Quantum math: transformations

Pre-script (dated 26 June 2020): Our ideas have evolved into a full-blown realistic (or classical) interpretation of all things quantum-mechanical. In addition, I note the dark force has amused himself by removing some material. So no use to read this. Read my recent papers instead. 🙂

Original post:

We’ve come a very long way. Now we’re ready for the Big Stuff. We’ll look at the rules for transforming amplitudes from one ‘base’ to ‘another’. [In quantum mechanics, however, we’ll talk about a ‘representation’, rather than a ‘base’, as we’ll reserve the latter term for a ‘base’ state.] In addition, we’ll look at how physicists model how amplitudes evolve over time using the so-called Hamiltonian matrix. So let’s go for it.

Transformations: how should we think about them?

In my previous post, I presented the following hypothetical set-up: we have an S-filter and a T-filter in series, but the T-filter at the angle α with respect to the first. In case you forgot: these ‘filters’ are modified Stern-Gerlach apparatuses, designed to split a particle beam according to the angular momentum in the direction of the gradient of the magnetic field, in which we may place masks to filter out one or more states.

The idea is illustrated in the hypothetical example below. The unpolarized beam goes through S, but we have masks blocking all particles with zero or negative spin in the z-direction, i.e. with respect to S. Hence, all particles entering the T-filter are in the +S state. Now, we assume the set-up of the T-filter is such that it filters out all particles with positive or negative spin. Hence, only particles with zero spin go through. So we’ve got something like this:

However, we need to be careful as what we are saying here. The T-apparatus is tilted, so the gradient of the magnetic field is different. To be precise, it’s got the same tilt as the T-filter itself (α). Hence, it will be filtering out all particles with positive or negative spin with respect to T. So, unlike what you might think at first, some fraction of the particles in the +S state will get through the T-filter, and come out in the 0T state. In fact, we know how many, because we have formulas for situations like this. To be precise, in this case, we should apply the following formula:

〈 0T | +S 〉 = −(1/√2)·sinα

This is a real-valued amplitude. As usual, we get the following probability by taking the absolute square, so P = |−(1/√2)·sinα|²= (1/2)·sin²α, which gives us the following graph of P:

The probability varies between 0 (for α = 0 or π) and 1/2 = 0.5 (for α = π/2 or 3π/2). Now, this graph may or may not make sense to you, so you should think about it. You’ll admit it makes sense to find P = 0 for α = 0, but what about the non-zero values?

Think about what this would mean in classical terms: we’ve got a beam of particles whose angular momentum is ‘up’ in the z-direction. To be precise, this means that J_z = +ħ. [Angular momentum and the quantum of action have the same dimension: the joule·second.] So that’s the maximum value out of the three permitted values, which are +ħ, 0 and –ħ. Note that the particles here must be bosons. So you may think we’re talking photons, in practice but… Well… No. As I’ll explain in a later post, the photon is a spin-one particle but it’s quite particular, because it has no ‘zero spin’-state. Don’t worry about it here – but it’s really quite remarkable. So, instead of thinking of a photon, you should think of some composite matter particle obeying Bose-Einstein statistics. These are not so rare as you may think: all matter-particles that contain an even number of fermions – like elementary particles – have integer spin – but… Well… Their spin number is usually zero – not one. So… Well… Feynman’s particle here is somewhat theoretical – but it doesn’t matter. Let’s move on. 🙂

Let’s look at another transformation formula. More in particular, let’s look at the formula we (should) get for 〈 0T | −S 〉 as a function of α. So we change the set-up of the S-filter to ensure all particles entering T have negative spin. The formula is:

〈 0T | −S 〉 = +(1/√2)·sinα

That gives the same probabilities: |+(1/√2)·sinα|²= (1/2)·sin²α. Adding |〈 0T | +S 〉|² and |〈 0T | −S 〉|²gives us a total probability equal to sin²α, which is equal to 1 if α = π/2 or 3π/2. We may be tempted to interpret this as follows: if a particle is in the +S or −S state before entering the T-apparatus, and the T-apparatus is tilted at an angle α = π/2 or 3π/2 with respect to the S-apparatus, then this particle will come out of the T-apparatus in the 0T-state. No ambiguity here: P = 1.

Is this strange? Well… Let’s think about what it means to tilt the T-apparatus. You’ll have to admit that, if the apparatus is tilted at the angle π/2 or 3π/2, it’s going to measure the angular momentum in the x-direction. [The y-direction is the common axis of both apparatuses here.] So… Well… It’s pretty plausible, isn’t it? If all of the angular momentum is in the positive or negative z-direction, then it’s not going to have any angular momentum in the x-direction, right? And not having any angular momentum in the x-direction effectively corresponds to being in the 0T-state, right?

Oh ! Is it that easy?

Well… No! Not at all! The reasoning above shows how easy it is to be led astray. We forgot to normalize. Remember, if we integrate the probability density function over its domain, i.e. α ∈ [0, 2π], then we have to get one, as all probabilities have to add up to one. The definite integral of (1/2)·sin²α over [0, 2π] is equal to π/2 (the definite integral of the sine or cosine squared over a full cycle is equal to π), so we need to multiply this function by 2/π to get the actual probability density function, i.e. (1/π)·sin²α. It’s got the same shape, obviously, but it gives us maximum probabilities equal to 1/π ≈ 0.32 for α = π/2 or 3π/2, instead of 1/2 = 0.5.

Likewise, the sin²α function we got when adding |〈 0T | +S 〉|² and |〈 0T | −S 〉|²should also be normalized. One really needs to keep one’s wits about oneself here. What we’re saying here is that we have a particle that is either in the +S or the −S state, so let’s say that the chance is 50/50 to be in either of the two states. We then have these probabilities |〈 0T | +S 〉|² and |〈 0T | −S 〉|², which we calculated as (1/π)·sin²α. So the total combined probability is equal to 0.5·(1/π)·sin²α + 0.5·(1/π)·sin²α = (1/π)·sin²α. So we’re now weighing the two (1/π)·sin²α functions – and it doesn’t matter if the weights are 50/50 or 75/25 or whatever, as long as the two weights add up to one. The bottom line is: we get the same (1/π)·sin²α function for P, and the same maximum probability 1/π ≈ 0.32 for α = π/2 or 3π/2.

So we don’t get unity: P ≠ 1 for α = π/2 or 3π/2. Why not? Think about it. The classical analysis made sense, didn’t it? If the angular momentum is all in the z-direction (or in one of the two z-directions, I should say), then we cannot have any of it in the x-direction, can it? Well… The surprising answer is: yes, we can. The remarkable thing is that, in quantum physics, we actually never have all of the angular momentum in one direction. As I explained in my post on spin and angular momentum, the classical concepts of angular momentum, and the related magnetic moment, have their limits in quantum mechanics. In quantum physics, we find that the magnitude of a vector quantity, like angular momentum, or the related magnetic moment, is generally not equal to the maximum value of the component of that quantity in any direction. The general rule is that the maximum value of any component of J in whatever direction – i.e. +ħ in the example we’re discussing here – is smaller than the magnitude of J – which I calculated in the mentioned post as |J| = J = +√2·ħ ≈ 1.414·ħ, so that’s almost 1.5 times ħ! So it’s quite a bit smaller! The upshot is that we cannot associate any precise and unambiguous direction with quantities like the angular momentum J or the magnetic moment μ. So the answer is: the angular momentum can never be all in the z-direction, so we can always have some of it in the x-direction, and so that explains the amplitudes and probabilities we’re having here.

Huh?

Yep. I know. We never seem to get out of this ‘weirdness’, but then that’s how quantum physics is like. Feynman warned us upfront:

“Because atomic behavior is so unlike ordinary experience, it is very difficult to get used to, and it appears peculiar and mysterious to everyone—both to the novice and to the experienced physicist. Even the experts do not understand it the way they would like to, and it is perfectly reasonable that they should not, because all of direct, human experience and of human intuition applies to large objects. We know how large objects will act, but things on a small scale just do not act that way. So we have to learn about them in a sort of abstract or imaginative fashion and not by connection with our direct experience.”

As I see it, quantum physics is about explaining all sorts of weird stuff, like electron interference and tunneling and what have you, so it shouldn’t surprise us that the framework is as weird as the stuff it’s trying to explain. 🙂 So… Well… All we can do is to try to go along with it, isn’t it? And so that’s what we’ll do here. 🙂

Transformations: the formulas

We need to distinguish various cases here. The first case is the case explained above: the T-apparatus shares the same y-axis – along which the particles move – but it’s tilted. To be precise, we should say that it’s rotated about the common y-axis by the angle α. That implies we can relate the x’, y’, z’ coordinate system of T to the x, y, z coordinate system of S through the following equations: z′ = z·cosα + x·sinα, x′ = x·cosα − z·sinα, and y′ = y. Then the transformation amplitudes are:

We used the formula for 〈 0T | +S 〉 and 〈 0T | −S 〉 above, and you can play with the formulas above by imagining the related set-up of the S and T filters, such as the one below:

If you do your homework (just check what formula and what set-up this corresponds to), you should find the following graph for the amplitude and the probability as a function of α: the graph is zero for α = π, but is non-zero everywhere else. As with the other example, you should think about this. It makes sense—sort of, that is. 🙂

OK. Next case. Now we’re going to rotate the T-apparatus around the z-axis by some angle β. To illustrate what we’re doing here, we need to take a ‘top view’ of our apparatus, as shown below, which shows a rotation over 90°. More in general, for any angle β, the coordinate transformation is given by z′ = z, x′ = x·cosβ + y·sinβ, y′ = y·cosβ − x·sinβ. [So it’s quite similar to case 1: we’re only rotating the thing in a different plane.]

The transformation amplitudes are now given by:

As you can see, we get complex-valued transformation amplitudes, unlike our first case, which yielded real-valued transformation amplitudes. That’s just the way it is. Nobody says transformation amplitudes have to be real-valued. On the contrary, one would expect them to be complex numbers. 🙂 Having said that, the combined set of transformation formulas is, obviously, rather remarkable. The amplitude to go from the +S state to, say, the 0T state is zero. Also, when our particle has zero spin when coming out of S, it will always have zero spin when and if it goes through T. In fact, the absolute value of those e^±iβ functions is also equal to one, so they are also associated with probabilities that are equal to one: |e^±iβ|² = 1² = 1. So… Well… Those formulas are simple and weird at the same time, aren’t they? They sure give us plenty of stuff to think about, I’d say.

So what’s next? Well… Not all that much. We’re sort of done, really. Indeed, it’s just a property of space that we can get any rotation of T by combining the two rotations above. As I only want to introduce the basic concepts here, I’ll refer you to Feynman for the details of how exactly that’s being done. [He illustrates it for spin-1/2 particles in particular.] I’ll just wrap up here by generalizing our results from base states to any state.

Transformations: generalization

We mentioned a couple of times already that the base states are like a particular coordinate system: we will usually describe a state in terms of base states indeed. More in particular, choosing S as our representation, we’ll say:

The state φ is defined by the three numbers:

C₊ = 〈 +S | φ 〉,

C₀ = 〈 0S | φ 〉,

C₋ = 〈 −S | φ 〉.

Now, the very same state can, of course, also be described in the ‘T system’, so then our numbers – i.e. the ‘components’ of φ – would be equal to:

C’₊ = 〈 +T | φ 〉, C’₀ = 〈 0T | φ 〉, and C’₋ = 〈 −T | φ 〉.

So how can we go from the unprimed ‘coordinates’ to the primed ones? The trick is to use the second of the three quantum math ‘Laws’ which I introduced in my previous post:

Capture

Just replace χ in [II] by +T, 0T and/or –T. More in general, if we denote +T, 0T or –T by jT, we can re-write this ‘Law’ as:

So the 〈 jT | iS 〉 amplitudes are those nine transformation amplitudes. Now, we can represent those nine amplitudes in a nice three-by-three matrix and, yes, we’ll call that matrix the transformation matrix. So now you know what that is.

To conclude, I should note that it’s only because we’re talking spin-one particles here that we have three base states here and, hence, three ‘components’, which we denoted by C₊, C₋ and C₀, which transform the way they do when going from one representation to another, and so that is very much like what vectors do when we move to a different coordinate system, which is why spin-one particles are often referred to as ‘vector particles‘. [I am just mentioning this in case you’d come across the term and wonder why they’re being called that way. Now you know.] In fact, if we have three base states, in respect to whatever representation, and we define some state φ in terms of them, then we can always re-define that state in terms of the following ‘special’ set of components:

The set is ‘special’ because one can show (you can do that yourself that by using those transformation laws) that these components transform exactly the way as x, y, z transform to x′, y′, z′. But so I’ll leave at this.

[…]

Oh… What about the Hamiltonian? Well… I’ll save that for my next posts, as my posts have become longer and longer, and so it’s probably a good idea to separate them out. 🙂

Post scriptum: transformations for spin-1/2 particles

You should actually really check out that chapter of Feynman. The transformation matrices for spin-1/2 particles look different because… Well… Because there’s only two base states for spin-1/2 particles. It’s a pretty technical chapter, but then spin-1/2 particles are the ones that make up the world. 🙂

Quantum math: the rules – all of them! :-)

Original post:

In my previous post, I made no compromise, and used all of the rules one needs to calculate quantum-mechanical stuff:

However, I didn’t explain them. These rules look simple enough, but let’s analyze them now. They’re simple and not at the same time, indeed.

[I] The first equation uses the Kronecker delta, which sounds fancy but it’s just a simple shorthand: δ_ij= δ_jiis equal to 1 if i = j, and zero if i ≠ j, with i and j representing base states. Equation (I) basically says that base states are all different. For example, the angular momentum in the x-direction of a spin-1/2 particle – think of an electron or a proton – is either +ħ/2 or −ħ/2, not something in-between, or some mixture. So 〈 +x | +x 〉 = 〈 −x | −x 〉 = 1 and 〈 +x | −x 〉 = 〈 −x | +x 〉 = 0.

We’re talking base states here, of course. Base states are like a coordinate system: we settle on an x-, y- and z-axis, and a unit, and any point is defined in terms of an x-, y– and z-number. It’s the same here, except we’re talking ‘points’ in four-dimensional spacetime. To be precise, we’re talking constructs evolving in spacetime. To be even more precise, we’re talking amplitudes with a temporal as well as a spatial frequency, which we’ll often represent as:

a·e^−i·θ= e^{−i·(ω·t − k ∙x)} = a·e^{−(i/ħ)·(E·t − p∙x)}

The coefficient in front (a) is just a normalization constant, ensuring all probabilities add up to one. It may not be a constant, actually: perhaps it just ensure our amplitude stays within some kind of envelope, as illustrated below.

As for the ω = E/ħ and k = p/ħ identities, these are the de Broglie equations for a matter-wave, which the young Comte jotted down as part of his 1924 PhD thesis. He was inspired by the fact that the E·t − p∙x factor is an invariant four-vector product (E·t − p∙x = p_μx_μ) in relativity theory, and noted the striking similarity with the argument of any wave function in space and time (ω·t − k ∙x) and, hence, couldn’t resist equating both. Louis de Broglie was inspired, of course, by the solution to the blackbody radiation problem, which Max Planck and Einstein had convincingly solved by accepting that the ω = E/ħ equation holds for photons. As he wrote it:

“When I conceived the first basic ideas of wave mechanics in 1923–24, I was guided by the aim to perform a real physical synthesis, valid for all particles, of the coexistence of the wave and of the corpuscular aspects that Einstein had introduced for photons in his theory of light quanta in 1905.” (Louis de Broglie, quoted in Wikipedia)

Looking back, you’d of course want the phase of a wavefunction to be some invariant quantity, and the examples we gave our previous post illustrate how one would expect energy and momentum to impact its temporal and spatial frequency. But I am digressing. Let’s look at the second equation. However, before we move on, note that minus sign in the exponent of our wavefunction: a·e^−i·θ. The phase turns counter-clockwise. That’s just the way it is. I’ll come back to this.

[II] The φ and χ symbols do not necessarily represent base states. In fact, Feynman illustrates this law using a variety of examples including both polarized as well as unpolarized beams, or ‘filtered’ as well as ‘unfiltered’ states, as he calls it in the context of the Stern-Gerlach apparatuses he uses to explain what’s going on. Let me summarize his argument here.

I discussed the Stern-Gerlach experiment in my post on spin and angular momentum, but the Wikipedia article on it is very good too. The principle is illustrated below: a inhomogeneous magnetic field – note the direction of the gradient ∇B = (∂B/∂x, ∂B/∂y, ∂B/∂z) – will split a beam of spin-one particles into three beams. [Matter-particles with spin one are rather rare (Lithium-6 is an example), but three states (rather than two only, as we’d have when analyzing spin-1/2 particles, such as electrons or protons) allow for more play in the analysis. 🙂 In any case, the analysis is easily generalized.]

The splitting of the beam is based, of course, on the quantized angular momentum in the z-direction (i.e. the direction of the gradient): its value is either ħ, 0, or −ħ. We’ll denote these base states as +, 0 or −, and we should note they are defined in regard to an apparatus with a specific orientation. If we call this apparatus S, then we can denote these base states as +S, 0S and −S respectively.

The interesting thing in Feynman’s analysis is the imagined modified Stern-Gerlach apparatus, which – I am using Feynman‘s words here 🙂 – “puts Humpty Dumpty back together.” It looks a bit monstruous, but it’s easy enough to understand. Quoting Feynman once more: “It consists of a sequence of three high-gradient magnets. The first one (on the left) is just the usual Stern-Gerlach magnet and splits the incoming beam of spin-one particles into three separate beams. The second magnet has the same cross section as the first, but is twice as long and the polarity of its magnetic field is opposite the field in magnet $1$ . The second magnet pushes in the opposite direction on the atomic magnets and bends their paths back toward the axis, as shown in the trajectories drawn in the lower part of the figure. The third magnet is just like the first, and brings the three beams back together again, so that leaves the exit hole along the axis.”

Now, we can use this apparatus as a filter by inserting blocking masks, as illustrated below.

But let’s get back to the lesson. What about the second ‘Law’ of quantum math? Well… You need to be able to imagine all kinds of situations now. The rather simple set-up below is one of them: we’ve got two of these apparatuses in series now, S and T, with T tilted at the angle α with respect to the first.

I know: you’re getting impatient. What about it? Well… We’re finally ready now. Let’s suppose we’ve got three apparatuses in series, with the first and the last one having the very same orientation, and the one in the middle being tilted. We’ll denote them by S, T and S’ respectively. We’ll also use masks: we’ll block the 0 and − state in the S-filter, like in that illustration above. In addition, we’ll block the + and − state in the T apparatus and, finally, the 0 and − state in the S’ apparatus. Now try to imagine what happens: how many particles will get through?

[…]

Just try to think about it. Make some drawing or something. Please!

[…]

OK… The answer is shown below. Despite the filtering in S, the +S particles that come out do have an amplitude to go through the 0T-filter, and so the number of atoms that come out will be some fraction (α) of the number of atoms (N) that came out of the +S-filter. Likewise, some other fraction (β) will make it through the +S’-filter, so we end up with βαN particles.

Now, I am sure that, if you’d tried to guess the answer yourself, you’d have said zero rather than βαN but, thinking about it, it makes sense: it’s not because we’ve got some angular momentum in one direction that we have none in the other. When everything is said and done, we’re talking components of the total angular momentum here, don’t we? Well… Yes and no. Let’s remove the masks from T. What do we get?

[…]

Come on: what’s your guess? N?

[…] You’re right. It’s N. Perfect. It’s what’s shown below.

Now, that should boost your confidence. Let’s try the next scenario. We block the 0 and − state in the S-filter once again, and the + and − state in the T apparatus, so the first two apparatuses are the same as in our first example. But let’s change the S’ apparatus: let’s close the + and − state there now. Now try to imagine what happens: how many particles will get through?

[…]

Come on! You think it’s a trap, isn’t it? It’s not. It’s perfectly similar: we’ve got some other fraction here, which we’ll write as γαN, as shown below.

Next scenario: S has the 0 and − gate closed once more, and T is fully open, so it has no masks. But, this time, we set S’ so it filters the 0-state with respect to it. What do we get? Come on! Think! Please!

[…]

The answer is zero, as shown below.

Does that make sense to you? Yes? Great! Because many think it’s weird: they think the T apparatus must ‘re-orient’ the angular momentum of the particles. It doesn’t: if the filter is wide open, then “no information is lost”, as Feynman puts it. Still… Have a look at it. It looks like we’re opening ‘more channels’ in the last example: the S and S’ filter are the same, indeed, and T is fully open, while it selected for 0-state particles before. But no particles come through now, while with the 0-channel, we had γαN.

Hmm… It actually is kinda weird, won’t you agree? Sorry I had to talk about this, but it will make you appreciate that second ‘Law’ now: we can always insert a ‘wide-open’ filter and, hence, split the beams into a complete set of base states − with respect to the filter, that is − and bring them back together provided our filter does not produce any unequal disturbances on the three beams. In short, the passage through the wide-open filter should not result in a change of the amplitudes. Again, as Feynman puts it: the wide-open filter should really put Humpty-Dumpty back together again. If it does, we can effectively apply our ‘Law’:

For an example, I’ll refer you to my previous post. This brings me to the third and final ‘Law’.

[III] The amplitude to go from state φ to state χ is the complex conjugate of the amplitude to to go from state χ to state φ:

〈 χ | φ 〉 = 〈 φ | χ 〉*

This is probably the weirdest ‘Law’ of all, even if I should say, straight from the start, we can actually derive it from the second ‘Law’, and the fact that all probabilities have to add up to one. Indeed, a probability is the absolute square of an amplitude and, as we know, the absolute square of a complex number is also equal to the product of itself and its complex conjugate:

|z|²= |z|·|z| = z·z*

[You should go through the trouble of reviewing the difference between the square and the absolute square of a complex number. Just write z as a + ib and calculate (a + ib)²= a² + 2abi + b², as opposed to |z|²= a² + b². Also check what it means when writing z as r·e^iθ= r·(cosθ + i·sinθ).]

Let’s applying the probability rule to a two-filter set-up, i.e. the situation with the S and the tilted T filter which we described above, and let’s assume we’ve got a pure beam of +S particles entering the wide-open T filter, so our particles can come out in either of the three base states with respect to T. We can then write:

〈 +T | +S 〉²+ 〈 0T | +S 〉²+ 〈 −T | +S 〉²= 1

⇔ 〈 +T | +S 〉〈 +T | +S 〉* + 〈 0T | +S 〉〈 0T | +S 〉* + 〈 −T | +S 〉〈 −T | +S 〉* = 1

Of course, we’ve got two other such equations if we start with a 0S or a −S state. Now, we take the 〈 χ | φ 〉 = ∑ 〈 χ | i 〉〈 i | φ 〉 ‘Law’, and substitute χ and φ for +S, and all i states for the base states with regard to T. We get:

〈 +S | +S 〉 = 1 = 〈 +S | +T 〉〈 +T | +S 〉 + 〈 +S | 0T 〉〈 0T | +S 〉 + 〈 +S | –T 〉〈 −T | +S 〉

These equations are consistent only if:

〈 +S | +T 〉 = 〈 +T | +S 〉*,

〈 +S | 0T 〉 = 〈 0T | +S 〉*,

〈 +S | −T 〉 = 〈 −T | +S 〉*,

which is what we wanted to prove. One can then generalize to any state φ and χ. However, proving the result is one thing. Understanding it is something else. One can write down a number of strange consequences, which all point to Feynman‘s rather enigmatic comment on this ‘Law’: “If this Law were not true, probability would not be ‘conserved’, and particles would get ‘lost’.” So what does that mean? Well… You may want to think about the following, perhaps. It’s obvious that we can write:

|〈 φ | χ 〉|²= 〈 φ | χ 〉〈 φ | χ 〉* = 〈 χ | φ 〉*〈 χ | φ 〉 = |〈 χ | φ 〉|²

This says that the probability to go from the φ-state to the χ-state is the same as the probability to go from the χ-state to the φ-state.

Now, when we’re talking base states, that’s rather obvious, because the probabilities involved are either 0 or 1. However, if we substitute for +S and −T, or some more complicated states, then it’s a different thing. My guts instinct tells me this third ‘Law’ – which, as mentioned, can be derived from the other ‘Laws’ – reflects the principle of reversibility in spacetime, which you may also interpret as a causality principle, in the sense that, in theory at least (i.e. not thinking about entropy and/or statistical mechanics), we can reverse what’s happening: we can go back in spacetime.

In this regard, we should also remember that the complex conjugate of a complex number in polar form, i.e. a complex number written as r·e^iθ, is equal to r·e^−iθ, so the argument in the exponent gets a minus sign. Think about what this means for our a·e^−i·θ= e^{−i·(ω·t − k ∙x)} = a·e^{−(i/ħ)·(E·t − p∙x)}function. Taking the complex conjugate of this function amounts to reversing the direction of t and x which, once again, evokes that idea of going back in spacetime.

I feel there’s some more fundamental principle here at work, on which I’ll try to reflect a bit more. Perhaps we can also do something with that relationship between the multiplicative inverse of a complex number and its complex conjugate, i.e. z⁻¹= z*/|z|². I’ll check it out. As for now, however, I’ll leave you to do that, and please let me know if you’ve got any inspirational ideas on this. 🙂

So… Well… Goodbye as for now. I’ll probably talk about the Hamiltonian in my next post. I think we really did a good job in laying the groundwork for the really hardcore stuff, so let’s go for that now. 🙂

Post Scriptum: On the Uncertainty Principle and other rules

After writing all of the above, I realized I should add some remarks to make this post somewhat more readable. First thing: not all of the rules are there—obviously! Most notably, I didn’t say anything about the rules for adding or multiplying amplitudes, but that’s because I wrote extensively about that already, and so I assume you’re familiar with that. [If not, see my page on the essentials.]

Second, I didn’t talk about the Uncertainty Principle. That’s because I didn’t have to. In fact, we don’t need it here. In general, all popular accounts of quantum mechanics have an excessive focus on the position and momentum of a particle, while the approach in this and my previous post is quite different. Of course, it’s Feynman’s approach to QM really. Not ‘mine’. 🙂 All of the examples and all of the theory he presents in his introductory chapters in the Third Volume of Lectures, i.e. the volume on QM, are related to things like:

What is the amplitude for a particle to go from spin state +S to spin state −T?
What is the amplitude for a particle to be scattered, by a crystal, or from some collision with another particle, in the θ direction?
What is the amplitude for two identical particles to be scattered in the same direction?
What is the amplitude for an atom to absorb or emit a photon? [See, for example, Feynman’s approach to the blackbody radiation problem.]
What is the amplitude to go from one place to another?

In short, you read Feynman, and it’s only at the very end of his exposé, that he starts talking about the things popular books start with, such as the amplitude of a particle to be at point (x, t) in spacetime, or the Schrödinger equation, which describes the orbital of an electron in an atom. That’s where the Uncertainty Principle comes in and, hence, one can really avoid it for quite a while. In fact, one should avoid it for quite a while, because it’s now become clear to me that simply presenting the Uncertainty Principle doesn’t help all that much to truly understand quantum mechanics.

Truly understanding quantum mechanics involves understanding all of these weird rules above. To some extent, that involves dissociating the idea of the wavefunction with our conventional ideas of time and position. From the questions above, it should be obvious that ‘the’ wavefunction does actually not exist: we’ve got a wavefunction for anything we can and possibly want to measure. That brings us to the question of the base states: what are they?

Feynman addresses this question in a rather verbose section of his Lectures titled: What are the base states of the world? I won’t copy it here, but I strongly recommend you have a look at it. 🙂

I’ll end here with a final equation that we’ll need frequently: the amplitude for a particle to go from one place (r₁) to another (r₂). It’s referred to as a propagator function, for obvious reasons—one of them being that physicists like fancy terminology!—and it looks like this:

The shape of the e^{(i/ħ)·(p∙r₁₂)}function is now familiar to you. Note the r₁₂in the argument, i.e. the vector pointing from r₁ to r₂. The p∙r₁₂ dot product equals |p|∙|r₁₂|·cosθ = p∙r₁₂·cosθ, with θ the angle between p and r₁₂. If the angle is the same, then cosθ is equal to 1. If the angle is π/2, then it’s 0, and the function reduces to 1/r₁₂. So the angle θ, through the cosθ factor, sort of scales the spatial frequency. Let me try to give you some idea of how this looks like by assuming the angle between p and r₁₂ is the same, so we’re looking at the space in the direction of the momentum only and |p|∙|r₁₂|·cosθ = p∙r₁₂. Now, we can look at the p/ħ factor as a scaling factor, and measure the distance x in units defined by that scale, so we write: x = p∙r₁₂/ħ. The function then reduces to (ħ/p)·e^i∙x/x = (ħ/p)·cos(x)/x + i·(ħ/p)·sin(x)/x, and we just need to square this to get the probability. All of the graphs are drawn hereunder: I’ll let you analyze them. [Note that the graphs do not include the ħ/p factor, which you may look at as yet another scaling factor.] You’ll see – I hope! – that it all makes perfect sense: the probability quickly drops off with distance, both in the positive as well as in the negative x-direction, while it’s going to infinity when very near. [Note that the absolute square, using cos(x)/x and sin(x)/xyields the same graph as squaring 1/x—obviously!]

Working with amplitudes

Note (22 May 2020): I am in the process of reviewing many of my posts—including the one below—as a result of my realist interpretation of quantum mechanics, which is based on the ring current model of an electron. It is a delicate exercise because I would like to keep track of the old posts. Perhaps I should copy all of them to some new site. Any case, the point is: I would significantly rephrase some of the below because I now believe spin is very real in quantum mechanics as well.

Original post:

Don’t worry: I am not going to introduce the Hamiltonian matrix—not yet, that is. But this post is going a step further than my previous ones, in the sense that it will be more abstract. At the same time, I do want to stick to real physical examples so as to illustrate what we’re doing when working with those amplitudes. The example that I am going to use involves spin. So let’s talk about that first.

Spin, angular momentum and the magnetic moment

You know spin: it allows experienced pool players to do the most amazing tricks with billiard balls, making a joke of what a so-called elastic collision is actually supposed to look like. So it should not come as a surprise that spin complicates the analysis in quantum mechanics too. We dedicated several posts to that (see, for example, my post on spin and angular momentum in quantum physics) and I won’t repeat these here. Let me just repeat the basics:

1. Classical and quantum-mechanical spin do share similarities: the basic idea driving the quantum-mechanical spin model is that of a electric charge – positive or negative – spinning about its own axis (this is often referred to as intrinsic spin) as well as having some orbital motion (presumably around some other charge, like an electron in orbit with a nucleus at the center). This intrinsic spin, and the orbital motion, give our charge some angular momentum (J) and, because it’s an electric charge in motion, there is a magnetic moment (μ). To put things simply: the classical and quantum-mechanical view of things converge in their analysis of atoms or elementary particles as tiny little magnets. Hence, when placed in an external magnetic field, there is some interaction – a force – and their potential and/or kinetic energy changes. The whole system, in fact, acquires extra energy when placed in an external magnetic field.

Note: The formula for that magnetic energy is quite straightforward, both in classical as well as in quantum physics, so I’ll quickly jot it down: U = −μ•B = −|μ|·|B|·cosθ = −μ·B·cosθ. So it’s just the scalar product of the magnetic moment and the magnetic field vector, with a minus sign in front so as to get the direction right. [θ is the angle between the μ and B vectors and determines whether U as a whole is positive or negative.

2. The classical and quantum-mechanical view also diverge, however. They diverge, first, because of the quantum nature of spin in quantum mechanics. Indeed, while the angular momentum can take on any value in classical mechanics, that’s not the case in quantum mechanics: in whatever direction we measure, we get a discrete set of values only. For example, the angular momentum of a proton or an electron is either −ħ/2 or +ħ/2, in whatever direction we measure it. Therefore, they are referred to as spin-1/2 particles. All elementary fermions, i.e. the particles that constitute matter (as opposed to force-carrying particles, like photons), have spin 1/2.

Note: Spin-1/2 particles include, remarkably enough, neutrons too, which has the same kind of magnetic moment that a rotating negative charge would have. The neutron, in other words, is not exactly ‘neutral’ in the magnetic sense. One can explain this by noting that a neutron is not ‘elementary’, really: it consists of three quarks, just like a proton, and, therefore, it may help you to imagine that the electric charges inside are, somehow, distributed unevenly—although physicists hate such simplifications. I am noting this because the famous Stern-Gerlach experiment, which established the quantum nature of particle spin, used silver atoms, rather than protons or electrons. More in general, we’ll tend to forget about the electric charge of the particles we’re describing, assuming, most of the time, or tacitly, that they’re neutral—which helps us to sort of forget about classical theory when doing quantum-mechanical calculations!

3. The quantum nature of spin is related to another crucial difference between the classical and quantum-mechanical view of the angular momentum and the magnetic moment of a particle. Classically, the angular momentum and the magnetic moment can have any direction.

Note: I should probably briefly remind you that J is a so-called axial vector, i.e. a vector product (as opposed to a scalar product) of the radius vector r and the (linear) momentum vector p = m·v, with v the velocity vector, which points in the direction of motion. So we write: J = r×p = r×m·v = |r|·|p|·sinθ·n. The n vector is the unit vector perpendicular to the plane containing r and p (and, hence, v, of course) given by the right-hand rule. I am saying this to remind you that the direction of the magnetic moment and the direction of motion are not the same: the simple illustration below may help to see what I am talking about.]

Back to quantum mechanics: the image above doesn’t work in quantum mechanics. We do not have an unambiguous direction of the angular momentum and, hence, of the magnetic moment. That’s where all of the weirdness of the quantum-mechanical concept of spin comes out, really. I’ll talk about that when discussing Feynman’s ‘filters’ – which I’ll do in a moment – but here I just want to remind you of the mathematical argument that I presented in the above-mentioned post. Just like in classical mechanics, we’ll have a maximum (and, hence, also a minimum) value for J, like +ħ, 0 and +ħ for a Lithium-6 nucleus. [I am just giving this rather special example of a spin-1 article so you’re reminded we can have particles with an integer spin number too!] So, when we measure its angular momentum in any direction really, it will take on one of these three values: +ħ, 0 or +ħ. So it’s either/or—nothing in-between. Now that leads to a funny mathematical situation: one would usually equate the maximum value of a quantity like this to the magnitude of the vector, which is equal to the (positive) square root of J² = J•J = J_x² + J_y² + J_z², with J_x, J_y and J_z the components of J in the x-, y- and z-direction respectively. But we don’t have continuity in quantum mechanics, and so the concept of a component of a vector needs to be carefully interpreted. There’s nothing definite there, like in classical mechanics: all we have is amplitudes, and all we can do is calculate probabilities, or expected values based on those amplitudes.

Huh? Yes. In fact, the concept of the magnitude of a vector itself becomes rather fuzzy: all we can do really is calculate its expected value. Think of it: in the classical world, we have a J² = J•J product that’s independent of the direction of J. For example, if J is all in the x-direction, then J_yand J_zwill be zero, and J² = J_x². If it’s all in the y-direction, then J_xand J_zwill be zero and all of the magnitude of J will be in the y-direction only, so we write: J² = J_y². Likewise, if J does not have any z-component, then our J•J product will only include the x- and y-components: J•J = J_x² + J_y². You get the idea: the J² = J•J product is independent of the direction of J exactly because, in classical mechanics, J actually has a precise and unambiguous magnitude and direction and, therefore, actually has a precise and unambiguous component in each direction. So we’d measure J_x, J_y, and J_zand, regardless of the actual direction of J, we’d find its magnitude |J| = J = +√J² = +(J_x² + J_y² + J_z²)^1/2.

In quantum mechanics, we just don’t have quantities like that. We say that J_x,J_yand J_zhave an amplitude to take on a value that’s equal to +ħ, 0 or +ħ (or whatever other value is allowed by the spin number of the system). Now that we’re talking spin numbers, please note that this characteristic number is usually denoted by j, which is a bit confusing, but so be it. So j can be 0, 1/2, 1, 3/2, etcetera, and the number of ‘permitted values’ is 2j + 1 values, with each value being separated by an amount equal to ħ. So we have 1, 2, 3, 4, 5 etcetera possible values for J_x,J_yand J_zrespectively. But let me get back to the lesson. We just can’t do the same thing in quantum mechanics. For starters, we can’t measure J_x, J_y, and J_zsimultaneously: our Stern-Gerlach apparatus has a certain orientation and, hence, measures one component of J only. So what can we do?

Frankly, we can only do some math here. The wave-mechanical approach does allow to think of the expected value of J² = J•J = J_x² + J_y² + J_z² value, so we write:

E[J²] = E[J•J] = E[J_x² + J_y² + J_z²] = ?

[Feynman’s use of the 〈 and 〉 brackets to denote an expected value is hugely confusing, because these brackets are also used to denote an amplitude. So I’d rather use the more commonly used E[X] notation.] Now, it is a rather remarkable property, but the expected value of the sum of two or more random variables is equal to the sum of the expected values of the variables, even if those variables may not be independent. So we can confidently use the linearity property of the expected value operator and write:

E[J_x²+ J_y² + J_z²] = E[J_x²] + E[J_x²] + E[J_x²]

Now we need something else. It’s also just part of the quantum-mechanical approach to things and so you’ll just have to accept it. It sounds rather obvious but it’s actually quite deep: if we measure the x-, y- or z-component of the angular momentum of a random particle, then each of the possible values is equally likely to occur. So that means, in our case, that the +ħ, 0 or +ħ values are equally likely, so their likelihood is one into three, i.e. 1/3. Again, that sounds obvious but it’s not. Indeed, please note, once again, that we can’t measure J_x, J_y, and J_zsimultaneously, so the ‘or’ in x-, y- or z-component is an exclusive ‘or’. Of course, I must add this equipartition of likelihoods is valid only because we do not have a preferred direction for J: the particles in our beam have random ‘orientations’. Let me give you the lingo for this: we’re looking at an unpolarized beam. You’ll say: so what? Well… Again, think about what we’re doing here: we may of may not assume that the J_x, J_y, and J_zvariables are related. In fact, in classical mechanics, they surely are: they’re determined by the magnitude and direction of J. Hence, they are not random at all ! But let me continue, so you see what comes out.

Because the +ħ, 0 and +ħ values are equally, we can write: E[J_x²] = ħ²/3 + 0/3 + (−ħ)²/3 = [ħ² + 0 + (−ħ)²]/3 = 2ħ²/3. In case you wonder, that’s just the definition of the expected value operator: E[X] = p₁x₁+ p₂x₂+ … = ∑p_ix_i, with p_i the likelihood of the possible value x_i. So we take a weighted average with the respective probabilities as the weights. However, in this case, with an unpolarized beam, the weighted average becomes a simple average.

Now, E[J_y²] and E[J_z²] are – rather unsurprisingly – also equal to 2ħ²/3, so we find that E[J²] = E[J_x²] + E[J_x²] + E[J_x²] = 3·(2ħ²/3) = 2ħ²and, therefore, we’d say that the magnitude of the angular momentum is equal to |J| = J = +√2·ħ ≈ 1.414·ħ. Now that value is not equal to the maximum value of our x-, y-, z-component of J, or the component of J in whatever direction we’d want to measure it. That maximum value is ħ, without the √2 factor, so that’s some 40% less than the magnitude we’ve just calculated!

Now, you’ve probably fallen asleep by now but, what this actually says, is that the angular momentum, in quantum mechanics, is never completely in any direction. We can state this in another way: it implies that, in quantum mechanics, there’s no such thing really as a ‘definite’ direction of the angular momentum.

[…]

OK. Enough on this. Let’s move on to a more ‘real’ example. Before I continue though, let me generalize the results above:

[I] A particle, or a system, will have a characteristic spin number: j. That number is always an integer or a half-integer, and it determines a discrete set of possible values for the component of the angular momentum J in any direction.

[II] The number of values is equal to 2j + 1, and these values are separated by ħ, which is why they are usually measured in units of ħ, i.e. Planck’s reduced constant: ħ ≈ 1×10⁻³⁴J·s, so that’s tiny but real. 🙂 [It’s always good to remind oneself that we’re actually trying to describe reality.] For example, the permitted values for a spin-3/2 particle are +3ħ/2, +ħ/2, −ħ/2 and −3ħ/2 or, measured in units of ħ, +3/2, +1/2, −1/2 and −3/2. When discussing spin-1/2 particles, we’ll often refer to the two possible states as the ‘up’ and the ‘down’ state respectively. For example, we may write the amplitude for an electron or a proton to have a angular momentum in the x-direction equal to +ħ/2 or −ħ/2 as 〈+x〉 and 〈−x〉 respectively. [Don’t worry too much about it right now: you’ll get used to the notation quickly.]

[III] The classical concepts of angular momentum, and the related magnetic moment, have their limits in quantum mechanics. The magnitude of a vector quantity like angular momentum is generally not equal to the maximum value of the component of that quantity in any direction. The general rule is:

J²= j·(j+1)ħ² > j²·ħ²

So the maximum value of any component of J in whatever direction (i.e. j·ħ) is smaller than the magnitude of J (i.e. √[ j·(j+1)]·ħ). This implies we cannot associate any precise and unambiguous direction with quantities like the angular momentum J or the magnetic moment μ. As Feynman puts it:

“That the energy of an atom [or a particle] in a magnetic field can have only certain discrete energies is really not more surprising than the fact that atoms in general have only certain discrete energy levels—something we mentioned often in Volume I. Why should the same thing not hold for atoms in a magnetic field? It does. But it is the attempt to correlate this with the idea of an oriented magnetic moment that brings out some of the strange implications of quantum mechanics.”

**A real example: the disintegration of a muon in a magnetic field**

I talked about muon integration before, when writing a much more philosophical piece on symmetries in Nature and time reversal in particular. I used the illustration below. We’ve got an incoming muon that’s being brought to rest in a block of material, and then, as muons do, it disintegrates, emitting an electron and two neutrinos. As you can see, the decay direction is (mostly) in the direction of the axial vector that’s associated with the spin direction, i.e. the direction of the grey dashed line. However, there’s some angular distribution of the decay direction, as illustrated by the blue arrows, that are supposed to visualize the decay products, i.e. the electron and the neutrinos.

This disintegration process is very interesting from a more philosophical side. The axial vector isn’t ‘real’: it’s a mathematical concept—a pseudovector. A pseudo- or axial vector is the product of two so-called true vectors, aka as polar vectors. Just look back at what I wrote about the angular momentum: the J in the J = r×p = r×m·v formula is such vector, and its direction depends on the spin direction, which is clockwise or counter-clockwise, depending from what side you’re looking at it. Having said that, who’s to judge if the product of two ‘true’ vectors is any less ‘true’ than the vectors themselves? 🙂

The point is: the disintegration process does not respect what is referred to as P-symmetry. That’s because our mathematical conventions (like all of these right-hand rules that we’ve introduced) are unambiguous, and they tell us that the pseudovector in the mirror image of what’s going on, has the opposite direction. It has to, as per our definition of a vector product. Hence, our fictitious muon in the mirror should send its decay products in the opposite direction too! So… Well… The mirror image of our muon decay process is actually something that’s not going to happen: it’s physically impossible. So we’ve got a process in Nature here that doesn’t respect ‘mirror’ symmetry. Physicists prefer to call it ‘P-symmetry’, for parity symmetry, because it involves a flip of sign of all space coordinates, so there’s a parity inversion indeed. So there’s processes in Nature that don’t respect it but, while that’s all very interesting, it’s not what I want to write about. [Just check that post of mine if you’d want to read more.] Let me, therefore, use another illustration—one that’s more to the point in terms of what we do want to talk about here:

So we’ve got the same muon here – well… A different one, of course! 🙂 – entering that block (A) and coming to a grinding halt somewhere in the middle, and then it disintegrates in a few micro-seconds, which is an eternity at the atomic or sub-atomic scale. It disintegrates into an electron and two neutrinos, as mentioned above, with some spread in the decay direction. [In case you wonder where we can find muons… Well… I’ll let you look it up yourself.] So we have:

Now it turns out that the presence of a magnetic field (represented by the B arrows in the illustration above) can drastically change the angular distribution of decay directions. That shouldn’t surprise us, of course, but how does it work, exactly? Well… To simplify the analysis, we’ve got a polarized beam here: the spin direction of all muons before they enter the block and/or the magnetic field, i.e. at time t = 0, is in the +x-direction. So we filtered them just, before they entered the block. [I will come back to this ‘filtering’ process.] Now, if the muon’s spin would stay that way, then the decay products – and the electron in particular – would just go straight, because all of the angular momentum is in that direction. However, we’re in the quantum-mechanical world here, and so things don’t stay the same. In fact, as we explained, there’s no such things as a definite angular momentum: there’s just an amplitude to be in the +x state, and that amplitude changes in time and in space.

