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

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

Original post:

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

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

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

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

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

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

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

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

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

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

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

dx_μ/ds = u_μ = p_μ/m₀

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

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

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

The four-force vector f_μ

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

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

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

The force and the tensor

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

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

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

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

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

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

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

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

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

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

Relativistic transformations of fields and the electromagnetic tensor

Original post:

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

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

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

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

The invariance of physics and the use of vector equations

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

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

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

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

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

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

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

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

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

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

The components of L are:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Electrodynamics in relativistic notation

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

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

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

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

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

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

Four-vectors and invariance under Lorentz transformations

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

The electromagnetic tensor

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

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

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

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

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

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

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

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

F_ij = ∇_iA_j − ∇_jA_i

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

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

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

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

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

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

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

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

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

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

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

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

The Lorentz transformation of the electric and magnetic fields

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

The components of L are:

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

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

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

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

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

So we get:

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

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

So… Done!

So what?

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

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

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

So that gives us four equations only:

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

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

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

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

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

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

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

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

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

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

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

Z = V/I ⇔ V = I∗Z

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

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

we preferred the complex or vector notation, writing:

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

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

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

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

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

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

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

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

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

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

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

This law is written and illustrated as follows:

KCR

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

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

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

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

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

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

The related formulas are:

I = Ɛ/Z_eff
V_n= Ɛ_eff− I_n∗Z_eff

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

Z = R + i·X

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

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

Ɛ = I∗Z = I∗(R + i·X)

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

Ɛ = I∗Z = I₀·eⁱ^(ωt)∗(R + i·X)

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

Ɛ = I₀·R·cos(ωt) − I₀·X·sin(ωt)

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

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

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

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

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

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

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

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

Done! 🙂

Electric circuits (1): the circuit elements

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

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

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

The impedance concept

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

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

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

A = A(φ) = A₀·cos(φ) = A₀·cos(ωt + θ)

As usual, Wikipedia has a nice animation for it:

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

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

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

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

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

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

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

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

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

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

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

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

V = I∗Z = I∗|Z|eⁱ^θ

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

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

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

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

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

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

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

Done!

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

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

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

Resistors

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

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

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

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

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

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

Z = R = V/I = V₀/I₀

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

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

Capacitors (condensers)

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

C = Q/V

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

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

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

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

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

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

I = dQ/dt = d(C·V)/dt = C·dV/dt

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

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

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

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

I = C·dV/dt ⇔ I = C·dV/dt

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

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

I = C·dV/dt = C·(iω)·V

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

Z = V/I = 1/(iωC)

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

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

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

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

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

V = V₀·cos(ωt)
I = I₀·cos(ωt + π/2)

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

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

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

Inductors

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

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

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

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

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

V = V₀·cos(ωt)
I = I₀·cos(ωt − π/2)

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

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

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

L = −Ɛ/(dI/dt)

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

V = −Ɛ = L·(dI/dt)

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

V = L·dI/dt⇔ V = L·dI/dt

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

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

So we get what we want and have to get:

V = L·dI/dt = iωL·I

Now, Z = V/I, so Z = iωL indeed!

Summary of conclusions

Let’s summarize what we found:

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

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

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

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

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

Addendum: Why is V = − Ɛ?

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

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

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

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

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

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

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

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

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

So this change in the magnetic flux is what it causing the ‘back emf’, and so the integral on the left is, effectively, equal to Ɛ, not minus Ɛ but +Ɛ. Now, the second integral is equal to V, because that’s the voltage V between the two terminals a and b. So the whole integral is equal to 0 = Ɛ + V and, therefore, we have that:

V = − Ɛ = L·dI/dt

Re-visiting the speed of light, Planck’s constant, and the fine-structure constant

Note: I have published a paper that is very coherent and fully explains what the fine-structure constant actually is. There is nothing magical about it. It’s not some God-given number. It’s a scaling constant – and then some more. But not God-given. Check it out: The Meaning of the Fine-Structure Constant. No ambiguity. No hocus-pocus.

Jean Louis Van Belle, 23 December 2018

Original post:

A brother of mine sent me a link to an article he liked. Now, because we share some interest in physics and math and other stuff, I looked at it and…

Well… I was disappointed. Despite the impressive credentials of its author – a retired physics professor – it was very poorly written. It made me realize how much badly written stuff is around, and I am glad I am no longer wasting my time on it. However, I do owe my brother some explanation of (a) why I think it was bad, and of (b) what, in my humble opinion, he should be wasting his time on. 🙂 So what it is all about?

The article talks about physicists deriving the speed of light from “the electromagnetic properties of the quantum vacuum.” Now, it’s the term ‘quantum‘, in ‘quantum vacuum’, that made me read the article.

Indeed, deriving the theoretical speed of light in empty space from the properties of the classical vacuum – aka empty space – is a piece of cake: it was done by Maxwell himself as he was figuring out his equations back in the 1850s (see my post on Maxwell’s equations and the speed of light). And then he compared it to the measured value, and he saw it was right on the mark. Therefore, saying that the speed of light is a property of the vacuum, or of empty space, is like a tautology: we may just as well put it the other way around, and say that it’s the speed of light that defines the (properties of the) vacuum!

Indeed, as I’ll explain in a moment: the speed of light determines both the electric as well as the magnetic constants μ₀and ε₀, which are the (magnetic) permeability and the (electric) permittivity of the vacuum respectively. Both constants depend on the units we are working with (i.e. the units for electric charge, for distance, for time and for force – or for inertia, if you want, because force is defined in terms of overcoming inertia), but so they are just proportionality coefficients in Maxwell’s equations. So once we decide what units to use in Maxwell’s equations, then μ₀and ε₀ are just proportionality coefficients which we get from c. So they are not separate constants really – I mean, they are not separate from c – and all of the ‘properties’ of the vacuum, including these constants, are in Maxwell’s equations.

In fact, when Maxwell compared the theoretical value of c with its presumed actual value, he didn’t compare c‘s theoretical value with the speed of light as measured by astronomers (like that 17th century Ole Roemer, to which our professor refers: he had a first go at it by suggesting some specific value for it based on his observations of the timing of the eclipses of one of Jupiter’s moons), but with c‘s value as calculated from the experimental values of μ₀and ε₀! So he knew very well what he was looking at. In fact, to drive home the point, it may also be useful to note that the Michelson-Morley experiment – which accurately measured the speed of light – was done some thirty years later. So Maxwell had already left this world by then—very much in peace, because he had solved the mystery all 19th century physicists wanted to solve through his great unification: his set of equations covers it all, indeed: electricity, magnetism, light, and even relativity!

I think the article my brother liked so much does a very lousy job in pointing all of that out, but that’s not why I wouldn’t recommend it. It got my attention because I wondered why one would try to derive the speed of light from the properties of the quantum vacuum. In fact, to be precise, I hoped the article would tell me what the quantum vacuum actually is. Indeed, as far as I know, there’s only one vacuum—one ’empty space’: empty is empty, isn’t it? 🙂 So I wondered: do we have a ‘quantum’ vacuum? And, if so, what is it, really?

Now, that is where the article is really disappointing, I think. The professor drops a few names (like the Max Planck Institute, the University of Paris-Sud, etcetera), and then, promisingly, mentions ‘fleeting excitations of the quantum vacuum’ and ‘virtual pairs of particles’, but then he basically stops talking about quantum physics. Instead, he wanders off to share some philosophical thoughts on the fundamental physical constants. What makes it all worse is that even those thoughts on the ‘essential’ constants are quite off the mark.

So… This post is just a ‘quick and dirty’ thing for my brother which, I hope, will be somewhat more thought-provoking than that article. More importantly, I hope that my thoughts will encourage him to try to grind through better stuff.

On Maxwell’s equations and the properties of empty space

Let me first say something about the speed of light indeed. Maxwell’s four equations may look fairly simple, but that’s only until one starts unpacking all those differential vector equations, and it’s only when going through all of their consequences that one starts appreciating their deep mathematical structure. Let me quickly copy how another blogger jotted them down: 🙂

As I showed in my above-mentioned post, the speed of light (i.e. the speed with which an electromagnetic pulse or wave travels through space) is just one of the many consequences of the mathematical structure of Maxwell’s set of equations. As such, the speed of light is a direct consequence of the ‘condition’, or the properties, of the vacuum indeed, as Maxwell suggested when he wrote that “we can scarcely avoid the inference that light consists in the transverse undulations of the same medium which is the cause of electric and magnetic phenomena”.

Of course, while Maxwell still suggests light needs some ‘medium’ here – so that’s a reference to the infamous aether theory – we now know that’s because he was a 19th century scientist, and so we’ve done away with the aether concept (because it’s a redundant hypothesis), and so now we also know there’s absolutely no reason whatsoever to try to “avoid the inference.” 🙂 It’s all OK, indeed: light is some kind of “transverse undulation” of… Well… Of what?

We analyze light as traveling fields, represented by two vectors, E and B, whose direction and magnitude varies both in space as well as in time. E and B are field vectors, and represent the electric and magnetic field respectively. An equivalent formulation – more or less, that is (see my post on the Liénard-Wiechert potentials) – for Maxwell’s equations when only one (moving) charge is involved is:

This re-formulation, which is Feynman’s preferred formula for electromagnetic radiation, is interesting in a number of ways. It clearly shows that, while we analyze the electric and magnetic field as separate mathematical entities, they’re one and the same phenomenon really, as evidenced by the B = –e_r‘×E/c equation, which tells us the magnetic field from a single moving charge is always normal (i.e. perpendicular) to the electric field vector, and also that B‘s magnitude is 1/c times the magnitude of E, so |B| = B = |E|/c = E/c. In short, B is fully determined by E, or vice versa: if we have one of the two fields, we have the other, so they’re ‘one and the same thing’ really—not in a mathematical sense, but in a real sense.

Also note that E and B‘s magnitude is just the same if we’re using natural units, so if we equate c with 1. Finally, as I pointed out in my post on the relativity of electromagnetic fields, if we would switch from one reference frame to another, we’ll have a different mix of E and B, but that different mix obviously describes the same physical reality. More in particular, if we’d be moving with the charges, the magnetic field sort of disappears to re-appear as an electric field. So the Lorentz force F = F_electric + F_magnetic= qE + qv×B is one force really, and its ‘electric’ and ‘magnetic’ component appear the way they appear in our reference frame only. In some other reference frame, we’d have the same force, but its components would look different, even if they, obviously, would and should add up to the same. [Well… Yes and no… You know there’s relativistic corrections to be made to the forces to, but that’s a minor point, really. The force surely doesn’t disappear!]

All of this reinforces what you know already: electricity and magnetism are part and parcel of one and the same phenomenon, the electromagnetic force field, and Maxwell’s equations are the most elegant way of ‘cutting it up’. Why elegant? Well… Click the Occam tab. 🙂

Now, after having praised Maxwell once more, I must say that Feynman’s equations above have another advantage. In Maxwell’s equations, we see two constants, the electric and magnetic constant (denoted by μ₀and ε₀ respectively), and Maxwell’s equations imply that the product of the electric and magnetic constant is the reciprocal of c²: μ₀·ε₀= 1/c². So here we see ε₀and c only, so no μ₀, so that makes it even more obvious that the magnetic and electric constant are related one to another through c.

[…] Let me digress briefly: why do we have c² in μ₀·ε₀= 1/c², instead of just c? That’s related to the relativistic nature of the magnetic force: think about that B = E/c relation. Or, better still, think about the Lorentz equation F = F_electric + F_magnetic= qE + qv×B = q[E + (v/c)×(E×e_r‘)]: the 1/c factor is there because the magnetic force involves some velocity, and any velocity is always relative—and here I don’t mean relative to the frame of reference but relative to the (absolute) speed of light! Indeed, it’s the v/c ratio (usually denoted by β = v/c) that enters all relativistic formulas. So the left-hand side of the μ₀·ε₀= 1/c² equation is best written as (1/c)·(1/c), with one of the two 1/c factors accounting for the fact that the ‘magnetic’ force is a relativistic effect of the ‘electric’ force, really, and the other 1/c factor giving us the proper relationship between the magnetic and the electric constant. To drive home the point, I invite you to think about the following:

μ₀ is expressed in (V·s)/(A·m), while ε₀is expressed in (A·s)/(V·m), so the dimension in which the μ₀·ε₀product is expressed is [(V·s)/(A·m)]·[(A·s)/(V·m)] = s²/m², so that’s the dimension of 1/c².
Now, this dimensional analysis makes it clear that we can sort of distribute 1/c² over the two constants. All it takes is re-defining the fundamental units we use to calculate stuff, i.e. the units for electric charge, for distance, for time and for force – or for inertia, as explained above. But so we could, if we wanted, equate both μ₀ as well as ε₀with 1/c.
Now, if we would then equate c with 1, we’d have μ₀ = ε₀= c = 1. We’d have to define our units for electric charge, for distance, for time and for force accordingly, but it could be done, and then we could re-write Maxwell’s set of equations using these ‘natural’ units.

In any case, the nitty-gritty here is less important: the point is that μ₀and ε₀are also related through the speed of light and, hence, they are ‘properties’ of the vacuum as well. [I may add that this is quite obvious if you look at their definition, but we’re approaching the matter from another angle here.]

In any case, we’re done with this. On to the next!

On quantum oscillations, Planck’s constant, and Planck units

The second thought I want to develop is about the mentioned quantum oscillation. What is it? Or what could it be? An electromagnetic wave is caused by a moving electric charge. What kind of movement? Whatever: the charge could move up or down, or it could just spin around some axis—whatever, really. For example, if it spins around some axis, it will have a magnetic moment and, hence, the field is essentially magnetic, but then, again, E and B are related and so it doesn’t really matter if the first cause is magnetic or electric: that’s just our way of looking at the world: in another reference frame, one that’s moving with the charges, the field would essential be electric. So the motion can be anything: linear, rotational, or non-linear in some irregular way. It doesn’t matter: any motion can always be analyzed as the sum of a number of ‘ideal’ motions. So let’s assume we have some elementary charge in space, and it moves and so it emits some electromagnetic radiation.

So now we need to think about that oscillation. The key question is: how small can it be? Indeed, in one of my previous posts, I tried to explain some of the thinking behind the idea of the ‘Great Desert’, as physicists call it. The whole idea is based on our thinking about the limit: what is the smallest wavelength that still makes sense? So let’s pick up that conversation once again.

The Great Desert lies between the 10³² and 10⁴³Hz scale. 10³² Hz corresponds to a photon energy of E_γ = h·f = (4×10⁻¹⁵ eV·s)·(10³² Hz) = 4×10¹⁷ eV = 400,000 tera-electronvolt (1 TeV = 10¹²eV). I use the γ (gamma) subscript in my E_γ symbol for two reasons: (1) to make it clear that I am not talking the electric field E here but energy, and (2) to make it clear we are talking ultra-high-energy gamma-rays here.

In fact, γ-rays of this frequency and energy are theoretical only. Ultra-high-energy gamma-rays are defined as rays with photon energies higher than 100 TeV, which is the upper limit for very-high-energy gamma-rays, which have been observed as part of the radiation emitted by so-called gamma-ray bursts (GRBs): flashes associated with extremely energetic explosions in distant galaxies. Wikipedia refers to them as the ‘brightest’ electromagnetic events know to occur in the Universe. These rays are not to be confused with cosmic rays, which consist of high-energy protons and atomic nuclei stripped of their electron shells. Cosmic rays aren’t rays really and, because they consist of particles with a considerable rest mass, their energy is even higher. The so-called Oh-My-God particle, for example, which is the most energetic particle ever detected, had an energy of 3×10²⁰ eV, i.e. 300 million TeV. But it’s not a photon: its energy is largely kinetic energy, with the rest mass m₀ counting for a lot in the m in the E = m·c² formula. To be precise: the mentioned particle was thought to be an iron nucleus, and it packed the equivalent energy of a baseball traveling at 100 km/h!

But let me refer you to another source for a good discussion on these high-energy particles, so I can get get back to the energy of electromagnetic radiation. When I talked about the Great Desert in that post, I did so using the Planck-Einstein relation (E = h·f), which embodies the idea of the photon being valid always and everywhere and, importantly, at every scale. I also discussed the Great Desert using real-life light being emitted by real-life atomic oscillators. Hence, I may have given the (wrong) impression that the idea of a photon as a ‘wave train’ is inextricably linked with these real-life atomic oscillators, i.e. to electrons going from one energy level to the next in some atom. Let’s explore these assumptions somewhat more.

Let’s start with the second point. Electromagnetic radiation is emitted by any accelerating electric charge, so the atomic oscillator model is an assumption that should not be essential. And it isn’t. For example, whatever is left of the nucleus after alpha or beta decay (i.e. a nuclear decay process resulting in the emission of an α- or β-particle) it likely to be in an excited state, and likely to emit a gamma-ray for about 10⁻¹² seconds, so that’s a burst that’s about 10,000 times shorter than the 10^–8seconds it takes for the energy of a radiating atom to die out. [As for the calculation of that 10^–8sec decay time – so that’s like 10 nanoseconds – I’ve talked about this before but it’s probably better to refer you to the source, i.e. one of Feynman’s Lectures.]

However, what we’re interested in is not the energy of the photon, but the energy of one cycle. In other words, we’re not thinking of the photon as some wave train here, but what we’re thinking about is the energy that’s packed into a space corresponding to one wavelength. What can we say about that?

As you know, that energy will depend both on the amplitude of the electromagnetic wave as well as its frequency. To be precise, the energy is (1) proportional to the square of the amplitude, and (2) proportional to the frequency. Let’s look at the first proportionality relation. It can be written in a number of ways, but one way of doing it is stating the following: if we know the electric field, then the amount of energy that passes per square meter per second through a surface that is normal to the direction in which the radiation is going (which we’ll denote by S – the s from surface – in the formula below), must be proportional to the average of the square of the field. So we write S ∝ 〈E²〉, and so we should think about the constant of proportionality now. Now, let’s not get into the nitty-gritty, and so I’ll just refer to Feynman for the derivation of the formula below:

S = ε₀c·〈E²〉

So the constant of proportionality is ε₀c. [Note that, in light of what we wrote above, we can also write this as S = (1/μ₀·c)·〈(c·B)²〉 = (c/μ₀)·〈B²〉, so that underlines once again that we’re talking one electromagnetic phenomenon only really.] So that’s a nice and rather intuitive result in light of all of the other formulas we’ve been jotting down. However, it is a ‘wave’ perspective. The ‘photon’ perspective assumes that, somehow, the amplitude is given and, therefore, the Planck-Einstein relation only captures the frequency variable: E_γ = h·f.

Indeed, ‘more energy’ in the ‘wave’ perspective basically means ‘more photons’, but photons are photons: they have a definite frequency and a definite energy, and both are given by that Planck-Einstein relation. So let’s look at that relation by doing a bit of dimensional analysis:

Energy is measured in electronvolt or, using SI units, joule: 1 eV ≈ 1.6×10⁻¹⁹J. Energy is force times distance: 1 joule = 1 newton·meter, which means that a larger force over a shorter distance yields the same energy as a smaller force over a longer distance. The oscillations we’re talking about here involve very tiny distances obviously. But the principle is the same: we’re talking some moving charge q, and the power – which is the time rate of change of the energy – that goes in or out at any point of time is equal to dW/dt = F·v, with W the work that’s being done by the charge as it emits radiation.
I would also like to add that, as you know, forces are related to the inertia of things. Newton’s Law basically defines a force as that what causes a mass to accelerate: F = m·a = m·(dv/dt) = d(m·v)/dt = dp/dt, with p the momentum of the object that’s involved. When charges are involved, we’ve got the same thing: a potential difference will cause some current to change, and one of the equivalents of Newton’s Law F = m·a = m·(dv/dt) in electromagnetism is V = L·(dI/dt). [I am just saying this so you get a better ‘feel’ for what’s going on.]
Planck’s constant is measured in electronvolt·seconds (eV·s) or in, using SI units, in joule·seconds (J·s), so its dimension is that of (physical) action, which is energy times time: [energy]·[time]. Again, a lot of energy during a short time yields the same energy as less energy over a longer time. [Again, I am just saying this so you get a better ‘feel’ for these dimensions.]
The frequency f is the number of cycles per time unit, so that’s expressed per second, i.e. in herz (Hz) = 1/second = s⁻¹.

So… Well… It all makes sense: [x joule] = [6.626×10⁻³⁴ joule]·[1 second]×[f cycles]/[1 second]. But let’s try to deepen our understanding even more: what’s the Planck-Einstein relation really about?

To answer that question, let’s think some more about the wave function. As you know, it’s customary to express the frequency as an angular frequency ω, as used in the wave function A(x, t) = A₀·sin(kx − ωt). The angular frequency is the frequency expressed in radians per second. That’s because we need an angle in our wave function, and so we need to relate x and t to some angle. The way to think about this is as follows: one cycle takes a time T (i.e. the period of the wave) which is equal to T = 1/f. Yes: one second divided by the number of cycles per second gives you the time that’s needed for one cycle. One cycle is also equivalent to our argument ωt going around the full circle (i.e. 2π), so we write: ω·T = 2π and, therefore:

ω = 2π/T = 2π·f

Now we’re ready to play with the Planck-Einstein relation. We know it gives us the energy of one photon really, but what if we re-write our equation E_γ = h·f as E_γ/f = h? The dimensions in this equation are:

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

⇔ x = 6.626×10⁻³⁴ joule per cycle

So that means that the energy per cycle is equal to 6.626×10⁻³⁴ joule, i.e. the value of Planck’s constant.

Let me rephrase truly amazing result, so you appreciate it—perhaps: regardless of the frequency of the light (or our electromagnetic wave, in general) involved, the energy per cycle, i.e. per wavelength or per period, is always equal to 6.626×10⁻³⁴ joule or, using the electronvolt as the unit, 4.135667662×10⁻¹⁵ eV. So, in case you wondered, that is the true meaning of Planck’s constant!

Now, if we have the frequency f, we also have the wavelength λ, because the velocity of the wave is the frequency times the wavelength: c = λ·f and, therefore, λ = c/f. So if we increase the frequency, the wavelength becomes smaller and smaller, and so we’re packing the same amount of energy – admittedly, 4.135667662×10⁻¹⁵ eV is a very tiny amount of energy – into a space that becomes smaller and smaller. Well… What’s tiny, and what’s small? All is relative, of course. 🙂 So that’s where the Planck scale comes in. If we pack that amount of energy into some tiny little space of the Planck dimension, i.e. a ‘length’ of 1.6162×10⁻³⁵ m, then it becomes a tiny black hole, and it’s hard to think about how that would work.

[…] Let me make a small digression here. I said it’s hard to think about black holes but, of course, it’s not because it’s ‘hard’ that we shouldn’t try it. So let me just mention a few basic facts. For starters, black holes do emit radiation! So they swallow stuff, but they also spit stuff out. More in particular, there is the so-called Hawking radiation, as Roger Penrose and Stephen Hawking discovered.

Let me quickly make a few remarks on that: Hawking radiation is basically a form of blackbody radiation, so all frequencies are there, as shown below: the distribution of the various frequencies depends on the temperature of the black body, i.e. the black hole in this case. [The black curve is the curve that Lord Rayleigh and Sir James Jeans derived in the late 19th century, using classical theory only, so that’s the one that does not correspond to experimental fact, and which led Max Planck to become the ‘reluctant’ father of quantum mechanics. In any case, that’s history and so I shouldn’t dwell on this.]

The interesting thing about blackbody radiation, including Hawking radiation, is that it reduces energy and, hence, the equivalent mass of our blackbody. So Hawking radiation reduces the mass and energy of black holes and is therefore also known as black hole evaporation. So black holes that lose more mass than they gain through other means are expected to shrink and ultimately vanish. Therefore, there’s all kind of theories that say why micro black holes, like that Planck scale black hole we’re thinking of right now, should be much larger net emitters of radiation than large black holes and, hence, whey they should shrink and dissipate faster.

Hmm… Interesting… What do we do with all of this information? Well… Let’s think about it as we continue our trek on this long journey to reality over the next year or, more probably, years (plural). 🙂

The key lesson here is that space and time are intimately related because of the idea of movement, i.e. the idea of something having some velocity, and that it’s not so easy to separate the dimensions of time and distance in any hard and fast way. As energy scales become larger and, therefore, our natural time and distance units become smaller and smaller, it’s the energy concept that comes to the fore. It sort of ‘swallows’ all other dimensions, and it does lead to limiting situations which are hard to imagine. Of course, that just underscores the underlying unity of Nature, and the mysteries involved.

So… To relate all of this back to the story that our professor is trying to tell, it’s a simple story really. He’s talking about two fundamental constants basically, c and h, pointing out that c is a property of empty space, and h is related to something doing something. Well… OK. That’s really nothing new, and surely not ground-breaking research. 🙂

Now, let me finish my thoughts on all of the above by making one more remark. If you’ve read a thing or two about this – which you surely have – you’ll probably say: this is not how people usually explain it. That’s true, they don’t. Anything I’ve seen about this just associates the 10⁴³Hz scale with the 10²⁸eV energy scale, using the same Planck-Einstein relation. For example, the Wikipedia article on micro black holes writes that “the minimum energy of a microscopic black hole is 10¹⁹ GeV [i.e. 10²⁸eV], which would have to be condensed into a region on the order of the Planck length.” So that’s wrong. I want to emphasize this point because I’ve been led astray by it for years. It’s not the total photon energy, but the energy per cycle that counts. Having said that, it is correct, however, and easy to verify, that the 10⁴³Hz scale corresponds to a wavelength of the Planck scale: λ = c/f = (3×10⁸m/s)/(10⁴³ s⁻¹) = 3×10⁻³⁵m. The confusion between the photon energy and the energy per wavelength arises because of the idea of a photon: it travels at the speed of light and, hence, because of the relativistic length contraction effect, it is said to be point-like, to have no dimension whatsoever. So that’s why we think of packing all of its energy in some infinitesimally small place. But you shouldn’t think like that. The photon is dimensionless in our reference frame: in its own ‘world’, it is spread out, so it is a wave train. And it’s in its ‘own world’ that the contradictions start… 🙂

OK. Done!

My third and final point is about what our professor writes on the fundamental physical constants, and more in particular on what he writes on the fine-structure constant. In fact, I could just refer you to my own post on it, but that’s probably a bit too easy for me and a bit difficult for you 🙂 so let me summarize that post and tell you what you need to know about it.

