The wavefunction as an oscillation of spacetime

Post scriptum note added on 11 July 2016: This is one of the more speculative posts which led to my e-publication analyzing the wavefunction as an energy propagation. With the benefit of hindsight, I would recommend you to immediately the more recent exposé on the matter that is being presented here, which you can find by clicking on the provided link.

Original post:

You probably heard about the experimental confirmation of the existence of gravitational waves by Caltech’s LIGO Lab. Besides further confirming our understanding of the Universe, I also like to think it confirms that the elementary wavefunction represents a propagation mechanism that is common to all forces. However, the fundamental question remains: what is the wavefunction? What are those real and imaginary parts of those complex-valued wavefunctions describing particles and/or photons? [In case you think photons have no wavefunction, see my post on it: it’s fairly straightforward to re-formulate the usual description of an electromagnetic wave (i.e. the description in terms of the electric and magnetic field vectors) in terms of a complex-valued wavefunction. To be precise, in the mentioned post, I showed an electromagnetic wave can be represented as the  sum of two wavefunctions whose components reflect each other through a rotation by 90 degrees.]

So what? Well… I’ve started to think that the wavefunction may not only describe some oscillation in spacetime. I’ve started to think the wavefunction—any wavefunction, really (so I am not talking gravitational waves only)—is nothing but an oscillation of spacetime. What makes them different is the geometry of those wavefunctions, and the coefficient(s) representing their amplitude, which must be related to their relative strength—somehow, although I still have to figure out how exactly.

Huh? Yes. Maxwell, after jotting down his equations for the electric and magnetic field vectors, wrote the following back in 1862: “The velocity of transverse undulations in our hypothetical medium, calculated from the electromagnetic experiments of MM. Kohlrausch and Weber, agrees so exactly with the velocity of light calculated from the optical experiments of M. Fizeau, that we can scarcely avoid the conclusion that light consists in the transverse undulations of the same medium which is the cause of electric and magnetic phenomena.”

We now know there is no medium – no aether – but physicists still haven’t answered the most fundamental question: what is it that is oscillating? No one has gone beyond the abstract math. I dare to say now that it must be spacetime itself. In order to prove this, I’ll have to study Einstein’s general theory of relativity. But this post will already cover some basics.

The quantum of action and natural units

We can re-write the quantum of action in natural units, which we’ll call Planck units for the time being. They may or may not be the Planck units you’ve heard about, so just think of them as being fundamental, or natural—for the time being, that is. You’ll wonder: what’s natural? What’s fundamental? Well… That’s the question we’re trying to explore in this post, so… Well… Just be patient… 🙂 We’ll denote those natural units as FP, lP, and tP, i.e. the Planck force, Planck distance and Planck time unit respectively. Hence, we write:

ħ = FPlP∙tP

Note that FP, lP, and tP are expressed in our old-fashioned SI units here, i.e. in newton (N), meter (m) and seconds (s) respectively. So FP, lP, and tP have a numerical value as well as a dimension, just like ħ. They’re not just numbers. If we’d want to be very explicit, we could write: FP = FP [force], or FP = FP N, and you could do the same for land tP. However, it’s rather tedious to mention those dimensions all the time, so I’ll just assume you understand the symbols we’re using do not represent some dimensionless number. In fact, that’s what distinguishes physical constants from mathematical constants.

Dimensions are also distinguishes physics equations from purely mathematical ones: an equation in physics will always relate some physical quantities and, hence, when you’re dealing with physics equations, you always need to wonder about the dimensions. [Note that the term ‘dimension’ has many definitions… But… Well… I suppose you know what I am talking about here, and I need to move on. So let’s do that.] Let’s re-write that ħ = FPlP∙tP formula as follows: ħ/tP = FPlP.

FPlP is, obviously, a force times a distance, so that’s energy. Please do check the dimensions on the left-hand side as well: [ħ/tP] = [[ħ]/[tP] = (N·m·s)/s = N·m. In short, we can think of EP = FPlP = ħ/tP as being some ‘natural’ unit as well. But what would it correspond to—physically? What is its meaning? We may be tempted to call it the quantum of energy that’s associated with our quantum of action, but… Well… No. While it’s referred to as the Planck energy, it’s actually a rather large unit, and so… Well… No. We should not think of it as the quantum of energy. We have a quantum of action but no quantum of energy. Sorry. Let’s move on.

In the same vein, we can re-write the ħ = FPlP∙tP as ħ/lP = FP∙tP. Same thing with the dimensions—or ‘same-same but different’, as they say in Asia: [ħ/lP] = [FP∙tP] = N·m·s)/m = N·s. Force times time is momentum and, hence, we may now be tempted to think of pP = FP∙tP = ħ/lP as the quantum of momentum that’s associated with ħ, but… Well… No. There’s no such thing as a quantum of momentum. Not now in any case. Maybe later. 🙂 But, for now, we only have a quantum of action. So we’ll just call ħ/lP = FP∙tP the Planck momentum for the time being.

So now we have two ways of looking at the dimension of Planck’s constant:

  1. [Planck’s constant] = N∙m∙s = (N∙m)∙s = [energy]∙[time]
  2. [Planck’s constant] = N∙m∙s = (N∙s)∙m = [momentum]∙[distance]

In case you didn’t get this from what I wrote above: the brackets here, i.e. the [ and ] symbols, mean: ‘the dimension of what’s between the brackets’. OK. So far so good. It may all look like kids stuff – it actually is kids stuff so far – but the idea is quite fundamental: we’re thinking here of some amount of action (h or ħ, to be precise, i.e. the quantum of action) expressing itself in time or, alternatively, expressing itself in spaceIn the former case, some amount of energy is expended during some time. In the latter case, some momentum is expended over some distance.

Of course, ideally, we should try to think of action expressing itself in space and time simultaneously, so we should think of it as expressing itself in spacetime. In fact, that’s what the so-called Principle of Least Action in physics is all about—but I won’t dwell on that here, because… Well… It’s not an easy topic, and the diversion would lead us astray. 🙂 What we will do, however, is apply the idea above to the two de Broglie relations: E = ħω and p = ħk. I assume you know these relations by now. If not, just check one of my many posts on them. Let’s see what we can do with them.

The de Broglie relations

We can re-write the two de Broglie relations as ħ = E/ω and ħ = p/k. We can immediately derive an interesting property here:

ħ/ħ = 1 = (E/ω)/(p/k) ⇔ E/p = ω/k

So the ratio of the energy and the momentum is equal to the wave velocity. What wave velocity? The group of the phase velocity? We’re talking an elementary wave here, so both are the same: we have only one E and p, and, hence, only one ω and k. The E/p = ω/k identity underscores the following point: the de Broglie equations are a pair of equations here, and one of the key things to learn when trying to understand quantum mechanics is to think of them as an inseparable pair—like an inseparable twin really—as the quantum of action incorporates both a spatial as well as a temporal dimension. Just think of what Minkowski wrote back in 1907, shortly after he had re-formulated Einstein’s special relativity theory in terms of four-dimensional spacetime, and just two years before he died—unexpectely—from an appendicitis: “Henceforth space by itself, and time by itself, are doomed to fade away into mere shadows, and only a kind of union of the two will preserve an independent reality.”

So we should try to think of what that union might represent—and that surely includes looking at the de Broglie equations as a pair of matter-wave equations. Likewise, we should also think of the Uncertainty Principle as a pair of equations: ΔpΔx ≥ ħ/2 and ΔEΔt ≥ ħ/2—but I’ll come back to those later.

The ω in the E = ħω equation and the argument (θ = kx – ωt) of the wavefunction is a frequency in time (or temporal frequency). It’s a frequency expressed in radians per second. You get one radian by dividing one cycle by 2π. In other words, we have 2π radians in one cycle. So ω is related the frequency you’re used to, i.e. f—the frequency expressed in cycles per second (i.e. hertz): we multiply f by 2π to get ω. So we can write: E = ħω = ħ∙2π∙f = h∙f, with h = ħ∙2π (or ħ = h/2π).

Likewise, the k in the p = ħk equation and the argument (θ = kx – ωt) of the wavefunction is a frequency in space (or spatial frequency). Unsurprisingly, it’s expressed in radians per meter.

At this point, it’s good to properly define the radian as a unit in quantum mechanics. We often think of a radian as some distance measured along the circumference, because of the way the unit is constructed (see the illustration below) but that’s right and wrong at the same time. In fact, it’s more wrong than right: the radian is an angle that’s defined using the length of the radius of the unit circle but, when everything is said and done, it’s a unit used to measure some anglenot a distance. That should be obvious from the 2π rad = 360 degrees identity. The angle here is the argument of our wavefunction in quantum mechanics, and so that argument combines both time (t) as well as distance (x): θ = kx – ωt = k(x – c∙t). So our angle (the argument of the wavefunction) integrates both dimensions: space as well as time. If you’re not convinced, just do the dimensional analysis of the kx – ωt expression: both the kx and ωt yield a dimensionless number—or… Well… To be precise, I should say: the kx and ωt products both yield an angle expressed in radians. That angle connects the real and imaginary part of the argument of the wavefunction. Hence, it’s a dimensionless number—but that does not imply it is just some meaningless number. It’s not meaningless at all—obviously!

Circle_radians

Let me try to present what I wrote above in yet another way. The θ = kx – ωt = (p/ħ)·x − (E/ħ)·t equation suggests a fungibility: the wavefunction itself also expresses itself in time and/or in space, so to speak—just like the quantum of action. Let me be precise: the p·x factor in the (p/ħ)·x term represents momentum (whose dimension is N·s) being expended over a distance, while the E·t factor in the (E/ħ)·t term represents energy (expressed in N·m) being expended over some time. [As for the minus sign in front of the (E/ħ)·t term, that’s got to do with the fact that the arrow of time points in one direction only while, in space, we can go in either direction: forward or backwards.] Hence, the expression for the argument tells us that both are essentially fungible—which suggests they’re aspects of one and the same thing. So that‘s what Minkowski intuition is all about: spacetime is one, and the wavefunction just connects the physical properties of whatever it is that we are observing – an electron, or a photon, or whatever other elementary particle – to it.

Of course, the corollary to thinking of unified spacetime is thinking of the real and imaginary part of the wavefunction as one—which we’re supposed to do as a complex number is… Well… One complex number. But that’s easier said than actually done, of course. One way of thinking about the connection between the two spacetime ‘dimensions’ – i.e. t and x, with x actually incorporating three spatial dimensions in space in its own right (see how confusing the term ‘dimension’ is?) – and the two ‘dimensions’ of a complex number is going from Cartesian to polar coordinates, and vice versa. You now think of Euler’s formula, of course – if not, you should – but let me insert something more interesting here. 🙂 I took it from Wikipedia. It illustrates how a simple sinusoidal function transforms as we go from Cartesian to polar coordinates.

Cartesian_to_polar

Interesting, isn’t it? Think of the horizontal and vertical axis in the Cartesian space as representing time and… Well… Space indeed. 🙂 The function connects the space and time dimension and might, for example, represent the trajectory of some object in spacetime. Admittedly, it’s a rather weird trajectory, as the object goes back and forth in some box in space, and accelerates and decelerates all of the time, reversing its direction in the process… But… Well… Isn’t that how we think of a an electron moving in some orbital? 🙂 With that in mind, look at how the same movement in spacetime looks like in polar coordinates. It’s also some movement in a box—but both the ‘horizontal’ and ‘vertical’ axis (think of these axes as the real and imaginary part of a complex number) are now delineating our box. So, whereas our box is a one-dimensional box in spacetime only (our object is constrained in space, but time keeps ticking), it’s a two-dimensional box in our ‘complex’ space. Isn’t it just nice to think about stuff this way?

As far as I am concerned, it triggers the same profound thoughts as that E/p = ω/k relation. The  left-hand side is a ratio between energy and momentum. Now, one way to look at energy is that it’s action per time unit. Likewise, momentum is action per distance unit. Of course, ω is expressed as some quantity (expressed in radians, to be precise) per time unit, and k is some quantity (again, expressed in radians) per distance unit. Because this is a physical equation, the dimension of both sides of the equation has to be the same—and, of course, it is the same: the action dimension in the numerator and denominator of the ratio on the left-hand side of the E/p = ω/k equation cancel each other. But… What? Well… Wouldn’t it be nice to think of the dimension of the argument of our wavefunction as being the dimension of action, rather than thinking of it as just some mathematical thing, i.e. an angle. I like to think the argument of our wavefunction is more than just an angle. When everything is said and done, it has to be something physical—if onlyh because the wavefunction describes something physical. But… Well… I need to do some more thinking on this, so I’ll just move on here. Otherwise this post risks becoming a book in its own right. 🙂

Let’s get back to the topic we were discussing here. We were talking about natural units. More in particular, we were wondering: what’s natural? What does it mean?

Back to Planck units

Let’s start with time and distance. We may want to think of lP and tP as the smallest distance and time units possible—so small, in fact, that both distance and time become countable variables at that scale.

Huh? Yes. I am sure you’ve never thought of time and distance as countable variables but I’ll come back to this rather particular interpretation of the Planck length and time unit later. So don’t worry about it now: just make a mental note of it. The thing is: if tP and lP are the smallest time and distance units possible, then the smallest cycle we can possibly imagine will be associated with those two units: we write: ωP = 1/tP and kP = 1/lP. What’s the ‘smallest possible’ cycle? Well… Not sure. You should not think of some oscillation in spacetime as for now. Just think of a cycle. Whatever cycle. So, as for now, the smallest cycle is just the cycle you’d associate with the smallest time and distance units possible—so we cannot write ωP = 2/tP, for example, because that would imply we can imagine a time unit that’s smaller than tP, as we can associate two cycles with tP now.

OK. Next step. We can now define the Planck energy and the Planck momentum using the de Broglie relations:

EP = ħ∙ωP = ħ/tP  and pP = ħ∙kP = ħ/lP

You’ll say that I am just repeating myself here, as I’ve given you those two equations already. Well… Yes and no. At this point, you should raise the following question: why are we using the angular frequency (ωP = 2π·fP) and the reduced Planck constant (ħ = h/2π), rather than fP or h?

That’s a great question. In fact, it begs the question: what’s the physical constant really? We have two mathematical constants – ħ and h – but they must represent the same physical reality. So is one of the two constants more real than the other? The answer is unambiguously: yes! The Planck energy is defined as EP = ħ/tP =(h/2π)/tP, so we cannot write this as EP = h/tP. The difference is that 1/2π factor, and it’s quite fundamental, as it implies we’re actually not associating a full cycle with tP and lP but a radian of that cycle only.

Huh? Yes. It’s a rather strange observation, and I must admit I haven’t quite sorted out what this actually means. The fundamental idea remains the same, however: we have a quantum of action, ħ (not h!), that can express itself as energy over the smallest distance unit possible or, alternatively, that expresses itself as momentum over the smallest time unit possible. In the former case, we write it as EP = FPlP = ħ/tP. In the latter, we write it as pP = FP∙tP = ħ/lP. Both are aspects of the same reality, though, as our particle moves in space as well as in time, i.e. it moves in spacetime. Hence, one step in space, or in time, corresponds to one radian. Well… Sort of… Not sure how to further explain this. I probably shouldn’t try anyway. 🙂

The more fundamental question is: with what speed is is moving? That question brings us to the next point. The objective is to get some specific value for lP and tP, so how do we do that? How can we determine these two values? Well… That’s another great question. 🙂

The first step is to relate the natural time and distance units to the wave velocity. Now, we do not want to complicate the analysis and so we’re not going to throw in some rest mass or potential energy here. No. We’ll be talking a theoretical zero-mass particle. So we’re not going to consider some electron moving in spacetime, or some other elementary particle. No. We’re going to think about some zero-mass particle here, or a photon. [Note that a photon is not just a zero-mass particle. It’s similar but different: in one of my previous posts, I showed a photon packs more energy, as you get two wavefunctions for the price of one, so to speak. However, don’t worry about the difference here.]