How exactly? Well… We don’t know, but we can apply some clever tricks here. The first thing to note is that our magnetic field will add to the energy of our muon. So, as I explained in my previous post, the magnetic field adds to the E in the exponent of our complex-valued wavefunction a·e^{−(i/ħ)(E·t − p∙x)}. In our example, we’ve got a magnetic field in the z-direction only, so that U = −μ•B reduces to U = −μ_z·B, and we can re-write our wavefunction as:

a·e^{−(i/ħ)[(E+U)·t − p∙x]} = a·e^{−(i/ħ)(E·t − p∙x)}·e^{(i/ħ)(μ_z·B·t)}

Of course, the magnetic field only acts from t = 0 to when the muon disintegrates, which we’ll denote by the point t = τ. So what we get is that the probability amplitude of a particle that’s been in a uniform magnetic field changes by a factor e⁽ⁱ^/ħ)(μ^_z^·B·τ). Note that it’s a factor indeed: we use it to multiply. You should also note that this is a complex exponential, so it’s a periodic function, with its real and imaginary part oscillating between zero and one. Finally, we know that μ_z can take on only certain values: for a spin-1/2 particle, they are plus or minus some number, which we’ll simply denote as μ, so that’s without the subscript, so our factor becomes:

e^{(i/ħ)(±μ·B·t)}

[The plus or minus sign needs to be explained here, so let’s do that quickly: we have two possible states for a spin-1/2 particle, one ‘up’, and the other ‘down’. But then we also know that the phase of our complex-valued wave function turns clockwise, which is why we have a minus sign in the exponent of our e^−iθexpression. In short, for the ‘up’ state, we should take the positive value, i.e. +μ, but the minus sign in the exponent of our e^−iθfunction makes it negative again, so our factor is e^{−(i/ħ)(μ·B·t)}for the ‘up’ state, and e^{+(i/ħ)(μ·B·t)}for the ‘down’ state.]

OK. We get that, but that doesn’t get us anywhere—yet. We need another trick first. One of the most fundamental rules in quantum-mechanics is that we can always calculate the amplitude to go from one state, say φ (read: ‘phi’), to another, say χ (read: ‘khi’), if we have a complete set of so-called base states, which we’ll denote by the index i or j (which you shouldn’t confuse with the imaginary unit, of course), using the following formula:

〈 χ | φ 〉 = ∑ 〈 χ | i 〉〈 i | φ 〉

I know this is a lot to swallow, so let me start with the notation. You should read 〈 χ | φ 〉 from right to left: it’s the amplitude to go from state φ to state χ. This notation is referred to as the bra-ket notation, or the Dirac notation. [Dirac notation sounds more scientific, doesn’t it?] The right part, i.e. | φ 〉, is the bra, and the left part, i.e. 〈 χ | is the ket. In our example, we wonder what the amplitude is for our muon staying in the +x state. Because that amplitude is time-dependent, we can write it as A₊(τ) = 〈 +x at time t = τ | +x at time t = 0 〉 = 〈 +x at t = τ | +x at t = 0 〉or, using a very misleading shorthand, 〈 +x | +x 〉. [The shorthand is misleading because the +x in the ket obviously means something else than the +x in the bra.]

But let’s apply the rule. We’ve got two states with respect to each coordinate axis only here. For example, in respect to the z-axis, the spin values are +z and −z respectively. [As mentioned above, we actually mean that the angular momentum in this direction is either +ħ/2 or −ħ/2, aka as ‘up’ or ‘down’ respectively, but then quantum theorists seem to like all kinds of symbols better, so we’ll use the +z and −z notations for these two base states here. So now we can use our rule and write:

A₊(t) = 〈 +x | +x 〉 = 〈 +x | +z 〉〈 +z | +x 〉 + 〈 +x | −z 〉〈 −z | +x 〉

You’ll say this doesn’t help us any further, but it does, because there is another set of rules, which are referred to as transformation rules, which gives us those 〈 +z | +x 〉 and 〈 −z | +x 〉 amplitudes. They’re real numbers, and it’s the same number for both amplitudes.

〈 +z | +x 〉 = 〈 −z | +x 〉 = 1/√2

This shouldn’t surprise you too much: the square root disappears when squaring, so we get two equal probabilities – 1/2, to be precise – that add up to one which – you guess it – they have to add up to because of the normalization rule: the sum of all probabilities has to add up to one, always. [I can feel your impatience, but just hang in here for a while, as I guide you through what is likely to be your very first quantum-mechanical calculation.] Now, the 〈 +z | +x 〉 = 〈 −z | +x 〉 = 1/√2 amplitudes are the amplitudes at time t = 0, so let’s be somewhat less sloppy with our notation and write 〈 +z | +x 〉 as C₊(0) and 〈 −z | +x 〉 as C₋(0), so we write:

〈 +z | +x 〉 = C₊(0) = 1/√2

〈 −z | +x 〉 = C₋(0) = 1/√2

Now we know what happens with those amplitudes over time: that e^{(i/ħ)(±μ·B·t)} factor kicks in, and so we have:

C₊(t) = C₊(0)·e^{−(i/ħ)(μ·B·t)} = e^{−(i/ħ)(μ·B·t)}/√2

C₋(t) = C₋(0)·e^{+(i/ħ)(μ·B·t)} = e^{+(i/ħ)(μ·B·t)}/√2

As for the plus and minus signs, see my remark on the tricky ± business in regard to μ. To make a long story somewhat shorter :-), our expression for A₊(t) = 〈 +x at t | +x 〉 now becomes:

A₊(t) = 〈 +x | +z 〉·C₊(t) + 〈 +x | −z 〉·C₋(t)

Now, you wouldn’t be too surprised if I’d just tell you that the 〈 +x | +z 〉 and 〈 +x | −z 〉 amplitudes are also real-valued and equal to 1/√2, but you can actually use yet another rule we’ll generalize shortly: the amplitude to go from state φ to state χ is the complex conjugate of the amplitude to to go from state χ to state φ, so we write 〈 χ | φ 〉 = 〈 φ | χ 〉*, and therefore:

〈 +x | +z 〉 = 〈 +z | +x 〉* = (1/√2)* = (1/√2)

〈 +x | −z 〉 = 〈 −z | +x 〉* = (1/√2)* = (1/√2)

So our expression for A₊(t) = 〈 +x at t | +x 〉 now becomes:

A₊(t) = e^{−(i/ħ)(μ·B·t)}/2 + e^{(i/ħ)(μ·B·t)}/2

That’s the sum of a complex-valued function and its complex conjugate, and we’ve shown more than once (see my page on the essentials, for example) that such sum reduces to the sum of the real parts of the complex exponentials. [You should not expect any explanation of Euler’s e^iθ = cosθ + i·sinθ rule at this level of understanding.] In short, we get the following grand result:

The big question, of course: what does this actually mean? 🙂 Well… Just square this thing and you get the probabilities shown below. [Note that the period of a squared cosine function is π, instead of 2π, which you can easily verify using an online graphing tool.]

Because you’re tired of this business, you probably don’t realize what we’ve just done. It’s spectacular and mundane at the same time. Let me quote Feynman to summarize the results:

“We find that the chance of catching the decay electron in the electron counter varies periodically with the length of time the muon has been sitting in the magnetic field. The frequency depends on the magnetic moment $μ$ . The magnetic moment of the muon has, in fact, been measured in just this way.”

As far as I am concerned, the key result is that we’ve learned how to work with those mysterious amplitudes, and the wavefunction, in a practical way, thereby using all of the theoretical rules of the quantum-mechanical approach to real-life physical situations. I think that’s a great leap forward, and we’ll re-visit those rules in a more theoretical and philosophical démarche in the next post. As for the example itself, Feynman takes it much further, but I’ll just copy the Grand Master here:

Huh? Well… I am afraid I have to leave it at this, as I discussed the precession of ‘atomic’ magnets elsewhere (see my post on precession and diamagnetism), which gives you the same formula: ω_p= μ·B/J (just substitute J for ±ħ/2). However, the derivation above approaches it from an entirely different angle, which is interesting. Of course, all fits. 🙂 However, I’ll let you do your own homework now. I hope to see you tomorrow for the mentioned theoretical discussion. Have a nice evening, or weekend – or whatever ! 🙂

Quantum mechanics and forces: the classical limit

Original post:

This post continues along the same line as the two previous ones: the objective is to get a better ‘feel’ for those weird amplitudes by showing how they are being affected by a potential − such as an electric potential, or a gravitational field, or a magnetic field − and how they relate to a classical analysis of a situation. It’s what Feynman tries to do too. The best example of such ‘familiarization exercises’, by far, is the comparison between a classical analysis of a force field, and the quantum-mechanical one, so let’s quickly go through that.

The classical situation is depicted below. We’ve got a particle moving in the x-direction and entering a region where the potential − again, think of an electrostatic field, for example, as it probably inspires the V (for voltage, I assume) – varies in the y-direction. As you know, the potential (energy) is measured by doing work against the force, and the work done by the force is the W = ∫F•ds integral along the path. Conversely, one can differentiate both sides of that W = ∫F•ds equation to get the force, so we can write: F = dW/ds. That explains the F = −∂V/∂y expression in the illustration below. Again, think of a positively charged particle that’s being attracted from a high or positive voltage (high V) to a lower or negative voltage.

As a result of the force, the particle is being accelerated transversely, i.e. in the y-direction, and so it adds some momentum in the y-direction (p_y) to the initial momentum p, which is all in the x-direction (p = p_x). Of course, the force only acts as the particle is passing through the area, so that’s during a time interval that’s equal to w/v, with w being the ‘width’ of the area. Now, the vertical momentum builds from 0 to p_y, and we can calculate p_y using Newton’s Law: F = m·a = m·(dv/dt) = d(m·v)/dt = dp/dt. It goes as follows:

We can then use a simple trigonometric formula to find the angle of deflection δθ:

Done! Now we need to show we get the same result with our quantum-mechanical approach or, at the very least, show how our quantum-mechanical approach actually works.

Of course, the assumption is that everything is on a very large scale as compared with the wavelength of our probability amplitudes. We then know that the presence of a potential V will just add to the energy E that we are to use in our wavefunction. Hence, in any small region, the amplitude will vary as:

a·e^{−(i/ħ)[(E_int+ p²/(2m) + V)·t − p∙x]}

Now, in the previous post, we explained that the energy conservation principle does not have any impact on the temporal variation of the wavefunction. Hence, E_int+ p²/(2m) + V effectively remains constant, and the effect of a changing V is on p, and works out through the −p∙x term in the exponent of our complex-valued wave function. To be specific, in a region where V is larger, p will be smaller and, therefore, we’ll have a larger wavelength λ = h/p. [Sorry if you can’t follow here: please do check out what I wrote in my two previous posts. It’s not that difficult.]

The illustration below shows how it works. The wavelength is the distance between successive wave nodes, which you should think of as surfaces where the phase of the amplitude is zero, or as wavefronts. So we’ve got a change in the angle of the wave nodes as well. How does that come about? Look at the two paths a and b: there’s a difference in potential between them, which we denote by ΔV, and which can be approximated as ΔV = (∂V/∂y)·D, with D the separation between the two paths.

Now, if E_int+ p²/(2m) + V is a constant, then ΔV must be equal to −Δ[p²/(2m)]. Now what can we do with that? We’re talking a differential really, so we should apply the Δf(x) = [df(x)/dx]·Δx formula, which gives us:

Δ[p²/(2m)] = (p/m)·Δp = −ΔV ⇔ Δp = −(m/p)·ΔV

[Note that the d[f(x)] = [df(x)/dx]·dx formula is an interesting one: you’ll surely have come across something like Δ[1/λ], and wondered what you could do with that. Now you know: Δ[1/λ] = Δλ/λ².]

Now, we’re interested in Δx, so how can we get some equation for that? Here we need to use the wavenumber k, and the derivation is not so easy—unfortunately! The (p/m)·Δp = −ΔV equation tells us that the wave number will be different along the two paths, and we can write the difference as Δk = Δ(p/ħ) = Δp/ħ. Now, we also now that the wavenumber is expressed in radians per unit distance. In addition, we also know that the wavenumber is the derivative of the phase with respect to distance, so we can approximately calculate the amount by which the phase on path b is ‘ahead’ of the phase on path a as the wave leaves the area, as:

Δ(phase) = Δk·w

Now, Δk = Δp/ħ, so Δ(phase) = Δp·w/ħ = −[m/(p·ħ)]·ΔV·w. Now, along path b, we will have some wavenumber k which, as per the definition of k, is equal to the derivative of the phase with respect to distance, so the ratio of the change in phase along path and the incremental distance Δx should equal the wave number k along path b, so we can write:

Δ(phase)/Δx = k ⇔ Δx = k·Δ(phase) = (λ/2π)·Δ(phase) = (ħ/p)·Δ(phase)

Combining that with Δ(phase) = −[m/(p·ħ)]·ΔV·w identity, we get:

Δx = −(ħ/p)·[m/(p·ħ)]·ΔV·w = −[m/p²]·ΔV·w

Phew! Just hang in there. We’re not done yet. I need to refer to the illustration now to show that Δx can also be approximated by D·δθ, so we write: Δx = D·δθ. [Again: it’s just plain triangle geometry, and the small angle approximation.] Combining both formulas for Δx yields:

δθ = −[m/p²]·ΔV·w/D

Now we can replace p/m by v and use our ΔV = (∂V/∂y)·D approximation to replace ΔV/D by (∂V/∂y), so we get:

δθ = −[1/(p·v)]·(∂V/∂y)·w = −[w/(p·v)]·(∂V/∂y)

So we get the very same equation for δθ as the one we got when applying classical mechanics. That’s a great result—also I have to admit that the differential analysis here was very painful and, hence, difficult to reproduce. In any case, what you should be interested in, is the result: as Feynman summarizes it,

“We get the same particle motions we get from Newton’s $F= m\cdota law$ , provided we assume that a potential contributes a phase to the probability amplitude equal to $V\cdott/ħ$ . Hence, in the classical limit, quantum mechanics agrees with Newtonian mechanics.”

However, he also cautions that “the result we have just got is correct only if the potential variations are slow and smooth”, which actually defines what is referred to as the “classical limit”.

Let me, to conclude this post, present another example out of Feynman’s ‘illustrations’ of what a field or a potential does with amplitudes. Trust me: we’ll need it later. As Feynman is really succinct here, I will just copy it. The quote describes what happens in a magnetic field, as opposed to the electrostatic field which I mentioned above. Just read it: it’s easy. [If it’s too small, just click on it, and the text size will be OK.] However, while it’s easy to understand, the consequences, on which I’ll write a separate post, are quite deep.

So… Well… That’s really it for today. 🙂

Potential energy and amplitudes: energy conservation and tunneling effects

Original post:

This post is intended to help you think about, and work with, those mysterious amplitudes. More in particular, I’ll explore how potential differences change amplitudes. But let’s first recapitulate the basics.

In my previous post, I explained why the young French Comte Louis de Broglie, when writing his PhD thesis back in 1924, i.e. before Schrödinger, Born, Heisenberg and others had published their work, boldly proposed to the ω·t − k·x argument in the wavefunction of a particle with the relativistic invariant product of the momentum and position four-vectors p_μ= (E, p) = (E, p_x, p_y, p_z,) and x_μ= (t, x) = (t, x, y, z), provided the energy and momentum are re-scaled in terms of ħ. Hence, he wrote:

θ = ω·t − k·x = (p_μx_μ)/ħ = (E∙t − p∙x)/ħ = (E/ħ)∙t − (p/ħ)∙x

As it’s usually instructive to do a quick dimensional analysis, let’s do one here too. Energy is expressed in joule, and dividing it by the quantum of action, which is expressed in joule·seconds (J·s) gives us the dimension of an (angular) frequency indeed, which, in turns, yields a pure number. Likewise, linear momentum can be expressed in newton·seconds which, when divided by joule·seconds (J·s), yields a quantity expressed per meter. Hence, the dimension of p/ħ is m^–1, which again yields a pure number when multiplied with the dimension of the coordinates x, y or z.

In the mentioned post, I also gave an unambiguous answer to the question as to what energy concept should be used in the equation: it is the total energy of the particle we are trying to describe, so that includes its kinetic energy, its rest mass energy and, finally, its potential energy in whatever force field it may find itself, such as a gravitational and/or electromagnetic force field. Now, while we know that, when talking potential energy, we have some liberty in choosing the zero point of our energy scale, this issue is easily overcome by noting that we are always talking about the amplitude to go from one state to another, or to go from one point in spacetime to another. Hence, what matters is the potential difference, really.

Feynman, in his description of the conservation of energy in a quantum-mechanical context, distinguishes:

The rest energy m₀∙c², which he describes as the rest energy ‘of the parts of the particle’. [One should remember he wrote this before the existence of quarks and the associated theory of matter was confirmed.]
The energy ‘over and above’ the rest energy, which includes both the kinetic energy, i.e. m∙v²/2 = p²/(2m), as well as the ‘binding and/or excitation energy’, which he refers to as ‘internal energy’.
Finally, there is the potential energy, which we’ll denote by U.

In my previous post, I also gave you the relativistically correct formula for the energy of a particle with momentum p:

However, we will follow Feynman in his description, who uses the non-relativistic formula E_p= E_int+ p²/(2m) + U. This is quite OK if we assume that the classical velocity of our particle does not approach the speed of light, so that covers a rather large number of real-life situations. Also, to make the example more real, we will assume the potential energy is electrostatic, and given by the formula U = q·Φ, with Φ the electrostatic potential (so just think of a number expressed in volt). Of course, q·Φ will be negative if the signs of q (i.e. the electric charge of our particle) and Φ are opposite, and positive if both have the same sign, as opposites attract and like repel when it comes to electric charge.

The illustration below visualizes the situation for Φ₂< Φ₁. For example, we may assume Φ₁ is zero, that Φ₂is negative, and that our particle is positively charged, so U₂= qΦ₂< 0. So it’s all rather simple really: we have two areas with a potential equal to U₁= qΦ₁and U₂= qΦ₂< 0 respectively. Hence, we need to use E₁= E_int+ p₁²/(2m) + U₁ to substitute ω₁for E₁/ħ in the first area, and then E₂= E_int+ p₂²/(2m) + U₂to substitute ω₂for E₂/ħ in the second area, which U₂– U₁< 0.

The corresponding amplitudes, or wavefunctions, are:

Ψ₁(θ₁) = Ψ₁(x, t) = a·e⁻ⁱ^θ^₁ = a·e^{−i[(E_int+ p₁²/(2m) + U₁)·t − p₁∙x]/ħ}
Ψ₂(θ₂) = Ψ₂(x, t) = a·e⁻ⁱ^θ^₂ = a·e^{−i[(E_int+ p₂²/(2m) + U₂)·t − p₂∙x]/ħ}

Now how should we think about these two equations? We are definitely talking different wavefunctions. However, having said that, there is no reason to assume the different potentials would have an impact on the temporal frequency. Therefore, we can boldly equate ω₁and ω₂and, therefore, write that:

E_int+ p₁²/(2m) + U₁= E_int+ p₂²/(2m) + U₂⇔ p₁²/(2m) − p₂²/(2m) = U₂– U₁< 0

⇒ p₁²− p₂²< 0 ⇔ p₂> p₁

What this says is that the kinetic energy, and/or the momentum, of our particle is greater in the second area, which is what we would classically expect, as a positive charged particle will pick up speed – and, therefore, momentum and kinetic energy – as it moves from an area with zero potential to an area with negative potential. However, the λ = h/p relation then implies that λ₂= h/p₂is smaller than λ₁= h/p₂, which is what is illustrated by the dashed lines in the illustration above – which represent surfaces of equal phase, or wavefronts – and also by the second diagram in the illustration, which shows the real part of the complex-valued amplitude and compares the wavelengths λ₁and λ₂. [As you know, the imaginary part is just like the real part but with a phase shift equal to π/2. Ideally, we should show both, but you get the idea.]

To sum it all up, the classical statement energy conservation principle is equivalent to the quantum-mechanical statement that the temporal frequency f or ω, i.e. the time-rate of change of the phase of the wavefunction, does not change – as long as the conditions do not change with time, of course – but that the spatial frequency, i.e. the wave number k or the wavelength λ – changes as the potential energy and/or kinetic energy change.

Tunneling

The p₁²/(2m) − p₂²/(2m) = U₂– U₁equation may be re-written to illustrate the quantum-mechanical effect of tunneling, i.e. the penetration of a potential barrier. Indeed, we can re-write p₁²/(2m) − p₂²/(2m) = U₂– U₁as

p₂² = 2m·[p₁²/(2m) − (U₂– U₁)]

and, importantly, try to analyze what happens if U₂– U₁ is larger than p₁²/(2m), so we get a negative value for p₂². Just imagine that Φ₁ is zero again, and that our particle is positively charged, but that Φ₂is also positive (instead of negative, as in the example above), so our particle is being repelled. In practical terms, it means that our particle just doesn’t have enough energy to “climb the potential hill”. Quantum-mechanically, however, the amplitude is still given by that equation above, and we have a purely imaginary number for p₂, as the square root of a negative number is a purely imaginary number, just like √−4 = 2i. So let’s denote p₂ as i·p’ and let’s analyze what happens by breaking our a·eⁱ^θ^₂ function up in two separate parts by writing: a·e⁻ⁱ^θ^₂ = a·e⁻ⁱ^[^{(E₂/ħ)∙t − (i·p’/ħ)x]} = a·e⁻ⁱ^{(E₂/ħ)∙t}·e^{i²·p’·x/ħ} = a·e⁻ⁱ^{(E₂/ħ)∙t}·e^{−p’·x/ħ}.

Now, the e^{−p’·x/ħ} factor in our formula for a·e⁻ⁱ^θ^₂ is a real-valued exponential function, and it’s a decreasing function, with the same shape as the general e^−x function, which I depict below.

This e^{−p’·x/ħ} basically ‘kills’ our wavefunction as we move in the positive x-direction, past the potential barrier, which is what is illustrated below.

However, the story doesn’t finish here. We may imagine that the region with the prohibitive potential is rather small—like a few wavelengths only—and that, past that region, we’ve got another region where p₂² = 2m·[p₁²/(2m) − (U₂– U₁)] is not negative. That’s the situation that’s depicted below, which also shows what might happen: the amplitude decays exponentially, but does not reach zero and, hence, there is a possibility that a particle might make it through the barrier, and that it will be found on the other side, with a real-valued and positive momentum and, hence, with a regular wavefunction.

Feynman gives a very interesting example of this: alpha-decay. Alpha decay is a type of radioactive decay in which an atomic nucleus emits an α-particle (so that’s a helium nucleus, really), thereby transforming or ‘decaying’ into an atom with a reduced mass and atomic number. The Wikipedia article on it hais not bad, but Feynman’s explanation is more to the point, especially when you’ve understood all of the above. The graph below illustrates the basic idea as it shows the potential energy U of an α-particle as a function of the distance from the center. As Feynman puts it: “If one tried to shoot an α-particle with the energy E into the nucleus, it would feel an electrostatic repulsion from the nucleus and would, classically, get no closer than the distance r₁, where its total energy is equal to U. Closer in, however, the potential energy is much lower because of the strong attraction of the short-range nuclear forces. How is it then that in radioactive decay we find α-particles which started out inside the nucleus coming out with the energy E? Because they start out with the energy E inside the nucleus and “leak” through the potential barrier.”

As for the numbers involved, the mean life of an α-particle in the uranium nucleus is as long as 4.5 billion years, according to Feynman, whereas the oscillations inside the nucleus are in the range of 10²² cycles per second! So how can one get a number like 10⁹ years from 10⁻²² seconds? The answer, as Feynman notes, is that that exponential gives a factor of about e⁻⁴⁵. So that gives the very small but definite probability of leakage. Once the α-particle is in the nucleus, there is almost no amplitude at all for finding it outside. However, if you take many nuclei and wait long enough, you’ll find one. 🙂

Now, that should be it for today, but let me end this post with something I should have told you a while ago, but then I didn’t, because I thought it would distract you from the essentials. If you’ve read my previous post carefully, you’ll note that I wrote the wavefunction as Ψ(θ) = a·eⁱ^θ, rather as a·e⁻ⁱ^θ, with the minus sign in front of the complex exponent. So why is that?

There is a long and a short answer to that. I’ll give the short answer. You’ll remember that the phase of our wavefunction is like the hand of a stopwatch. Now we could imagine a stopwatch going counter-clockwise, and we could actually make one. Now, there is no arbitrariness here: it’s one way or the other, depending on our other conventions, and the phase of our complex-valued wavefunction does actually turn clockwise if we write things the way we’re writing them, rather than anti-clockwise. That’s a direction that’s actually not as per the usual mathematical convention: an angle in the unit circle is usually measured counter-clockwise. If you’d want it that way, we can fix easily by reversing the signs inside of the bracket, so we could write θ = k·x − ω·t, which is actually what you’ll often see. But so there’s only way to get it right: there’s a direction to it, and if we use the θ = ω·t − k·x, then we need the minus sign in the Ψ(θ) = a·e⁻ⁱ^θ equation.

It’s just one of those things that is easy to state, but actually gives us a lot of food for thought. Hence, I’ll probably come back to this one day. As for now, however, I think you’ve had enough. Or I’ve had enough, at least. 🙂 I hope this was not too difficult, and that you enjoyed it.

Amplitudes, wavefunctions and relativity – or the de Broglie equation re-visited

Original post:

My previous posts were rather technical and, hence, I thought I’d re-visit a topic on which I’ve written before – but represent it from another angle: the de Broglie equation. You know it by heart: it associates a wavelength (λ) with the momentum (p) of a particle: λ = h/p. It’s a simple relationship: the wavelength and the momentum are inversely proportional, and the constant of proportionality is Planck’s constant. It’s an equation you’ll find in all of the popular accounts of quantum mechanics. However, I am of the opinion that the equation may actually not help novices to understand what quantum mechanics is all about—at least not in an initial approach.

One barrier to a proper understanding is that the de Broglie relation is always being presented as the twin of the Planck-Einstein relation for photons, which relates the energy (E) of a photon to its frequency (n): E = h∙ν = ħ∙ω [i]. It’s only natural, then, to try to relate the two equations, as momentum and energy are obviously related one to anotyher. But how exactly? What energy concept should we use? Potential energy? Kinetic energy? Should we include the equivalent energy of the rest mass?

One quickly gets into trouble here. For example, one can try the kinetic energy, K.E. = m∙v²/2 and use the definition of momentum (p = m∙v) to write E = p²/(2m), and then relate the frequency ν to the wavelength λ using the general rule that the traveling speed of a wave is equal to the product of its wavelength and its frequency (v = λ∙ν). But if E = p²/(2m) and ν = v/λ, we get:

p²/(2m) = h∙v/λ ⇔ λ = 2∙h/p

So that is almost right, but not quite: that factor 2 should not be there. In fact, it’s easy to see that we’d get de Broglie’s equation if we’d use E = m∙v² rather than E = m∙v²/2. But E = m∙v²? How could we possibly justify the use of that formula? There’s something weird here—something deep, but I will probably die before I figure out exactly what. 🙂

Note: I should make a reference to the argument of the wavefunction here: E·t −p·x. [The argument of the wavefunction has a 1/ħ factor in front, but we assume we measure both E as well as p in units of ħ here.] So that’s an invariant quantity, i.e. it doesn’t change under a relativistic transformation of the reference frame. Now, if we measure time and distance in equivalent units, so c = 1, then we can show that E/p = 1/v. [Remember: if c would not be one, we’d write: E·β = p·c, with β = v/c, i.e. the relative velocity of our particle, as measured as a ratio of the speed of light.] As E·t − p·x, is an invariant quantity, it’s some constant that’s characteristic of the particle. But x and t change as the clock is tick. Well… Yes. But if we believe the particle is somehow real, and its velocity is v, then the ratio of the real position x and the time t should be equal to v = x/t. Hence, for these very special positions x, i.e. the real position of the particle, we can equate E·t −p·x to E·t −p·v·t = E·t −m·v·v·t = (E − m∙v²)·t. So there we have the m∙v² factor. There must be something very deep about it, but, as mentioned above, I will probably die before I figure out exactly what. 🙂

The second problem is the interpretation of λ. Of course, λ is just a length in space, which we can relate to the spatial frequency or wavenumber k = 2π/λ [ii]. And, of course, the frequency and the wavelength are, once again, related through the traveling speed of the wave: v = λ∙ν. But then, when you think about it, it’s actually not that simple: the wavefunction of a particle is much more complicated. For starters, you should think of it as a wave packet, or wavetrain, i.e. a composite wave: a sum of a potentially infinite number of elementary waves. So we do not have a simple periodic phenomenon here: we need to distinguish the so-called group velocity from the phase velocity, and we’re also talking a complex-valued wavefunction, so it’s all quite different from what we’re used to.

But back to the energy concept. We have Einstein’s E = m∙c² = m₀∙γ∙c² equation, of course, from which the relativistically correct momentum-energy relationship can be derived [iii]:

E_pis the energy of a particle with momentum p, and the relationship establishes a one-to-one relationship between the energy and the momentum of a particle [iv], with the rest mass (or rest energy) E₀² = m₀∙c² appearing as a constant (the rest mass does not depend on the reference frame – per definition). However, you can try this formula too, but it will not give you the de Broglie relation. In short, it doesn’t help us in terms of understanding what the de Broglie relation is all about.

So how did this young nobleman, back in 1924, as he was writing his PhD thesis, get this λ = h/p or – using the wavenumber – the k = p/ħ equation?

Well… The relativity theory had been around for quite a while and, amongst other things (including the momentum-energy relationship above), it had also established the invariance of the four-vector product p_μx_μ = E∙t – p∙x = p_μ‘x_μ‘ = E’∙t’ – p’∙x’.

Now, any regular sinusoidal wave is associated with a phase θ = ωt – k∙x, and then quantum theory had associated a complex-valued wavefunction Ψ(θ) = Ψ(ωt – k∙x) with a particle, and so the young count, Louis de Broglie, saw the mathematical similarity between the E∙t – p∙x and ωt – k∙x expressions, and then just took the bold step of substituting ω and k for E/ħ and p/ħ respectively in the Ψ(ωt – k∙x) function. That’s it really: as the laws of physics should look the same, regardless of our frame of reference, he realized the argument of the wavefunction needed to be some invariant quantity, and so that’s what he gets through this substitution.

Of course, the substitution makes sense: it has to. To show why, let’s consider the limiting situation of a particle with zero momentum – so it’s at rest, really – but assuming that the probability of finding it at some point in space, at some point of time is equally distributed over space and time. So we have the rather nonsensical but oft-used wavefunction:

Ψ(θ) = Ψ(x, t) = a·e^iθ = a·e^{i(ωt – k∙x)}

Taking the absolute square of this yields a constant probability equal to a²indeed. The equation is non-sensical or – to put it more politely – a limiting case only because it assumes perfect knowledge about the particle’s momentum p (we said it was zero, exactly) and, hence, about its energy. So we don’t need to worry about any Δ here: Δp = ΔE = 0 and E_p= E₀. As mentioned, we know that doesn’t make much sense, and that a particle in free space will actually be represented by a wave train with a group velocity v (i.e. the classical speed of the particle) and a phase velocity, and so that’s where the modeling of uncertainty comes in, but let’s just go along with the example now. [Note that, while p = 0, that does not imply that E₀ is equal to zero. Indeed, there’s energy in the rest mass and possibly potential energy too!]

Now, if we substitute ω and k for E/ħ and p/ħ respectively – note that we did away with the bold-face k and x, so we’ve reduced the analysis to one dimension (x) only – we get:

Ψ(θ) = Ψ(x, t) = a·e^iθ = a·e^{i(E₀t – p∙x)/ħ}= a·e^i(E₀/ħ)t

What this means is that the phase θ = (E₀/ħ)·t does not depend on x: it only varies in time. Hence, the diagram below – don’t look at the x’ and t’ right now: that’s another reference frame that we’ll introduce in a moment – shows equal-phase lines parallel to the x-axis and, because of the θ = (E₀/ħ)·t equation, they’re equally spaced in time, i.e. in the t-coordinate.

Of course, a particle at rest in one reference frame – let’s say S – will appear to be moving in another – which we’ll denote as S’, so we have the primed coordinates x’ and t’, which are related to the x and t by the Lorentz transformation rules. It’s easy to see that the points of equal phase have a different spacing along the t’-axis, so the frequency in time must be different. Indeed, we’ll write that frequency as ω’ = E_p‘/ħ in the S’ reference frame.

Likewise, we see that the phase now does vary in space, so the probability amplitude does vary in space now, as the particle’s momentum in the primed reference frame is no longer zero: if we write it as p’, then we can write that θ = (E₀/ħ)·t = (E_p‘/ħ)·t − (p/ħ)∙x = (E_p‘/ħ)·t − (p/ħ)∙x. Therefore, our wavefunction becomes:

Ψ(θ) = Ψ(x, t) = a·e^iθ = a·e^i(E₀/ħ)t= Ψ(x’, t’) = a·e^{i(E_p‘·t’ − p’∙x’)/ħ}

We could introduce yet another reference frame, and we’d get similar results. The point is: the k in our Ψ(θ) = Ψ(x, t) = a·e^iθ = a·e^{i(ωt – k∙x)}equation is, effectively, equal to k = p/ħ, and that identity holds in any reference frame.

Now, none of what I wrote above actually proves the de Broglie relation: it merely explains it. When everything is said and done, the de Broglie relation is a hypothesis, but it is an important one—and it does fit into the overall quantum-mechanical or wave-mechanical approach, that physicists take for granted nowadays.

So that’s it, really. I have nothing more to write but I should, perhaps, just remind you about what I said about a ‘particle wave’: we should look at it as some composite wave. Indeed, there will be uncertainty, and the uncertainty in E implies a frequency range Δω = Δ(E/ħ) = ΔE/ħ. Likewise, the momentum will be unknown, and so we’ll have a spread in the wavenumber k as well. We write: Δk = Δ(p/ħ) = Δp/ħ. We can try to reduce this uncertainty, but the Uncertainty Principle gives us the limits: the Δp·Δx and/or the ΔE·Δt products cannot be smaller than ħ/2.

So we’ll have a potentially infinite number of waves with slightly different values for ω and k, whose sum may be visualized as a complex-valued traveling wavetrain, like the lump below.

All of the component waves necessarily need to travel at the same speed, and this speed, which is the ratio of ω and k, will be equal to the so-called phase velocity of the wave (v_p), which we can calculate as v_p= ω/k = (E/ħ)/(p/ħ) = E/p = (m·c²)/(m·v) = c²/v. This speed is superluminal (c²/v = c/β, with β = v/c < 1), but that is not in contradiction with special relativity because the phase velocity carries no ‘signal’ or ‘information’. As for the group velocity (v_g), we can effectively see that this is equal to the classical velocity of our ‘particle’ by noting that:

v_g= dω/dk = d(E/ħ)/d(p/ħ) = dE/dp = d[(p²/(2m)]/dp = p/m = v.

You may think there’s some cheating here, as we equate E with the kinetic energy only. You’re right: the total energy should also include potential energy and rest energy, so E is a sum, but then rest energy and potential energy are treated as constants and, hence, it’s only the kinetic energy that matters when taking the derivative with respect to p, and so that’s why get the result we get, which makes perfect sense.

I wanted this to be a very short post, and so I will effectively end it here. I hope you enjoyed it. It actually sets the stage for a more interesting discussion, and that’s a discussion on how a change in potential energy effectively changes the phase and, hence, the amplitude. But so that’s for next time. I also need to devote a separate post on a discussion of the wave-mechanical framework in general, with a particular focus on the math behind. So… Well… Yes, the next posts are likely to be somewhat more technical again.

[i] I should make a note on notations here, and also insert some definitions. I will denote a frequency by ν (nu), rather than by f, so as to not cause confusion with any function f. A frequency is expressed in cycles per second, while the angular frequency ω is expressed in radians per second. One cycle covers 2π radians and, therefore, we can write: ν = ω/2π. Hence, h∙ν = h∙ω/2π = ħ∙ω. Both ν as well as ω measure the time-rate of change of the phase, as opposed to k, i.e. the spatial frequency of the wave, which depends on the speed of wave. I will also use the symbol v for the speed of a wave, although that is hugely confusing, because I will also use it to denote the classical velocity of the particle. However, I find the conventional use of the symbol of c even more confusing, because this symbol is also used for the speed of light, and the speed of a wave is not necessarily equal to the speed of light. In fact, both the group as well as the phase velocity of a particle wave are very different from the speed of light. The speed of a wave and the speed of light only coincide for electromagnetic waves and, even then, it should be noted that photons also have amplitudes to travel faster or slower than the speed of light.

[ii] Note that the de Broglie relation can be re-written as k = p/ħ, with ħ = h/2π. Indeed, λ = 2π/k and, hence, we get: λ = h/p ⇔ 2π/k = h/p ⇔ p = ħ∙k ⇔ k = p/ħ.

[iii] One gets the equation by equating m to m = m₀γ (γ is the Lorentz factor) in the E² = (mc²)² equation, and re-arranging.

[iv] Note that this energy formula does not include any potential energy: it is (equivalent) rest mass plus kinetic energy only.

Atomic magnets: precession and diamagnetism

This and the next posts will further build on the concepts introduced in my previous post on particle spin. This post in particular will focus on some of the math we’ll need to understand what quantum mechanics is all about. The first topic is about the quantum-mechanical equivalent of the phenomenon of precession. The other topics are… Well… You’ll see… 🙂

The Larmor frequency

The motion of a spinning object in a force field is quite complicated. In our post on gyroscopes, we introduced the concepts of precession and nutation. The concept of precession is illustrated below for the Earth as well as for a spinning top. In both cases, the external force is just gravity.

Nutation is an additional movement: on top of the precessional movement, a spinning object may wobble, as illustrated below.

There seems to be no analog for nutation in quantum mechanics. In fact, the terms nutation and precession seem to be used interchangeably in quantum physics, although they are very different in classical physics. But let’s not complicate things and, hence, talk about the phenomenon of precession only.

We will not re-explain the phenomenon of precession here but just remind you that the phenomenon can be described in terms of (a) the angle between the symmetry axis and the momentum vector, which we’ll denote by θ, and (b) the angular velocity of the precession, which we’ll denote by ω_p= dφ/dt, as shown below. The J in the illustration below is the angular momentum of the object. Hence, if we’d imagine it to be an electron, then J would be the spin angular momentum only, not its orbital angular momentum—although the analysis would obviously be valid for the orbital and/or total angular momentum as well.

OK. Let’s look at what’s going on. The angular displacement – which is also, rather confusingly, referred to as the angle of precession – in the time interval Δt is, obviously, equal to Δφ = ω_p·Δt. Now, looking at the geometry of the situation, and using the small-angle approximation for the sine, one can also see that ΔJ ≈ (J·sinθ)·(ω_p·Δt). In fact, going to the limit (i.e. for infinitesimally small Δφ and ΔJ), we can write:

dJ/dt = ω_p·J·sinθ

But the angular momentum cannot change if there’s no torque. In fact, the time rate of change of the angular momentum is equal to the torque. [You should look this up but, if you don’t want to do that, note that this is just the equivalent, for rotational motion, of the F = dp/dt law for linear motion.] Now, in my post on magnetic dipoles, I showed that the torque τ on a loop of current with magnetic moment μ in an external magnetic field B is equal to τ = μ×B. So the magnitude of the torque is equal to |τ| = |μ|·|B|·sinθ = μ·B·sinθ. Therefore, ω_p·J·sinθ = μ·B·sinθ and, hence,

ω_p= μ·B/J

However, from the general μ/J = –g·(q_e/2m) equation we derived in our previous post, we know that μ/J – for an atomic magnet, that is – must be equal to μ/J = g·q_e/2m. So we get the formula we wanted to get here:

ω_p= g·(q_e/2m)·B

This equation says that the angular velocity of the precession is proportional to the magnitude of the external magnetic field, and that the constant of proportionality is equal to g·(q_e/2m). It’s good to do the math and actually calculate the precession frequency f_p = ω_p/2π. It’s easy. We had calculated q_e/2m already: it was equal to 1.6×10⁻¹⁹ C divided by 2·9.1×10⁻³¹kg, so that’s 0.0879×10¹² C/kg or 0.0879×10¹²(C·m)/(N·s²), more or less. 🙂 Now, g is dimensionless, and B is expressed in tesla: 1 T = (N·s)/(C·m), so we get the s⁻¹ dimension we want for a frequency. For g = 2 (so we look at the spin of the electron itself only), we get:

f_p = ω_p/2π = 2·0.0879×10¹²/2π ≈ 28×10⁹= 28 gigacycles per tesla = 28 GHz/T

This is a number expressed per unit of the magnetic field strength B. Note that you’ll often see this number expressed as 1.4 megacycles per gauss, using the older gauss unit for magnetic field strength: 1 tesla = 10,000 gauss. For a nucleus, we get a somewhat less impressive number because the proton (or neutron) mass is so much bigger: it’s a number expressed in megacycles per tesla, indeed, and for a proton (i.e. a hydrogen nucleus), it’s about 42.58 MHz/T.

Now, you may wonder about the numbers here. Are they astronomical? Maybe. Maybe not. It’s probably good to note that the strength of the magnetic field in medical MRI systems (magnetic resonance imaging systems) is only 1.5 to 3 tesla, so it’s a rather large unit. You should also note that the clock speed of the CPU in your laptop – so that’s the speed at which it executes instructions – is measured in GHz too, so perhaps it’s not so astronomic. I’ll let you judge. 🙂

So… Well… That’s all nice. The key question, of course, is whether or not this classical view of the electron spinning around a proton is accurate, quantum-mechanically, that is. I’ll let Feynman answer that question provisionally:

“According to the classical theory, then, the electron orbits—and spins—in an atom should precess in a magnetic field. Is it also true quantum-mechanically? It is essentially true, but the meaning of the “precession” is different. In quantum mechanics one cannot talk about the direction of the angular momentum in the same sense as one does classically; nevertheless, there is a very close analogy—so close that we continue to call it precession.”

To distinguish classical and quantum-mechanical precession, quantum-mechanical precession is usually referred to as Larmor precession, and the frequencies above are often referred to as Larmor frequencies. However, I should note that, technically speaking, the term Larmor frequency is actually reserved for the frequency I’ll describe in the next section. I should also note that the ω_p= g·(q_e/2m)·B is usually written, quite simply, as ω_p= γ·B. Of course, the gamma is not the Lorentz factor here, but the so-called gyromagnetic ratio (aka as the magnetogyric ratio): γ = g·(q_e/2m). Oh—just so you know: Sir Joseph Larmor was a British physicists and, yes, he developed all of the stuff we’re talking about here. 🙂

At this point, you may wonder if and why all of the above is relevant. Well… There’s more than one answer to this question, but I’d recommend you start with reading the Wikipedia article on NMR spectroscopy. 🙂 And then you should also read Feynman’s exposé on the Rabi atomic or molecular beam method for determining the precession frequency. It’s really fascinating stuff, but you are sufficiently armed now to read those things for yourself, and so I’ll just move on. Indeed, there’s something else I need to talk about here, and that’s Larmor’s Theorem.