The fine-structure constant

The fine-structure constant α is a dimensionless constant which also illustrates the underlying unity of Nature, but in a way that’s much more fascinating than the two or three things the professor mentions. Indeed, it’s quite incredible how this number (α = 0.00729735…, but you’ll usually see it written as its reciprocal, which is a number that’s close to 137.036…) links charge with the relative speeds, radii, and the mass of fundamental particles and, therefore, how this number also these concepts with each other. And, yes, the fact that it is, effectively, dimensionless, unlike h or c, makes it even more special. Let me quickly sum up what the very same number α all stands for:

(1) α is the square of the electron charge expressed in Planck units: α = e_P².

(2) α is the square root of the ratio of (a) the classical electron radius and (b) the Bohr radius: α = √(r_e /r). You’ll see this more often written as r_e = α²r. Also note that this is an equation that does not depend on the units, in contrast to equation 1 (above), and 4 and 5 (below), which require you to switch to Planck units. It’s the square of a ratio and, hence, the units don’t matter. They fall away.

(3) α is the (relative) speed of an electron: α = v/c. [The relative speed is the speed as measured against the speed of light. Note that the ‘natural’ unit of speed in the Planck system of units is equal to c. Indeed, if you divide one Planck length by one Planck time unit, you get (1.616×10⁻³⁵m)/(5.391×10⁻⁴⁴s) = c m/s. However, this is another equation, just like (2), that does not depend on the units: we can express v and c in whatever unit we want, as long we’re consistent and express both in the same units.]

(4) α is also equal to the product of (a) the electron mass (which I’ll simply write as m_e here) and (b) the classical electron radius r_e (if both are expressed in Planck units): α = m_e·r_e. Now I think that’s, perhaps, the most amazing of all of the expressions for α. [If you don’t think that’s amazing, I’d really suggest you stop trying to study physics. :-)]

Also note that, from (2) and (4), we find that:

(5) The electron mass (in Planck units) is equal m_e = α/r_e= α/α²r = 1/αr. So that gives us an expression, using α once again, for the electron mass as a function of the Bohr radius r expressed in Planck units.

Finally, we can also substitute (1) in (5) to get:

(6) The electron mass (in Planck units) is equal to m_e = α/r_e = e_P²/r_e. Using the Bohr radius, we get m_e = 1/αr = 1/e_P²r.

So… As you can see, this fine-structure constant really links all of the fundamental properties of the electron: its charge, its radius, its distance to the nucleus (i.e. the Bohr radius), its velocity, its mass (and, hence, its energy),…

So… Why is what it is?

Well… We all marvel at this, but what can we say about it, really? I struggle how to interpret this, just as much – or probably much more 🙂 – as the professor who wrote the article I don’t like (because it’s so imprecise, and that’s what made me write all what I am writing here).

Having said that, it’s obvious that it points to a unity beyond these numbers and constants that I am only beginning to appreciate for what it is: deep, mysterious, and very beautiful. But so I don’t think that professor does a good job at showing how deep, mysterious and beautiful it all is. But then that’s up to you, my brother and you, my imaginary reader, to judge, of course. 🙂

[…] I forgot to mention what I mean with ‘Planck units’. Well… Once again, I should refer you to one of my other posts. But, yes, that’s too easy for me and a bit difficult for you. 🙂 So let me just note we get those Planck units by equating not less than five fundamental physical constants to 1, notably (1) the speed of light, (2) Planck’s (reduced) constant, (3) Boltzmann’s constant, (4) Coulomb’s constant and (5) Newton’s constant (i.e. the gravitational constant). Hence, we have a set of five equations here (c = ħ = k_B = k_e = G = 1), and so we can solve that to get the five Planck units, i.e. the Planck length unit, the Planck time unit, the Planck mass unit, the Planck energy unit, the Planck charge unit and, finally (oft forgotten), the Planck temperature unit. Of course, you should note that all mass and energy units are directly related because of the mass-energy equivalence relation E = mc², which simplifies to E = m if c is equated to 1. [I could also say something about the relation between temperature and (kinetic) energy, but I won’t, as it would only further confuse you.]

OK. Done! 🙂

Addendum: How to think about space and time?

If you read the argument on the Planck scale and constant carefully, then you’ll note that it does not depend on the idea of an indivisible photon. However, it does depend on that Planck-Einstein relation being valid always and everywhere. Now, the Planck-Einstein relation is, in its essence, a fairly basic result from classical electromagnetic theory: it incorporates quantum theory – remember: it’s the equation that allowed Planck to solve the black-body radiation problem, and so it’s why they call Planck the (reluctant) ‘Father of Quantum Theory’ – but it’s not quantum theory.

So the obvious question is: can we make this reflection somewhat more general, so we can think of the electromagnetic force as an example only. In other words: can we apply the thoughts above to any force and any movement really?

The truth is: I haven’t advanced enough in my little study to give the equations for the other forces. Of course, we could think of gravity, and I developed some thoughts on how gravity waves might look like, but nothing specific really. And then we have the shorter-range nuclear forces, of course: the strong force, and the weak force. The laws involved are very different. The strong force involves color charges, and the way distances work is entirely different. So it would surely be some different analysis. However, the results should be the same. Let me offer some thoughts though:

We know that the relative strength of the nuclear force is much larger, because it pulls like charges (protons) together, despite the strong electromagnetic force that wants to push them apart! So the mentioned problem of trying to ‘pack’ some oscillation in some tiny little space should be worse with the strong force. And the strong force is there, obviously, at tiny little distances!
Even gravity should become important, because if we’ve got a lot of energy packed into some tiny space, its equivalent mass will ensure the gravitational forces also become important. In fact, that’s what the whole argument was all about!
There’s also all this talk about the fundamental forces becoming one at the Planck scale. I must, again, admit my knowledge is not advanced enough to explain how that would be possible, but I must assume that, if physicists are making such statements, the argument must be fairly robust.

So… Whatever charge or whatever force we are talking about, we’ll be thinking of waves or oscillations—or simply movement, but it’s always a movement in a force field, and so there’s power and energy involved (energy is force times distance, and power is the time rate of change of energy). So, yes, we should expect the same issues in regard to scale. And so that’s what’s captured by h.

As we’re talking the smallest things possible, I should also mention that there are also other inconsistencies in the electromagnetic theory, which should (also) have their parallel for other forces. For example, the idea of a point charge is mathematically inconsistent, as I show in my post on fields and charges. Charge, any charge really, must occupy some space. It cannot all be squeezed into one dimensionless point. So the reasoning behind the Planck time and distance scale is surely valid.

In short, the whole argument about the Planck scale and those limits is very valid. However, does it imply our thinking about the Planck scale is actually relevant? I mean: it’s not because we can imagine how things might look like – they may look like those tiny little black holes, for example – that these things actually exist. GUT or string theorists obviously think they are thinking about something real. But, frankly, Feynman had a point when he said what he said about string theory, shortly before his untimely death in 1988: “I don’t like that they’re not calculating anything. I don’t like that they don’t check their ideas. I don’t like that for anything that disagrees with an experiment, they cook up an explanation—a fix-up to say, ‘Well, it still might be true.'”

It’s true that the so-called Standard Model does not look very nice. It’s not like Maxwell’s equations. It’s complicated. It’s got various ‘sectors’: the electroweak sector, the QCD sector, the Higgs sector,… So ‘it looks like it’s got too much going on’, as a friend of mine said when he looked at a new design for mountainbike suspension. 🙂 But, unlike mountainbike designs, there’s no real alternative for the Standard Model. So perhaps we should just accept it is what it is and, hence, in a way, accept Nature as we can see it. So perhaps we should just continue to focus on what’s here, before we reach the Great Desert, rather than wasting time on trying to figure out how things might look like on the other side, especially because we’ll never be able to test our theories about ‘the other side.’

On the other hand, we can see where the Great Desert sort of starts (somewhere near the 10³² Hz scale), and so it’s only natural to think it should also stop somewhere. In fact, we know where it stops: it stops at the 10⁴³ Hz scale, because everything beyond that doesn’t make sense. The question is: is there actually there? Like fundamental strings or whatever you want to call it. Perhaps we should just stop where the Great Desert begins. And what’s the Great Desert anyway? Perhaps it’s a desert indeed, and so then there is absolutely nothing there. 🙂

Hmm… There’s not all that much one can say about it. However, when looking at the history of physics, there’s one thing that’s really striking. Most of what physicists can think of, in the sense that it made physical sense, turned out to exist. Think of anti-matter, for instance. Paul Dirac thought it might exist, that it made sense to exist, and so everyone started looking for it, and Carl Anderson found in a few years later (in 1932). In fact, it had been observed before, but people just didn’t pay attention, so they didn’t want to see it, in a way. […] OK. I am exaggerating a bit, but you know what I mean. The 1930s are full of examples like that. There was a burst of scientific creativity, as the formalism of quantum physics was being developed, and the experimental confirmations of the theory just followed suit.

In the field of astronomy, or astrophysics I should say, it was the same with black holes. No one could really imagine the existence of black holes until the 1960s or so: they were thought of a mathematical curiosity only, a logical possibility. However, the circumstantial evidence now is quite large and so… Well… It seems a lot of what we can think of actually has some existence somewhere. 🙂

So… Who knows? […] I surely don’t. And so I need to get back to the grind and work my way through the rest of Feynman’s Lectures and the related math. However, this was a nice digression, and so I am grateful to my brother he initiated it. 🙂

Self-inductance, mutual inductance, and the power and energy of inductors

As Feynman puts it, studying physics is not always about the ‘the great and esoteric heights’. In fact, you usually have to come down from then fairly quickly – studying physics is similar to mountain climbing in that regard 🙂 – and study ‘relatively low-level subjects’, such as electrical circuits, which is what we’ll do in this and the next post.

As I’ve introduced some key concepts in a previous post already, let me recapitulate the basics, which include the concept of the electromotive force, which is basically the voltage, i.e. the potential difference, that’s produced in a loop or coil of wire as the magnetic flux changes. I also talked about the impedance in a AC circuit. Finally, we also discussed the power and energies involved. Important results from this previous discussion include (but are not limited to):

A constant speed AC generator will create an alternating current with the emf, i.e. the voltage, varying as V₀·sin(ωt).
If we only have resistors as circuit elements, and the resistance in the circuit adds up to R, then the electric current in the circuit will be equal to I = Ɛ/R = V/R = (V₀/R)·sin(ωt). So that’s Ohm’s Law, basically.
The power that’s produced and consumed in an AC circuit is the product of the voltage and the current, so P = Ɛ·I = V·I. We also showed this electrical power is equal to the mechanical power dW/dt that makes the generator run.
Finally, we explained the concept of impedance (denoted by Z) using Euler’s formula: Z = |Z|eⁱ^θ, mentioning that, if other circuit elements than resistors are involved, such as inductors, then it’s quite likely that the current signal will lag the voltage signal, with the phase factor θ telling us by how much.

It’s now time to introduce those ‘other’ circuit elements. So let’s start with inductors here, and the concept of inductance itself. There’s a lot of stuff to them, and so let’s go over it point by point.

The concept of self-inductance

In its simplest form, an inductor is just a coil, but they come in all sizes and shapes. If you want to see how they might look like, just google some images of micro-electronic inductors, and then, just to see the range of applications, some images of inductors used for large-scale industrial applications. If you do so, you’re likely to see images of transformers too, because transformers work on the principle of mutual inductance, and so they involve two coils, i.e. two inductors.

Contrary to what you might expect, the concept of self-inductance (or inductance tout court) is quite simply: a changing current will cause a changing magnetic field and, hence, some emf. Now, it turns out that the induced emf is proportional to the change in current. So we’ve got another constant of proportionality here, so it’s like how we defined resistance, or capacitance. So, in many ways, the inductance is just another proportionality coefficient. If we denote it by L – the symbol is said to honor the Russian phyicist Heinrich Lenz, whom you know from Lenz’ Law – then we define it as:

L = −Ɛ/(dI/dt)

The dI/dt factor is, obviously, the time rate of change of the current, and the negative sign indicates that the emf opposes the change in current, so it will tend to cause an opposing current. That’s why the emf involved is often referred to as a ‘back emf’. So that’s Lenz’ Law basically. As you might expect, the physicists came up with yet another derived unit, the Henry, to honor yet another physicist, Joseph Henry, an American scientist who was a contemporary of Michael Faraday and independently discovered pretty much the same as Faraday: one henry (H) equals one volt·second per ampere: 1 H = 1 V·s/A.

The concept of mutual inductance

Feynman introduces the topic of inductance with a two-coil set-up, as shown below, noting that a current in coil 1 will induce some emf in coil 2, which he denotes by M₁₂. Conversely, a current in coil 2 will induce some emf in coil 1, which he denoted by M₂₁. M₁₂ and M₂₁ are also constants: they depend on the geometry of the coils, including the length of the solenoid (l), its surface area (S) and the number of loop turns of the coils (N₁ and N₂).

The next step in the analysis is then to acknowledge that each coil should also produce a ‘back emf’ in itself, which we can denote by M₁₁ and M₂₂ respectively, but then these constants are, of course, equal to the self-inductance of the coils so, taking into account the convention in regard to the sign of the self-inductance, we write:

L₁ = − M₁₁ and L₁ = − M₂₂

You will now wonder: what’s the total emf in each coil, taking into account that we do not only have mutual inductance but also self-inductance? Frankly, when I was a kid, and my father tried to tell me one or two things about this, it confused me very much. I could not imagine what happened in one coil, let alone in two coils. I had this vision of a current producing some ‘back-current’, and then the ‘back-current’ producing ‘back-current’ again, and so I could not imagine how one could solve this problem. So the image in my head was very much like that baking powder box which Feynman evokes when talking about the method of images to find the electric fields in situations with an easy geometry, so that’s the picture of a baking powder box which has on its label a picture of a baking powder box which has… Well… Etcetera. Unfortunately, my father didn’t push us to study math and, therefore, I knew that one could solve such problems mathematically – we’re talking a converging series here – but I did not know how, and that’s why I found it all very confusing.

Now I understand there’s one current only, and one potential difference only, and that the formulas do not involve some infinite series of terms. But… Well… I am not ashamed to say these problems are still testing the (limited) agility of my mind. The first thing to ‘get’ is that we’re talking a back emf, and so that’s not a current but a potential difference. In fact, as I explained in my post on the electromotive force, the term ‘force’ in emf is actually misleading, and may lead to that same erroneous vision that I had as a kid: forces generating counter-forces, that generate counter-forces, that generate counter-forces, etcetera. It’s not like that: we have some current – one current – in a coil and we’ll have some voltage – one voltage – across the coil. If the coil would be a resistor instead of a coil, we’d find that the ratio of this voltage and the current would be some constant R = V/I. Now here we’re talking a coil indeed, so that’s a different circuit element, and we find some other ratio, L = −V/(dI/dt) = −Ɛ/(dI/dt). Why the minus sign? Well… As said, the induced emf will be such that it will tend to counter the current, and current flows from positive to negative as per our convention.

But… Yes? So how does it work when we put this coil in some circuit, and how does the resistance of the inductor come into play? Relax. We’ve just been talking ideal circuit elements so far, and we’ve discussed only two: the resistor and the inductor. We’ll talk about voltage sources (or generators) and capacitors too, and then we’ll link all of these ideal circuit elements. In short, we’ll analyze some real-life electrical circuit soon, but first you need to understand the basics. Let me just note that an ideal inductor appears a zero-resistance conductor in a direct current (DC) circuit, so it’s a short-circuit really! Please try to mentally separate out those ‘ideal’ circuit components. Otherwise you’ll never be able to make sense of it all!

In fact, there’s a good reason why Feynman starts with explaining mutual inductance before discussing a little circuit like the one below, which has an inductor and a voltage source. The two-coil situation above is effectively easier to understand, although it may not look like that at first. So let’s analyze that two-coil situation in more detail first. In other words, let me try to understand the situation that I didn’t understand as a kid. 🙂

Because of the law of superposition, we should add fluxes and changes in fluxes and, hence, we should also add the electromotive forces, i.e. the induced voltages. So, what we have here is that the total emf in coil 2 should be written as:

Ɛ₂ = M₂₁·(dI₁/dt) + M₂₂·(dI₂/dt) = M₂₁·(dI₁/dt) – L₂·(dI₂/dt)

What we’re saying here is that the emf, i.e. the voltage across the coil, will indeed depend on the change in current in the other coil, but also on the change in current of the coil itself. Likewise, the total emf in coil 1 should be written as:

Ɛ₁ = M₁₂·(dI₂/dt) + M₁₁·(dI₁/dt) = M₁₂·(dI₂/dt) – L₁·(dI₁/dt)

Of course, this does reduce to the simple L = −Ɛ/(dI/dt) if there’s one coil only. But so you see where it comes from and, while we do not have some infinite series 🙂 we do have a system of two equations here, and so let me say one or two things about it.

The first thing to note is that it is not so difficult to show that M₂₁is equal to M₁₂, so we can simplify and write that M₂₁ = M₁₂= M. Now, while I said ‘not so difficult’, I didn’t mean it’s easy and, because I don’t want this post to become too long, I’ll refer you to Feynman for the proof of this M₂₁ = M₁₂= M equation. It’s a general proof for any two coils or ‘circuits’ of arbitrary shape and it’s really worth the read. However, I have to move on.

The second thing to note is that this coefficient M, which is referred to as the mutual inductance now (so singular instead of plural) depends on the ‘circuit geometry’ indeed. For a simple solenoid, Feynman calculates it as

M = −(1/ε₀c²)·(N₁·N₂)·S/l,

with l the length of the solenoid, S its surface area (S), and N₁ and N₂the respective number of loop turns of the two coils. So, yes, only ‘geometry’ comes into play. [Note that’s quite obvious from the formula because a switch of the subscripts of N₁and N₂makes no difference, of course!]. Now, it’s interesting to note that M is the same for, let’s say, N₁= 100 and N₂ = 10 and for N₁= 20 and N₂ = 50. In fact, because you’re familiar with what transformers do, i.e. transforming voltages, you may think that’s counter-intuitive. It’s not. The thing with the number of coils does not imply that Ɛ₁ and Ɛ₂ remain the same. Our set of equations is Ɛ₁ = M·(dI₂/dt) – L₁·(dI₁/dt) and Ɛ₂= M·(dI₁/dt) – L₂·(dI₂/dt), and so L₁and L₂clearly do vary as N₁and N₂ vary! So… Well… Yes. We’ve got a set of two equations with two independent variables (I₁and I₂) and two dependent variables (Ɛ₁ and Ɛ₂). Of course, we could also phrase the problem the other way around: given two voltages, what are the currents? 🙂

Of course, that makes us think of the power that goes in and out of a transformer. Indeed, you’ll remember that power is voltage times current. So what’s going on here in regard to that?

Well… There’s a thing with transformers, or with two-coil systems like this in general, that is referred to as coupling. The geometry of the situation will determine how much flux from one coil is linked with the flux of the other coil. If most, or all of it, is linked, we say the two coils are ‘tightly coupled’ or, in the limit, that they are fully coupled. There’s a measure for that, and it’s called the coefficient of coupling. Let’s first explore that concept of power once more.

Inductance, energy and electric power

It’s easy to see that we need electric power to get some current going. Now, as we pointed out in our previous post, the power is equal to the voltage times the current. It’s also equal, of course, to the amount of work done per second, i.e. the time rate of change of the energy W, so we write:

dW/dt = Ɛ·I

Now, we defined the self-inductance as L = −Ɛ/(dI/dt) and, therefore, we know that Ɛ = −L·(dI/dt), so we have:

dW/dt = −L·I·(dI/dt)

What is this? A differential equation? Yes and no. We’ve got not one but two functions of time here, W and I, and, while their derivatives with respect to time do appear in the equation, what we need to do is just integrate the two sides over time. We get: W = −(1/2)·L·I². Just check it by taking the time derivative of both sides. Of course, we can add any constant, to both sides in fact, but that’s just a matter of choosing some reference point. We’ll chose our constant to be zero, and also think about the energy that’s stored in the coil, i.e. U, which we define as:

U = −W = −(1/2)·L·I²

Huh? What’s going on here? Well… It’s not an easy discussion, but let’s try to make sense of it. We have some changing current in the coil here but, obviously, some kind of inertia also: the coil itself opposes the change in current through the ‘back emf’. It requires energy, or power, to overcome the inertia. We may think of applying some voltage to offset the ‘back emf’, so we may effectively think of that little circuit with an inductor and a voltage source. The voltage V we’d need to apply to offset the inertia would, obviously, be equal to the ‘back emf’, but with its sign reversed, so we have:

V = − Ɛ = L·(dI/dt)

Now, it helps to think of what a current really is: it’s about electric charges that are moving at some velocity v because some force is being applied to them. As in any system, the power that’s being delivered is the dot product of the force and the velocity vectors (that ensures we only take the tangential component of the force into account), so if we have n moving charges, the power that is being delivered to the circuit is (F·v)·n. What is F? It’s obviously, qE, as the electric field is the force per unit charge, so E = F/q. But so we’re talking some circuit here and we need to think of the power being delivered to some infinitesimal element ds in the coil, and so that’s (F·v)·n·ds, which can be written as: (F·ds)·n·v. And then we integrate over the whole coil to find:

Now, you may or may not remember that the emf (Ɛ) is actually defined as the line integral ∫ E·ds line, taken around the entire coil and, hence, noting that E = F/q, and that the current I is equal to I = q·n·v, we got our power equation. Indeed, the integrand or kernel of our integral becomes F·n·v·ds = q·E·n·v·ds = I·E·ds. Hence, we get our power formula indeed: P = V·I, with V the potential difference, i.e. the voltage across the coil.

I am getting too much into the weeds here. The point is: we’ve got a full and complete analog to the concept of inertia in mechanics here: instead of some force F causing some mass m to change its velocity according to Newton’s Law, i.e. F = m·a = m·(dv/dt), we here have a potential difference V causing some current I to change according to the V = L·(dI/dt) law.

This is very confusing but, remember, the same equations must have the same solutions! So, in an electric circuit, the inductance is really like what the mass is in a mechanics. Now, in mechanics, we’ll say that our mass has some momentum p = m·v, and we’ll also say that its kinetic energy is equal to (1/2)m·v². We can do the same for our circuit: potential energy is continuously being converted into kinetic energy which, for our inductor, we write as U = (1/2)·L·I².

Just think about by playing with one of the many online graphing tools. The graph below, for example, assumes the current builds up to some maximum. As it reaches its maximum, the stored energy will also max out. Now, you should not worry about the units here, or the scale of the graphs. The assumption is that I builds up from 0 to 1, and that L = 1, so that makes U what it is. Using a different constant for L, and/or different units for I, will change the scale of U too, but not its general shape, and that shape gives you the general idea.

The example above obviously assumes some direct current, so it’s a DC circuit: the current builds up, but then stabilizes at some maximum that we can find by applying Ohm’s Law to the resistance of the circuit: I = V/R. Resistance? But we were talking an ideal inductor? We are. If there’s no other resistance in the circuit, we’ll have a short-circuit, so the assumption is that we do have some resistance in the circuit and, therefore, we should also think of some energy loss to heat from the current in the resistance, but that’s not our worry here.

The illustration below is, perhaps, more interesting. Here we are, obviously, applying an alternating current, and so the current goes in one and then in the other direction, so I > 0, and then I < 0, etcetera. We’re assuming some nice sinusoidal curve for the current here (i.e. the blue curve), and so we get what we get for U (i.e. the red curve): the energy goes up and down between zero and some maximum amplitude that’s determined by the maximum current.

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

Let’s now get back to that coupling constant.

The coupling constant

We can apply our reasoning to two coils. Indeed, we know that Ɛ₁ = M·(dI₂/dt) – L₁·(dI₁/dt) and Ɛ₂= M·(dI₁/dt) – L₂·(dI₂/dt). So the power in the two-coils system is dW/dt = Ɛ₁·I₁ + Ɛ₂·I₂, so we have:

dW/dt = M·I₁(dI₂/dt) – L₁·I₁·(dI₁/dt) + M·I₂·(dI₁/dt) – L₂·I₂·(dI₂/dt)

= – L₁·I₁·(dI₁/dt) – L₂·I₂·(dI₂/dt) + M·I₁(dI₂/dt)·I₂·(dI₁/dt)

Integrating both sides, and equating U with −W once more, yields:

U = (1/2)·L₁·I₁²+ (1/2)·L₂·I₂²+ M·I₁·I₂

[Again, you should just take the time derivative to verify this. If you don’t forget to apply the product rule for the M·I₁·I₂ term, you’ll see I am not writing too much nonsense here.] Now, there’s an interesting algebraic transformation of this expression, and an equally interesting explanation why we’d re-write the expression as we do. Let me copy it from Feynman so I’ll be using his fancier L and M symbols now. 🙂

So what? Well… Taking into account that inequality above, we can write the relation between M and the self-inductances L₁ and L₂ using some constant k, which varies between 0 and 1 and which we’ll refer to as the coupling constant:

We refer to k as the coupling constant, for rather obvious reasons: if it’s near zero, the mutual inductance will be very small, and if it’s near one, then the coils are said to be ‘tightly coupled’, and the ‘mutual flux linkage’ is then maximized. As you can imagine, there’s a whole body of literature out there relating this coupling constant to the behavior of transformers or other circuits where mutual inductance plays a role.

The formula for self-inductance

We gave the formula for the mutual inductance of two coils that are arranged as one solenoid on top of the other (cf. the illustration I started with):

M = −(1/ε₀c²)·(N₁·N₂)·S/l

It’s a very easy calculation, so let me quickly copy it from Feynman:

You’ll say: where is the M here? This is a formula for the emf! It is, but M is the constant of proportionality in front, remember? So there you go. 🙂

Now, you would think that getting a formula for the self-inductance L of some solenoid would be equally straightforward. It turns out that that is not the case. Feynman needs two full pages and… Well… By now, you should now how ‘dense’ his writing really is: if it weren’t so dense, you’d be reading Feynman yourself, rather than my ‘explanations’ of him. 🙂 So… Well.. If you want to see how it works, just click on the link here and scroll down to the last two pages of his exposé on self-inductance. I’ll limit myself to just jotting down the formula he does obtain when he’s through the whole argument:

See why he uses a fancier L than ‘my’ L? ‘His’ L is the length of the solenoid. 🙂 And, yes, r is the radius of the coil and n the number of turns per unit length in the winding. Also note this formula is valid only if L >> R, so the effects at the end of the solenoid can be neglected. OK. Done. 🙂

Well… That’s it for today! I am sorry to say but the next post promises to be as boring as this one because… Well… It’s on electric circuits again. 😦

Reconciling the wave-particle duality in electromagnetism

As I talked about Feynman’s equation for electromagnetic radiation in my previous post, I thought I should add a few remarks on wave-particle duality, but then I didn’t do it there, because my post would have become way too long. So let me add those remarks here. In fact, I’ve written about this before, and so I’ll just mention the basic ideas without going too much in detail. Let me first jot down the formula once again, as well as illustrate the geometry of the situation:

The gist of the matter is that light, in classical theory, is a traveling electromagnetic field caused by an accelerating electric charge and that, because light travels at speed c, it’s the acceleration at the retarded time t – r/c, i.e. a‘ = a(t – r/c), that enters the formula. You’ve also seen the diagrams that accompany this formula:

The two diagrams above show that the curve of the electric field in space is a “reversed” plot of the acceleration as a function of time. As I mentioned before, that’s quite obvious from the mathematical behavior of a function with argument like the argument above, i.e. a function F(t – r/c). When we write t – r/c, we basically measure distance units in seconds, instead of in meter. So we basically use c as the scale for both time as well as distance. I explained that in a previous post, so please have a look there if you’d want so see how that works.