Now, you know that the wave velocity for a zero-mass particle and/or a photon is equal to the speed of light. To be precise, the wave velocity of a photon is the speed of light and, hence, the speed of any zero-mass particle must be the same—as per the definition of mass in Einstein’s special relativity theory. So we write: lP/tP = c ⇔ lP = c∙tP and tP = lP/c. In fact, we also get this from dividing EP by pP, because we know that E/p = c, for any photon (and for any zero-mass particle, really). So we know that EP/pP must also equal c. We can easily double-check that by writing: EP/pP = (ħ/tP)/(ħ/lP) = lP/tP = c. Substitution in ħ = FPlP∙tP yields ħ = c∙FP∙tP2 or, alternatively, ħ = FPlP2/c. So we can now write FP as a function of lP and/or tP:

FP = ħ∙c/lP2 = ħ/(c∙tP2)

We can quickly double-check this by dividing FP = ħ∙c/lP2 by FP = ħ/(c∙tP2). We get: 1 = c2∙tP2/lP2 ⇔ lP2/tP2 = c2 ⇔ lP/tP = c.

Nice. However, this does not uniquely define FP, lP, and tP. The problem is that we’ve got only two equations (ħ = FPlP∙tP and lP/tP = c) for three unknowns (FP, lP, and tP). Can we throw in one or both of the de Broglie equations to get some final answer?

I wish that would help, but it doesn’t—because we get the same ħ = FPlP∙tP equation. Indeed, we’re just re-defining the Planck energy (and the Planck momentum) by that EP = ħ/tP (and pP = ħ/lP) equation here, and so that does not give us a value for EP (and pP). So we’re stuck. We need some other formula so we can calculate the third unknown, which is the Planck force unit (FP). What formula could we possibly choose?

Well… We got a relationship by imposing the condition that lP/tP = c, which implies that if we’d measure the velocity of a photon in Planck time and distance units, we’d find that its velocity is one, so c = 1. Can we think of some similar condition involving ħ? The answer is: we can and we can’t. It’s not so simple. Remember we were thinking of the smallest cycle possible? We said it was small because tP and lP were the smallest units we could imagine. But how do we define that? The idea is as follows: the smallest possible cycle will pack the smallest amount of action, i.e. h (or, expressed per radian rather than per cycle, ħ).

Now, we usually think of packing energy, or momentum, instead of packing action, but that’s because… Well… Because we’re not good at thinking the way Minkowski wanted us to think: we’re not good at thinking of some kind of union of space and time. We tend to think of something moving in space, or, alternatively, of something moving in time—rather than something moving in spacetime. In short, we tend to separate dimensions. So that’s why we’d say the smallest possible cycle would pack an amount of energy that’s equal to EP = ħ∙ωP = ħ/tP, or an amount of momentum that’s equal to pP = ħ∙kP = ħ/lP. But both are part and parcel of the same reality, as evidenced by the E = m∙c2 = m∙cc = p∙c equality. [This equation only holds for a zero-mass particle (and a photon), of course. It’s a bit more complicated when we’d throw in some rest mass, but we can do that later. Also note I keep repeating my idea of the smallest cycle, but we’re talking radians of a cycle, really.]

So we have that mass-energy equivalence, which is also a mass-momentum equivalence according to that E = m∙c2 = m∙cc = p∙c formula. And so now the gravitational force comes into play: there’s a limit to the amount of energy we can pack into a tiny space. Or… Well… Perhaps there’s no limit—but if we pack an awful lot of energy into a really tiny speck of space, then we get a black hole.

However, we’re getting a bit ahead of ourselves here, so let’s first try something else. Let’s throw in the Uncertainty Principle.

The Uncertainty Principle

As mentioned above, we can think of some amount of action expressing itself over some time or, alternatively, over some distance. In the former case, some amount of energy is expended over some time. In the latter case, some momentum is expended over some distance. That’s why the energy and time variables, and the momentum and distance variables, are referred to as complementary. It’s hard to think of both things happening simultaneously (whatever that means in spacetime), but we should try! Let’s now look at the Uncertainty relations once again (I am writing uncertainty with a capital U out of respect—as it’s very fundamental, indeed!):

ΔpΔx ≥ ħ/2 and ΔEΔt ≥ ħ/2.

Note that the ħ/2 factor on the right-hand side quantifies the uncertainty, while the right-hand side of the two equations (ΔpΔx and ΔEΔt) are just an expression of that fundamental uncertainty. In other words, we have two equations (a pair), but there’s only one fundamental uncertainty, and it’s an uncertainty about a movement in spacetime. Hence, that uncertainty expresses itself in both time as well as in space.

Note the use of ħ rather than h, and the fact that the  1/2 factor makes it look like we’re splitting ħ over ΔpΔx and ΔEΔt respectively—which is actually a quite sensible explanation of what this pair of equations actually represent. Indeed, we can add both relations to get the following sum:

ΔpΔx + ΔEΔt ≥ ħ/2 + ħ/2 = ħ

Interesting, isn’t it? It explains that 1/2 factor which troubled us when playing with the de Broglie relations.

Let’s now think about our natural units again—about lP, and tP in particular. As mentioned above, we’ll want to think of them as the smallest distance and time units possible: so small, in fact, that both distance and time become countable variables, so we count x and t as 0, 1, 2, 3 etcetera. We may then imagine that the uncertainty in x and t is of the order of one unit only, so we write Δx = lP and Δt = tP. So we can now re-write the uncertainty relations as:

  • Δp·lP = ħ/2
  • ΔE·tP = ħ/2

Hey! Wait a minute! Do we have a solution for the value of lP and tP here? What if we equate the natural energy and momentum units to ΔE and Δp here? Well… Let’s try it. First note that we may think of the uncertainty in t, or in x, as being equal to plus or minus one unit, i.e. ±1. So the uncertainty is two units really. [Frankly, I just want to get rid of that 1/2 factor here.] Hence, we can re-write the ΔpΔx = ΔEΔt = ħ/2 equations as:

  • ΔpΔx = pPlP = FP∙tPlP = ħ
  • ΔEΔt = EP∙tP = FPlP∙tP = ħ

Hmm… What can we do with this? Nothing much, unfortunately. We’ve got the same problem: we need a value for FP (or for pP, or for EP) to get some specific value for lP and tP, so we’re stuck once again. We have three variables and two equations only, so we have no specific value for either of them. 😦

What to do? Well… I will give you the answer now—the answer you’ve been waiting for, really—but not the technicalities of it. There’s a thing called the Schwarzschild radius, aka as the gravitational radius. Let’s analyze it.

The Schwarzschild radius and the Planck length

The Schwarzschild radius is just the radius of a black hole. Its formal definition is the following: it is the radius of a sphere such that, if all the mass of an object were to be compressed within that sphere, the escape velocity from the surface of the sphere would equal the speed of light (c). The formula for the Schwartzschild radius is the following:

RS = 2m·G/c2

G is the gravitational constant here: G ≈ 6.674×10−11 N⋅m2/kg2. [Note that Newton’s F = m·Law tells us that 1 kg = 1 N·s2/m, as we’ll need to substitute units later.]

But what is the mass (m) in that RS = 2m·G/c2 equation? Using equivalent time and distance units (so = 1), we wrote the following for a zero-mass particle and for a photon respectively:

  • E = m = p = ħ/2 (zero-mass particle)
  • E = m = p = ħ (photon)

How can a zero-mass particle, or a photon, have some mass? Well… Because it moves at the speed of light. I’ve talked about that before, so I’ll just refer you to my post on that. Of course, the dimension of the right-hand side of these equations (i.e. ħ/2 or ħ) symbol has to be the same as the dimension on the left-hand side, so the ‘ħ’ in the E = ħ equation (or E = ħ/2 equation) is a different ‘ħ’ in the p = ħ equation (or p = ħ/2 equation). So we must be careful here. Let’s write it all out, so as to remind ourselves of the dimensions involved:

  • E [N·m] = ħ [N·m·s/s] = EP = FPlP∙tP/tP
  • p [N·s] = ħ [N·m·s/m] = pP = FPlP∙tP/lP

Now, let’s check this by cheating. I’ll just give you the numerical values—even if we’re not supposed to know them at this point—so you can see I am not writing nonsense here:

  • EP = 1.0545718×10−34 N·m·s/(5.391×10−44 s) = (1.21×1044 N)·(1.6162×10−35 m) = 1.9561×10N·m
  • pP =1.0545718×10−34 N·m·s/(1.6162×10−35 m) = (1.21×1044 N)·(5.391×10−44 s) = 6.52485 N·s

You can google the Planck units, and you’ll see I am not taking you for a ride here. 🙂

The associated Planck mass is mP = EP/c2 = 1.9561×10N·m/(2.998×10m/s)2 = 2.17651×108 N·s2/m = 2.17651×108 kg. So let’s plug that value into RS = 2m·G/cequation. We get:

RS = 2m·G/c= [(2.17651×108 kg)·(6.674×10−11 N⋅m2/kg)/(8.988×1016 m2·s−2)

= 1.6162×1035 kg·N⋅m2·kg−2·m2·s−2 = 1.6162×1035 kg·N⋅m2·kg−2·m2·s−2 = 1.6162×1035 m = lP

Bingo! You can look it up: 1.6162×1035 m is the Planck length indeed, so the Schwarzschild radius is the Planck length. We can now easily calculate the other Planck units:

  • t= lP/c = 1.6162×1035 m/(2.998×10m/s) = 5.391×10−44 s
  • F= ħ/(tPlP)= (1.0545718×10−34 N·m·s)/[(1.6162×1035 m)·(5.391×10−44 s) = 1.21×10−44 N

Bingo again! 🙂

[…] But… Well… Look at this: we’ve been cheating all the way. First, we just gave you that formula for the Schwarzschild radius. It looks like an easy formula but its derivation involves a profound knowledge of general relativity theory. So we’d need to learn about tensors and what have you. The formula is, in effect, a solution to what is known as Einstein’s field equations, and that’s pretty complicated stuff.

However, my crime is much worse than that: I also gave you those numerical values for the Planck energy and momentum, rather than calculating them. I just couldn’t calculate them with the knowledge we have so far. When everything is said and done, we have more than three unknowns. We’ve got five in total, including the Planck charge (qP) and, hence, we need five equations. Again, I’ll just copy them from Wikipedia, because… Well… What we’re discussing here is way beyond the undergraduate physics stuff that we’ve been presenting so far. The equations are the following. Just have a look at them and move on. 🙂

Planck units

Finally, I should note one more thing: I did not use 2m but m in Schwarzschild’s formula. Why? Well… I have no good answer to that. I did it to ensure I got the result we wanted to get. It’s that 1/2 factor again. In fact, the E = m = p = ħ/2 is the correct formula to use, and all would come out alright if we did that and defined the magnitude of the uncertainty as one unit only, but so we used the E = m = p = ħ formula instead, i.e. the equation that’s associated with a photon. You can re-do the calculations as an exercise: you’ll see it comes out alright.

Just to make things somewhat more real, let me note that the Planck energy is very substantial: 1.9561×10N·m ≈ 2×10J is equivalent to the energy that you’d get out of burning 60 liters of gasoline—or the mileage you’d get out of 16 gallons of fuel! In short, it’s huge,  and so we’re packing that into a unimaginably small space. To understand how that works, you can think of the E = h∙f ⇔ h = E/f relation once more. The h = E/f ratio implies that energy and frequency are directly proportional to each other, with h the coefficient of proportionality. Shortening the wavelength, amounts to increasing the frequency and, hence, the energy. So, as you think of our cycle becoming smaller and smaller, until it becomes the smallest cycle possible, you should think of the frequency becoming unimaginably large. Indeed, as I explained in one of my other posts on physical constants, we’re talking the the 1043 Hz scale here. However, we can always choose our time unit such that we measure the frequency as one cycle per time unit. Because the energy per cycle remains the same, it means the quantum of action (ħ = FPlP∙tP) expresses itself over extremely short time spans, which means the EP = FPlP product becomes huge, as we’ve shown above. The rest of the story is the same: gravity comes into play, and so our little blob in spacetime becomes a tiny black hole. Again, we should think of both space and time: they are joined in ‘some kind of union’ here, indeed, as they’re intimately connected through the wavefunction, which travels at the speed of light.

The wavefunction as an oscillation in and of spacetime

OK. Now I am going to present the big idea I started with. Let me first ask you a question: when thinking about the Planck-Einstein relation (I am talking about the E = ħ∙ω relation for a photon here, rather than the equivalent de Broglie equation for a matter-particle), aren’t you struck by the fact that the energy of a photon depends on the frequency of the electromagnetic wave only? I mean… It does not depend on its amplitude. The amplitude is mentioned nowhere. The amplitude is fixed, somehow—or considered to be fixed.

Isn’t that strange? I mean… For any other physical wave, the energy would not only depend on the frequency but also on the amplitude of the wave. For a photon, however, it’s just the frequency that counts. Light of the same frequency but higher intensity (read: more energy) is not a photon with higher amplitude, but just more photons. So it’s the photons that add up somehow, and so that explains the amplitude of the electric and magnetic field vectors (i.e. E and B) and, hence, the intensity of the light. However, every photon considered separately has the same amplitude apparently. We can only increase its energy by increasing the frequency. In short, ω is the only variable here.

Let’s look at that angular frequency once more. As you know, it’s expressed in radians per second but, if you multiply ω by 2π, you get the frequency you’re probably more acquainted with: f = 2πω = f cycles per second. The Planck-Einstein relation is then written as E = h∙f. That’s easy enough. But what if we’d change the time unit here? For example, what if our time unit becomes the time that’s needed for a photon to travel one meter? Let’s examine it.

Let’s denote that time unit by tm, so we write: 1 tm = 1/c s ⇔ tm1 = c s1, with c ≈ 3×108. The frequency, as measured using our new time unit, changes, obviously: we have to divide its former value by c now. So, using our little subscript once more, we could write: fm = f/c. [Why? Just add the dimension to make things more explicit: f s1 = f/c tm1 = f/c tm1.] But the energy of the photon should not depend on our time unit, should it?

Don’t worry. It doesn’t: the numerical value of Planck’s constant (h) would also change, as we’d replace the second in its dimension (N∙m∙s) by c times our new time unit tm. However, Planck’s constant remains what it is: some physical constant. It does not depend on our measurement units: we can use the SI units, or the Planck units (FP, lP, and tP), or whatever unit you can think of. It doesn’t matter: h (or ħ = h/2π) is what is—it’s the quantum of action, and so that’s a physical constant (as opposed to a mathematical constant) that’s associated with one cycle.

Now, I said we do not associate the wavefunction of a photon with an amplitude, but we do associate it with a wavelength. We do so using the standard formula for the velocity of a wave: c = f∙λ ⇔ λ = c/f. We can also write this using the angular frequency and the wavenumber: c = ω/k, with k = 2π/λ. We can double-check this, because we know that, for a photon, the following relation holds: E/p = c. Hence, using the E = ħ∙ω and p = ħ∙k relations, we get: (ħ∙ω)/(ħ∙k) = ω/k = c. So we have options here: h can express itself over a really long wavelength, or it can do so over an extremely short wavelength. We re-write p = ħ∙k as p = E/c = ħ∙2π/λ = h/λ ⇔ E = h∙c/λ ⇔ h∙c = E∙λ. We know this relationship: the energy and the wavelength of a photon (or an electromagnetic wave) are inversely proportional to each other.

Once again, we may want to think of the shortest wavelength possible. As λ gets a zillion times smaller, E gets a zillion times bigger. Is there a limit? There is. As I mentioned above, the gravitational force comes into play here: there’s a limit to the amount of energy we can pack into a tiny space. If we pack an awful lot of energy into a really tiny speck of space, then we get a black hole. In practical terms, that implies our photon can’t travel, as it can’t escape from the black hole it creates. That’s what that calculation of the Schwarzschild radius was all about.

We can—in fact, we should—now apply the same reasoning to the matter-wave. Instead of a photon, we should try to think of a zero-mass matter-particle. You’ll say: that’s a contradiction. Matter-particles – as opposed to force-carrying particles, like photons (or bosons in general) – must have some rest mass, so they can’t be massless. Well… Yes. You’re right. But we can throw the rest mass in later. I first want to focus on the abstract principles, i.e. the propagation mechanism of the matter-wave.