Larmor’s Theorem

We’ve been talking single electrons only so far. Now, you may fear that things become quite complicated when many electrons are involved and… Well… That’s true, of course. And then you may also think that things become even more complicated when external fields are involved, like that external magnetic field we introduced above, and that led our electrons to precess at extraordinary frequencies. Well… That’s not true. Here we get some help: Larmor proved a theorem that basically says that, if we can work out the motions of the electrons without the external field, the solution for the motions with the external field is the no-field solution with an added rotation about the axis of the field. More specifically, for an external magnetic field, the added rotation will have an angular frequency equal to:

ω_L= (q_e/2m)·B

So that’s the same formula as we found for the angular velocity of the precession if g = 1, so that’s very easy to remember. The ω_L frequency, which is the precession frequency for g = 1, is referred to as the Larmor frequency. The proof of the above is remarkably easy, but… Well… I don’t want to copy Feynman here, so I’ll just refer you to the relevant Lecture on it. 🙂

Diamagnetism

I guess it’s about time we relate all of what we learned so far to properties of matter we can relate to, and so that’s what I’ll do here. We’re not going to talk about ferromagnetism here, i.e. the mechanism through which iron, nickel and cobalt and most of their alloys become permanent magnets. That’s quite peculiar and so we will not discuss it here. Here we’ll talk about the very weak quantum-mechanical magnetic effect – a thousand to a million times less than the effects in ferromagnetic materials – that occurs in all materials when placed in an external magnetic field.

While the effect is there in all materials, it’s stronger for some than for others. In fact, it’s usually so weak it is hard to detect, and so it’s usually demonstrated using elements for which the diamagnetic effect is somewhat stronger, like bismuth or antimony. The effect is demonstrated by suspending a piece of material in a non-uniform field, as illustrated below. The diamagnetic effect will cause a small displacement of the material, away from the high-field region, i.e. away from the pointed pole.

I should immediately add that some materials, like aluminium, will actually be attracted to the pointed pole, but that’s because of yet another effect that not all materials share: paramagnetism. I’ll talk about that in another post, together with ferromagnetism. So… Diamagnetism: what is it?

The illustration below shows our spinning electron (q) once again. It also shows a magnetic field B but, unlike our analysis above, or the analysis in our previous post, we assume the external magnetic field is not just there. We assume it changes, because it’s been turned on or off—hopefully slowly: if not, we’d have eddy-current forces causing potentially strong impulses.

But so we’ve got some change in the magnetic flux , and so we know, because of Faraday or Maxwell – you choose 🙂 – that we’ll have some circulation of E, i.e. the electric field. The magnetic flux is B times the surface area, and the circulation is the average tangential component E times the length of the path. Because our model of the orbiting electron is so nice and symmetric, we can write Faraday’s Law here as:

E·2π·r = −d(B·π·r²)/dt ⇔ E = −(r/2)·dB/dt

A field implies a force and, therefore, a torque on the electron. The torque is equal to the force times the lever arm, so it’s equal to (−q_e·E)·r = −q_e·E·r. Of course, the torque is also equal to the rate of the change of the angular momentum, so dJ/dt must equal:

dJ/dt = −q_e·E·r = q_e·(r/2)·(dB/dt)·r = (q_e·r²/2)·(dB/dt)

Now, the assumption is that the field goes from zero to B, so ΔB = B. Therefore, ΔJ must be equal to:

ΔJ = (q_e·r²/2)·B

You should, in fact, derive this more formally, by integrating—but let’s keep things as simple as we can. 🙂 What does this formula say, really? It’s the extra angular momentum from the ‘twist’ that’s given to the electrons as the field is turned on. Now, this added angular momentum makes an extra magnetic moment which, because it is an orbital motion, is just q_e $/2m$ times the angular momentum that’s already there. But more angular momentum means the magnetic moment has changed, according to the μ = (q_e/2m)·J formula we derived in our previous post, so we have:

Δμ = –(q_e/2m)·ΔJ

The minus sign is there because of Lenz’ law: the added momentum is opposite to the magnetic field—and, yes, I know: it’s hard to keep track of all of the conventions involved here. In any case, we get the following grand equation:

So we found that the induced magnetic moment is directly proportional to the magnetic field $B,$ and opposing it. Now that is what explains why our piece of bismuth does what it does in that non-uniform magnetic field. Of course, you’ll say: why is stronger for bismuth than for other materials? And what about aluminium, or paramagnetism in general? Well… Good questions, but we’ll tackle them in the next posts. 🙂

Let me conclude this post by copying Feynman’s little exposé on why the phenomenon of diamagnetism is so particular. In fact, he notes that, because we’re talking a piece of material here that can’t spin – so it’s held in place, so to say – we should have “no magnetic effects whatsoever”. The reasoning is as follows:

This is very interesting indeed. This classical theorem basically says that the energy of a system should not be affected by the presence of a magnetic field. However, we know magnetic effects, such as the diamagnetic effect, are there, so these effects are referred to as ‘quantum-mechanical’ effects indeed: they cannot be explained using classical theory only, even if all of what we wrote above used classical theory only.

I should also note another point: why do we need a non-homogeneous field? Well… The situation is comparable to what we wrote on the Stern-Gerlach experiment. If we would have a homogeneous magnetic field, then we would only have a torque on all of the atomic magnets, but no net force in one or the other direction. There’s something else here too: you may think that the forces pointing towards and away from the pointed tip should cancel each other out, so there should actually be no net movement of the material at all! Feynman’s analysis works for one atom, indeed, but does it still make sense if we look at the whole piece of material? It does, because we’re talking an induced magnetic moment that’s opposing the field, regardless of the orientation of the magnetic moment of the individual atoms in the piece of material. So, even if the individual atoms have opposite momenta, the extra induced magnetic moment will point in the same direction for all. So that solves that issue. However, it does not address Feynman’s own critical remark in regard to the supposed ‘impossibility’ of diamagnetism in classical mechanics.

But I’ll let you think about this, and sign off for today. 🙂 I hope you enjoyed this post.

Spin and angular momentum in quantum mechanics

Note: A few years after writing the post below, I published a paper on the anomalous magnetic moment which makes (some of) what is written below irrelevant. It gives a clean classical explanation for it. Have a look by clicking on the link here !

Original blog post:

Feynman starts his Volume of Lectures on quantum mechanics (so that’s Volume III of the whole series) with the rules we already know, so that’s the ‘special’ math involving probability amplitudes, rather than probabilities. However, these introductory chapters assume theoretical zero-spin particles, which means they don’t have any angular momentum. While that makes it much easier to understand the basics of quantum math, real elementary particles do have angular momentum, which makes the analysis much more complicated. Therefore, Feynman makes it very clear, after his introductory chapters, that he expects all prospective readers of his third volume to first work their way through chapter 34 and 35 of the second volume, which discusses the angular momentum of elementary particles from both a classical as well as a quantum-mechanical perspective. So that’s what we will do here. I have to warn you, though: while the mentioned two chapters are more generous with text than other textbooks on quantum mechanics I’ve looked at, the matter is still quite hard to digest. By way of introduction, Feynman writes the following:

“The behavior of matter on a small scale—as we have remarked many times—is different from anything that you are used to and is very strange indeed. Understanding of these matters comes very slowly, if at all. One never gets a comfortable feeling that these quantum-mechanical rules are ‘natural’. Of course they are, but they are not natural to our own experience at an ordinary level. The attitude that we are going to take with regard to this rule about angular momentum is quite different from many of the other things we have talked about. We are not going to try to ‘explain’ it but tell you what happens.”

I personally feel it’s not all as mysterious as Feynman claims it to be, but I’ll let you judge for yourself. So let’s just go for it and see what comes out. 🙂

Atomic magnets and the g-factor

When discussing electromagnetic radiation, we introduced the concept of atomic oscillators. It was a very useful model to help us understand what’s supposed to be going on. Now we’re going to introduce atomic magnets. It is based on the classical idea of an electron orbiting around a proton. Of course, we know this classical idea is wrong: we don’t have nice circular electron orbitals, and our discussion on the radius of an the electron in our previous post makes it clear that the idea of the electron itself is rather fuzzy. Nevertheless, the classical concepts used to analyze rotation are also used, mutatis mutandis, in quantum mechanics. Mutatis mutandis means: with necessary alterations. So… Well… Let’s go for it. 🙂 The basic idea is the following: an electron in a circular orbit is a circular current and, hence, it causes a magnetic field, i.e. a magnetic flux through the area of the loop—as illustrated below.

As such, we’ll have a magnetic (dipole) moment, and you may want to review my post(s) on that topic so as to ensure you understand what follows. The magnetic moment (μ) is the product of the current (I) and the area of the loop (π·r²), and its conventional direction is given by the μ vector in the illustration below, which also shows the other relevant scalar and/or vector quantities, such as the velocity v and the orbital angular momentum J. The orbital angular momentum is to be distinguished from the spin angular momentum, which results from the spin around its own axis. So the spin angular momentum – which is often referred to as the spin tout court – is not depicted below, and will only be discussed in a few minutes.

Let me interject something on notation here. Feynman’s always uses J, for whatever momentum. That’s not so usual. Indeed, if you’d google a bit, you’ll see the convention is to use S and L respectively to distinguish spin and orbital angular momentum respectively. If we’d use S and L, we can write the total angular momentum as J = S + L, and the illustration below shows how the S and L vectors are to be added. It looks a bit complicated, so you can skip this for now and come back to it later. But just try to visualize things:

The L vector is moving around, so that assumes the orbital plane is moving around too. That happens when we’d put our atomic system in a magnetic field. We’ll come back to that. In what follows, we’ll assume the orbital plane is not moving.
The S vector here is also moving, which also assumes the axis of rotation is not steady. What’s going on here is referred to as precession, and we discussed it when presenting the math one needs to understand gyroscopes.
Adding S and L yields J, the total angular momentum. Unsurprisingly, this vector wiggles around too. Don’t worry about the magnitudes of the vectors here. Also, in case you’d wonder why the axis of symmetry for the movement of the J vector happens to be the J_z axis, the answer is simple: we chose the coordinate system so as to ensure that was the case.

But I am digressing. I just inserted the illustration above to give you an inkling of where we’re going with this. Indeed, what’s shown above will make it easier for you to see how we can generalize the analysis that we’ll do now, which is an analysis of the orbital angular momentum and the related magnetic moment only. Let me copy the illustration we started with once more, so you don’t have to scroll up to see what we’re talking about.

So we have a charge orbiting around some center. It’s a classical analysis, and so it’s really like a planet around the Sun, except that we should remember that likes repel, and opposites attract, so we’ve got a minus sign in the force law here.

Let’s go through the math. The magnetic moment is the current times the area of the loop. As the velocity is constant, the current is just the charge q times the frequency of rotation. The frequency of rotation is, of course, the velocity (i.e. the distance traveled per second) divided by the circumference of the orbit (i.e. 2π·r). Hence, we write: I = (q_e·v)/(2π·r) and, therefore: μ = (q_e/·v)·π·r²)/(2π·r) = q_e·v·r/2. Note that, as per the convention, current is defined as a flow of positive charges, so the illustration above actually assumes we’re talking some proton in orbit, so q = q_e would be the elementary charge +1. If we’d be talking an electron, then its charge is to be denoted as –q_e(minus q_e, i.e. −1), and we’d need to reverse the direction of μ, which we’ll do in a moment. However, to simplify the discussion, you should just think of some positive charge orbiting the way it does in the illustration above.

OK. That’s all there’s to say about the magnetic moment—for the time being, that is. Let’s think about the angular momentum now. It’s orbital angular momentum here, and so that’s the type of angular momentum we discussed in our post on gyroscopes. We denoted it as L indeed – i.e. not as J, but that’s just a matter of conventions – and we noted that L could be calculated as the vector cross product of the position vector r and the momentum vector p, as shown in the animation below, which also shows the torque vector τ.

The angular momentum L changes in the animation above. In our J case above, it doesn’t. Also, unlike what’s happening with the angular momentum of that swinging ball above, the magnitude of our J doesn’t change. It remains constant, and it’s equal to |J| = J = |r×p| = |r|·|p|·sinθ = r·p = r·m·v. One should note this is a non-relativistic formula, but as the relative velocity of an electron v/c is equal to the fine-structure constant, so that’s α ≈ 0.0073 (see my post on the fine-structure constant if you wonder where this formula comes from), it’s OK to not include the Lorentz factor in our formulas as for now.

Now, as I mentioned already, the illustration we’re using to explain μ and J is somewhat unreal because it assumes a positive charge q, and so μ and J point in the same direction in this case, which is not the case if we’d be talking an actual atomic system with an electron orbiting around a proton. But let’s go along with it as for now and so we’ll put the required minus sign in later. We can combine the J = r·m·v and μ = q·v·r/2 formulas to write:

μ = (q/2m)·J or μ/J = (q/2m) (electron orbit)

In other words, the ratio of the magnetic moment and the angular moment depends on (1) the charge (q) and (2) the mass of the charge, and on those two variables only. So the ratio does not depend on the velocity v nor on the radius r. It can be noted that the q/2m factor is often referred to as the gyromagnetic factor (not to be confused with the g-factor, which we’ll introduce shortly). It’s good to do a quick dimensional check of this relation: the magnetic moment is expressed in ampère per second times the loop area, so that’s (C/s)·m². On the right-hand side, we have the dimension of the gyromagnetic factor, which is C/kg, times the dimension of the angular momentum, which is m·kg·m/s, so we have the same units on both sides: C·m²/s, which is often written as joule per tesla (J/T): the joule is the energy unit (1 J = 1 N·m), and the tesla measures the strength of the magnetic field (1 T = 1 (N·s)/(C·m). OK. So that works out.

So far, so good. The story is a little bit different for the spin angular momentum and the spin magnetic moment. The formula is the following:

μ = (q/m)·J (electron spin)

This formula says that the μ/J ratio is twice what it is for the orbital motion of the electron. Why is that? Feynman says “the reasons are pure quantum-mechanical—there is no classical explanation.” So I’d suggest we just leave that question open for the moment and see if we’re any wiser once we’ve worked ourselves through all of his Lectures on quantum physics. 🙂 Let’s just go along with it as for now.

Now, we can write both formulas – i.e. the formula for the spin and the orbital angular momentum – in a more general way using the format below:

μ = –g·(q_e/2m_e)·J

Why the minus sign? Well… I wanted to get the sign right this time. Our model assumed some positive charge in orbit, but so we want a formula for a atomic system, and so our circling charge should be an electron. So the formula above is the formula for a electron, and the direction of the magnetic moment and of the angular motion will be opposite for electrons: it just replaces q by –q_e. The format above also applies to any atomic system: as Feynman writes, “For an isolated atom, the direction of the magnetic moment will always be exactly opposite to the direction of the angular momentum.” So the $g-factor will be$ characteristic of the state of the atom. It will be $1$ for a pure orbital moment, $2$ for a pure spin moment, or some other number in-between for a complicated system like an atom, indeed.

You may have one last question: why q_e/2m instead of q_e/m in the middle? Well… If we’d take q_e/m, then g would be 1/2 for the orbital angular momentum, and the initial idea with g was that it would be some integer (we’ll quickly see that’s an idea only). So… Well… It’s just one more convention. Of course, conventions are not always respected so sometimes you’ll see the expression above written without the minus sign, so you may see it as μ = g·(q_e/2m_e)·J. In that case, the g-factor for our example involving the spin angular momentum and the spin magnetic moment, will obviously have to be written as minus 2.

Of course, it’s easy to see that the formula for the spin of a proton will look the same, except that we should take the mass of the proton in the formula, so that’s m_pinstead of m_e. Having said that, the elementary charge remains what it is, but so we write it without the minus sign here. To make a long story short, the formula for the proton is:

μ = g·(q_e/2m_p)·J

OK. That’s clear enough. For electrons, the g-factor is referred to as the Landé g-factor, while the g-factor for protons or, more generally, for any spinning nucleus, is referred to as the nuclear g-factor. Now, you may or may not be surprised, but there’s a g-factor for neutrons too, despite the fact that they do not carry a net charge: the explanation for it must have something to do with the quarks that make up the neutron but that’s a complicated matter which we will not get into here. Finally, there is a g-factor for a whole atom, or a whole atomic system, and that’s referred to as… Well… Just the g-factor. 🙂 It’s, obviously, a number that’s characteristic of the state of the atom.

So… This was a big step forward. We’ve got all of the basics on that ‘magical’ spin number here, and so I hope it’s somewhat less ‘magical’ now. 🙂 Let me just copy the values of the g-factor for some elementary particles. It also shows how hard physicists have been trying to narrow down the uncertainty in the measurement. Quite impressive! The table comes from the Wikipedia article on it. I hope the explanations above will now enable you to read and understand that. 🙂

Let’s now move on to the next topic.

Spin numbers and states

Of course, we’re talking quantum mechanics and, therefore, J can only take on a finite number of values. While that’s weird – as weird as other quantum-mechanical things, such as the boson-fermion math, for example – it should not surprise us. As we will see in a moment, the values of J will determine the magnetic energy our system will acquire when we put in some magnetic field and, as Feynman writes: “That the energy of an atom in the magnetic field can have only certain discrete energies is really not more surprising than the fact that atoms in general have only certain discrete energy levels. Why should the same thing not hold for atoms in a magnetic field? It does. It is just correlation of this with the idea of an oriented magnetic moment that brings out some of the strange implications of quantum mechanics.” Yep. That’s true. We’ll talk about that later.

Of course, you’ll probably want some ‘easier’ explanation. I am afraid I can’t give you that. All I can say is that, perhaps, you should think of our discussion on the fine-structure constant, which made it clear that the various radii of the electron, its velocity and its mass and/or energy are all related one to another and, hence, that they can only take on certain values. Indeed, of all the relations we discussed, there’s two you should always remember. The first relationship is the U= (e²/r) = α/r. So that links the energy (which we can express in equivalent mass units), the electron charge and its radius. The second thing you should remember is that the Bohr radius and the classical electron radius are also related through α: α r_e/r = α². So you may want to think of the different values for J as being associated with different ‘orbitals’, so to speak. But that’s a very crude way of thinking about it, so I’d say: just accept the fact and see where it leads us. You’ll see, in a few moments from now, that the whole thing is not unlike the quantum-mechanical explanation of the blackbody radiation problem, which assumes that the permitted energy levels (or states) are equally spaced and h·f apart, with f the frequency of the light that’s being absorbed and/or emitted. So the atom takes up energies only h·f at a time. Here we’ve got something similar: the energy levels that we’ll associate with the discrete values of J – or J‘s components , I should say – will also be equally spaced. Let me show you how it works, as that will make things somewhat more clear.

If we have an object with a given total angular momentum J in classical mechanics, then any of its components x, y or z, could take on any value from +J to −J. That’s not the case here. The rule is that the ‘system’ – the atom, the nucleus, or anything really – will have a characteristic number, which is referred to as the ‘spin’ of the system and, somewhat confusingly, it’s denoted by j (as you can, it’s extremely important, indeed, to distinguish capital letters (like J) from small letters (like j) if you want to keep track of what we’re explaining here). Now, if we have that characteristic spin number j, then any component of J (think of the z-direction, for example) can take on only (one of) the following values:

Note that we will always have 2j + 1 values. For example, if j = 3/2, we’ll have 2·(3/2) + 1 = 4 permitted values, and in the extreme case where j is zero, we’ll still have 2·0 + 1 = 1 permitted value: zero itself. So that basically says we have no angular momentum. […] OK. That should be clear enough, but let’s pause here for a moment and analyze this—just to make sure we ‘get’ this indeed. What’s being written here? What are those numbers? Let’s do a quick dimensional analysis first. Because j, j − 1, j − 2, etcetera are pure numbers, it’s only the dimension of ħ that we need to look at. We know ħ: it’s the Planck constant h, which is expressed in joule·second, i.e. J·s = N·m·s, divided by 2π. That makes sense, because we get the same dimension for the angular momentum. Indeed, the L or J = r·m·v formula also gives us the dimension of physical action, i.e. N·m·s. Just check it: [r]·[m]·[v] = m·kg·m/s = m·(N·s²/m)·m/s = N·m·s. Done!

So we’ve got some kind of unit of action once more here, even if it’s not h but ħ = h/2π. That makes it a quantum of action expressed for a radian, so that’s a unit of length, rather than for a full cycle. Just so you know, ħ = h/2π is 1×10⁻³⁴ J·s ≈ 6.6×10⁻¹⁶ eV·s, and we could chose to express the components of J in terms of h by multiplying the whole thing with 2π. That would boil down to saying that our unit length is not unity but the unit circle, which is 2π times unity. Huh? Just think about it: h is a fundamental unit linked to one full cycle of something, so it all makes sense. Before we move on, you may want to compare the value of h or ħ with the energy of a photon, which is 1.6 to 3.2 eV in the visible light spectrum, but you should note that energy does not have the time dimension, and a second is an eternity in quantum physics, so the comparison is a bit tricky. So… […] Well… Let’s just move on. What about those coefficients? What constraints are there?

Well… The constraint is that the difference between +j and −j must be some integer, so +j−(−j) = 2j must be an integer. That implies that the spin number j is always an integer or a half-integer, depending on whether j is even or odd. Let’s do a few examples:

A lithium (Li-7) nucleus has spin j = 3/2 and, therefore, the permitted values for the angular momentum around any axis (the z-axis, for example) are: 3/2, 3/2−1=1/2, 3/2−2=−1/2, and −3/2—all times ħ of course! Note that the difference between +j and –j is 3, and each ‘step’ between those two levels is ħ, as we’d like it to be.
The nucleus of the much rarer Lithium-6 isotope is one of the few stable nuclei that has spin j = 1, so the permitted values are 1, 0 and −1. Again, all needs to be multiplied with ħ to get the actual value for the J-component that we’re looking at. So each step is ‘one’ again, and the total difference (between +j and –j) is 2.]
An electron is a spin-1/2 particle, and so there are only two permitted values: +ħ/2 and −ħ/2. So there is just one ‘step’ and it’s equal to the whole difference between +j and –j. In fact, this is the most common situation, because we’ll be talking elementary fermions most of the time.
Photons are an example of spin-1 ‘particles’, and ‘particles’ with integer spin are referred to as bosons. In this regard, you may heard of superfluid Helium-4, which is caused by Bose-Einstein condensation near the zero temperature point, and demonstrates the integer spin number of Helium-4, so it resembles Lithium-6 in this regard.

The four ‘typical’ examples makes it clear that the actual situations that we’ll be analyzing will usually be quite simple: we’ll only have 2, 3 or 4 permitted values only. As mentioned, there is this fundamental dichotomy between fermions and bosons. Fermions have half-integer spin, and all elementary fermions, such as protons, neutrons, electrons, neutrinos and quarks are spin-1/2 particles. [Note that a proton and a neutron are, strictly speaking, not elementary, as their constituent parts are quarks.] Bosons have integer spin, and the bosons we know of are spin-one particles, (except for the newly discovered Higgs boson, which is an actual spin-zero particle). The photon is an example, but the helium nucleus (He-4) also has spin one, which – as mentioned above – gives rise to superfluidity when its cooled near the absolute zero point.

In any case, to make a long story short, in practice, we’ll be dealing almost exclusively with spin-1, spin-1/2 particles and, occasionally, with spin-3/2 particles. In addition, to analyze simple stuff, we’ll often pretend particles do not have any spin, so our ‘theoretical’ particles will often be spin zero. That’s just to simplify stuff.

We now need to learn how to do a bit of math with all of this. Before we do so, let me make some additional remarks on these permitted values. Regardless of whether or not J is ‘wobbling’ or moving or not – let me be clear: J is not moving in the analysis above, but we’ll discuss the phenomenon of precession in the next post, and that will involve a J like that J circling around the J_z axis, so I am just preparing the terrain here – J‘s magnitude will always be some constant, which we denoted by |J| = J.

Now there’s something really interesting here, which again distinguishes classical mechanics from quantum mechanics. As mentioned, in classical mechanics, any of J‘s components J_x, J_y or J_z, could take on any value from +J to −J and, therefore, the maximum value of any component of J – say J_z – would be equal to J. To be precise, J would be the value of the component of J in the direction of J itself. So, in classical mechanics, we’d write: |J| = +√(J·J) = +√J² $=$ J, and it would be the maximum value of any component of J. But so we said that, if the spin number of J is j, then the maximum value of any component of J was equal to j·ħ. So, naturally, one would think that J = |J| = +√(J·J) = +√J²= j·ħ.

However, that’s not the case in quantum mechanics: the maximum value of any component of J is not J = j·ħ but the square root of j·(j+1)·ħ.

Huh? Yes. Let me spell it out: |J| = +√(J·J) = +√J²≠ jħ. Indeed, quantum math has many particularities, and this is one of them. The magnitude of J is not equal to the largest possible value of any component of J:

J‘s magnitude is not jħ but √(j(j+1)ħ).

As for the proof of this, let me simplify my life and just copy Feynman here:

The formula can be easily generalized for j ≠ 3/2. Also note that we used a fact that we didn’t mention as yet: all possible values of the z-component (or of whatever component) of J are equally likely.

Now, the result is fascinating, but the implications are even better. Let me paraphrase Feynman as he relates them:

From what we have so far, we can get another interesting and somewhat surprising conclusion. In certain classical calculations the quantity that appears in the final result is the square of the magnitude of the angular momentum $J$ —in other words, $J \cdot J$ = J². It turns out that it is often possible to guess at the correct quantum-mechanical formula by using the classical calculation and the following simple rule: Replace J² $= J \cdot J$ by $j(j+1)ħ$ . This rule is commonly used, and usually gives the correct result.
The second implication is the one we announced already: although we would think classically that the largest possible value of the any component of $J$ is just the magnitude of $J,$ quantum-mechanically the maximum of any component of J is always less than that, because j $ħ is always less than \sqrt(j(j+1)ħ). For example, for j = 3/2 = 1.5, we have j(j+1) = (3/2)\cdot(5/2) = 15/4 = 3.75. Now, the square root of this value is \sqrt3.75 \approx 1.9365, so the magnitude of J is about 30% larger than the maximum value of any of J ‘s components. That’s a pretty spectacular difference, obviously!$

$The second point is quite deep: it implies that the$ angular momentum is ‘never completely along any direction’. Why? Well… Think of it: “any of J‘s components” also includes the component in the direction of J itself! But if the maximum value of that component is 30% less than the magnitude of J, what does that mean really? All we can say is that it implies that the concept of the direction of the magnitude itself is actually quite fuzzy in quantum mechanics! Of course, that’s got to do with the Uncertainty Principle, and so we’ll come back to this later.

In fact, if you look at the math, you may think: what’s that business with those average or expected values? A magnitude is a magnitude, isn’t it? It’s supposed to be calculated from the actual values of J_x, J_y and J_z, not from some average that’s based on the (equal) likelihoods of the permitted values. You’re right. Feynman’s derivation here is quantum-mechanical from the start and, therefore, we get a quantum-mechanical result indeed: the magnitude of J is calculated as the magnitude of a quantum-mechanical variable in the derivation above, not as the magnitude of a classical variable.

[…] OK. On to the next.

The magnetic energy of atoms

Before we start talking about this topic, we should, perhaps, relate the angular momentum to the magnetic moment once again. We can do that using the μ = (q/2m)·J and/or μ = (q/m)·J formula (so that’s the simple formulas for the orbital and spin angular momentum respectively) or, else, by using the more general μ = – g·(q/2m)·J formula.

Let’s use the simpler μ = (q_e/2m)·J formula, which is the one for the orbital angular momentum. What’s q_e/2m? It should be equal to 1.6×10⁻¹⁹ C divided by 2·9.1×10⁻³¹kg, so that’s about 0.0879×10¹² C/kg, or 0.0879×10¹²(C·m)/(N·s²). Now we multiply by ħ/2 ≈ 0.527×10⁻³⁴ J·s. We get something like 0.0463×10⁻²²m²·C/s or J/T. These numbers are ridiculously small, so they’re usually measured in terms of a so-called natural unit: the Bohr magneton, which I’ll explain in a moment but so here we’re interested in its value only, which is μ_B = 9.274×10⁻²⁴J/T. Hence, μ/μ_B = 0.5 = 1/2. What a nice number!

Hmm… This cannot be a coincidence… […] You’re right. It isn’t. To get the full picture, we need to include the spin angular momentum, so we also need to see what the μ = (q/m)·J will yield. That’s easy, of course, as it’s twice the value of (q/2m)·J, so μ/μ_B = 1, and so the total is equal to 3/2. So the magnetic moment of an electron has the same value (when expressed in terms of the Bohr magneton) as the spin (when expressed in terms of ħ). Now that’s just sweet!

Yes, it is. All our definitions and formulas were formulated so as to make it sweet. Having said that, we do have a tiny little problem. If we use the general μ = −g·(q/2m)·J to write the result we found for the spin of the electron only (so we’re not looking at the orbital momentum here), then we’d write: μ = 2·(q/2m)·J = (q/m)·J and, hence, the g-factor here is −2. Yes. We know that. You told me so already. What’s the issue? Well… The problem is: experiments reveal the actual value of g is not exactly −2: it’s −2.00231930436182(52) instead, with the last two digits (in brackets) the uncertainty in the current measurements. Just check it for yourself on the NIST website. 🙂 [Please do check it: it brings some realness to this discussion.]

Hmm…. The accuracy of the measurement suggests we should take it seriously, even if we’re talking a difference of 0.1% only. We should. It can be explained, of course: it’s something quantum-mechanical. However, we’ll talk about this later. As for now, just try to understand the basics here. It’s complicated enough already, and so we’ll stay away from the nitty-gritty as long as we can.

Let’s now get back to the magnetic energy of our atoms. From our discussion on the torque on a magnetic dipole in an external magnetic field, we know that our magnetic atoms will have some extra magnetic energy when placed in an external field. So now we have an external magnetic field B, and we derived the formula for the energy is

U_mag= −μ·B·cosθ = −μ·B

I won’t explain the whole thing once again, but it might help to visualize the situation, which we do below. The loop here is not circular but square, and it’s a current-carrying wire instead of an electron in orbit, but I hope you get the point.

We need to chose some coordinate system to calculate stuff and so we’ll just choose our z-axis along the direction of the external magnetic field B so as to simplify those calculations. If we do that, we can just take the z-component of μ and then combine the interim result with our general μ = – g·(q/2m)·J formula, so we write:

U_mag= −μ_z·B = g·(q/2m)·J_z·B

Now, we know that the maximum value of J_z is equal to j·ħ, and so the maximum value of U_magwill be equal g(q/2m)jħB. Let’s now simplify this expression by choosing some natural unit, and that’s the unit we introduced already above: the Bohr magneton. It’s equal to (q_eħ)/(2m_e) and its value is μ_B ≈ 9.274×10⁻²⁴J/T. So we get the result we wanted, and that is:

Let me make a few remarks here. First on that magneton: you should note there’s also something which is known as the nuclear magneton which, you guessed it, is calculated using the proton charge and the proton mass: μ_N = (q_pħ)/(2m_p) ≈ 5.05×10⁻²⁷J/T. My second remark is a question: what does that formula mean, really? Well… Let me quote Feynman on that. The formula basically says the following:

“The energy of an atomic system is changed when it is put in a magnetic field by an amount that is proportional to the field, and proportional to J_z. We say that the energy of an atomic system is ‘split’ into $2 j + 1$ ‘levels’ by a magnetic field. For instance, an atom whose energy is U₀ outside a magnetic field and whose $j$ is $3/2$ , will have four possible energies when placed in a field. We can show these energies by an energy-level diagram like that drawn below. Any particular atom can have only one of the four possible energies in any given field B. That is what quantum mechanics says about the behavior of an atomic system in a magnetic field.”

Of course, the simplest ‘atomic’ system is a single electron, which has spin 1/2 only (like most fermions really: the example in the diagram above, with spin 3/2, would be that Li-7 system or something similar). If the spin is 1/2, then there are only two energy levels, with J_z = ±ħ/2 and, as we mentioned already, the g-factor for an electron is −2 (again, the use of minus signs (or not) is quite confusing: I am sorry for that), and so our formula above becomes very simple:

U_mag= ± μ_B·B

The graph above becomes the graph below, and we can now speak more loosely and say that the electron either has its spin ‘up’ (so that’s along the field), or ‘down’ (so that’s opposite the field).

By now, you’re probably tired of the math and you’ll wonder: how can we prove all of this permitted value business? Well… That question leads me to the last topic of my post: the Stern-Gerlach experiment.

The Stern-Gerlach experiment

Here again, I can just copy straight of out of Feynman, and so I hope you’ll forgive me if I just do that, as I don’t think there’s any advantage to me trying to summarize what he writes on it:

“The fact that the angular momentum is quantized is such a surprising thing that we will talk a little bit about it historically. It was a shock from the moment it was discovered (although it was expected theoretically). It was first observed in an experiment done in 1922 by Stern and Gerlach. If you wish, you can consider the experiment of Stern-Gerlach as a direct justification for a belief in the quantization of angular momentum. Stern and Gerlach devised an experiment for measuring the magnetic moment of individual silver atoms. They produced a beam of silver atoms by evaporating silver in a hot oven and letting some of them come out through a series of small holes. This beam was directed between the pole tips of a special magnet, as shown in the illustration below. Their idea was the following. If the silver atom has a magnetic moment $μ$ , then in a magnetic field $B$ it has an energy $-μ z B$ , where $z$ is the direction of the magnetic field. In the classical theory, μ_z would be equal to the magnetic moment times the cosine of the angle between the moment and the magnetic field, so the extra energy in the field would be

ΔU = −μ·B·cosθ

Of course, as the atoms come out of the oven, their magnetic moments would point in every possible direction, so there would be all values of $θ$ . Now if the magnetic field varies very rapidly with $z$ —if there is a strong field gradient—then the magnetic energy will also vary with position, and there will be a force on the magnetic moments whose direction will depend on whether cosine $θ$ is positive or negative. The atoms will be pulled up or down by a force proportional to the derivative of the magnetic energy; from the principle of virtual work,

Stern and Gerlach made their magnet with a very sharp edge on one of the pole tips in order to produce a very rapid variation of the magnetic field. The beam of silver atoms was directed right along this sharp edge, so that the atoms would feel a vertical force in the inhomogeneous field. A silver atom with its magnetic moment directed horizontally would have no force on it and would go straight past the magnet. An atom whose magnetic moment was exactly vertical would have a force pulling it up toward the sharp edge of the magnet. An atom whose magnetic moment was pointed downward would feel a downward push. Thus, as they left the magnet, the atoms would be spread out according to their vertical components of magnetic moment. In the classical theory all angles are possible, so that when the silver atoms are collected by deposition on a glass plate, one should expect a smear of silver along a vertical line. The height of the line would be proportional to the magnitude of the magnetic moment. The abject failure of classical ideas was completely revealed when Stern and Gerlach saw what actually happened. They found on the glass plate two distinct spots. The silver atoms had formed two beams.

That a beam of atoms whose spins would apparently be randomly oriented gets split up into two separate beams is most miraculous. How does the magnetic moment know that it is only allowed to take on certain components in the direction of the magnetic field? Well, that was really the beginning of the discovery of the quantization of angular momentum, and instead of trying to give you a theoretical explanation, we will just say that you are stuck with the result of this experiment just as the physicists of that day had to accept the result when the experiment was done. It is an experimental fact that the energy of an atom in a magnetic field takes on a series of individual values. For each of these values the energy is proportional to the field strength. So in a region where the field varies, the principle of virtual work tells us that the possible magnetic force on the atoms will have a set of separate values; the force is different for each state, so the beam of atoms is split into a small number of separate beams. From a measurement of the deflection of the beams, one can find the strength of the magnetic moment.”

I should note one point which Feynman hardly addresses in the analysis above: why do we need a non-homogeneous field? Well… Think of it. The individual silver atoms are not like electrons in some electric field. They are tiny little magnets, and magnets do not behave like electrons. Remember we said there’s no such thing as a magnetic charge? So that applies here. If the silver atoms are tiny magnets, with a magnetic dipole moment, then the only thing they will do is turn, so as to minimize their energy U = −μBcosθ.

That energy is minimized when μ and B are at right angles of each other, so as to make the cosθ factor zero, which happens when θ = π/2. Hence, in a homogeneous magnetic field, we will have a torque on the loop of current – think of our electron(s) in orbit here – but no net force pulling it in this or that direction as a whole. So the atoms would just rotate but not move in our classical analysis here.

To make the atoms themselves move towards or away one of the poles (with or without a sharp tip), the magnetic field must be non-homogeneous, so as to ensure that the force that’s pulling on one side of the loop of current is slightly different from the force that’s pulling (in the opposite direction) on the other side of the loop of current. So that’s why the field has to be non-homogeneous (or inhomogeneous as Feynman calls it), and so that’s why one pole needs to have a sharply pointed tip.

As for the force formula, it’s crucial to remember that energy (or work) is force times distance. To be precise, it’s the ∫F∙ds integral. This integral will have a minus sign in front when we’re doing work against the force, so that’s when we’re increasing the potential energy of an object. Conversely, we’ll just take the positive value when we’re converting potential energy into kinetic energy. So that explains the F = −∂U/∂z formula above. In fact, in the analysis above, Feynman assumes the magnetic moment doesn’t turn at all. That’s pretty obvious from the F_z = −∂U/∂z = −μ∙cosθ∙∂B/∂z formula, in which μ is clearly being treated as a constant. So the F_z in this formula is a net force in the z-direction, and it’s crucially dependent on the variation of the magnetic field in the z-direction. If the field would not be varying, ∂B/∂z would be zero and, therefore, we would not have any net force in the z-direction. As mentioned above, we would only have a torque.

Well… This sort of covers all of what we wanted to cover today. 🙂 I hope you enjoyed it.

Taking the magic out of God’s number: some additional reflections

Note: I have published a paper that is very coherent and fully explains this so-called God-given number. There is nothing magical about it. It is just a scaling constant. Check it out: The Meaning of the Fine-Structure Constant. No ambiguity. No hocus-pocus.

Jean Louis Van Belle, 23 December 2018

Original post:

In my previous post, I explained why the fine-structure constant α is not a ‘magical’ number, even if it relates all fundamental properties of the electron: its mass, its energy, its charge, its radius, its photon scattering cross-section (i.e. the Bohr radius, or the size of the atom really) and, finally, the coupling constant for photon-electron interactions. The key to such understanding of α was the model of an electron as a tiny ball of charge. As such, we have two energy formulas for it. One is the energy that’s needed to assemble the charge from infinitely dispersed infinitesimal charges, which we denoted as U_elec. The other formula is the energy of the field of the tiny ball of charge, which we denoted as E_elec.

The formula for E_elec is calculated using the formula for the field momentum of a moving charge and, using the m = E/c²mas-energy equivalence relationship, is equivalent to the electromagnetic mass. We went through the derivation in our previous post, so let me just jot down the result:

The second formula depends on what ball of charge we’re thinking of, because the formulas for a charged sphere and a spherical shell of charge are different: both have the same structure as the relationship above (so the energy is also proportional to the square of the electron charge and inversely proportional to the radius a), but the constant of proportionality is different. For a sphere of charge, we write:

For a spherical shell of charge we write:

shell

To compare the formulas, you need to note that the square of the electron charge e in the formula for the field energy is equal to e²= q_e²/4πε₀= k_e·q_e². So we multiply the square of the actual electron charge by the Coulomb constant k_e= 1/4πε₀. As you can see, the three formulas have exactly the same form then. It’s just the proportionality constant that’s different: it’s 2/3, 3/5 and 1/2 respectively. It’s interesting to quickly reflect on the dimensions here: [k_e] ≈ 9×10⁹N·m²/C², so e² is expressed in N·m². That makes the units come out alright, as we divide by a (so that’s in meter) and so we get the energy in joule (which is newton·meter). In fact, now that we’re here, let’s quickly calculate the value of e²: it’s that k_e·q_e² product, so it’s equal to 2.3×10⁻²⁸N·m². We can quickly check this value because we know that the classical electron radius is equal to:

So we divide 2.3×10⁻²⁸N·m²by m_ec_²≈ 8.2×10⁻¹⁴J, so we get r₀≈ 2.82×10⁻¹⁵m. So we’re spot on! Why did I do this check? Not really to check what I wrote. It’s more to show what’s going on. We’ve got yet another formula relating the energy and the radius of an electron here, so now we have three. In fact we have more because the formula for U_elec depends on the finer details of our model for the electron (sphere versus shell, uniform versus non-uniform distribution):

E_elec= (2/3)·(e²/a): This is the formula for the energy of the field, so we may all it is external energy.
U_elec= (3/5)·(e²/a), or U_elec= (1/2)·(e²/a): This is the energy needed to assemble our electron, so we might, perhaps, call it its internal energy. The first formula assumes our electron is a uniformly charged sphere. The second assumes all charges sit on the surface of the sphere. If we drop the assumption of the charge having to be uniformly distributed, we’ll find yet another formula.
m_ec²= e²/r₀: This is the energy associated with the so-called classical electron radius (r₀) and the electron’s rest mass (m_e).

In our previous posts, we assumed the last equation was the right one. Why? Because it’s the one that’s been verified experimentally. The discrepancies between the various proportionality coefficients – i.e. the difference between 2/3 and 1, basically – are to be explained because of the binding forces within the electron, without which the electron would just ‘explode’, as the French physicist and polymath Henri Poincaré famously put it. Indeed, if the electron is a little ball of negative charge, the repulsive forces between its parts should rip it apart. So we will not say anything more about this. You can have fun yourself by googling all the various theories that try to model these binding forces. [I may do the same some day, but now I’ve got other priorities: I want to move to Feynman’s third volume of Lectures, which is devoted to quantum physics only, so I look very much forward to that.]

In this post, I just wanted to reflect once more on what constants are really fundamental and what constants are somewhat less fundamental. From all what I wrote in my previous post, I said there were three:

The fine-structure constant α, which is a dimensionless number.
Planck’s constant h, whose dimension is joule·second, so that’s the dimension of action.
The speed of light c, whose dimension is that of a velocity.

The three are related through the following expression:

This is an interesting expression. Let’s first check its dimension. We already explained that e_² is expressed in N·m². That’s rather strange, because it means the dimension of e itself is N^1/2·m: what’s the square root of a force of one newton? In fact, to interpret the formula above, it’s probably better to re-write e²as e²= q_e²/4πε₀= k_e·q_e². That shows you how the electron charge and Coulomb’s constant are related. Of course, they are part and parcel of one and the same force law: Coulomb’s law. We don’t need anything else, except for relativity theory, because we need to explain the magnetic force as well—and that we can do because magnetism is just a relativistic effect. Think of the field momentum indeed: the magnetic field comes into play only when we start to move our electron. The relativity effect is captured by c in that formula for α above. As for ħ, ħ = h/2π comes with the E = h·f equation, which links us to the electron’s Compton wavelength λ through the de Broglie relation λ = h/p.

The point is: we should probably not look at α as a ‘fundamental physical constant’. It’s e² that’s the third fundamental constant, besides h and c. Indeed, it’s from e² that all the rest follows: the electron’s internal energy, its external energy, and its radius, and then all the rest by combining stuff with other stuff.

Now, we took the magic out of α by doing what we did in the previous posts, and that’s to combine stuff with other stuff, and so now you may think I am putting the magic back in with that formula for α, which seems to define α in terms of the three mentioned ‘fundamental’ constants. That’s not the case: this relation comes out of all of the other relationships we found, and so it’s nothing new really. It’s actually not a definition of α: it just does what it does, and that’s to relate α to the ‘fundamental’ physical constants behind.

So… No new magic. In fact, I want to close this post by taking away even more of the magic. If you read my previous post, I said that α was ‘God’s cut-off factor’ 🙂 ensuring our energy functions do not blow up, but I also said it was impossible to say why he chose 0.00729735256 as the cut-off factor. The question is actually easily answered by thinking about those two formulas we had for the internal and external energy respectively. Let’s re-write them in natural units and, temporarily, two different subscripts for α, so we write:

E_elec= α_e/r₀: This is the formula for the energy of the field.
U_elec= α_u/r₀: This is the energy needed to assemble our electron.

Both energies are determined by the above-mentioned laws, i.e. Coulomb’s Law and the theory of relativity, so α has got nothing to do what that. However, both energies have to be the same, and so α_ehas to be equal to α_u. In that sense, α is, quite simply, a proportionality constant that achieves that equality. Now that explains why we can derive α from the three other constants which, as mentioned above, are probably more fundamental. In fact, we’ve got only three degrees of freedom here, so if we chose c, h and e as ‘fundamental’, then α isn’t any more.