So it’s pretty straightforward, really. However, having said that, when I see a diagram like the one above, so all of these diagrams plotting an E or B wave in space, I can’t help thinking it’s somewhat misleading: after all, we’re talking something traveling at the speed of light here and, therefore, its length – in our frame of reference – should be zero. And it is, obviously. Electromagnetic radiation comes packed in point-like, dimensionless photons: the length of something that travels at the speed of light must be zero.

Now, I don’t claim to know what’s going on exactly, but my thinking on it may not be far off the mark. We know that light is emitted and absorbed by atoms, as electrons go from one energy level to another, and the energy of the photons of light corresponds to the difference between those energy levels (i.e. a few electron-volt only, typically: it’s given by the E = h·ν relation). Therefore, we can look at a photon as a transient electromagnetic wave. It’s a very short pulse: the decay time for one such pulse of sodium light, i.e. one photon of sodium light, is 3.2×10^–8seconds. However, taking into account the frequency of sodium light (500 THz), that still makes for some 16 million oscillations, and a wave-train with a length of almost 10 meter. [Yes. Quite incredible, isn’t it?] So the photon could look like the transient wave I depicted below, except… Well… This wavetrain is traveling at the speed of light and, hence, we will not see it as a ten-meter long wave-train. Why not? Well… Because of the relativistic length contraction, it will effectively appear as a point-like particle to us.

So relativistic length contraction is why the wave and particle duality can be easily reconciled in electromagnetism: we can think of light as an irregular beam of point-like photons indeed, as one atomic oscillator after the other releases a photon, in no particularly organized way. So we can think of photons as transient wave-trains, but we should remind ourselves that they are traveling at the speed of light, so they’ll look point-like to us.

Is such view consistent with the results of the famous – of should I say infamous? – double-slit experiment. Well… Maybe. As I mentioned in one of my posts, it is rather remarkable that is actually hard to find actual double-slit experiments that use actual detectors near the slits, and even harder to find such experiments involving photons! Indeed, experiments involving detectors near the slits are usually experiments with ‘real’ particles, such as electrons, for example. Now, a lot of advances have been made in the set-up of these experiments over the past five years, and one of these experiments is a 2010 experiment of an Italian team which suggests that it’s the interaction between the detector and the electron wave that may cause the interference pattern to disappear. Now that throws some doubts on the traditional explanation of the results of the double-slit experiment.

The idea is shown below. The electron is depicted as an incoming plane wave which effectively breaks up as it goes through the slits. The slit on the left has no ‘filter’ (which you may think of as a detector) and, hence, the plane wave goes through as a cylindrical wave. The slit on the right-hand side is covered by a ‘filter’ made of several layers of ‘low atomic number material’, so the electron goes through but, at the same time, the barrier creates a spherical wave as it goes through. The researchers note that “the spherical and cylindrical wave do not have any phase correlation, and so even if an electron passed through both slits, the two different waves that come out cannot create an interference pattern on the wall behind them.” [I hope I don’t have to remind you that, while being represented as ‘real’ waves here, the ‘waves’ are, obviously, complex-valued psi functions.]

In fact, to be precise, the experimenters note that there still was an interference effect if the filter was thin enough. Let me quote the reason for that: “The thicker the filter, the greater the probability for inelastic scattering. When the electron suffers inelastic scattering, it is localized. This means that its wavefunction collapses and, after the measurement act, it propagates roughly as a spherical wave from the region of interaction, with no phase relation at all with other elastically or inelastically scattered electrons. If the filter is made thick enough, the interference effects cancels out almost completely.”

This does not solve the ‘mystery’ of the double-slit experiment, but it throws doubt on how it’s usually being explained. The mystery in such experiments is that, when we put detectors, it is either the detector at A or the detector at B that goes off. They should never go off together—”at half strength, perhaps”, as Feynman puts it. But so there are doubts here now. Perhaps the electron does go through both slits at the same time! And so that’s why I used italics when writing “even if an electron passed through both slits”: the electron, or the photon in a similar set-up, is not supposed to do that according to the traditional explanation of the results of the double-slit experiment! It’s one or the other, and the wavefunction collapses or reduces as it goes through.

However, that’s where these so-called ‘weak measurement’ experiments now come in, like this 2010 experiment: it does not prove but indicates that interaction does not have to be that way. They strongly suggest that it is not all or nothing, that our observations should not necessarily destroy the wavefunction. So, who knows, perhaps we will be able, one day, to show that the wavefunction does go through both slits, as it should (otherwise the interference pattern cannot be explained), and then we will have resolved the paradox.

I am pretty sure that, when that’s done, physicists will also be able to relate the image of a photon as a transient electromagnetic wave (cf. the diagram above), being emitted by an atomic oscillator for a few nanoseconds only (we gave the example for sodium light, for which the decay time was 3.2×10^–8seconds) with the image of a photon as a particle that can be represented by a complex-valued probability amplitude function (cf. the diagram below). I look forward to that day. I think it will come soon.

Here I should add two remarks. First, a lot has been said about the so-called indivisibility of a photon, but inelastic scattering implies that photons are not monolithic: the photon loses energy to the electron and, hence, its wavelength changes. Now, you’ll say: the scattered photon is not the same photon as the incident photon, and you’re right. But… Well. Think about it. It does say something about the presumed oneness of a photon.

The other remark is on the mathematics of interference. Photons are bosons and, therefore, we have to add their amplitudes to get the interference effect. So you may try to think of an amplitude function, like Ψ = (1/√2π)·eⁱ^θor whatever, and think it’s just a matter of ‘splitting’ this function before it enters the two slits and then ‘putting it back together’, so to say, after our photon has gone through the slits. [For the detailed math of interference in quantum mechanics, see my page on essentials.] Well… No. It’s not that simple. The illustration with that plane wave entering the slits, and the cylindrical and/or spherical wave coming out, makes it obvious that something happens to our wave as it goes through the slit. As I said a couple of times already, the two-slit experiment is interesting, but the interference phenomenon – or diffraction as it’s called – involving one slit only is at least as interesting. So… Well… The analysis is not that simple. Not at all, really. 🙂

The Liénard–Wiechert potentials and the solution for Maxwell’s equations

In my post on gauges and gauge transformations in electromagnetics, I mentioned the full and complete solution for Maxwell’s equations, using the electric and magnetic (vector) potential Φ and A. Feynman frames it nicely, so I should print it and put it on the kitchen door, so I can look at it everyday. 🙂

I should print the wave equation we derived in our previous post too. Hmm… Stupid question, perhaps, but why is there no wave equation above? I mean: in the previous post, we said the wave equation was the solution for Maxwell’s equation, didn’t we? The answer is simple, of course: the wave equation is a solution for waves originating from some source and traveling through free space, so that’s a special case. Here we have everything. Those integrals ‘sweep’ all over space, and so that’s real space, which is full of moving charges and so there’s waves everywhere. So the solution above is far more general and captures it all: it’s the potential at every point in space, and at every point in time, taking into account whatever else is there, moving or not moving. In fact, it is the general solution of Maxwell’s equations.

How do we find it? Well… I could copy Feynman’s 21st Lecture but I won’t do that. The solution is based on the formula for Φ and A for a small blob of charge, and then the formulas above just integrate over all of space. That solution for a small blob of charge, i.e. a point charge really, was first deduced in 1898, by a French engineer: Alfred-Marie Liénard. However, his equations did not get much attention, apparently, because a German physicist, Emil Johann Wiechert, worked on the same thing and found the very same equations just two years later. That’s why they are referred to as the Liénard-Wiechert potentials, so they both get credit for it, even if both of them worked it out independently. These are the equations:

Now, you may wonder why I am mentioning them, and you may also wonder how we get those integrals above, i.e. our general solution for Maxwell’s equations, from them. You can find the answer to your second question in Feynman’s 21st Lecture. 🙂 As for the first question, I mention them because one can derive two other formulas for E and B from them. It’s the formulas that Feynman uses in his first Volume, when studying light:

Now you’ll probably wonder how we can get these two equations from the Liénard-Wiechert potentials. They don’t look very similar, do they? No, they don’t. Frankly, I would like to give you the same answer as above, i.e. check it in Feynman’s 21st Lecture, but the truth is that the derivation is so long and tedious that even Feynman says one needs “a lot of paper and a lot of time” for that. So… Well… I’d suggest we just use all of those formulas and not worry too much about where they come from. If we can agree on that, we’re actually sort of finished with electromagnetism. All the chapters that follow Feynman’s 21st Lecture are applications indeed, so they do not add all that much to the core of the classical theory of electromagnetism.

So why did I write this post? Well… I am not sure. I guess I just wanted to sum things up for myself, so I can print it all out and put it on the kitchen door indeed. 🙂 Oh, and now that I think of it, I should add one more formula, and that’s the formula for spherical waves (as opposed to the plane waves we discussed in my previous post). It’s a very simple formula, and entirely what you’d expect to see:

The S function is the source function, and you can see that the formula is a Coulomb-like potential, but with the retarded argument. You’ll wonder: what is ψ? Is it E or B or what? Well… You can just substitute: ψ can be anything. Indeed, Feynman gives a very general solution for any type of spherical wave here. 🙂

So… That’s it, folks. That’s all there is to it. I hope you enjoyed it. 🙂

Addendum: Feynman’s equation for electromagnetic radiation

I talked about Feynman’s formula for electromagnetic radiation before, but it’s probably good to quickly re-explain it here. Note that it talks about the electric field only, as the magnetic field is so tiny and, in any case, if we have E then we can find B. So the formula is:

The geometry of the situation is depicted below. We have some charge q that, we assume, is moving through space, and so it creates some field E at point P. The e_r‘vector is the unit vector from P to Q, so it points at the charge. Well… It points to where the charge was at the time just a little while ago, i.e. at the time t – r‘/c. Why? Well… We don’t know where q is right now, because the field needs some time travel, we don’t know q right now, i.e. q at time t. It might be anywhere. Perhaps it followed some weird trajectory during the time r‘/c, like the trajectory below.

So our e_r‘vector moves as the charge moves, and so it will also have velocity and, likely, some acceleration, but what we measure for its velocity and acceleration, i.e. the d(e_r‘)/dt and d²(e_r‘)/dt² in that Feynman equation, is also the retarded velocity and the retarded acceleration. But look at the terms in the equation. The first two terms have a 1/r’² in them, so these two effects diminish with the square of the distance. The first term is just Coulomb’s Law (note that the minus sign in front takes care of the fact that like charges repel and so the E vector will point in the other way). Well… It is and it isn’t, because of the retarded time argument, of course. And so we have the second term, which sort of compensates for that. Indeed, the d(e_r‘)/dt is the time rate of change of e_r‘ and, hence, if r‘/c = Δt, then (r‘/c)·d(e_r‘)/dt is a first-order approximation of Δe_r‘.

As Feynman puts it: “The second term is as though nature were trying to allow for the fact that the Coulomb effect is retarded, if we might put it very crudely. It suggests that we should calculate the delayed Coulomb field but add a correction to it, which is its rate of change times the time delay that we use. Nature seems to be attempting to guess what the field at the present time is going to be, by taking the rate of change and multiplying by the time that is delayed.” In short, the first two terms can be written as E = −(q/4πε₀)/r‘²·[e_r‘ + Δe_r‘] and, hence, it’s a sort of modified Coulomb Law that sort of tries to guess what the electrostatic field at P should be based on (a) what it is right now, and (b) how q’s direction and velocity, as measured now, would change it.

Now, the third term has a 1/c² factor in front but, unlike the other two terms, this effect does not fall off with distance. So the formula below fully describes electromagnetic radiation, indeed, because it’s the only important term when we get ‘far enough away’, with ‘far enough’ meaning that the parts that go as the square of the distance have fallen off so much that they’re no longer significant.

Of course, you’re smart, and so you’ll immediately note that, as r increases, that unit vector keeps wiggling but that effect will also diminish. You’re right. It does, but in a fairly complicated way. The acceleration of e_r‘ has two components indeed. One is the transverse or tangential piece, because the end of e_r‘ goes up and down, and the other is a radial piece because it stays on a sphere and so it changes direction. The radial piece is the smallest bit, and actually also varies as the inverse square of $r$ when $r$ is fairly large. The tangential piece, however, varies only inversely as the distance, so as 1/r. So, yes, the wigglings of e_r‘ look smaller and smaller, inversely as the distance, but the tangential piece is and remains significant, because it does not vary as 1/r² but as 1/r only. That’s why you’ll usually see the law of radiation written in an even simpler way:

This law reduces the whole effect to the component of the acceleration that is perpendicular to the line of sight only. It assumes the distance is huge as compared to the distance over which the charge is moving and, therefore, that r‘ and r can be equated for all practical purposes. It also notes that the tangential piece is all that matters, and so it equates d²(e_r‘)/dt²with a_x/r. The whole thing is probably best illustrated as below: we have a generator driving charges up and down in G – so it’s an antenna really – and so we’ll measure a strong signal when putting the radiation detector D in position 1, but we’ll measure nothing in position 3. [The detector is, of course, another antenna, but with an amplifier for the signal.] But so here I am starting to talk about electromagnetic radiation once more, which was not what I wanted to do here, if only because Feynman does a much better job at that than I could ever do. 🙂

Traveling fields: the wave equation and its solutions

Original post:

We’ve climbed a big mountain over the past few weeks, post by post, 🙂 slowly gaining height, and carefully checking out the various routes to the top. But we are there now: we finally fully understand how Maxwell’s equations actually work. Let me jot them down once more:

As for how real or unreal the E and B fields are, I gave you Feynman’s answer to it, so… Well… I can’t add to that. I should just note, or remind you, that we have a fully equivalent description of it all in terms of the electric and magnetic (vector) potential Φ and A, and so we can ask the same question about Φ and A. They explain real stuff, so they’re real in that sense. That’s what Feynman’s answer amounts to, and I am happy with it. 🙂

What I want to do here is show how we can get from those equations to some kind of wave equation: an equation that describes how a field actually travels through space. So… Well… Let’s first look at that very particular wave function we used in the previous post to prove that electromagnetic waves propagate with speed c, i.e. the speed of light. The fields were very simple: the electric field had a y-component only, and the magnetic field a z-component only. Their magnitudes, i.e. their magnitude where the field had reached, as it fills the space traveling outwards, were given in terms of J, i.e. the surface current density going in the positive y-direction, and the geometry of the situation is illustrated below.

The fields were, obviously, zero where the fields had not reached as they were traveling outwards. And, yes, I know that sounds stupid. But… Well… It’s just to make clear what we’re looking at here. 🙂

We also showed how the wave would look like if we would turn off its First Cause after some time T, so if the moving sheet of charge would no longer move after time T. We’d have the following pulse traveling through space, a rectangular shape really:

We can imagine more complicated shapes for the pulse, like the shape shown below. J goes from one unit to two units at time t = t₁ and then to zero at t = t₂. Now, the illustration on the right shows the electric field as a function of x at the time t shown by the arrow. We’ve seen this before when discussing waves: if the speed of travel of the wave is equal to c, then x is equal to x = c·t, and the pattern is as shown below indeed: it mirrors what happened at the source x/c seconds ago. So we write:

This idea of using the retarded time t’ = t − x/c in the argument of a wave function f – or, what amounts to the same, using x − c/t – is key to understanding wave functions. I’ve explained this in very simple language in a post for my kids and, if you don’t get this, I recommend you check it out. What we’re doing, basically, is converting something expressed in time units into something expressed in distance units, or vice versa, using the velocity of the wave as the scale factor, so time and distance are both expressed in the same unit, which may be seconds, or meter.

To see how it works, suppose we add some time Δt to the argument of our wave function f, so we’re looking at f[x−c(t+Δt)] now, instead of f(x−ct). Now, f[x−c(t+Δt)] = f(x−ct−cΔt), so we’ll get a different value for our function—obviously! But it’s easy to see that we can restore our wave function F to its former value by also adding some distance Δx = cΔt to the argument. Indeed, if we do so, we get f[x+Δx−c(t+Δt)] = f(x+cΔt–ct−cΔt) = f(x–ct). You’ll say: t − x/c is not the same as x–ct. It is and it isn’t: any function of x–ct is also a function of t − x/c, because we can write:

Here, I need to add something about the direction of travel. The pulse above travel in the positive x-direction, so that’s why we have x minus ct in the argument. For a wave traveling in the negative x-direction, we’ll have a wave function y = F(x+ct). In any case, I can’t dwell on this, so let me move on.

Now, Maxwell’s equations in free or empty space, where are there no charges nor currents to interact with, reduce to:

Now, how can we relate this set of complicated equations to a simple wave function? Let’s do the exercise for our simple E_y and B_z wave. Let’s start by writing out the first equation, i.e. ∇·E = 0, so we get:

Now, our wave does not vary in the y and z direction, so none of the components, including E_y and E_zdepend on y or z. It only varies in the x-direction, so ∂E_y/∂y and ∂E_z/∂z are zero. Note that the cross-derivatives ∂E_y/∂z and ∂E_z/∂y are also zero: we’re talking a plane wave here, the field varies only with x. However, because ∇·E = 0, ∂E_x/∂x must be zero and, hence, E_x must be zero.

Huh? What? How is that possible? You just said that our field does vary in the x-direction! And now you’re saying it doesn’t it? Read carefully. I know it’s complicated business, but it all makes sense. Look at the function: we’re talking E_y, not E_x. E_y does vary as a function of x, but our field does not have an x-component, so E_x = 0. We have no cross-derivative ∂E_y/∂x in the divergence of E (i.e. in ∇·E = 0).

Huh? What? Let me put it differently. E has three components: E_x, E_y and E_z, and we have three space coordinates: x, y and z, so we have nine cross-derivatives. What I am saying is that all derivatives with respect to y and z are zero. That still leaves us with three derivatives: ∂E_x/∂x, ∂E_y/∂x, and ∂E_y/∂x. So… Because all derivatives in respect to y and z are zero, and because of the ∇·E = 0 equation, we know that ∂E_x/∂x must be zero. So, to make a long story short, I did not say anything about ∂E_y/∂x or ∂E_z/∂x. These may still be whatever they want to be, and they may vary in more or in less complicated ways. I’ll give an example of that in a moment.

Having said that, I do agree that I was a bit quick in writing that, because ∂E_x/∂x = 0, E_x must be zero too. Looking at the math only, E_x is not necessarily zero: it might be some non-zero constant. So… Yes. That’s a mathematical possibility. The static field from some charged condenser plate would be an example of a constant E_x field. However, the point is that we’re not looking at such static fields here: we’re talking dynamics here, and we’re looking at a particular type of wave: we’re talking a so-called plane wave. Now, the wave front of a plane wave is… Well… A plane. 🙂 So E_x is zero indeed. It’s a general result for plane waves: the electric field of a plane wave will always be at right angles to the direction of propagation.

Hmm… I can feel your skepticism here. You’ll say I am arbitrarily restricting the field of analysis… Well… Yes. For the moment. It’s not a reasonable restriction though. As I mentioned above, the field of a plane wave may still vary in both the y- and z-directions, as shown in the illustration below (for which the credit goes to Wikipedia), which visualizes the electric field of circularly polarized light. In any case, don’t worry too much about. Let’s get back to the analysis. Just note we’re talking plane waves here. We’ll talk about non-plane waves i.e. incoherent light waves later. 🙂

So we have plane waves and, therefore, a so-called transverse E field which we can resolve in two components: E_yand E_z. However, we wanted to study a very simply E_yfield only. Why? Remember the objective of this lesson: it’s just to show how we go from Maxwell’s equations to the wave function, and so let’s keep the analysis simple as we can for now: we can make it more general later. In fact, if we do the analysis now for non-zero E_yand zero E_z, we can do a similar analysis for non-zero E_zand zero E_y, and the general solution is going to be some superposition of two such fields, so we’ll have a non-zero E_yand E_z. Capito? 🙂 So let me write out Maxwell’s second equation, and use the results we got above, so I’ll incorporate the zero values for the derivatives with respect to y and z, and also the assumption that E_z is zero. So we get:

[By the way: note that, out of the nine derivatives, the curl involves only the (six) cross-derivatives. That’s linked to the neat separation between the curl and the divergence operator. Math is great! :-)]

Now, because of the flux rule (∇×E = –∂B/∂t), we can (and should) equate the three components of ∇×E above with the three components of –∂B/∂t, so we get:

[In case you wonder what it is that I am trying to do, patience, please! We’ll get where we want to get. Just hang in there and read on.] Now, ∂B_x/∂t = 0 and ∂B_y/∂t = 0 do not necessarily imply that B_x and B_yare zero: there might be some magnets and, hence, we may have some constant static field. However, that’s a matter of choosing a reference point or, more simply, assuming that empty space is effectively empty, and so we don’t have magnets lying around and so we assume that B_x and B_yare effectively zero. [Again, we can always throw more stuff in when our analysis is finished, but let’s keep it simple and stupid right now, especially because the B_x = B_y= 0 is entirely in line with the E_x = E_z= 0 assumption.]

The equations above tell us what we know already: the E and B fields are at right angles to each other. However, note, once again, that this is a more general result for all plane electromagnetic waves, so it’s not only that very special caterpillar or butterfly field that we’re looking at it. [If you didn’t read my previous post, you won’t get the pun, but don’t worry about it. You need to understand the equations, not the silly jokes.]

OK. We’re almost there. Now we need Maxwell’s last equation. When we write it out, we get the following monstrously looking set of equations:

However, because of all of the equations involving zeroes above 🙂 only ∂B_z/∂x is not equal to zero, so the whole set reduced to only simple equation only:

Simplifying assumptions are great, aren’t they? 🙂 Having said that, it’s easy to be confused. You should watch out for the denominators: a ∂x and a ∂t are two very different things. So we have two equations now involving first-order derivatives:

∂B_z/∂t = −∂E_y/∂x
−c²∂B_z/∂x = −∂E_y/∂t

So what? Patience, please! 🙂 Let’s differentiate the first equation with respect to x and the second with respect to t. Why? Because… Well… You’ll see. Don’t complain. It’s simple. Just do it. We get:

∂[∂B_z/∂t]/∂x = −∂²E_y/∂x²
∂[−c²∂B_z/∂x]/∂t = −∂²E_y/∂x²

So we can equate the left-hand sides of our two equations now, and what we get is a differential equation of the second order that we’ve encountered already, when we were studying wave equations. In fact, it is the wave equation for one-dimensional waves:

In case you want to double-check, I did a few posts on this, but, if you don’t get this, well… I am sorry. You’ll need to do some homework. More in particular, you’ll need to do some homework on differential equations. The equation above is basically some constraint on the functional form of E_y. More in general, if we see an equation like:

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

So any function ψ like that will work. You can check it out by doing the necessary derivatives and plug them into the wave equation. [In case you wonder how you should go about this, Feynman actually does it for you in his Lecture on this topic, so you may want to check it there.]

In fact, the functions f(x − c/t) and g(x + c/t) themselves will also work as possible solutions. So we can drop one or the other, which amounts to saying that our ‘shape’ has to travel in some direction, rather than in both at the same time. 🙂 Indeed, from all of my explanations above, you know what f(x − c/t) represents: it’s a wave that travels in the positive x-direction. Now, it may be periodic, but it doesn’t have to be periodic. The f(x − c/t) function could represent any constant ‘shape’ that’s traveling in the positive x-direction at speed c. Likewise, the g(x + c/t) function could represent any constant ‘shape’ that’s traveling in the negative x-direction at speed c. As for super-imposing both…

Well… I suggest you check that post I wrote for my son, Vincent. It’s on the math of waves, but it doesn’t have derivatives and/or differential equations. It just explains how superimposition and all that works. It’s not very abstract, as it revolves around a vibrating guitar string. So, if you have trouble with all of the above, you may want to read that first. 🙂 The bottom line is that we can get any wavefunction we want by superimposing simple sinusoidals that are traveling in one or the other direction, and so that’s what’s the more general solution really says. Full stop. So that’s what’s we’re doing really: we add very simple waves to get very more complicated waveforms. 🙂

Now, I could leave it at this, but then it’s very easy to just go one step further, and that is to assume that E_zand, therefore, B_yare not zero. It’s just a matter of super-imposing solutions. Let me just give you the general solution. Just look at it for a while. If you understood all that I’ve said above, 20 seconds or so should be sufficient to say: “Yes, that makes sense. That’s the solution in two dimensions.” At least, I hope so! 🙂

OK. I should really stop now. But… Well… Now that we’ve got a general solution for all plane waves, why not be even bolder and think about what we could possibly say about three-dimensional waves? So then E_xand, therefore, B_xwould not necessarily be zero either. After all, light can behave that way. In fact, light is likely to be non-polarized and, hence, E_xand, therefore, B_xare most probably not equal to zero!

Now, you may think the analysis is going to be terribly complicated. And you’re right. It would be if we’d stick to our analysis in terms of x, y and z coordinates. However, it turns out that the analysis in terms of vector equations is actually quite straightforward. I’ll just copy the Master here, so you can see His Greatness. 🙂

But what solution does an equation like (20.27) have? We can appreciate it’s actually three equations, i.e. one for each component, and so… Well… Hmm… What can we say about that? I’ll quote the Master on this too:

“How shall we find the general wave solution? The answer is that all the solutions of the three-dimensional wave equation can be represented as a superposition of the one-dimensional solutions we have already found. We obtained the equation for waves which move in the $x$ -direction by supposing that the field did not depend on $y$ and $z$ . Obviously, there are other solutions in which the fields do not depend on $x$ and $z$ , representing waves going in the $y$ -direction. Then there are solutions which do not depend on $x$ and $y$ , representing waves travelling in the $z$ -direction. Or in general, since we have written our equations in vector form, the three-dimensional wave equation can have solutions which are plane waves moving in any direction at all. Again, since the equations are linear, we may have simultaneously as many plane waves as we wish, travelling in as many different directions. Thus the most general solution of the three-dimensional wave equation is a superposition of all sorts of plane waves moving in all sorts of directions.”