Using natural units, we know our particle will move in spacetime with velocity Δx/Δt = 1/1 = 1. Of course, it has to have some energy to move, or some momentum. We also showed that, if it’s massless, and the elementary wavefunction is ei[(p/ħ)x – (E/ħ)t), then we know the energy, and the momentum, has to be equal to ħ/2. Where does it get that energy, or momentum? Not sure. I like to think it borrows it from spacetime, as it breaks some potential barrier between those two points, and then it gives it back. Or, if it’s at point x = t = 0, then perhaps it gets it from some other massless particle moving from x = t = −1. In both cases, we’d like to think our particle keeps moving. So if the first description (borrowing) is correct, it needs to keep borrowing and returning energy in some kind of interaction with spacetime itself. If it’s the second description, it’s more like spacetime bumping itself forward.

In both cases, however, we’re actually trying to visualize (or should I say: imagine?) some oscillation of spacetime itself, as opposed to an oscillation in spacetime.

Huh? Yes. The idea is the following here: we like to think of the wavefunction as the dependent variable: both its real as well as its imaginary part are a function of x and t, indeed. But what if we’d think of x and t as dependent variables? In that case, the real and imaginary part of the wavefunction would be the independent variables. It’s just a matter of perspective. We sort of mirror our function: we switch its domain for its range, and its range for its domain, as shown below. It all makes sense, doesn’t it? Space and time appear as separate dimensions to us, but they’re intimately connected through c, ħ and the other fundamental physical constants. Likewise, the real and imaginary part of the wavefunction appear as separate dimensions, but they’re intimately connected through π and Euler’s number, i.e. through mathematical constants. That cannot be a coincidence: the mathematical and physical ‘space’ reflect each other through the wavefunction, just like the domain and range of a function reflect each other through that function. So physics and math must meet in some kind of union—at least in our mind, they do!

dependent independent

So, yes, we can—and probably should—be looking at the wavefunction as an oscillation of spacetime, rather than as an oscillation in spacetime only. As mentioned in my introduction, I’ll need to study general relativity theory—and very much in depth—to convincingly prove that point, but I am sure it can be done.

You’ll probably think I am arrogant when saying that—and I probably am—but then I am very much emboldened by the fact some nuclear scientist told me a photon doesn’t have any wavefunction: it’s just those E and B vectors, he told me—and then I found out he was dead wrong, as I showed in my previous post! So I’d rather think more independently now. I’ll keep you guys posted on progress—but it will probably take a while to figure it all out. In the meanwhile, please do let me know your ideas on this. 🙂

Let me wrap up this little excursion with two small notes:

  • We have this E/c = p relation. The mass-energy equivalence relation implies momentum must also have an equivalent mass. If E = m∙c2, then p = m∙c ⇔ m = p/c. It’s obvious, but I just thought it would be useful to highlight this.
  • When we studied the ammonia molecule as a two-state system, our space was not a continuum: we allowed just two positions—two points in space, which we defined with respect to the system. So x was a discrete variable. We assumed time to be continuous, however, and so we got those nice sinusoids as a solution to our set of Hamiltonian equations. However, if we look at space as being discrete, or countable, we should probably think of time as being countable as well. So we should, perhaps, think of a particle being at point x = t = 0 first, and, then, being at point x = t = 1. Instead of the nice sinusoids, we get some boxcar function, as illustrated below, but probably varying between 0 and 1—or whatever other normalized values. You get the idea, I hope. 🙂

boxcar1

Post Scriptum on the Principle of Least Action: As noted above, the Principle of Least Action is not very intuitive, even if Feynman’s exposé of it is not as impregnable as it may look at first. To put it simply, the Principle of Least Action says that the average kinetic energy less the average potential energy is as little as possible for the path of an object going from one point to another. So we have a path or line integral here. In a gravitation field, that integral is the following:

Least action

The integral is not all that important. Just note its dimension is the dimension of action indeed, as we multiply energy (the integrand) with time (dt). We can use the Principle of Least Action to re-state Newton’s Law, or whatever other classical law. Among other things, we’ll find that, in the absence of any potential, the trajectory of a particle will just be some straight line.

In quantum mechanics, however, we have uncertainty, as expressed in the ΔpΔx ≥ ħ/2 and ΔEΔt ≥ ħ/2 relations. Now, that uncertainty may express itself in time, or in distance, or in both. That’s where things become tricky. 🙂 I’ve written on this before, but let me copy Feynman himself here, with a more exact explanation of what’s happening (just click on the text to enlarge):

Feynman

The ‘student’ he speaks of above, is himself, of course. 🙂

Too complicated? Well… Never mind. I’ll come back to it later. 🙂

All what you ever wanted to know about the photon wavefunction…

Post scriptum note added on 11 July 2016: This is one of the more speculative posts which led to my e-publication analyzing the wavefunction as an energy propagation. With the benefit of hindsight, I would recommend you to immediately read the more recent exposé on the matter that is being presented here, which you can find by clicking on the provided link.

Original post:

This post is, essentially, a continuation of my previous post, in which I juxtaposed the following images:

Animation 5d_euler_f

Both are the same, and then they’re not. The illustration on the right-hand side is a regular quantum-mechanical wavefunction, i.e. an amplitude wavefunction. You’ve seen that one before. In this case, the x-axis represents time, so we’re looking at the wavefunction at some particular point in space. ]You know we can just switch the dimensions and it would all look the same.] The illustration on the left-hand side looks similar, but it’s not an amplitude wavefunction. The animation shows how the electric field vector (E) of an electromagnetic wave travels through space. Its shape is the same. So it’s the same function. Is it also the same reality?

Yes and no. And I would say: more no than yes—in this case, at least. Note that the animation does not show the accompanying magnetic field vector (B). That vector is equally essential in the electromagnetic propagation mechanism according to Maxwell’s equations, which—let me remind you—are equal to:

  1. B/∂t = –∇×E
  2. E/∂t = ∇×B

In fact, I should write the second equation as ∂E/∂t = c2∇×B, but then I assume we measure time and distance in equivalent units, so c = 1.

You know that E and B are two aspects of one and the same thing: if we have one, then we have the other. To be precise, B is always orthogonal to in the direction that’s given by the right-hand rule for the following vector cross-product: B = ex×E, with ex the unit vector pointing in the x-direction (i.e. the direction of propagation). The reality behind is illustrated below for a linearly polarized electromagnetic wave.

E and b

The B = ex×E equation is equivalent to writing B= i·E, which is equivalent to:

B = i·E = ei(π/2)·ei(kx − ωt) = cos(kx − ωt + π/2) + i·sin(kx − ωt + π/2)

= −sin((kx − ωt) + i·cos(kx − ωt)

Now, E and B have only two components: Eand Ez, and Band Bz. That’s only because we’re looking at some ideal or elementary electromagnetic wave here but… Well… Let’s just go along with it. 🙂 It is then easy to prove that the equation above amounts to writing:

  1. B= cos(kx − ωt + π/2) = −sin(kx − ωt) = −Ez
  2. B= sin(kx − ωt + π/2) = cos(kx − ωt) = Ey

We should now think of Ey and Eas the real and imaginary part of some wavefunction, which we’ll denote as ψE = ei(kx − ωt). So we write:

E = (Ey, Ez) = Ey + i·E= cos(kx − ωt) + i∙sin(kx − ωt) = ReE) + i·ImE) = ψE = ei(kx − ωt)

What about B? We just do the same, so we write:

B = (By, Bz) = By + i·B= ψB = i·E = i·ψE = −sin(kx − ωt) + i∙sin(kx − ωt) = − ImE) + i·ReE)

Now we need to prove that ψE and ψB are regular wavefunctions, which amounts to proving Schrödinger’s equation, i.e. ∂ψ/∂t = i·(ħ/m)·∇2ψ, for both ψE and ψB. [Note I use the Schrödinger’s equation for a zero-mass spin-zero particle here, which uses the ħ/m factor rather than the ħ/(2m) factor.] To prove that ψE and ψB are regular wavefunctions, we should prove that:

  1. Re(∂ψE/∂t) =  −(ħ/m)·Im(∇2ψE) and Im(∂ψE/∂t) = (ħ/m)·Re(∇2ψE), and
  2. Re(∂ψB/∂t) =  −(ħ/m)·Im(∇2ψB) and Im(∂ψB/∂t) = (ħ/m)·Re(∇2ψB).

Let’s do the calculations for the second pair of equations. The time derivative on the left-hand side is equal to:

∂ψB/∂t = −iω·iei(kx − ωt) = ω·[cos(kx − ωt) + i·sin(kx − ωt)] = ω·cos(kx − ωt) + iω·sin(kx − ωt)

The second-order derivative on the right-hand side is equal to:

2ψ= ∂2ψB/∂x= i·k2·ei(kx − ωt) = k2·cos(kx − ωt) + i·k2·sin(kx − ωt)

So the two equations for ψare equivalent to writing:

  1. Re(∂ψB/∂t) =   −(ħ/m)·Im(∇2ψB) ⇔ ω·cos(kx − ωt) = k2·(ħ/m)·cos(kx − ωt)
  2. Im(∂ψB/∂t) = (ħ/m)·Re(∇2ψB) ⇔ ω·sin(kx − ωt) = k2·(ħ/m)·sin(kx − ωt)

So we see that both conditions are fulfilled if, and only if, ω = k2·(ħ/m).

Now, we also demonstrated in that post of mine that Maxwell’s equations imply the following:

  1. ∂By/∂t = –(∇×E)y = ∂Ez/∂x = ∂[sin(kx − ωt)]/∂x = k·cos(kx − ωt) = k·Ey
  2. ∂Bz/∂t = –(∇×E)z = – ∂Ey/∂x = – ∂[cos(kx − ωt)]/∂x = k·sin(kx − ωt) = k·Ez

Hence, using those B= −Eand B= Eequations above, we can also calculate these derivatives as:

  1. ∂By/∂t = −∂Ez/∂t = −∂sin(kx − ωt)/∂t = ω·cos(kx − ωt) = ω·Ey
  2. ∂Bz/∂t = ∂Ey/∂t = ∂cos(kx − ωt)/∂t = −ω·[−sin(kx − ωt)] = ω·Ez

In other words, Maxwell’s equations imply that ω = k, which is consistent with us measuring time and distance in equivalent units, so the phase velocity is  = 1 = ω/k.

So far, so good. We basically established that the propagation mechanism for an electromagnetic wave, as described by Maxwell’s equations, is fully coherent with the propagation mechanism—if we can call it like that—as described by Schrödinger’s equation. We also established the following equalities:

  1. ω = k
  2. ω = k2·(ħ/m)

The second of the two de Broglie equations tells us that k = p/ħ, so we can combine these two equations and re-write these two conditions as:

ω/k = 1 = k·(ħ/m) = (p/ħ)·(ħ/m) = p/m ⇔ p = m

What does this imply? The p here is the momentum: p = m·v, so this condition implies must be equal to 1 too, so the wave velocity is equal to the speed of light. Makes sense, because we actually are talking light here. 🙂 In addition, because it’s light, we also know E/p = = 1, so we have – once again – the general E = p = m equation, which we’ll need!

OK. Next. Let’s write the Schrödinger wave equation for both wavefunctions:

  1. ∂ψE/∂t = i·(ħ/mE)·∇2ψE, and
  2. ∂ψB/∂t = i·(ħ/mB)·∇2ψB.

Huh? What’s mE and mE? We should only associate one mass concept with our electromagnetic wave, shouldn’t we? Perhaps. I just want to be on the safe side now. Of course, if we distinguish mE and mB, we should probably also distinguish pE and pB, and EE and EB as well, right? Well… Yes. If we accept this line of reasoning, then the mass factor in Schrödinger’s equations is pretty much like the 1/c2 = μ0ε0 factor in Maxwell’s (1/c2)·∂E/∂t = ∇×B equation: the mass factor appears as a property of the medium, i.e. the vacuum here! [Just check my post on physical constants in case you wonder what I am trying to say here, in which I explain why and how defines the (properties of the) vacuum.]

To be consistent, we should also distinguish pE and pB, and EE and EB, and so we should write ψand ψB as:

  1. ψE = ei(kEx − ωEt), and
  2. ψB = ei(kBx − ωBt).

Huh? Yes. I know what you think: we’re talking one photon—or one electromagnetic wave—so there can be only one energy, one momentum and, hence, only one k, and one ω. Well… Yes and no. Of course, the following identities should hold: kE = kB and, likewise, ω= ωB. So… Yes. They’re the same: one k and one ω. But then… Well… Conceptually, the two k’s and ω’s are different. So we write:

  1. pE = EE = mE, and
  2. pB = EB = mB.

The obvious question is: can we just add them up to find the total energy and momentum of our photon? The answer is obviously positive: E = EE + EB, p = pE + pB and m = mE + mB.

Let’s check a few things now. How does it work for the phase and group velocity of ψand ψB? Simple:

  1. vg = ∂ωE/∂kE = ∂[EE/ħ]/∂[pE/ħ] = ∂EE/∂pE = ∂pE/∂pE = 1
  2. vp = ωE/kE = (EE/ħ)/(pE/ħ) = EE/pE = pE/pE = 1

So we’re fine, and you can check the result for ψby substituting the subscript E for B. To sum it all up, what we’ve got here is the following:

  1. We can think of a photon having some energy that’s equal to E = p = m (assuming c = 1), but that energy would be split up in an electric and a magnetic wavefunction respectively: ψand ψB.
  2. Schrödinger’s equation applies to both wavefunctions, but the E, p and m in those two wavefunctions are the same and not the same: their numerical value is the same (pE =EE = mE = pB =EB = mB), but they’re conceptually different. They must be: if not, we’d get a phase and group velocity for the wave that doesn’t make sense.

Of course, the phase and group velocity for the sum of the ψand ψwaves must also be equal to c. This is obviously the case, because we’re adding waves with the same phase and group velocity c, so there’s no issue with the dispersion relation.

So let’s insert those pE =EE = mE = pB =EB = mB values in the two wavefunctions. For ψE, we get:

ψ= ei[kEx − ωEt) ei[(pE/ħ)·x − (EE/ħ)·t] 

You can do the calculation for ψyourself. Let’s simplify our life a little bit and assume we’re using Planck units, so ħ = 1, and so the wavefunction simplifies to ψei·(pE·x − EE·t). We can now add the components of E and B using the summation formulas for sines and cosines:

1. B+ Ey = cos(pB·x − EB·t + π/2) + cos(pE·x − EE·t) = 2·cos[(p·x − E·t + π/2)/2]·cos(π/4) = √2·cos(p·x/2 − E·t/2 + π/4)

2. B+ Ez = sin(pB·x − EB·t+π/2) + sin(pE·x − EE·t) = 2·sin[(p·x − E·t + π/2)/2]·cos(π/4) = √2·sin(p·x/2 − E·t/2 + π/4)

Interesting! We find a composite wavefunction for our photon which we can write as:

E + B = ψ+ ψ= E + i·E = √2·ei(p·x/2 − E·t/2 + π/4) = √2·ei(π/4)·ei(p·x/2 − E·t/2) = √2·ei(π/4)·E

What a great result! It’s easy to double-check, because we can see the E + i·E = √2·ei(π/4)·formula implies that 1 + should equal √2·ei(π/4). Now that’s easy to prove, both geometrically (just do a drawing) or formally: √2·ei(π/4) = √2·cos(π/4) + i·sin(π/4ei(π/4) = (√2/√2) + i·(√2/√2) = 1 + i. We’re bang on! 🙂

We can double-check once more, because we should get the same from adding E and B = i·E, right? Let’s try:

E + B = E + i·E = cos(pE·x − EE·t) + i·sin(pE·x − EE·t) + i·cos(pE·x − EE·t) − sin(pE·x − EE·t)

= [cos(pE·x − EE·t) – sin(pE·x − EE·t)] + i·[sin(pE·x − EE·t) – cos(pE·x − EE·t)]

Indeed, we can see we’re going to obtain the same result, because the −sinθ in the real part of our composite wavefunction is equal to cos(θ+π/2), and the −cosθ in its imaginary part is equal to sin(θ+π/2). So the sum above is the same sum of cosines and sines that we did already.