The underlying deep question behind it all is why those two energies should be equal. Why would our electron have some internal energy if it’s elementary? The answer to that question is: because it has some non-zero radius, and it has some non-zero radius because we don’t want our formula for the field energy (or the field momentum) to blow up. Now, if it has some radius, then it has to have some internal energy.

You’ll say: that makes sense, but it doesn’t answer the question. Why would it have internal energy, with or without a zero radius? If an electron is an elementary particle, then it’s really elementary, isn’t? And so then we shouldn’t try to ‘assemble’ it from an infinite number of infinitesimally small charges. You’re right, and here we can also note that the fact that the electron doesn’t blow up is firm evidence it’s very elementary indeed.

I should also note that Feynman actually doesn’t talk about the energy that’s needed to assemble a charge: he gets his U_elec= (1/2)·(e²/a) by calculating the external field energy for a spherical shell of charge, and he sticks to it—presumably because it’s the same field for a uniform or non-uniform sphere of charge. He only notes there has to be some radius because, if not, the formula he uses blows up, indeed. So – who knows? – perhaps he doesn’t quite believe that formula for the internal energy is relevant either.

So perhaps there is no internal energy indeed. Perhaps there’s just the energy of the field. So… Well… I can’t say much about this… Except… Well… Perhaps just one more thing. Let me note something that, I hope, you noticed as well: the k_e·q_e²is the numerator in Coulomb’s Law itself. You also know that energy equals force times distance. So if we divide both sides by r₀, we get Coulomb’s Law itself F_elec= k_e·q_e²/r₀². The only thing is: what’s the distance? It’s one charge only, and there is no distance between one charge, is there? Well… Yes and no. I have been thinking that the requirement of the internal and external energies being equal resembles the statement that the forces between two charges are equal and opposite. That ties in with the idea of the internal energy itself: remember we were basically talking forces between infinitesimally small elements of charge within the electron itself? So r₀ is, perhaps, some average distance or so. There must be some way of thinking of it like that. But… Well… Which one exactly?

This kind of reflection may not make sense. Who knows? I obviously need to think all of this through and so this post is, indeed, just a bunch of reflections for which I will have more time later—hopefully. 🙂 Perhaps we’re all just pushing the matter too far. Perhaps we should just accept that the external energy has that 2/3 factor but that the actual energy of the electron should also include the equivalent energy of some binding force that holds the electron together. Well… In any case. That’s all I am going to do on this extremely complicated matter. It’s time to move indeed! So the point to take home here is probably just this:

When calculating the radius of an electron using classical theory, we get in trouble: not only do we find different radii, but the radii that we find do not respect the E = m_ec²law. It’s only the m_ec²= e²/r₀ that’s relativistically correct.
That suggests the electron also has some non-electric mass, which are referred to as ‘binding forces’ or ‘Poincaré stresses’, but which remain to be explained convincingly.
All of this shouldn’t surprise us: for all we know, the electron is something fuzzy. 🙂

So my next posts will focus on the ‘essentials’ preparing for Feynman’s Volume on quantum mechanics. Those ‘essentials’ will still involve some classical stuff but, as you will see, even more contradictions, that – hopefully! – will then be solved in the quantum-mechanical picture of it all. 🙂

Taking the magic out of God’s number

Jean Louis Van Belle, 23 December 2018

Original post:

I think the post scriptum to my previous post is interesting enough to separate it out as a piece of its own, so let me do that here. You’ll remember that we were trying to find some kind of a model for the electron, picturing it like a tiny little ball of charge, and then we just applied the classical energy formulas to it to see what comes out of it. The key formula is the integral that gives us the energy that goes into assembling a charge. It was the following thing:

This is a double integral which we simplified in two stages, so we’re looking at an integral within an integral really, but we can substitute the integral over the ρ(2)·dV₂product by the formula we got for the potential, so we write that as Φ(1), and so the integral above becomes:

Now, this integral integrates the ρ(1)·Φ(1)·dV₁product over all of space, so that’s over all points in space, and so we just dropped the index and wrote the whole thing as the integral of ρ·Φ·dV over all of space:

We then established that this integral was mathematically equivalent to the following equation:

So this integral is actually quite simple: it just integrates E•E = E² over all of space. The illustration below shows E as a function of the distance r for a sphere of radius R filled uniformly with charge.

So the field (E) goes as r for r ≤ R and as 1/r²for r ≥ R. So, for r ≥ R, the integral will have (1/r²)² = 1/r⁴in it. Now, you know that the integral of some function is the surface under the graph of that function. Look at the 1/r⁴function below: it blows up between 1 and 0. That’s where the problem is: there needs to be some kind of cut-off, because that integral will effectively blow up when the radius of our little sphere of charge gets ‘too small’. So that makes it clear why it doesn’t make sense to use this formula to try to calculate the energy of a point charge. It just doesn’t make sense to do that.

In fact, the need for a ‘cut-off factor’ so as to ensure our energy function doesn’t ‘blow up’ is not because of the exponent in the 1/r⁴expression: the need is also there for any 1/r relation, as illustrated below. All 1/rⁿfunction have the same pivot point, as you can see from the simple illustration below. So, yes, we cannot go all the way to zero from there when integrating: we have to stop somewhere.

So what’s the ‘cut-off point’? What’s ‘too small’ a radius? Let’s look at the formula we got for our electron as a shell of charge (so the assumption here is that the charge is uniformly distributed on the surface of a sphere with radius a):

So we’ve got an even simpler formula here: it’s just a 1/r relation (a is r in this formula), not 1/r⁴. Why is that? Well… It’s just the way the math turns out: we’re integrating over volumes and so that involves an r³ factor and so it all simplifies to 1/r, and so that gives us this simple inversely proportional relationship between U and r, i.e a, in this case. 🙂 I copied the detail of Feynman’s calculation in my previous post, so you can double-check it. It’s quite wonderful, really. Look at it again: we have a very simple inversely proportional relationship between the radius of our electron and its energy as a sphere of charge. We could write it as:

U_elect = α/a, with α = e²/2

Still… We need the cut-off point’. Also note that, as I pointed out, we don’t necessarily need to assume that the charge in our little ball of charge (i.e. our electron) sits on the surface only: if we’d assume it’s a uniformly charged sphere of charge, we’d just get another constant of proportionality: our 1/2 factor would become a 3/5 factor, so we’d write: U_elect = (3/5)·e²/a. But we’re not interested in finding the right model here. We know the U_elect = (3/5)·e²/a gives us a value for a that differs with a 2/5 factor as the classical electron radius. That’s not so bad and so let’s go along with it. 🙂

We’re going to look at the simple structure of this relation, and all of its implications. The simple equation above says that the energy of our electron is (a) proportional to the square of its charge and (b) inversely proportional to its radius. Now, that is a very remarkable result. In fact, we’ve seen something like this before, and we were astonished. We saw it when we were discussing the wonderful properties of that magical number, the fine-structure constant, which we also denoted by α. However, because we used α already, I’ll denote the fine-structure constant as α_ehere, so you don’t get confused. You’ll remember that the fine-structure constant is a God-like number indeed: it links all of the fundamental properties of the electron, i.e. its charge, its radius, its distance to the nucleus (i.e. the Bohr radius), its velocity, its mass (and, hence, its energy), its de Broglie wavelength. Whatever: all these physical constants are all related through the fine-structure constant.

In my various posts on this topic, I’ve repeatedly said that, but I never showed why it’s true, and so it was a very magical number indeed. I am going to take some of the magic out now. Not too much but… Well… You can judge for yourself how much of the magic remains after I am done here. 🙂

So, at this stage of the argument, α can be anything, and α_ecannot, of course. It’s just that magical number out there, which relates everything to everything: it’s the God-given number we don’t understand, or didn’t understand, I should say. Past tense. Indeed, we’re going to get some understanding here because we know that one of the many expressions involving α_ewas the following one:

m_e = α_e/r_e

This says that the mass of the electron is equal to the ratio of the fine-structure constant and the electron radius. [Note that we express everything in natural units here, so that’s Planck units. For the detail of the conversion, please see the relevant section on that in my one of my posts on this and other stuff.] In fact, the U = (3/5)·e²/a and m_e = α_e/r_erelations looks exactly the same, because one of the other equations involving the fine-structure constant was: α_e = e_P². So we’ve got the square of the charge here as well! Indeed, as I’ll explain in a moment, the difference between the two formulas is just a matter of units.

Now, mass is equivalent to energy, of course: it’s just a matter of units, so we can equate m_e with E_e (this amounts to expressing the energy of the electron in a kg unit—bit weird, but OK) and so we get:

E_e = α_e/r_e

So there we have: the fine-structure constant α_e is Nature’s ‘cut-off’ factor, so to speak. Why? Only God knows. 🙂 But it’s now (fairly) easy to see why all the relations involving α_e are what they are. As I mentioned already, we also know that α_e is the square of the electron charge expressed in Planck units, so we have:

α_e = e_P² and, therefore, E_e = e_P²/r_e

Now, you can check for yourself: it’s just a matter of re-expressing everything in standard SI units, and relating e_P² to e², and it should all work: you should get the E_elect = (2/3)·e²/a expression. So… Well… At least this takes some of the magic out the fine-structure constant. It’s still a wonderful thing, but so you see that the fundamental relationship between (a) the energy (and, hence, the mass), (b) the radius and (c) the charge of an electron is not something God-given. What’s God-given are Maxwell’s equations, and so the E_e = α_e/r_e= e_P²/r_e is just one of the many wonderful things that you can get out of them.

So we found God’s ‘cut-off factor’ 🙂 It’s equal to α_e ≈ 0.0073 = 7.3×10⁻³. So 7.3 thousands of… What? Well… Nothing. It’s just a pure ratio between the energy and the radius of an electron (if both are expressed in Planck units, of course). And so it determines the electron charge (again, expressed in Planck units). Indeed, we write:

e_P = √α_e

Really? Yes. Just do all these formulas:

e_P = √α_e≈ √0.0073·1.9×10⁻¹⁸coulomb ≈ 1.6 ×10⁻¹⁹C

Just re-check it with all the known decimals: you’ll see it’s bang on. Let’s look at the E_e= m_e = α_e/r_eratio once again. What’s the meaning of it? Let’s first calculate the value of r_e and m_e, i.e. the electron radius and electron mass expressed in Planck units. It’s equal to the classical electron radius divided by the Planck length, and then the same for the mass, so we get the following thing:

r_e ≈ (2.81794×10⁻¹⁵m)/(1.6162×10⁻³⁵m) = 1.7435×10²⁰

m_e ≈ (9.1×10⁻³¹kg)/(2.17651×10⁻⁸kg) = 4.18×10⁻²³

α_e = (4.18×10⁻²³)·(1.7435×10²⁰) ≈ 0.0073

It works like a charm, but what does it mean? Well… It’s just a ratio between two physical quantities, and the scale you use to measure those quantities matters very much. We’ve explained that the Planck mass is a rather large unit at the atomic scale and, therefore, it’s perhaps not quite appropriate to use it here. In fact, out of the many interesting expressions for α_e, I should highlight the following one:

α_e = e²/(ħ·c) ≈ (1.60217662×10⁻¹⁹C)²/(4πε₀·[(1.054572×10⁻³⁴N·m·s)·(2.998×10⁸m/s)]) ≈ 0.0073 once more 🙂

Note that the elementary charge e is actually equal to q_e/4πε₀, which is what I am using in the formula. I know that’s confusing, but it what it is. As for the units, it’s a bit tedious to write it all out, but you’ll get there. Note that ε₀≈ 8.8542×10⁻¹²C²/(N·m²) so… Well… All the units do cancel out, and we get a dimensionless number indeed, which is what α_e is.

The point is: this expression links α_e to the the de Broglie relation (p = h/λ), with λ the wavelength that’s associated with the electron. Of course, because of the Uncertainty Principle, we know we’re talking some wavelength range really, so we should write the de Broglie relation as Δp = h·Δ(1/λ). Now, that, in turn, allows us to try to work out the Bohr radius, which is the other ‘dimension’ we associate with an electron. Of course, now you’ll say: why would you do that. Why would you bring in the de Broglie relation here?

Well… We’re talking energy, and so we have the Planck-Einstein relation first: the energy of some particle can always be written as the product of h and some frequency f: E = h·f. The only thing that de Broglie relation adds is the Uncertainty Principle indeed: the frequency f will be some frequency range, associated with some momentum range, and so that’s what the Uncertainty Principle really says. I can’t dwell too much on that here, because otherwise this post would become a book. 🙂 For more detail, you can check out one of my many posts on the Uncertainty Principle. In fact, the one I am referring to here has Feynman’s calculation of the Bohr radius, so I warmly recommend you check it out. The thrust of the argument is as follows:

If we assume that (a) an electron takes some space – which I’ll denote by r 🙂 – and (b) that it has some momentum p because of its mass m and its velocity v, then the ΔxΔp = ħ relation (i.e. the Uncertainty Principle in its roughest form) suggests that the order of magnitude of r and p should be related in the very same way. Hence, let’s just boldly write r ≈ ħ/p and see what we can do with that.
We know that the kinetic energy of our electron equals mv²/2, which we can write as p²/2m so we get rid of the velocity factor.Well… Substituting our p ≈ ħ/r conjecture, we get K.E. = ħ²/2mr². So that’s a formula for the kinetic energy. Next is potential.
The formula for the potential energy is U = q₁q₂/4πε₀r₁₂. Now, we’re actually talking about the size of an atom here, so one charge is the proton (+e) and the other is the electron (–e), so the potential energy is U = P.E. = –e²/4πε₀r, with r the ‘distance’ between the proton and the electron—so that’s the Bohr radius we’re looking for!
We can now write the total energy (which I’ll denote by E, but don’t confuse it with the electric field vector!) as E = K.E. + P.E. = ħ²/2mr²– e²/4πε₀r. Now, the electron (whatever it is) is, obviously, in some kind of equilibrium state. Why is that obvious? Well… Otherwise our hydrogen atom wouldn’t or couldn’t exist. 🙂 Hence, it’s in some kind of energy ‘well’ indeed, at the bottom. Such equilibrium point ‘at the bottom’ is characterized by its derivative (in respect to whatever variable) being equal to zero. Now, the only ‘variable’ here is r (all the other symbols are physical constants), so we have to solve for dE/dr = 0. Writing it all out yields: dE/dr = –ħ²/mr³+ e²/4πε₀r²= 0 ⇔ r = 4πε₀ħ²/me²
We can now put the values in: r = 4πε₀h²/me²= [(1/(9×10⁹) C²/N·m²)·(1.055×10^–34J·s)²]/[(9.1×10^–31kg)·(1.6×10^–19C)²] = 53×10^–12m = 53 pico-meter (pm)

Done. We’re right on the spot. The Bohr radius is, effectively, about 53 trillionths of a meter indeed!

Phew!

Yes… I know… Relax. We’re almost done. You should now be able to figure out why the classical electron radius and the Bohr radius can also be related to each other through the fine-structure constant. We write:

m_e = α/r_e= α/α²r = 1/αr

So we get that α/r_e= 1/αr and, therefore, we get r_e/r = α², which explains why α is also equal to the so-called junction number, or the coupling constant, for an electron-photon coupling (see my post on the quantum-mechanical aspects of the photon-electron interaction). It gives a physical meaning to the probability (which, as you know, is the absolute square of the probability amplitude) in terms of the chance of a photon actually ‘hitting’ the electron as it goes through the atom. Indeed, the ratio of the Thomson scattering cross-section and the Bohr size of the atom should be of the same order as r_e/r, and so that’s α².

[Note: To be fully correct and complete, I should add that the coupling constant itself is not α² but √α = e_P. Why do we have this square root? You’re right: the fact that the probability is the absolute square of the amplitude explains one square root (√α² = α), but not two. The thing is: the photon-electron interaction consists of two things. First, the electron sort of ‘absorbs’ the photon, and then it emits another one, that has the same or a different frequency depending on whether or not the ‘collision’ was elastic or not. So if we denote the coupling constant as j, then the whole interaction will have a probability amplitude equal to j². In fact, the value which Feynman uses in his wonderful popular presentation of quantum mechanics (The Strange Theory of Light and Matter), is −α ≈ −0.0073. I am not quite sure why the minus sign is there. It must be something with the angles involved (the emitted photon will not be following the trajectory of the incoming photon) or, else, with the special arithmetic involved in boson-fermion interactions (we add amplitudes when bosons are involved, but subtract amplitudes when it’s fermions interacting. I’ll probably find out once I am true through Feynman’s third volume of Lectures, which focus on quantum mechanics only.]

Finally, the last bit of unexplained ‘magic’ in the fine-structure constant is that the fine-structure constant (which I’ve started to write as α again, instead of α_e) also gives us the (classical) relative speed of an electron, so that’s its speed as it orbits around the nucleus (according to the classical theory, that is), so we write

α = v/c = β

I should go through the motions here – I’ll probably do so in the coming days – but you can see we must be able to get it out somehow from all what we wrote above. See how powerful our U_elect ∼ e²/a relation really is? It links the electron, charge, its radius and its energy, and it’s all we need to all the rest out of it: its mass, its momentum, its speed and – through the Uncertainty Principle – the Bohr radius, which is the size of the atom.

We’ve come a long way. This is truly a milestone. We’ve taken the magic out of God’s number—to some extent at least. 🙂

You’ll have one last question, of course: if proportionality constants are all about the scale in which we measure the physical quantities on either side of an equation, is there some way the fine-structure constant would come out differently? That’s the same as asking: what if we’d measure energy in units that are equivalent to the energy of an electron, and the radius of our electron just as… Well… What if we’d equate our unit of distance with the radius of the electron, so we’d write r_e= 1? What would happen to α? Well… I’ll let you figure that one out yourself. I am tired and so I should go to bed now. 🙂

[…] OK. OK. Let me tell you. It’s not that simple here. All those relationships involving α, in one form or the other, are very deep. They relate a lot of stuff to a lot of stuff, and we can appreciate that only when doing a dimensional analysis. A dimensional analysis of the E_e = α_e/r_e= e_P²/r yields [e_P²/r] = C²/m on the right-hand side and [E_e] = J = N·mon the left-hand side. How can we reconcile both? The coulomb is an SI base unit , so we can’t ‘translate’ it into something with N and m. [To be fully correct, for some reason, the ampère (i.e. coulomb per second) was chosen as an SI base unit, but they’re interchangeable in regard to their place in the international system of units: they can’t be reduced.] So we’ve got a problem. Yes. That’s where we sort of ‘smuggled’ the 4πε₀factor in when doing our calculations above. That ε₀constant is, obviously, not ‘as fundamental’ as c or α (just think of the c⁻²= ε₀μ₀relationship to understand what I mean here) but, still, it was necessary to make the dimensions come out alright: we need the reciprocal dimension of ε₀, i.e. (N·m²)/C², to make the dimensional analysis work. We get: (C²/m)·(N·m²)/C² = N·m = J, i.e. joule, so that’s the unit in which we measure energy or – using the E = mc² equivalence – mass, which is the aspect of energy emphasizing its inertia.

So the answer is: no. Changing units won’t change alpha. So all that’s left is to play with it now. Let’s try to do that. Let me first plot that E_e= m_e = α_e/r_e= 0.00729735256/r_e:

Unsurprisingly, we find the pivot point of this curve is at the intersection of the diagonal and the curve itself, so that’s at the (0.00729735256, 0.00729735256) point, where slopes are ± 1, i.e. plus or minus unity. What does this show? Nothing much. What? I can hear you: I should be excited because… Well… Yes! Think of it. If you would have to chose a cut-off point, you’d chose this one, wouldn’t you? 🙂 Sure, you’re right. How exciting! Let me show you. Look at it! It proves that God thinks in terms of logarithms. He has chosen α such that ln(E) = ln(α/r) = lnα – lnr = lnα – lnr = 0, so ln α = lnr and, therefore, α = r. 🙂

Huh? Excuse me?

I am sorry. […] Well… I am not, of course… 🙂 I just wanted to illustrate the kind of exercise some people are tempted to do. It’s no use. The fine-structure constant is what it is: it sort of summarizes an awful lot of formulas. It basically shows what Maxwell’s equation imply in terms of the structure of an atom defined as a negative charge orbiting around some positive charge. It shows we can get calculate everything as a function of something else, and that’s what the fine-structure constant tells us: it relates everything to everything. However, when everything is said and done, the fine-structure constant shows us two things:

Maxwell’s equations are complete: we can construct a complete model of the electron and the atom, which includes: the electron’s energy and mass, its velocity, its own radius, and the radius of the atom. [I might have forgotten one of the dimensions here, but you’ll add it. :-)]
God doesn’t want our equations to blow up. Our equations are all correct but, in reality, there’s a cut-off factor that ensures we don’t go to the limit with them.

So the fine-structure constant anchors our world, so to speak. In other words: of all the worlds that are possible, we live in this one.

[…] It’s pretty good as far as I am concerned. Isn’t it amazing that our mind is able to just grasp things like that? I know my approach here is pretty intuitive, and with ‘intuitive’, I mean ‘not scientific’ here. 🙂 Frankly, I don’t like the talk about physicists “looking into God’s mind.” I don’t think that’s what they’re trying to do. I think they’re just trying to understand the fundamental unity behind it all. And that’s religion enough for me. 🙂

So… What’s the conclusion? Nothing much. We’ve sort of concluded our description of the classical world… Well… Of its ‘electromagnetic sector’ at least. 🙂 That sector can be summarized in Maxwell’s equations, which describe an infinite world of possible worlds. However, God fixed three constants: h, c and α. So we live in a world that’s defined by this Trinity of fundamental physical constants. Why is it not two, or four?

My guts instinct tells me it’s because we live in three dimensions, and so there’s three degrees of freedom really. But what about time? Time is the fourth dimension, isn’t it? Yes. But time is symmetric in the ‘electromagnetic’ sector: we can reverse the arrow of time in our equations and everything still works. The arrow of time involves other theories: statistics (physicists refer to it as ‘statistical mechanics‘) and the ‘weak force’ sector, which I discussed when talking about symmetries in physics. So… Well… We’re not done. God gave us plenty of other stuff to try to understand. 🙂

The classical explanation for the electron’s mass and radius

Feynman’s 28th Lecture in his series on electromagnetism is one of the more interesting but, at the same time, it’s one of the few Lectures that is clearly (out)dated. In essence, it talks about the difficulties involved in applying Maxwell’s equations to the elementary charges themselves, i.e. the electron and the proton. We already signaled some of these problems in previous posts. For example, in our post on the energy in electrostatic fields, we showed how our formulas for the field energy and/or the potential of a charge blow up when we use it to calculate the energy we’d need to assemble a point charge. What comes out is infinity: ∞. So our formulas tell us we’d need an infinite amount of energy to assemble a point charge.

Well… That’s no surprise, is it? The idea itself is impossible: how can one have a finite amount of charge in something that’s infinitely small? Something that has no size whatsoever? It’s pretty obvious we get some division by zero there. 🙂 The mathematical approach is often inconsistent. Indeed, a lot of blah-blah in physics is obviously just about applying formulas to situations that are clearly not within the relevant area of application of the formula. So that’s why I went through the trouble (in my previous post, that is) of explaining you how we get these energy and potential formulas, and that’s by bringing charges (note the plural) together. Now, we may assume these charges are point charges, but that assumption is not so essential. What I tried to say when being so explicit was the following: yes, a charge causes a field, but the idea of a potential makes sense only when we’re thinking of placing some other charge in that field. So point charges with ‘infinite energy’ should not be a problem. Feynman admits as much when he writes:

“If the energy can’t get out, but must stay there forever, is there any real difficulty with an infinite energy? Of course, a quantity that comes out infinite may be annoying, but what really matters is only whether there are any observable physical effects.”

So… Well… Let’s see. There’s another, more interesting, way to look at an electron: let’s have a look at the field it creates. A electron – stationary or moving – will create a field in Maxwell’s world, which we know inside out now. So let’s just calculate it. In fact, Feynman calculates it for the unit charge (+1), so that’s a positron. It eases the analysis because we don’t have to drag any minus sign along. So how does it work? Well…

We’ll have an energy flux density vector – i.e. the Poynting vector S – as well as a momentum density vector g all over space. Both are related through the g = S/c² equation which, as I explained in my previous post, is probably best written as cg = S/c, because we’ve got units then, on both sides, that we can readily understand, like N/m²(so that’s force per unit area) or J/m³ (so that’s energy per unit volume). On the other hand, we’ll need something that’s written as a function of the velocity of our positron, so that’s v, and so it’s probably best to just calculate g, the momentum, which is measured in N·s or kg·(m/s²)·s (both are equivalent units for the momentum p = mv, indeed) per unit volume (so we need to add a 1/ m³ to the unit). So we’ll have some integral all over space, but I won’t bother you with it. Why not? Well… Feynman uses a rather particular volume element to solve the integral, and so I want you to focus on the solution. The geometry of the situation, and the solution for g, i.e. the momentum of the field per unit volume, is what matters here.

So let’s look at that geometry. It’s depicted below. We’ve got a radial electric field—a Coulomb field really, because our charge is moving at a non-relativistic speed, so v << c and we can approximate with a Coulomb field indeed. Maxwell’s equations imply that B = v×E/c², so g = ε₀E×B is what it is in the illustration below. Note that we’d have to reverse the direction of both E and B for an electron (because it’s negative), but g would be the same. It is directed obliquely toward the line of motion and its magnitude is g = (ε₀v/c²)·E²·sinθ. Don’t worry about it: Feynman integrates this thing for you. 🙂 It’s not that difficult, but still… To solve it, he uses the fact that the fields are symmetric about the line of motion, which is indicated by the little arrow around the v-axis, with the Φ symbol next to it (it symbolizes the potential). [The ‘rather particular volume element’ is a ring around the v-axis, and it’s because of this symmetry that Feynman picks the ring. Feynman’s Lectures are not only great to learn physics: they’re a treasure drove of mathematical tricks too. :-)]

As said, I don’t want to bother you with the technicalities of the integral here. This is the result:

What does this say? It says that the momentum of the field – i.e. the electromagnetic momentum, integrated over all of space – is proportional to the velocity v of our charge. That makes sense: when v = 0, we’ll have an electrostatic field all over space and, hence, some inertia, but it’s only when we try to move our charge that Newton’s Law comes into play: then we’ll need some force to overcome that inertia. It all works through the Poynting formula: S = E×B/μ₀. If nothing’s moving, then B = 0, and so we’ll have some E and, therefore, we’ll have field energy alright, but the energy flow will be zero. But when we move the charge, we’re moving the field, and so then B ≠ 0 and so it’s through B that the E in our S equation start kicking in. Does that make sense? Think about it: it’s good to try to visualize things in your mind. 🙂

The constants in the proportionality constant (2e²)/(3ac²) of our p ∼ v formula above are:

e² = q_e²/(4πε₀), with q_ethe electron charge (without the minus sign) and ε₀ our ubiquitous electric constant. [Note that, unlike Feynman, I prefer to not write e in italics, so as to not confuse it with Euler’s number e ≈ 2.71828 etc. However, I know I am not always consistent in my notation. We don’t need Euler’s number in this post, so e or e is always an expression for the electron charge, not Euler’s number. Stupid remark, perhaps, but I don’t want you to be confused.]
a is the radius of our charge—see we got away from the idea of a point charge? 🙂
c² is just c², i.e. our weird constant (the square of the speed of light) which seems to connect everything to everything. Indeed, think about stuff like this: S/g = c²= 1/(ε₀μ₀).

Now, p = mv, so that formula for p basically says that our elementary charge (as mentioned, g is the same for a positron or an electron: E and B will be reversed, but g is not) has an electromagnetic mass m_elecequal to:

That’s an amazing result. We don’t need to give our electron any rest mass: just its charge and its movement will do! Super! So we don’t need any Higgs fields here! 🙂 The electromagnetic field will do!

Well… Maybe. Let’s explore what we’ve got here.

First, let’s compare that radius a in our formula to what’s found in experiments. Huh? Did someone ever try to measure the electron radius? Of course. There are all these scattering experiments in which electrons get fired at atoms. They can fly through or, else, hit something. Therefore, one can some statistical analysis and determine what is referred to as a cross-section. A cross-section is denoted by the same symbol as the standard deviation: σ (sigma). In any case… So there’s something that’s referred to as the classical electron radius, and it’s equal to the so-called Thomsom scattering length. Thomson scattering, as opposed to Compton scattering, is elastic scattering, so it preserves kinetic energy (unlike Compton scattering, where energy gets absorbed and changes frequencies). So… Well… I won’t go into too much detail but, yes, this is the electron radius we need. [I am saying this rather explicitly because there are two other numbers around: the so-called Bohr radius and, as you might imagine, the Compton scattering cross-section.]

The Thomson scattering length is 2.82 femtometer (so that’s 2.82×10⁻¹⁵m), more or less that is :-), and it’s usually related to the observed electron mass m_ethrough the fine-structure constant α. In fact, using Planck units, we can write: r_e·m_e= α, which is an amazing formula but, unfortunately, I can’t dwell on it here. Using ordinary m, s, C and what have you units, we can write r_eas:

That’s good, because if we equate m_eand m_elecand switch m_elecand a in our formula for m_elec, we get:

So, frankly, we’re spot on! Well… Almost. The two numbers differ by 1/3. But who cares about a 1/3 factor indeed? We’re talking rather fuzzy stuff here – scattering cross-sections and standard deviations and all that – so… Yes. Well done! Our theory works!

Well… Maybe. Physicists don’t think so. They think the 1/3 factor is an issue. It’s sad because it really makes a lot of sense. In fact, the Dutch physicist Hendrik Lorentz – whom we know so well by now 🙂 – had also worked out that, because of the length contraction effect, our spherical charge would contract into an ellipsoid and… Well… He worked it all out, and it was not a problem: he found that the momentum was altered by the factor (1−v²/c²)^−1/2, so that’s the ubiquitous Lorentz factor γ! He got this formula in the 1890s already, so that’s long before the theory of relativity had been developed. So, many years before Planck and Einstein would come up with their stuff, Hendrik Antoon Lorentz had the correct formulas already: the mass, or everything really, all should vary with that γ-factor. 🙂

Why bother about the 1/3 factor? [I should note it’s actually referred to as the 4/3 problem in physics.] Well… The critics do have a point: if we assume that (a) an electron is not a point charge – so if we allow it to have some radius a – and (b) that Maxwell’s Laws apply, then we should go all the way. The energy that’s needed to assemble an electron should then, effectively, be the same as the value we’d get out of those field energy formulas. So what do we get when we apply those formulas? Well… Let me quickly copy Feynman as he does the calculation for an electron, not looking at it as a point particle, but as a tiny shell of charge, i.e. a sphere with all charge sitting on the surface:

Let me enlarge the formula:

Now, if we combine that with our formula for m_elec above, then we get:

So that formula does not respect Einstein’s universal mass-energy equivalence formula E = mc². Now, you will agree that we really want Einstein’s mass-energy equivalence relation to be respected by all, so our electron should respect it too. 🙂 So, yes, we’ve got a problem here, and it’s referred to as the 4/3 problem (yes, the ratio got turned around).

Now, you may think it got solved in the meanwhile. Well… No. It’s still a bit of a puzzle today, and the current-day explanation is not really different from what the French scientist Henri Poincaré proposed as a ‘solution’ to the problem back in the 1890s. He basically told Lorentz the following: “If the electron is some little ball of charge, then it should explode because of the repulsive forces inside. So there should be some binding forces there, and so that energy explains the ‘missing mass’ of the electron.” So these forces are effectively being referred to as Poincaré stresses, and the non-electromagnetic energy that’s associated with them – which, of course, has to be equal to 1/3 of the electromagnetic energy (I am sure you see why) 🙂 – adds to the total energy and all is alright now. We get:

U = mc² = (m_elec+ m_Poincaré)c²

So… Yes… Pretty ad hoc. Worse, according to the Wikipedia article on electromagnetic mass, that’s still where we are. And, no, don’t read Feynman’s overview of all of the theories that were around then (so that’s in the 1960s, or earlier). As I said, it’s the one Lecture you don’t want to waste time on. So I won’t do that either.

In fact, let me try to do something else here, and that’s to de-construct the whole argument really. 🙂 Before I do so, let me highlight the essence of what was written above. It’s quite amazing really. Think of it: we say that the mass of an electron – i.e. its inertia, or the proportionality factor in Newton’s F = m·a law of motion – is the energy in the electric and magnetic field it causes. So the electron itself is just a hook for the force law, so to say. There’s nothing there, except for the charge causing the field. But so its mass is everywhere and, hence, nowhere really. Well… I should correct that: the field strength falls of as 1/r²and, hence, the energy flow and momentum density that’s associated with it, falls of as 1/r⁴, so it falls of very rapidly and so the bulk of the energy is pretty near the charge. 🙂

[Note: You’ll remember that the field that’s associated with electromagnetic radiation falls of as 1/r, not as 1/r², which is why there is an energy flux there which is never lost, which can travel independently through space. It’s not the same here, so don’t get confused.]

So that’s something to note: the m_elec= (2c⁻²/3)·(e²/a) has the radius a in it, but that radius is only the hook, so to say. That’s fine, because it is not inconsistent with the idea of the Thomson scattering cross-section, which is the area that one can hit. Now, you’ll wonder how one can hit an electron: you can readily imagine an electron beam aimed at nuclei, but how would one hit electrons? Well… You can shoot photons at them, and see if they bounce back elastically or non-elastically. The cross-section area that bounces them off elastically must be pretty ‘hard’, and the cross-section that deflects them non-elastically somewhat less so. 🙂

OK… But… Yes? Hey! How did we get that electron radius in that formula?

Good question! Brilliant, in fact! You’re right: it’s here that the whole argument falls apart really. We did a substitution. That radius a is the radius of a spherical shell of charge with an energy that’s equal to U_elec= (1/2)·(e²/a), so there’s another way of stating the inconsistency: the equivalent energy of m_elec= (2c⁻²/3)·(e²)/a) is equal to E = m_elec·c²= (2/3)·(e²/a) and that’s not the same as U_elec= (1/2)·(e²/a). If we take the ratio of U_elecand m_elec·c²=, we get the same factor: (1/2)/(2/3) = 3/4. But… Your question is superb! Look at it: putting it the way we put it reveals the inconsistency in the whole argument. We’re mixing two things here:

We first calculate the momentum density, and the momentum, that’s caused by the unit charge, so we get some energy which I’ll denote as E_elec= m_elec·c²
Now, we then assume this energy must be equal to the energy that’s needed to assemble the unit charge from an infinite number of infinitesimally small charges, thereby also assuming the unit charge is a uniformly charged sphere of charge with radius a.
We then use this radius a to simplify our formula for E_elec= m_elec·c²

Now that is not kosher, really! First, it’s (a) a lot of assumptions, both implicit as well as explicit, and then (b) it’s, quite simply, not a legit mathematical procedure: calculating the energy in the field, or calculating the energy we need to assemble a uniformly charged sphere of radius a are two very different things.

Well… Let me put it differently. We’re using the same laws – it’s all Maxwell’s equations, really – but we should be clear about what we’re doing with them, and those two things are very different. The legitimate conclusion must be that our a is wrong. In other words, we should not assume that our electron is spherical shell of charge. So then what? Well… We could easily imagine something else, like a uniform or even a non-uniformly charged sphere. Indeed, if we’re just filling empty space with infinitesimally small charge ‘elements’, then we may want to think the density at the ‘center’ will be much higher, like what’s going on when planets form: the density of the inner core of our own planet Earth is more than four times the density of its surface material. [OK. Perhaps not very relevant here, but you get the idea.] Or, conversely, taking into account Poincaré’s objection, we may want to think all of the charge will be on the surface, just like on a perfect conductor, where all charge is surface charge!

Note that the field outside of a uniformly charged sphere and the field of a spherical shell of charge is exactly the same, so we would not find a different number for E_elec= m_elec·c², but we surely would find a different number for U_elec. You may want to look up some formulas here: you’ll find that the energy of a uniformly distributed sphere of charge (so we do not assume that all of the charge sits on the surface here) is equal to (3/5)·(e²/a). So we’d already have much less of a problem, because the 3/4 factor in the U_elec= (3/4)·m_elec·c² becomes a (5/3)·(2/3) = 10/9 factor. So now we have a discrepancy of some 10% only. 🙂

You’ll say: 10% is 10%. It’s huge in physics, as it’s supposed to be an exact science. Well… It is and it isn’t. Do you realize we haven’t even started to talk about stuff like spin? Indeed, in modern physics, we think of electrons as something that also spins around one or the other axis, so there’s energy there too, and we didn’t include that in our analysis.

In short, Feynman’s approach here is disappointing. Naive even, but then… Well… Who knows? Perhaps he didn’t do this Lecture himself. Perhaps it’s just an assistant or so. In fact, I should wonder why there’s still physicists wasting time on this! I should also note that naively comparing that a radius with the classical electron radius also makes little or no sense. Unlike what you’d expect, the classical electron radius r_e and the Thomson scattering cross-section σ_eare not related like you might think they are, i.e. like σ_e= π·r_e² or σ_e= π·(r_e/2)² or σ_e= r_e² or σ_e= π·(2·r_e)² or whatever circular surface calculation rule that might make sense here. No. The Thomson scattering cross-section is equal to:

σ_e= (8π/3)·r_e² = (2π/3)·(2·r_e)² = (2/3)·π·(2·r_e)² ≈ 66.5×10⁻³⁰m²= 66.5 (fm)²

Why? I am not sure. I must assume it’s got to do with the standard deviation and all that. The point is, we’ve got a 2/3 factor here too, so do we have a problem really? I mean… The a we got was equal to a = (2/3)·r_e, wasn’t it? It was. But, unfortunately, it doesn’t mean anything. It’s just a coincidence. In fact, looking at the Thomson scattering cross-section, instead of the Thomson scattering radius, makes the ‘problem’ a little bit worse. Indeed, applying the π·r² rule for a circular surface, we get that the radius would be equal to (8/3)^1/2·r_e ≈ 1.633·r_e, so we get something that’s much larger rather than something that’s smaller here.

In any case, it doesn’t matter. The point is: this kind of comparisons should not be taken too seriously. Indeed, when everything is said and done, we’re comparing three very different things here:

The radius that’s associated with the energy that’s needed to assemble our electron from infinitesimally small charges, and so that’s based on Coulomb’s law and the model we use for our electron: is it a shell or a sphere of charge? If it’s a sphere, do we want to think of it as something that’s of uniform of non-uniform density.
The second radius is associated with the field of an electron, which we calculate using Poynting’s formula for the energy flow and/or the momentum density. So that’s not about the internal structure of the electron but, of course, it would be nice if we could find some model of an electron that matches this radius.
Finally, there’s the radius that’s associated with elastic scattering, which is also referred to as hard scattering because it’s like the collision of two hard spheres indeed. But so that’s some value that has to be established experimentally and so it involves judicious choices because there’s probabilities and standard deviations involved.

So should we worry about the gaps between these three different concepts? In my humble opinion: no. Why? Because they’re all damn close and so we’re actually talking about the same thing. I mean: isn’t terrific that we’ve got a model that brings the first and the second radius together with a difference of 10% only? As far as I am concerned, that shows the theory works. So what Feynman’s doing in that (in)famous chapter is some kind of ‘dimensional analysis’ which confirms rather than invalidates classical electromagnetic theory. So it shows classical theory’s strength, rather than its weakness. It actually shows our formula do work where we wouldn’t expect them to work. 🙂

The thing is: when looking at the behavior of electrons themselves, we’ll need a different conceptual framework altogether. I am talking quantum mechanics here. Indeed, we’ll encounter other anomalies than the ones we presented above. There’s the issue of the anomalous magnetic moment of electrons, for example. Indeed, as I mentioned above, we’ll also want to think as electrons as spinning around their own axis, and so that implies some circulation of E that will generate a permanent magnetic dipole moment… […] OK, just think of some magnetic field if you don’t have a clue what I am saying here (but then you should check out my post on it). […] The point is: here too, the so-called ‘classical result’, so that’s its theoretical value, will differ from the experimentally measured value. Now, the difference here will be 0.0011614, so that’s about 0.1%, i.e. 100 times smaller than my 10%. 🙂

Personally, I think that’s not so bad. 🙂 But then physicists need to stay in business, of course. So, yes, it is a problem. 🙂

Post scriptum on the math versus the physics

The key to the calculation of the energy that goes into assembling a charge was the following integral:

This is a double integral which we simplified in two stages, so we’re looking at an integral within an integral really, but we can substitute the integral over the ρ(2)·dV₂product by the formula we got for the potential, so we write that as Φ(1), and so the integral above becomes:

We then established that this integral was mathematically equivalent to the following equation:

What’s ‘too small’? Let’s look at the formula we got for our electron as a spherical shell of charge:

So we’ve got an even simpler formula here: it’s just a 1/r relation. Why is that? Well… It’s just the way the math turns it out. I copied the detail of Feynman’s calculation above, so you can double-check it. It’s quite wonderful, really. We have a very simple inversely proportional relationship between the radius of our electron and its energy as a sphere of charge. We could write it as:

U_elect = α/a , with α = e²/2

But – Hey! Wait a minute! We’ve seen something like this before, haven’t we? We did. We did when we were discussing the wonderful properties of that magical number, the fine-structure constant, which we also denoted by α. 🙂 However, because we used α already, I’ll denote the fine-structure constant as α_ehere, so you don’t get confused. As you can see, the fine-structure constant links all of the fundamental properties of the electron: its charge, its radius, its distance to the nucleus (i.e. the Bohr radius), its velocity, and its mass (and, hence, its energy). So, at this stage of the argument, α can be anything, and α_ecannot, of course. It’s just that magical number out there, which relates everything to everything: it’s the God-given number we don’t understand. 🙂 Having said that, it seems like we’re going to get some understanding here because we know that, one the many expressions involving α_ewas the following one:

m_e = α_e/r_e

This says that the mass of the electron is equal to the ratio of the fine-structure constant and the electron radius. [Note that we express everything in natural units here, so that’s Planck units. For the detail of the conversion, please see the relevant section on that in my one of my posts on this and other stuff.] Now, mass is equivalent to energy, of course: it’s just a matter of units, so we can equate m_e with E_e (this amounts to expressing the energy of the electron in a kg unit—bit weird, but OK) and so we get:

E_e = α_e/r_e

So there we have: the fine-structure constant α_e is Nature’s ‘cut-off’ factor, so to speak. Why? Only God knows. 🙂 But it’s now (fairly) easy to see why all the relations involving α_e are what they are. For example, we also know that α_e is the square of the electron charge expressed in Planck units, so we have:

α = e_P² and, therefore, E_e = e_P²/r_e

Now, you can check for yourself: it’s just a matter of re-expressing everything in standard SI units, and relating e_P² to e², and it should all work: you should get the U_elect = (1/2)·e²/a expression. So… Well… At least this takes some of the magic out the fine-structure constant. It’s still a wonderful thing, but so you see that the fundamental relationship between (a) the energy (and, hence, the mass), (b) the radius and (c) the charge of an electron is not something God-given. What’s God-given are Maxwell’s equations, and so the E_e = α_e/r_e= e_P²/r_e is just one of the many wonderful things that you can get out of them. 🙂

Field energy and field momentum

This post goes to the heart of the E = mc², equation. It’s kinda funny, because Feynman just compresses all of it in a sub-section of his Lectures. However, as far as I am concerned, I feel it’s a very crucial section. Pivotal, I’d say, which would fit with its place in all of the 115 Lectures that make up the three volumes, which is sort of mid-way, which is where we are here. So let’s get go for it. 🙂

Let’s first recall what we wrote about the Poynting vector S, which we calculate from the magnetic and electric field vectors E and B by taking their cross-product:

This vector represents the energy flow, per unit area and per unit time, in electrodynamical situations. If E and/or B are zero (which is the case in electrostatics, for example, because we don’t have magnetic fields in electrostatics), then S is zero too, so there is no energy flow then. That makes sense, because we have no moving charges, so where would the energy go to?