It’s the same thing once more: we add very simple waves to get very more complicated waveforms. 🙂

You must have fallen asleep by now or, else, be watching something else. Feynman must have felt the same. After explaining all of the nitty-gritty above, Feynman wakes up his students. He does so by appealing to their imagination:

“Try to imagine what the electric and magnetic fields look like at present in the space in this lecture room. First of all, there is a steady magnetic field; it comes from the currents in the interior of the earth—that is, the earth’s steady magnetic field. Then there are some irregular, nearly static electric fields produced perhaps by electric charges generated by friction as various people move about in their chairs and rub their coat sleeves against the chair arms. Then there are other magnetic fields produced by oscillating currents in the electrical wiring—fields which vary at a frequency of $6060$ cycles per second, in synchronism with the generator at Boulder Dam. But more interesting are the electric and magnetic fields varying at much higher frequencies. For instance, as light travels from window to floor and wall to wall, there are little wiggles of the electric and magnetic fields moving along at $186,000$ miles per second. Then there are also infrared waves travelling from the warm foreheads to the cold blackboard. And we have forgotten the ultraviolet light, the x-rays, and the radiowaves travelling through the room.

Flying across the room are electromagnetic waves which carry music of a jazz band. There are waves modulated by a series of impulses representing pictures of events going on in other parts of the world, or of imaginary aspirins dissolving in imaginary stomachs. To demonstrate the reality of these waves it is only necessary to turn on electronic equipment that converts these waves into pictures and sounds.

If we go into further detail to analyze even the smallest wiggles, there are tiny electromagnetic waves that have come into the room from enormous distances. There are now tiny oscillations of the electric field, whose crests are separated by a distance of one foot, that have come from millions of miles away, transmitted to the earth from the Mariner II space craft which has just passed Venus. Its signals carry summaries of information it has picked up about the planets (information obtained from electromagnetic waves that travelled from the planet to the space craft).

There are very tiny wiggles of the electric and magnetic fields that are waves which originated billions of light years away—from galaxies in the remotest corners of the universe. That this is true has been found by “filling the room with wires”—by building antennas as large as this room. Such radiowaves have been detected from places in space beyond the range of the greatest optical telescopes. Even they, the optical telescopes, are simply gatherers of electromagnetic waves. What we call the stars are only inferences, inferences drawn from the only physical reality we have yet gotten from them—from a careful study of the unendingly complex undulations of the electric and magnetic fields reaching us on earth.

There is, of course, more: the fields produced by lightning miles away, the fields of the charged cosmic ray particles as they zip through the room, and more, and more. What a complicated thing is the electric field in the space around you! Yet it always satisfies the three-dimensional wave equation.”

So… Well… That’s it for today, folks. 🙂 We have some more gymnastics to do, still… But we’re really there. Or here, I should say: on top of the peak. What a view we have here! Isn’t it beautiful? It took us quite some effort to get on top of this thing, and we’re still trying to catch our breath as we struggle with what we’ve learned so far, but it’s really worthwhile, isn’t it? 🙂

Maxwell’s equations and the speed of light

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

Original post:

We know how electromagnetic waves travel through space: they do so because of the mechanism described in Maxwell’s equation: a changing magnetic field causes a changing electric field, and a changing magnetic field causes a (changing) electric field, as illustrated below.

So we need some First Cause to get it all started 🙂 i.e. some current, i.e. some moving charge, but then the electromagnetic wave travels, all by itself, through empty space, completely detached from the cause. You know that by now – indeed, you’ve heard this a thousand times before – but, if you’re reading this, you want to know how it works exactly. 🙂

In my post on the Lorentz gauge, I included a few links to Feynman’s Lectures that explain the nitty-gritty of this mechanism from various angles. However, they’re pretty horrendous to read, and so I just want to summarize them a bit—if only for myself, so as to remind myself what’s important and not. In this post, I’ll focus on the speed of light: why do electromagnetic waves – light – travel at the speed of light?

You’ll immediately say: that’s a nonsensical question. It’s light, so it travels at the speed of light. Sure, smart-arse! Let me be more precise: how can we relate the speed of light to Maxwell’s equations? That is the question here. Let’s go for it.

Feynman deals with the matter of the speed of an electromagnetic wave, and the speed of light, in a rather complicated exposé on the fields from some infinite sheet of charge that is suddenly set into motion, parallel to itself, as shown below. The situation looks – and actually is – very simple, but the math is rather messy because of the rather exotic assumptions: infinite sheets and infinite acceleration are not easy to deal with. 🙂 But so the whole point of the exposé is just to prove that the speed of propagation (v) of the electric and magnetic fields is equal to the speed of light (c), and it does a marvelous job at that. So let’s focus on that here only. So what I am saying is that I am going to leave out most of the nitty-gritty and just try to get to that v = c result as fast as I possibly can. So, fasten your seat belt, please.

Most of the nitty-gritty in Feynman’s exposé is about how to determine the direction and magnitude of the electric and magnetic fields, i.e. E and B. Now, when the nitty-gritty business is finished, the grand conclusion is that both E and B travel out in both the positive as well as the negative x-direction at some speed v and sort of ‘fill’ the entire space as they do. Now, the region they are filling extends infinitely far in both the y- and z-direction but, because they travel along the x-axis, there are no fields (yet) in the region beyond x = ± v·t (t = 0 is the moment when the sheet started moving, and it moves in the positive y-direction). As you can see, the sheet of charge fills the yz-plane, and the assumption is that its speed goes from zero to u instantaneously, or very very quickly at least. So the E and B fields move out like a tidal wave, as illustrated below, and thereby ‘fill’ the space indeed, as they move out.

The magnitude of E and B is constant, but it’s not the same constant, and part of the exercise here is to determine the relationship between the two constants. As for their direction, you can see it in the first illustration: B points in the negative z-direction for x > 0 and in the positive z-direction for x < 0, while E‘s direction is opposite to u‘s direction everywhere, so E points in the negative y-direction. As said, you should just take my word for it, because the nitty-gritty on this – which we do not want to deal with here – is all in Feynman and so I don’t want to copy that.

The crux of the argument revolves around what happens at the wavefront itself, as it travels out. Feynman relates flux and circulation there. It’s the typical thing to do: it’s at the wavefront itself that the fields change: before they were zero, and now they are equal to that constant. The fields do not change anywhere else, so there’s no changing flux or circulation business to be analyzed anywhere else. So we define two loops at the wavefront itself: Γ₁ and Γ₂. They are normal to each other (cf. the top and side view of the situation below), because the E and B fields are normal to each other. And so then we use Maxwell’s equations to check out what happens with the flux and circulation there and conclude what needs to be concluded. 🙂

We start with rectangle Γ₂. So one side is in the region where there are fields, and one side is in the region where the fields haven’t reached yet. There is some magnetic flux through this loop, and it is changing, so there is an emf around it, i.e. some circulation of E. The flux changes because the area in which B exists increases at speed v. Now, the time rate of change of the flux is, obviously, the width of the rectangle L times the rate of change of the area, so that’s (B·L·v·Δt)/Δt = B·L·v, with Δt some differential time interval co-defining how slow or how fast the field changes. Now, according to Faraday’s Law (see my previous post), this will be equal to minus the line integral of $E around Γ 2, which is E\cdotL. So E\cdotL = B\cdotL\cdotv and, hence, we find:$ $E = v\cdotB.$

$Interesting! To satisfy Faraday’s equation (which is just one of Maxwell’s equations in integral rather than in differential form), E must equal B times v, with v the speed of propagation of our ‘tidal’ wave. Now let’s look at Γ 1 . There we should apply:$

$Now the line integral is just B\cdotL, and the right-hand side is E\cdotL\cdotv, so, not forgetting that c 2 in front—i.e. the square of the speed of light, as you know!—we get:$ $c 2 B = E\cdotv, or E = (c 2 /v)\cdotB.$

Now, the E = v·B and E = (c²/v)·B equations must both apply (we’re talking one wave and one and the same phenomenon) and, obviously, that’s only possible if v = c²/v, i.e. if v = c. So the wavefront must travel at the speed of light! Waw ! That’s fast. 🙂 Yes. […] Jokes aside, that’s the result we wanted here: we just proved that the speed of travel of an electromagnetic wave must be equal to the speed of light.

As an added bonus, we also showed the mechanism of travel. It’s obvious from the equations we used to prove the result: it works through the derivatives of the fields with respect to time, i.e. ∂E/∂t and ∂B/∂t.

Done! Great! Enjoy the view!

Well… Yes and no. If you’re smart, you’ll say: we got this result because of the c² factor in that equation, so Maxwell had already put it in, so to speak. Waw! You really are a smart-arse, aren’t you? 🙂

The thing is… Well… The answer is: no. Maxwell did not put it in. Well… Yes and no. Let me explain. Maxwell’s first equation was the electric flux law ∇·E = σ/ε₀: the flux of E through a closed surface is proportional to the charge inside. So that’s basically an other way of writing Coulomb’s Law, and ε₀ was just some constant in it, the electric constant. So it’s a constant of proportionality that depends on the unit in which we measure electric charge. The only reason that it’s there is to make the units come out alright, so if we’d measure charge not in coulomb (C) in a unit equal to 1 C/ε₀, it would disappear. If we’d do that, our new unit would be equivalent to the charge of some 700,000 protons. You can figure that magical number yourself by checking the values of the proton charge and ε₀. 🙂

OK. And then Faraday came up with the exact laws for magnetism, and they involved current and some other constant of proportionality, and Maxwell formalized that by writing ∇×B = μ₀j, with μ₀ the magnetic constant. It’s not a flux law but a circulation law: currents cause circulation of B. We get the flux rule from it by integrating it. But currents are moving charges, and so Maxwell knew magnetism was related to the same thing: electric charge. So Maxwell knew the two constants had to be related. In fact, when putting the full set of equations together – there are four, as you know – Maxwell figured out that μ₀times ε₀would have to be equal to the reciprocal of c², with c the speed of propagation of the wave. So Maxwell knew that, whatever the unit of charge, we’d get two constants of proportionality, and electric and a magnetic constant, and that μ₀·ε₀would be equal to 1/c². However, while he knew that, at the time, light and electromagnetism were considered to be separate phenomena, and so Maxwell did not say that c was the speed of light: the only thing his equations told him was that c is the speed of propagation of that ‘electromagnetic’ wave that came out of his equations.

The rest is history. In 1856, the great Wilhelm Eduard Weber – you’ve seen his name before, didn’t you? – did a whole bunch of experiments which measured the electric constant rather precisely, and Maxwell jumped on it and calculated all the rest, i.e. μ₀, and so then he took the reciprocal of the square root of μ₀·ε₀and – Bang! – he had c, the speed of propagation of the electromagnetic wave he was thinking of. Now, c was some value of the order of 3×10⁸ m/s, and so that happened to be the same as the speed of light, which suggested that Maxwell’s c and the speed of light were actually one and the same thing!

Now, I am a smart-arse too 🙂 and, hence, when I first heard this story, I actually wondered how Maxwell could possibly know the speed of light at the time: Maxwell died many years before the Michelson-Morley experiment unequivocally established the value of the speed of light. [In case, you wonder: the Michelson-Morley experiment was done in 1887. So I check it. The fact is that the Michelson-Morley experiment concluded that the speed of light was an absolute value and that, in the process of doing so, they got a rather precise value for it, but the value of c itself has already been established, more or less, that is, by a Danish astronomer, Ole Römer, in 1676 ! He did so by carefully observing the timing of the repeating eclipses of Io, one of Jupiter’s moons. Newton mentioned his results in his Principia, which he wrote in 1687, duly noting that it takes about seven to eight minutes for light to travel from the Sun to the Earth. Done! The whole story is fascinating, really, so you should check it out yourself. 🙂

In any case, to make a long story short, Maxwell was puzzled by this mysterious coincidence, but he was bold enough to immediately point to the right conclusion, tentatively at least, and so he told the Cambridge Philosophical Society, in the very same year, i.e. 1856, 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.”

So… Well… Maxwell still suggests light needs some medium here, so the ‘medium’ is a reference to the infamous aether theory, but that’s not the point: what he says here is what we all take for granted now: light is an electromagnetic wave. So now we know there’s absolute no reason whatsoever to avoid the ‘inference’, but… Well… 160 years ago, it was quite a big deal to suggest something like that. 🙂

So that’s the full story. I hoped you like it. Don’t underestimate what you just did: understanding an argument like this is like “climbing a great peak”, as Feynman puts it. So it is “a great moment” indeed. 🙂 The only thing left is, perhaps, to explain the ‘other’ flux rules I used above. Indeed, you know Faraday’s Law:

But that other one? Well… As I explained in my previous post, Faraday’s Law is the integral form of Maxwell’s second equation: −∂B/∂t = ∇×E. The ‘other’ flux rule above – so that’s the one with the c² in front and without a minus sign, is the integral form of Maxwell’s fourth equation: c²∇×B = j/ε₀+ ∂E/∂t, taking into account that we’re talking a wave traveling in free space, so there are no charges and currents (it’s just a wave in empty space—whatever that means) and, hence, the Maxwell equation reduces to c²∇×B = ∂E/∂t. Now, I could take you through the same gymnastics as I did in my previous post but, if I were you, I’d just apply the general principle that ”the same equations must yield the same solutions” and so I’d just switch E for B and vice versa in Faraday’s equation. 🙂

So we’re done… Well… Perhaps one more thing. We’ve got these flux rules above telling us that the electromagnetic wave will travel all by itself, through empty space, completely detached from its First Cause. But… […] Well… Again you may think there’s some trick here. In other words, you may think the wavefront has to remain connected to the First Cause somehow, just like the whip below is connected to some person whipping it. 🙂

There’s no such connection. The whip is not needed. 🙂 If we’d switch off the First Cause after some time T, so our moving sheet stops moving, then we’d have the pulse below traveling through empty space. As Feynman puts it: “The fields have taken off: they are freely propagating through space, no longer connected in any way with the source. The caterpillar has turned into a butterfly!“

Now, the last question is always the same: what are those fields? What’s their reality? Here, I should refer you to one of the most delightful sections in Feynman’s Lectures. It’s on the scientific imagination. I’ll just quote the introduction to it, but I warmly recommend you go and check it out for yourself: it has no formulas whatsoever, and so you should understand all of it without any problem at all. 🙂

“I have asked you to imagine these electric and magnetic fields. What do you do? Do you know how? How do I imagine the electric and magnetic field? What do I actually see? What are the demands of scientific imagination? Is it any different from trying to imagine that the room is full of invisible angels? No, it is not like imagining invisible angels. It requires a much higher degree of imagination to understand the electromagnetic field than to understand invisible angels. Why? Because to make invisible angels understandable, all I have to do is to alter their properties a little bit—I make them slightly visible, and then I can see the shapes of their wings, and bodies, and halos. Once I succeed in imagining a visible angel, the abstraction required—which is to take almost invisible angels and imagine them completely invisible—is relatively easy. So you say, “Professor, please give me an approximate description of the electromagnetic waves, even though it may be slightly inaccurate, so that I too can see them as well as I can see almost invisible angels. Then I will modify the picture to the necessary abstraction.”

I’m sorry I can’t do that for you. I don’t know how. I have no picture of this electromagnetic field that is in any sense accurate. I have known about the electromagnetic field a long time—I was in the same position 25 years ago that you are now, and I have had 25 years more of experience thinking about these wiggling waves. When I start describing the magnetic field moving through space, I speak of the $E$ and $B$ fields and wave my arms and you may imagine that I can see them. I’ll tell you what I see. I see some kind of vague shadowy, wiggling lines—here and there is an $E and a$ $B$ written on them somehow, and perhaps some of the lines have arrows on them—an arrow here or there which disappears when I look too closely at it. When I talk about the fields swishing through space, I have a terrible confusion between the symbols I use to describe the objects and the objects themselves. I cannot really make a picture that is even nearly like the true waves. So if you have some difficulty in making such a picture, you should not be worried that your difficulty is unusual.

Our science makes terrific demands on the imagination. The degree of imagination that is required is much more extreme than that required for some of the ancient ideas. The modern ideas are much harder to imagine. We use a lot of tools, though. We use mathematical equations and rules, and make a lot of pictures. What I realize now is that when I talk about the electromagnetic field in space, I see some kind of a superposition of all of the diagrams which I’ve ever seen drawn about them. I don’t see little bundles of field lines running about because it worries me that if I ran at a different speed the bundles would disappear, I don’t even always see the electric and magnetic fields because sometimes I think I should have made a picture with the vector potential and the scalar potential, for those were perhaps the more physically significant things that were wiggling.

Perhaps the only hope, you say, is to take a mathematical view. Now what is a mathematical view? From a mathematical view, there is an electric field vector and a magnetic field vector at every point in space; that is, there are six numbers associated with every point. Can you imagine six numbers associated with each point in space? That’s too hard. Can you imagine even one number associated with every point? I cannot! I can imagine such a thing as the temperature at every point in space. That seems to be understandable. There is a hotness and coldness that varies from place to place. But I honestly do not understand the idea of a number at every point.

So perhaps we should put the question: Can we represent the electric field by something more like a temperature, say like the displacement of a piece of jello? Suppose that we were to begin by imagining that the world was filled with thin jello and that the fields represented some distortion—say a stretching or twisting—of the jello. Then we could visualize the field. After we “see” what it is like we could abstract the jello away. For many years that’s what people tried to do. Maxwell, Ampère, Faraday, and others tried to understand electromagnetism this way. (Sometimes they called the abstract jello “ether.”) But it turned out that the attempt to imagine the electromagnetic field in that way was really standing in the way of progress. We are unfortunately limited to abstractions, to using instruments to detect the field, to using mathematical symbols to describe the field, etc. But nevertheless, in some sense the fields are real, because after we are all finished fiddling around with mathematical equations—with or without making pictures and drawings or trying to visualize the thing—we can still make the instruments detect the signals from Mariner II and find out about galaxies a billion miles away, and so on.

The whole question of imagination in science is often misunderstood by people in other disciplines. They try to test our imagination in the following way. They say, “Here is a picture of some people in a situation. What do you imagine will happen next?” When we say, “I can’t imagine,” they may think we have a weak imagination. They overlook the fact that whatever we are allowed to imagine in science must be consistent with everything else we know: that the electric fields and the waves we talk about are not just some happy thoughts which we are free to make as we wish, but ideas which must be consistent with all the laws of physics we know. We can’t allow ourselves to seriously imagine things which are obviously in contradiction to the known laws of nature. And so our kind of imagination is quite a difficult game. One has to have the imagination to think of something that has never been seen before, never been heard of before. At the same time the thoughts are restricted in a strait jacket, so to speak, limited by the conditions that come from our knowledge of the way nature really is. The problem of creating something which is new, but which is consistent with everything which has been seen before, is one of extreme difficulty.”

Isn’t that great? I mean: Feynman, one of the greatest physicists of all time, didn’t write what he wrote above when he was a undergrad student or so. No. He did so in 1964, when he was 45 years old, at the height of his scientific career! And it gets better, because Feynman then starts talking about beauty. What is beauty in science? Well… Just click and check what Feynman thinks about it. 🙂

Oh… Last thing. So what is the magnitude of the E and B field? Well… You can work it out yourself, but I’ll give you the answer. The geometry of the situation makes it clear that the electric field has a y-component only, and the magnetic field a z-component only. Their magnitudes are given in terms of J, i.e. the surface current density going in the positive y-direction:

An introduction to electric circuits

In my previous post,I introduced electric motors, generators and transformers. They all work because of Faraday’s flux rule: a changing magnetic flux will produce some circulation of the electric field. The formula for the flux rule is given below:

It is a wonderful thing, really, but not easy to grasp intuitively. It’s one of these equations where I should quote Feynman’s introduction to electromagnetism: “The laws of Newton were very simple to write down, but they had a lot of complicated consequences and it took us a long time to learn about them all. The laws of electromagnetism are not nearly as simple to write down, which means that the consequences are going to be more elaborate and it will take us quite a lot of time to figure them all out.”

Now, among Maxwell’s Laws, this is surely the most complicated one! However, that shouldn’t deter us. 🙂 Recalling Stokes’ Theorem helps to appreciate what the integral on the left-hand side represents:

We’ve got a line integral around some closed loop Γ on the left and, on the right, we’ve got a surface integral over some surface S whose boundary is Γ. The illustration below depicts the geometry of the situation. You know what it all means. If not, I am afraid I have to send you back to square one, i.e. my posts on vector analysis. Yep. Sorry. Can’t keep copying stuff and make my posts longer and longer. 🙂

To understand the flux rule, you should imagine that the loop Γ is some loop of electric wire, and then you just replace C by E, the electric field vector. The circulation of E, which is caused by the change in magnetic flux, is referred to as the electromotive force (emf), and it’s the tangential force (E·ds) per unit charge in the wire integrated over its entire length around the loop, which is denoted by Γ here, and which encloses a surface S.

Now, you can go from the line integral to the surface integral by noting Maxwell’s Law: −∂B/∂t = ∇×E. In fact, it’s the same flux rule really, but in differential form. As for (∇×E)_n, i.e. the component of ∇×E that is normal to the surface, you know that any vector multiplied with the normal unit vector will yield its normal component. In any case, if you’re reading this, you should already be acquainted with all of this. Let’s explore the concept of the electromotive force, and then apply it our first electric circuit. 🙂

Indeed, it’s now time for a small series on circuits, and so we’ll start right here and right now, but… Well… First things first. 🙂

The electromotive force: concept and units

The term ‘force’ in ‘electromotive force’ is actually somewhat misleading. There is a force involved, of course, but the emf is not a force. The emf is expressed in volts. That’s consistent with its definition as the circulation of E: a force times a distance amounts to work, or energy (one joule is one newton·meter), and because E is the force on a unit charge, the circulation of E is expressed in joule per coulomb, so that’s a voltage: 1 volt = 1 joule/coulomb. Hence, on the left-hand side of Faraday’s equation, we don’t have any dimension of time: it’s energy per unit charge, so it’s x joule per coulomb . Full stop.

On the right-hand side, however, we have the time rate of change of the magnetic flux. through the surface S. The magnetic flux is a surface integral, and so it’s a quantity expressed in [B]·m², with [B] the measurement unit for the magnetic field strength. The time rate of change of the flux is then, of course, expressed in [B]·m²per second, i.e. [B]·m²/s. Now what is the unit for the magnetic field strength B, which we denoted by [B]?

Well… [B] is a bit of a special unit: it is not measured as some force per unit charge, i.e. in newton per coulomb, like the electric field strength E. No. [B] is measured in (N/C)/(m/s). Why? Because the magnetic force is not F = qE but F = qv×B. Hence, so as to make the units come out alright, we need to express B in (N·s)/(C·m), which is a unit known as the tesla (1 T = N·s/C·m), so as to honor the Serbian-American genius Nikola Tesla. [I know it’s a bit of short and dumb answer, but the complete answer is quite complicated: it’s got to do with the relativity of the magnetic force, which I explained in another post: both the v in F = qv×B equation as well as the m/s unit in [B] should make you think: whose velocity? In which reference frame? But that’s something I can’t summarize in two lines, so just click the link if you want to know more. I need to get back to the lesson.]

Now that we’re talking units, I should note that the unit of flux also got a special name, the weber, so as to honor one of Germany’s most famous physicists, Wilhelm Eduard Weber: as you might expect, 1 Wb = 1 T·m². But don’t worry about these strange names. Besides the units you know, like the joule and the newton, I’ll only use the volt, which got its name to honor some other physicist, Alessandro Volta, the inventor of the electrical battery. Or… Well… I might mention the watt as well at some point… 🙂

So how does it work? On one side, we have something expressed per second – so that’s per unit time – and on the other we have something that’s expressed per coulomb – so that’s per unit charge. The link between the two is the power, so that’s the time rate of doing work. It’s expressed in joule per second. So… Well… Yes. Here we go: in honor of yet another genius, James Watt, the unit of power got its own special name too: the watt. 🙂 In the argument below, I’ll show that the power that is being generated by a generator, and that is being consumed in the circuit (through resistive heating, for example, or whatever else taking energy out of the circuit) is equal to the emf times the current. For the moment, however, I’ll just assume you believe me. 🙂

We need to look at the whole circuit now, indeed, in which our little generator (i.e. our loop or coil of wire) is just one of the circuit elements. The units come out alright: the power = emf·current product is expressed in volt·coulomb/second = (joule/coulomb)·(coulomb/second) = joule/second. So, yes, it looks OK. But what’s going on really? How does it work, literally?

A short digression: on Ohm’s Law and electric power

Well… Let me first recall the basic concepts involved which, believe it or not, are probably easiest to explain by briefly recalling Ohm’s Law, which you’ll surely remember from your high-school physics classes. It’s quite simple really: we have some resistance in a little circuit, so that’s something that resists the passage of electric current, and then we also have a voltage source. Now, Ohm’s Law tells us that the ratio of (i) the voltage V across the resistance (so that’s between the two points marked as + and −) and (ii) the current I will be some constant. It’s the same as saying that V and I are inversely proportional to each other. The constant of proportionality is referred to as the resistance itself and, while it’s often looked at as a property of the circuit itself, we may embody it in a circuit element itself: a resistor, as shown below.

So we write R = V/I, and the brief presentation above should remind you of the capacity of a capacitor, which was just another constant of proportionality. Indeed, instead of feeding a resistor (so all energy gets dissipated away), we could charge a capacitor with a voltage source, so that’s a energy storage device, and then we find that the ratio between (i) the charge on the capacitor and (ii) the voltage across the capacitor was a constant too, which we defined as the capacity of the capacitor, and so we wrote C = Q/V. So, yes, another constant of proportionality (there are many in electricity!).

In any case, the point is: to increase the current in the circuit above, you need to increase the voltage, but increasing both amounts to increasing the power that’s being consumed in the circuit, because the power is voltage times current indeed, so P = V·I (or v·i, if I use the small letters that are used in the two animations below). For example, if we’d want to double the current, we’d need to double the voltage, and so we’re quadrupling the power: (2·V)·(2·I) = 2²·V·I. So we have a square-cube law for the power, which we get by substituting V for R·I or by substituting I for V/R, so we can write the power P as P = V²/R = I²·R. This square-cube law says exactly the same: if you want to double the voltage or the current, you’ll actually have to double both and, hence, you’ll quadruple the power. Now let’s look at the animations below (for which credit must go to Wikipedia).