So our electromagnetic wavefunction, i.e. the wavefunction for the photon, is equal to:

ψ = ψ+ ψ= √2·ei(p·x/2 − E·t/2 + π/4) = √2·ei(π/4)·ei(p·x/2 − E·t/2) 

What about the √2 factor in front, and the π/4 term in the argument itself? No sure. It must have something to do with the way the magnetic force works, which is not like the electric force. Indeed, remember the Lorentz formula: the force on some unit charge (q = 1) will be equal to F = E + v×B. So… Well… We’ve got another cross-product here and so the geometry of the situation is quite complicated: it’s not like adding two forces Fand Fto get some combined force F = Fand F2.

In any case, we need the energy, and we know that its proportional to the square of the amplitude, so… Well… We’re spot on: the square of the √2 factor in the √2·cos product and √2·sin product is 2, so that’s twice… Well… What? Hold on a minute! We’re actually taking the absolute square of the E + B = ψ+ ψ= E + i·E = √2·ei(p·x/2 − E·t/2 + π/4) wavefunction here. Is that legal? I must assume it is—although… Well… Yes. You’re right. We should do some more explaining here.

We know that we usually measure the energy as some definite integral, from t = 0 to some other point in time, or over the cycle of the oscillation. So what’s the cycle here? Our combined wavefunction can be written as √2·ei(p·x/2 − E·t/2 + π/4) = √2·ei(θ/2 + π/4), so a full cycle would correspond to θ going from 0 to 4π here, rather than from 0 to 2π. So that explains the √2 factor in front of our wave equation.

Bingo! If you were looking for an interpretation of the Planck energy and momentum, here it is.:-) And, while everything that’s written above is not easy to understand, it’s close to the ‘intuitive’ understanding to quantum mechanics that we were looking for, isn’t it? The quantum-mechanical propagation model explains everything now. 🙂 I only need to show one more thing, and that’s the different behavior of bosons and fermions:

  1. The amplitudes of identitical bosonic particles interfere with a positive sign, so we have Bose-Einstein statistics here. As Feynman writes it: (amplitude direct) + (amplitude exchanged).
  2. The amplitudes of identical fermionic particles interfere with a negative sign, so we have Fermi-Dirac statistics here: (amplitude direct) − (amplitude exchanged).

I’ll think about it. I am sure it’s got something to do with that B= i·E formula or, to put it simply, with the fact that, when bosons are involved, we get two wavefunctions (ψand ψB) for the price of one. The reasoning should be something like this:

I. For a massless particle (i.e. a zero-mass fermion), our wavefunction is just ψ = ei(p·x − E·t). So we have no √2 or √2·ei(π/4) factor in front here. So we can just add any number of them – ψ1 + ψ2 + ψ3 + … – and then take the absolute square of the amplitude to find a probability density, and we’re done.

II. For a photon (i.e. a zero-mass boson), our wavefunction is √2·ei(π/4)·ei(p·x − E·t)/2, which – let’s introduce a new symbol – we’ll denote by φ, so φ = √2·ei(π/4)·ei(p·x − E·t)/2. Now, if we add any number of these, we get a similar sum but with that √2·ei(π/4) factor in front, so we write: φ1 + φ2 + φ3 + … = √2·ei(π/4)·(ψ1 + ψ2 + ψ3 + …). If we take the absolute square now, we’ll see the probability density will be equal to twice the density for the ψ1 + ψ2 + ψ3 + … sum, because

|√2·ei(π/4)·(ψ1 + ψ2 + ψ3 + …)|2 = |√2·ei(π/4)|2·|ψ1 + ψ2 + ψ3 + …)|2 2·|ψ1 + ψ2 + ψ3 + …)|2

So… Well… I still need to connect this to Feynman’s (amplitude direct) ± (amplitude exchanged) formula, but I am sure it can be done.

Now, we haven’t tested the complete √2·ei(π/4)·ei(p·x − E·t)/2 wavefunction. Does it respect Schrödinger’s ∂ψ/∂t = i·(1/m)·∇2ψ or, including the 1/2 factor, the ∂ψ/∂t = i·[1/2m)]·∇2ψ equation? [Note we assume, once again, that ħ = 1, so we use Planck units once more.] Let’s see. We can calculate the derivatives as:

  • ∂ψ/∂t = −√2·ei(π/4)·ei∙[p·x − E·t]/2·(i·E/2)
  • 2ψ = ∂2[√2·ei(π/4)·ei∙[p·x − E·t]/2]/∂x= √2·ei(π/4)·∂[√2·ei(π/4)·ei∙[p·x − E·t]/2·(i·p/2)]/∂x = −√2·ei(π/4)·ei∙[p·x − E·t]/2·(p2/4)

So Schrödinger’s equation becomes:

i·√2·ei(π/4)·ei∙[p·x − E·t]/2·(i·E/2) = −i·(1/m)·√2·ei(π/4)·ei∙[p·x − E·t]/2·(p2/4) ⇔ 1/2 = 1/4!?

That’s funny ! It doesn’t work ! The E and m and p2 are OK because we’ve got that E = m = p equation, but we’ve got problems with yet another factor 2. It only works when we use the 2/m coefficient in Schrödinger’s equation.

So… Well… There’s no choice. That’s what we’re going to do. The Schrödinger equation for the photon is ∂ψ/∂t = i·(2/m)·∇2ψ !

It’s a very subtle point. This is all great, and very fundamental stuff! Let’s now move on to Schrödinger’s actual equation, i.e. the ∂ψ/∂t = i·(ħ/2m)·∇2ψ equation.

Post scriptum on the Planck units:

If we measure time and distance in equivalent units, say seconds, we can re-write the quantum of action as:

1.0545718×10−34 N·m·s = (1.21×1044 N)·(1.6162×10−35 m)·(5.391×10−44 s)

⇔ (1.0545718×10−34/2.998×108) N·s2 = (1.21×1044 N)·(1.6162×10−35/2.998×108 s)(5.391×10−44 s)

⇔ (1.21×1044 N) = [(1.0545718×10−34/2.998×108)]/[(1.6162×10−35/2.998×108 s)(5.391×10−44 s)] N·s2/s2

You’ll say: what’s this? Well… Look at it. We’ve got a much easier formula for the Planck force—much easier than the standard formulas you’ll find on Wikipedia, for example. If we re-interpret the symbols ħ and so they denote the numerical value of the quantum of action and the speed of light in standard SI units (i.e. newton, meter and second)—so ħ and c become dimensionless, or mathematical constants only, rather than physical constants—then the formula above can be written as:

FP newton = (ħ/c)/[(lP/c)·tP] newton ⇔ FP = ħ/(lP·tP)

Just double-check it: 1.0545718×10−34/(1.6162×10−35·5.391×10−44) = 1.21×1044. Bingo!

You’ll say: what’s the point? The point is: our model is complete. We don’t need the other physical constants – i.e. the Coulomb, Boltzmann and gravitational constant – to calculate the Planck units we need, i.e. the Planck force, distance and time units. It all comes out of our elementary wavefunction! All we need to explain the Universe – or, let’s be more modest, quantum mechanics – is two numerical constants (c and ħ) and Euler’s formula (which uses π and e, of course). That’s it.

If you don’t think that’s a great result, then… Well… Then you’re not reading this. 🙂

The quantum of time and distance

Post scriptum note added on 11 July 2016: This is one of the more speculative posts which led to my e-publication analyzing the wavefunction as an energy propagation. With the benefit of hindsight, I would recommend you to immediately the more recent exposé on the matter that is being presented here, which you can find by clicking on the provided link. In fact, I actually made some (small) mistakes when writing the post below.

Original post:

In my previous post, I introduced the elementary wavefunction of a particle with zero rest mass in free space (i.e. the particle also has zero potential). I wrote that wavefunction as ei(kx − ωt) ei(x/2 − t/2) = cos[(x−t)/2] + i∙sin[(x−t)/2], and we can represent that function as follows:

5d_euler_f

If the real and imaginary axis in the image above are the y- and z-axis respectively, then the x-axis here is time, so here we’d be looking at the shape of the wavefunction at some fixed point in space.

Now, we could also look at its shape at some fixed in point in time, so the x-axis would then represent the spatial dimension. Better still, we could animate the illustration to incorporate both the temporal as well as the spatial dimension. The following animation does the trick quite well:

Animation

Please do note that space is one-dimensional here: the y- and z-axis represent the real and imaginary part of the wavefunction, not the y- or z-dimension in space.

You’ve seen this animation before, of course: I took it from Wikipedia, and it actually represents the electric field vector (E) for a circularly polarized electromagnetic wave. To get a complete picture of the electromagnetic wave, we should add the magnetic field vector (B), which is not shown here. We’ll come back to that later. Let’s first look at our zero-mass particle denuded of all properties, so that’s not an electromagnetic wave—read: a photon. No. We don’t want to talk charges here.

OK. So far so good. A zero-mass particle in free space. So we got that ei(x/2 − t/2) = cos[(x−t)/2] + i∙sin[(x−t)/2] wavefunction. We got that function assuming the following:

  1. Time and distance are measured in equivalent units, so = 1. Hence, the classical velocity (v) of our zero-mass particle is equal to 1, and we also find that the energy (E), mass (m) and momentum (p) of our particle are numerically the same. We wrote: E = m = p, using the p = m·v (for = c) and the E = m∙c2 formulas.
  2. We also assumed that the quantum of energy (and, hence, the quantum of mass, and the quantum of momentum) was equal to ħ/2, rather than ħ. The de Broglie relations (k = p/ħ and ω = E/ħ) then gave us the rather particular argument of our wavefunction: kx − ωt = x/2 − t/2.

The latter hypothesis (E = m = p = ħ/2) is somewhat strange at first but, as I showed in that post of mine, it avoids an apparent contradiction: if we’d use ħ, then we would find two different values for the phase and group velocity of our wavefunction. To be precise, we’d find for the group velocity, but v/2 for the phase velocity. Using ħ/2 solves that problem. In addition, using ħ/2 is consistent with the Uncertainty Principle, which tells us that ΔxΔp = ΔEΔt = ħ/2.

OK. Take a deep breath. Here I need to say something about dimensions. If we’re saying that we’re measuring time and distance in equivalent units – say, in meter, or in seconds – then we are not saying that they’re the same. The dimension of time and space is fundamentally different, as evidenced by the fact that, for example, time flows in one direction only, as opposed to x. To be precise, we assumed that x and t become countable variables themselves at some point in time. However, if we’re at t = 0, then we’d count time as t = 1, 2, etcetera only. In contrast, at the point x = 0, we can go to x = +1, +2, etcetera but we may also go to x = −1, −2, etc.

I have to stress this point, because what follows will require some mental flexibility. In fact, we often talk about natural units, such as Planck units, which we get from equating fundamental constants, such as c, or ħ, to 1, but then we often struggle to interpret those units, because we fail to grasp what it means to write = 1, or ħ = 1. For example, writing = 1 implies we can measure distance in seconds, or time in meter, but it does not imply that distance becomes time, or vice versa. We still need to keep track of whether or not we’re talking a second in time, or a second in space, i.e. c meter, or, conversely, whether we’re talking a meter in space, or a meter in time, i.e. 1/c seconds. We can make the distinction in various ways. For example, we could mention the dimension of each equation between brackets, so we’d write: t = 1×10−15 s [t] ≈ 299.8×10−9 m [t]. Alternatively, we could put a little subscript (like t, or d), so as to make sure it’s clear our meter is a a ‘light-meter’, so we’d write: t = 1×10−15 s ≈ 299.8×10−9 mt. Likewise, we could add a little subscript when measuring distance in light-seconds, so we’d write x = 3×10m ≈ 1 sd, rather than x = 3×10m [x] ≈ 1 s [x].

If you wish, we could refer to the ‘light-meter’ as a ‘time-meter’ (or a meter of time), and to the light-second as a ‘distance-second’ (or a second of distance). It doesn’t matter what you call it, or how you denote it. In fact, you will never hear of a meter of time, nor will you ever see those subscripts or brackets. But that’s because physicists always keep track of the dimensions of an equation, and so they know. They know, for example, that the dimension of energy combines the dimensions of both force as well as distance, so we write: [energy] = [force]·[distance]. Read: energy amounts to applying a force over a distance. Likewise, momentum amounts to applying some force over some time, so we write: [momentum] = [force]·[time]. Using the usual symbols for energy, momentum, force, distance and time respectively, we can write this as [E] = [F]·[x] and [p] = [F]·[t]. Using the units you know, i.e. joulenewton, meter and seconds, we can also write this as: 1 J = 1 N·m and 1…

Hey! Wait a minute! What’s that N·s unit for momentum? Momentum is mass times velocity, isn’t it? It is. But it amounts to the same. Remember that mass is a measure for the inertia of an object, and so mass is measured with reference to some force (F) and some acceleration (a): F = m·⇔ m = F/a. Hence, [m] = kg = [F/a] = N/(m/s2) = N·s2/m. [Note that the m in the brackets is symbol for mass but the other m is a meter!] So the unit of momentum is (N·s2/m)·(m/s) = N·s = newton·second.

Now, the dimension of Planck’s constant is the dimension of action, which combines all dimensions: force, time and distance. We write: ħ ≈ 1.0545718×10−34 N·m·s (newton·meter·second). That’s great, and I’ll show why in a moment. But, at this point, you should just note that when we write that E = m = p = ħ/2, we’re just saying they are numerically the same. The dimensions of E, m and p are not the same. So what we’re really saying is the following:

  1. The quantum of energy is ħ/2 newton·meter ≈ 0.527286×10−34 N·m.
  2. The quantum of momentum is ħ/2 newton·second ≈ 0.527286×10−34 N·s.

What’s the quantum of mass? That’s where the equivalent units come in. We wrote: 1 kg = 1 N·s2/m. So we could substitute the distance unit in this equation (m) by sd/= sd/(3×108). So we get: 1 kg = 3×108 N·s2/sd. Can we scrap both ‘seconds’ and say that the quantum of mass (ħ/2) is equal to the quantum of momentum? Think about it.

[…]

The answer is… Yes and no—but much more no than yes! The two sides of the equation are only numerically equal, but we’re talking a different dimension here. If we’d write that 1 kg = 0.527286×10−34 N·s2/sd = 0.527286×10−34 N·s, you’d be equating two dimensions that are fundamentally different: space versus time. To reinforce the point, think of it the other way: think of substituting the second (s) for 3×10m. Again, you’d make a mistake. You’d have to write 0.527286×10−34 N·(mt)2/m, and you should not assume that a time-meter is equal to a distance-meter. They’re equivalent units, and so you can use them to get some number right, but they’re not equal: what they measure, is fundamentally different. A time-meter measures time, while a distance-meter measure distance. It’s as simple as that. So what is it then? Well… What we can do is remember Einstein’s energy-mass equivalence relation once more: E = m·c2 (and m is the mass here). Just check the dimensions once more: [m]·[c2] = (N·s2/m)·(m2/s2) = N·m. So we should think of the quantum of mass as the quantum of energy, as energy and mass are equivalent, really.

Back to the wavefunction

The beauty of the construct of the wavefunction resides in several mathematical properties of this construct. The first is its argument:

θ = kx − ωt, with k = p/ħ and ω = E/ħ

Its dimension is the dimension of an angle: we express in it in radians. What’s a radian? You might think that a radian is a distance unit because… Well… Look at how we measure an angle in radians below:

Circle_radians

But you’re wrong. An angle’s measurement in radians is numerically equal to the length of the corresponding arc of the unit circle but… Well… Numerically only. 🙂 Just do a dimensional analysis of θ = kx − ωt = (p/ħ)·x − (E/ħ)·t. The dimension of p/ħ is (N·s)/(N·m·s) = 1/m = m−1, so we get some quantity expressed per meter, which we then multiply by x, so we get a pure number. No dimension whatsoever! Likewise, the dimension of E/ħ is (N·m)/(N·m·s) = 1/s = s−1, which we then multiply by t, so we get another pure number, which we then add to get our argument θ. Hence, Planck’s quantum of action (ħ) does two things for us:

  1. It expresses p and E in units of ħ.
  2. It sorts out the dimensions, ensuring our argument is a dimensionless number indeed.