I also made it clear we should think of S as something physical, by comparing it to the heat flow vector h, which we presented when discussing vector analysis and vector operators. The heat flow out of a surface element da is the area times the component of h perpendicular to da, so that’s (h•n)·da = h_n·da. Likewise, we can write (S•n)·da = S_n·da. The units of S and h are also the same: joule per second and per square meter or, using the definition of the watt (1 W = 1 J/s), in watt per square meter. In fact, if you google a bit, you’ll find that both h and S are referred to as a flux density:

The heat flow vector h is the heat flux density vector, from which we get the heat flux through an area through the (h•n)·da = h_n·da product.
The energy flow S is the energy flux density vector, from which we get the energy flux through the (S•n)·da = S_n·da product.

So that should be enough as an introduction to what I want to talk about here. Let’s first look at the energy conservation principle once again.

Local energy conservation

In a way, you can look at my previous post as being all about the equation below, which we referred to as the ‘local’ energy conservation law:

Of course, it is not the complete energy conservation law. The local energy is not only in the field. We’ve got matter as well, and so that’s what I want to discuss here: we want to look at the energy in the field as well as the energy that’s in the matter. Indeed, field energy is conserved, and then it isn’t: if the field is doing work on matter, or matter is doing work on the field, then… Well… Energy goes from one to the other, i.e. from the field to the matter or from the matter to the field. So we need to include matter in our analysis, which we didn’t do in our last post. Feynman gives the following simple example: we’re in a dark room, and suddenly someone turns on the light switch. So now the room is full of field energy—and, yes, I just mean it’s not dark anymore. :-). So that means some matter out there must have radiated its energy out and, in the process, it must have lost the equivalent mass of that energy. So, yes, we had matter losing energy and, hence, losing mass.

Now, we know that energy and momentum are related. Respecting and incorporating relativity theory, we’ve got two equivalent formulas for it:

E²− p²c² = m₀²c⁴
pc = E·(v/c) ⇔ p = v·E/c²= m·v

The E = mc² and m = ·m₀·(1−v²/c²)^−1/2 formulas connect both expressions. So we can look at it in either of two ways. We could use the energy conservation law, but Feynman prefers the conservation of momentum approach, so let’s see where he takes us. If the field has some energy (and, hence, some equivalent mass) per unit volume, and if there’s some flow, so if there’s some velocity (which there is: that’s what our previous post was all about), then it will have a certain momentum per unit volume. [Remember: momentum is mass times velocity.] That momentum will have a direction, so it’s a vector, just like p = mv. We’ll write it as g, so we define g as:

g is the momentum of the field per unit volume.

What units would we express it in? We’ve got a bit of choice here. For example, because we’re relating everything to energy here, we may want to convert our kilogram into eV/c²or J/c²units, using the mass-energy equivalence relation E = mc². Hmm… Let’s first keep the kg as a measure of inertia though. So we write: [g] = [m]·[v]/m³= (kg·m/s)/m³. Hmm… That doesn’t show it’s energy, so let’s replace the kg with a unit that’s got newton and meter in it, cf. the F = ma law. So we write: [g] = (kg·m/s)/m³= (kg/s)/m²= [(N·s²/m)/s]/m²= N·s/m³. Well… OK. The newton·second is the unit of momentum indeed, and we can re-write it including the joule (1 J = 1 N·m), so then we get [g] = (J·s/m⁴), so what’s that? Well… Nothing much. However, I do note it happens to be the dimension of S/c², so that’s [S/c²] = [J/(s·m²)]·(s²/m²) = (J·s/m⁴). 🙂 Let’s continue the discussion.

Now, momentum is conserved, and each component of it is conserved. So let’s look at the x-direction. We should have something like:

If you look at this carefully, you’ll probably say: “OK. I understood the thing with the dark room and light switch. Mass got converted into field energy, but what’s that second term of the left?”

Good. Smart. Right remark. Perfect. […] Let me try to answer the question. While all of the quantities above are expressed per unit volume, we’re actually looking at the same infinitesimal volume element here, so the example of the light switch is actually an example of a ‘momentum outflow’, so it’s actually an example of that second term of the left-hand side of the equation above kicking in! 🙂

Indeed, the first term just sort of reiterates the mass-energy equivalence: the energy that’s in the matter can become field energy, so to speak, in our infinitesimal volume element itself, and vice versa. But if it doesn’t, then it should get out and, hence, become ‘momentum outflow’. Does that make sense? No?

Hmm… What to say? You’ll need to look at that equation a couple of times more, I guess. But I need to move on, unfortunately. [Don’t get put off when I say things like this: I am basically talking to myself, so it means I’ll need to re-visit this myself. :-/]

Let’s look at all of the three terms:

The left-hand side (i.e. the time rate-of-change of the momentum of matter) is easy. It’s just the force on it, which we know is equal to F = $q (E + v \times B). Do we know that? OK\dots I’ll admit it. Sometimes it’s easy to forget where we are in an analysis like this, but so we’re looking at the electromagnetic force here. 🙂 As we’re talking infinitesimals here and, therefore, charge density rather than discrete charges, we should re-write this as the force per unit volume which is$ $ρ E + j \times B . [This is an interesting formula which I didn’t use before, so you should double-check it. :-)]$
The first term on the right-hand side should be equally obvious, or… Well… Perhaps somewhat less so. But with all my rambling on the Uncertainty Principle and/or the wave-particle duality, it should make sense. If we scrap the second term on the right-hand side, we basically have an equation that is equivalent to the E = mc² equation. No? Sorry. Just look at it, again and again. You’ll end up understanding it. 🙂
So it’s that second term on the right-hand side. What the hell does that say? Well… I could say: it’s the local energy or momentum conservation law. If the energy or momentum doesn’t stay in, it has to go out. 🙂 But that’s not very satisfactory as an answer, of course. However, please just go along with this ‘temporary’ answer for a while.

So what is that second term on the right-hand side? As we wrote it, it’s an x-component – or, let’s put it differently, it is or was part of the x-component of the momentum density – but, frankly, we should probably allow it to go out in any direction really, as the only constraint on the left-hand side is a per second rate of change of something. Hence, Feynman suggest to equate it to something like this:

What a, b and c? The components of some vector? Not sure. We’re stuck. This piece really requires very advanced math. In fact, as far as I know, this is the only time where Feynman says: “Sorry. This is too advanced. I’ll just give you the equation. Sorry.” So that’s what he does. He explains the philosophy of the argument, which is the following:

On the left-hand side, we’ve got the time rate-of-change of momentum, so that obeys the F = dp/dt = d(mv)/dt law, with the force F, $per unit volume, being equal to F (unit volume) =$ $ρ E + j \times B .$
$On the right-hand side, we’ve got something that can be written as:$

So we’d need to find a way to $ρ E + j \times B$ in terms of $E$ and $B only -$ eliminating $ρ$ and $j$ by using Maxwell’s equations or whatever other trick – and then juggle terms and make substitutions to get it into a form that looks like the formula above, i.e. the right-hand side of that equation. But so Feynman doesn’t show us how it’s being done. He just mentions some theorem in physics, which says that the energy that’s flowing through a unit area per unit time divided by c² – so that’s E/c²per unit area and per unit time – must be equal to the momentum per unit volume in the space, so we write:

g = S/c²

He illustrates the general theorem that’s used to get the equation above by giving two examples:

OK. Two good examples. However, it’s still frustrating to not see how we get the g = S/c² in the specific context of the electromagnetic force, so let’s do a dimensional analysis at least. In my previous post, I showed that the dimension of S must be J/(m²·s), so [S/c²] = [J/(m²·s)]/(m²/s²) = [N·m/(m²·s)]·(s²/m²) = [N·s/m³]. Now, we know that the unit of mass is 1 kg = N/(m/s²). That’s just the force law: a force of 1 newton will give a mass of 1 kg an acceleration of 1 m/s per second, so 1 N = 1 kg·(m/s²). So the [N·s/m³] dimension is equal to [kg·(m/s²)·s/m³] = [(kg·(m/s)/m³] = [(kg·(m/s)]/m³ $, which is the dimension of momentum (p = m v) per unit volume, indeed. So, yes, the dimensional analysis works out, and it’s also in line with the p = v\cdot E / c 2$ $= m\cdot v equation, but\dots Oh\dots We did a dimensional analysis already, where we also showed that [g] = [S / c 2] = (J\cdots/m 4). Well\dots In any case\dots It’s a bit frustrating to not see the detail here, but let us note the the Grand Result once again:$

The Poynting vector S gives us the energy flow as well as the $momentum density$ $g$ = S/c².

But what does it all mean, really? Let’s go through Einstein’s illustration of the principle. That will help us a lot. Before we do, however, I’d like to note something. I’ve always wondered a bit about that dichotomy between energy and momentum. Energy is force times distance: 1 joule is 1 newton × 1 meter indeed (1 J = 1 N·m). Momentum is force times time, as we can express it in N·s. Planck’s constant h combines all three in the dimension of action, which is force times distance times time: h ≈ 6.6×10⁻³⁴ N·m·s, indeed. I like that unity. In this regard, you should, perhaps, quickly review that post in which I explain that h is the energy per cycle, i.e. per wavelength or per period, of a photon, regardless of its wavelength. So it’s really something very fundamental.

We’ve got something similar here: energy and momentum coming together, and being shown as one aspect of the same thing: some oscillation. Indeed, just see what happens with the dimensions when we ‘distribute’ the 1/c²factor on the right-hand side over the two sides, so we write: $c$ $\cdot$ $g$ = S/c and work out the dimensions:

$[$ $c$ $\cdot$ $g$ $]$ = (m/s)·(N·s)/m³= N/m²= J/m³.
[S/c] = (s/m)·(N·m)/(s·m²) = N/m²= J/m³.

Isn’t that nice? Both sides of the equation now have a dimension like ‘the force per unit area’, or ‘the energy per unit volume’. To get that, we just re-scaled g and S, by c and 1/c respectively. As far as I am concerned, this shows an underlying unity we probably tend to mask with our ‘related but different’ energy and momentum concepts. It’s like E and B: I just love it we can write them together in our Poynting formula S = ε₀c²E×B. In fact, let me show something else here, which you should think about. You know that c²= 1/(ε₀μ₀), so we can write S also as S = E×B/μ₀. That’s nice, but what’s nice too is the following:

S/c = $c$ $\cdot$ $g$ = ε₀cE×B = E×B/μ₀c
S/g = c²= 1/(ε₀μ₀)

So, once again, Feynman may feel the Poynting vector is sort of counter-intuitive when analyzing specific situations but, as far as I am concerned, I feel the Poyning vector makes things actually easier to understand. Instead of two E and B vectors, and two concepts to deal with ‘energy’ (i.e. energy and momentum), we’re sort of unifying things here. In that regard – i.e in regard of feeling we’re talking the same thing really – I’d really highlight the S/g = c²= 1/(ε₀μ₀) equation. Indeed, the universal constant c acts just like the fine-structure constant here: it links everything to everything. 🙂

And, yes, it’s also about time we introduce the so-called principle of least action to explain things, because action, as a concept, combines force, distance and time indeed, so it’s a bit more promising than just energy, of just momentum. Having said that, you’ll see in the next section that it’s sometimes quite useful to have the choice between one formula or the other. But… Well… Enough talk. Let’s look at Einstein’s car.

Einstein’s car

Einstein’s car is a wonderful device: it rolls without any friction and it moves with a little flashlight. That’s all it needs. It’s pictured below. 🙂 So the situation is the following: the flashlight shoots some light out from one side, which is then stopped at the opposite end of the car. When the light is emitted, there must be some recoil. In fact, we know it’s going to be equal to 1/c times the energy because all we need to do is apply the pc = E·(v/c) formula for v = c, so we know that p = E/c. Of course, this momentum now needs to move Einstein’s car. It’s frictionless, so it should work, but still… The car has some mass M, and so that will determine its recoil velocity: v = p/M. We just apply the general p = mv formula here, and v is not equal to c here, of course! Of course, then the light hits the opposite end of the car and delivers the same momentum, so that stops the car again. However, it did move over some distance x = vt. So we could flash our light again and get to wherever we want to get. [Never mind the infinite accelerations involved!] So… Well… Great! Yes, but Einstein didn’t like this car when he first saw it. In fact, he still doesn’t like it, because he knows it won’t take you very far. 🙂

The problem is that we seem to be moving the center of gravity of this car by fooling around on the inside only. Einstein doesn’t like that. He thinks it’s impossible. And he’s right of course. The thing is: the center of gravity did not change. What happened here is that we’ve got some blob of energy, and so that blob has some equivalent mass (which we’ll denote by U/c²), and so that equivalent mass moved all the way from one side to the other, i.e. over the length of the car, which we denote by L. In fact, it’s stuff like this that inspired the whole theory of the field energy and field momentum, and how it interacts with matter.

What happens here is like switching the light on in the dark room: we’ve got matter doing work on the field, and so matter loses mass, and the field gains it, through its momentum and/or energy. To calculate how much, we could integrate S/c or $c$ $\cdot$ $g$ over the volume of our blob, and we’d get something in joule indeed, but there’s a simpler way here. The momentum conservation says that the momentum of our car and the momentum of our blob must be equal, so if T is the time that was needed for our blob to go to the other side – and so that’s, of course, also the time during which our car was rolling – then M·v = M·x/T must be equal to (U/c²)·c = (U/c²)·L/T. The 1/T factor on both sides cancel, so we write: M·x = (U/c²)·L. Now, what is x? Yes. In case you were wondering, that’s what we’re looking for here. 🙂 Here it is:

x = vT = vL/c = (p/M)·(L/c) = [U/c)/M]·(L/c) = (U/c²)·(L/M)

So what’s next? Well… Now we need to show that the center-of-mass actually did not move with this ‘transfer’ of the blob. I’ll leave the math to you here: it should all work out. And you can also think through the obvious questions:

Where is the energy and, hence, the mass of our blob after it stops the car? Hint: think about excited atoms and imagine they might radiate some light back. 🙂
As the car did move a little bit, we should be able to move it further and further away from its center of gravity, until the center of gravity is no longer in the car. Hint: think about batteries and energy levels going down while shooting light out. It just won’t happen. 🙂

Now, what about a blob of light going from the top to the bottom of the car? Well… That involves the conservation of angular momentum: we’ll have more mass on the bottom, but on a shorter lever-arm, so angular momentum is being conserved. It’s a very good question though, and it led Einstein to combine the center-of-gravity theorem with the angular momentum conservation theorem to explain stuff like this.

It’s all fascinating, and one can think of a great many paradoxes that, at first, seem to contradict the Grand Principles we used here, which means that they would contradict all that we have learned so far. However, a careful analysis of those paradox reveals that they are paradoxes indeed: propositions which sound true but are, in the end, self-contradictory. In fact, when explaining electromagnetism over his various Lectures, Feynman tasks his readers with a rather formidable paradox when discussing the laws of induction, he solves it here, ten chapters later, after describing what we described above. You can busy yourself with it but… Well… I guess you’ve got something better to do. If so, just take away the key lesson: there’s momentum in the field, and it’s also possible to build up angular momentum in a magnetic field and, if you switch it off, the angular momentum will be given back, somehow, as it’s stored energy.

That’s also why the seemingly irrelevant circulation of S we discussed in my previous post, where we had a charge next to an ordinary magnet, and where we found that there was energy circulating around, is not so queer. The energy is there, in the circulating field, and it’s real. As real as can be. 🙂

The energy of fields and the Poynting vector

For some reason, I always thought that Poynting was a Russian physicist, like Minkowski. He wasn’t. I just looked it up. Poynting was an Englishman, born near Manchester, and he teached in Birmingham. I should have known. Poynting is a very English name, isn’t it? My confusion probably stems from the fact that it was some Russian physicist, Nikolay Umov, who first proposed the basic concepts we are going to discuss here, i.e. the speed and direction of energy itself, or its movement. And as I am double-checking, I just learned that Hermann Minkowski is generally considered to be German-Jewish, not Russian. Makes sense. With Einstein and all that. His personal life story is actually quite interesting. You should check it out. 🙂

Let’s go for it. We’ve done a few posts on the energy in the fields already, but all in the contexts of electrostatics. Let me first walk you through the ideas we presented there.

The basic concepts: force, work, energy and potential

1. A charge q causes an electric field E, and E‘s magnitude E is a simple function of the charge (q) and its distance (r) from the point that we’re looking at, which we usually write as P = (x, y, z). Of course, the origin of our reference frame here is q. The formula is the simple inverse-square law that you (should) know: E ∼ q/r², and the proportionality constant is just Coulomb’s constant, which I think you wrote as k_e in your high-school days and which, as you know, is there so as to make sure the units come out alright. So we could just write E = k_e·q/r². However, just to make sure it does not look like a piece of cake 🙂 physicists write the proportionality constant as 1/4πε₀, so we get:

Now, the field is the force on any unit charge (+1) we’d bring to P. This led us to think of energy, potential energy, because… Well… You know: energy is measured by work, so that’s some force acting over some distance. The potential energy of a charge increases if we move it against the field, so we wrote:

Well… We actually gave the formula below in that post, so that’s the work done per unit charge. To interpret it, you just need to remember that F = qE, which is equivalent to saying that E is the force per unit charge.

As for the F•ds or E•ds product in the integrals, that’s a vector dot product, which we need because it’s only the tangential component of the force that’s doing work, as evidenced by the formula F•ds = |F|·|ds|·cosθ = F_t·ds, and as depicted below.

Now, this allowed us to describe the field in terms of the (electric) potential Φ and the potential differences between two points, like the points a and b in the integral above. We have to chose some reference point, of course, some P₀ defining zero potential, which is usually infinitely far away. So we wrote our formula for the work that’s being done on a unit charge, i.e. W(unit) as:

2. The world is full of charges, of course, and so we need to add all of their fields. But so now you need a bit of imagination. Let’s reconstruct the world by moving all charges out, and then we bring them back one by one. So we take q₁ now, and we bring it back into the now-empty world. Now that does not require any energy, because there’s no field to start with. However, when we take our second charge q₂, we will be doing work as we move it against the field or, if it’s an opposite charge, we’ll be taking energy out of the field. Huh? Yes. Think about it. All is symmetric. Just to make sure you’re comfortable with every step we take, let me jot down the formula for the force that’s involved. It’s just the Coulomb force of course:

F₁is the force on charge q₁, and F₂is the force on charge q₂. Now, q₁and q₂. may attract or repel each other but the forces will always be equal and opposite. The e₁₂ vector makes sure the directions and signs come out alright, as it’s the unit vector from q₂to q₁(not from q₁to q₂, as you might expect when looking at the order of the indices). So we would need to integrate this for r going from infinity to… Well… The distance between q₁and q₂ – wherever they end up as we put them back into the world – so that’s what’s denoted by r₁₂. Now I hate integrals too, but this is an easy one. Just note that ∫ r⁻²dr = 1/r and you’ll be able to figure out that what I’ll write now makes sense (if not, I’ll do a similar integral in a moment): the work done in bringing two charges together from a large distance (infinity) is equal to:

So now we should bring in q₃and then q₄, of course. That’s easy enough. Bringing the first two charges into that world we had emptied took a lot of time, but now we can automate processes. Trust me: we’ll be done in no time. 🙂 We just need to sum over all of the pairs of charges q_i and q_j. So we write the total electrostatic energy U as the sum of the energies of all possible pairs of charges:

Huh? Can we do that? I mean… Every new charge that we’re bringing in here changes the field, doesn’t it? It does. But it’s the magic of the superposition principle at work here. Our third charge q₃is associated with two pairs in this formula. Think of it: we’ve got the q₁q₃and the q₂q₃combination, indeed. Likewise, our fourh charge q₄is to be paired up with three charges now: q₁, q₁ and q₃. This formula takes care of it, and the ‘all pairs’ mention under the summation sign (Σ) reminds us we should watch we don’t double-count pairs: the q₁q₃and q₃q₁combination, for example, count for one pair only, obviously. So, yes, we write ‘all pairs’ instead of the usual i, j subscripts. But then, yes, this formula takes care of it. We’re done!

Well… Not really, of course. We’ve still got some way to go before I can introduce the Poynting vector. 🙂 However, to make sure you ‘get’ the energy formula above, let me insert an extremely simple diagram so you’ve got a bit of a visual of what we’re talking about.

3. Now, let’s take a step back. We just calculated the (potential) energy of the world (U), which is great. But perhaps we should also be interested in the world’s potential Φ, rather than its potential energy U. Why? Well, we’ll want to know what happens when we bring yet another charge in—from outer space or so. 🙂 And so then it’s easier to know the world’s potential, rather than its energy, because we can calculate the field from it using the E = −∇Φ formula. So let’s de- and re-construct the world once again 🙂 but now we’ll look at what happens with the field and the potential.

We know our first charge created a field with a field strength we calculated as:

So, when bringing in our second charge, we can use our Φ(P) integral to calculate the potential:

[Let me make a note here, just for the record. You probably think I am being pretty childish when talking about my re-construction of the world in terms of bringing all charges out and then back in again but, believe me, there will be a lot of confusion when we’ll start talking about the energy of one charge, and that confusion can be avoided, to a large extent, when you realize that the idea (I mean the concept itself, really—not its formula) of a potential involves two charges really. Just remember: it’s the first charge that causes the field (and, of course, any charge causes a field), but calculating a potential only makes sense when we’re talking some other charge. Just make a mental note of it. You’ll be grateful to me later.]

Let’s now combine the integral and the formula for E above. Because you hate integrals as much as I do, I’ll spell it out: the antiderivative of the Φ(P) integral is ∫ q/(4πε₀r²)·dr. Now, let’s bring q/4πε₀out for a while so we can focus on solving ∫(1/r²)dr. Now, ∫(1/r²)dr is equal to –1/r + k, and so the whole antiderivative is –q/4πε₀r + k. Now, we integrate from r = ∞ to r, and so the definite integral is [–q/(4πε₀)]·[1/∞ − 1/r] = [–q/(4πε₀)]·[0 − 1/r] = q/(4πε₀r). Let me present this somewhat nicer:

You’ll say: so what? Well… We’re done! The only thing we need to do now is add up the potentials of all of the charges in the world. So the formula for the potential Φ at a point which we’ll simply refer to as point 1, is:

Note that our index j starts at 2, otherwise it doesn’t make sense: we’d have a division by zero for the q₁/r₁₁ term. Again, it’s an obvious remark, but not thinking about it can cause a lot of confusion down the line.

4. Now, I am very sorry but I have to inform you that we’ll be talking charge densities and all that shortly, rather than discrete charges, so I have to give you the continuum version of this formula, i.e. the formula we’ll use when we’ve got charge densities rather than individual charges. That sum above then becomes an infinite sum (i.e. an integral), and q_j becomes a variable which we write as ρ(2). [That’s totally in line with our index j starts at 2, rather than from 1.] We get:

Just look at this integral, and try to understand it: we’re integrating over all of space – so we’re integrating the whole world, really 🙂 – and the ρ(2)·dV₂product in the integral is just the charge of an infinitesimally small volume of our world. So the whole integral is just the (infinite) sum of the contributions to the potential (at point 1) of all (infinitesimally small) charges that are around indeed. Now, there’s something funny here. It’s just a mathematical thing: we don’t need to worry about double-counting here. Why? We’re not having products of volumes here. Just make a mental note of it because it will be different in a moment.

Now we’re going to look at the continuum version for our energy formula indeed. Which energy formula? That electrostatic energy formula, which said that the total electrostatic energy U as the sum of the energies of all possible pairs of charges:

Its continuum version is the following monster:

Hmm… What kind of integral is that? We’ve got two variables here: dV₂ and dV₁. Yes. And we’ve also got a 1/2 factor now, because we do not want to double-count and, unfortunately, there is no convenient way of writing an integral like this that keeps track of the pairs. It’s a so-called double integral, but I’ll let you look up the math yourself. In any case, we can simplify this integral so you don’t need to worry about it too much. How do we simplify it? Well… Just look at that integral we got for Φ(1): we calculated the potential at point 1 by integrating the ρ(2)·dV₂product over all of space, so the integral above can be written as:

But so this integral integrates the ρ(1)·Φ(1)·dV₁product over all of space, so that’s over all points in space. So we can just drop the index and write the whole thing as the integral of ρ·Φ·dV over all of space:

5. It’s time for the hat-trick now. The equation above is mathematically equivalent to the following equation:

Huh? Yes. Let me make two remarks here. First on the math, the E = −∇Φ formula allows you to the integrand of the integral above as E•E = (−∇Φ)•(−∇Φ) = (∇Φ)•(∇Φ). And then you may or may not remember that, when substituting E = −∇Φ in Maxwell’s first equation (∇•E = ρ/ε₀), we got the following equality: ρ = ε₀·∇•(∇Φ) = ε₀·∇²Φ, so we can write ρΦ as ε₀·Φ·∇²Φ. However, that still doesn’t show the two integrals are the same thing. The proof is actually rather involved, and so I’ll refer to that post I referred to, so you can check the proof there.

The second remark is much more fundamental. The two integrals are mathematically equivalent, but are they also physically? What do I mean with that? Well… Look at it. The second integral implies that we can look at (ε₀/2)·E•E = ε₀E²/2 as an energy density, which we’ll denote by u, so we write:

Just to make sure you ‘get’ what we’re talking about here: u is the energy density in the little cube dV in the rather simplistic (and, therefore, extremely useful) illustration below (which, just like most of what I write above, I got from Feynman).

Now the question: what is the reality of that formula? Indeed, what we did when calculating U amounted to calculating the Universe with some number U – and that’s kinda nice, of course! – but then what? Is u = ε₀E²/2 anything real? Well… That’s what this post is about. So we’re finished with the introduction now. 🙂

Energy density and energy flow in electrodynamics

Before giving you any more formulas, let me answer the question: there is no doubt, in the classical theory of electromagnetism at least, that the energy density u is something very real. It has to be because of the charge conservation law. Charges cannot just disappear in space, to then re-appear somewhere else. The charge conservation law is written as ∇•j = −∂ρ/∂t, and that makes it clear it’s a local conservation law. Therefore, charges can only disappear and re-appear through some current. We write dQ₁/dt = ∫ (j•n)·da = −dQ₂/dt, and here’s the simple illustration that comes with it:

So we do not allow for any ‘non-local’ interactions here! Therefore, we say that, if energy goes away from a region, it’s because it flows away through the boundaries of that region. So that’s what the Poynting formulas are all about, and so I want to be clear on that from the outset.

Now, to get going with the discussion, I need to give you the formula for the energy density in electrodynamics. Its shape won’t surprise you:

However, it’s just like the electrostatic formula: it takes quite a bit of juggling to get this from our electrodynamic equations, so, if you want to see how it’s done, I’ll refer you to Feynman. Indeed, I feel the derivation doesn’t matter all that much, because the formula itself is very intuitive: it’s really the thing everyone knows about a wave, electromagnetic or not: the energy in it is proportional to the square of its amplitude, and so that’s E•E = E² and B•B = B². Now, you also know that the magnitude of B is 1/c of that of E, so cB = E, and so that explains the extra c² factor in the second term.

The second formula is also very intuitive. Let me write it down:

Just look at it: u is the energy density, so that’s the amount of energy per unit volume at a given point, and so whatever flows out of that point must represent its time rate of change. As for the –∇•S expression… Well… Sorry, I can’t keep re-explaining things: the ∇• operator is the divergence, and so it give us the magnitude of a (vector) field’s source or sink at a given point. ∇•C is a scalar, and if it’s positive in a region, then that region is a source. Conversely, if it’s negative, then it’s a sink. To be precise, the divergence represents the volume density of the outward flux of a vector field from an infinitesimal volume around a given point. So, in this case, it gives us the volume density of the flux of S. As you can see, the formula has exactly the same shape as ∇•j = −∂ρ/∂t.

So what is S? Well… Think about the more general formula for the flux out of some closed surface, which we get from integrating over the volume enclosed. It’s just Gauss’ Theorem:

Just replace C by E, and think about what it meant: the flux of E was the field strength multiplied by the surface area, so it was the total flow of E. Likewise, S represents the flow of (field) energy. Let me repeat this, because it’s an important result:

S represents the flow of field energy.

Huh? What flow? Per unit area? Per second? How do you define such ‘flow’? Good question. Let’s do a dimensional analysis:

E is measured in newton per coulomb, so [E•E] = [E²] = N²/C².
B is measured in (N/C)/(m/s). [Huh? Well… Yes. I explained that a couple of times already. Just check it in my introduction to electric circuits.] So we get [B•B] = [B²] = (N²/C²)·(s²/m²) but the dimension of our c² factor is (m²/s²) so we’re left with N²/C². That’s nice, because we need to add in the same units.
Now we need to look at ε₀. That constant usually ‘fixes’ our units, but can we trust it to do the same now? Let’s see… One of the many ways in which we can express its dimension is [ε₀] = C²/(N·m²), so if we multiply that with N²/C², we find that u is expressed in N/m². Wow! That’s kinda neat. Why? Well… Just multiply with m/m and its dimension becomes N·m/m³= J/m³, so that’s joule per cubic meter, so… Yes: u has got the right unit for something that’s supposed to measure energy density!
OK. Now, we take the time rate of change of u, and so both the right and left of our ∂u/∂t = −∇•S formula are expressed in (J/m³)/s, which means that the dimension of S itself must be J/(m²·s). Just check it by writing it all out: ∇•S = ∂S_x/∂x + ∂S_y/∂x + ∂S_z/∂z, and so that’s something per meter so, to get the dimension of S itself, we need to go from cubic meter to square meter. Done! Let me highlight the grand result:

S is the energy flow per unit area and per second.

Now we’ve got its magnitude and its dimension, but what is its direction? Indeed, we’ve been writing S as a vector, but… Well… What’s its direction indeed?

Well… Hmm… I referred you to Feynman for that derivation of that u = ε₀E²/2 + ε₀c²B²/2 formula energy for u, and so the direction of S – I should actually say, its complete definition – comes out of that derivation as well. So… Well… I think you should just believe what I’ll be writing here for S:

So it’s the vector cross product of E and B with ε₀c²thrown in. It’s a simple formula really, and because I didn’t drag you through the whole argument, you should just quickly do a dimensional analysis again—just to make sure I am not talking too much nonsense. 🙂 So what’s the direction? Well… You just need to apply the usual right-hand rule:

OK. We’re done! This S vector, which – let me repeat it – represents the energy flow per unit area and per second, is what is referred to as Poynting’s vector, and it’s a most remarkable thing, as I’ll show now. Let’s think about the implications of this thing.

Poynting’s vector in electrodynamics

The S vector is actually quite similar to the heat flow vector h, which we presented when discussing vector analysis and vector operators. The heat flow out of a surface element da is the area times the component of h perpendicular to da, so that’s (h•n)·da = h_n·da. Likewise, we can write (S•n)·da = S_n·da. The units of S and h are also the same: joule per second and per square meter or, using the definition of the watt (1 W = 1 J/s), in watt per square meter. In fact, if you google a bit, you’ll find that both h and S are referred to as a flux density:

The heat flow vector h is the heat flux density vector, from which we get the heat flux through an area through the (h•n)·da = h_n·da product.
The energy flow S is the energy flux density vector, from which we get the energy flux through the (S•n)·da = S_n·da product.

The big difference, of course, is that we get h from a simpler vector equation:

h = −κ∇T ⇔ (h_x, h_y, h_z) = −κ(∂T_x/∂x, ∂T_y/∂y,∂T_z/∂x)

The vector equation for S is more complicated:

So it’s a vector product. Note that S will be zero if E = 0 and/or if B = 0. So S = 0 in electrostatics, i.e. when there are no moving charges and only steady currents. Let’s examine Feynman’s examples.

The illustration below shows the geometry of the E, B and S vectors for a light wave. It’s neat, and totally in line with what we wrote on the radiation pressure, or the momentum of light. So I’ll refer you to that post for an explanation, and to Feynman himself, of course.

OK. The situation here is rather simple. Feynman gives a few others examples that are not so simple, like that of a charging capacitor, which is depicted below.

The Poynting vector points inwards here, toward the axis. What does it mean? It means the energy isn’t actually coming down the wires, but from the space surrounding the capacitor.

What? I know. It’s completely counter-intuitive, at first that is. You’d think it’s the charges. But it actually makes sense. The illustration below shows how we should think of it. The charges outside of the capacitor are associated with a weak, enormously spread-out field that surrounds the capacitor. So if we bring them to the capacitor, that field gets weaker, and the field between the plates gets stronger. So the field energy which is way out moves into the space between the capacitor plates indeed, and so that’s what Poynting’s vector tells us here.

Hmm… Yes. You can be skeptic. You should be. But that’s how it works. The next illustration looks at a current-carrying wire itself. Let’s first look at the B and E vectors. You’re familiar with the magnetic field around a wire, so the B vector makes sense, but what about the electric field? Aren’t wires supposed to be electrically neutral? It’s a tricky question, and we handled it in our post on the relativity of fields. The positive and negative charges in a wire should cancel out, indeed, but then it’s the negative charges that move and, because of their movement, we have the relativistic effect of length contraction, so the volumes are different, and the positive and negative charge density do not cancel out: the wire appears to be charged, so we do have a mix of E and B! Let me quickly give you the formula: E = (2πε₀)·(λ/r), with λ the (apparent) charge per unit length, so it’s the same formula as for a long line of charge, or for a long uniformly charged cylinder.

So we have a non-zero E and B and, hence, a non-zero Poynting vector S, whose direction is radially inward, so there is a flow of energy into the wire, all around. What the hell? Where does it go? Well… There’s a few possibilities here: the charges need kinetic energy to move, or as they increase their potential energy when moving towards the terminals of our capacitor to increase the charge on the plates or, much more mundane, the energy may be radiated out again in the form of heat. It looks crazy, but that’s how it is really. In fact, the more you think about, the more logical it all starts to sound. Energy must be conserved locally, and so it’s just field energy going in and re-appearing in some other form. So it does make sense. But, yes, it’s weird, because no one bothered to teach us this in school. 🙂

The ‘craziest’ example is the one below: we’ve got a charge and a magnet here. All is at rest. Nothing is moving… Well… I’ll correct that in a moment. 🙂 The charge (q) causes a (static) Coulomb field, while our magnet produces the usual magnetic field, whose shape we (should) recognize: it’s the usual dipole field. So E and B are not changing. But so when we calculate our Poynting vector, we see there is a circulation of S. The E×B product is not zero. So what’s going on here?

Well… There is no net change in energy with time: the energy just circulates around and around. Everything which flows into one volume flows out again. As Feynman puts it: “It is like incompressible water flowing around.” What’s the explanation? Well… Let me copy Feynman’s explanation of this ‘craziness’:

“Perhaps it isn’t so terribly puzzling, though, when you remember that what we called a “static” magnet is really a circulating permanent current. In a permanent magnet the electrons are spinning permanently inside. So maybe a circulation of the energy outside isn’t so queer after all.”

So… Well… It looks like we do need to revise some of our ‘intuitions’ here. I’ll conclude this post by quoting Feynman on it once more:

“You no doubt get the impression that the Poynting theory at least partially violates your intuition as to where energy is located in an electromagnetic field. You might believe that you must revamp all your intuitions, and, therefore have a lot of things to study here. But it seems really not necessary. You don’t need to feel that you will be in great trouble if you forget once in a while that the energy in a wire is flowing into the wire from the outside, rather than along the wire. It seems to be only rarely of value, when using the idea of energy conservation, to notice in detail what path the energy is taking. The circulation of energy around a magnet and a charge seems, in most circumstances, to be quite unimportant. It is not a vital detail, but it is clear that our ordinary intuitions are quite wrong.”

Well… That says it all, I guess. As far as I am concerned, I feel the Poyning vector makes things actually easier to understand. Indeed, the E and B vectors were quite confusing, because we had two of them, and the magnetic field is, frankly, a weird thing. Just think about the units in which we’re measuring B: (N/C)/(m/s). I can’t imagine what a unit like that could possible represent, so I must assume you can’t either. But so now we’ve got this Poynting vector that combines both E and B, and which represents the flow of the field energy. Frankly, I think that makes a lot of sense, and it’s surely much easier to visualize than E and/or B. [Having said that, of course, you should note that E and B do have their value, obviously, if only because they represent the lines of force, and so that’s something very physical too, of course. I guess it’s a matter of taste, to some extent, but so I’d tend to soften Feynman’s comments on the supposed ‘craziness’ of S.

In any case… The next thing I should discuss is field momentum. Indeed, if we’ve got flow, we’ve got momentum. But I’ll leave that for my next post. This topic can’t be exhausted in one post only, indeed. 🙂 So let me conclude this post. I’ll do with a very nice illustration I got from the Wikipedia article on the Poynting vector. It shows the Poynting vector around a voltage source and a resistor, as well as what’s going on in-between. [Note that the magnetic field is given by the field vector H, which is related to B as follows: B = μ₀(H + M), with M the magnetization of the medium. B and H are obviously just proportional in empty space, with μ₀ as the proportionality constant.]

Re-visiting relativity and four-vectors: the proper time, the tensor and the four-force

Original post:

My previous post explained how four-vectors transform from one reference frame to the other. Indeed, a four-vector is not just some one-dimensional array of four numbers: it represent something—a physical vector that… Well… Transforms like a vector. 🙂 So what vectors are we talking about? Let’s see what we have:

We knew the position four-vector already, which we’ll write as x_μ= (ct, x, y, z) = (ct, x).
We also proved that A_μ= (Φ, A_x, A_y, A_z) = (Φ, A) is a four-vector: it’s referred to as the four-potential.
We also know the momentum four-vector from the Lectures on special relativity. We write it as p_μ= (E, p_x, p_y, p_z) = (E, p), with E = γm₀, p = γm₀v, and γ = (1−v²/c²)^−1/2 or, for c = 1, γ = (1−v²)^−1/2

To show that it’s not just a matter of adding some fourth t-component to a three-vector, Feynman gives the example of the four-velocity vector. We have v_x= dx/dt, v_y= dy/dt and v_z= dz/dt, but a v_μ= (d(ct)/dt, dx/dt, dy/dt, dz/dt) = (c, dx/dt, dy/dt, dz/dt) ‘vector’ is, obviously, not a four-vector. [Why obviously? The inner product v_μv_μ is not invariant.] In fact, Feynman ‘fixes’ the problem by noting that ct, x, y and z have the ‘right behavior’, but the d/dt operator doesn’t. The d/dt operator is not an invariant operator. So how does he fix it then? He tries the (1−v²/c²)^−1/2·d/dt operator and, yes, it turns out we do get a four-vector then. In fact, we get that four-velocity vector u_μ that we were looking for:[Note we assume we’re using equivalent time and distance units now, so c = 1 and v/c reduces to a new variable v.]

Now how do we know this is four-vector? How can we prove this one? It’s simple. We can get it from our p_μ= (E, p) by dividing it by m₀, which is an invariant scalar in four dimensions too. Now, it is easy to see that a division by an invariant scalar does not change the transformation properties. So just write it all out, and you’ll see that p_μ/m₀ = u_μ and, hence, that u_μ is a four-vector too. 🙂

We’ve got an interesting thing here actually: division by an invariant scalar, or applying that (1−v²/c²)^−1/2·d/dt operator, which is referred to as an invariant operator, on a four-vector will give us another four-vector. Why is that? Let’s switch to compatible time and distance units so c = 1 so to simplify the analysis that follows.

The invariant (1−v²)^−1/2·d/dt operator and the proper time s

Why is the (1−v²)^−1/2·d/dt operator invariant? Why does it ‘fix’ things? Well… Think about the invariant spacetime interval (Δs)²= Δt²− Δx²− Δy²− Δz² going to the limit (ds)²= dt²− dx²− dy²− dz² . Of course, we can and should relate this to an invariant quantity s = ∫ ds. Just like Δs, this quantity also ‘mixes’ time and distance. Now, we could try to associate some derivative d/ds with it because, as Feynman puts it, “it should be a nice four-dimensional operation because it is invariant with respect to a Lorentz transformation.” Yes. It should be. So let’s relate ds to dt and see what we get. That’s easy enough: dx = v_x·dt, dy = v_y·dt, dz = v_z·dt, so we write:

(ds)²= dt²− v_x²·dt²− v_y²·dt²− v_z²·dt²⇔ (ds)²= dt²·(1 − v_x²− v_y²− v_z²) = dt²·(1 − v²)

and, therefore, ds = dt·(1−v²)^1/2. So our operator d/ds is equal to (1−v²)^−1/2·d/dt, and we can apply it to any four-vector, as we are sure that, as an invariant operator, it’s going to give us another four-vector. I’ll highlight the result, because it’s important:

The d/ds = (1−v²)^−1/2·d/dt operator is an invariant operator for four-vectors.

For example, if we apply it to x_μ= (t, x, y, z), we get the very same four-velocity vector μ_μ:

dx_μ/ds = u_μ = p_μ/m₀

Now, if you’re somewhat awake, you should ask yourself: what is this s, really, and what is this operator all about? Our new function s = ∫ ds is not the distance function, as it’s got both time and distance in it. Likewise, the invariant operator d/ds = (1−v²)^−1/2·d/dt has both time and distance in it (the distance is implicit in the v² factor). Still, it is referred to as the proper time along the path of a particle. Now why is that? If it’s got distance and time in it, why don’t we call it the ‘proper distance-time’ or something?

Well… The invariant quantity s actually is the time that would be measured by a clock that’s moving along, in spacetime, with the particle. Just think of it: in the reference frame of the moving particle itself, Δx, Δyand Δz must be zero, because it’s not moving in its own reference frame. So the (Δs)²= Δt²− Δx²− Δy²− Δz² reduces to (Δs)²= Δt², and so we’re only adding time to s. Of course, this view of things implies that the proper time itself is fixed only up to some arbitrary additive constant, namely the setting of the clock at some event along the ‘world line’ of our particle, which is its path in four-dimensional spacetime. But… Well… In a way, s is the ‘genuine’ or ‘proper’ time coming with the particle’s reference frame, and so that’s why Einstein called it like that. You’ll see (later) that it plays a very important role in general relativity theory (which is a topic we haven’t discussed yet: we’ve only touched special relativity, so no gravity effects).

OK. I know this is simple and complicated at the same time: the math is (fairly) easy but, yes, it may be difficult to ‘understand’ this in some kind of intuitive way. But let’s move on.

The four-force vector f_μ

We know the relativistically correct equation for the motion of some charge q. It’s just Newton’s Law F = dp/dt = d(mv)/dt. The only difference is that we are not assuming that m is some constant. Instead, we use the p = γm₀v formula to get:

How can we get a four-vector for the force? It turns out that we get it when applying our new invariant operator to the momentum four-vector p_μ= (E, p), so we write: f_μ= dp_μ/ds. But p_μ= m₀u_μ = m₀dx_μ/ds, so we can re-write this as f_μ= d(m₀·dx_μ/ds)/ds, which gives us a formula which is reminiscent of the Newtonian F = ma equation:

What is this thing? Well… It’s not so difficult to verify that the x, y and z-components are just our old-fashioned F_x, F_y and F_z, so these are the components of F. The t-component is (1−v²)^−1/2·dE/dt. Now, dE/dt is the time rate of change of energy and, hence, it’s equal to the rate of doing work on our charge, which is equal to F•v. So we can write f_μas:

The force and the tensor

We will now derive that formula which we ended the previous post with. We start with calculating the spacelike components of f_μfrom the Lorentz formula F = q(E + v×B). [The terminology is nice, isn’t it? The spacelike components of the four-force vector! Now that sounds impressive, doesn’t it? But so… Well… It’s really just the old stuff we know already.] So we start with f_x = F_x, and write it all out:

What a monster! But, hey! We can ‘simplify’ this by substituting stuff by (1) the t-, x-, y- and z-components of the four-velocity vector u_μand (2) the components of our tensor F_μν = [F_ij] = [∇_iA_j − ∇_jA_i] with i, j = t, x, y, z. We’ll also pop in the diagonal F_xx = 0 element, just to make sure it’s all there. We get:

Looks better, doesn’t it? 🙂 Of course, it’s just the same, really. This is just an exercise in symbolism. Let me insert the electromagnetic tensor we defined in our previous post, just as a reminder of what that F_μν matrix actually is:

If you read my previous post, this matrix – or the concept of a tensor – has no secrets for you. Let me briefly summarize it, because it’s an important result as well. The tensor is (a generalization of) the cross-product in four-dimensional space. We take two vectors: a_μ = (a_t, a_x, a_y, a_z) and b_μ = (b_t, b_x, b_y, b_z) and then we take cross-products of their components just like we did in three-dimensional space, so we write T_ij = a_ib_j − a_jb_i. Now, it’s easy to see that this combination implies that T_ij = − T_ji and that T_ii= 0, which is why we only have six independent numbers out of the 16 possible combinations, and which is why we’ll get a so-called anti-symmetric matrix when we organize them in a matrix. In three dimensions, the very same definition of the cross-product T_ij gives us 9 combinations, and only 3 independent numbers, which is why we represented our ‘tensor’ as a vector too! In four-dimensional space we can’t do that: six things cannot be represented by a four-vector, so we need to use this matrix, which is referred to as a tensor of the second rank in four dimensions. [When you start using words like that, you’ve come a long way, really. :-)]

[…] OK. Back to our four-force. It’s easy to get a similar one-liner for f_y and f_z too, of course, as well as for f_t. But… Yes, f_t… Is it the same thing really? Let me quickly copy Feynman’s calculation for f_t:

It does: remember that v×B and v are orthogonal, and so their dot product is zero indeed. So, to make a long story short, the four equations – one for each component of the four-force vector f_μ– can be summarized in the following elegant equation:

Writing this all requires a few conventions, however. For example, F_μν is a 4×4 matrix and so u_ν has to be written as a 1×4 vector. And the formula for the f_x and f_t component also make it clear that we also want to use the +−−− signature here, so the convention for the signs in the u_νF_μν product is the same as that for the scalar product a_μb_μ. So, in short, you really need to interpret what’s being written here.