They show how energy is being used in an electric circuit in terms of power. [Note that the little moving pluses are in line with the convention that a current is defined as the movement of positive charges, so we write I = dQ/dt instead of I = −dQ/dt. That also explains the direction of the field line E, which has been added to show that the power source effectively moves charges against the field and, hence, against the electric force.] What we have here is that, on one side of the circuit, some generator or voltage source will create an emf pushing the charges, and then some load will consume their energy, so they lose their push. So power, i.e. energy per unit time, is supplied, and is then consumed.

Back to the emf…

Now, I mentioned that the emf is a ratio of two terms: the numerator is expressed in joule, and the denominator is expressed in coulomb. So you might think we’ve got some trade-off here—something like: if we double the energy of half of the individual charges, then we still get the same emf. Or vice versa: we could, perhaps, double the number of charges and load them with only half the energy. One thing is for sure: we can’t both.

Hmm… Well… Let’s have a look at this line of reasoning by writing it down more formally.

The time rate of change of the magnetic flux generates some emf, which we can and should think of as a property of the loop or the coil of wire in which it is being generated. Indeed, the magnetic flux through it depends on its orientation, its size, and its shape. So it’s really very much like the capacity of a capacitor or the resistance of a conductor. So we write: emf = Δ(flux)/Δt. [In fact, the induced emf tries to oppose the change in flux, so I should add the minus sign, but you get the idea.]
For a uniform magnetic field, the flux is equal to the field strength B times the surface area S. [To be precise, we need to take the normal component of B, so the flux is B·S = B·S·cosθ.] So the flux can change because of a change in B or because of a change in S, or because of both.
The emf = Δ(flux)/Δt formula makes it clear that a very slow change in flux (i.e. the same Δ(flux) over a much larger Δt) will generate little emf. In contrast, a very fast change (i.e. the the same Δ(flux) over a much smaller Δt) will produce a lot of emf. So, in that sense, emf is not like the capacity or resistance, because it’s variable: it depends on Δ(flux), as well as on Δt. However, you should still think of it as a property of the loop or the ‘generator’ we’re talking about here.
Now, the power that is being produced or consumed in the circuit in which our ‘generator’ is just one of the elements, is equal to the emf times the current. The power is the time rate of change of the energy, and the energy is the work that’s being done in the circuit (which I’ll denote by ΔU), so we write: emf·current = ΔU/Δt.
Now, the current is equal to the time rate of change of the charge, so I = ΔQ/Δt. Hence, the emf is equal to emf = (ΔU/Δt)/I = (ΔU/Δt)/(ΔQ/Δt) = ΔU/ΔQ. From this, it follows that: emf = Δ(flux)/Δt = ΔU/ΔQ, which we can re-write as:

Δ(flux) = ΔU·Δt/ΔQ

What this says is the following. For a given amount of change in the magnetic flux (so we treat Δ(flux) as constant in the equation above), we could do more work on the same charge (ΔQ) – we could double ΔU by moving the same charge over a potential difference that’s twice as large, for example – but then Δt must be cut in half. So the same change in magnetic flux can do twice as much work if the change happens in half of the time.

Now, does that mean the current is being doubled? We’re talking the same ΔQ and half the Δt, so… Well? No. The Δt here measures the time of the flux change, so it’s not the dt in I = dQ/dt. For the current to change, we’d need to move the same charge faster, i.e. over a larger distance over the same time. We didn’t say we’d do that above: we only said we’d move the charge across a larger potential difference: we didn’t say we’d change the distance over which they are moved.

OK. That makes sense. But we’re not quite finished. Let’s first try something else, to then come back to where we are right now via some other way. 🙂 Can we change ΔQ? Here we need to look at the physics behind. What’s happening really is that the change in magnetic flux causes an induced current which consists of the free electrons in the Γ loop. So we have electrons moving in and out of our loop, and through the whole circuit really, but so there’s only so many free electrons per unit length in the wire. However, if we would effectively double the voltage, then their speed will effectively increase proportionally, so we’ll have more of them passing through per second. Now that effect surely impacts the current. It’s what we wrote above: all other things being the same, including the resistance, then we’ll also double the current as we double the voltage.

So where is that effect in the flux rule? The answer is: it isn’t there. The circulation of E around the loop is what it is: it’s some energy per unit charge. Not per unit time. So our flux rule gives us a voltage, which tells us that we’re going to have some push on the charges in the wire, but it doesn’t tell us anything about the current. To know the current, we must know the velocity of the moving charges, which we can calculate from the push if we also get some other information (such as the resistance involved, for instance), but so it’s not there in the formula of the flux rule. You’ll protest: there is a Δt on the right-hand side! Yes, that’s true. But it’s not the Δt in the v = Δs/Δt equation for our charges. Full stop.

Hmm… I may have lost you by now. If not, please continue reading. Let me drive the point home by asking another question. Think about the following: we can re-write that Δ(flux) = ΔU·Δt/ΔQ equation above as Δ(flux) = (ΔU/ΔQ)·Δt equation. Now, does that imply that, with the same change in flux, i.e. the same Δ(flux), and, importantly, for the same Δt, we could double both ΔU as well as ΔQ? I mean: (2·ΔU)/(2·ΔQ) = ΔU/ΔQ and so the equation holds, mathematically that is. […] Think about it.

You should shake your head now, and rightly so, because, while the Δ(flux) = (ΔU/ΔQ)·Δt equation suggests that would be possible, it’s totally counter-intuitive. We’re changing nothing in the real world (what happens there is the same change of flux in the same amount of time), but so we’d get twice the energy and twice the charge ?! Of course, we could also put a 3 there, or 20,000, or minus a million. So who decides on what we get? You get the point: it is, indeed, not possible. Again, what we can change is the speed of the free electrons, but not their number, and to change their speed, you’ll need to do more work, and so the reality is that we’re always looking at the same ΔQ, so if we want a larger ΔU, then we’ll need a larger change in flux, or we a shorter Δt during which that change in flux is happening.

So what can we do? We can change the physics of the situation. We can do so in many ways, like we could change the length of the loop, or its shape. One particularly interesting thing to do would be to increase the number of loops, so instead of one loop, we could have some coil with, say, N turns, so that’s N of these Γ loops. So what happens then? In fact, contrary to what you might expect, the ΔQ still doesn’t change as it moves into the coil and then from loop to loop to get out and then through the circuit: it’s still the same ΔQ. But the work that can be done by this current becomes much larger. In fact, two loops give us twice the emf of one loop, and N loops give us N times the emf of one loop. So then we can make the free electrons move faster, so they cover more distance in the same time (and you know work is force times distance), or we can move them across a larger potential difference over the same distance (and so then we move them against a larger force, so it also implies we’re doing more work). The first case is a larger current, while the second is a larger voltage. So what is it going to be?

Think about the physics of the situation once more: to make the charges move faster, you’ll need a larger force, so you’ll have a larger potential difference, i.e. a larger voltage. As for what happens to the current, I’ll explain that below. Before I do, let me talk some more basics.

In the exposé below, we’ll talk about power again, and also about load. What is load? Think about what it is in real life: when buying a battery for a big car, we’ll want a big battery, so we don’t look at the voltage only (they’re all 12-volt anyway). We’ll look at how many ampères it can deliver, and for how long. The starter motor in the car, for example, can suck up like 200 A, but for a very short time only, of course, as the car engine itself should kick in. So that’s why the capacity of batteries is expressed in ampère-hours.

Now, how do we get such large currents, such large loads? Well… Use Ohm’s Law: to get 200 A at 12 V, the resistance of the starter motor will have to as low as 0.06 ohm. So large currents are associated with very low resistance. Think practical: a 240-volt 60 watt light-bulb will suck in 0.25 A, and hence, its internal resistance, is about 960 Ω. Also think of what goes on in your house: we’ve got a lot of resistors in parallel consuming power there. The formula for the total resistance is 1/R_total= 1/R₁+ 1/R₂+ 1/R₃+ … So more appliances is less resistance, so that’s what draws in the larger current.

The point is: when looking at circuits, emf is one thing, but energy and power, i.e. the work done per second, are all that matters really. And so then we’re talking currents, but our flux rule does not say how much current our generator will produce: that depends on the load. OK. We really need to get back to the lesson now.

A circuit with an AC generator

The situation is depicted below. We’ve got a coil of wire of, let’s say, N turns of wire, and we’ll use it to generate an alternating current (AC) in a circuit.

The coil is really like the loop of wire in that primitive electric motor I introduced in my previous post, but so now we use the motor as a generator. To simplify the analysis, we assume we’ll rotate our coil of wire in a uniform magnetic field, as shown by the field lines B.

Now, our coil is not a loop, of course: the two ends of the coil are brought to external connections through some kind of sliding contacts, but that doesn’t change the flux rule: a changing magnetic flux will produce some emf and, therefore, some current in the coil.

OK. That’s clear enough. Let’s see what’s happening really. When we rotate our coil of wire, we change the magnetic flux through it. If S is the area of the coil, and θ is the angle between the magnetic field and the normal to the plane of the coil, then the flux through the coil will be equal to B·S·cosθ. Now, if we rotate the coil at a uniform angular velocity ω, then θ varies with time as θ = ω·t. Now, each turn of the coil will have an emf equal to the rate of change of the flux, i.e. d(B·S·cosθ)/dt. We’ve got N turns of wire, and so the total emf, which we’ll denote by Ɛ (yep, a new symbol), will be equal to:

Now, that’s just a nice sinusoidal function indeed, which will look like the graph below.

When no current is being drawn from the wire, this Ɛ will effectively be the potential difference between the two wires. What happens really is that the emf produces a current in the coil which pushes some charges out to the wire, and so then they’re stuck there for a while, and so there’s a potential difference between them, which we’ll denote by V, and that potential difference will be equal to Ɛ. It has to be equal to Ɛ because, if it were any different, we’d have an equalizing counter-current, of course. [It’s a fine point, so you should think about it.] So we can write:

So what happens when we do connect the wires to the circuit, so we’ve got that closed circuit depicted above (and below)?

Then we’ll have a current I going through the circuit, and Ohm’s Law then tells us that the ratio between (i) the voltage across the resistance in this circuit (we assume the connections between the generator and the resistor itself are perfect conductors) and (ii) the current will be some constant, so we have R = V/I and, therefore:

[To be fully complete, I should note that, when other circuit elements than resistors are involved, like capacitors and inductors, we’ll have a phase difference between the voltage and current functions, and so we should look at the impedance of the circuit, rather than its resistance. For more detail, see the addendum below this post.]

OK. Let’s now look at the power and energy involved.

Energy and power in the AC circuit

You’ll probably have many questions about the analysis above. You should. I do. The most remarkable thing, perhaps, is that this analysis suggests that the voltage doesn’t drop as we connect the generator to the circuit. It should. Why not? Why do the charges at both ends of the wire simply discharge through the circuit? In real life, there surely is such tendency: sudden large changes in loading will effectively produce temporary changes in the voltage. But then it’s like Feynman writes: “The emf will continue to provide charge to the wires as current is drawn from them, attempting to keep the wires always at the same potential difference.”

So how much current is drawn from them? As I explained above, that depends not on the generator but on the circuit, and more in particular on the load, so that’s the resistor in this case. Again, the resistance is the (constant) ratio of the voltage and the current: R = V/I. So think about increasing or decreasing the resistance. If the voltage remains the same, it implies the current must decrease or increase accordingly, because R = V/I implies that I = V/R. So the current is inversely proportional to R, as I explained above when discussing car batteries and lamps and loads. 🙂

Now, I still have to prove that the power provided by our generator is effectively equal to P = Ɛ·I but, if it is, it implies the power that’s being delivered will be inversely proportional to R. Indeed, when Ɛ and/or V remain what they are as we insert a larger resistance in the circuit, then P = Ɛ·I = Ɛ²/R, and so the power that’s being delivered would be inversely proportional to R. To be clear, we’d have a relation between P and R like the one below.

This is somewhat weird. Why? Well… I also have to show you that the power that goes into moving our coil in the magnetic field, i.e. the rate of mechanical work required to rotate the coil against the magnetic forces, is equal to the electric power Ɛ·I, i.e. the rate at which electrical energy is being delivered by the emf of the generator. However, I’ll postpone that for a while and, hence, I’ll just ask you, once again, to take me on my word. 🙂 Now, if that’s true, so if the mechanical power equals the electric power, then that implies that a larger resistance will reduce the mechanical power we need to maintain the angular velocity ω. Think of a practical example: if we’d double the resistance (i.e. we halve the load), and if the voltage stays the same, then the current would be halved, and the power would also be halved. And let’s think about the limit situations: as the resistance goes to infinity, the power that’s being delivered goes to zero, as the current goes to zero, while if the resistance goes to zero, both the current as well as the power would go to infinity!

Well… We actually know that’s also true in real-life: actual generators consume more fuel when the load increases, so when they deliver more power, and much less fuel, so less power, when there’s no load at all. You’ll know that, at least when you’re living in a developing country with a lot of load shedding! 🙂 And the difference is huge: no or just a little load will only consume 10% of what you need when fully loading it. It’s totally in line with what I wrote on the relationship between the resistance and the current that it draws in. So, yes, it does make sense:

An emf does produce more current if the resistance in the circuit is low (so i.e. when the load is high), and the stronger currents do represent greater mechanical forces.

That’s a very remarkable thing. It means that, if we’d put a larger load on our little AC generator, it should require more mechanical work to keep the coil rotating at the same angular velocity ω. But… What changes? The change in flux is the same, the Δt is the same, and so what changes really? What changes is the current going through the coil, and it’s not a change in that ΔQ factor above, but a change in its velocity v.

Hmm… That all looks quite complicated, doesn’t it? It does, so let’s get back to the analysis of what we have here, so we’ll simply assume that we have some dynamic equilibrium obeying that formula above, and so I and R are what they are, and we relate them to Ɛ according to that equation above, i.e.:

Now let me prove those formulas on the power of our generator and in the circuit. We have all these charges in our coil that are receiving some energy. Now, the rate at which they receive energy is F·v.

Huh? Yes. Let me explain: the work that’s being done on a charge along some path is the line integral ∫ F·ds along this path. But the infinitesimal distance ds is equal to v·dt, as ds/dt = v (note that we write s and v as vectors, so the dot product with F gives us the component of F that is tangential to the path). So ∫ F·ds = ∫ (F·v)dt. So the time rate of change of the energy, which is the power, is F·v. Just take the time derivative of the integral. 🙂

Now let’s assume we have n moving charges per unit length of our coil (so that’s in line with what I wrote about ΔQ above), then the power being delivered to any element ds of the coil is (F·v)·n·ds, which can be written as: (F·ds)·n·v. [Why? Because v and ds have the same direction: the direction of both vectors is tangential to the wire, always.] Now all we need to do to find out how much power is being delivered to the circuit by our AC generator is integrate this expression over the coil, so we need to find:

However, the emf (Ɛ) is defined as the line integral ∫ E·ds line, taken around the entire coil, and E = F/q, and the current I is equal to I = q·n·v. So the power from our little AC generator is indeed equal to:

Power = Ɛ·I

So that’s done. Now I need to make good on my other promise, and that is to show that Ɛ·I product is equal to the mechanical power that’s required to rotate the coil in the magnetic field. So how do we do that?

We know there’s going to be some torque because of the current in the coil. It’s formula is given by τ = μ×B. What magnetic field? Well… Let me refer you to my post on the magnetic dipole and its torque: it’s not the magnetic field caused by the current, but the external magnetic field, so that’s the B we’ve been talking about here all along. So… Well… I am not trying to fool you here. 🙂 However, the magnetic moment μ was not defined by that external field, but by the current in the coil and its area. Indeed, μ‘s magnitude was the current times the area, so that’s N·I·S in this case. Of course, we need to watch out because μ is a vector itself and so we need the angle between μ and B to calculate that vector cross product τ = μ×B. However, if you check how we defined the direction of μ, you’ll see it’s normal to the plane of the coil and, hence, the angle between μ and B is the very same θ = ω·t that we started our analysis with. So, to make a long story short, the magnitude of the torque τ is equal to:

τ = (N·I·S)·B·sinθ

Now, we know the torque is also equal to the work done per unit of distance traveled (around the axis of rotation, that is), so τ = dW/dθ. Now dθ = d(ω·t) = ω·dt. So we can now find the work done per unit of time, so that’s the power once more:

dW/dt = ω·τ = ω·(N·I·S)·B·sinθ

But so we found that Ɛ = N·S·B·ω·sinθ, so… Well… We find that:

dW/dt = Ɛ·I

Now, this equation doesn’t sort out our question as to how much power actually goes in and out of the circuit as we put some load on it, but it is what we promised to do: I showed that the mechanical work we’re doing on the coil is equal to the electric energy that’s being delivered to the circuit. 🙂

It’s all quite mysterious, isn’t it? It is. And we didn’t include other stuff that’s relevant here, such as the phenomenon of self-inductance: the varying current in the coil will actually produce its own magnetic field and, hence, in practice, we’d get some “back emf” in the circuit. This “back emf” is opposite to the current when it is increasing, and it is in the direction of the current when it is decreasing. In short, the self-inductance effect causes a current to have ‘inertia’: the inductive effects try to keep the flow constant, just as mechanical inertia tries to keep the velocity of an object constant. But… Well… I left that out. I’ll take about next time because…

[…] Well… It’s getting late in the day, and so I must assume this is sort of ‘OK enough’ as an introduction to what we’ll be busying ourselves with over the coming week. You take care, and I’ll talk to you again some day soon. 🙂

Perhaps one little note, on a question that might have popped up when you were reading all of the above: so how do actual generators keep the voltage up? Well… Most AC generators are, indeed, so-called constant speed devices. You can download some manuals from the Web, and you’ll find things like this: don’t operate at speeds above 4% of the rated speed, or more than 1% below the rated speed. Fortunately, the so-called engine governor will take car of that. 🙂

Addendum: The concept of impedance

In one of my posts on oscillators, I explain the concept of impedance, which is the equivalent of resistance, but for AC circuits. Just like resistance, impedance also sort of measures the ‘opposition’ that a circuit presents to a current when a voltage is applied, but it’s a complex ratio, as opposed to R = V/I. It’s literally a complex ratio because the impedance has 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|eⁱ^θ

The illustration below (credit goes to Wikipedia, once again) explains what’s going on. It’s a pretty generic view of the same AC circuit. The truth is: if we apply an alternating current, then the current and the voltage will both go up and down, but the current signal will usually lag the voltage signal, and the phase factor θ tells us by how much. Hence, using complex-number notation, we write:

V = I∗Z = I∗|Z|eⁱ^θ

Now, while that resembles the V = R·I formula, 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 to make it clear we’re not talking a vector cross product A×B here, but a product of two complex numbers. The bold-face for V and I implies they’re like vectors, or like complex numbers: so they have a phase too and, hence, we can write them as:

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

To be fully complete – you may skip all of this if you want, but it’s not that difficult, nor very long – it all works out as follows. We write:

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

Now, this equation must hold for all t, so we can equate the magnitudes and phases and, hence, we get: |V| = |I||Z| and so we get the formula we need, i.e. the phase difference between our function for the voltage and our function for the current.

θ_V= θ_I+ θ

Of course, you’ll say: voltage and current are something real, isn’t it? So what’s this about complex numbers? You’re right. I’ve used the complex notation only to simplify the calculus, so it’s only the real part of those complex-valued functions that counts.

Oh… And also note that, as mentioned above, we do not have such lag or phase difference when only resistors are involved. So we don’t need the concept of impedance in the analysis above. With this addendum, I just wanted to be as complete as I can be. 🙂

Induced currents

In my two previous posts, I presented all of the ingredients of the meal we’re going to cook now, most notably:

The formula for the torque on a loop of a current in a magnetic field, and its energy: (i) τ = μ×B, and (ii) U_mech= −μ·B.
The Biot-Savart Law, which gives you the magnetic field that’s produced by wires carrying currents:

Both ingredients are, obviously, relevant to the design of an electromagnetic motor, i.e. an ‘engine that can do some work’, as Feynman calls it. 🙂 Its principle is illustrated below.

The two formulas above explain how and why the coil go around, and the coil can be made to keep going by arranging that the connections to the coil are reversed each half-turn by contacts mounted on the shaft. Then the torque is always in the same direction. That’s how a small direct current (DC) motor is made. My father made me make a couple of these thirty years ago, with a magnet, a big nail and some copper coil. I used sliding contacts, and they were the most difficult thing in the whole design. But now I found a very nice demo on YouTube of a guy whose system to ‘reverse’ the connections is wonderfully simple: he doesn’t use any sliding contacts. He just removes half of the insulation on the wire of the coil on one side. It works like a charm, but I think it’s not so sustainable, as it spins so fast that the insulation on the other side will probably come off after a while! 🙂

Now, to make this motor run, you need current and, hence, 19th century physicists and mechanical engineers also wondered how one could produce currents by changing the magnetic field. Indeed, they could use Alessandro Volta’s ‘voltaic pile‘ to produce currents but it was not very handy: it consisted of alternating zinc and copper discs, with pieces of cloth soaked in salt water in-between!

Now, while the Biot-Savart Law goes back to 1820, it took another decade to find out how that could be done. Initially, people thought magnetic fields should just cause some current, but that didn’t work. Finally, Faraday unequivocally established the fundamental principle that electric effects are only there when something is changing. So you’ll get a current in a wire by moving it in a magnetic field, or by moving the magnet or, if the magnetic field is caused by some other current, by changing the current in that wire. It’s referred to as the ‘flux rule’, or Faraday’s Law. Remember: we’ve seen Gauss’ Law, then Ampère’s Law, and then that Biot-Savart Law, and so now it’s time for Faraday’s Law. 🙂 Faraday’s Law is Maxwell’s third equation really, aka as the Maxwell-Faraday Law of Induction:

∇×E = −∂B/∂t

Now you’ll wonder: what’s flux got to do with this formula? ∇×E is about circulation, not about flux! Well… Let me copy Feynman’s answer:

So… There you go. And, yes, you’re right, instead of writing Faraday’s Law as ∇×E = −∂B/∂t, we should write it as:

That’s a easier to understand, and it’s also easier to work with, as we’ll see in a moment. So the point is: whenever the magnetic flux changes, there’s a push on the electrons in the wire. That push is referred to as the electromotive force, abbreviated as emf or EMF, and so it’s that line and/or surface integral above indeed. Let me paraphrase Feynman so you fully understand what we’re talking about here:

When we move our wire in a magnetic field, or when we move a magnet near the wire, or when we change the current in a nearby wire, there will be some net push on the electrons in the wire in one direction along the wire. There may be pushes in different directions at different places, but there will be more push in one direction than another. What counts is the push integrated around the complete circuit. We call this net integrated push the electromotive force (abbreviated emf) in the circuit. More precisely, the emf is defined as the tangential force per unit charge in the wire integrated over length, once around the complete circuit.

So that’s the integral. 🙂 And that’s how we can turn that motor above into a generator: instead of putting a current through the wire to make it turn, we can turn the loop, by hand or by a waterwheel or by whatever. Now, when the coil rotates, its wires will be moving in the magnetic field and so we will find an emf in the circuit of the coil, and so that’s how the motor becomes a generator.

Now, let me quickly interject something here: when I say ‘a push on the electrons in the wire’, what electrons are we talking about? How many? Well… I’ll answer that question in very much detail in a moment but, as for now, just note that the emf is some quantity expressed per coulomb or, as Feynman puts it above, per unit charge. So we’ll need to multiply it with the current in the circuit to get the power of our little generator.

OK. Let’s move on. Indeed, all I can do here is mention just a few basics, so we can move on to the next thing. If you really want to know all of the nitty-gritty, then you should just read Feynman’s Lecture on induced currents. That’s got everything. And, no, don’t worry: contrary to what you might expect, my ‘basics’ do not amount to a terrible pile of formulas. In fact, it’s all easy and quite amusing stuff, and I should probably include a lot more. But then… Well… I always need to move on… If not, I’ll never get to the stuff that I really want to understand. 😦

The electromotive force

We defined the electromotive force above, including its formula:

What are the units? Let’s see… We know B was measured not in newton per coulomb, like the electric field E, but in N·s/C·m, because we had to multiply the magnetic field strength with the velocity of the charge to find the force per unit charge, cf. the F/q = v×B equation. Now what’s the unit in which we’d express that surface integral? We must multiply with m², so we get N·m·s/C. Now let’s simplify that by noting that one volt is equal to 1 N·m/C. [The volt has a number of definitions, but the one that applies here is that it’s the potential difference between two points that will impart one joule (i.e. 1 N·m) of energy to a unit of charge (i.e. 1 C) that passes between them.] So we can measure the magnetic flux in volt-seconds, i.e. V·s. And then we take the derivative in regard to time, so we divide by s, and so we get… Volt! The emf is measured in volt!

Does that make sense? I guess so: the emf causes a current, just like a potential difference, i.e. a voltage, and, therefore, we can and should look at the emf as a voltage too!

But let’s think about it some more, though. In differential form, Faraday’s Law, is just that ∇×E = −∂B/∂t equation, so that’s just one of Maxwell’s four equations, and so we prefer to write it as the “flux rule”. Now, the “flux rule” says that the electromotive force (abbreviated as emf or EMF) on the electrons in a closed circuit is equal to the time rate of change of the magnetic flux it encloses. As mentioned above, we measure magnetic flux in volt-seconds (i.e. V·s), so its time rate of change is measured in volt (because the time rate of change is a quantity expressed per second), and so the emf is measured in volt, i.e. joule per coulomb, as 1 V = 1 N·m/C = 1 J/C. What does it mean?

The time rate of change of the magnetic flux can change because the surface covered by our loop changes, or because the field itself changes, or by both. Whatever the cause, it will change the emf, or the voltage, and so it will make the electrons move. So let’s suppose we have some generator generating some emf. The emf can be used to do some work. We can charge a capacitor, for example. So how would that work?