In fact, I’d say the ħ in the (p/ħ)·x term in the argument is a different ħ than the ħ in the (E/ħ)·t term. Huh? What? Yes. Think of the distinction I made between s and sd, or between m and mt. Both were numerically the same: they captured a magnitude, but they measured different things. We’ve got the same thing here:

  1. The meter (m) in ħ ≈ 1.0545718×10−34 N·m·s in (p/ħ)·x is the dimension of x, and so it gets rid of the distance dimension. So the m in ħ ≈ 1.0545718×10−34 m·s goes, and what’s left measures p in terms of units equal to 1.0545718×10−34 N·s, so we get a pure number indeed.
  2. Likewise, the second (s) in ħ ≈ 1.0545718×10−34 N·m·s in (E/ħ)·t is the dimension of t, and so it gets rid of the time dimension. So the s in ħ ≈ 1.0545718×10−34 N·m·s goes, and what’s left measures E in terms of units equal to 1.0545718×10−34 N·m, so we get another pure number.
  3. Adding both gives us the argument θ: a pure number that measures some angle.

That’s why you need to watch out when writing θ = (p/ħ)·x − (E/ħ)·t as θ = (p·x − E·t)/ħ or – in the case of our elementary wavefunction for the zero-mass particle – as θ = (x/2 − t/2) = (x − t)/2. You can do it – in fact, you should do when trying to calculate something – but you need to be aware that you’re making abstraction of the dimensions. That’s quite OK, as you’re just calculating something—but don’t forget the physics behind!

You’ll immediately ask: what are the physics behind here? Well… I don’t know. Perhaps nobody knows. As Feynman once famously said: “I think I can safely say that nobody understands quantum mechanics.” But then he never wrote that, and I am sure he didn’t really mean that. And then he said that back in 1964, which is 50 years ago now. 🙂 So let’s try to understand it at least. 🙂

Planck’s quantum of action – 1.0545718×10−34 N·m·s – comes to us as a mysterious quantity. A quantity is more than a a number. A number is something like π or e, for example. It might be a complex number, like eiθ, but that’s still a number. In contrast, a quantity has some dimension, or some combination of dimensions. A quantity may be a scalar quantity (like distance), or a vector quantity (like a field vector). In this particular case (Planck’s ħ or h), we’ve got a physical constant combining three dimensions: force, time and distance—or space, if you want.  It’s a quantum, so it comes as a blob—or a lump, if you prefer that word. However, as I see it, we can sort of project it in space as well as in time. In fact, if this blob is going to move in spacetime, then it will move in space as well as in time: t will go from 0 to 1, and x goes from 0 to ± 1, depending on what direction we’re going. So when I write that E = p = ħ/2—which, let me remind you, are two numerical equations, really—I sort of split Planck’s quantum over E = m and p respectively.

You’ll say: what kind of projection or split is that? When projecting some vector, we’ll usually have some sine and cosine, or a 1/√2 factor—or whatever, but not a clean 1/2 factor. Well… I have no answer to that, except that this split fits our mathematical construct. Or… Well… I should say: my mathematical construct. Because what I want to find is this clean Schrödinger equation:

∂ψ/∂t = i·(ħ/2m)·∇2ψ = i·∇2ψ for m = ħ/2

Now I can only get this equation if (1) E = m = p and (2) if m = ħ/2 (which amounts to writing that E = p = m = ħ/2). There’s also the Uncertainty Principle. If we are going to consider the quantum vacuum, i.e. if we’re going to look at space (or distance) and time as count variables, then Δx and Δt in the ΔxΔp = ΔEΔt = ħ/2 equations are ± 1 and, therefore, Δp and ΔE must be ± ħ/2. In any case, I am not going to try to justify my particular projection here. Let’s see what comes out of it.

The quantum vacuum

Schrödinger’s equation for my zero-mass particle (with energy E = m = p = ħ/2) amounts to writing:

  1. Re(∂ψ/∂t) = −Im(∇2ψ)
  2. Im(∂ψ/∂t) = Re(∇2ψ)

Now that reminds of the propagation mechanism for the electromagnetic wave, which we wrote as ∂B/∂t = –∇×and E/∂t = ∇×B, also assuming we measure time and distance in equivalent units. However, we’ll come back to that later. Let’s first study the equation we have, i.e.

ei(kx − ωt) = ei(ħ·x/2 − ħ·t/2)/ħ = ei(x/2 − t/2) = cos[(x−t)/2] + i∙sin[(x−t)/2]

Let’s think some more. What is that ei(x/2 − t/2) function? It’s subject to conceiving time and distance as countable variables, right? I am tempted to say: as discrete variables, but I won’t go that far—not now—because the countability may be related to a particular interpretation of quantum physics. So I need to think about that. In any case… The point is that x can only take on values like 0, 1, 2, etcetera. And the same goes for t. To make things easy, we’ll not consider negative values for x right now (and, obviously, not for t either). But you can easily check it doesn’t make a difference: if you think of the propagation mechanism – which is what we’re trying to model here – then x is always positive, because we’re moving away from some source that caused the wave. In any case, we’ve got a infinite set of points like:

  • ei(0/2 − 0/2) ei(0) = cos(0) + i∙sin(0)
  • ei(1/2 − 0/2) = ei(1/2) = cos(1/2) + i∙sin(1/2)
  • ei(0/2 − 1/2) = ei(−1/2) = cos(−1/2) + i∙sin(−1/2)
  • ei(1/2 − 1/2) = ei(0) = cos(0) + i∙sin(0)

In my previous post, I calculated the real and imaginary part of this wavefunction for x going from 0 to 14 (as mentioned, in steps of 1) and for t doing the same (also in steps of 1), and what we got looked pretty good:

graph real graph imaginary

I also said that, if you wonder how the quantum vacuum could possibly look like, you should probably think of these discrete spacetime points, and some complex-valued wave that travels as illustrated above. In case you wonder what’s being illustrated here: the right-hand graph is the cosine value for all possible x = 0, 1, 2,… and t = 0, 1, 2,… combinations, and the left-hand graph depicts the sine values, so that’s the imaginary part of our wavefunction. Taking the absolute square of both gives 1 for all combinations. So it’s obvious we’d need to normalize and, more importantly, we’d have to localize the particle by adding several of these waves with the appropriate contributions. But so that’s not our worry right now. I want to check whether those discrete time and distance units actually make sense. What’s their size? Is it anything like the Planck length (for distance) and/or the Planck time?

Let’s see. What are the implications of our model? The question here is: if ħ/2 is the quantum of energy, and the quantum of momentum, what’s the quantum of force, and the quantum of time and/or distance?

Huh? Yep. We treated distance and time as countable variables above, but now we’d like to express the difference between x = 0 and x = 1 and between t = 0 and t = 1 in the units we know, this is in meter and in seconds. So how do we go about that? Do we have enough equations here? Not sure. Let’s see…

We obviously need to keep track of the various dimensions here, so let’s refer to that discrete distance and time unit as tand lP respectively. The subscript (P) refers to Planck, and the refers to a length, but we’re likely to find something else than Planck units. I just need placeholder symbols here. To be clear: tand lP are expressed in meter and seconds respectively, just like the actual Planck time and distance, which are equal to 5.391×10−44 s (more or less) and  1.6162×10−35 m (more or less) respectively. As I mentioned above, we get these Planck units by equating fundamental physical constants to 1. Just check it: (1.6162×10−35 m)/(5.391×10−44 s) = ≈ 3×10m/s. So the following relation must be true: lP = c·tP, or lP/t= c.

Now, as mentioned above, there must be some quantum of force as well, which we’ll write as FP, and which is – obviously – expressed in newton (N). So we have:

  1. E = ħ/2 ⇒ 0.527286×10−34 N·m = FP·lN·m
  2. p = ħ/2 ⇒ 0.527286×10−34 N·s = FP·tN·s

Let’s try to divide both formulas: E/p = (FP·lN·m)/(FP·tN·s) = lP/tP m/s = lP/tP m/s = c m/s. That’s consistent with the E/p = equation. Hmm… We found what we knew already. My model is not fully determined, it seems. 😦

What about the following simplistic approach? E is numerically equal to 0.527286×10−34, and its dimension is [E] = [F]·[x], so we write: E = 0.527286×10−34·[E] = 0.527286×10−34·[F]·[x]. Hence, [x] = [E]/[F] = (N·m)/N = m. That just confirms what we already know: the quantum of distance (i.e. our fundamental unit of distance) can be expressed in meter. But our model does not give that fundamental unit. It only gives us its dimension (meter), which is stuff we knew from the start. 😦

Let’s try something else. Let’s just accept that Planck length and time, so we write:

  • lP = 1.6162×10−35 m
  • t= 5.391×10−44 s

Now, if the quantum of action is equal to ħ N·m·s = FP·lP·tP N·m·s = 1.0545718×10−34 N·m·s, and if the two definitions of land tP above hold, then 1.0545718×10−34 N·m·s = (FN)×(1.6162×10−35 m)×(5.391×10−44 s) ≈ FP  8.713×10−79 N·m·s ⇔ FP ≈ 1.21×1044 N.

Does that make sense? It does according to Wikipedia, but how do we relate this to our E = p = m = ħ/2 equations? Let’s try this:

  1. EP = (1.0545718×10−34 N·m·s)/(5.391×10−44 s) = 1.956×109 J. That corresponds to the regular Planck energy.
  2. pP = (1.0545718×10−34 N·m·s)/(1.6162×10−35 m) = 0.6525 N·s. That corresponds to the regular Planck momentum.

Is EP = pP? Let’s substitute: 1.956×109 N·m = 1.956×109 N·(s/c) = 1.956×109/2.998×10N·s = 0.6525 N·s. So, yes, it comes out alright. In fact, I omitted the 1/2 factor in the calculations, but it doesn’t matter: it does come out alright. So I did not prove that the difference between my x = 0 and x = 1 points (or my t = 0 and t  = 1 points) is equal to the Planck length (or the Planck time unit), but I did show my theory is, at the very least, compatible with those units. That’s more than enough for now. And I’ll come surely come back to it in my next post. 🙂

Post Scriptum: One must solve the following equations to get the fundamental Planck units:

Planck units

We have five fundamental equations for five fundamental quantities respectively: tP, lP, FP, mP, and EP respectively, so that’s OK: it’s a fully determined system alright! But where do the expressions with G, kB (the Boltzmann constant) and ε0 come from? What does it mean to equate those constants to 1? Well… I need to think about that, and I’ll get back to you on it. 🙂

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 μand 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 μand 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 μand 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: 🙂

god-said-maxwell-equation

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:

E

B

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 = –er×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/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 = Felectric + Fmagnetic = 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 μand ε0 respectively), and Maxwell’s equations imply that the product of the electric and magnetic constant is the reciprocal of c2: μ0·ε= 1/c2. So here we see εand 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·ε= 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 = Felectric + Fmagnetic = qE + qv×B = q[E + (v/c)×(E×er)]: 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·ε= 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 εis expressed in (A·s)/(V·m), so the dimension in which the μ0·εproduct 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 εwith 1/c.
  • Now, if we would then equate c with 1, we’d have μ0 = ε= 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 μand εare 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 1043 Hz scale. 1032 Hz corresponds to a photon energy of Eγ = h·f = (4×10−15 eV·s)·(1032 Hz) = 4×1017 eV = 400,000 tera-electronvolt (1 TeV = 1012 eV). I use the γ (gamma) subscript in my Eγ 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 10–8 seconds it takes for the energy of a radiating atom to die out. [As for the calculation of that 10–8 sec 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〉 = (c0)·〈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: Eγ = 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 ≈ 1.6×10−19 J. 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 Eγ = h·f as Eγ/f = h? The dimensions in this equation are:

[x joule]·[1 second]/[cyles] = [6.626×10−34 joule]·[1 second]

⇔ = 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: = 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.]

600px-Black_body

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 1043 Hz scale with the 1028 eV 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. 1028 eV], 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 1043 Hz scale corresponds to a wavelength of the Planck scale: λ = c/= (3×10m/s)/(1043 s−1) = 3×10−35 m. 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 = eP2.

(2) α is the square root of the ratio of (a) the classical electron radius and (b) the Bohr radius: α = √(re /r). You’ll see this more often written as re = 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 (1.616×10−35 m)/(5.391×10−44 s) = 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 me here) and (b) the classical electron radius re (if both are expressed in Planck units): α = me·re. Now 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 me = α/r= 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 me = α/r = eP2/re. Using the Bohr radius, we get me = 1/αr = 1/eP2r.

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 (ħ = kB = ke = 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. 🙂

The Uncertainty Principle and the stability of atoms

The Model of the Atom

In one of my posts, I explained the quantum-mechanical model of an atom. Feynman sums it up as follows:

“The electrostatic forces pull the electron as close to the nucleus as possible, but the electron is compelled to stay spread out in space over a distance given by the Uncertainty Principle. If it were confined in too small a space, it would have a great uncertainty in momentum. But that means it would have a high expected energy—which it would use to escape from the electrical attraction. The net result is an electrical equilibrium not too different from the idea of Thompson—only is it the negative charge that is spread out, because the mass of the electron is so much smaller than the mass of the proton.”

This explanation is a bit sloppy, so we should add the following clarification: “The wave function Ψ(r) for an electron in an atom does not describe a smeared-out electron with a smooth charge density. The electron is either here, or there, or somewhere else, but wherever it is, it is a point charge.” (Feynman’s Lectures, Vol. III, p. 21-6)

The two quotes are not incompatible: it is just a matter of defining what we really mean by ‘spread out’. Feynman’s calculation of the Bohr radius of an atom in his introduction to quantum mechanics clears all confusion in this regard:

Bohr radius

It is a nice argument. One may criticize he gets the right thing out because he puts the right things in – such as the values of e and m, for example 🙂 − but it’s nice nevertheless!

Mass as a Scale Factor for Uncertainty

Having complimented Feynman, the calculation above does raise an obvious question: why is it that we cannot confine the electron in “too small a space” but that we can do so for the nucleus (which is just one proton in the example of the hydrogen atom here). Feynman gives the answer above: because the mass of the electron is so much smaller than the mass of the proton.

Huh? What’s the mass got to do with it? The uncertainty is the same for protons and electrons, isn’t it?

Well… It is, and it isn’t. 🙂 The Uncertainty Principle – usually written in its more accurate σxσp ≥ ħ/2 expression – applies to both the electron and the proton – of course! – but the momentum is the product of mass and velocity (p = m·v), and so it’s the proton’s mass that makes the difference here. To be specific, the mass of a proton is about 1836 times that of an electron. Now, as long as the velocities involved are non-relativistic—and they are non-relativistic in this case: the (relative) speed of electrons in atoms is given by the fine-structure constant α = v/c ≈ 0.0073, so the Lorentz factor is very close to 1—we can treat the m in the p = m·v identity as a constant and, hence, we can also write: Δp = Δ(m·v) = m·Δv. So all of the uncertainty of the momentum goes into the uncertainty of the velocity. Hence, the mass acts likes a reverse scale factor for the uncertainty. To appreciate what that means, let me write ΔxΔp = ħ as:

ΔxΔv = ħ/m

It is an interesting point, so let me expand the argument somewhat. We actually use a more general mathematical property of the standard deviation here: the standard deviation of a variable scales directly with the scale of the variable. Hence, we can write: σ(k·x) = k·σ(x), with k > 0. So the uncertainty is, indeed, smaller for larger masses. Larger masses are associated with smaller uncertainties in their position x. To be precise, the uncertainty is inversely proportional to the mass and, hence, the mass number effectively acts like a reverse scale factor for the uncertainty.