A more important question, perhaps, is: what can we do with it? Well… Feynman’s evaluation of the usefulness of this formula is rather succinct: “Although it is nice to see that the equations can be written that way, this form is not particularly useful. It’s usually more convenient to solve for particle motions by using the F = q(E + v×B) = (1−v²)^−1/2·d(m₀v)/dt equations, and that’s what we will usually do.”

Having said that, this formula really makes good on the promise I started my previous post with: we wanted a formula, some mathematical construct, that effectively presents the electromagnetic force as one force, as one physical reality. So… Well… Here it is! 🙂

Well… That’s it for today. Tomorrow we’ll talk about energy and about a very mysterious concept—the electromagnetic mass. That should be fun! So I’ll c u tomorrow! 🙂

Relativistic transformations of fields and the electromagnetic tensor

Original post:

We’re going to do a very interesting piece of math here. It’s going to bring a lot of things together. The key idea is to present a mathematical construct that effectively presents the electromagnetic force as one force, as one physical reality. Indeed, we’ve been saying repeatedly that electromagnetism is one phenomenon only but we’ve been writing it always as something involving two vectors: he electric field vector E and the magnetic field vector B. Of course, Lorentz’ force law F = q(E + v×B) makes it clear we’re talking one force only but… Well… There is a way of writing it all up that is much more elegant.

I have to warn you though: this post doesn’t add anything to the physics we’ve seen so far: it’s all math, really and, to a large extent, math only. So if you read this blog because you’re interested in the physics only, then you may just as well skip this post. Having said that, the mathematical concept we’re going to present is that of the tensor and… Well… You’ll have to get to know that animal sooner or later anyway, so you may just as well give it a try right now, and see whatever you can get out of this post.

The concept of a tensor further builds on the concept of the vector, which we liked so much because it allows us to write the laws of physics as vector equations, which do not change when going from one reference frame to another. In fact, we’ll see that a tensor can be described as a ‘special’ vector cross product (to be precise, we’ll show that a tensor is a ‘more general’ cross product, really). So the tensor and vector concepts are very closely related, but then… Well… If you think about it, the concept of a vector and the concept of a scalar are closely related, too! So we’re just moving up the value chain, so to speak: from scalar fields to vector fields to… Well… Tensor fields! And in quantum mechanics, we’ll introduce spinors, and so we also have spinor fields! Having said that, don’t worry about tensor fields. Let’s first try to understand tensors tout court. 🙂

So… Well… Here we go. Let me start with it all by reminding you of the concept of a vector, and why we like to use vectors and vector equations.

The invariance of physics and the use of vector equations

What’s a vector? You may think, naively, that any one-dimensional array of numbers is a vector. But… Well… No! In math, we may, effectively, refer to any one-dimensional array of numbers as a ‘vector’, perhaps, but in physics, a vector does represent something real, something physical, and so a vector is only a vector if it transforms like a vector under the transformation rules that apply when going from one another frame of reference, i.e. one coordinate system, to another. Examples of vectors in three dimensions are: the velocity vector v, or the momentum vector p = m·v, or the position vector r.

Needless to say, the same can be said of scalars: mathematicians may define a scalar as just any real number, but it’s not in physics. A scalar in physics refers to something real, i.e. a scalar field, like the temperature (T) inside of a block of material. In fact, think about your first vector equation: it may have been the one determining the heat flow (h), i.e. h = −κ·∇T = (−κ·∂T/∂x, −κ·∂T/∂y, −κ·∂T/∂z). It immediately shows how scalar and vector fields are intimately related.

Now, when discussing the relativistic framework of physics, we introduced vectors in four dimensions, i.e. four-vectors. The most basic four-vector is the spacetime four-vector R = (ct, x, y, z), which is often referred to as an event, but it’s just a point in spacetime, really. So it’s a ‘point’ with a time as well as a spatial dimension, so it also has t in it, besides x, y and z. It is also known as the position four-vector but, again, you should think of a ‘position’ that includes time! Of course, we can re-write R as R = (ct, r), with r = (x, y, z), so here we sort of ‘break up’ the four-vector in a scalar and a three-dimensional vector, which is something we’ll do from time to time, indeed. 🙂

We also have a displacement four-vector, which we can write as ΔR = (c·Δt, Δr). There are other four-vectors as well, including the four-velocity, the four-momentum and the four-force four-vectors, which we’ll discuss later (in the last section of this post).

So it’s just like using three-dimensional vectors in three-dimensional physics, or ‘Newtonian’ physics, I should say: the use of four-vectors is going to allow us to write the laws of physics using vector equations, but in four dimensions, rather than three, so we get the ‘Einsteinian’ physics, the real physics, so to speak—or the relativistically correct physics, I should say. And so these four-dimensional vector equations will also not change when going from one reference frame to another, and so our four-vector will be vectors indeed, i.e. they will transform like a vector under the transformation rules that apply when going from one another frame of reference, i.e. one coordinate system, to another.

What transformation? Well… In Newtonian or Galilean physics, we had translations and rotations and what have you, but what we are interested in right now are ‘Einsteinian’ transformations of coordinate systems, so these have to ensure that all of the laws of physics that we know of, including the principle of relativity, still look the same. You’ve seen these transformation rules. We don’t call them the ‘Einsteinian’ transformation rules, but the Lorentz transformation rules, because it was a Dutch physicist (Hendrik Lorentz) who first wrote them down. So these rules are very different from the Newtonian or Galilean transformation rules which everyone assumed to be valid until the Michelson-Morley experiment unequivocally established that the speed of light did not respect the Galilean transformation rules. Very different? Well… Yes. In their mathematical structure, that is. Of course, when velocities are low, i.e. non-relativistic, then they yield the same result, approximately, that is. However, I explained that in my post on special relativity, and so I won’t dwell on that here.

Let me just jot down both sets of rules assuming that the two reference frames move with respect to each other along the x- axis only, so the y- and z-component of u is zero.

The Galilean or Newtonian rules are the simple rules on the right. Going from one reference frame to another (let’s call them S and S’ respectively) is just a matter of adding or subtracting speeds: if my car goes 100 km/h, and yours goes 120 km/h, then you will see my car falling behind at a speed of (minus) 20 km/h. That’s it. We could also rotate our reference frame, and our Newtonian vector equations would still look the same. As Feynman notes, smilingly, it’s what a lot of armchair philosophers think relativity theory is all about, but so it’s got nothing to do with it. It’s plain wrong!

In any case, back to vectors and transformations. The key to the so-called invariance of the laws of physics is the use of vectors and vector operators that transform like vectors. For example, if we defined A and B as (A_x, A_y, A_z) and (B_x, B_y, B_z), then we knew that the so-called inner product A•B would look the same in all rotated coordinate systems, so we can write: A•B = A’•B’. So we know that if we have a product like that on both sides of an equation, we’re fine: the equation will have the same form in all rotated coordinate systems. Also, the gradient, i.e. our vector operator ∇ = (∂/∂_x, ∂/∂_y, ∂/∂_z), when applied to a scalar function, gave three quantities that also transform like a vector under rotation. We also defined a vector cross product, which yielded a vector (as opposed to the inner product, i.e. the vector dot product, which yields a scalar):

So how does this thing behave under a Galilean transformation? Well… You may or may not remember that we used this cross-product to define the angular momentum L, which was a cross product of the radius vector r and the momentum vector p = mv, as illustrated below. The animation also gives the torque τ, which is, loosely speaking, a measure of the turning force: it’s the cross product of r and F, i.e. the force on the lever-arm.

The components of L are:

Now, we find that these three numbers, or objects if you want, transform in exactly the same way as the components of a vector. However, as Feynman points out, that’s a matter of ‘luck’ really. It’s something ‘special’. Indeed, you may or may not remember that we distinguished axial vectors from polar vectors. L is an axial vector, while r and p are polar vectors, and so we find that, in three dimensions, the cross product of two polar vectors will always yields an axial vector. Axial vectors are sometimes referred to as pseudovectors, which suggests that they are ‘not so real’ as… Well… Polar vectors, which are sometimes referred to as ‘true’ vectors. However, it doesn’t matter when doing these Newtonian or Galilean transformations: pseudo or true, both vectors transform like vectors. 🙂

But so… Well… We’re actually getting a bit of a heads-up here: if we’d be mixing (or ‘crossing’) polar and axial vectors, or mixing axial vectors only, so if we’d define something involving L and p (rather than r and p), or something involving L and τ, then we may not be so lucky, and then we’d have to carefully examine our cross-product, or whatever other product we’d want to define, because its components may not behave like a vector.

Huh? Whatever other product we’d want to define? Why are you saying that? Well… We actually can think of other products. For example, if we have two vectors a = (a_x, a_y, a_z) and b = (b_x, b_y, b_z), then we’ll have nine possible combinations of their components, which we can write as T_ij = a_ib_j. So that’s like L_xy, L_yz and L_zx really. Now, you’ll say: “No. It isn’t. We don’t have nine combinations here. Just three numbers.” Well… Think about it: we actually do have nine L_ij combinations too here, as we can write: L_ij = r_i·p_j – r_j·p_i. It just happens that, with this definition, only three of these combinations L_ij are independent. That’s because the other six numbers are either zero or the opposite. Indeed, it’s easy to verify that L_ij = –L_ji , and L_ii = 0. So… Well… It turns out that the three components of our L = r×p ‘vector’ are actually a subset of a set of nine L_ij numbers. So… Well… Think about it. We cannot just do whatever we want with our ‘vectors’. We need to watch out.

In fact, I do not want to get too much ahead of myself, but I can already tell you that the matrix with these nine T_ij = a_ib_j combinations is what is referred to as the tensor. To be precise, it’s referred to as a tensor of the second rank in three dimensions. The ‘second rank’, aka as ‘degree’ or ‘order’ refers to the fact that we’ve got two indices, and the ‘three dimensions’ is because we’re using three-dimensional vectors. We’ll soon see that the electromagnetic tensor is also of the second rank, but it’s a tensor in four dimensions. In any case, I should not get ahead of myself. Just note what I am saying here: the tensor is like a ‘new’ product of two vectors, a new type of ‘cross’ product really (because we’re mixing the components, so to say), but it doesn’t yield a vector: it yields a matrix. For three-dimensional vectors, we get a 3×3 matrix. For four-vectors, we’ll get a 4×4 matrix. And so the full truth about our angular momentum vector L, is the following:

There is a thing which we call the angular momentum tensor. It’s a 3×3 matrix, so it has nine elements which are defined as: L_ij = r_i·p_j – r_j·p_i. Because of this definition, it’s an antisymmetric tensor of the second order in three dimensions, so it’s got only three independent components.
The three independent elements are the components of our ‘vector’ L, and picking them out and calling these three components a ‘vector’ is actually a ‘trick’ that only works in three dimensions. They really just happen to transform like a vector under rotation or under whatever Galilean transformation! [By the way, do you know understand why I was saying that we can look at a tensor as a ‘more general’ cross product?]
In fact, in four dimensions, we’ll use a similar definition and define 16 elements F_ij as F_ij = ∇_iA_j − ∇_jA_i, using the two four-vectors ∇_μand A_μ (so we have 4×4 = 16 combinations indeed), out of which only six will be independent for the very same reason: we have an antisymmetric vector combination here, F_ij = −F_ji and F_ii = 0. 🙂 However, because we cannot represent six independent things by four things, we do not get some other four-vector, and so that’s why we cannot apply the same ‘trick’ in four dimensions.

However, here I am getting way ahead of myself and so… Well… Yes. Back to the main story line. 🙂 So let’s try to move to the next level of understanding, which is… Well…

Because of guys like Maxwell and Einstein, we now know that rotations are part of the Newtonian world, in which time and space are neatly separated, and that things are not so simple in Einstein’s world, which is the real world, as far as we know, at least! Under a Lorentz transformation, the new ‘primed’ space and time coordinates are a mixture of the ‘unprimed’ ones. Indeed, the new x’ is a mixture of x and t, and the new t’ is a mixture of x and t as well. [Yes, please scroll all the way up and have a look at the transformation on the left-hand side!]

So you don’t have that under a Galilean transformation: in the Newtonian world, space and time are neatly separated, and time is absolute, i.e. it is the same regardless of the reference frame. In Einstein’s world – our world – that’s not the case: time is relative, or local as Hendrik Lorentz termed it quite appropriately, and so it’s space-time – i.e. ‘some kind of union of space and time’ as Minkowski termed it – that transforms.

So that’s why physicists use four-vectors to keep track of things. These four-vectors always have three space-like components, but they also include one so-called time-like component. It’s the only way to ensure that the laws of physics are unchanged when moving with uniform velocity. Indeed, any true law of physics we write down must be arranged so that the invariance of physics (as a “fact of Nature”, as Feynman puts it) is built in, and so that’s why we use Lorentz transformations and four-vectors.

In the mentioned post, I gave a few examples illustrating how the Lorentz rules work. Suppose we’re looking at some spaceship that is moving at half the speed of light (i.e. 0.5c) and that, inside the spaceship, some object is also moving at half the speed of light, as measured in the reference frame of the spaceship, then we get the rather remarkable result that, from our point of view (i.e. our reference frame as observer on the ground), that object is not going as fast as light, as Newton or Galileo – and most present-day armchair philosophers 🙂 – would predict (0.5c + 0.5c = c). We’d see it move at a speed equal to v = 0.8c. Huh? How do we know that? Well… We can derive a velocity formula from the Lorentz rules:

So now you can just put in the numbers now: v_x = (0.5c + 0.5c)/(1 + 0.5·0.5) = 0.8c. See?

Let’s do another example. Suppose we’re looking at a light beam inside the spaceship, so something that’s traveling at speed c itself in the spaceship. How does that look to us? The Galilean transformation rules say its speed should be 1.5c, but that can’t be true of course, and the Lorentz rules save us once more: v_x = (0.5c + c)/(1 + 0.5·1) = c, so it turns out that the speed of light does not depend on the reference frame: it looks the same – both to the man in the ship as well as to the man on the ground. As Feynman puts it: “This is good, for it is, in fact, what the Einstein theory of relativity was designed to do in the first place—so it had better work!” 🙂

So let’s now apply relativity to electromagnetism. Indeed, that’s what this post is all about! However, before I do so, let me re-write the Lorentz transformation rules for c = 1. We can equate the speed of light to one, indeed, when measure time and distance in equivalent units. It’s just a matter of ditching our seconds for meters (so our time unit becomes the time that light needs to travel a distance of one meter), or ditching our meters for seconds (so our distance unit becomes the distance that light travels in one second). You should be familiar with this procedure. If not, well… Check out my posts on relativity. So here’s the same set of rules for c = 1:

They’re much easier to remember and work with, and so that’s good, because now we need to look at how these rules work with four-vectors and the various operations and operators we’ll be defining on them. Let’s look at that step by step.

Electrodynamics in relativistic notation

Let me copy the Universal Set of Equations and Their Solution once more:

The solution for Maxwell’s equations is given in terms of the (electric) potential Φ and the (magnetic) vector potential A. I explained that in my post on this, so I won’t repeat myself too much here either. The only point you should note is that this solution is the result of a special choice of Φ and A, which we referred to as the Lorentz gauge. We’ll touch upon this condition once more, so just make a mental note of it.

Now, E and B do not correspond to four-vectors: they depend on x, y, z and t, but they have three components only: E_x, E_y, E_z, and B_x, B_y, and B_z respectively. So we have six independent terms here, rather than four things that, somehow, we could combine into some four-vector. [Does this ring a bell? It should. :-)] Having said that, it turns out that we can combine Φ and A into a four-vector, which we’ll refer to as the four-potential and which we’ll will write as:

A_μ= (Φ, A) = (Φ, A_x, A_y, A_z) = (A_t, A_x, A_y, A_z) with A_t = Φ.

So that’s a four-vector just like R = (ct, x, y, z).

How do we know that A_μis a four-vector? Well… Here I need to say a few things about those Lorentz transformation rules and, more importantly, about the required condition of invariance under a Lorentz transformation. So, yes, here we need to dive into the math.

Four-vectors and invariance under Lorentz transformations

When you were in high-school, you learned how to rotate your coordinate frame. You also learned that the distance of a point from the origin does not change under a rotation, so you’d write r’²= x’²+ y’²+ z’²= r²= x²+ y²+ z², and you’d say that r² is an invariant quantity under a rotation. Indeed, transformations leave certain things unchanged. From the Lorentz transformation rules itself, it is easy to see that

c·t’²– x’²– y’²–z ‘²= c·t²–x²– y² – z², or,

if c = 1, that t’²– x’²– y’²– z’²= t²– x²– y² – z²,

is an invariant under a Lorentz transformation. We found the same for the so-called spacetime interval Δs² = Δr²– cΔt², which we write as Δs² = Δr²– Δt² as we chose our time or distance units such that c = 1. [Note that, from now on, we’ll assume that’s the case, so c = 1 everywhere. We can always change back to our old units when we’re done with the analysis.] Indeed, such invariance allowed us to define spacelike, timelike and lightlike intervals using the so-called light cone emanating from a single event and traveling in all directions.

You should note that, for four-vectors, we do not have a simple sum of three terms. Indeed, we don’t write x²+ y²+ z² but t²– x²– y² – z². So we’ve got a +−−− thing here or, it’s just another convention, we could also work with a −+++ sum of terms. The convention is referred to as the signature, and we will use the so-called metric signature here, which is +−−−. Let’s continue the story. Now, all four-vectors a_μ= (a_t, a_x, a_y, a_z) have this property that:

a_t‘²– a_x‘²– a_y‘²– a_z‘²= a_t²– a_x²– a_y² – a_z².

[The primed quantities are, obviously, the quantities as measured in the other reference frame.] So. Well… Yes. 🙂 But… Well… Hmm… We can say that our four-potential vector is a four-vector, but so we still have to prove that. So we need to prove that Φ’²– A_x‘²– A_y‘²– A_z‘²= Φ²– A_x²– A_y² – A_z² for our four-potential vector A_μ= (Φ, A). So… Yes… How can we do that? The proof is not so easy, but you need to go through it as it will introduce some more concepts and ideas you need to understand.

In my post on the Lorentz gauge, I mentioned that Maxwell’s equations can be re-written in terms of Φ and A, rather than in terms of E and B. The equations are:

The expression look rather formidable, but don’t panic: just look at it. Of course, you need to be familiar with the operators that are being used here, so that’s the Laplacian ∇² and the divergence operator ∇• that’s being applied to the scalar Φ and the vector A. I can’t re-explain this. I am sorry. Just check my posts on vector analysis. You should also look at the third equation: that’s just the Lorentz gauge condition, which we introduced when deriving these equations from Maxwell’s equations. Having said that, it’s the first and second equation which describe Φ and A as a function of the charges and currents in space, and so that’s what matters here. So let’s unfold the first equation. It says the following:

In fact, if we’d be talking free or empty space, i.e. regions where there are no charges and currents, then the right-hand side would be zero and this equation would then represent a wave equation, so some potential Φ that is changing in time and moving out at the speed c. Here again, I am sorry I can’t write about this here: you’ll need to check one of my posts on wave equations. If you don’t want to do that, you should believe me when I say that, if you see an equation like this:

then the function Ψ(x, t) must be some function

Now, that’s a function representing a wave traveling at speed c, i.e. the phase velocity. Always? Yes. Always! It’s got to do with the x − ct and/or x + ct argument in the function. But, sorry, I need to move on here.

The unfolding of the equation with Φ makes it clear that we have four equations really. Indeed, the second equation is three equations: one for A_x, one for A_y, and one for A_z respectively. The four quantities on the right-hand side of these equations are ρ, j_x, j_y and j_z respectively, divided by ε₀, which is a universal constant which does not change when going from one coordinate system to another. Now, the quantities ρ, j_x, j_y and j_z transform like a four-vector. How do we know that? It’s just the charge conservation law. We used it when solving the problem of the fields around a moving wire, when we demonstrated the relativity of the electric and magnetic field. Indeed, the relevant equations were:

You can check that against the Lorentz transformation rules for c = 1. They’re exactly the same, but so we chose t = 0, so the rules are even simpler. Hence, the (ρ, j_x, j_y, j_z) vector is, effectively, a four-vector, and we’ll denote it by j_μ= (ρ, j). I now need to explain something else. [And, yes, I know this is becoming a very long story but… Well… That’s how it is.]

It’s about our operators ∇, ∇•, ∇× and ∇², so that’s the gradient, the divergence, curl and Laplacian operator respectively: they all have a four-dimensional equivalent. Of course, that won’t surprise you. 😦 Let me just jot all of them down, so we’re done with that, and then I’ll focus on the four-dimensional equivalent of the Laplacian ∇•∇ = ∇², which is referred to as the D’Alembertian, and which is denoted by □², because that’s the one we need to prove that our four-potential vector is a real four-vector. [I know: □²is a tiny symbol for a pretty monstrous thing, but I can’t help it: my editor tool is pretty limited.]

Now, we’re almost there. Just hang in for a little longer. It should be obvious that we can re-write those two equations with Φ, A, ρ and j, as:

Just to make sure, let me remind you that A_μ= (Φ, A) and that j_μ= (ρ, j). Now, our new D’Alembertian operator is just an operator—a pretty formidable operator but, still, it’s an operator, and so it doesn’t change when the coordinate system changes, so the conclusion is that, IF j_μ= (ρ, j) is a four-vector – which it is – and, therefore, transforms like a four-vector, THEN the quantities Φ, A_x, A_y, and A_z must also transform like a four-vector, which means they are (the components of) a four-vector.

So… Well… Think about it, but not too long, because it’s just an intermediate result we had to prove. So that’s done. But we’re not done here. It’s just the beginning, actually. Let me repeat our intermediate result:

A_μ= (Φ, A) is a four-vector. We call it the four-potential vector.

OK. Let’s continue. Let me first draw your attention to that expression with the D’Alembertian above. Which expression? This one:

What about it? Well… You should note that the physics of that equation is just the same as Maxwell’s equations. So it’s one equation only, but it’s got it all.

It’s quite a pleasure to re-write it in such elegant form. Why? Think about it: it’s a four-vector equation: we’ve got a four-vector on the left-hand side, and a four-vector on the right-hand side. Therefore, this equation is invariant under a transformation. So, therefore, it directly shows the invariance of electrodynamics under the Lorentz transformation.

Huh? Yes. You may think about this a little longer. 🙂

To wrap this up, I should also note that we can also express the gauge condition using our new four-vector notation. Indeed, we can write it as:

It’s referred to as the Lorentz condition and it is, effectively, a condition for invariance, i.e. it ensures that the four-vector equation above does stay in the form it is in for all reference frames. Note that we’re re-writing it using the four-dimensional equivalent of the divergence operator ∇•, but so we don’t have a dot between ∇_μ and A_μ. In fact, the notation is pretty confusing, and it’s easy to think we’re talking some gradient, rather than the divergence. So let me therefore highlight the meaning of both once again. It looks the same, but it’s two very different things: the gradient operates on a scalar, while the divergence operates on a (four-)vector. Also note the +−−− signature is only there for the gradient, not for the divergence!

You’ll wonder why they didn’t use some • or ∗ symbol, and the answer: I don’t know. I know it’s hard to keep inventing symbols for all these different ‘products’ – the ⊗ symbol, for example, is reserved for tensor products, which we won’t get into – but… Well… I think they could have done something here. 😦

In any case… Let’s move on. Before we do, please note that we can also re-write our conservation law for electric charge using our new four-vector notation. Indeed, you’ll remember that we wrote that conservation law as:

Using our new four-vector operator ∇_μ, we can re-write that as ∇_μj_μ= 0. So all of electrodynamics can be summarized in the two equations only—Maxwell’s law and the charge conservation law:

OK. We’re now ready to discuss the electromagnetic tensor. [I know… This is becoming an incredibly long and incredibly complicated piece but, if you get through it, you’ll admit it’s really worth it.]

The electromagnetic tensor

The whole analysis above was done in terms of the Φ and A potentials. It’s time to get back to our field vectors E and B. We know we can easily get them from Φ and A, using the rules we mentioned as solutions:

These two equations should not look as yet another formula. They are essential, and you should be able to jot them down anytime anywhere. They should be on your kitchen door, in your toilet and above your bed. 🙂 For example, the second equation gives us the components of the magnetic field vector B:

Now, look at these equations. The $x$ -component is equal to a couple of terms that involve only $y$ – and $z$ -components. The y-component is equal to something involving only x and $z.$ Finally, the $z$ -component only involves x and y. Interesting. Let’s define a ‘thing’ we’ll denote by F_zy and define as:

So now we can write: B_x = F_zy, B_y = F_xz, and B_z = F_xy. Now look at our equation for E. It turns out the components of E are equal to things like F_xt, F_ytand F_zt! Indeed, F_xt = ∂A_x/∂t − ∂A_t/∂x = E_x!

But… Well… No. 😦 The sign is wrong! E_x = −∂A_x/∂t−∂A_t/∂x, so we need to modify our definition of F_xt. When the t-component is involved, we’ll define our ‘F-things’ as:

So we’ve got a plus instead of a minus. It looks quite arbitrary but, frankly, you’ll have to admit it’s sort of consistent with our +−−− signature for our four-vectors and, in just a minute, you’ll see it’s fully consistent with our definition of the four-dimensional vector operator ∇_μ= (∂/∂t, −∂/∂x, −∂/∂y, −∂/∂z). So… Well… Let’s go along with it.

What about the F_xx, F_yy, F_zzand F_ttterms? Well… F_xx = ∂A_x/∂x − ∂A_x/∂x = 0, and it’s easy to see that F_yy and F_zz are zero too. But F_tt? Well… It’s a bit tricky but, applying our definitions carefully, we see that F_tt must be zero too. In any case, the F_tt = 0 will become obvious as we will be arranging these ‘F-things’ in a matrix, which is what we’ll do now. [Again: does this ring a bell? If not, it should. :-)]

Indeed, we’ve got sixteen possible combinations here, which Feynman denotes as F_μν, which is somewhat confusing, because F_μν usually denotes the 4×4 matrix representing all of these combinations. So let me use the subscripts i and j instead, and define F_ij as:

F_ij = ∇_iA_j − ∇_jA_i

with ∇_i being the t-, x-, y- or z-component of ∇_μ = (∂/∂t, −∂/∂x, −∂/∂y, −∂/∂z) and, likewise, A_i being the t-, x-, y- or z-component of A_μ = (Φ, A_x, A_y, A_z). Just check it: F_zy = −∂A_y/∂z + ∂A_z/∂y = ∂A_z/∂y − ∂A_y/∂z = B_x, for example, and F_xt = −∂Φ/∂x − ∂A_x/∂t = E_x. So the +−−− convention works. [Also note that it’s easier now to see that F_tt = ∂Φ/∂t − ∂Φ/∂t = 0.]

We can now arrange the F_ij in a matrix. This matrix is antisymmetric, because F_ij = – F_ji, and its diagonal elements are zero. [For those of you who love math: note that the diagonal elements of an antisymmetric matrix are always zero because of the F_ij = – F_ji constraint: just use k = i = j in the constraint.]

Now that matrix is referred to as the electromagnetic tensor and it’s depicted below (we plugged c back in, remember that B’s magnitude is 1/c times E’s magnitude).

So… Well… Great ! We’re done! Well… Not quite. 🙂

We can get this matrix in a number of ways. The least complicated way is, of course, just to calculate all F_ij components and them put them in a [F_ij] matrix using the i as the row number and the j as the column number. You need to watch out with the conventions though, and so i and j start on t and end on z. 🙂

The other way to do it is to write the ∇_μ = (∂/∂t, −∂/∂x, −∂/∂y, −∂/∂z) operator as a 4×1 column vector, which you then multiply with the four-vector A_μ written as a 4×1 row vector. So ∇_μA_μis then a 4×4 matrix, which we combine with its transpose, i.e. (∇_μA_μ)^T, as shown below. So what’s written below is (∇_μA_μ) − (∇_μA_μ)^T.

If you google, you’ll see there’s more than one way to go about it, so I’d recommend you just go through the motions and double-check the whole thing yourself—and please do let me know if you find any mistake! In fact, the Wikipedia article on the electromagnetic tensor denotes the matrix above as F^μν, rather than as F_μν, which is the same tensor but in its so-called covariant form, but so I’ll refer you to that article as I don’t want to make things even more complicated here! As said, there’s different conventions around here, and so you need to double-check what is what really. 🙂

Where are we heading with all of this? The next thing is to look at the Lorentz transformation of these F_ij = ∇_iA_j − ∇_jA_icomponents, because then we know how our E and B fields transform. Before we do so, however, we should note the more general results and definitions which we obtained here:

1. The F_μν matrix (a matrix is just a multi-dimensional array, of course) is a so-called tensor. It’s a tensor of the second rank, because it has two indices in it. We think of it as a very special ‘product’ of two vectors, not unlike the vector cross product a × b, whose components were also defined by a similar combination of the components of a and b. Indeed, we wrote:

So one should think of a tensor as “another kind of cross product” or, preferably, and as Feynman puts it, as a “generalization of the cross product”.

2. In this case, the four-vectors are ∇_μ = (∂/∂t, −∂/∂x, −∂/∂y, −∂/∂z) and A_μ = (Φ, A_x, A_y, A_z). Now, you will probably say that ∇_μ is an operator, not a vector, and you are right. However, we know that ∇_μ behaves like a vector, and so this is just a special case. The point is: because the tensor is based on four-vectors, the F_μν tensor is referred to as a tensor of the second rank in four dimensions. In addition, because of the F_ij = – F_ji result, F_μν is an asymmetric tensor of the second rank in four dimensions.

3. Now, the whole point is to examine how tensors transform. We know that the vector dot product, aka the inner product, remains invariant under a Lorentz transformation, both in three as well as in four dimensions, but what about the vector cross product, and what about the tensor? That’s what we’ll be looking at now.

The Lorentz transformation of the electric and magnetic fields

Cross products are complicated, and tensors will be complicated too. Let’s recall our example in three dimensions, i.e. the angular momentum vector L, which was a cross product of the radius vector r and the momentum vector p = mv, as illustrated below (the animation also gives the torque τ, which is, loosely speaking, a measure of the turning force).

The components of L are:

Now, this particular definition ensures that L_ijturns out to be an antisymmetric object:

So it’s a similar situation here. We have nine possible combinations, but only three independent numbers. So it’s a bit like our tensor in four dimensions: 16 combinations, but only 6 independent numbers.

Now, it so happens that that these three numbers, or objects if you want, transform in exactly the same way as the components of a vector. However, as Feynman points out, that’s a matter of ‘luck’ really. In fact, Feynman points out that, when we have two vectors a = (a_x, a_y, a_z) and b = (b_x, b_y, b_z), we’ll have nine products T_ij = a_ib_j which will also form a tensor of the second rank (cf. the two indices) but which, in general, will not obey the transformation rules we got for the angular momentum tensor, which happened to be an antisymmetric tensor of the second rank in three dimensions.

To make a long story short, it’s not simple in general, and surely not here: with E and B, we’ve got six independent terms, and so we cannot represent six things by four things, so the transformation rules for E and B will differ from those for a four-vector. So what are they then?

Well… Feynman first works out the rules for the general antisymmetric vector combination G_ij = a_ib_j− a_jb_i, with a_iand b_j the t-, x-, y- or z-component of the four-vectors a_μ= (a_t, a_x, a_y, a_z) and b_μ= (b_t, b_x, b_y, b_z) respectively. The idea is to first get some general rules, and then replace G_ij = a_ib_j− a_jb_i by F_ij = ∇_iA_j − ∇_jA_i, of course! So let’s apply the Lorentz rules, which – let me remind you – are the following ones:

So we get:

The rest is all very tedious: you just need to plug these things into the various G_ij = a_ib_j− a_jb_i formulas. For example, for G’_tx, we get:

Hey! That’s just G’_tx, so we find that G’_tx= G_tx! What about the rest? Well… That yields something different. Let me shorten the story by simply copying Feynman here:

So… Done!

So what?

Well… Now we just substitute. In fact, there are two alternative formulations of the Lorentz transformations of E and B. They are given below (note the units are such that c = 1):

In addition, there is a third equivalent formulation which is more practical, and also simpler, even if it puts the c‘s back in. It re-defines the field components, distinguishing only two:

The ‘parallel’ components E_|| and B_||along the x-direction ( because they are parallel to the relative velocity of the S and S’ reference frames), and
The ‘perpendicular’ or ‘total transverse’ components E_⊥ and B_⊥, which are the vector sums of the y- and z-components.

So that gives us four equations only:

And, yes, we are done now. This is the Lorentz transformation of the fields. I am sure it has left you totally exhausted. Well… If not… […] It sure left me totally exhausted. 🙂

To lighten things up, let me insert an image of how the transformed field E actually looks like. The first image is the reference frame of a charge itself: we have a simple Coulomb field. The second image shows the charge flying by. Its electric field is ‘squashed up’. To be precise, it’s just like the scale of x is squashed up by a factor ((1−v²/c²)^1/2. Let me refer you to Feynman for the detail of the calculations here.

OK. So that’s it. You may wonder: what about that promise I made? Indeed, when I started this post, I said I’d present a mathematical construct that presents the electromagnetic force as one force only, as one physical reality, but so we’re back writing all of it in terms of two vectors—the electric field vector E and the magnetic field vector B. Well… What can I say? I did present the mathematical construct: it’s the electromagnetic tensor. So it’s that antisymmetric matrix really, which one can combine with a transformation matrix embodying the Lorentz transformation rules. So, I did what I promised to do. But you’re right: I am re-presenting stuff in the old style once again.

The second objection that you may have—in fact, that you should have, is that all of this has been rather tedious. And you’re right. The whole thing just re-emphasizes the value of using the four-potential vector. It’s obviously much easier to take that vector from one reference frame to another – so we just apply the Lorentz transformation rules to A_μ= (Φ, A) and get A_μ‘ = (Φ’, A’) from it – and then calculate E’ and B’ from it, rather than trying to remember those equations above. However, that’s not the point, or…

Well… It is and it isn’t. We wanted to get away from those two vectors E and B, and show that electromagnetism is really one phenomenon only, and so that’s where the concept of the electromagnetic tensor came in. There were two objectives here: the first objective was to introduce you to the concept of tensors, which we’ll need in the future. The second objective was to show you that, while Lorentz’ force law – F = q(E + v×B) makes it clear we’re talking one force only, there is a way of writing it all up that is much more elegant.

I’ve introduced the concept of tensors here, so the first objective should have been achieved. As for the second objective, I’ll discuss that in my next post, in which I’ll introduce the four-velocity vector μ_μas well as the four-force vector f_μ. It will explain the following beautiful equation of motion:

Now that looks very elegant and unified, doesn’t it? 🙂

[…] Hmm… No reaction. I know… You’re tired now, and you’re thinking: yet another way of representing the same thing? Well… Yes! So…

OK… Enough for today. Let’s follow up tomorrow.

Electric circuits (2): Kirchoff’s rules, and the energy in a circuit

My previous post was long and tedious, and all it did was presenting the three (passive) circuit elements as well as the concept of impedance. It show the inner workings of these little devices are actually quite complicated. Fortunately, the conclusions were very neat and short: for all circuit elements, we have a very simple relationship between (a) the voltage across the terminals of the element (V) and (b) the current that’s going through the circuit element (I). We found they are always in some ratio, which is referred to as the impedance, which we denoted by Z:

Z = V/I ⇔ V = I∗Z

So it’s a ‘simple’ ratio, indeed. But… Well… Simple and not simple. It’s a ratio of two complex numbers and, therefore, it’s a complex number itself. That’s why I use the ∗ symbol when re-writing the Z = V/I formula as V = I∗Z, so it’s clear we’re talking a product of two complex numbers). This ‘complexity’ is best understood by thinking of the voltage and the current as phase vectors (or phasors as engineers call them). Indeed, instead of using the sinusoidal functions we are used to, so that’s

V = V₀·cos(ωt + θ_V),
I = I₀·cos(ωt + θ_I), and
Z = Z₀·cos(ωt + θ) = (V₀/I₀)·cos(ωt + θ_V − θ_I),

we preferred the complex or vector notation, writing:

V = |V|eⁱ^{(ωt +}^θ_V⁾= V₀·eⁱ^{(ωt +}^θ_V⁾
I = |I|eⁱ^{(ωt +}^θ_I⁾= I₀·eⁱ^{(ωt +}^θ_I⁾
Z = |ZIeⁱ^{(ωt +}^θ⁾= Z₀·eⁱ^{(ωt +}^θ⁾= (V₀/I₀)·eⁱ^{(ωt + θ_V − θ_I}⁾

For the three circuit elements, we found the following solution for Z in terms of the previously defined properties of the respective circuit elements, i.e. their resistance (R), capacitance (C), and inductance (L) respectively:

For a resistor, we have Z(resistor) = Z_R= R
For an capacitor, we have Z(capacitor) = Z_C= 1/iωC = –i/(ωC)
For an inductor, we have Z(inductance) = Z_L= iωL

We also explained what these formulas meant, using graphs like the ones below:

The graph on the left-hand side gives you the ratio of the peak voltage and peak current for the three devices as a function of C, L, R and ω respectively.
The graph on the right-hand side shows you the relationship between the phase of the voltage and the current for a capacitor and for an inductor. [For a resistor, the phases are the same, so no need for a graph. Also note that the lag of the phase of the current vis-á-vis the voltage phase is 90 degrees for an inductor, while it’s 270 degrees for a capacitor (which amounts to the current leading the voltage with a 90° phase difference).]

The inner workings of our circuit elements are all wonderful and mysterious, and so we spent a lot of time writing about them. That’s finished now. The summary about describes all of them in very simple terms, relating the voltage and current phasor through the concept of impedance, which is just a ratio—albeit a complex ratio.

As the graphs above suggest, we can build all kinds of crazy circuits now, and the idea of resonance as we’ve learned it when studying the behavior of waves will be particularly relevant when discussing circuits that are designed to filter certain frequencies or, the opposite, to amplify some. We won’t go that far in this post, however, as I just want to explain the basic rules one needs to know when looking at a circuit, i.e. Kirchoff’s circuit laws. There are two of them:

1. Kirchoff’s Voltage Law (KCL): The sum of the voltage drops around any closed path is zero.

The principle is illustrated below. It doesn’t matter whether or not we have other circuits feeding into this one: Kirchoff’s Voltage Law (KCL) remains valid.

We can write this law using the concept of circulation once again or, what you’ll probably like more, just using plain summation:

2. Kirchoff’s Current Law (KCL): The sum of the currents into any node is zero.

This law is written and illustrated as follows:

KCR

This Law requires some definition of a node, of course. Feynman defines a node as any set of terminals such as a, b, c, d in the illustration above which are connected. So it’s a set of connected terminals.

Now, I’ll refer you Feynman for some practical examples. The circuit below is one of them. It looks complicated but it all boils down to solving a set of linear equations. So… Well… That’s it, really. We’re done! We should do the exercises, of course, but then we’re too lazy for that, I guess. 🙂 So we’re done!

Well… Almost. I also need to mention how one can reduce complicated circuits by combining parallel impedances, using the following formula:

And then another powerful idea is the idea of equivalent circuits. The rules for this are as follows:

Any two-terminal network of passive elements is equivalent to (and, hence, can be replaced by) an effective impedance (Z_eff).
Any two-terminal network of passive elements is equivalent to (and, hence, can be replaced by) a generator in series with an impedance.

These two principles are illustrated below: (a) is equivalent to (b) in each diagram.

The related formulas are:

I = Ɛ/Z_eff
V_n= Ɛ_eff− I_n∗Z_eff

Last but not least, I need to say something about the energy in circuits. As we noted in our previous post, the impedance will consist of a real and an imaginary part. We write:

Z = R + i·X

This gives rise to the following powerful equivalence: any impedance is equivalent to a series combination of a pure resistance and a pure reactance, as illustrated below (the ≡ sign stands for equivalence):

Of course, because this post risks becoming too short 🙂 I need to insert some more formulas now. If Z = R + i·X is the impedance of the whole circuit, then the whole circuit can be summarized in the following equation:

Ɛ = I∗Z = I∗(R + i·X)

Now, if we bring the analysis back to the real parts of this equation, then we may write our current as I = I₀·cos(ωt). This implies we chose a t = 0 point so θ_I= 0. [Note that this is somewhat different than what we usually do: we usually chose our t = 0 point such that θ_V= 0, but it doesn’t matter.] The real emf is then going to be the real part of Ɛ = I∗Z = I∗(R + i·X), so we’ll write it as Ɛ (no bold-face), and it’s going to be the real part of that expression above, which we can also write as:

Ɛ = I∗Z = I₀·eⁱ^(ωt)∗(R + i·X)

So Ɛ is the real part of this Ɛ and, you should check, it’s going to be equal to:

Ɛ = I₀·R·cos(ωt) − I₀·X·sin(ωt)

The two terms in this equation represent the voltage drops across the resistance R and the reactance X in that illustration above. […] Now that I think of it, in line with the -or and -ance convention for circuit elements and their properties, should we, perhaps, say resistor and reactor in this case? 🙂 […] OK. That’s a bad joke. [I don’t seem to have good ones, isn’t it?] 🙂

Jokes aside, we see that the voltage drop across the resistance is in phase with the current (because it’s a simple cosine function of ωt as well), while the voltage drop across the purely reactive part is out of phase with the current (as you know, the sine and cosine are the same function, but with a phase difference of π/2 indeed).

You’ll wonder where are we going with this, so let me wrap it all up. You know the power is the emf times the current, and so let’s integrate this thing over one cycle to get the average rate (and then I mean a time rate of change) of the energy that gets lost in the circuit. So we need to solve the following integral:

This may look like a monster, but if you look back at your notes from your math classes, you should be able to figure it out:

The first integral is (1/2)I₀²·R..
The second integral is zero.

So what? Well… Look at it! It means that the (average) energy loss in a circuit with impedance Z = R + i·X only depends on the real part of Z, which is equal to I₀²·R/2. That’s, of course, how we want it to be: ideal inductances and capacitors store energy when being powered, and give whatever they stored when ‘charging’ back to the circuit when the current reverses direction.

So it’s a nice result, because it’s consistent with everything. Hmm… Let’s double-check though… Is it also consistent with the power equation for a resistor which, remember, is written as: P = V·I = I·R·I = I²·R. […] Well… What about the 1/2 factor?

Well… Think about it. I is a sine or a cosine here, and so we want the average value of its square, so that’s 〈cos²(ωt)〉 = 1/2.