More charge on the capacitor will increase the voltage V of the capacitor, i.e. the potential difference V = Φ₁− Φ₂between the two plates. Now, we know that the increase of the voltage V will be proportional to the increase of the charge Q, and that the constant of proportionality is defined by the capacity C of the capacitor: C = Q/V. [How do we know that? Well… Have a look at my post on capacitors.] Now, if our capacitor has an enormous capacity, then its voltage won’t increase very rapidly. However, it’s clear that, no matter how large the capacity, its voltage will increase. It’s just a matter of time. Now, its voltage cannot be higher than the emf provided by our ‘generator’, because it will then want to discharge through the same circuit!

So we’re talking power and energy here, and so we need to put some load on our generator. Power is the rate of doing work, so it’s the time rate of change of energy, and it’s expressed in joule per second. The energy of our capacitor is U = (1/2)·Q²/C = (1/2)·C·V². [How do we know that? Well… Have a look at my post on capacitors once again. :-)] So let’s take the time derivative of U assuming some constant voltage V. We get: dU/dt = d[(1/2)·Q²/C]/dt = (Q/C)·dQ/dt = V·dQ/dt. So that’s the power that the generator would need to supply to charge the generator. As I’ll show in a moment, the power supplied by a generator is, indeed, equal to the emf times the current, and the current is the time rate of change of the charge, so I = dQ/dt.

So, yes, it all works out: the power that’s being supplied by our generator will be used to charge our capacitor. Now, you may wonder: what about the current? Where is the current in Faraday’s Law? The answer is: Faraday’s Law doesn’t have the current. It’s just not there. The emf is expressed in volt, and so that’s energy per coulomb, so it’s per unit charge. How much power an generator can and will deliver depends on its design, and the circuit and load that we will be putting on it. So we can’t say how many coulomb we will have. It all depends. But you can imagine that, if the loop would be bigger, or if we’d have a coil with many loops, then our generator would be able to produce more power, i.e. it would be able to move more electrons, so the mentioned power = (emf)×(current) product would be larger. 🙂

Finally, to conclude, note Feynman’s definition of the emf: the tangential force per unit charge in the wire integrated over length around the complete circuit. So we’ve got force times distance here, but per unit charge. Now, force times distance is work, or energy, and so… Yes, emf is joule per coulomb, definitely! 🙂

[…] Don’t worry too much if you don’t quite ‘get’ this. I’ll come back to it when discussing electric circuits, which I’ll do in my next posts.

Self-inductance and Lenz’s rule

We talked about motors and generators above. We also have transformers, like the one below. What’s going on here is that an alternating current (AC) produces a continuously varying magnetic field, which generates an alternating emf in the second coil, which produces enough power to light an electric bulb.

Now, the total emf in coil (b) is the sum of the emf’s of the separate turns of coil, so if we wind (b) with many turns, we’ll get a larger emf, so we can ‘transform’ the voltage to some other voltage. From your high-school classes, you should know how that works.

The thing I want to talk about here is something else, though. There is an induction effect in coil (a) itself. Indeed, the varying current in coil (a) produces a varying magnetic field inside itself, and the flux of this field is continually changing, so there is a self-induced emf in coil (a). The effect is called self-inductance, and so it’s the emf acting on a current itself when it is building up a magnetic field or, in general, when its field is changing in any way. It’s a most remarkable phenomenon, and so let me paraphrase Feynman as he describes it:

“When we gave “the flux rule” that the emf is equal to the rate of change of the flux linkage, we didn’t specify the direction of the emf. There is a simple rule, called Lenz’s rule, for figuring out which way the emf goes: the emf tries to oppose any flux change. That is, the direction of an induced emf is always such that if a current were to flow in the direction of the emf, it would produce a flux of B that opposes the change in $B$ that produces the emf. In particular, if there is a changing current in a single coil (or in any wire), there is a “back” emf in the circuit. This emf acts on the charges flowing in the coil to oppose the change in magnetic field, and so in the direction to oppose the change in current. It tries to keep the current constant; it is opposite to the current when the current is increasing, and it is in the direction of the current when it is decreasing. A current in a self-inductance has “inertia,” because the inductive effects try to keep the flow constant, just as mechanical inertia tries to keep the velocity of an object constant.”

Hmm… That’s something you need to read a couple of times to fully digest it. There’s a nice demo on YouTube, showing an MIT physics video demonstrating this effect with a metal ring placed on the end of an electromagnet. You’ve probably seen it before: the electromagnet is connected to a current, and the ring flies into the air. The explanation is that the induced currents in the ring create a magnetic field opposing the change of field through it. So the ring and the coil repel just like two magnets with opposite poles. The effect is no longer there when a thin radial cut is made in the ring, because then there can be no current. The nice thing about the video is that it shows how the effect gets much more dramatic when an alternating current is applied, rather than a DC current. And it also shows what happens when you first cool the ring in liquid nitrogen. 🙂

You may also notice the sparks when the electromagnet is being turned on. Believe it or not, that’s also related to a “back emf”. Indeed, when we disconnect a large electromagnet by opening a switch, the current is supposed to immediately go to zero but, in trying to do so, it generates a large “back emf”: large enough to develop an arc across the opening contacts of the switch. The high voltage is also not good for the insulation of the coil, as it might damage it. So that’s why large electromagnets usually include some extra circuit, which allows the “back current” to discharge less dramatically. But I’ll refer you to Feynman for more details, as any illustration here would clutter the exposé.

Eddy currents

I like educational videos, and so I should give you a few references here, but there’s so many of this that I’ll let you google a few yourself. The most spectacular demonstration of eddy currents is those that appear in a superconductor: even back in the 1970s, when Feynman wrote his Lectures, the effect of magnetic levitation was well known. Feynman illustrates the effect with the simple diagram below: when bringing a magnet near to a perfect conductor, such as tin below 3.8°K, eddy currents will create opposing fields, so that no magnetic flux enters the superconducting material. The effect is also referred to as the Meisner effect, after the German physicist Walther Meisner, although it was discovered much earlier (in 1911) by a Dutch physicist in Leiden, Heike Kamerlingh Onnes, who got a Nobel Prize for it.

Of course, we have eddy currents in less dramatic situations as well. The phenomenon of eddy currents is usually demonstrated by the braking of a sheet of metal as it swings back and forth between the poles of an electromagnet, as illustrated below (left). The illustration on the right shows how eddy-current effect can be drastically reduced by cutting slots in the plate, so that’s like making a radial cut in our jumping ring. 🙂

The Faraday disc

The Faraday disc is interesting, not only from a historical point of view – the illustration below is a 19th century model, so Michael Faraday may have used himself – but also because it seems to contradict the “flux of rule”: as the disc rotates through a steady magnetic field, it will produce some emf, but so there’s no change in the flux. How is that possible?

The answer, of course, is that we are ‘cheating’ here: the material is moving, so we’re actually moving the ‘wire’, or the circuit if you want, so here we need to combine two equations:

If we do that, you’ll see it all makes sense. 🙂 Oh… That Faraday disc is referred to as a homopolar generator, and it’s quite interesting. You should check out what happened to the concept in the Wikipedia article on it. The Faraday disc was apparently used as a source for power pulses in the 1950s. The thing below could store 500 mega-joules and deliver currents up to 2 mega-ampère, i.e. 2 million amps! Fascinating, isn’t it? 🙂

Magnetic dipoles and their torque and energy

We studied the magnetic dipole in very much detail in one of my previous posts but, while we talked about an awful lot of stuff there, we actually managed to not talk about the torque on a it, when it’s placed in the magnetic field of other currents, or some other magnetic field tout court. Now, that’s what drives electric motors and generators, of course, and so we should talk about it, which is what I’ll do here. Let me first remind you of the concept of torque, and then we’ll apply it to a loop of current. 🙂

The concept of torque

The concept of torque is easy to grasp intuitively, but the math involved is not so easy. Let me sum up the basics (for the detail, I’ll refer you to my posts on spin and angular momentum). In essence, for rotations in space (i.e. rotational motion), the torque is what the force is for linear motion:

It’s the torque (τ) that makes an object spin faster or slower around some axis, just like the force would accelerate or decelerate that very same object when it would be moving along some curve.
There’s also a similar ‘law of Newton’ for torque: you’ll remember that the force equals the time rate of change of a vector quantity referred to as (linear) momentum: F = dp/dt = d(mv)/dt = ma (the mass times the acceleration). Likewise, we have a vector quantity that is referred to as angular momentum (L), and we can write: τ (i.e. the Greek tau) = dL/dt.
Finally, instead of linear velocity, we’ll have an angular velocity ω (omega), which is the time rate of change of the angle θ that defines how far the object has gone around (as opposed to the distance in linear dynamics, describing how far the object has gone along). So we have ω = dθ/dt. This is actually easy to visualize because we know that θ, expressed in radians, is actually the length of the corresponding arc on the unit circle. Hence, the equivalence with the linear distance traveled is easily ascertained.

There are many more similarities, like an angular acceleration: α = dω/dt = d²θ/dt², and we should also note that, just like the force, the torque is doing work – in its conventional definition as used in physics – as it turns an object instead of just moving it, so we can write:

ΔW = τ·Δθ

So it’s all the same-same but different once more 🙂 and so now we also need to point out some differences. The animation below does that very well, as it relates the ‘new’ concepts – i.e. torque and angular momentum – to the ‘old’ concepts – i.e. force and linear momentum. It does so using the vector cross product, which is really all you need to understand the math involved. Just look carefully at all of the vectors involved, which you can identify by their colors, i.e. red-brown (r), light-blue (τ), dark-blue (F), light-green (L), and dark-green (p).

So what do we have here? We have vector quantities once again, denoted by symbols in bold-face. Having said that, I should note that τ, L and ω are ‘special’ vectors: they are referred to as axial vectors, as opposed to the polar vectors F, p and v. To put it simply: polar vectors represent something physical, and axial vectors are more like mathematical vectors, but that’s a very imprecise and, hence, essential non-correct definition. 🙂 Axial vectors are directed along the axis of spin – so that is, strangely enough, at right angles to the direction of spin, or perpendicular to the ‘plane of the twist’ as Feynman calls it – and the direction of the axial vector is determined by a convention which is referred to as the ‘right-hand screw rule’. 🙂

Now, I know it’s not so easy to visualize vector cross products, so it may help to first think of torque (also known, for some obscure reason, as the moment of the force) as a twist on an object or a plane. Indeed, the torque’s magnitude can be defined in another way: it’s equal to the tangential component of the force, i.e. F·sin(Δθ), times the distance between the object and the axis of rotation (we’ll denote this distance by r). This quantity is also equal to the product of the magnitude of the force itself and the length of the so-called lever arm, i.e. the perpendicular distance from the axis to the line of action of the force (this lever arm length is denoted by r₀). So, we can define τ without the use of the vector cross-product, and in not less than three different ways actually. Indeed, the torque is equal to:

The product of the tangential component of the force times the distance r: τ = r·F_t= r·F·sin(Δθ);
The product of the length of the lever arm times the force: τ = r₀·F;
The work done per unit of distance traveled: τ = ΔW/Δθ or τ = dW/dθ in the limit.

Phew! Yeah. I know. It’s not so easy… However, I regret to have to inform you that you’ll need to go even further in your understanding of torque. More specifically, you really need to understand why and how we define the torque as a vector cross product, and so please do check out that post of mine on the fundamentals of ‘torque math’. If you don’t want to do that, then just try to remember the definition of torque as an axial vector, which is:

τ = (τ_yz, τ_zx, τ_xy) = (τ_x, τ_y, τ_z) with

τ_x = τ_yz = yF_z – zF_y (i.e. the torque about the x-axis, i.e. in the yz-plane),

τ_y = τ_zx = zF_x – xF_z (i.e. the torque about the y-axis, i.e. in the zx-plane), and

τ_z = τ_xy = xF_y – yF_x (i.e. the torque about the z-axis, i.e. in the xy-plane).

The angular momentum L is defined in the same way:

L = (L_yz, L_zx, L_xy) = (L_x, L_y, L_z) with

L_x = L_yz = yp_z – zp_y (i.e. the angular momentum about the x-axis),

L_y = L_zx = zp_x – xp_z(i.e. the angular momentum about the y-axis), and

L_z = L_xy = xp_y – yp_x (i.e. the angular momentum about the z-axis).

Let’s now apply the concepts to a loop of current.

The forces on a current loop

The geometry of the situation is depicted below. I know it looks messy but let me help you identifying the moving parts, so to speak. 🙂 We’ve got a loop with current and so we’ve got a magnetic dipole with some moment μ. From my post on the magnetic dipole, you know that μ‘s magnitude is equal to |μ| = μ = (current)·(area of the loop) = I·a·b.

Now look at the B vectors, i.e. the magnetic field. Please note that these vectors represent some external magnetic field! So it’s not like what we did in our post on the dipole: we’re not looking at the magnetic field caused by our loop, but at how it behaves in some external magnetic field. Now, because it’s kinda convenient to analyze, we assume that the direction of our external field B is the direction of the z-axis, so that’s what you see in this illustration: the B vectors all point north. Now look at the force vectors, remembering that the magnetic force is equal to:

F_magnetic = qv×B

So that gives the F₁, F₂, F₃, and F₄ vectors (so that’s the force on the first, second, third and fourth leg of the loop respectively) the magnitude and direction they’re having. Now, it’s easy to see that the opposite forces, i.e. the F₁–F₂ and F₃–F₄pair respectively, create a torque. The torque because of F₁and F₂is a torque which will tend to rotate the loop about the y-axis, so that’s a torque in the xz-plane, while the torque because of F₃and F₄will be some torque about the x-axis and/or the z-axis. As you can see, the torque is such that it will try to line up the moment vector μ with the magnetic field B. In fact, the geometry of the situation above is such that F₃and F₄have already done their job, so to speak: the moment vector μ is already lined up with the xz-plane, so there’s not net torque in that plane. However, that’s just because of the specifics of the situation here: the more general situation is that we’d have some torque about all three axes, and so we need to find that vector τ.

If we’d be talking some electric dipole, the analysis would be very straightforward, because the electric force is just F_electric = qE, which we can also write as E = F_electric =/q, so the field is just the force on one unit of electric charge, and so it’s (relatively) easy to see that we’d get the following formula for the torque vector:

τ = p×E

Of course, the p is the electric dipole moment here, not some linear momentum. [And, yes, please do try to check this formula. Sorry I can’t elaborate on it, but the objective of this blog is not substitute for a textbook!]

Now, all of the analogies between the electric and magnetic dipole field, which we explored in the above-mentioned post of mine, would tend to make us think that we can write τ here as:

τ = μ×B

Well… Yes. It works. Now you may want to know why it works 🙂 and so let me give you the following hint. Each charge in a wire feels that F_magnetic = qv×B force, so the total magnetic force on some volume ΔV, which I’ll denote by ΔF for a while, is the sum of the forces on all of the individual charges. So let’s assume we’ve got N charges per unit volume, then we’ve got N·ΔV charges in our little volume ΔV, so we write: ΔF = N·ΔV·q·v×B. You’re probably confused now: what’s the v here? It’s the (drift) velocity of the (free) electrons that make up our current I. Indeed, the protons don’t move. 🙂 So N·q·v is just the current density j, so we get: ΔF = j×BΔV, which implies that the force per unit volume is equal to j×B. But we need to relate it to the current in our wire, not the current density. Relax. We’re almost there. The ΔV in a wire is just its cross-sectional area A times some length, which I’ll denote by ΔL, so ΔF = j×BΔV becomes ΔF = j×BAΔL. Now, jA is the vector current I, so we get the simple result we need here: ΔF = I×BΔL, i.e. the magnetic force per unit length on a wire is equal to ΔF/ΔL = I×B.

Let’s now get back to our magnetic dipole and calculate F₁and F₂. The length of ‘wire’ is the length of the leg of the loop, i.e. b, so we can write:

F₁= −F₂ = b·I×B

So the magnitude of these forces is equal F₁= F₂ = I·B·b. Now, The length of the moment or lever arm is, obviously, equal to a·sinθ, so the magnitude of the torque is equal to the force times the lever arm (cf. the τ = r₀·F formula above) and so we can write:

τ = I·B·b·a·sinθ

But I·a·b is the magnitude of the magnetic moment μ, so we get:

τ = μ·B·sinθ

Now that’s consistent with the definition of the vector cross product:

τ = μ×B = |μ|·|B|·sinθ·n = μ·B·sinθ·n

Done! Now, electric motors and generators are all about work and, therefore, we also need to briefly talk about energy here.

The energy of a magnetic dipole

Let me remind you that we could also write the torque as the work done per unit of distance traveled, i.e. as τ = ΔW/Δθ or τ = dW/dθ in the limit. Now, the torque tries to line up the moment with the field, and so the energy will be lowest when μ and B are parallel, so we need to throw in a minus sign when writing:

τ = −dU/dθ ⇔ dU = −τ·dθ

We should now integrate over the [0, θ] interval to find U, also using our τ = μ·B·sinθ formula. That’s easy, because we know that d(cosθ)/dθ = −sinθ, so that integral yields:

U = 1 − μ·B·cosθ + a constant

If we choose the constant to be zero, and if we equate μ·B with 1, we get the blue graph below:

The μ·B in the U = 1 − μ·B·cosθ formula is just a scaling factor, obviously, so it determines the minimum and maximum energy. Now, you may want to limit the relevant range of θ to [0, π], but that’s not necessary: the energy of our loop of current does go up and down as shown in the graph. Just think about it: it all makes perfect sense!

Now, there is, of course, more energy in the loop than this U energy because energy is needed to maintain the current in the loop, and so we didn’t talk about that here. Therefore, we’ll qualify this ‘energy’ and call it the mechanical energy, which we’ll abbreviate by U_mech. In addition, we could, and will, choose some other constant of integration, so that amounts to choosing some other reference point for the lowest energy level. Why? Because it then allows us to write U_mechas a vector dot product, so we get:

U_mech= −μ·B·cosθ = −μ·B

The graph is pretty much the same, but it now goes from −μ·B to +μ·B, as shown by the red graph in the illustration above.

Finally, you should note that the U_mech= −μ·B formula is similar to what you’ll usually see written for the energy of an electric dipole: U= −p·E. So that’s all nice and good! However, you should remember that the electrostatic energy of an electric dipole (i.e. two opposite charges separated by some distance d) is all of the energy, as we don’t need to maintain some current to create the dipole moment!

Now, Feynman does all kinds of things with these formulas in his Lectures on electromagnetism but I really think this is all you need to know about it—for the moment, at least. 🙂

The magnetic field of circuits: the Law of Biot and Savart

Pre-script (dated 26 June 2020): This post got mutilated by the removal of some material by the dark force. You should be able to follow the main story line, however. If anything, the lack of illustrations might actually help you to think things through for yourself. In any case, we now have different views on these concepts as part of our realist interpretation of quantum mechanics, so we recommend you read our recent papers instead of these old blog posts.

Original post:

We studied the magnetic dipole in very much detail in one of my previous posts. While we talked about an awful lot of stuff there, we actually managed to not talk about the torque on a it, when it’s placed in the magnetic field of other currents. Now, that’s what drives electric motors and generators, of course, and so we should talk about it, which is what I’ll do in my next post. Before doing so, however, I need to give you one or two extra formulas generalizing some of the results we obtained in our previous posts on magnetostatics. So that’s what I do under this heading: the magnetic field of circuits. The idea is simple: loops of current are not always nice squares or circles. Their shape might be quite irregular, indeed, like the loop below.

Of course, the same general formula should apply. So we can find the magnetic vector potential with the following integral:

Just to make sure, let me re-insert its equivalent for electrostatics, so you can see they’re (almost) the same:

But we’re talking a wire here, so how can we relate the current density j and the volume element dV to that? It’s easy: the illustration below shows that we can simply write:

j·dV = j·S·ds = I·ds

Therefore, we can write our integral for the vector potential as:

Of course, you should note the subtle change from a volume integral to a line integral, so it’s not all that straightforward, but we’re good to go. Now, in electrostatics, we actually had a fairly simple integral for the electric field itself:

To be clear, E(1) is the field of a known charge distribution, which is represented by ρ(2), at point (1). The integral is almost the same as the one for Φ, but we’re talking vectors here (E and e₁₂) rather than scalars (ρ and Φ), and you should also note the square in the denominator of the integral. 🙂

As you might expect, there is a similar integral for B, which we find by… Well… We just need to calculate B, so that’s the curl of A:

How do we do that? It’s not so easy, so let me just copy the master himself:

So this integral gives B directly in terms of the known currents. The geometry involved is easy but, just in case, Feynman illustrates it, quite simply, as follows:

Now, there’s one more step to take, and then we’re done. If we’re talking a circuit of small wire, then we can replace j·dV by I·ds once more, and, hence, we get the Biot-Savart Law in its final form:

Note the minus sign: it appears because we reversed the order of the vector cross product, and also note we actually have three integrals here, one for each component of B, so that’s just like that integral for A.

So… That’s it. 🙂 I’ll conclude by two small remarks:

The law is named after Jean-Baptiste Biot and Félix Savart, two incredible Frenchmen (it’s really worthwhile checking their biographies on Wikipedia), who jotted it down in 1820, so that’s almost 200 years ago. Isn’t that amazing?
You see we sort of got rid of the vector potential with this formula. So the question is: “What is the advantage of the vector potential if we can find $B$ directly with a vector integral? After all, A also involves three integrals!” I’ll let Feynman reply to that question:

Because of the cross product, the integrals for B are usually more complicated. Also, since the integrals for $A$ are like those of electrostatics, we may already know them. Finally, we will see that in more advanced theoretical matters (in relativity, in advanced formulations of the laws of mechanics, like the principle of least action to be discussed later, and in quantum mechanics), the vector potential plays an important role.

In fact, Feynman makes the point on the vector potential being relevant very explicit by just boldly stating two laws in quantum mechanics in which the magnetic and electric potential are used, not the magnetic or electric field. Indeed, it seems an external magnetic or electric field changes probability amplitudes. I’ll just jot down the two laws below, but leave it to you to decide whether or not you want to read the whole argument. qm2

The key point that Feynman is making is that Φ and A are equally ‘real’ or ‘unreal’ as E and B in terms of explaining physical realities. I get the point, but I don’t find it necessary to copy the whole argument here. Perhaps it’s sufficient to just quote Feynman’s introduction to it, which says it all, in my humble opinion, that is:

“There are many changes in what concepts are important when we go from classical to quantum mechanics. We have already discussed some of them in Volume I. In particular, the force concept gradually fades away, while the concepts of energy and momentum become of paramount importance. You remember that instead of particle motions, one deals with probability amplitudes which vary in space and time. In these amplitudes there are wavelengths related to momenta, and frequencies related to energies. The momenta and energies, which determine the phases of wave functions, are therefore the important quantities in quantum mechanics. Instead of forces, we deal with the way interactions change the wavelength of the waves. The idea of a force becomes quite secondary—if it is there at all. When people talk about nuclear forces, for example, what they usually analyze and work with are the energies of interaction of two nucleons, and not the force between them. Nobody ever differentiates the energy to find out what the force looks like. In this section we want to describe how the vector and scalar potentials enter into quantum mechanics. It is, in fact, just because momentum and energy play a central role in quantum mechanics that A and Φ provide the most direct way of introducing electromagnetic effects into quantum descriptions.”

OK. That’s sufficient really. Onwards!

Gravitational waves: how should we imagine them?

This post is not a post. It’s just a reminder for myself to look into gravitational waves at some point in time. We know how electromagnetic waves travel through space: they do so because of the mechanism described in Maxwell’s equation: a changing magnetic field causes a changing electric field, and a changing magnetic field causes a (changing) electric field, as illustrated below.

So… Electromagnetism is one phenomenon only, but we do analyze the E and B fields as separate things, as the equations below illustrate:

B is co-defined with E, and then all of the dynamics work themselves out through the ∂E/∂t and ∂E/∂t functions. Now, when talking gravity, we only have positive ‘charges’, referred to as masses, always attracting each other, but that doesn’t matter all that much: Coulomb’s and Newton’s Law have the same mathematical structure (as shown below), except for the minus sign, and so there must be some equivalent to the electromagnetic wave, explaining gravitational waves , using the same mathematical concepts, as a propagation of a force on a ‘unit charge’ (so that’s a unit mass in this case) using the very same concepts, i.e. the concepts of a field – two separate fields, I should say, just like E and B – and of its flux and circulation.

So we’d have an E_G and a B_G field, so to speak, or a G_E and a G_B field, and formulas for the flux and circulation of both, resembling Maxwell’s equations. In fact, they’d be the same, more or less. It’s a powerful idea, and I am sure the idea has been fully developed elsewhere. In fact, I quickly googled a few things, but it seems that the whole area is a pretty new area of research, both theoretically as well as experimentally—to my surprise!

Hmm… Interesting idea, but I’ll need to do a lot more analysis to be able to grind through this… One day, I will… The first thing I need to do, obviously, is to thoroughly study how these equations for E and B above can be derived from Maxwell’s equations. I’ll need some time for that, and then I can see if it’s consistent with Einstein’s special and general theories of relativity.

I’ll update you on progress. 🙂

The field of electric and magnetic dipoles

You’l surely remember the electric field of an electric dipole, as depicted below:

We distinguish two components:

A z-component along the axis of the dipole itself, and
A so-called transverse component, i.e. the component that is perpendicular to the axis of the dipole, or the z-axis.

Pythagoras’ rule then gives us the total field, which is equal to:

I’ll give you the formulas for both components in a moment, but let’s first introduce the concept of a magnetic dipole. Look at the magnetic field of a solenoid below, and imagine we reduce the solenoid to one loop of current only. What would we get?

We get a magnetic field that resembles the electric field from an electric dipole. Of course, it’s a magnetic field, and it’s not the field of an electric dipole but of a magnetic dipole which, in this case, is the field of a small loop of current. Feynman depicts the geometry of the situation, and the relevant variables, as follows:

Now, in my previous post, in which I presented the magnetic vector potential, I pointed out that the equations for the x-, y- and z-component of the vector potential A are Poisson equations and, therefore, they are mathematically identical to the Poisson equation for electrostatics. Now that’s why the x-, y- and z-component of the vector potential arising from a current density j is exactly the same as the electric potential Φ that would be produced by a charge density $ρ$ equal to j_x, j_y and j_z respectively, divided by c² $, and we use that fact to calculate them. Huh? Yes.$ $Let me just jot it down the equations once more, so you can appreciate what I am saying here:$

∇²Φ = –ρ/ε₀(Poisson equation for electrostatics)

∇²A = –j/ε₀c²(Poisson equation for magnetostatics)

⇔ A_x = –j_x/ε₀c², A_y = –j_y/ε₀c²and A_z = –j_z/ε₀c²

We didn’t go through all of the nitty-gritty of the exercise for a straight wire and a solenoid in our previous post, but we will do so now for our single loop of current. It’s kinda tedious stuff, so just hang in there and try to get through it. 🙂

We’ll start with A_x. To calculate A_x, we need j_x, so that’s the component of the current density that’s shown below. [As you can see, j_zis zero, and we’ll talk about j_y in a minute.]