Of course, you’ll say that the uncertainty still applies to both factors on the left-hand side of the equation, and so you’ll wonder: why can’t we keep Δx the same and multiply Δv with m, so its product yields ħ again? In other words, why can’t we have a uncertainty in velocity for the proton that is 1836 times larger than the uncertainty in velocity for the electron? The answer to that question should be obvious: the uncertainty should not be greater than the expected value. When everything is said and done, we’re talking a distribution of some variable here (the velocity variable, to be precise) and, hence, that distribution is likely to be the Maxwell-Boltzmann distribution we introduced in previous posts. Its formula and graph are given below:

Formula M-B distribution428px-Maxwell-Boltzmann_distribution_pdf

In statistics (and in probability theory), they call this a chi distribution with three degrees of freedom and a scale parameter which is equal to a = (kT/m)1/2. The formula for the scale parameter shows how the mass of a particle indeed acts as a reverse scale parameter. The graph above shows three graphs for a = 1, 2 and 5 respectively. Note the square root though: quadrupling the mass (keeping kT the same) amounts to going from a = 2 to a = 1, so that’s halving a. Indeed, [kT/(4m)]1/2 = (1/2)(kT/m)1/2. So we can’t just do what we want with Δv (like multiplying it with 1836, as suggested). In fact, the graph and the formulas show that Feynman’s assumption that we can equate p with Δp (i.e. his assumption that “the momenta must be of the order p = ħ/Δx, with Δx the spread in position”), more or less at least, is quite reasonable.

Of course, you are very smart and so you’ll have yet another objection: why can’t we associate a much higher momentum with the proton, as that would allow us to associate higher velocities with the proton? Good question. My answer to that is the following (and it might be original, as I didn’t find this anywhere else). When everything is said and done, we’re talking two particles in some box here: an electron and a proton. Hence, we should assume that the average kinetic energy of our electron and our proton is the same (if not, they would be exchanging kinetic energy until it’s more or less equal), so we write <melectron·v2electron/2> = <mproton·v2proton/2>. We can re-write this as mp/m= 1/1836 = <v2e>/<v2p> and, therefore, <v2e> = 1836·<v2p>. Now, <v2> ≠ <v>2 and, hence, <v> ≠ √<v2>. So the equality does not imply that the expected velocity of the electron is √1836 ≈ 43 times the expected velocity of the proton. Indeed, because of the particularities of the distribution, there is a difference between (a) the most probable speed, which is equal to √2·a ≈ 1.414·a, (b) the root mean square speed, which is equal to √<v2> = √3·a ≈ 1.732·a, and, finally, (c) the mean or expected speed, which is equal to <v> = 2·(2/π)1/2·a ≈ 1.596·a.

However, we are not far off. We could use any of these three values to roughly approximate Δv, as well as the scale parameter a itself: our answers would all be of the same order. However, to keep the calculations simple, let’s use the most probable speed. Let’s equate our electron mass with unity, so the mass of our proton is 1836. Now, such mass implies a scale factor (i.e. a) that’s √1836 ≈ 43 times smaller. So the most probable speed of the proton and, therefore, its spread, would be about √2/√1836 = √(2/1836) ≈ 0.033 that of the electron, so we write: Δvp ≈ 0.033·Δve. Now we can insert this in our ΔxΔv = ħ/m = ħ/1836 identity. We get: ΔxpΔvp = Δxp·√(2/1836)·Δve = ħ/1836. That, in turn, implies that √(2·1836)·Δxp = ħ/Δve, which we can re-write as: Δx= Δxe/√(2·1836) ≈ Δxe/60. In other words, the expected spread in the position of the proton is about 60 times smaller than the expected spread of the electron. More in general, we can say that the spread in position of a particle, keeping all else equal, is inversely proportional to (2m)1/2. Indeed, in this case, we multiplied the mass with about 1800, and we found that the uncertainty in position went down with a factor 1/60 = 1/√3600. Not bad as a result ! Is it precise? Well… It could be like √3·√m or 2·(2/π)1/2··√m depending on our definition of ‘uncertainty’, but it’s all of the same order. So… Yes. Not bad at all… 🙂

You’ll raise a third objection now: the radius of a proton is measured using the femtometer scale, so that’s expressed in 10−15 m, which is not 60 but a million times smaller than the nanometer (i.e. 10−9 m) scale used to express the Bohr radius as calculated by Feynman above. You’re right, but the 10−15 m number is the charge radius, not the uncertainty in position. Indeed, the so-called classical electron radius is also measured in femtometer and, hence, the Bohr radius is also like a million times that number. OK. That should settle the matter. I need to move on.

Before I do move on, let me relate the observation (i.e. the fact that the uncertainty in regard to position decreases as the mass of a particle increases) to another phenomenon. As you know, the interference of light beams is easy to observe. Hence, the interference of photons is easy to observe: Young’s experiment involved a slit of 0.85 mm (so almost 1 mm) only. In contrast, the 2012 double-slit experiment with electrons involved slits that were 62 nanometer wide, i.e. 62 billionths of a meter! That’s because the associated frequencies are so much higher and, hence, the wave zone is much smaller. So much, in fact, that Feynman could not imagine technology would ever be sufficiently advanced so as to actually carry out the double slit experiment with electrons. It’s an aspect of the same: the uncertainty in position is much smaller for electrons than it is for photons. Who knows: perhaps one day, we’ll be able to do the experiment with protons. 🙂 For further detail, I’ll refer you one of my posts on this.

What’s Explained, and What’s Left Unexplained?

There is another obvious question: if the electron is still some point charge, and going around as it does, why doesn’t it radiate energy? Indeed, the Rutherford-Bohr model had to be discarded because this ‘planetary’ model involved circular (or elliptical) motion and, therefore, some acceleration. According to classical theory, the electron should thus emit electromagnetic radiation, as a result of which it would radiate its kinetic energy away and, therefore, spiral in toward the nucleus. The quantum-mechanical model doesn’t explain this either, does it?

I can’t answer this question as yet, as I still need to go through all Feynman’s Lectures on quantum mechanics. You’re right. There’s something odd about the quantum-mechanical idea: it still involves a electron moving in some kind of orbital − although I hasten to add that the wavefunction is a complex-valued function, not some real function − but it does not involve any loss of kinetic energy due to circular motion apparently!

There are other unexplained questions as well. For example, the idea of an electrical point charge still needs to be re-conciliated with the mathematical inconsistencies it implies, as Feynman points out himself in yet another of his Lectures. :-/

Finally, you’ll wonder as to the difference between a proton and a positron: if a positron and an electron annihilate each other in a flash, why do we have a hydrogen atom at all? Well… The proton is not the electron’s anti-particle. For starters, it’s made of quarks, while the positron is made of… Well… A positron is a positron: it’s elementary. But, yes, interesting question, and the ‘mechanics’ behind the mutual destruction are quite interesting and, hence, surely worth looking into—but not here. 🙂

Having mentioned a few things that remain unexplained, the model does have the advantage of solving plenty of other questions. It explains, for example, why the electron and the proton are actually right on top of each other, as they should be according to classical electrostatic theory, and why they are not at the same time: the electron is still a sort of ‘cloud’ indeed, with the proton at its center.

The quantum-mechanical ‘cloud’ model of the electron also explains why “the terrific electrical forces balance themselves out, almost perfectly, by forming tight, fine mixtures of the positive and the negative, so there is almost no attraction or repulsion at all between two separate bunches of such mixtures” (Richard Feynman, Introduction to Electromagnetism, p. 1-1) or, to quote from one of his other writings, why we do not fall through the floor as we walk:

“As we walk, our shoes with their masses of atoms push against the floor with its mass of atoms. In order to squash the atoms closer together, the electrons would be confined to a smaller space and, by the uncertainty principle, their momenta would have to be higher on the average, and that means high energy; the resistance to atomic compression is a quantum-mechanical effect and not a classical effect. Classically, we would expect that if we were to draw all the electrons and protons closer together, the energy would be reduced still further, and the best arrangement of positive and negative charges in classical physics is all on top of each other. This was well known in classical physics and was a puzzle because of the existence of the atom. Of course, the early scientists invented some ways out of the trouble—but never mind, we have the right way out, now!”

So that’s it, then. Except… Well…

The Fine-Structure Constant

When talking about the stability of atoms, one cannot escape a short discussion of the so-called fine-structure constant, denoted by α (alpha). I discussed it another post of mine, so I’ll refer you there for a more comprehensive overview. I’ll just remind you of the basics:

(1) α is the square of the electron charge expressed in Planck units: α = eP2.

(2) α is the square root of the ratio of (a) the classical electron radius and (b) the Bohr radius: α = √(re /r). You’ll see this more often written as re = 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 (1.616×10−35 m)/(5.391×10−44 s) = 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) Finally, α is also equal to the product of (a) the electron mass (which I’ll simply write as me here) and (b) the classical electron radius re (if both are expressed in Planck units): α = me·re. [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.]

Note that, from (2) and (4), we also find that:

(5) The electron mass (in Planck units) is equal me = α/r= 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 me = α/r = eP2/re. Using the Bohr radius, we get me = 1/αr = 1/eP2r.

In addition, in the mentioned post, I also related α to the so-called coupling constant determining the strength of the interaction between electrons and photons. So… What a magical number indeed ! It suggests some unity that our little model of the atom above doesn’t quite capture. As far as I am concerned, it’s one of the many other ‘unexplained questions’, and one of my key objectives, as I struggle through Feynman’s Lectures, is to understand it all. 🙂 One of the issues is, of course, how to relate this coupling constant to the concept of a gauge, which I briefly discussed in my previous post. In short, I’ve still got a long way to go… 😦

Post Scriptum: The de Broglie relations and the Uncertainty Principle

My little exposé on mass being nothing but a scale factor in the Uncertainty Principle is a good occasion to reflect on the Uncertainty Principle once more. Indeed, what’s the uncertainty about, if it’s not about the mass? It’s about the position in space and velocity, i.e. it’s movement and time. Velocity or speed (i.e. the magnitude of the velocity vector) is, in turn, defined as the distance traveled divided by the time of travel, so the uncertainty is about time as well, as evidenced from the ΔEΔt = h expression of the Uncertainty Principle. But how does it work exactly?

Hmm… Not sure. Let me try to remember the context. We know that the de Broglie relation, λ = h/p, which associates a wavelength (λ) with the momentum (p) of a particle, is somewhat misleading, because we’re actually associating a (possibly infinite) bunch of component waves with a particle. So we’re talking some range of wavelengths (Δλ) and, hence, assuming all these component waves travel at the same speed, we’re also talking a frequency range (Δf). The bottom line is that we’ve got a wave packet and we need to distinguish the velocity of its phase (vp) versus the group velocity (vg), which corresponds to the classical velocity of our particle.

I think I explained that pretty well in one of my previous posts on the Uncertainty Principle, so I’d suggest you have a look there. The mentioned post explains how the Uncertainty Principle relates position (x) and momentum (p) as a Fourier pair, and it also explains that general mathematical property of Fourier pairs: the more ‘concentrated’ one distribution is, the more ‘spread out’ its Fourier transform will be. In other words, it is not possible to arbitrarily ‘concentrate’ both distributions, i.e. both the distribution of x (which I denoted as Ψ(x) as well as its Fourier transform, i.e. the distribution of p (which I denoted by Φ(p)). So, if we’d ‘squeeze’ Ψ(x), then its Fourier transform Φ(p) will ‘stretch out’.

That was clear enough—I hope! But how do we go from ΔxΔp = h to ΔEΔt = h? Why are energy and time another Fourier pair? To answer that question, we need to clearly define what energy and what time we are talking about. The argument revolves around the second de Broglie relation: E = h·f. How do we go from the momentum p to the energy E? And how do we go from the wavelength λ to the frequency f?

The answer to the first question is the energy-mass equivalence: E = mc2, always. This formula is relativistic, as m is the relativistic mass, so it includes the rest mass m0 as well as the equivalent mass of its kinetic energy m0v2/2 + … [Note, indeed, that the kinetic energy – defined as the excess energy over its rest energy – is a rapidly converging series of terms, so only the m0v2/2 term is mentioned.] Likewise, momentum is defined as p = mv, always, with m the relativistic mass, i.e. m = (1−v2/c2)−1/2·m0 = γ·m0, with γ the Lorentz factor. The E = mc2 and p = mv relations combined give us the E/c = m·c = p·c/v or E·v/c = p·c relationship, which we can also write as E/p = c2/v. However, we’ll need to write E as a function of p for the purpose of a derivation. You can verify that E− p2c= m02c4) and, hence, that E = (p2c+ m02c4)1/2.

Now, to go from a wavelength to a frequency, we need the wave velocity, and we’re obviously talking the phase velocity here, so we write: vp = λ·f. That’s where the de Broglie hypothesis comes in: de Broglie just assumed the Planck-Einstein relation E = h·ν, in which ν is the frequency of a massless photon, would also be valid for massive particles, so he wrote: E = h·f. It’s just a hypothesis, of course, but it makes everything come out alright. More in particular, the phase velocity vp = λ·f can now be re-written, using both de Broglie relations (i.e. h/p = λ and E/h = f) as vp = (E/h)·(p/h) = E/p = c2/v. Now, because v is always smaller than c for massive particles (and usually very much smaller), we’re talking a superluminal phase velocity here! However, because it doesn’t carry any signal, it’s not inconsistent with relativity theory.

Now what about the group velocity? To calculate the group velocity, we need the frequencies and wavelengths of the component waves. The dispersion relation assumes the frequency of each component wave can be expressed as a function of its wavelength, so f = f(λ). Now, it takes a bit of wave mechanics (which I won’t elaborate on here) to show that the group velocity is the derivative of f with respect to λ, so we write vg = ∂f/∂λ. Using the two de Broglie relations, we get: vg = ∂f/∂λ = ∂(E/h)/∂(p/h) = ∂E/∂p = ∂[p2c+ m02c4)1/2]/∂p. Now, when you write it all out, you should find that vg = ∂f/∂λ = pc2/E = c2/vp = v, so that’s the classical velocity of our particle once again.

Phew! Complicated! Yes. But so we still don’t have our ΔEΔt = h expression! All of the above tells us how we can associate a range of momenta (Δp) with a range of wavelengths (Δλ) and, in turn, with a frequency range (Δf) which then gives us some energy range (ΔE), so the logic is like:

Δp ⇒ Δλ ⇒ Δf ⇒ ΔE

Somehow, the same sequence must also ‘transform’ our Δx into Δt. I googled a bit, but I couldn’t find any clear explanation. Feynman doesn’t seem to have one in his Lectures either so, frankly, I gave up. What I did do in one of my previous posts, is to give some interpretation. However, I am not quite sure if it’s really the interpretation: there are probably several ones. It must have something to do with the period of a wave, but I’ll let you break your head over it. 🙂 As far as I am concerned, it’s just one of the other unexplained questions I have as I sort of close my study of ‘classical’ physics. So I’ll just make a mental note of it. [Of course, please don’t hesitate to send me your answer, if you’d have one!] Now it’s time to really dig into quantum mechanics, so I should really stay silent for quite a while now! 🙂

Fields and charges (II)

My previous posts was, perhaps, too full of formulas, without offering much reflection. Let me try to correct that here by tying up a few loose ends. The first loose end is about units. Indeed, I haven’t been very clear about that and so let me somewhat more precise on that now.

Note: In case you’re not interested in units, you can skip the first part of this post. However, please do look at the section on the electric constant εand, most importantly, the section on natural units—especially Planck units, as I will touch upon the topic of gauge coupling parameters there and, hence, on quantum mechanics. Also, the third and last part, on the theoretical contradictions inherent in the idea of point charges, may be of interest to you.]

The field energy integrals

When we wrote that down that u = ε0E2/2 formula for the energy density of an electric field (see my previous post on fields and charges for more details), we noted that the 1/2 factor was there to avoid double-counting. Indeed, those volume integrals we use to calculate the energy over all space (i.e. U = ∫(u)dV) count the energy that’s associated with a pair of charges (or, to be precise, charge elements) twice and, hence, they have a 1/2 factor in front. Indeed, as Feynman notes, there is no convenient way, unfortunately, of writing an integral that keeps track of the pairs so that each pair is counted just once. In fact, I’ll have to come back to that assumption of there being ‘pairs’ of charges later, as that’s another loose end in the theory.