Done! 🙂

Electric circuits (1): the circuit elements

OK. No escape. It’s part of physics. I am not going to go into the nitty-gritty of it all (because this is a blog about physics, not about engineering) but it’s good to review the basics, which are, essentially, Kirchoff’s rules. Just for the record, Gustav Kirchhoff was a German genius who formulated these circuit laws while he was still a student, when he was like 20 years old or so. He did it as a seminar exercise 170 years ago, and then turned it into doctoral dissertation. Makes me think of that Dire Straits song—That’s the way you do it—Them guys ain’t dumb. 🙂

So this post is, in essence, just an ‘explanation’ of Feynman’s presentation of Kirchoff’s rules, so I am writing this post basically for myself, so as to ensure I am not missing anything. To be frank, Feynman’s use of notation when working with complex numbers is confusing at times and so, yes, I’ll do some ‘re-writing’ here. The nice thing about Feynman’s presentation of electrical circuits is that he sticks to Maxwell’s Laws when describing all ideal circuit elements, so he keeps using line integrals of the electric field E around closed paths (that’s what a circuit is, indeed) to describe the so-called passive circuit elements, and he also recapitulates the idea of the electromotive force when discussing the so-called active circuit element, so that’s the generator. That’s nice, because it links it all with what we’ve learned so far, i.e. the fundamentals as expressed in Maxwell’s set of equations. Having said that, I won’t make that link here in this post, because I feel it makes the whole approach rather heavy.

OK. Let’s go for it. Let’s first recall the concept of impedance.

The impedance concept

There are three ideal (passive) circuit elements: the resistor, the capacitor and the inductor. Real circuit elements usually combine characteristics of all of them, even if they are designed to work like ideal circuit elements. Collectively, these ideal (passive) circuit elements are referred to as impedances, because… Well… Because they have some impedance. In fact, you should note that, if we reserve the terms ending with -ance for the property of the circuit elements, and those ending on -or for the objects themselves, then we should call them impedors. However, that term does not seem to have caught on.

You already know what impedance is. I explained it before, notably in my post on the intricacies related to self- and mutual inductance. Impedance basically extends the concept of resistance, as we know it from direct current (DC) circuits, to alternating current (AC) circuits. To put it simply, when AC currents are involved – so when the flow of charge periodically changes reverses direction – then it’s likely that, because of the properties of the circuit, the current signal will lag the voltage signal, and so we’ll have some phase difference telling us by how much. So, resistance is just a simple real number R – it’s the ratio between (1) the voltage that is being applied across the resistor and (2) the current through it, so we write R = V/I – and it’s got a magnitude only, but impedance is a ‘number’ that has both a magnitude as well as phase, so it’s a complex number, or a vector.

In engineering, such ‘numbers’ with a magnitude as well as a phase are referred to as phasors. A phasor represents voltages, currents and impedances as a phase vector (note the bold italics: they explain how we got the pha-sor term). It’s just a rotating vector really. So a phasor has a varying magnitude (A) and phase (φ) , which is determined by (1) some maximum magnitude A₀, (2) some angular frequency ω and (3) some initial phase (θ). So we can write the amplitude A as:

A = A(φ) = A₀·cos(φ) = A₀·cos(ωt + θ)

As usual, Wikipedia has a nice animation for it:

In case you wonder why I am using a cosine rather than a sine function, the answer is that it doesn’t matter: the sine and the cosine are the same function except for a π/2 phase difference: just rotate the animation above by 90 degrees, or think about the formula: sinφ = cos(φ−π/2). 🙂

So A = A₀·cos(ωt + θ) is the amplitude. It could be the voltage, or the current, or whatever real variable. The phase vector itself is represented by a complex number, i.e. a two-dimensional number, so to speak, which we can write as all of the following:

A = A₀·e^iφ = A₀·cosφ + i·A₀·sinφ = A₀·cos(ωt+θ) + i·A₀·sin(ωt+θ)

= A₀·e^i(ωt+θ)= A₀·e^iθ·e^iωt= A₀·e^iωtwith A₀= A₀·e^iθ

That’s just Euler’s formula, and I am afraid I have to refer you to my page on the essentials if you don’t get this. I know what you are thinking: why do we need the vector notation? Why can’t we just be happy with the A = A₀·cos(ωt+θ) formula? The truthful answer is: it’s just to simplify calculations: it’s easier to work with exponentials than with cosines or sines. For example, writing e^{i(ωt + θ)}= e^iθ·e^iωtis easier than writing cos(ωt + θ) = … […] Well? […] Hmm… 🙂

See! You’re stuck already. You’d have to use the cos(α+β) = cosα·cosβ − sinα·sinβ formula: you’d get the same results (just do it for the simple calculation of the impedance below) but it takes a lot more time, and it’s easier to make mistake. Having said why complex number notation is great, I also need to warn you. There are a few things you have to watch out for. One of these things is notation. The other is the kind of mathematical operations we can do: it’s usually alright but we need to watch out with the i² = –1 thing when multiplying complex numbers. However, I won’t talk about that here because it would only confuse you even more. 🙂

Just for the notation, let me note that Feynman would write A₀as A₀ with the little hat or caret symbol (∧) on top of it, so as to indicate the complex coefficient is not a variable. So he writes A₀as Â₀ = A₀·e^iθ. However, I find that confusing and, hence, I prefer using bold-type for any complex number, variable or not. The disadvantage is that we need to remember that the coefficient in front of the exponential is not a variable: it’s a complex number alright, but not a variable. Indeed, do look at that A₀= A₀·e^iθ equality carefully: A₀is a specific complex number that captures the initial phase θ. So it’s not the magnitude of the phasor itself, i.e. |A| = A₀. In fact, magnitude, amplitude, phase… We’re using a lot confusing terminology here, and so that’s why you need to ‘get’ the math.

The impedance is not a variable either. It’s some constant. Having said that, this constant will depend on the angular frequency ω. So… Well… Just think about this as you continue to read. 🙂 So the impedance is some number, just like resistance, but it’s a complex number. We’ll denote it by Z and, using Euler’s formula once again, we’ll write it as:

Z = |Z|eⁱ^θ= V/I = |V|eⁱ^{(ωt +}^θ_V⁾/|I|eⁱ^{(ωt +}^θ_I⁾= [|V|/|I|]·eⁱ⁽^θ_V⁻^θ_I⁾

So, as you can see, it is, literally, some complex ratio, just like R = V/I was some real ratio: it is a complex ratio because it has a magnitude and a direction, obviously. Also please do note that, as I mentioned already, the impedance is, in general, some function of the frequency ω, as evidenced by the ωt term in the exponential, but so we’re not looking at ω as a variable: V and I are variables and, as such, they depend on ω, but so you should look at ω as some parameter. I know I should, perhaps, not be so explicit on what’s going on, but I want to make sure you understand.

So what’s going on? The illustration below (credit goes to Wikipedia, once again) explains. It’s a pretty generic view of a very simple AC circuit. So we don’t care what the impedance is: it might be an inductor or a capacitor, or a combination of both, but we don’t care: we just call it an impedance, or an impedor if you want. 🙂 The point is: if we apply an alternating current, then the current and the voltage will both go up and down, but the current signal will lag the voltage signal, and some phase factor θ tells us by how much, so θ will be the phase difference.

Now, we’re dividing one complex number by another in that Z = V/I formula above, and dividing one complex number by another is not all that straightforward, so let me re-write that formula for Z above as:

V = I∗Z = I∗|Z|eⁱ^θ

Now, while that V = I∗Z formula resembles the V = I·R formula, you should note the bold-face type for V and I, and the ∗ symbol I am using here for multiplication. The bold-face for V and I implies they’re vectors, or complex numbers. As for the ∗ symbol, that’s to make it clear we’re not talking a vector cross product A×B here, but a product of two complex numbers. [It’s obviously not a vector dot product either, because a vector dot product yields a real number, not some other vector.]

Now we write V and I as you’d expect us to write them:

V = |V|eⁱ^{(ωt +}^θ_V⁾= V₀·eⁱ^{(ωt +}^θ_V⁾
I = |I|eⁱ^{(ωt +}^θ_I⁾= I₀·eⁱ^{(ωt +}^θ_I⁾

θ_V and θ_Iare, obviously, the so-called initial phase of the voltage and the current respectively. These ‘initial’ phases are not independent: we’re talking a phase difference really, between the voltage and the current signal, and it’s determined by the properties of the circuit. In fact, that’s the whole point here: the impedance is a property of the circuit and determines how the current signal varies as a function of the voltage signal. In fact, we’ll often choose the t = 0 point such that θ_Vand so then we need to find θ_I. […] OK. Let’s get on with it. Writing out all of the factors in the V = I∗Z = I∗|Z|eⁱ^θ equation yields:

V = |V|eⁱ^{(ωt +}^θ_V⁾= I∗Z = |I|eⁱ^{(ωt +}^θ_I⁾∗|Z|eⁱ^θ= |I||Z|eⁱ^{(ωt +}^θ_I^+ θ)

Now, this equation must hold for all values of t, so we can equate the magnitudes and phases and, hence, the following equalities must hold:

|V| = |I||Z| ⇔ |Z| = |V|/|I|
ωt + θ_V = ωt + θ_I+ θ ⇔ θ = θ_V − θ_I

Done!

Of course, you’ll complain once again about those complex numbers: voltage and current are something real, isn’t it? And so what is really about this complex numbers? Well… I can just say what I said already. You’re right. I’ve used the complex notation only to simplify the calculus, so it’s only the real part of those complex-valued functions that counts.

OK. We’re done with impedance. We can now discuss the impedors, including resistors (for which we won’t have such lag or phase difference, but the concept of impedance applies nevertheless).

Before I start, however, you should think about what I’ve done above: I explained the concept of impedance, but I didn’t do much with it. The real-life problem will usually be that you get the voltage as a function of time, and then you’ll have to calculate the impedance of a circuit and, then, the current as a function of time. So I just showed the fundamental relations but, in real life, you won’t know what θ and θ_I could possibly be. Well… Let me correct that statement: we’ll give you formulas for θ as we discuss the various circuit elements and their impedance below, and so then you can use these formulas to calculate θ_I. 🙂

Resistors

Let’s start with what seems to be the easiest thing: a resistor. A real resistor is actually not easy to understand, because it requires us to understand the properties of real materials. Indeed, it may or may not surprise you, but the linear relation between the voltage and the current for real materials is only approximate. Also, the way resistors dissipate energy is not easy to understand. Indeed, unlike inductors and capacitors, i.e. the other two passive components of an electrical circuit, a resistor does not store but dissipates energy, as shown below.

It’s a nice animation (credit for it has to go to Wikipedia once more), as it shows how energy is being used in an electric circuit. Note that the little moving pluses are in line with the convention that a current is defined as the movement of positive charges, so we write I = dQ/dt instead of I = −dQ/dt. That also explains the direction of the field line E, which has been added to show that the charges move with the field that is being generated by the power source (which is not shown here). So, what we have here is that, on one side of the circuit, some generator or voltage source will create an emf pushing the charges, and so the animation shows how some load – i.e. the resistor in this case – will consume their energy, so they lose their push (as shown by the change in color from yellow to black). So power, i.e.energy per unit time, is supplied, and is then consumed.

To increase the current in the circuit above, you need to increase the voltage, but increasing both amounts to increasing the power that’s being consumed in the circuit. Electric power is voltage times current, so P = V·I (or v·i, if I use the small letters that are used in the two animations below). Now, Ohm’s Law (I = V/R) says that, if we’d want to double the current, we’d need to double the voltage, and so we’re quadrupling the power then: P₂ = V₂·I₂= (2·V₁)·(2·I₁) = 4·V₁·I₁= 2²·P₁. So we have a square-cube law for the power, which we get by substituting V for R·I or by substituting I for V/R, so we can write the power P as P = V²/R = I²·R. This square-cube law says exactly the same: if you want to double the voltage or the current, you’ll actually have to double both and, hence, you’ll quadruple the power.

But back to the impedance: Ohm’s Law is the Z = V/I law for resistors, but we can simplify it because we know the voltage across the resistor and the current that’s going through are in phase. Hence, θ_V and θ_Iare identical and, therefore, the θ = θ_V− θ_Iin Z = |Z|eⁱ^θis equal to zero and, hence, Z = |Z|. Now, |Z| = |V|/|I| = V₀/I₀. So the impedance Z is just some real number R = V₀/I₀, which we can also write as:

R = V₀/I₀= (V₀·eⁱ^{(ωt + α}⁾)/(I₀·eⁱ^{(ωt + α}⁾) = V(t)/I(t), with α = θ_V = θ_I

The equation above goes from R = V₀/I₀to R = V(t)/I(t) = V/I. It’s note the same thing: the second equation says that, at any point in time, the voltage and the current will be proportional to each other, with R or its reciprocal as the proportionality constant. In any case, we have our formula for Z here:

Z = R = V/I = V₀/I₀

So that’s simple. Before we move to the next, let me note that the resistance of a real resistor may depend on its temperature, so in real-life applications one will want to keep its temperature as stable as possible. That’s why real-life resistors have power ratings and recommended operating temperatures. The image below illustrates how so-called heat-sink resistors can be mounted on a heat sink with a simple spring clip so as to ensure the dissipated heat is transported away. These heat-sink resistors are rather small (10 by 15 mm only) but are rated for 35 watt – so that’s quite a lot for such small thing – if correctly mounted.

As mentioned, the linear relation between the voltage and the current is only approximate, and the observed relation is also there only for frequencies that are not ‘too high’ because, if the frequency becomes very high, the free electrons will start radiating energy away, as they produce electromagnetic radiation. So one always needs to look at the tolerances of real-life resistors, which may be ± 5%, ± 10%, or whatever. In any case… On to the next.

Capacitors (condensers)

We talked at length about capacitors (aka condensers) in our post explaining capacitance or, the more widely used term, capacity: the capacity of a capacitor is the observed proportionality between (1) the voltage (V) across and (2) the charge (Q) on the capacitor, so we wrote it as:

C = Q/V

Now, it’s easy to confuse the C here with the C for coulomb, which I’ll also use in a moment, and so… Well… Just don’t! 🙂 The meaning of the symbol is usually obvious from the context.

As for the explanation of this relation, it’s quite simple: a capacitor consists of two separate conductors in space, with positive charge on one, and an equal and opposite (i.e. negative) charge on the other. Now, the logic of the superposition of fields implies that, if we double the charges, we will also double the fields, and so the work one needs to do to carry a unit charge from one conductor to the other is also doubled! So that’s why the potential difference between the conductors is proportional to the charge.

The C = Q/V formula actually measures the ability of the capacitor to store electric charge and, therefore, to store energy, so that’s why the term capacity is really quite appropriate. I’ll let you google a few illustrations like the one below, that shows how a capacitor is actually being charged in a circuit. Usually, some resistance will be there in the circuit, so as to limit the current when it’s connected to the voltage source and, therefore, as you can see, the R times C factor (R·C) determines how fast or how slow the capacitor charges and/or discharges. Also note that the current is equal to the time rate of change of the charge: I = dQ/dt.

In the above-mentioned post, we also give a few formulas for the capacity of specific types of condensers. For example, for a parallel-plate condenser, the formula was C = ε₀A/d. We also mentioned its unit, which is is coulomb/volt, obviously, but – in honor of Michael Faraday, who gave us Faraday’s Law, and many other interesting formulas – it’s referred to as the farad: 1 F = 1 C/V. The C here is coulomb, of course. Sorry we have to use C to denote two different things but, as I mentioned, the meaning of the symbol is usually clear from the context.

We also talked about how dielectrics actually work in that post, but we did not talk about the impedance of a capacitor, so let’s do that now. The calculation is pretty straightforward. Its interpretation somewhat less so. But… Well… Let’s go for it.

It’s the current that’s charging the condenser (sorry I keep using both terms interchangeably), and we know that the current is the time rate of change of the charge (I = dQ/dt). Now, you’ll remember that, in general, we’d write a phasor A as A = A₀·e^iωtwith A₀= A₀·e^iθ, so A₀is a complex coefficient incorporating the initial phase, which we wrote as θ_Vand θ_Ifor the voltage and for the current respectively. So we’ll represent the voltage and the current now using that notation, so we write: V = V₀·e^iωtand I = I₀·e^iωt. So let’s now use that C = Q/V by re-writing it as Q = C·V and, because C is some constant, we can write:

I = dQ/dt = d(C·V)/dt = C·dV/dt

Now, what’s dV/dt? Oh… You’ll say: V is the magnitude of V, so it’s equal to |V| = |V₀·e^iωt| = |V₀|·|e^iωt| = |V₀| = |V₀·e^iθ| = |V₀|·|e^iθ| = |V₀| = V₀. So… Well… What? V₀ is some constant here! It’s the maximum amplitude of V, so… Well… It’s time derivative is zero: dV₀/dt = 0.

Yes. Indeed. We did something very wrong here! You really need to watch out with this complex-number notation, and you need to think about what you’re doing. V is not the magnitude of V but its (varying) amplitude. So it’s the real voltage V that varies with time: it’s equal to V₀·cos(ωt + θ_V), which is the real part of our phasor V. Huh? Yes. Just hang in for a while. I know it’s difficult and, frankly, Feynman doesn’t help us very much here. Let’s take one step back and so – you will see why I am doing this in a moment – let’s calculate the time derivative of our phasor V, instead of the time derivative of our real voltage V. So we calculate dV/dt, which is equal to:

dV/dtd(V₀·e^iωt)/dt = V₀·d(e^iωt)/dt = V₀·(iω)·e^iωt = iω·V₀·e^iωt = iω·V

Remarkable result, isn’t it? We take the time derivative of our phasor, and the result is the phasor itself multiplied with iω. Well… Yes. It’s a general property of exponentials, but still… Remarkable indeed! We’d get the same with I, but we don’t need that for the moment. What we do need to do is go from our I = C·dV/dt relation, which connects the real parts of I and V one to another, to the I = C·dV/dt relation, which relates the (complex) phasors. So we write:

I = C·dV/dt ⇔ I = C·dV/dt

Can we do that? Just like that? We just replace I and V by I and V? Yes, we can. Why? Well… We know that I is the real part of I and so we can write I = Re(I)+ Im(I)·i = I + Im(I)·i, and then we can write the right-hand side of the equation as C·dV/dt = Re(C·dV/dt)+ Im(C·dV/dt)·i. Now, two complex numbers are equal if, and only if, their real and imaginary parts are the same, so… Well… Write it all out, if you want, using Euler’s formula, and you’ll see it all makes sense indeed.

So what do we get? The I = C·dV/dt gives us:

I = C·dV/dt = C·(iω)·V

That implies that I/V = C·(iω) and, hence, we get – finally! – what we need to get:

Z = V/I = 1/(iωC)

This is a grand result and, while I am sorry I made you suffer for it, I think it did a good job here because, if you’d check Feynman on it, you’ll see he – or, more probably, his assistants, – just skate over this without bothering too much about mathematical rigor. OK. All that’s left now is to interpret this ‘number’ Z = 1/(iωC). It is a purely imaginary number, and it’s a constant indeed, albeit a complex constant. It can be re-written as:

Z = 1/(iωC) = i^-1/(ωC) = –i/(ωC) = (1/ωC)·e^−i·π/2

[Sorry. I can’t be more explicit here. It’s just of the wonders of complex numbers: i^-1= –i. Just check one my posts on complex numbers for more detail.] Now, a –i factor corresponds to a rotation of minus 90 degrees, and so that gives you the true meaning of what’s usually said about a circuit with a capacitor: the voltage across the capacitor will lag the current with a phase difference equal to π/2, as shown below. Of course, as it’s the voltage driving the current, we should say it’s the current that is lagging with a phase difference of 3π/2, rather than stating it the other way around! Indeed, i^-1= –i = –1·i = i²·i = i³, so that amounts to three ‘turns’ of the phase in the counter-clockwise direction, which is the direction in which our ωt angle is ‘turning’.

It is a remarkable result, though. The illustration above assumes the maximum amplitude of the voltage and the current are the same, so |Z| = |V|/|I| = 1, but what if they are not the same? What are the real bits then? I can hear you, indeed: “To hell with the bold-face letters: what’s V and I? What’s the real thing?”

Well… V and I are the real bits of V = |V|eⁱ^(ωt+^θ_V⁾= V₀·eⁱ^(ωt+^θ_V⁾and of I = |I|eⁱ^(ωt+^θ_I⁾= I₀·eⁱ^{(ωt+θ_V−θ}⁾ = I₀·eⁱ^(ωt−^θ⁾= I₀·eⁱ^(ωt+^π/2⁾respectively so, assuming θ_V= 0 (as mentioned above, that’s just a matter of choosing a convenient t = 0 point), we get:

V = V₀·cos(ωt)
I = I₀·cos(ωt + π/2)

So the π/2 phase difference is there (you need to watch out with the signs, of course: θ = −π/2, but so it’s the current that seems to lead here) but the V₀/I₀ratio doesn’t have to be one, so the real voltage and current could look like something below, where the maximum amplitude of the current is only half of the maximum amplitude of the voltage.

So let’s analyze this quickly: the V₀/I₀ratio is equal to |Z| = |V|/|I| = V₀/I₀= 1/ωC = (1/ω)(1/C) (note that it’s not equal to V/I = V(t)/I(t), which is a ratio that doesn’t make sense because I(t) goes through zero as the current switches direction). So what? Well… It means the ratio is inversely proportional to both the frequency ω as well as the capacity C, as shown below. Think about this: if ω goes to zero, V₀/I₀goes to ∞, which means that, for a given voltage, the current must go to zero. That makes sense, because we’re talking DC current when ω → 0, and the capacitor charges itself and then that’s it: no more currents. Now, if C goes to zero, so we’re talking capacitors with hardly any capacity, we’ll also get tiny currents. Conversely, for large C, we’ll get huge currents, as the capacitor can take pretty much any charge you throw at it, so that makes for small V₀/I₀ratios. The most interesting thing to consider is ω going to infinity, as the V₀/I₀ratio is also quite small then. What happens? The capacitor doesn’t get the time to charge, and so it’s always in this state where it has large currents flowing in and out of it, as it can’t build the voltage that would counter the electromotive force that’s being supplied by the voltage source.

OK. That’s it. Le’s discuss the last (passive) element.

Inductors

We’ve spoiled the party a bit with that illustration above, as it gives the phase difference for an inductor already:

Z = iωL = ωL·e^i·π/2, with L the inductance of the coil

So, again assuming that θ_V= 0, we can calculate I as:

I = |I|eⁱ^(ωt+^θ_I⁾= I₀·eⁱ^{(ωt+θ_V−θ}⁾ = I₀·eⁱ^(ωt−^θ⁾= I₀·eⁱ^(ωt−^π/2⁾

Of course, you’ll want to relate this, once again, to the real voltage and the real current, so let’s write the real parts of our phasors:

V = V₀·cos(ωt)
I = I₀·cos(ωt − π/2)

Just to make sure you’re not falling asleep as you’re reading, I’ve made another graph of how things could look like. So now’s it’s the current signal that’s lagging the voltage signal with a phase difference equal to θ = π/2.

Also, to be fully complete, I should show you how the V₀/I₀ratio now varies with L and ω. Indeed, here also we can write that |Z| = |V|/|I| = V₀/I₀, but so here we find that V₀/I₀ = ωL, so we have a simple linear proportionality here! For example, for a given voltage V₀, we’ll have smaller currents as ω increases, so that’s the opposite of what happens with our ideal capacitors. I’ll let you think about that… 🙂

Now how do we get that Z = iωL formula? In my post on inductance, I explained what an inductor is: a coil of wire, basically. Its defining characteristic is that a changing current will cause a changing magnetic field in it and, hence, some change in the flux of the magnetic field. Now, Faraday’s Law tells us that that will cause some circulation of the electric field in the coil, which amounts to an induced potential difference which is referred to as the electromotive force (emf). Now, it turns out that the induced emf is proportional to the change in current. So we’ve got another constant of proportionality here, so it’s like how we defined resistance, or capacitance. So, in many ways, the inductance is just another proportionality coefficient. If we denote it by L – the symbol is said to honor the Russian phyicist Heinrich Lenz, whom you know from Lenz’ Law – then we define it as:

L = −Ɛ/(dI/dt)

The dI/dt factor is, obviously, the time rate of change of the current, and the negative sign indicates that the emf opposes the change in current, so it will tend to cause an opposing current. However, the power of our voltage source will ensure the current does effectively change, so it will counter the ‘back emf’ that’s being generated by the inductor. To be precise, the voltage across the terminals of our inductor, which we denote by V, will be equal and opposite to Ɛ, so we write:

V = −Ɛ = L·(dI/dt)

Now, this very much resembles the I = C·dV/dt relation we had for capacitors, and it’s completely analogous indeed: we just need to switch the I and V, and C and L symbols. So we write:

V = L·dI/dt⇔ V = L·dI/dt

Now, dI/dt is a similar time derivative as dV/dt. We calculate it as:

dI/dtd(I₀·e^iωt)/dt = I₀·d(e^iωt)/dt = I₀·(iω)·e^iωt = iω·I₀·e^iωt = iω·I

So we get what we want and have to get:

V = L·dI/dt = iωL·I

Now, Z = V/I, so Z = iωL indeed!

Summary of conclusions

Let’s summarize what we found:

For a resistor, we have Z(resistor) = Z_R= R = V/I = V₀/I₀
For an capacitor, we have Z(capacitor) = Z_C= 1/(iωC) = –i/(ωC)
For an inductor, we have Z(inductance) = Z_L= iωL

Note that the impedance of capacitors decreases as frequency increases, while for inductors, it’s the other way around. We explained that by making you think of the currents: for a given voltage, we’ll have large currents for high frequencies, and, hence, a small V₀/I₀ratio. Can you think of what happens with an inductor? It’s not so easy, so I’ll refer you to the addendum below for some more explanation.

Let me also note that, as you can see, the impedance of (ideal) inductors and capacitors is a pure imaginary number, so that’s a complex number which has no real part. In engineering, the imaginary part of the impedance is referred to as the reactance, so engineers will say that ideal capacitors and inductors have a purely imaginary reactive impedance.

However, in real life, the impedance will usually have both a real as well as an imaginary part, so it will be some kind of mix, so to speak. The real part is referred to as the ‘resistance’ R, and the ‘imaginary’ part is referred to as the ‘reactance’ X. The formula for both is given below:

But here I have to end my post on circuit elements. It’s become quite long, so I’ll discuss Kirchoff’s rules in my next post.

Addendum: Why is V = − Ɛ?

Inductors are not easy to understand—intuitively, that is. That’s why I spent so much time writing on them in my other post on them, to which I should be referring you here. But let me recapitulate the key points. The key idea is that we’re pumping energy into an inductor when applying a current and, as you know, the time rate of change is power: P = dW/dt, so we’re talking power here too, which is voltage times current: P = dW/dt = V·I. The illustration below shows what happens when an alternating current is applied to the circuit with the inductor. So the assumption is that the current goes in one and then in the other direction, so I > 0, and then I < 0, etcetera. We’re also assuming some nice sinusoidal curve for the current here (i.e. the blue curve), and so we get what we get for U (i.e. the red curve), which is the energy that’s stored in the inductor really, as it tries to resist the changing current: the energy goes up and down between zero and some maximum amplitude that’s determined by the maximum current.

So, yes, building up current requires energy from some external source, which is used to overcome the ‘back emf’ in the inductor, and that energy is stored in the inductor itself. [If you still wonder why it’s stored in the inductor, think about the other question: where else would it be stored?] How is stored? Look at the graph and think: it’s stored as kinetic energy of the charges, obviously. That explains why the energy is zero when the current is zero, and why the energy maxes out when the current maxes out. So, yes, it all makes sense! 🙂

Let me give another example. The graph below assumes the current builds up to some maximum. As it reaches its maximum, the stored energy will also max out. This example assumes direct current, so it’s a DC circuit: the current builds up, but then stabilizes at some maximum that we can find by applying Ohm’s Law to the resistance of the circuit: I = V/R. Resistance? But we were talking an ideal inductor? We are. If there’s no other resistance in the circuit, we’ll have a short-circuit, so the assumption is that we do have some resistance in the circuit and, therefore, we should also think of some energy loss to heat from the current in the resistance. If not, well… Your power source will obviously soon reach its limits. 🙂

So what’s going on then? We have some changing current in the coil but, obviously, some kind of inertia also: the coil itself opposes the change in current through the ‘back emf’. Now, it requires energy, or power, to overcome the inertia, so that’s the power that comes from our voltage source: it will offset the ‘back emf’, so we may effectively think of a little circuit with an inductor and a voltage source, as shown below.

But why do we write V = − Ɛ? Our voltage source can have any voltage, can’t it? Yes. Sure. But so the coil will always provide an emf that’s exactly the opposite of this voltage. Think of it: we have some voltage that’s being applied across the terminals of the inductor, and so we’ll have some current. A current that’s changing. And it’s that current will generate an emf that’s equal to Ɛ = –L·(dI/dt). So don’t think of Ɛ as some constant: it’s the self-inductance coefficient L that’s constant, but I (and, hence, dI/dt) and V are variable.

The point is: we cannot have any potential difference in a perfect conductor, which is what the terminals are: any potential difference, i.e. any electric field really, would cause huge currents. In other words, the voltage V and the emf Ɛ have to cancel each other out, all of the time. If not, we’d have huge currents in the wires re-establishing the V = −Ɛ equality.

Let me use Feynman’s argument here. Perhaps that will work better. 🙂 Our ideal inductor is shown below: it’s shielded by some metal box so as to ensure it does not interact with the rest of the circuit. So we have some current I, which we assume to be an AC current, and we know some voltage is needed to cause that current, so that’s the potential difference V between the terminals.

The total circulation of E – around the whole circuit – can be written as the sum of two parts:

Now, we know circulation of E can only be caused by some changing magnetic field, which is what’s going on in the inductor:

So this change in the magnetic flux is what it causing the ‘back emf’, and so the integral on the left is, effectively, equal to Ɛ, not minus Ɛ but +Ɛ. Now, the second integral is equal to V, because that’s the voltage V between the two terminals a and b. So the whole integral is equal to 0 = Ɛ + V and, therefore, we have that:

V = − Ɛ = L·dI/dt

Re-visiting the speed of light, Planck’s constant, and the fine-structure constant

Note: I have published a paper that is very coherent and fully explains what the fine-structure constant actually is. There is nothing magical about it. It’s not some God-given number. It’s a scaling constant – and then some more. But not God-given. Check it out: The Meaning of the Fine-Structure Constant. No ambiguity. No hocus-pocus.

Jean Louis Van Belle, 23 December 2018

Original post:

A brother of mine sent me a link to an article he liked. Now, because we share some interest in physics and math and other stuff, I looked at it and…

Well… I was disappointed. Despite the impressive credentials of its author – a retired physics professor – it was very poorly written. It made me realize how much badly written stuff is around, and I am glad I am no longer wasting my time on it. However, I do owe my brother some explanation of (a) why I think it was bad, and of (b) what, in my humble opinion, he should be wasting his time on. 🙂 So what it is all about?

The article talks about physicists deriving the speed of light from “the electromagnetic properties of the quantum vacuum.” Now, it’s the term ‘quantum‘, in ‘quantum vacuum’, that made me read the article.

Indeed, deriving the theoretical speed of light in empty space from the properties of the classical vacuum – aka empty space – is a piece of cake: it was done by Maxwell himself as he was figuring out his equations back in the 1850s (see my post on Maxwell’s equations and the speed of light). And then he compared it to the measured value, and he saw it was right on the mark. Therefore, saying that the speed of light is a property of the vacuum, or of empty space, is like a tautology: we may just as well put it the other way around, and say that it’s the speed of light that defines the (properties of the) vacuum!

Indeed, as I’ll explain in a moment: the speed of light determines both the electric as well as the magnetic constants μ₀and ε₀, which are the (magnetic) permeability and the (electric) permittivity of the vacuum respectively. Both constants depend on the units we are working with (i.e. the units for electric charge, for distance, for time and for force – or for inertia, if you want, because force is defined in terms of overcoming inertia), but so they are just proportionality coefficients in Maxwell’s equations. So once we decide what units to use in Maxwell’s equations, then μ₀and ε₀ are just proportionality coefficients which we get from c. So they are not separate constants really – I mean, they are not separate from c – and all of the ‘properties’ of the vacuum, including these constants, are in Maxwell’s equations.

In fact, when Maxwell compared the theoretical value of c with its presumed actual value, he didn’t compare c‘s theoretical value with the speed of light as measured by astronomers (like that 17th century Ole Roemer, to which our professor refers: he had a first go at it by suggesting some specific value for it based on his observations of the timing of the eclipses of one of Jupiter’s moons), but with c‘s value as calculated from the experimental values of μ₀and ε₀! So he knew very well what he was looking at. In fact, to drive home the point, it may also be useful to note that the Michelson-Morley experiment – which accurately measured the speed of light – was done some thirty years later. So Maxwell had already left this world by then—very much in peace, because he had solved the mystery all 19th century physicists wanted to solve through his great unification: his set of equations covers it all, indeed: electricity, magnetism, light, and even relativity!

I think the article my brother liked so much does a very lousy job in pointing all of that out, but that’s not why I wouldn’t recommend it. It got my attention because I wondered why one would try to derive the speed of light from the properties of the quantum vacuum. In fact, to be precise, I hoped the article would tell me what the quantum vacuum actually is. Indeed, as far as I know, there’s only one vacuum—one ’empty space’: empty is empty, isn’t it? 🙂 So I wondered: do we have a ‘quantum’ vacuum? And, if so, what is it, really?

Now, that is where the article is really disappointing, I think. The professor drops a few names (like the Max Planck Institute, the University of Paris-Sud, etcetera), and then, promisingly, mentions ‘fleeting excitations of the quantum vacuum’ and ‘virtual pairs of particles’, but then he basically stops talking about quantum physics. Instead, he wanders off to share some philosophical thoughts on the fundamental physical constants. What makes it all worse is that even those thoughts on the ‘essential’ constants are quite off the mark.

So… This post is just a ‘quick and dirty’ thing for my brother which, I hope, will be somewhat more thought-provoking than that article. More importantly, I hope that my thoughts will encourage him to try to grind through better stuff.

On Maxwell’s equations and the properties of empty space

Let me first say something about the speed of light indeed. Maxwell’s four equations may look fairly simple, but that’s only until one starts unpacking all those differential vector equations, and it’s only when going through all of their consequences that one starts appreciating their deep mathematical structure. Let me quickly copy how another blogger jotted them down: 🙂

As I showed in my above-mentioned post, the speed of light (i.e. the speed with which an electromagnetic pulse or wave travels through space) is just one of the many consequences of the mathematical structure of Maxwell’s set of equations. As such, the speed of light is a direct consequence of the ‘condition’, or the properties, of the vacuum indeed, as Maxwell suggested when he wrote that “we can scarcely avoid the inference that light consists in the transverse undulations of the same medium which is the cause of electric and magnetic phenomena”.

Of course, while Maxwell still suggests light needs some ‘medium’ here – so that’s a reference to the infamous aether theory – we now know that’s because he was a 19th century scientist, and so we’ve done away with the aether concept (because it’s a redundant hypothesis), and so now we also know there’s absolutely no reason whatsoever to try to “avoid the inference.” 🙂 It’s all OK, indeed: light is some kind of “transverse undulation” of… Well… Of what?

We analyze light as traveling fields, represented by two vectors, E and B, whose direction and magnitude varies both in space as well as in time. E and B are field vectors, and represent the electric and magnetic field respectively. An equivalent formulation – more or less, that is (see my post on the Liénard-Wiechert potentials) – for Maxwell’s equations when only one (moving) charge is involved is:

This re-formulation, which is Feynman’s preferred formula for electromagnetic radiation, is interesting in a number of ways. It clearly shows that, while we analyze the electric and magnetic field as separate mathematical entities, they’re one and the same phenomenon really, as evidenced by the B = –e_r‘×E/c equation, which tells us the magnetic field from a single moving charge is always normal (i.e. perpendicular) to the electric field vector, and also that B‘s magnitude is 1/c times the magnitude of E, so |B| = B = |E|/c = E/c. In short, B is fully determined by E, or vice versa: if we have one of the two fields, we have the other, so they’re ‘one and the same thing’ really—not in a mathematical sense, but in a real sense.

Also note that E and B‘s magnitude is just the same if we’re using natural units, so if we equate c with 1. Finally, as I pointed out in my post on the relativity of electromagnetic fields, if we would switch from one reference frame to another, we’ll have a different mix of E and B, but that different mix obviously describes the same physical reality. More in particular, if we’d be moving with the charges, the magnetic field sort of disappears to re-appear as an electric field. So the Lorentz force F = F_electric + F_magnetic= qE + qv×B is one force really, and its ‘electric’ and ‘magnetic’ component appear the way they appear in our reference frame only. In some other reference frame, we’d have the same force, but its components would look different, even if they, obviously, would and should add up to the same. [Well… Yes and no… You know there’s relativistic corrections to be made to the forces to, but that’s a minor point, really. The force surely doesn’t disappear!]

All of this reinforces what you know already: electricity and magnetism are part and parcel of one and the same phenomenon, the electromagnetic force field, and Maxwell’s equations are the most elegant way of ‘cutting it up’. Why elegant? Well… Click the Occam tab. 🙂

Now, after having praised Maxwell once more, I must say that Feynman’s equations above have another advantage. In Maxwell’s equations, we see two constants, the electric and magnetic constant (denoted by μ₀and ε₀ respectively), and Maxwell’s equations imply that the product of the electric and magnetic constant is the reciprocal of c²: μ₀·ε₀= 1/c². So here we see ε₀and c only, so no μ₀, so that makes it even more obvious that the magnetic and electric constant are related one to another through c.

[…] Let me digress briefly: why do we have c² in μ₀·ε₀= 1/c², instead of just c? That’s related to the relativistic nature of the magnetic force: think about that B = E/c relation. Or, better still, think about the Lorentz equation F = F_electric + F_magnetic= qE + qv×B = q[E + (v/c)×(E×e_r‘)]: the 1/c factor is there because the magnetic force involves some velocity, and any velocity is always relative—and here I don’t mean relative to the frame of reference but relative to the (absolute) speed of light! Indeed, it’s the v/c ratio (usually denoted by β = v/c) that enters all relativistic formulas. So the left-hand side of the μ₀·ε₀= 1/c² equation is best written as (1/c)·(1/c), with one of the two 1/c factors accounting for the fact that the ‘magnetic’ force is a relativistic effect of the ‘electric’ force, really, and the other 1/c factor giving us the proper relationship between the magnetic and the electric constant. To drive home the point, I invite you to think about the following:

μ₀ is expressed in (V·s)/(A·m), while ε₀is expressed in (A·s)/(V·m), so the dimension in which the μ₀·ε₀product is expressed is [(V·s)/(A·m)]·[(A·s)/(V·m)] = s²/m², so that’s the dimension of 1/c².
Now, this dimensional analysis makes it clear that we can sort of distribute 1/c² over the two constants. All it takes is re-defining the fundamental units we use to calculate stuff, i.e. the units for electric charge, for distance, for time and for force – or for inertia, as explained above. But so we could, if we wanted, equate both μ₀ as well as ε₀with 1/c.
Now, if we would then equate c with 1, we’d have μ₀ = ε₀= c = 1. We’d have to define our units for electric charge, for distance, for time and for force accordingly, but it could be done, and then we could re-write Maxwell’s set of equations using these ‘natural’ units.

In any case, the nitty-gritty here is less important: the point is that μ₀and ε₀are also related through the speed of light and, hence, they are ‘properties’ of the vacuum as well. [I may add that this is quite obvious if you look at their definition, but we’re approaching the matter from another angle here.]

In any case, we’re done with this. On to the next!

On quantum oscillations, Planck’s constant, and Planck units

The second thought I want to develop is about the mentioned quantum oscillation. What is it? Or what could it be? An electromagnetic wave is caused by a moving electric charge. What kind of movement? Whatever: the charge could move up or down, or it could just spin around some axis—whatever, really. For example, if it spins around some axis, it will have a magnetic moment and, hence, the field is essentially magnetic, but then, again, E and B are related and so it doesn’t really matter if the first cause is magnetic or electric: that’s just our way of looking at the world: in another reference frame, one that’s moving with the charges, the field would essential be electric. So the motion can be anything: linear, rotational, or non-linear in some irregular way. It doesn’t matter: any motion can always be analyzed as the sum of a number of ‘ideal’ motions. So let’s assume we have some elementary charge in space, and it moves and so it emits some electromagnetic radiation.

So now we need to think about that oscillation. The key question is: how small can it be? Indeed, in one of my previous posts, I tried to explain some of the thinking behind the idea of the ‘Great Desert’, as physicists call it. The whole idea is based on our thinking about the limit: what is the smallest wavelength that still makes sense? So let’s pick up that conversation once again.

The Great Desert lies between the 10³² and 10⁴³Hz scale. 10³² Hz corresponds to a photon energy of E_γ = h·f = (4×10⁻¹⁵ eV·s)·(10³² Hz) = 4×10¹⁷ eV = 400,000 tera-electronvolt (1 TeV = 10¹²eV). I use the γ (gamma) subscript in my E_γ symbol for two reasons: (1) to make it clear that I am not talking the electric field E here but energy, and (2) to make it clear we are talking ultra-high-energy gamma-rays here.

In fact, γ-rays of this frequency and energy are theoretical only. Ultra-high-energy gamma-rays are defined as rays with photon energies higher than 100 TeV, which is the upper limit for very-high-energy gamma-rays, which have been observed as part of the radiation emitted by so-called gamma-ray bursts (GRBs): flashes associated with extremely energetic explosions in distant galaxies. Wikipedia refers to them as the ‘brightest’ electromagnetic events know to occur in the Universe. These rays are not to be confused with cosmic rays, which consist of high-energy protons and atomic nuclei stripped of their electron shells. Cosmic rays aren’t rays really and, because they consist of particles with a considerable rest mass, their energy is even higher. The so-called Oh-My-God particle, for example, which is the most energetic particle ever detected, had an energy of 3×10²⁰ eV, i.e. 300 million TeV. But it’s not a photon: its energy is largely kinetic energy, with the rest mass m₀ counting for a lot in the m in the E = m·c² formula. To be precise: the mentioned particle was thought to be an iron nucleus, and it packed the equivalent energy of a baseball traveling at 100 km/h!

But let me refer you to another source for a good discussion on these high-energy particles, so I can get get back to the energy of electromagnetic radiation. When I talked about the Great Desert in that post, I did so using the Planck-Einstein relation (E = h·f), which embodies the idea of the photon being valid always and everywhere and, importantly, at every scale. I also discussed the Great Desert using real-life light being emitted by real-life atomic oscillators. Hence, I may have given the (wrong) impression that the idea of a photon as a ‘wave train’ is inextricably linked with these real-life atomic oscillators, i.e. to electrons going from one energy level to the next in some atom. Let’s explore these assumptions somewhat more.

Let’s start with the second point. Electromagnetic radiation is emitted by any accelerating electric charge, so the atomic oscillator model is an assumption that should not be essential. And it isn’t. For example, whatever is left of the nucleus after alpha or beta decay (i.e. a nuclear decay process resulting in the emission of an α- or β-particle) it likely to be in an excited state, and likely to emit a gamma-ray for about 10⁻¹² seconds, so that’s a burst that’s about 10,000 times shorter than the 10^–8seconds it takes for the energy of a radiating atom to die out. [As for the calculation of that 10^–8sec decay time – so that’s like 10 nanoseconds – I’ve talked about this before but it’s probably better to refer you to the source, i.e. one of Feynman’s Lectures.]

However, what we’re interested in is not the energy of the photon, but the energy of one cycle. In other words, we’re not thinking of the photon as some wave train here, but what we’re thinking about is the energy that’s packed into a space corresponding to one wavelength. What can we say about that?