What are those + and − signs next to those two loop legs? Well… Think about what we wrote above: we need to think of j_xdivided by c² $as some charge density, and the associated Φ will then equal A x . It’s an easy principle: the same equations have the same solutions. We’re only switching symbols. So look at the geometry of the situation: we have a uniform current and, hence, a uniform current density, in each leg of the loop, so j x$ is some constant over the two legs shown here. Therefore, the electrostatic equivalent is that of two charged rods with some charge density λ.

Now, it’s easy to see that λ = j_x/c² = I/c²and… So… Well… Hmm… […] So how do we calculate the electrostatic potential Φ from two charged rods? Well… We haven’t calculated that in any of our posts, but if R is ‘large enough’ (and it should be, because we’re interested in the fields only at distances that are large compared to the size of the loop), then Φ will be the potential of a dipole with (opposite) charges that are equal to the (opposite) charges of the rods. Now, λ is a charge density, so we need to multiply it with the length of the rods a to get the charge. And then we need to multiply the charge with the distance between the two charges to get the dipole moment, so we write:

p = λab = Iab/c²

OK. That’s a nice result because we can now use that to calculate the potential. Indeed, you may or may not remember the formula for the potential of an electric dipole, but we wrote it as follows in one of my posts:

Φ = −p·∇φ₀= −(1/4πε₀)p·∇(1/R) = −(1/4πε₀)(p·e_r)/R²

∇φ₀is the gradient of φ₀, and φ₀is the potential of a unit point of charge: φ₀ = 1/4πε₀r. In any case, don’t worry too much about this formula right now, because I’ll give you some easier version of it later on. So let’s just move on. To calculate that vector product p·e_r = |p|·|e_r |·cosθ = p·cosθ, we need to know the angle θ between them, and the direction of the dipole moment. That’s simple enough: the vector p points in the negative y-direction, so cosθ = –y/R, where y is a coordinate of P. So we have:

Now we do the grand substitution, the hat-trick really: Φ becomes A_x, and λ = j_x/c² = I/c², and so we get:

Huh? Yes. It’s a great trick really. Brilliant! 🙂 You’ll get it. Just think about what happened for a while.

Next thing is j_y. So what about it? Well… As you may have suspected, it’s a perfectly symmetrical situation, so we can just repeat the same reasoning but swap x for y and y for x. We get:

Note the minus sign has disappeared, but you can explain that yourself. And, of course, we only have the vector potential here, so we still need to calculate B from it, using the B = ∇×A equation, which is three equations involving all possible cross-derivatives really:

Phew! It looks quite horrible indeed! Feynman says: “It’s a little more complicated than electrostatics, but the same idea.” But… Well… I guess it’s a bit hard to not take this statement with a pinch of salt. For starters, there is no intuitive physical interpretation of the magnetic vector potential A, as opposed to the electric potential Φ. More importantly, calculating E = −∇Φ involves only three derivatives, and that’s a helluva lot easier than calculating not less than six cross-derivatives and keep track of their order and all of that. In any case, nothing much we can do about, so let’s grind through it. And to be fully honest: A_z is zero, so the ∂A_z/∂y and ∂A_z/∂x derivatives above are also zero, so that leaves only four cross-derivatives to be calculated.

Let’s first define the magnetic dipole moment, however. We talked a lot about it, but what is it? We said it’s similar to the electric dipole moment p = q·d, whose magnitude was λab = Iab/c²in this example. So what have we got here? No charge, but a current, so that’s charge per second. No distance either, but some area a·b. And then the 1/c²factor that, somehow, always slips in because we’ve got it in Maxwell’s equation too. So, we said the fields look the same, and just like Φ was proportional to p, we now see that A is proportional to I·a·b, so it’s quite similar really and so we’ll just define the magnetic dipole moment as:

μ = (current)·(area of the loop) = I·a·b

Now, the area of the loop doesn’t have to be some square: it could also be some circle or some triangle or whatever. You should just change the formula to calculate the area accordingly. You’ll say: where’s the 1/c²factor? It’s true: we didn’t put it in. Why? I could say: that’s the convention, but that’s a bit too easy as an answer, I guess. I am not quite sure but the best answer I have is that the 1/c²factor has nothing to do with the physicality of the situation: it’s part of Maxwell’s equations, or the electromagnetic force equations, if you want. So it’s typical of the magnetic force but it’s got nothing to do with our little loop of current, so to speak. So that’s why we just leave it out in our definition of μ.

OK. That should be it. However, I should also note that we can (and, hence, we will) make a vector quantity out of μ, so then we write it in boldface: μ. The symmetry of the situation implies that μ should be some vector that’s normal to the plane of the loop because, if not, all directions would be equally likely candidates. But up or down? As you may expect, we’ll use the right-hand rule again, as illustrated below: if your fingers point in the direction of current, your thumb will point in μ‘s direction. The illustration below also gives the direction of A, because we can now combine the definition of μ and our formulas for A_x, A_y and A_z and write:

Note that the formula for A above is quite similar to the formula for Φ below, but not quite the same: in the formula for A we’ve got a vector cross product – so that’s very different from the vector dot product below – and then we’ve also got that 1/c²factor. So watch out: it’s the “same-same but different”, as they say in Asia. 🙂[…] OK. Sooooo much talk, and still I did not calculate B to effectively show that the magnetic field of a magnetic dipole looks like the electric field of an electric dipole indeed. So let me no longer postpone those tedious calculations and effectively do that B = ∇×A cross-product using our new formula for A. Well… Calculating B_x and B_y is relatively simple because A‘s z-component is zero, so ∂A_z/∂x = ∂A_z/∂y = 0. So for B_x, we get:

For B_y and B_z, we get:

The … in the formulas above obviously stands for μ/4πε₀c². Now, let’s compare this with the components of the electric field (E) of an electric dipole. Let’s re-write the formula for the electric potential above as:

Note that this formula assumes the z-axis points in the same direction as the dipole, and that the dipole is centered at the origin of the coordinate system, so it’s the same coordinate system as the one for the magnetic dipole above. Therefore, we have that cosθ = z/r, and so the formula above is equivalent to:

Now, let’s find the electric field by using the E = (E_x, E_y, E_z) = −∇Φ = (−∂Φ/∂x, Φ, −∂Φ/∂y, −∂Φ/∂z) equation. For E_z, we get:

which we can re-write as:

Huh? Yes. Note that z also appears in r, because r = (x²+y²+z²)^1/2, and so we need to apply the product (or quotient) rule for derivatives: (u·v)’ = u’·v + u·v’. Note that this simplifies to E = (1/4πε₀)·(2p/r³) for the field at distance r from the dipole in the direction along its axis.

OK, so we’re done with E_z. For E_xand E_y, we get:

So… Well… Yes! We’ve checked it: the formulas for E_x, E_y, and E_z have exactly the same shape as those for B_x, B_y, and B_z, except for that 1/c² factor and, of course, the fact that we switched p for μ, so we may have some other number there too. [Oh – before I forget – I promised to give you the formula for the transverse component. The transverse component is, obviously, equal to (E_x² + E_y²)^1/2. So it’s just Pythagoras’ rule once more. Let me refer you to Feynman for the explicit formula, i.e. the formula in terms of r and θ, or in terms of x and y, as I don’t need it here and there’s too much clutter already in this post.]

So… Well… Yes, that’s all very interesting! Indeed, as Feynman notes, it is quite remarkable that, starting with completely different laws, $\nabla • E =ρ/ϵ 0$ and $\nabla \times B = j / ϵ 0 c 2$ , we end up with the same kind of field.

Is it? Well… Yes and no. You should, of course, note that the sources whose configuration we summarizing here by the dipole moments are physically quite different—in one case, it’s a circulating current; in the other, a pair of charges—one above and one below the plane of the loop for the corresponding field. So… Well… It’s the same-same but different indeed! 🙂

Now, to conclude this post, we should, perhaps, have a look at the units and order of magnitude of what we’re talking about here. The unit of charge and the unit of current are related: 1 ampere is a current of 1 coulomb per second. The coulomb is a rather large unit of charge, as it’s equivalent to the charge of some 6.241×10¹⁸ protons. Now, that’s less than a mole, which is about 6×10²³, but it’s still a quite respectable number. 🙂 Having said that, a current of one ampere is not so exceptional in everyday life, and so a magnetic dipole moment of 1 A·m² (i.e. the equivalent of an electric dipole moment of 1 C·m) would not be exceptional either.

The really interesting thing about the magnetic force is that 1/c² factor, which causes its magnitude to be so much smaller than that of the electric force. Indeed, the electric field at distance r from the dipole in the direction along its axis is given by E = (1/4πε₀)·(2p/r³). Assuming p = 1 C·m, and noting that 1/4πε₀≈ 9×10⁹N·m²/C² we get E ≈ (18×10⁹)·r⁻³N/C. So we’ve got an incredibly large (18×10⁹) factor in front here! [Note how the units come out alright: the electric field is effectively measured in newton per coulomb, indeed.]

What do we get for the magnetic field (along the z-axis once more) from a magnetic dipole with moment μ = 1 A·m²? The formula is now B = (1/4πε₀c²)·(2μ/r³). Now, 1/4πε₀c² ≈ (1×10⁻⁷)N·s²/C², so B ≈ (2×10⁻⁷)·r⁻³N·s/C·m. So we’ve got an incredibly small (2×10⁻⁷) factor in front here! [As for the unit, 1 N·s/C·m is the unit of the magnetic field indeed. It’s referred to as the tesla (1 T = 1 N·s/C·m), and it makes the force equation come out alright: the magnetic force on some charge q is, indeed, F_magnetic = qv×B, with v expressed in m/s, and so that’s why we need to multiply the N/C factor with an s/m factor in the unit for B.]

So look at this! The point to note is the relative strength – I should say weakness – of the magnetic force as compared to the electric force: they differ with a factor equal to c²≈ 9×10¹⁶. That’s quite incredible, especially because our electric motors and generators work through the magnetic force, as I’ll show in my next posts. In fact, looking at the F_magnetic = qv×B, and comparing it to the electric force (F_electric = q·E), we may say that the magnitude of the magnetic force, as compared to the magnitude of the electric force, is weaker because of two factors:

First, there is the relative velocity of the charge that we are looking, which is equal to β = v/c. The electrostatic Coulomb force has no such factor.
Second, there is the 1/c factor, which we also encountered when discussing radiation pressure, or the momentum of light, which is equal to p = E/c, with the symbol p denoting momentum here – not some dipole moment – and with E denoting the energy of the light photons – not the electric field involved!

Both combine to give us the actual relative strength of the magnetic versus the electric force, so we can write that relative strength as:

|F_magnetic|/|F_electric| = F_B/F_E = (v/c)·(1/c) = v/c²

Now, I am tempted to now write a lot about that 1/c factor, but I can’t do that here, as this post has become way too long already. Let me just add a small addendum here. It’s, perhaps, a not-so-easy piece 🙂 but I warmly recommend trying to work through it.

Addendum: On the momentum of light and radiation pressure

Look at the illustration below: it represents a beam of light, i.e. electromagnetic radiation, originating at some source S and hitting some charge q. What charge? Whatever charge the electromagnetic radiation is affecting, so you may want to think of an electron or a proton in a mirror or a piece of glass, or whatever other surface that is absorbing a photon fully or partially—so that’s any charge that is absorbing the energy of the light.

I’ve written a post on this before. It concluded a series of posts on electromagnetic radiation by noting that we usually look at the electric field only when discussing radiation. The electric field is given by the following formula for E:

I know. It looks like a monster, but the first term is just the Coulomb effect. For the explanation of the second and term term, I need to refer you to the mentioned section in Feynman’s Lecture on it. [Sorry for that, but I am still struggling somewhat with this equation myself. If and when I fully ‘get it’ myself, I’ll jot down a summary of my understanding of it. Now now, however.]

The point is: when discussing light, like interference and what have you, we usually forget about the B vector, for which the formula is the following:

So we define B by referring to E. Of course, this again shows the electric and magnetic force are one and the same phenomenon, really. The cross-product gives you both (i) the magnitude of B, which is equal to B = E/c (the magnitude of the unit vectors e_r’and n in the vector cross-product are obviously equal to one, and sinθ is equal to one too), and (ii) its direction: it’s perpendicular to both E as well as to e_r’, which is the unit vector from the point P where E is measured to the charge q that is producing the field.

The B = E/c equation tells us why the magnetic field is never looked at when discussing electromagnetic radiation: because it’s so tiny. It’s so tiny because of that 1/c factor, and so the B vector in the illustration above is surely not to scale: if all was drawn to scale, you just wouldn’t see it, because it’s 1/c times smaller than E. However, it’s there, and the illustration shows how the magnetic force resulting from it looks like, if we forget about scale, that is. 🙂

The magnetic force is, obviously, F = qv×B, and you need to apply the usual right-hand screw rule to find the direction of the force. [Please note that the order of magnitude of the force is the same as that of B so… Well… Again I need to warn you the illustration is not to scale and, hence, somewhat misleading. But I can’t think of an alternative to it.] As you can see, the magnetic force – as tiny as it is – is oriented in the direction of propagation, and it is what is responsible for the so-called radiation pressure.

Indeed, there is a ‘pushing momentum’ here, and we can calculate how strong it is. In fact, we should check how strong it is in order to see if it could, potentially, make a space ship like the one below. 🙂

Sorry. That was just to add a lighter note. 🙂 I know posts like these are rather tedious, so I just wanted to lighten it all up. 🙂 So… Well… Back to our F = qv×B equation. Because of that B = –e_r’×E/c formula, we can substitute B for E/c, and so we get the following formula for the magnitude of the magnetic force:

F = q·v·E/c

Now, the charge q times the electric field is the electric force on the charge, and the force on the charge times the velocity is equal to dW/dt, so that’s the time rate of change of the work that’s being done on the charge.

Huh? I know: not easy to see what’s being done here. Think of it like this: work equals force times distance, so W = qE·s in this case. Hence, dW/dt = d(qE·s)/dt = qE·ds/dt = qE·ds/dt. [Note that, if you’d check Feynman’s analysis, you’ll find that it’s based on average values here for some reason, so he writes 〈F〉 and 〈E〉, but I am not quite sure why, so I simplified here. I’ll try to figure the why of these averages later, and let you know.]

Again, think physical, rather than mathematical: what’s the charge here? It’s whatever charge the electromagnetic radiation is hitting, so you should think of an electron or a proton in a mirror or a piece of glass or whatever other surface. Now, because of the energy conservation principle, dW/dt must also be equal to the energy that is being absorbed from the light per second. So F = (dW/dt)/c.

Now, Newton’s Law (he had many, of course, so let me be precise: I am talking Newton’s Second Law of Motion) tells us that the force is the rate of change of the momentum: F = m·a = m·dv/dt = d(m·v)/dt. Hence, integrating both sides of the F = (dW/dt)/c equation gives us the associated momentum of the light: p = W/c.

In short, we know that light carries energy, but so we also know that light also carries momentum, and that momentum is due to the magnetic force, and it’s equal to 1 $/c$ times the energy. To be clear: it’s real pressure, so when light is emitted from a source there is a recoil effect, and when light hits some charge, it’s the same thing: the momentum of the light is being conserved as the charge that’s absorbing the energy picks it up. To be fully complete: we also have Newton’s Third Law of Motion coming into play: for every action, there is an equal and opposite reaction.

So… Well… That’s about it, I think. Just note that you’ll usually see this momentum written as $p = E/c, with E denoting the energy of the radiation, not the electric field (sorry for the switch in symbols). In this equation, E is given by the Planck-Einstein relation E = h \cdot f, with h Planck’s constant and f the frequency of the light.$

Interesting! So it all makes sense! Isn’t it wonderful how all these equations come together in the end? 🙂 However, I shouldn’t digress even more, and so I’ll leave it to you to further reflect on this. 🙂

Magnetostatics: the vector potential

This and the next posts are supposed to wrap up a few loose ends on magnetism. One of these loose ends is the (magnetic) vector potential, which we introduced in our post on gauge transformations, but then we didn’t do much with it. Another topic I neglected so far is that of the magnetic dipole moment (as opposed to the electric dipole moment), which is an extremely important concept both in classical as well as in quantum mechanics. So let’s do the vector potential here, and the magnetic dipole moment in the next. 🙂

Let’s go for it. Let me recall the basics which, as usual, are just Maxwell’s equations. You’ll remember that the electrostatic field was curl-free: ∇×E = 0, everywhere. Therefore, we can apply the following mathematical theorem: if the curl of a vector field is zero (everywhere), then the vector field can be represented as the gradient of some scalar function:

if ∇×C = 0, then there is some Ψ for which C = ∇Ψ

Substituting C for E, and taking into account our conventions on charge and the direction of flow, we wrote:

E = –∇Φ

Φ (phi) is referred to as the electric potential. Combining E = –∇Φ with Gauss’ Law – ∇•E = ρ/ε₀ − we got Poisson’s equation:

∇²Φ = −ρ/ε₀

So that equation sums up all of electrostatics. Really: that’s it! 🙂

Now, the two equations for magnetostatics are: ∇•B = 0 and c²∇×B = j/ε₀. Let me say something more about them:

The ∇•B = 0 equation is true, always, unlike the ∇×E = 0 expression, which is true for electrostatics only (no moving charges).
The ∇•B = 0 equation says the divergence of B is zero, always. Now, you can verify for yourself that the divergence of the curl of a vector field is always zero, so div (curl A) = ∇•(∇×A) = 0, always. Therefore, there’s another theorem that we can apply. It says the following: if the divergence of a vector field, say D, is zero – so if ∇•D = 0, then $D will be the$ the curl of some other vector field $C, so we can write: D = \nabla \times C . Applying this to \nabla • B = 0, we can write:$

If ∇•B = 0, then there is an A such that B = ∇×A

We can also write this as follows: ∇·B = ∇·(∇×A) = 0 and, hence, B = ∇×A. Now, it’s this vector field A that is referred to as the (magnetic) vector potential, and so that’s what we want to talk about here. As a start, it may be good to write all of the components of our B = ∇×A vector:

Note that we have no ‘time component’ because we assume the fields are static, so they do not change with time. Now, because that’s a relatively simple situation, you may wonder whether we really simplified anything with this vector potential. B is a vector with three components, and so is A. The answer to that question is somewhat subtle, and similar to what we did for electrostatics: it’s mathematically convenient to use A, and then calculate the derivatives above to find B. So the number of components doesn’t matter really: it’s just more convenient to first get A using our data on the currents j, and then we get B from A.

That’s it really. Let me show you how it works. The whole argument is somewhat lengthy, but it’s not difficult, and once it’s done, it’s done. So just carry on and please bear with me 🙂

First, we need to put some constraints on A, because the B = ∇×A equation does not fully define A. It’s like the scalar potential Φ: any Φ’ = Φ + C was as good a choice as Φ (with C any constant), so we needed a reference point Φ = 0, which we usually took at infinity. With the vector potential A, we have even more latitude: we can not only add a constant but any field which is the gradient of some scalar field, so any A’ = A + ∇Ψ will do. Why? Just write it all out: ∇×(A + ∇Ψ) = ∇×A + ∇×(∇Ψ). But the curl of the gradient of a scalar field (or a scalar function) is always zero (you can check my post on vector calculus on this), so ∇×(∇Ψ) = 0 and so ∇×(A + ∇Ψ) = ∇×A + ∇×(∇Ψ) = ∇×A + 0 = ∇×A = B.

So what constraints should we put on our choice of A? The choice is, once again, based on mathematical convenience: in magnetostatics, we’ll choose A such that ∇•A = 0. Can we do that? Yes. The A’ = A + ∇Ψ flexibility allows us to make ∇•A’ anything we wish, and so A and A’ will have the same curl, but they don’t need to have the same divergence. So we can choose an A’ so ∇•A’ = 0, and then we denote A’ by A. 🙂 So our ‘definition’ of the vector potential A is now:

B = ∇×A and ∇•A = 0

I have to make two points here:

First, you should note that, in my post on gauges, I mentioned that the choice is different when the time derivatives of E and B are not equal to zero, so when we’re talking changing currents and charge distributions, so that’s dynamics. However, that’s not a concern here.
To be fully complete, I should note that the ‘definition’ above does still not uniquely determine A. For a unique specification, we also need some reference point, or say how the field behaves on some boundary, or at large distances. It is usually convenient to choose a field which goes to zero at large distances, just like our electric potential.

Phew! We’ve said so many things about A now, but nothing that has any relevance to how we’d calculate A. 😦 So we are we heading here?

Fortunately, we can go a bit faster now. The c²∇×B = j/ε₀ equation and our B = ∇×A give us:

c²∇×(∇×A) = j/ε₀

Now, there’s this other vector identity, which you surely won’t remember either—but trust me: I am not lying: ∇×(∇×A) = ∇(∇•A) − ∇²A. So, now you see why we choose A such that ∇•A = 0 ! It allows us to write:

c²∇×(∇×A) = − c²∇²A = j/ε₀⇔ ∇²A = –j/ε₀c²

Now, the three components of ∇²A = –j/ε₀c²are, of course:

As you can see, each of these three equations is mathematically identical to that Poisson equation: ∇²Φ = − ρ/ε₀. So all that we learned about solving for potentials when ρ is known can now be used to solve for each component of A when j is known. Now, to calculate Φ, we used the following integral:

Simply substituting symbols then gives us the solution for A_x:

We have a similar integral for A_y and A_z, of course, and we can combine the three equations in vector form:

Finally, and just in case you wonder what is what, there’s the illustration below (taken from Feynman’s Lecture on this topic here) that, hopefully, will help you to make sense of it all.

At this point, you’re probably tired of these formulas (or asleep) or (if you’re not asleep) wondering what they mean really, so let’s do two examples. Of course, you won’t be surprised that we’ll be talking a straight wire and a solenoid respectively once again. 🙂

The magnetic field of a straight wire

We already calculated the magnetic field of a straight wire, using Ampère’s Law and the symmetry of the situation, in our previous post on magnetostatics. We got the following formula:

Do we get the same using those formulas for A and then doing our derivations to get B? We should, and we do, but I’ll be lazy here and just refer you to the relevant section in Feynman’s Lecture on it, because the solenoid stuff is much more interesting. 🙂

The magnetic field of a solenoid

In the mentioned post on magnetostatics, we also derived a formula for the magnetic field inside a solenoid. We got:

solenoid formula 2 with $n$ the number of turns per unit length of the solenoid, and I the current going through it. However, in the mentioned post, we assumed that the magnetic field outside of the solenoid was zero, for all practical purposes, but it is not. It is very weak but not zero, as shown below. In fact, it’s fairly strong at very short distances from the solenoid! Calculating the vector potential allows us to calculate its exact value, everywhere. So let’s go for it.

The relevant quantities are shown in the illustration below. So we’ve got a very long solenoid here once again, with n turns of wire per unit length and, therefore, a circumferential current on the surface of n·I per unit length (the slight pitch of the winding is being neglected).

Now, just like that surface charge density ρ in electrostatics, we have a ‘surface current density’ J here, which we define as J = n·I. So we’re going from a scalar to a vector quantity, and the components of J are:

J_x = –J·sinϕ, J_y = –J·cosϕ, J_z = 0

So how do we do this? As should be clear from the whole development above, the principle is that the x-component of the vector potential arising from a current density j is the same as the electric potential Φ that would be produced by a charge density $ρ$ equal to j_x divided by c² $, and similarly for the y- and z-components. Huh? Yes. Just read it a couple of times and think about it: we should imagine some cylinder with a surface charge ρ = -(J/ c 2) \cdotsinϕ to calculate A x . And then we equate ρ with -(J/ c 2) \cdotcosϕ and zero respectively to find A y and A z .$

Now, that sounds pretty easy but Feynman’s argument is quite convoluted here, so I’ll just skip it (click the link here if you’d want to see it) and give you the final result, i.e. the magnitude of A:

Of course, you need to interpret the result above with the illustration, which shows that A is always perpendicular to r’. [In case you wonder why we write r’ (so r with a prime) and not r, that’s to make clear we’re talking the distance from the z-axis, so it’s not the distance from the origin.]

Now, you may think that c² $in the denominator explains the very weak field, but it doesn’t: it’s the inverse proportionality to r’ that makes the difference!$ Indeed, you should compare the formula above with the result we get for the vector potential inside of the solenoid, which is equal to:

The illustration below shows the quantities involved. Note that we’re talking a uniform magnetic field here, along the z-axis, which has the same direction as B₀and, hence, is pointing towards you as you look at the illustration, which is why you don’t see the B₀ field lines and/or the z-axis: they’re perpendicular to your computer screen, so to speak.

As for the direction of A, it’s shown on the illustration, of course, but let me remind you of the right-hand rule for the vector cross product a×b once again, so you can make sense of the direction of A = (1/2)B₀×r’ indeed:

Also note the magnitude this formula implies: a×b = |a|·|b|·sinθ·n, with θ the angle between a and b, and n the normal unit vector in the direction given by that right-hand rule above. Now, unlike a vector dot product, the magnitude of the vector cross product is not zero for perpendicular vectors. In fact, when θ = π/2, which is the case for B₀and r’, then sinθ = 1, and, hence, we can write:

|A| = A = (1/2)|B₀||r’| = (1/2)·B₀·r’

Now, just substitute B₀for B₀= n·I/ε₀c², which is the field inside the solenoid, then you get:

A = (1/2)·n·I·r’/ε₀c²

You should compare this formula with the formula for A outside the solenoid, so you can draw the right conclusions. Note that both formulas incorporate the same (1/2)·n·I/ε₀c²factor. The difference, really, is that inside the solenoid, A is proportional to r’ (as shown in the illustration: if r’ doubles, triples etcetera, then A will double, triple etcetera too) while, outside of the solenoid, A is inversely proportional to r’. In addition, outside the solenoid, we have the a²factor, which doesn’t matter inside. Indeed, the radius of the solenoid (i.e. a) changes the flux, which is the product of B and the cross-section area π·a², but not B itself.

Let’s do a quick check to see if the formula makes sense. We do not want A to be larger outside of the solenoid than inside, obviously, so the a²/r’ factor should be smaller than r’ for r’ > a. Now, a²/r’ < r’ if a²< r’², and because a an r’ are both positive real numbers, that’s the case if r’ > a indeed. So we’ve got something that resembles the electric field inside and outside of a uniformly charged sphere, except that A decreases as 1/r’ rather than as 1/r’², as shown below.