U 6

U 7

Now, we also said that that εfactor in the second integral (i.e. the one with the vector dot product EE =|E||E|cos(0) = E2) is there to make the units come out alright. Now, when I say that, what does it mean really? I’ll explain. Let me first make a few obvious remarks:

  1. Densities are always measures in terms per unit volume, so that’s the cubic meter (m3). That’s, obviously, an astronomical unit at the atomic or molecular scale.
  2. Because of historical reasons, the conventional unit of charge is not the so-called elementary charge +e (i.e. the charge of a proton), but the coulomb. Hence, the charge density ρ is expressed in Coulomb per cubic meter (C/m3). The coulomb is a rather astronomical unit too—at the atomic or molecular scale at least: 1 e ≈ 1.6022×10−19 C. [I am rounding here to four digits after the decimal point.]
  3. Energy is in joule (J) and that’s, once again, a rather astronomical unit at the lower end of the scales. Indeed, theoretical physicists prefer to use the electronvolt (eV), which is the energy gained (or lost) when an electron (so that’s a charge of –e, i.e. minus e) moves across a potential difference of one volt. But so we’ll stick to the joule as for now, not the eV, because the joule is the SI unit that’s used when defining most electrical units, such as the ampere, the watt and… Yes. The volt. Let’s start with that one.

The volt

The volt unit (V) measures both potential (energy) as well as potential difference (in both cases, we mean electric potential only, of course). Now, from all that you’ve read so far, it should be obvious that potential (energy) can only be measured with respect to some reference point. In physics, the reference point is infinity, which is so far away from all charges that there is no influence there. Hence, any charge we’d bring there (i.e. at infinity) will just stay where it is and not be attracted or repelled by anything. We say the potential there is zero: Φ(∞) = 0. The choice of that reference point allows us, then, to define positive or negative potential: the potential near positive charges will be positive and, vice versa, the potential near negative charges will be negative. Likewise, the potential difference between the positive and negative terminal of a battery will be positive.

So you should just note that we measure both potential as well as potential difference in volt and, hence, let’s now answer the question of what a volt really is. The answer is quite straightforward: the potential at some point r = (x, y, z) measures the work done when bringing one unit charge (i.e. +e) from infinity to that point. Hence, it’s only natural that we define one volt as one joule per unit charge:

1 volt = 1 joule/coulomb (1 V = 1 J/C).

Also note the following:

  1. One joule is the energy energy transferred (or work done) when applying a force of one newton over a distance of one meter, so one volt can also be measured in newton·meter per coulomb: 1 V = 1 J/C = N·m/C.
  2. One joule can also be written as 1 J = 1 V·C.

It’s quite easy to see why that energy = volt-coulomb product makes sense: higher voltage will be associated with higher energy, and the same goes for higher charge. Indeed, the so-called ‘static’ on our body is usually associated with potential differences of thousands of volts (I am not kidding), but the charges involved are extremely small, because the ability of our body to store electric charge is minimal (i.e. the capacitance (aka capacity) of our body). Hence, the shock involved in the discharge is usually quite small: it is measured in milli-joules (mJ), indeed.

The farad

The remark on ‘static’ brings me to another unit which I should mention in passing: the farad. It measures the capacitance (formerly known as the capacity) of a capacitor (formerly known as a condenser). A condenser consists, quite simply, of two separated conductors: it’s usually illustrated as consisting of two plates or of thin foils (e.g. aluminum foil) separated by an insulating film (e.g. waxed paper), but one can also discuss the capacity of a single body, like our human body, or a charged sphere. In both cases, however, the idea is the same: we have a ‘positive’ charge on one side (+q), and a ‘negative’ charge on the other (–q). In case of a single object, we imagine the ‘other’ charge to be some other large object (the Earth, for instance, but it can also be a car or whatever object that could potentially absorb the charge on our body) or, in case of the charged sphere, we could imagine some other sphere of much larger radius. The farad will then measure the capacity of one or both conductors to store charge.

Now, you may think we don’t need another unit here if that’s the definition: we could just express the capacity of a condensor in terms of its maximum ‘load’, couldn’t we? So that’s so many coulomb before the thing breaks down, when the waxed paper fails to separate the two opposite charges on the aluminium foil, for example. No. It’s not like that. It’s true we can not continue to increase the charge without consequences. However, what we want to measure with the farad is another relationship. Because of the opposite charges on both sides, there will be a potential difference, i.e. a voltage difference. Indeed, a capacitor is like a little battery in many ways: it will have two terminals. Now, it is fairly easy to show that the potential difference (i.e. the voltage) between the two plates will be proportional to the charge. Think of it as follows: if we double the charges, we’re doubling the fields, right? So then we need to do twice the amount of work to carry the unit charge (against the field) from one plate to the other. Now, because the distance is the same, that means the potential difference must be twice what it was.

Now, while we have a simple proportionality here between the voltage and the charge, the coefficient of proportionality will depend on the type of conductors, their shape, the distance and the type of insulator (aka dielectric) between them, and so on and so on. Now, what’s being measured in farad is that coefficient of proportionality, which we’ll denote by C(the proportionality coefficient for the charge), CV ((the proportionality coefficient for the voltage) or, because we should make a choice between the two, quite simply, as C. Indeed, we can either write (1) Q = CQV or, alternatively, V = CVQ, with C= 1/CV. As Feynman notes, “someone originally wrote the equation of proportionality as Q = CV”, so that’s what it will be: the capacitance (aka capacity) of a capacitor (aka condenser) is the ratio of the electric charge Q (on each conductor) to the potential difference V between the two conductors. So we know that’s a constant typical of the type of condenser we’re talking about. Indeed, the capacitance is the constant of proportionality defining the linear relationship between the charge and the voltage means doubling the voltage, and so we can write:

C = Q/V

Now, the charge is measured in coulomb, and the voltage is measured in volt, so the unit in which we will measure C is coulomb per volt (C/V), which is also known as the farad (F):

1 farad = 1 coulomb/volt (1 F = 1 C/V)

[Note the confusing use of the same symbol C for both the unit of charge (coulomb) as well as for the proportionality coefficient! I am sorrry about that, but so that’s convention!].

To be precise, I should add that the proportionality is generally there, but there are exceptions. More specifically, the way the charge builds up (and the way the field builds up, at the edges of the capacitor, for instance) may cause the capacitance to vary a little bit as it is being charged (or discharged). In that case, capacitance will be defined in terms of incremental changes: C = dQ/dV.

Let me conclude this section by giving you two formulas, which are also easily proved but so I will just give you the result:

  1. The capacity of a parallel-plate condenser is C = ε0A/d. In this formula, we have, once again, that ubiquitous electric constant ε(think of it as just another coefficient of proportionality), and then A, i.e. the area of the plates, and d, i.e. the separation between the two plates.
  2. The capacity of a charged sphere of radius r (so we’re talking the capacity of a single conductor here) is C = 4πε0r. This may remind you of the formula for the surface of a sphere (A = 4πr2), but note we’re not squaring the radius. It’s just a linear relationship with r.

I am not giving you these two formulas to show off or fill the pages, but because they’re so ubiquitous and you’ll need them. In fact, I’ll need the second formula in this post when talking about the other ‘loose end’ that I want to discuss.

Other electrical units

From your high school physics classes, you know the ampere and the watt, of course:

  1. The ampere is the unit of current, so it measures the quantity of charge moving or circulating per second. Hence, one ampere is one coulomb per second: 1 A = 1 C/s.
  2. The watt measures power. Power is the rate of energy conversion or transfer with respect to time. One watt is one joule per second: 1 W = 1 J/s = 1 N·m/s. Also note that we can write power as the product of current and voltage: 1 W = (1 A)·(1 V) = (1 C/s)·(1 J/C) = 1 J/s.

Now, because electromagnetism is such well-developed theory and, more importantly, because it has so many engineering and household applications, there are many other units out there, such as:

  • The ohm (Ω): that’s the unit of electrical resistance. Let me quickly define it: the ohm is defined as the resistance between two points of a conductor when a (constant) potential difference (V) of one volt, applied to these points, produces a current (I) of one ampere. So resistance (R) is another proportionality coefficient: R = V/I, and 1 ohm (Ω) = 1 volt/ampere (V/A). [Again, note the (potential) confusion caused by the use of the same symbol (V) for voltage (i.e. the difference in potential) as well as its unit (volt).] Now, note that it’s often useful to write the relationship as V = R·I, so that gives the potential difference as the product of the resistance and the current.
  • The weber (Wb) and the tesla (T): that’s the unit of magnetic flux (i.e. the strength of the magnetic field) and magnetic flux density (i.e. one tesla = one weber per square meter) respectively. So these have to do with the field vector B, rather than E. So we won’t talk about it here.
  • The henry (H): that’s the unit of electromagnetic inductance. It’s also linked to the magnetic effect. Indeed, from Maxwell’s equations, we know that a changing electric current will cause the magnetic field to change. Now, a changing magnetic field causes circulation of E. Hence, we can make the unit charge go around in some loop (we’re talking circulation of E indeed, not flux). The related energy, or the work that’s done by a unit of charge as it travels (once) around that loop, is – quite confusingly! – referred to as electromotive force (emf). [The term is quite confusing because we’re not talking force but energy, i.e. work, and, as you know by now, energy is force times distance, so energy and force are related but not the same.] To ensure you know what we’re talking about, let me note that emf is measured in volts, so that’s in joule per coulomb: 1 V = 1 J/C. Back to the henry now. If the rate of change of current in a circuit (e.g. the armature winding of an electric motor) is one ampere per second, and the resulting electromotive force (remember: emf is energy per coulomb) is one volt, then the inductance of the circuit is one henry. Hence, 1 H = 1 V/(1 A/s) = 1 V·s/A.     

The concept of impedance

You’ve probably heard about the so-called impedance of a circuit. That’s a complex concept, literally, because it’s a complex-valued ratio. I should refer you to the Web for more details, but let me try to summarize it because, while it’s complex, that doesn’t mean it’s complicated. 🙂 In fact, I think it’s rather easy to grasp after all you’ve gone through already. 🙂 So let’s give it a try.

When we have a simple direct current (DC), then we have a very straightforward definition of resistance (R), as mentioned above: it’s a simple ratio between the voltage (as measured in volt) and the current (as measured in ampere). Now, with alternating current (AC) circuits, it becomes more complicated, and so then it’s the concept of impedance that kicks in. Just like resistance, impedance also sort of measures the ‘opposition’ that a circuit presents to a current when a voltage is applied, but we have a complex ratio—literally: it’s a ratio with a magnitude and a direction, or a phase as it’s usually referred to. Hence, one will often write the impedance (denoted by Z) using Euler’s formula:

Z = |Z|eiθ

Now, if you don’t know anything about complex numbers, you should just skip all of what follows and go straight to the next section. However, if you do know what a complex number is (it’s an ‘arrow’, basically, and if θ is a variable, then it’s a rotating arrow, or a ‘stopwatch hand’, as Feynman calls it in his more popular Lectures on QED), then you may want to carry on reading.

The illustration below (credit goes to Wikipedia, once again) is, probably, the most generic view of an AC circuit that one can jot down. If we apply an alternating current, both the current as well as the voltage will go up and down. However, the current signal will lag the voltage signal, and the phase factor θ tells us by how much. Hence, using complex-number notation, we write:

V = IZ = I∗|Z|eiθ

General_AC_circuit

Now, while that resembles the V = R·I formula I mentioned when discussing resistance, you should note the bold-face type for V and I, and the ∗ symbol I am using here for multiplication. First the ∗ symbol: that’s a convention Feynman adopts in the above-mentioned popular account of quantum mechanics. I like it, because it makes it very clear we’re not talking a vector cross product A×B here, but a product of two complex numbers. Now, that’s also why I write V and I in bold-face: they have a phase too and, hence, we can write them as:

  • = |V|ei(ωt + θV)
  • = |I|ei(ωt + θI)

This works out as follows:

IZ = |I|ei(ωt + θI)∗|Z|eiθ = |I||Z|ei(ωt + θ+ θ) = |V|ei(ωt + θV)

Indeed, because the equation must hold for all t, we can equate the magnitudes and phases and, hence, we get: |V| = |I||Z| and θ= θI + θ. But voltage and current is something real, isn’t it? Not some complex number? You’re right. The complex notation is used mainly to simplify the calculus, but it’s only the real part of those complex-valued functions that count. [In any case, because we limit ourselves to complex exponentials here, the imaginary part (which is the sine, as opposed to the real part, which is the cosine) is the same as the real part, but with a lag of its own (π/2 or 90 degrees, to be precise). Indeed: when writing Euler’s formula out (eiθ = cos(θ) + isin(θ), you should always remember that the sine and cosine function are basically the same function: they differ only in the phase, as is evident from the trigonometric identity sin(θ+π/) = cos(θ).]

Now, that should be more than enough in terms of an introduction to the units used in electromagnetic theory. Hence, let’s move on.

The electric constant ε0

Let’s now look at  that energy density formula once again. When looking at that u = ε0E2/2 formula, you may think that its unit should be the square of the unit in which we measure field strength. How do we measure field strength? It’s defined as the force on a unit charge (E = F/q), so it should be newton per coulomb (N/C). Because the coulomb can also be expressed in newton·meter/volt (1 V = 1 J/C = N·m/C and, hence, 1 C = 1 N·m/V), we can express field strength not only in newton/coulomb but also in volt per meter: 1 N/C = 1 N·V/N·m = 1 V/m. How do we get from N2/C2 and/or V2/mto J/m3?

Well… Let me first note there’s no issue in terms of units with that ρΦ formula in the first integral for U: [ρ]·[Φ] = (C/m3)·V = [(N·m/V)/m3)·V = (N·m)/m3 = J/m3. No problem whatsoever. It’s only that second expression for U, with the u = ε0E2/2 in the integrand, that triggers the question. Here, we just need to accept that we need that εfactor to make the units come out alright. Indeed, just like other physical constants (such as c, G, or h, for example), it has a dimension: its unit is either C2/N·m2 or, what amounts to the same, C/V·m. So the units come out alright indeed if, and only if, we multiply the N2/C2 and/or V2/m2 units with the dimension of ε0:

  1. (N2/C2)·(C2/N·m2) = (N2·m)·(1/m3) = J/m3
  2. (V2/m2)·(C/V·m) = V·C/m3 = (V·N·m/V)/m= N·m/m3 = J/m3

Done! 

But so that’s the units only. The electric constant also has a numerical value:

ε0 = 8.854187817…×10−12 C/V·m ≈ 8.8542×10−12 C/V·m

This numerical value of εis as important as its unit to ensure both expressions for U yield the same result. Indeed, as you may or may not remember from the second of my two posts on vector calculus, if we have a curl-free field C (that means ×= 0 everywhere, which is the case when talking electrostatics only, as we are doing here), then we can always find some scalar field ψ such that C = ψ. But so here we have E = –ε0Φ, and so it’s not the minus sign that distinguishes the expression from the C = ψ expression, but the εfactor in front.

It’s just like the vector equation for heat flow: h = –κT. Indeed, we also have a constant of proportionality here, which is referred to as the thermal conductivity. Likewise, the electric constant εis also referred to as the permittivity of the vacuum (or of free space), for similar reasons obviously!

Natural units

You may wonder whether we can’t find some better units, so we don’t need the rather horrendous 8.8542×10−12 C/V·m factor (I am rounding to four digits after the decimal point). The answer is: yes, it’s possible. In fact, there are several systems in which the electric constant (and the magnetic constant, which we’ll introduce later) reduce to 1. The best-known are the so-called Gaussian and Lorentz-Heaviside units respectively.

Gauss defined the unit of charge in what is now referred to as the statcoulomb (statC), which is also referred to as the franklin (Fr) and/or the electrostatic unit of charge (esu), but I’ll refer you to the Wikipedia article on it in case you’d want to find out more about it. You should just note the definition of this unit is problematic in other ways. Indeed, it’s not so easy to try to define ‘natural units’ in physics, because there are quite a few ‘fundamental’ relations and/or laws in physics and, hence, equating this or that constant to one usually has implications on other constants. In addition, one should note that many choices that made sense as ‘natural’ units in the 19th century seem to be arbitrary now. For example:

  1. Why would we select the charge of the electron or the proton as the unit charge (+1 or –1) if we now assume that protons (and neutrons) consists of quarks, which have +2/3 or –1/3?
  2. What unit would we choose as the unit for mass, knowing that, despite all of the simplification that took place as a result of the generalized acceptance of the quark model, we’re still stuck with quite a few elementary particles whose mass would be a ‘candidate’ for the unit mass? Do we chose the electron, the u quark, or the d quark?