As you know, that energy will depend both on the amplitude of the electromagnetic wave as well as its frequency. To be precise, the energy is (1) proportional to the square of the amplitude, and (2) proportional to the frequency. Let’s look at the first proportionality relation. It can be written in a number of ways, but one way of doing it is stating the following: if we know the electric field, then the amount of energy that passes per square meter per second through a surface that is normal to the direction in which the radiation is going (which we’ll denote by S – the s from surface – in the formula below), must be proportional to the average of the square of the field. So we write S ∝ 〈E²〉, and so we should think about the constant of proportionality now. Now, let’s not get into the nitty-gritty, and so I’ll just refer to Feynman for the derivation of the formula below:

S = ε₀c·〈E²〉

So the constant of proportionality is ε₀c. [Note that, in light of what we wrote above, we can also write this as S = (1/μ₀·c)·〈(c·B)²〉 = (c/μ₀)·〈B²〉, so that underlines once again that we’re talking one electromagnetic phenomenon only really.] So that’s a nice and rather intuitive result in light of all of the other formulas we’ve been jotting down. However, it is a ‘wave’ perspective. The ‘photon’ perspective assumes that, somehow, the amplitude is given and, therefore, the Planck-Einstein relation only captures the frequency variable: E_γ = h·f.

Indeed, ‘more energy’ in the ‘wave’ perspective basically means ‘more photons’, but photons are photons: they have a definite frequency and a definite energy, and both are given by that Planck-Einstein relation. So let’s look at that relation by doing a bit of dimensional analysis:

Energy is measured in electronvolt or, using SI units, joule: 1 eV ≈ 1.6×10⁻¹⁹J. Energy is force times distance: 1 joule = 1 newton·meter, which means that a larger force over a shorter distance yields the same energy as a smaller force over a longer distance. The oscillations we’re talking about here involve very tiny distances obviously. But the principle is the same: we’re talking some moving charge q, and the power – which is the time rate of change of the energy – that goes in or out at any point of time is equal to dW/dt = F·v, with W the work that’s being done by the charge as it emits radiation.
I would also like to add that, as you know, forces are related to the inertia of things. Newton’s Law basically defines a force as that what causes a mass to accelerate: F = m·a = m·(dv/dt) = d(m·v)/dt = dp/dt, with p the momentum of the object that’s involved. When charges are involved, we’ve got the same thing: a potential difference will cause some current to change, and one of the equivalents of Newton’s Law F = m·a = m·(dv/dt) in electromagnetism is V = L·(dI/dt). [I am just saying this so you get a better ‘feel’ for what’s going on.]
Planck’s constant is measured in electronvolt·seconds (eV·s) or in, using SI units, in joule·seconds (J·s), so its dimension is that of (physical) action, which is energy times time: [energy]·[time]. Again, a lot of energy during a short time yields the same energy as less energy over a longer time. [Again, I am just saying this so you get a better ‘feel’ for these dimensions.]
The frequency f is the number of cycles per time unit, so that’s expressed per second, i.e. in herz (Hz) = 1/second = s⁻¹.

So… Well… It all makes sense: [x joule] = [6.626×10⁻³⁴ joule]·[1 second]×[f cycles]/[1 second]. But let’s try to deepen our understanding even more: what’s the Planck-Einstein relation really about?

To answer that question, let’s think some more about the wave function. As you know, it’s customary to express the frequency as an angular frequency ω, as used in the wave function A(x, t) = A₀·sin(kx − ωt). The angular frequency is the frequency expressed in radians per second. That’s because we need an angle in our wave function, and so we need to relate x and t to some angle. The way to think about this is as follows: one cycle takes a time T (i.e. the period of the wave) which is equal to T = 1/f. Yes: one second divided by the number of cycles per second gives you the time that’s needed for one cycle. One cycle is also equivalent to our argument ωt going around the full circle (i.e. 2π), so we write: ω·T = 2π and, therefore:

ω = 2π/T = 2π·f

Now we’re ready to play with the Planck-Einstein relation. We know it gives us the energy of one photon really, but what if we re-write our equation E_γ = h·f as E_γ/f = h? The dimensions in this equation are:

[x joule]·[1 second]/[f cyles] = [6.626×10⁻³⁴ joule]·[1 second]

⇔ x = 6.626×10⁻³⁴ joule per cycle

So that means that the energy per cycle is equal to 6.626×10⁻³⁴ joule, i.e. the value of Planck’s constant.

Let me rephrase truly amazing result, so you appreciate it—perhaps: regardless of the frequency of the light (or our electromagnetic wave, in general) involved, the energy per cycle, i.e. per wavelength or per period, is always equal to 6.626×10⁻³⁴ joule or, using the electronvolt as the unit, 4.135667662×10⁻¹⁵ eV. So, in case you wondered, that is the true meaning of Planck’s constant!

Now, if we have the frequency f, we also have the wavelength λ, because the velocity of the wave is the frequency times the wavelength: c = λ·f and, therefore, λ = c/f. So if we increase the frequency, the wavelength becomes smaller and smaller, and so we’re packing the same amount of energy – admittedly, 4.135667662×10⁻¹⁵ eV is a very tiny amount of energy – into a space that becomes smaller and smaller. Well… What’s tiny, and what’s small? All is relative, of course. 🙂 So that’s where the Planck scale comes in. If we pack that amount of energy into some tiny little space of the Planck dimension, i.e. a ‘length’ of 1.6162×10⁻³⁵ m, then it becomes a tiny black hole, and it’s hard to think about how that would work.

[…] Let me make a small digression here. I said it’s hard to think about black holes but, of course, it’s not because it’s ‘hard’ that we shouldn’t try it. So let me just mention a few basic facts. For starters, black holes do emit radiation! So they swallow stuff, but they also spit stuff out. More in particular, there is the so-called Hawking radiation, as Roger Penrose and Stephen Hawking discovered.

Let me quickly make a few remarks on that: Hawking radiation is basically a form of blackbody radiation, so all frequencies are there, as shown below: the distribution of the various frequencies depends on the temperature of the black body, i.e. the black hole in this case. [The black curve is the curve that Lord Rayleigh and Sir James Jeans derived in the late 19th century, using classical theory only, so that’s the one that does not correspond to experimental fact, and which led Max Planck to become the ‘reluctant’ father of quantum mechanics. In any case, that’s history and so I shouldn’t dwell on this.]

The interesting thing about blackbody radiation, including Hawking radiation, is that it reduces energy and, hence, the equivalent mass of our blackbody. So Hawking radiation reduces the mass and energy of black holes and is therefore also known as black hole evaporation. So black holes that lose more mass than they gain through other means are expected to shrink and ultimately vanish. Therefore, there’s all kind of theories that say why micro black holes, like that Planck scale black hole we’re thinking of right now, should be much larger net emitters of radiation than large black holes and, hence, whey they should shrink and dissipate faster.

Hmm… Interesting… What do we do with all of this information? Well… Let’s think about it as we continue our trek on this long journey to reality over the next year or, more probably, years (plural). 🙂

The key lesson here is that space and time are intimately related because of the idea of movement, i.e. the idea of something having some velocity, and that it’s not so easy to separate the dimensions of time and distance in any hard and fast way. As energy scales become larger and, therefore, our natural time and distance units become smaller and smaller, it’s the energy concept that comes to the fore. It sort of ‘swallows’ all other dimensions, and it does lead to limiting situations which are hard to imagine. Of course, that just underscores the underlying unity of Nature, and the mysteries involved.

So… To relate all of this back to the story that our professor is trying to tell, it’s a simple story really. He’s talking about two fundamental constants basically, c and h, pointing out that c is a property of empty space, and h is related to something doing something. Well… OK. That’s really nothing new, and surely not ground-breaking research. 🙂

Now, let me finish my thoughts on all of the above by making one more remark. If you’ve read a thing or two about this – which you surely have – you’ll probably say: this is not how people usually explain it. That’s true, they don’t. Anything I’ve seen about this just associates the 10⁴³Hz scale with the 10²⁸eV energy scale, using the same Planck-Einstein relation. For example, the Wikipedia article on micro black holes writes that “the minimum energy of a microscopic black hole is 10¹⁹ GeV [i.e. 10²⁸eV], which would have to be condensed into a region on the order of the Planck length.” So that’s wrong. I want to emphasize this point because I’ve been led astray by it for years. It’s not the total photon energy, but the energy per cycle that counts. Having said that, it is correct, however, and easy to verify, that the 10⁴³Hz scale corresponds to a wavelength of the Planck scale: λ = c/f = (3×10⁸m/s)/(10⁴³ s⁻¹) = 3×10⁻³⁵m. The confusion between the photon energy and the energy per wavelength arises because of the idea of a photon: it travels at the speed of light and, hence, because of the relativistic length contraction effect, it is said to be point-like, to have no dimension whatsoever. So that’s why we think of packing all of its energy in some infinitesimally small place. But you shouldn’t think like that. The photon is dimensionless in our reference frame: in its own ‘world’, it is spread out, so it is a wave train. And it’s in its ‘own world’ that the contradictions start… 🙂

OK. Done!

My third and final point is about what our professor writes on the fundamental physical constants, and more in particular on what he writes on the fine-structure constant. In fact, I could just refer you to my own post on it, but that’s probably a bit too easy for me and a bit difficult for you 🙂 so let me summarize that post and tell you what you need to know about it.

The fine-structure constant

The fine-structure constant α is a dimensionless constant which also illustrates the underlying unity of Nature, but in a way that’s much more fascinating than the two or three things the professor mentions. Indeed, it’s quite incredible how this number (α = 0.00729735…, but you’ll usually see it written as its reciprocal, which is a number that’s close to 137.036…) links charge with the relative speeds, radii, and the mass of fundamental particles and, therefore, how this number also these concepts with each other. And, yes, the fact that it is, effectively, dimensionless, unlike h or c, makes it even more special. Let me quickly sum up what the very same number α all stands for:

(1) α is the square of the electron charge expressed in Planck units: α = e_P².

(2) α is the square root of the ratio of (a) the classical electron radius and (b) the Bohr radius: α = √(r_e /r). You’ll see this more often written as r_e = α²r. Also note that this is an equation that does not depend on the units, in contrast to equation 1 (above), and 4 and 5 (below), which require you to switch to Planck units. It’s the square of a ratio and, hence, the units don’t matter. They fall away.

(3) α is the (relative) speed of an electron: α = v/c. [The relative speed is the speed as measured against the speed of light. Note that the ‘natural’ unit of speed in the Planck system of units is equal to c. Indeed, if you divide one Planck length by one Planck time unit, you get (1.616×10⁻³⁵m)/(5.391×10⁻⁴⁴s) = c m/s. However, this is another equation, just like (2), that does not depend on the units: we can express v and c in whatever unit we want, as long we’re consistent and express both in the same units.]

(4) α is also equal to the product of (a) the electron mass (which I’ll simply write as m_e here) and (b) the classical electron radius r_e (if both are expressed in Planck units): α = m_e·r_e. Now I think that’s, perhaps, the most amazing of all of the expressions for α. [If you don’t think that’s amazing, I’d really suggest you stop trying to study physics. :-)]

Also note that, from (2) and (4), we find that:

(5) The electron mass (in Planck units) is equal m_e = α/r_e= α/α²r = 1/αr. So that gives us an expression, using α once again, for the electron mass as a function of the Bohr radius r expressed in Planck units.

Finally, we can also substitute (1) in (5) to get:

(6) The electron mass (in Planck units) is equal to m_e = α/r_e = e_P²/r_e. Using the Bohr radius, we get m_e = 1/αr = 1/e_P²r.

So… As you can see, this fine-structure constant really links all of the fundamental properties of the electron: its charge, its radius, its distance to the nucleus (i.e. the Bohr radius), its velocity, its mass (and, hence, its energy),…

So… Why is what it is?

Well… We all marvel at this, but what can we say about it, really? I struggle how to interpret this, just as much – or probably much more 🙂 – as the professor who wrote the article I don’t like (because it’s so imprecise, and that’s what made me write all what I am writing here).

Having said that, it’s obvious that it points to a unity beyond these numbers and constants that I am only beginning to appreciate for what it is: deep, mysterious, and very beautiful. But so I don’t think that professor does a good job at showing how deep, mysterious and beautiful it all is. But then that’s up to you, my brother and you, my imaginary reader, to judge, of course. 🙂

[…] I forgot to mention what I mean with ‘Planck units’. Well… Once again, I should refer you to one of my other posts. But, yes, that’s too easy for me and a bit difficult for you. 🙂 So let me just note we get those Planck units by equating not less than five fundamental physical constants to 1, notably (1) the speed of light, (2) Planck’s (reduced) constant, (3) Boltzmann’s constant, (4) Coulomb’s constant and (5) Newton’s constant (i.e. the gravitational constant). Hence, we have a set of five equations here (c = ħ = k_B = k_e = G = 1), and so we can solve that to get the five Planck units, i.e. the Planck length unit, the Planck time unit, the Planck mass unit, the Planck energy unit, the Planck charge unit and, finally (oft forgotten), the Planck temperature unit. Of course, you should note that all mass and energy units are directly related because of the mass-energy equivalence relation E = mc², which simplifies to E = m if c is equated to 1. [I could also say something about the relation between temperature and (kinetic) energy, but I won’t, as it would only further confuse you.]

OK. Done! 🙂

Addendum: How to think about space and time?

If you read the argument on the Planck scale and constant carefully, then you’ll note that it does not depend on the idea of an indivisible photon. However, it does depend on that Planck-Einstein relation being valid always and everywhere. Now, the Planck-Einstein relation is, in its essence, a fairly basic result from classical electromagnetic theory: it incorporates quantum theory – remember: it’s the equation that allowed Planck to solve the black-body radiation problem, and so it’s why they call Planck the (reluctant) ‘Father of Quantum Theory’ – but it’s not quantum theory.

So the obvious question is: can we make this reflection somewhat more general, so we can think of the electromagnetic force as an example only. In other words: can we apply the thoughts above to any force and any movement really?

The truth is: I haven’t advanced enough in my little study to give the equations for the other forces. Of course, we could think of gravity, and I developed some thoughts on how gravity waves might look like, but nothing specific really. And then we have the shorter-range nuclear forces, of course: the strong force, and the weak force. The laws involved are very different. The strong force involves color charges, and the way distances work is entirely different. So it would surely be some different analysis. However, the results should be the same. Let me offer some thoughts though:

We know that the relative strength of the nuclear force is much larger, because it pulls like charges (protons) together, despite the strong electromagnetic force that wants to push them apart! So the mentioned problem of trying to ‘pack’ some oscillation in some tiny little space should be worse with the strong force. And the strong force is there, obviously, at tiny little distances!
Even gravity should become important, because if we’ve got a lot of energy packed into some tiny space, its equivalent mass will ensure the gravitational forces also become important. In fact, that’s what the whole argument was all about!
There’s also all this talk about the fundamental forces becoming one at the Planck scale. I must, again, admit my knowledge is not advanced enough to explain how that would be possible, but I must assume that, if physicists are making such statements, the argument must be fairly robust.

So… Whatever charge or whatever force we are talking about, we’ll be thinking of waves or oscillations—or simply movement, but it’s always a movement in a force field, and so there’s power and energy involved (energy is force times distance, and power is the time rate of change of energy). So, yes, we should expect the same issues in regard to scale. And so that’s what’s captured by h.

As we’re talking the smallest things possible, I should also mention that there are also other inconsistencies in the electromagnetic theory, which should (also) have their parallel for other forces. For example, the idea of a point charge is mathematically inconsistent, as I show in my post on fields and charges. Charge, any charge really, must occupy some space. It cannot all be squeezed into one dimensionless point. So the reasoning behind the Planck time and distance scale is surely valid.

In short, the whole argument about the Planck scale and those limits is very valid. However, does it imply our thinking about the Planck scale is actually relevant? I mean: it’s not because we can imagine how things might look like – they may look like those tiny little black holes, for example – that these things actually exist. GUT or string theorists obviously think they are thinking about something real. But, frankly, Feynman had a point when he said what he said about string theory, shortly before his untimely death in 1988: “I don’t like that they’re not calculating anything. I don’t like that they don’t check their ideas. I don’t like that for anything that disagrees with an experiment, they cook up an explanation—a fix-up to say, ‘Well, it still might be true.'”

It’s true that the so-called Standard Model does not look very nice. It’s not like Maxwell’s equations. It’s complicated. It’s got various ‘sectors’: the electroweak sector, the QCD sector, the Higgs sector,… So ‘it looks like it’s got too much going on’, as a friend of mine said when he looked at a new design for mountainbike suspension. 🙂 But, unlike mountainbike designs, there’s no real alternative for the Standard Model. So perhaps we should just accept it is what it is and, hence, in a way, accept Nature as we can see it. So perhaps we should just continue to focus on what’s here, before we reach the Great Desert, rather than wasting time on trying to figure out how things might look like on the other side, especially because we’ll never be able to test our theories about ‘the other side.’

On the other hand, we can see where the Great Desert sort of starts (somewhere near the 10³² Hz scale), and so it’s only natural to think it should also stop somewhere. In fact, we know where it stops: it stops at the 10⁴³ Hz scale, because everything beyond that doesn’t make sense. The question is: is there actually there? Like fundamental strings or whatever you want to call it. Perhaps we should just stop where the Great Desert begins. And what’s the Great Desert anyway? Perhaps it’s a desert indeed, and so then there is absolutely nothing there. 🙂

Hmm… There’s not all that much one can say about it. However, when looking at the history of physics, there’s one thing that’s really striking. Most of what physicists can think of, in the sense that it made physical sense, turned out to exist. Think of anti-matter, for instance. Paul Dirac thought it might exist, that it made sense to exist, and so everyone started looking for it, and Carl Anderson found in a few years later (in 1932). In fact, it had been observed before, but people just didn’t pay attention, so they didn’t want to see it, in a way. […] OK. I am exaggerating a bit, but you know what I mean. The 1930s are full of examples like that. There was a burst of scientific creativity, as the formalism of quantum physics was being developed, and the experimental confirmations of the theory just followed suit.

In the field of astronomy, or astrophysics I should say, it was the same with black holes. No one could really imagine the existence of black holes until the 1960s or so: they were thought of a mathematical curiosity only, a logical possibility. However, the circumstantial evidence now is quite large and so… Well… It seems a lot of what we can think of actually has some existence somewhere. 🙂

So… Who knows? […] I surely don’t. And so I need to get back to the grind and work my way through the rest of Feynman’s Lectures and the related math. However, this was a nice digression, and so I am grateful to my brother he initiated it. 🙂

Self-inductance, mutual inductance, and the power and energy of inductors

As Feynman puts it, studying physics is not always about the ‘the great and esoteric heights’. In fact, you usually have to come down from then fairly quickly – studying physics is similar to mountain climbing in that regard 🙂 – and study ‘relatively low-level subjects’, such as electrical circuits, which is what we’ll do in this and the next post.

As I’ve introduced some key concepts in a previous post already, let me recapitulate the basics, which include the concept of the electromotive force, which is basically the voltage, i.e. the potential difference, that’s produced in a loop or coil of wire as the magnetic flux changes. I also talked about the impedance in a AC circuit. Finally, we also discussed the power and energies involved. Important results from this previous discussion include (but are not limited to):

A constant speed AC generator will create an alternating current with the emf, i.e. the voltage, varying as V₀·sin(ωt).
If we only have resistors as circuit elements, and the resistance in the circuit adds up to R, then the electric current in the circuit will be equal to I = Ɛ/R = V/R = (V₀/R)·sin(ωt). So that’s Ohm’s Law, basically.
The power that’s produced and consumed in an AC circuit is the product of the voltage and the current, so P = Ɛ·I = V·I. We also showed this electrical power is equal to the mechanical power dW/dt that makes the generator run.
Finally, we explained the concept of impedance (denoted by Z) using Euler’s formula: Z = |Z|eⁱ^θ, mentioning that, if other circuit elements than resistors are involved, such as inductors, then it’s quite likely that the current signal will lag the voltage signal, with the phase factor θ telling us by how much.

It’s now time to introduce those ‘other’ circuit elements. So let’s start with inductors here, and the concept of inductance itself. There’s a lot of stuff to them, and so let’s go over it point by point.

The concept of self-inductance

In its simplest form, an inductor is just a coil, but they come in all sizes and shapes. If you want to see how they might look like, just google some images of micro-electronic inductors, and then, just to see the range of applications, some images of inductors used for large-scale industrial applications. If you do so, you’re likely to see images of transformers too, because transformers work on the principle of mutual inductance, and so they involve two coils, i.e. two inductors.

Contrary to what you might expect, the concept of self-inductance (or inductance tout court) is quite simply: a changing current will cause a changing magnetic field and, hence, some emf. Now, it turns out that the induced emf is proportional to the change in current. So we’ve got another constant of proportionality here, so it’s like how we defined resistance, or capacitance. So, in many ways, the inductance is just another proportionality coefficient. If we denote it by L – the symbol is said to honor the Russian phyicist Heinrich Lenz, whom you know from Lenz’ Law – then we define it as:

L = −Ɛ/(dI/dt)

The dI/dt factor is, obviously, the time rate of change of the current, and the negative sign indicates that the emf opposes the change in current, so it will tend to cause an opposing current. That’s why the emf involved is often referred to as a ‘back emf’. So that’s Lenz’ Law basically. As you might expect, the physicists came up with yet another derived unit, the Henry, to honor yet another physicist, Joseph Henry, an American scientist who was a contemporary of Michael Faraday and independently discovered pretty much the same as Faraday: one henry (H) equals one volt·second per ampere: 1 H = 1 V·s/A.

The concept of mutual inductance

Feynman introduces the topic of inductance with a two-coil set-up, as shown below, noting that a current in coil 1 will induce some emf in coil 2, which he denotes by M₁₂. Conversely, a current in coil 2 will induce some emf in coil 1, which he denoted by M₂₁. M₁₂ and M₂₁ are also constants: they depend on the geometry of the coils, including the length of the solenoid (l), its surface area (S) and the number of loop turns of the coils (N₁ and N₂).

The next step in the analysis is then to acknowledge that each coil should also produce a ‘back emf’ in itself, which we can denote by M₁₁ and M₂₂ respectively, but then these constants are, of course, equal to the self-inductance of the coils so, taking into account the convention in regard to the sign of the self-inductance, we write:

L₁ = − M₁₁ and L₁ = − M₂₂

You will now wonder: what’s the total emf in each coil, taking into account that we do not only have mutual inductance but also self-inductance? Frankly, when I was a kid, and my father tried to tell me one or two things about this, it confused me very much. I could not imagine what happened in one coil, let alone in two coils. I had this vision of a current producing some ‘back-current’, and then the ‘back-current’ producing ‘back-current’ again, and so I could not imagine how one could solve this problem. So the image in my head was very much like that baking powder box which Feynman evokes when talking about the method of images to find the electric fields in situations with an easy geometry, so that’s the picture of a baking powder box which has on its label a picture of a baking powder box which has… Well… Etcetera. Unfortunately, my father didn’t push us to study math and, therefore, I knew that one could solve such problems mathematically – we’re talking a converging series here – but I did not know how, and that’s why I found it all very confusing.

Now I understand there’s one current only, and one potential difference only, and that the formulas do not involve some infinite series of terms. But… Well… I am not ashamed to say these problems are still testing the (limited) agility of my mind. The first thing to ‘get’ is that we’re talking a back emf, and so that’s not a current but a potential difference. In fact, as I explained in my post on the electromotive force, the term ‘force’ in emf is actually misleading, and may lead to that same erroneous vision that I had as a kid: forces generating counter-forces, that generate counter-forces, that generate counter-forces, etcetera. It’s not like that: we have some current – one current – in a coil and we’ll have some voltage – one voltage – across the coil. If the coil would be a resistor instead of a coil, we’d find that the ratio of this voltage and the current would be some constant R = V/I. Now here we’re talking a coil indeed, so that’s a different circuit element, and we find some other ratio, L = −V/(dI/dt) = −Ɛ/(dI/dt). Why the minus sign? Well… As said, the induced emf will be such that it will tend to counter the current, and current flows from positive to negative as per our convention.

But… Yes? So how does it work when we put this coil in some circuit, and how does the resistance of the inductor come into play? Relax. We’ve just been talking ideal circuit elements so far, and we’ve discussed only two: the resistor and the inductor. We’ll talk about voltage sources (or generators) and capacitors too, and then we’ll link all of these ideal circuit elements. In short, we’ll analyze some real-life electrical circuit soon, but first you need to understand the basics. Let me just note that an ideal inductor appears a zero-resistance conductor in a direct current (DC) circuit, so it’s a short-circuit really! Please try to mentally separate out those ‘ideal’ circuit components. Otherwise you’ll never be able to make sense of it all!

In fact, there’s a good reason why Feynman starts with explaining mutual inductance before discussing a little circuit like the one below, which has an inductor and a voltage source. The two-coil situation above is effectively easier to understand, although it may not look like that at first. So let’s analyze that two-coil situation in more detail first. In other words, let me try to understand the situation that I didn’t understand as a kid. 🙂

Because of the law of superposition, we should add fluxes and changes in fluxes and, hence, we should also add the electromotive forces, i.e. the induced voltages. So, what we have here is that the total emf in coil 2 should be written as:

Ɛ₂ = M₂₁·(dI₁/dt) + M₂₂·(dI₂/dt) = M₂₁·(dI₁/dt) – L₂·(dI₂/dt)

What we’re saying here is that the emf, i.e. the voltage across the coil, will indeed depend on the change in current in the other coil, but also on the change in current of the coil itself. Likewise, the total emf in coil 1 should be written as:

Ɛ₁ = M₁₂·(dI₂/dt) + M₁₁·(dI₁/dt) = M₁₂·(dI₂/dt) – L₁·(dI₁/dt)

Of course, this does reduce to the simple L = −Ɛ/(dI/dt) if there’s one coil only. But so you see where it comes from and, while we do not have some infinite series 🙂 we do have a system of two equations here, and so let me say one or two things about it.

The first thing to note is that it is not so difficult to show that M₂₁is equal to M₁₂, so we can simplify and write that M₂₁ = M₁₂= M. Now, while I said ‘not so difficult’, I didn’t mean it’s easy and, because I don’t want this post to become too long, I’ll refer you to Feynman for the proof of this M₂₁ = M₁₂= M equation. It’s a general proof for any two coils or ‘circuits’ of arbitrary shape and it’s really worth the read. However, I have to move on.

The second thing to note is that this coefficient M, which is referred to as the mutual inductance now (so singular instead of plural) depends on the ‘circuit geometry’ indeed. For a simple solenoid, Feynman calculates it as

M = −(1/ε₀c²)·(N₁·N₂)·S/l,

with l the length of the solenoid, S its surface area (S), and N₁ and N₂the respective number of loop turns of the two coils. So, yes, only ‘geometry’ comes into play. [Note that’s quite obvious from the formula because a switch of the subscripts of N₁and N₂makes no difference, of course!]. Now, it’s interesting to note that M is the same for, let’s say, N₁= 100 and N₂ = 10 and for N₁= 20 and N₂ = 50. In fact, because you’re familiar with what transformers do, i.e. transforming voltages, you may think that’s counter-intuitive. It’s not. The thing with the number of coils does not imply that Ɛ₁ and Ɛ₂ remain the same. Our set of equations is Ɛ₁ = M·(dI₂/dt) – L₁·(dI₁/dt) and Ɛ₂= M·(dI₁/dt) – L₂·(dI₂/dt), and so L₁and L₂clearly do vary as N₁and N₂ vary! So… Well… Yes. We’ve got a set of two equations with two independent variables (I₁and I₂) and two dependent variables (Ɛ₁ and Ɛ₂). Of course, we could also phrase the problem the other way around: given two voltages, what are the currents? 🙂

Of course, that makes us think of the power that goes in and out of a transformer. Indeed, you’ll remember that power is voltage times current. So what’s going on here in regard to that?

Well… There’s a thing with transformers, or with two-coil systems like this in general, that is referred to as coupling. The geometry of the situation will determine how much flux from one coil is linked with the flux of the other coil. If most, or all of it, is linked, we say the two coils are ‘tightly coupled’ or, in the limit, that they are fully coupled. There’s a measure for that, and it’s called the coefficient of coupling. Let’s first explore that concept of power once more.

Inductance, energy and electric power

It’s easy to see that we need electric power to get some current going. Now, as we pointed out in our previous post, the power is equal to the voltage times the current. It’s also equal, of course, to the amount of work done per second, i.e. the time rate of change of the energy W, so we write:

dW/dt = Ɛ·I

Now, we defined the self-inductance as L = −Ɛ/(dI/dt) and, therefore, we know that Ɛ = −L·(dI/dt), so we have:

dW/dt = −L·I·(dI/dt)

What is this? A differential equation? Yes and no. We’ve got not one but two functions of time here, W and I, and, while their derivatives with respect to time do appear in the equation, what we need to do is just integrate the two sides over time. We get: W = −(1/2)·L·I². Just check it by taking the time derivative of both sides. Of course, we can add any constant, to both sides in fact, but that’s just a matter of choosing some reference point. We’ll chose our constant to be zero, and also think about the energy that’s stored in the coil, i.e. U, which we define as:

U = −W = −(1/2)·L·I²

Huh? What’s going on here? Well… It’s not an easy discussion, but let’s try to make sense of it. We have some changing current in the coil here but, obviously, some kind of inertia also: the coil itself opposes the change in current through the ‘back emf’. It requires energy, or power, to overcome the inertia. We may think of applying some voltage to offset the ‘back emf’, so we may effectively think of that little circuit with an inductor and a voltage source. The voltage V we’d need to apply to offset the inertia would, obviously, be equal to the ‘back emf’, but with its sign reversed, so we have:

V = − Ɛ = L·(dI/dt)

Now, it helps to think of what a current really is: it’s about electric charges that are moving at some velocity v because some force is being applied to them. As in any system, the power that’s being delivered is the dot product of the force and the velocity vectors (that ensures we only take the tangential component of the force into account), so if we have n moving charges, the power that is being delivered to the circuit is (F·v)·n. What is F? It’s obviously, qE, as the electric field is the force per unit charge, so E = F/q. But so we’re talking some circuit here and we need to think of the power being delivered to some infinitesimal element ds in the coil, and so that’s (F·v)·n·ds, which can be written as: (F·ds)·n·v. And then we integrate over the whole coil to find:

Now, you may or may not remember that the emf (Ɛ) is actually defined as the line integral ∫ E·ds line, taken around the entire coil and, hence, noting that E = F/q, and that the current I is equal to I = q·n·v, we got our power equation. Indeed, the integrand or kernel of our integral becomes F·n·v·ds = q·E·n·v·ds = I·E·ds. Hence, we get our power formula indeed: P = V·I, with V the potential difference, i.e. the voltage across the coil.

I am getting too much into the weeds here. The point is: we’ve got a full and complete analog to the concept of inertia in mechanics here: instead of some force F causing some mass m to change its velocity according to Newton’s Law, i.e. F = m·a = m·(dv/dt), we here have a potential difference V causing some current I to change according to the V = L·(dI/dt) law.

This is very confusing but, remember, the same equations must have the same solutions! So, in an electric circuit, the inductance is really like what the mass is in a mechanics. Now, in mechanics, we’ll say that our mass has some momentum p = m·v, and we’ll also say that its kinetic energy is equal to (1/2)m·v². We can do the same for our circuit: potential energy is continuously being converted into kinetic energy which, for our inductor, we write as U = (1/2)·L·I².

Just think about by playing with one of the many online graphing tools. The graph below, for example, assumes the current builds up to some maximum. As it reaches its maximum, the stored energy will also max out. Now, you should not worry about the units here, or the scale of the graphs. The assumption is that I builds up from 0 to 1, and that L = 1, so that makes U what it is. Using a different constant for L, and/or different units for I, will change the scale of U too, but not its general shape, and that shape gives you the general idea.

The example above obviously assumes some direct current, so it’s a DC circuit: the current builds up, but then stabilizes at some maximum that we can find by applying Ohm’s Law to the resistance of the circuit: I = V/R. Resistance? But we were talking an ideal inductor? We are. If there’s no other resistance in the circuit, we’ll have a short-circuit, so the assumption is that we do have some resistance in the circuit and, therefore, we should also think of some energy loss to heat from the current in the resistance, but that’s not our worry here.

The illustration below is, perhaps, more interesting. Here we are, obviously, applying an alternating current, and so the current goes in one and then in the other direction, so I > 0, and then I < 0, etcetera. We’re assuming some nice sinusoidal curve for the current here (i.e. the blue curve), and so we get what we get for U (i.e. the red curve): the energy goes up and down between zero and some maximum amplitude that’s determined by the maximum current.

So, yes, it is, after all, quite intuitive: building up a current does require energy from some external source, which is used to overcome the ‘back emf’ in the inductor, and that energy is stored in the inductor itself. [If you still wonder why it’s stored in the inductor, think about the other question: where else would it be stored?] How is stored? Look at the graph and think: it’s stored as kinetic energy of the charges, obviously. That explains why the energy is zero when the current is zero, and why the energy maxes out when the current maxes out. So, yes, it all makes sense! 🙂

Let’s now get back to that coupling constant.

The coupling constant

We can apply our reasoning to two coils. Indeed, we know that Ɛ₁ = M·(dI₂/dt) – L₁·(dI₁/dt) and Ɛ₂= M·(dI₁/dt) – L₂·(dI₂/dt). So the power in the two-coils system is dW/dt = Ɛ₁·I₁ + Ɛ₂·I₂, so we have:

dW/dt = M·I₁(dI₂/dt) – L₁·I₁·(dI₁/dt) + M·I₂·(dI₁/dt) – L₂·I₂·(dI₂/dt)

= – L₁·I₁·(dI₁/dt) – L₂·I₂·(dI₂/dt) + M·I₁(dI₂/dt)·I₂·(dI₁/dt)

Integrating both sides, and equating U with −W once more, yields:

U = (1/2)·L₁·I₁²+ (1/2)·L₂·I₂²+ M·I₁·I₂

[Again, you should just take the time derivative to verify this. If you don’t forget to apply the product rule for the M·I₁·I₂ term, you’ll see I am not writing too much nonsense here.] Now, there’s an interesting algebraic transformation of this expression, and an equally interesting explanation why we’d re-write the expression as we do. Let me copy it from Feynman so I’ll be using his fancier L and M symbols now. 🙂

So what? Well… Taking into account that inequality above, we can write the relation between M and the self-inductances L₁ and L₂ using some constant k, which varies between 0 and 1 and which we’ll refer to as the coupling constant:

We refer to k as the coupling constant, for rather obvious reasons: if it’s near zero, the mutual inductance will be very small, and if it’s near one, then the coils are said to be ‘tightly coupled’, and the ‘mutual flux linkage’ is then maximized. As you can imagine, there’s a whole body of literature out there relating this coupling constant to the behavior of transformers or other circuits where mutual inductance plays a role.

The formula for self-inductance

We gave the formula for the mutual inductance of two coils that are arranged as one solenoid on top of the other (cf. the illustration I started with):

M = −(1/ε₀c²)·(N₁·N₂)·S/l

It’s a very easy calculation, so let me quickly copy it from Feynman:

You’ll say: where is the M here? This is a formula for the emf! It is, but M is the constant of proportionality in front, remember? So there you go. 🙂

Now, you would think that getting a formula for the self-inductance L of some solenoid would be equally straightforward. It turns out that that is not the case. Feynman needs two full pages and… Well… By now, you should now how ‘dense’ his writing really is: if it weren’t so dense, you’d be reading Feynman yourself, rather than my ‘explanations’ of him. 🙂 So… Well.. If you want to see how it works, just click on the link here and scroll down to the last two pages of his exposé on self-inductance. I’ll limit myself to just jotting down the formula he does obtain when he’s through the whole argument:

See why he uses a fancier L than ‘my’ L? ‘His’ L is the length of the solenoid. 🙂 And, yes, r is the radius of the coil and n the number of turns per unit length in the winding. Also note this formula is valid only if L >> R, so the effects at the end of the solenoid can be neglected. OK. Done. 🙂

Well… That’s it for today! I am sorry to say but the next post promises to be as boring as this one because… Well… It’s on electric circuits again. 😦

Reconciling the wave-particle duality in electromagnetism

As I talked about Feynman’s equation for electromagnetic radiation in my previous post, I thought I should add a few remarks on wave-particle duality, but then I didn’t do it there, because my post would have become way too long. So let me add those remarks here. In fact, I’ve written about this before, and so I’ll just mention the basic ideas without going too much in detail. Let me first jot down the formula once again, as well as illustrate the geometry of the situation:

The gist of the matter is that light, in classical theory, is a traveling electromagnetic field caused by an accelerating electric charge and that, because light travels at speed c, it’s the acceleration at the retarded time t – r/c, i.e. a‘ = a(t – r/c), that enters the formula. You’ve also seen the diagrams that accompany this formula:

The two diagrams above show that the curve of the electric field in space is a “reversed” plot of the acceleration as a function of time. As I mentioned before, that’s quite obvious from the mathematical behavior of a function with argument like the argument above, i.e. a function F(t – r/c). When we write t – r/c, we basically measure distance units in seconds, instead of in meter. So we basically use c as the scale for both time as well as distance. I explained that in a previous post, so please have a look there if you’d want so see how that works.

So it’s pretty straightforward, really. However, having said that, when I see a diagram like the one above, so all of these diagrams plotting an E or B wave in space, I can’t help thinking it’s somewhat misleading: after all, we’re talking something traveling at the speed of light here and, therefore, its length – in our frame of reference – should be zero. And it is, obviously. Electromagnetic radiation comes packed in point-like, dimensionless photons: the length of something that travels at the speed of light must be zero.

Now, I don’t claim to know what’s going on exactly, but my thinking on it may not be far off the mark. We know that light is emitted and absorbed by atoms, as electrons go from one energy level to another, and the energy of the photons of light corresponds to the difference between those energy levels (i.e. a few electron-volt only, typically: it’s given by the E = h·ν relation). Therefore, we can look at a photon as a transient electromagnetic wave. It’s a very short pulse: the decay time for one such pulse of sodium light, i.e. one photon of sodium light, is 3.2×10^–8seconds. However, taking into account the frequency of sodium light (500 THz), that still makes for some 16 million oscillations, and a wave-train with a length of almost 10 meter. [Yes. Quite incredible, isn’t it?] So the photon could look like the transient wave I depicted below, except… Well… This wavetrain is traveling at the speed of light and, hence, we will not see it as a ten-meter long wave-train. Why not? Well… Because of the relativistic length contraction, it will effectively appear as a point-like particle to us.

So relativistic length contraction is why the wave and particle duality can be easily reconciled in electromagnetism: we can think of light as an irregular beam of point-like photons indeed, as one atomic oscillator after the other releases a photon, in no particularly organized way. So we can think of photons as transient wave-trains, but we should remind ourselves that they are traveling at the speed of light, so they’ll look point-like to us.

Is such view consistent with the results of the famous – of should I say infamous? – double-slit experiment. Well… Maybe. As I mentioned in one of my posts, it is rather remarkable that is actually hard to find actual double-slit experiments that use actual detectors near the slits, and even harder to find such experiments involving photons! Indeed, experiments involving detectors near the slits are usually experiments with ‘real’ particles, such as electrons, for example. Now, a lot of advances have been made in the set-up of these experiments over the past five years, and one of these experiments is a 2010 experiment of an Italian team which suggests that it’s the interaction between the detector and the electron wave that may cause the interference pattern to disappear. Now that throws some doubts on the traditional explanation of the results of the double-slit experiment.

The idea is shown below. The electron is depicted as an incoming plane wave which effectively breaks up as it goes through the slits. The slit on the left has no ‘filter’ (which you may think of as a detector) and, hence, the plane wave goes through as a cylindrical wave. The slit on the right-hand side is covered by a ‘filter’ made of several layers of ‘low atomic number material’, so the electron goes through but, at the same time, the barrier creates a spherical wave as it goes through. The researchers note that “the spherical and cylindrical wave do not have any phase correlation, and so even if an electron passed through both slits, the two different waves that come out cannot create an interference pattern on the wall behind them.” [I hope I don’t have to remind you that, while being represented as ‘real’ waves here, the ‘waves’ are, obviously, complex-valued psi functions.]

In fact, to be precise, the experimenters note that there still was an interference effect if the filter was thin enough. Let me quote the reason for that: “The thicker the filter, the greater the probability for inelastic scattering. When the electron suffers inelastic scattering, it is localized. This means that its wavefunction collapses and, after the measurement act, it propagates roughly as a spherical wave from the region of interaction, with no phase relation at all with other elastically or inelastically scattered electrons. If the filter is made thick enough, the interference effects cancels out almost completely.”

This does not solve the ‘mystery’ of the double-slit experiment, but it throws doubt on how it’s usually being explained. The mystery in such experiments is that, when we put detectors, it is either the detector at A or the detector at B that goes off. They should never go off together—”at half strength, perhaps”, as Feynman puts it. But so there are doubts here now. Perhaps the electron does go through both slits at the same time! And so that’s why I used italics when writing “even if an electron passed through both slits”: the electron, or the photon in a similar set-up, is not supposed to do that according to the traditional explanation of the results of the double-slit experiment! It’s one or the other, and the wavefunction collapses or reduces as it goes through.

However, that’s where these so-called ‘weak measurement’ experiments now come in, like this 2010 experiment: it does not prove but indicates that interaction does not have to be that way. They strongly suggest that it is not all or nothing, that our observations should not necessarily destroy the wavefunction. So, who knows, perhaps we will be able, one day, to show that the wavefunction does go through both slits, as it should (otherwise the interference pattern cannot be explained), and then we will have resolved the paradox.

I am pretty sure that, when that’s done, physicists will also be able to relate the image of a photon as a transient electromagnetic wave (cf. the diagram above), being emitted by an atomic oscillator for a few nanoseconds only (we gave the example for sodium light, for which the decay time was 3.2×10^–8seconds) with the image of a photon as a particle that can be represented by a complex-valued probability amplitude function (cf. the diagram below). I look forward to that day. I think it will come soon.

Here I should add two remarks. First, a lot has been said about the so-called indivisibility of a photon, but inelastic scattering implies that photons are not monolithic: the photon loses energy to the electron and, hence, its wavelength changes. Now, you’ll say: the scattered photon is not the same photon as the incident photon, and you’re right. But… Well. Think about it. It does say something about the presumed oneness of a photon.

The other remark is on the mathematics of interference. Photons are bosons and, therefore, we have to add their amplitudes to get the interference effect. So you may try to think of an amplitude function, like Ψ = (1/√2π)·eⁱ^θor whatever, and think it’s just a matter of ‘splitting’ this function before it enters the two slits and then ‘putting it back together’, so to say, after our photon has gone through the slits. [For the detailed math of interference in quantum mechanics, see my page on essentials.] Well… No. It’s not that simple. The illustration with that plane wave entering the slits, and the cylindrical and/or spherical wave coming out, makes it obvious that something happens to our wave as it goes through the slit. As I said a couple of times already, the two-slit experiment is interesting, but the interference phenomenon – or diffraction as it’s called – involving one slit only is at least as interesting. So… Well… The analysis is not that simple. Not at all, really. 🙂

Transformations: how should we think about them?

Transformations: the formulas

Transformations: generalization

Post scriptum: transformations for spin-1/2 particles

Spin, angular momentum and the magnetic moment

A real example: the disintegration of a muon in a magnetic field

The Larmor frequency

Larmor’s Theorem

Diamagnetism

Atomic magnets and the g-factor

Spin numbers and states

The magnetic energy of atoms

The Stern-Gerlach experiment

Local energy conservation

Einstein’s car

The basic concepts: force, work, energy and potential

Energy density and energy flow in electrodynamics

Poynting’s vector in electrodynamics

The invariant (1−v2)−1/2·d/dt operator and the proper time s

The four-force vector fμ

The force and the tensor

The invariance of physics and the use of vector equations

Electrodynamics in relativistic notation

Four-vectors and invariance under Lorentz transformations

The electromagnetic tensor

The Lorentz transformation of the electric and magnetic fields

The impedance concept

Resistors

Capacitors (condensers)

Inductors

Summary of conclusions

Addendum: Why is V = − Ɛ?

On Maxwell’s equations and the properties of empty space

On quantum oscillations, Planck’s constant, and Planck units

The fine-structure constant

Addendum: How to think about space and time?

**A real example: the disintegration of a muon in a magnetic field**

The invariant (1−v²)^−1/2·d/dt operator and the proper time s

The four-force vector f_μ