Therefore, the approach to ‘natural units’ has not been to redefine mass or charge or temperature, but the physical constants themselves. Obvious candidates are, of course, c and ħ, i.e. the speed of light and Planck’s constant. [You may wonder why physicists would select ħ, rather than h, as a ‘natural’ unit, but I’ll let you think about that. The answer is not so difficult.] That can be done without too much difficulty indeed, and so one can equate some more physical constants with one. The next candidate is the so-called Boltzmann constant (kB). While this constant is not so well known, it does pop up in a great many equations, including those that led Planck to propose his quantum of action, i.e.(see my post on Planck’s constant). When we do that – so when we equate c, ħ and kB with one (ħ = kB = 1), we still have a great many choices, so we need to impose further constraints. The next is to equate the gravitational constant with one, so then we have ħ = kB = G = 1.

Now, it turns out that the ‘solution’ of this ‘set’ of four equations (ħ = kB = G = 1) does, effectively, lead to ‘new’ values for most of our SI base units, most notably length, time, mass and temperature. These ‘new’ units are referred to as Planck units. You can look up their values yourself, and I’ll let you appreciate the ‘naturalness’ of the new units yourself. They are rather weird. The Planck length and time are usually referred to as the smallest possible measurable units of length and time and, hence, they are related to the so-called limits of quantum theory. Likewise, the Planck temperature is a related limit in quantum theory: it’s the largest possible measurable unit of temperature. To be frank, it’s hard to imagine what the scale of the Planck length, time and temperature really means. In contrast, the scale of the Planck mass is something we actually can imagine – it is said to correspond to the mass of an eyebrow hair, or a flea egg – but, again, its physical significance is not so obvious: Nature’s maximum allowed mass for point-like particles, or the mass capable of holding a single elementary charge. That triggers the question: do point-like charges really exist? I’ll come back to that question. But first I’ll conclude this little digression on units by introducing the so-called fine-structure constant, of which you’ve surely heard before.

The fine-structure constant

I wrote that the ‘set’ of equations ħ = kB = G = 1 gave us Planck units for most of our SI base units. It turns out that these four equations do not lead to a ‘natural’ unit for electric charge. We need to equate a fifth constant with one to get that. That fifth constant is Coulomb’s constant (often denoted as ke) and, yes, it’s the constant that appears in Coulomb’s Law indeed, as well as in some other pretty fundamental equations in electromagnetics, such as the field caused by a point charge q: E = q/4πε0r2. Hence, ke = 1/4πε0. So if we equate kwith one, then ε0 will, obviously, be equal to ε= 1/4π.

To make a long story short, adding this fifth equation to our set of five also gives us a Planck charge, and I’ll give you its value: it’s about 1.8755×10−18 C. As I mentioned that the elementary charge is 1 e ≈ 1.6022×10−19 C, it’s easy to that the Planck charge corresponds to some 11.7 times the charge of the proton. In fact, let’s be somewhat more precise and round, once again, to four digits after the decimal point: the qP/e ratio is about 11.7062. Conversely, we can also say that the elementary charge as expressed in Planck units, is about 1/11.7062 ≈ 0.08542455. In fact, we’ll use that ratio in a moment in some other calculation, so please jot it down.

0.08542455? That’s a bit of a weird number, isn’t it? You’re right. And trying to write it in terms of the charge of a u or d quark doesn’t make it any better. Also, note that the first four significant digits (8542) correspond to the first four significant digits after the decimal point of our εconstant. So what’s the physical significance here? Some other limit of quantum theory?

Frankly, I did not find anything on that, but the obvious thing to do is to relate is to what is referred to as the fine-structure constant, which is denoted by α. This physical constant is dimensionless, and can be defined in various ways, but all of them are some kind of ratio of a bunch of these physical constants we’ve been talking about:

Fine-structure constant formula

The only constants you have not seen before are μ0Rand, perhaps, ras well as m. However, these can be defined as a function of the constants that you did see before:

  1. The μ0 constant is the so-called magnetic constant. It’s something similar as ε0 and it’s referred to as the magnetic permeability of the vacuum. So it’s just like the (electric) permittivity of the vacuum (i.e. the electric constant ε0) and the only reason why you haven’t heard of this before is because we haven’t discussed magnetic fields so far. In any case, you know that the electric and magnetic force are part and parcel of the same phenomenon (i.e. the electromagnetic interaction between charged particles) and, hence, they are closely related. To be precise, μ= 1/ε0c2. That shows the first and second expression for α are, effectively, fully equivalent.
  2. Now, from the definition of ke = 1/4πε0, it’s easy to see how those two expressions are, in turn, equivalent with the third expression for α.
  3. The Rconstant is the so-called von Klitzing constant, but don’t worry about it: it’s, quite simply, equal to Rh/e2. Hene, substituting (and don’t forget that h = 2πħ) will demonstrate the equivalence of the fourth expression for α.
  4. Finally, the re factor is the classical electron radius, which is usually written as a function of me, i.e. the electron mass: re = e2/4πε0mec2. This very same equation implies that reme = e2/4πε0c2. So… Yes. It’s all the same really.

Let’s calculate its (rounded) value in the old units first, using the third expression:

  • The econstant is (roughly) equal to (1.6022×10–19 C)= 2.5670×10–38 C2. Coulomb’s constant k= 1/4πεis about 8.9876×10N·m2/C2. Hence, the numerator e2k≈ 23.0715×10–29 N·m2.
  • The (rounded) denominator is ħc = (1.05457×10–34 N·m·s)(2.998×108 m/s) = 3.162×10–26 N·m2.
  • Hence, we get α = kee2/ħc ≈ 7.297×10–3 = 0.007297.

Note that this number is, effectively, dimensionless. Now, the interesting thing is that if we calculate α using Planck units, we get an econstant that is (roughly) equal to 0.08542455= … 0.007297! Now, because all of the other constants are equal to 1 in Planck’s system of units, that’s equal to α itself. So… Yes ! The two values for α are one and the same in the two systems of units and, of course, as you might have guessed, the fine-structure constant is effectively dimensionless because it does not depend on our units of measurement. So what does it correspond to?

Now that would take me a very long time to explain, but let me try to summarize what it’s all about. In my post on quantum electrodynamics (QED) – so that’s the theory of light and matter basically and, most importantly, how they interact – I wrote about the three basic events in that theory, and how they are associated with a probability amplitude, so that’s a complex number, or an ‘arrow’, as Feynman puts it: something with (a) a magnitude and (b) a direction. We had to take the absolute square of these amplitudes in order to calculate the probability (i.e. some real number between 0 and 1) of the event actually happening. These three basic events or actions were:

  1. A photon travels from point A to B. To keep things simple and stupid, Feynman denoted this amplitude by P(A to B), and please note that the P stands for photon, not for probability. I should also note that we have an easy formula for P(A to B): it depends on the so-called space-time interval between the two points A and B, i.e. I = Δr– Δt= (x2–x1)2+(y2–y1)2+(z2–z1)– (t2–t1)2. Hence, the space-time interval takes both the distance in space as well as the ‘distance’ in time into account.
  2. An electron travels from point A to B: this was denoted by E(A to B) because… Well… You guessed it: the of electron. The formula for E(A to B) was much more complicated, but the two key elements in the formula was some complex number j (see below), and some other (real) number n.
  3. Finally, an electron could emit or absorb a photon, and the amplitude associated with this event was denoted by j, for junction.

Now, that junction number j is about –0.1. To be somewhat more precise, I should say it’s about –0.08542455.

–0.08542455? That’s a bit of a weird number, isn’t it? Hey ! Didn’t we see this number somewhere else? We did, but before you scroll up, let’s first interpret this number. It looks like an ordinary (real) number, but it’s an amplitude alright, so you should interpret it as an arrow. Hence, it can be ‘combined’ (i.e. ‘added’ or ‘multiplied’) with other arrows. More in particular, when you multiply it with another arrow, it amounts to a shrink to a bit less than one-tenth (because its magnitude is about 0.085 = 8.5%), and half a turn (the minus sign amounts to a rotation of 180°). Now, in that post of mine, I wrote that I wouldn’t entertain you on the difficulties of calculating this number but… Well… We did see this number before indeed. Just scroll up to check it. We’ve got a very remarkable result here:

j ≈ –0.08542455 = –√0.007297 = –√α = –e expressed in Planck units

So we find that our junction number j or – as it’s better known – our coupling constant in quantum electrodynamics (aka as the gauge coupling parameter g) is equal to the (negative) square root of that fine-structure constant which, in turn, is equal to the charge of the electron expressed in the Planck unit for electric charge. Now that is a very deep and fundamental result which no one seems to be able to ‘explain’—in an ‘intuitive’ way, that is.

I should immediately add that, while we can’t explain it, intuitively, it does make sense. A lot of sense actually. Photons carry the electromagnetic force, and the electromagnetic field is caused by stationary and moving electric charges, so one would expect to find some relation between that junction number j, describing the amplitude to emit or absorb a photon, and the electric charge itself, but… An equality? Really?

Well… Yes. That’s what it is, and I look forward to trying to understand all of this better. For now, however, I should proceed with what I set out to do, and that is to tie up a few loose ends. This was one, and so let’s move to the next, which is about the assumption of point charges.

Note: More popular accounts of quantum theory say α itself is ‘the’ coupling constant, rather than its (negative) square –√α = j = –e (expressed in Planck units). That’s correct: g or j are, technically speaking, the (gauge) coupling parameter, not the coupling constant. But that’s a little technical detail which shouldn’t bother you. The result is still what it is: very remarkable! I should also note that it’s often the value of the reciprocal (1/α) that is specified, i.e. 1/0.007297 ≈ 137.036. But so now you know what this number actually stands for. 🙂

Do point charges exist?

Feynman’s Lectures on electrostatics are interesting, among other things, because, besides highlighting the precision and successes of the theory, he also doesn’t hesitate to point out the contradictions. He notes, for example, that “the idea of locating energy in the field is inconsistent with the assumption of the existence of point charges.”

Huh?

Yes. Let’s explore the point. We do assume point charges in classical physics indeed. The electric field caused by a point charge is, quite simply:

E = q/4πε0r2

Hence, the energy density u is ε0E2/2 = q2/32πε0r4. Now, we have that volume integral U = (ε0/2)∫EEdV = ∫(ε0E2/2)dV integral. As Feynman notes, nothing prevents us from taking a spherical shell for the volume element dV, instead of an infinitesimal cube. This spherical shell would have the charge q at its center, an inner radius equal to r, an infinitesimal thickness dr, and, finally, a surface area 4πr(that’s just the general formula for the surface area of a spherical shell, which I also noted above). Hence, its (infinitesimally small) volume is 4πr2dr, and our integral becomes:

Integral

To calculate this integral, we need to take the limit of –q2/8πε0r for (a) r tending to zero (r→0) and for (b) r tending to infinity (r→∞). The limit for r = ∞ is zero. That’s OK and consistent with the choice of our reference point for calculating the potential of a field. However, the limit for r = 0 is zero is infinity! Hence, that U = (ε0/2)∫EEdV basically says there’s an infinite amount of energy in the field of a point charge! How is that possible? It cannot be true, obviously.

So… Where did we do wrong?

Your first reaction may well be that this very particular approach (i.e. replacing our infinitesimal cubes by infinitesimal shells) to calculating our integral is fishy and, hence, not allowed. Maybe you’re right. Maybe not. It’s interesting to note that we run into similar problems when calculating the energy of a charged sphere. Indeed, we mentioned the formula for the capacity of a charged sphere: C = 4πε0r. Now, there’s a similarly easy formula for the energy of a charged sphere. Let’s look at how we charge a condenser:

  • We know that the potential difference between two plates of a condenser represents the work we have to do, per unit charge, to transfer a charge (Q) from one plate to the other. Hence, we can write V = ΔU/ΔQ.
  • We will, of course, want to do a differential analysis. Hence, we’ll transfer charges incrementally, one infinitesimal little charge dQ at the time, and re-write V as V = dU/dQ or, what amounts to the same: dU = V·dQ.
  • Now, we’ve defined the capacitance of a condenser as C = Q/V. [Again, don’t be confused: C stands for capacity here, measured in coulomb per volt, not for the coulomb unit.] Hence, we can re-write dU as dU = Q·dQ/C.
  • Now we have to integrate dU going from zero charge to the final charge Q. Just do a little bit of effort here and try it. You should get the following result: U = Q2/2C. [We could re-write this as U = (C2/V2)/2C =  C·V2/2, which is a form that may be more useful in some other context but not here.]
  • Using that C = 4πε0r formula, we get our grand result. The energy of a charged sphere is:

U = Q2/8πε0r

From that formula, it’s obvious that, if the radius of our sphere goes to zero, its energy should also go to infinity! So it seems we can’t really pack a finite charge Q in one single point. Indeed, to do that, our formula says we need an infinite amount of energy. So what’s going on here?

Nothing much. You should, first of all, remember how we got that integral: see my previous post for the full derivation indeed. It’s not that difficult. We first assumed we had pairs of charges qi and qfor which we calculated the total electrostatic energy U as the sum of the energies of all possible pairs of charges:

U 3

And, then, we looked at a continuous distribution of charge. However, in essence, we still did the same: we counted the energy of interaction between infinitesimal charges situated at two different points (referred to as point 1 and 2 respectively), with a 1/2 factor in front so as to ensure we didn’t double-count (there’s no way to write an integral that keeps track of the pairs so that each pair is counted only once):

U 4

Now, we reduced this double integral by a clever substitution to something that looked a bit better:

U 6

Finally, some more mathematical tricks gave us that U = (ε0/2)∫EEdV integral.

In essence, what’s wrong in that integral above is that it actually includes the energy that’s needed to assemble the finite point charge q itself from an infinite number of infinitesimal parts. Now that energy is infinitely large. We just can’t do it: the energy required to construct a point charge is ∞.

Now that explains the physical significance of that Planck mass ! We said Nature has some kind of maximum allowable mass for point-like particles, or the mass capable of holding a single elementary charge. What’s going on is, as we try to pile more charge on top of the charge that’s already there, we add energy. Now, energy has an equivalent mass. Indeed, the Planck charge (q≈ 1.8755×10−18 C), the Planck length (l= 1.616×10−35 m), the Planck energy (1.956×109 J), and the Planck mass (2.1765×10−8 kg) are all related. Now things start making sense. Indeed, we said that the Planck mass is tiny but, still, it’s something we can imagine, like a flea’s egg or the mass of a hair of a eyebrow. The associated energy (E = mc2, so that’s (2.1765×10−8 kg)·(2.998×108 m/s)2 ≈ 19.56×108 kg·m2/s= 1.956×109 joule indeed.

Now, how much energy is that? Well… That’s about 2 giga-joule, obviously, but so what’s that in daily life? It’s about the energy you would get when burning 40 liter of fuel. It’s also likely to amount, more or less, to your home electricity consumption over a month. So it’s sizable, and so we’re packing all that energy into a Planck volume (lP≈ 4×10−105 m3). If we’d manage that, we’d be able to create tiny black holes, because that’s what that little Planck volume would become if we’d pack so much energy in it. So… Well… Here I just have to refer you to more learned writers than I am. As Wikipedia notes dryly: “The physical significance of the Planck length is a topic of theoretical research. Since the Planck length is so many orders of magnitude smaller than any current instrument could possibly measure, there is no way of examining it directly.”

So… Well… That’s it for now. The point to note is that we would not have any theoretical problems if we’d assume our ‘point charge’ is actually not a point charge but some small distribution of charge itself. You’ll say: Great! Problem solved! 

Well… For now, yes. But Feynman rightly notes that assuming that our elementary charges do take up some space results in other difficulties of explanation. As we know, these difficulties are solved in quantum mechanics, but so we’re not supposed to know that when doing these classical analyses. 🙂