Hmm… That’s all stuff to think about… The thing you should take home from all of this is the following:

A (uniform) magnetic field B in the z-direction corresponds to a vector potential A that rotates about the z-axis with magnitude A = B₀·r’/2 (with r’ the displacement from the z-axis, not from the origin—obviously!). So that gives you the A inside of a solenoid. The magnitude is A = (1/2)·n·I·r’/ε₀c², so A is proportional with r’.
Outside of the solenoid, A‘s magnitude (i.e. A) is inversely proportional to the distance r’, and it’s given by the formula: A = (1/2)·n·I·a²/ε₀c²·r’. That’s, of course, consistent with the magnetic field diminishing with distance there. But remember: contrary to what you’ve been taught or what you often read, it is not zero. It’s only near zero if r’ >> a.

Alright. Done. Next post. So that’s on the magnetic dipole moment 🙂

Ferroelectrics and ferromagnetics

Ferroelectricity and ferromagnetism are two different things, but they are analogous. Materials are ferroelectric if they have a spontaneous electric polarization that can be changed or reversed by the application of an external electric field. Ferromagnetism, in contrast, refers to materials which exhibit a permanent magnetic moment.

The materials are very different. In fact, most ferroelectric materials do not contain any iron and, hence, the ferro in the term is somewhat misleading. Ferroelectric materials are a special class of crystals, like barium or lead titanate (BaTiO₃or PbTiO₃). Lead zirconate titanate (LZT) is another example. These materials are also piezoelectric: when applying some mechanical stress, they will generate some voltage. In fact, the process goes both ways: when applying some voltage to them, it will also create mechanical deformation, as illustrated below (credit for this illustration goes to Wikipedia).

Ferroelectricity has to do with electric dipoles, while ferromagnetism has to do with magnetic dipoles. We’ve only discussed electric dipoles so far (see the section on dielectrics in my post on capacitors) and so we’re only in a position to discuss ferroelectricity right now, which is what I’ll do here. However, before doing so, let me briefly quote from the Wikipedia article on ferromagnetism, because that’s really concise and to the point on this:

“One of the fundamental properties of an electron (besides that it carries charge) is that it has a magnetic dipole moment, i.e. it behaves like a tiny magnet. This dipole moment comes from the more fundamental property of the electron that it has quantum mechanical spin. Due to its quantum nature, the spin of the electron can be in one of only two states; with the magnetic field either pointing “up” or “down” (for any choice of up and down). The spin of the electrons in atoms is the main source of ferromagnetism, although there is also a contribution from the orbital angular momentum of the electron about the nucleus.”

In short, ferromagnetism was discovered and known much before ferroelectricity was discovered and studied, but it’s actually more complicated, because it’s a quantum-mechanical thing really, unlike ferroelectricity, which we’ll discuss now. Before we start, let me note that, in many ways, this post is a continuation of the presentation on dielectrics, which I referred to above already, so you may want to check that discussion in that post I referred to if you have trouble following the arguments below.

Molecular dipoles

Let me first remind you of the basics. The (electric) dipole moment is the product of the distance between two equal but opposite charges q₊ and q₋. Usually, it’s written as a vector so as to also keep track of its direction and use it in vector equations, so we write p = qd, with d the vector going from the negative to the positive charge, as shown below.

Now, molecules like water molecules have a permanent dipole moment, as illustrated below. It’s because the center of ‘gravity’ of the positive and negative charges do not coincide, so that’s what makes the H₂O molecule polar, as opposed to the O₂ molecule, which is non-polar.

Now, if we place polar molecules in some electric field, we’d expect them to line up, to some extent at least, as shown below (the second illustration has more dipoles pointing vaguely north).

However, at ordinary temperatures and electric fields, the collisions of the molecules in their thermal motion keeps them from lining up too much. In fact, we can apply the principles of statistics mechanics to calculate how much exactly. You can check out the details in Feynman’s Lecture on it, but the result is that the net dipole moment per unit volume (so that’s the polarization) is equal to:

So the polarization is proportional to the number of molecules per unit volume (N), the square of their dipole moment (p₀) and, as we’d might expect, the electric field E, and inversely proportional to the temperature (T). In fact, the formula above is a sort of first-order approximation, in line with what we wrote on the electric susceptibility χ (chi) in our post on capacitors, where we also assumed the relation between P and E was linear, so we wrote: P = ε₀·χ·E. Now, engineers and physicists often use different symbols and definitions and so you may of may not have heard about another concept saying essentially the same thing: the dielectric constant, which is denoted by κ (kappa) and is, quite simply, equal to κ = 1 + χ. Combining the expression for P above, and the P = ε₀·χ·E = ε₀·(κ−1)·E expression, we get:

This doesn’t say anything new: it just states the dependence of χ on the temperature. Now, you can imagine this linear relationship has been verified experimentally. As it turns out, it’s sort of valid, but it is not as straightforward as you might imagine. There’s a nice post on this on the University of Cambridge’s Materials Science site. But this blog is about physics, not about materials science, so let’s move on. The only thing I should add to this section is a remark on the energy of dipoles.

You know charges in a field have energy, potential energy. You can look up the detail behind the formulas in one of my other posts on electromagnetism, so I’ll just remind you of them: the energy of a charge is, quite simply, the product of the charge (q) and the electric potential (Φ) at the location of the charge. Why? Well… The potential is the amount of work we’d do when bringing the unit charge there from some other (reference) point where Φ = 0. In short, the energy of the positive charge is q·Φ(1) and the energy of the negative charge is −q·Φ(2), with 1 and 2 denoting their respective location, as illustrated below.

So we have U = q·Φ(1) − q·Φ(2) = q·[Φ(1)−Φ(2)]. Now, we’re talking tiny little dipoles here, so we can approximate ΔΦ = Φ(1)−Φ(2) by ΔΦ = d•∇Φ = Δx·(∂Φ/∂x) + Δy·(∂Φ/∂y). Hence, also noting that E = −∇Φ and qd = p₀, we get:

U = q·Φ(1) − q·Φ(2) = qd•∇Φ = −p₀•E = −p₀·E·cosθ, with θ the angle between p₀and E

So the energy is lower when the dipoles are lined up with the field, which is what we would expect, of course. However, it’s an interesting thing so I just wanted to show you that. 🙂

Electrets, piezoelectricity and ferroelectricity

The analysis above was very general, so we actually haven’t started our discussion on ferroelectricity yet! All of the above is just a necessary introduction to the topic. So let’s move on. Ferroelectrics are solids, so let’s look at solids. Let me just copy Feynman’s introduction here, as it’s perfectly phrased:

“The first interesting fact about solids is that there can be a permanent polarization built in—which exists even without applying an electric field. An example occurs with a material like wax, which contains long molecules having a permanent dipole moment. If you melt some wax and put a strong electric field on it when it is a liquid, so that the dipole moments get partly lined up, they will stay that way when the liquid freezes. The solid material will have a permanent polarization which remains when the field is removed. Such a solid is called an electret. An electret has permanent polarization charges on its surface. It is the electrical analog of a magnet. It is not as useful, though, because free charges from the air are attracted to its surfaces, eventually cancelling the polarization charges. The electret is “discharged” and there are no visible external fields.”

Another example (i.e. other than wax) of an electret is the crystal lattice below. As you can see, all the dipoles are pointing in the same direction even with no applied electric field. Many crystals have such polarization but, again, we do not normally notice it because the external fields are discharged, just as for the electrets.

Now, this gives rise to the phenomena of pyroelectricity and piezoelectricity. Indeed, as Feynman explains: “If these internal dipole moments of a crystal are changed, external fields appear because there is not time for stray charges to gather and cancel the polarization charges. If the dielectric is in a condenser, free charges will be induced on the electrodes. The moments can also change when a dielectric is heated, because of thermal expansion. The effect is called pyroelectricity. Similarly, if we change the stresses in a crystal—for instance, if we bend it—again the moment may change a little bit, and a small electrical effect, called piezoelectricity, can be detected.”

But, still, piezoelectricity is not the same as ferroelectricity. In fact, there’s a hierarchy here:

Out of all crystals, some will be piezoelectric.
Among all piezoelectric crystals, some will also be pyroelectric.
Among the pyroelectric crystals, we can find some ferroelectric crystals.

The defining characteristic of ferroelectricity is that the built-in permanent moment can be reversed by the application of an external electric field. Feynman defines them as “nearly cubic crystals, whose moments can be turned in different directions, so we can detect a large change in the moment when an applied electric field is changed: all the moments flip over and we get a large effect.”

Because this is a blog, not a physics handbook, I’ll refer you to Feynman and/or the Wikipedia article on ferroelectricity for an explanation of the mechanism. Indeed, the objective of this post is to explain what it is, and so I don’t want to go off into the weeds. The two diagrams below, which I took from the mentioned Wikipedia article, illustrate the difference between your average dielectric material as opposed to a ferroelectric material. The first diagram shows you the linear relationship between P and E we discussed above: if we reverse the field, so E becomes negative, then the polarization will be reversed as well, but gradually, as shown below.

In contrast, the illustration below shows a hysteresis effect, which can be used as a memory function, and ferroelectric materials are indeed used for ferroelectric RAM (FeRAM) memory chips for computers! I’ll let you google that for yourself − it’s fun: just have a look at the following link, for example − because it’s about time I start wrapping up this post. 🙂

OK. That’s it for today. More tomorrow. 🙂

Magnetostatics

Original post:

Not all is exciting when studying physics. In fact, electromagnetism is, most of the time, a extremely boring subject-matter. But so we need to get through it, because we need the math and the formulas. So… Here we go…

When going from electrostatics to electrodynamics, one first needs to have a look at magnetostatics, to get familiar with (steady) electric currents. So let’s have a look at what they are. Of course, you already know what steady currents are. In that case, you should, perhaps, stop reading. But I’d recommend you go through it anyway. It’s always good to be explicit, so let’s be explicit.

Let me first make a very pedantic note. There are a couple of sections in Feynman’s Lectures in which he assumes that a steady current in a wire is uniformly distributed throughout the cross-section of the current-carrying wire: that assumption amounts to saying that the current density j is uniform. He uses that assumption, for example, when calculating the force per unit length of a current-carrying wire in a magnetic field (see Vol. II, section 13-3). He also uses it when calculating the magnetic field it creates itself (see Vol. II, section 14-3). This raises two questions:

Is the assumption true?
Does it matter?

My impression is that it’s a simplification that doesn’t matter. So the answer to both question would be negative. But let’s examine them. First note that, in previous posts, we repeatedly said that, if we place a charge Q on any conductor, all charges will spread out in some way on the surface, so we have an equipotential on the surface and no electric field inside of the conductor. The physics behind are easy to understand: if there were an electric field inside of the conductor, and the surface were not an equipotential, the charges would keep moving until it became zero.

Does it matter? Maybe. Maybe not. I discussed the electric field from a conductor in a previous post, so let me just recall some formulas here, first and foremost Gauss’ Law, which says that the electric flux from any closed surface S is equal to Q_inside/ε₀. Now, Q_insideis, obviously, the sum of the charges inside the volume enclosed by the surface, and the most remarkable thing about Gauss’ Law is that the charge distribution inside of the volume doesn’t matter. So if we’re talking a uniformly charged sphere or a thin spherical shell of charge, it’s the same. The illustration below shows the field for a uniformly charged sphere: E is proportional to r (to be precise: E = (ρ·r)/(3ε₀) for r ≤ R) inside the sphere, and outside E is proportional to 1/r² (to be precise: E = Q_inside/(4πε₀r²) for r ≥ R).

However, Gauss’ Law is a law that gives us the electric flux only, so we’re talking E only. We also have the magnetic field, i.e. the field vector B. So what’s the equivalent of Gauss’ Law for B? That’s Ampère’s Law, obviously, so let’s have a look at how Feynman derives that law.

Ampère’s Law

Feynman starts by defining the current through some surface S as the following integral:

The illustration below explains the logic behind. The vector j is like the heat flow vector h which we used when explaining the basics of vector calculus: it is some amount passing expressed per unit time and per unit area. As for the use of n, that’s the same normal unit vector we used for h as well: we then wrote that h·n = |h|·|n|·cosθ = h·cosθ was the component of the heat flow that’s perpendicular or normal (as mathematicians prefer to say) to the surface. So here we’ve got the same: j·n·dS is the amount of charge flowing across an infinitesimally small area dS in a unit time. So to get the electric current I, which is the total charge passing per unit time through a surface S, we need to integrate the normal component of the flow through all the surface elements, which is what the integral above is doing.

Note that I is not a vector but a scalar. We could, however, include the idea of the direction of flow by making I a vector, so then we write it in boldface: I. It is measured in coulomb per second, aka as ampere: 1 A = 1 C/s. Also note we don’t have any wires here: just surfaces and volumes. 🙂 Onwards!

The equations of magnetostatics are Maxwell’s third and fourth equation and, as we used Maxwell’s first and second equation to derive Gauss’ Law, we’ll use these two to derive Ampère’s Law: (1) ∇•B = 0 and (2) c²∇×B = j/ε₀.

You know these equations: the first one basically says there’s no flux of B: there’s no such thing as magnetic charges, in other words. The second one says that a current produces some circulation of B. You also know these equations are valid only for static fields: all electric charge densities are constant, and all currents are steady, so the electric and magnetic fields are not changing with time: ∂E/∂t = 0 = ∂B/∂t. Forget about c² for a moment (it’s just a constant) and note that ∇×B is referred to as the curl of B.

Now, as I pointed out in one of my posts on vector analysis, the divergence of the curl of a vector is always equal to zero, so ∇•(∇×B) = 0. However, because ∇×B = j/ε₀c², that means ∇•(j/ε₀c²) must also be equal to zero (we’re just taking the divergence of both sides of the equation here), and so we find that ∇•j must be equal to zero. What does that mean?Well… From the same post, you may or may not remember that the divergence of some vector field C (so that’s ∇•C) is the (net) flux out of an (infinitesimal) volume around the point we’re considering, so ∇•j = 0 implies that as much charge must be coming in as it going out, always and everywhere. So that means that, because of the charge conservation law (no charges are created or lost), we can only look at charges flowing in paths that close back on themselves, so we can only consider closed circuits. It’s a minor point – so don’t worry too much about it – but it does imply that we’re not looking at condensers, for example. Just remember: magnetostatics is about circulation, we have no flux, not of B, and not of j: our field, and our charges, circulate. 🙂

OK. Let’s get back to the lesson. We need to find Ampère’s Law, so we’d better get on with it. 🙂 To find Gauss’ Law, we used Gauss’ Theorem. To find Ampère’s Law, we’ll use… Stokes’ Theorem. [Sorry!] I need to refer you, once again, to that post on vector analysis for it. Here I can only remind you of the Theorem itself. It says that the line integral of the tangential component of a vector (field) around a closed loop is equal to the surface integral of the normal component of the curl of that vector over any surface which is bounded by the loop. […] I know that’s quite a mouthful, so let me jot down the equation:

Applying it to the magnetic field vector B, we get:

This is the illustration which goes with it.

Now, using our ∇×B = j/ε₀c² equation, we get:

Finally, we just plug in our I = ∫ j·n dS integral and we’re done. This is Ampère’s Law:

It basically says that the circulation of B around any closed curve is equal to the current I through the loop, divided by ε₀c². So what can we do with it? Well… We used Gauss’ Law to find the electric field in various circumstances, so let’s now use Ampère’s Law to find the magnetic field in various circumstances. 🙂

Before doing so, however, let me note that Ampère’s Law does not depend on any particular assumption in regard to the distribution of the charge densities j. So, frankly speaking, don’t worry too much about that assumption about a steady current in a wire: a current is a current in Ampère’s Law. 🙂

Wires

You know the magnetic field around a wire, as you’ll surely remember that right-hand rule for it from your high-school physics classes. Note, however, that it assumes you apply the usual convention: charge flows from positive to negative, because our unit of electric charge is obviously +1, not –1. So the electron flow actually goes the other way. 🙂

But so we’re past our high school days and we need to apply Ampère’s Law. The symmetry of the situation implies that that line integral of B·ds, taken along some closed circle around the wire, is, quite simply, the magnitude of B times the circumference r of our circle. Indeed, the symmetry of the situation implies that B at some distance r should be of the same magnitude everywhere, so we have:

But from Ampère’s Law we know that integral is equal to I/ε₀c² and, therefore, B·2π·r must equal I/ε₀c², and so we get the grand result we were looking for. The magnetic field outside of a (long) wire carrying the current I is:

As Feynman notes, we can write this in vector form to include the directions, remembering that B is at right angles both to I as well as to r, and remembering that the order matters, of course, because of the right-hand rule for a vector cross product. 🙂

Solenoids

Coils of wire, and solenoids, pop up almost everywhere when studying electromagnetism. Indeed, transformers, inductances, electrical motors: it’s all coils. So, yes, we can’t escape them. 😦 So let’s get on with it. As you know, a solenoid is a long coil of wire wound in a tight spiral. The illustrations below show a cross-section and its magnetic field.

Now, this is probably one of Feynman’s most intuitive arguments. Read: he’s cutting an awful lot of corners here. 🙂 I’ll just copy him:

We observe experimentally that when a solenoid is very long compared with its diameter, the field outside is very small compared with the field inside. Using just that fact, together with Ampère’s law, we can find the size of the field inside. Since the field stays inside (and has zero divergence), its lines must go along parallel to the axis, as shown above. That being the case, we can use Ampère’s law with the rectangular ‘curve’ Γ shown in the figure. This loop goes the distance $L$ inside the solenoid, where the field is, say, B₀, then goes at right angles to the field, and returns along the outside, where the field is negligible. The line integral of B for this curve is just B₀·L, and it must be $1/ε 0 c 2$ times the total current through Γ, which is $N\cdotI$ if there are $N$ turns of the solenoid in the length $L$ . We have:

Or, letting $n$ be the number of turns per unit length of the solenoid (that is, $n=N/L$ ), we get:

Oh… What happens to the lines of B when they get to the end of the solenoid? Well… They just spread out in some way and return to enter the solenoid at the other end. Hmm… He’s really cutting corners here, isn’t he? But the formula is right, and I’d rather keep it short—just like he seems to want to do here. 🙂 I’ll just insert an illustration showing another right-hand rule—the right-hand rule for solenoids: if the direction of the fingers of your right hand is the direction of current, then your thumb gives the direction of the magnetic field inside.

You may wonder: does it matter where the + and − ends of the coil are? Good question because, in practice, we’ll have something that’s very tightly wound, like the coil below, so when making an actual coil (click on this link for a nice video), we’ll have several rows and so we wind from right to left and then back from left to right and so on and so on. So if we’d have two rows of wire, the two ends of the wire would come out on the same side, and that’s OK.

Of course, the wire needs to be insulated. What you see on the picture (and in the video) is the use of so-called magnet wire, which has a polymer film insulation. So when making the electrical connections at both ends, after winding the coil, you need to get rid of the insulation, but then it often melts just by the heat of soldering. And now that we’re talking practical stuff, let me say something about the magnetic core you see in the illustration above.

A magnetic core is a material with high magnetic permeability as compared to the surrounding air, and this high permeability will cause the magnetic field to be concentrated in the core material. Now, there’s a phenomenon that’s called hysteresis, which means that the core material will tend to retain its magnetization when the applied field is removed. This is not very desirable in many applications, such as transformers or electric engines. That’s why so-called ‘soft’ magnetic materials with low hysteresis are often preferred. The so-called soft iron is such material: it’s literally softer because of a heat treatment increasing its ductility and reducing its hardness. Of course, for permanent magnets, a so-called ‘hard’ magnetic material will be used. But here we’re getting into engineering and that’s not what I want to write about in this blog.

I’ll just end by noting that a magnetic field has a so-called north (N) and south (S) pole. That convention refers to the Earth’s north and south pole, of course. However, since opposite poles (north and south) attract, the North Magnetic Pole is actually the south pole of the Earth’s magnetic field, and the South is the north. 🙂 So it’s better not to think too much of the Earth’s poles when discussing the poles of a magnet. By convention, a magnet’s north pole is where the field lines of a magnet emerge, and the south pole is where they enter, as shown below.

In any case… Folks: that’s it for today. I’ll continue tomorrow. 🙂

A post for Vincent: on the math of waves

Pre-scriptum (dated 26 June 2020): These posts on elementary math and physics for my kids (they are 21 and 23 now and no longer need such explanations) have not suffered much the attack by the dark force—which is good because I still like them. While my views on the true nature of light, matter and the force or forces that act on them have evolved significantly as part of my explorations of a more realist (classical) explanation of quantum mechanics, I think most (if not all) of the analysis in this post remains valid and fun to read. In fact, I find the simplest stuff is often the best. 🙂

Original post:

I wrote this post to just briefly entertain myself and my teenage kids. To be precise, I am writing this for Vincent, as he started to study more math this year (eight hours a week!), and as he also thinks he might go for engineering studies two years from now. So let’s see if he gets this and − much more importantly − if he likes the topic. If not… Well… Then he should get even better at golf than he already is, so he can make a living out of it. 🙂

To be sure, nothing what I write below requires an understanding of stuff you haven’t seen yet, like integrals, or complex numbers. There’s no derivatives, exponentials or logarithms either: you just need to know what a sine or a cosine is, and then it’s just a bit of addition and multiplication. So it’s just… Well… Geometry and waves as I would teach it to an interested teenager. So let’s go for it. And, yes, I am talking to you now, Vincent! 🙂

The animation below shows a repeating pulse. It is a periodic function: a traveling wave. It obviously travels in the positive x-direction, i.e. from left to right as per our convention. As you can see, the amplitude of our little wave varies as a function of time (t) and space (x), so it’s a function in two variables, like y = F(u, v). You know what that is, and you also know we’d refer to y as the dependent variable and to u and v as the independent variables.

Now, because it’s a wave, and because it travels in the positive x-direction, the argument of the wave function F will be x−ct, so we write:

y = F(x−ct)

Just to make sure: c is the speed of travel of this particular wave, so don’t think it’s the speed of light. This wave can be any wave: a water wave, a sound wave,… Whatever. Our dependent variable y is the amplitude of our wave, so it’s the vertical displacement − up or down − of whatever we’re looking at. As it’s a repeating pulse, y is zero most of the time, except when that pulse is pulsing. 🙂

So what’s the wavelength of this thing?

[…] Come on, Vincent. Think! Don’t just look at this!

[…] I got it, daddy! It’s the distance between two peaks, or between the center of two successive pulses— obviously! 🙂

[…] Good! 🙂 OK. That was easy enough. Now look at the argument of this function once again:

F = F(x−ct)

We are not merely acknowledging here that F is some function of x and t, i.e. some function varying in space and time. Of course, F is that too, so we can write: y = F = F(x, t) = F(x−ct), but it’s more than just some function: we’ve got a very special argument here, x−ct, and so let’s start our little lesson by explaining it.

The x−ct argument is there because we’re talking waves, so that is something moving through space and time indeed. Now, what are we actually doing when we write x−ct? Believe it or not, we’re basically converting something expressed in time units into something expressed in distance units. So we’re converting time into distance, so to speak. To see how this works, suppose we add some time Δt to the argument of our function y = F, so we’re looking at F[x−c(t+Δt)] now, instead of F(x−ct). Now, F[x−c(t+Δt)] = F(x−ct−cΔt), so we’ll get a different value for our function—obviously! But it’s easy to see that we can restore our wave function F to its former value by also adding some distance Δx = cΔt to the argument. Indeed, if we do so, we get F[x+Δx−c(t+Δt)] = F(x+cΔt–ct−cΔt) = F(x–ct). For example, if c = 3 m/s, then 2 seconds of time correspond to (2 s)×(3 m/s) = 6 meters of distance.

The idea behind adding both some time Δt as well as some distance Δx is that you’re traveling with the waveform itself, or with its phase as they say. So it’s like you’re riding on its crest or in its trough, or somewhere hanging on to it, so to speak. Hence, the speed of a wave is also referred to as its phase velocity, which we denote by v_p = c. Now, let me make some remarks here.

First, there is the direction of travel. The pulses above travel in the positive x-direction, so that’s why we have x minus ct in the argument. For a wave traveling in the negative x-direction, we’ll have a wave function y = F(x+ct). [And, yes, don’t be lazy, Vincent: please go through the Δx = cΔt math once again to double-check that.]

The second thing you should note is that the speed of a regular periodic wave is equal to to the product of its wavelength and its frequency, so we write: v_p = c = λ·f, which we can also write as λ = c/f or f = c/λ. Now, you know we express the frequency in oscillations or cycles per second, i.e. in hertz: one hertz is, quite simply, 1 s⁻¹, so the unit of frequency is the reciprocal of the second. So the m/s and the Hz units in the fraction below give us a wavelength λ equal to λ = (20 m/s)/(5/s) = 4 m. You’ll say that’s too simple but I just want to make sure you’ve got the basics right here.

The third thing is that, in physics, and in math, we’ll usually work with nice sinusoidal functions, i.e. sine or cosine functions. A sine and a cosine function are the same function but with a phase difference of 90 degrees, so that’s π/2 radians. That’s illustrated below: cosθ = sin(θ+π/2).

Now, when we converted time to distance by multiplying it with c, what we actually did was to ensure that the argument of our wavefunction F was expressed in one unit only: the meter, so that’s the distance unit in the international SI system of units. So that’s why we had to convert time to distance, so to speak.

The other option is to express all in seconds, so that’s in time units. So then we should measure distance in seconds, rather than meters, so to speak, and the corresponding argument is t–x/c, and our wave function would be written as y = G(t–x/c). Just go through the same Δx = cΔt math once more: G[t+Δt–(x+Δx)/c] = G(t+Δt–x/c−cΔt/c) = G(t–x/c).

In short, we’re talking the same wave function here, so F(x−ct) = G(t−x/c), but the argument of F is expressed in distance units, while the argument of G is expressed in time units. If you’d want to double-check what I am saying here, you can use the same 20 m/s wave example again: suppose the distance traveled is 100 m, so x = 100 m and x/c = (100 m)/(20 m/s) = 5 seconds. It’s always important to check the units, and you can see they come out alright in both cases! 🙂

Now, to go from F or G to our sine or cosine function, we need to do yet another conversion of units, as the argument of a sinusoidal function is some angle θ, not meters or seconds. In physics, we refer to θ as the phase of the wave function. So we need degrees or, more common now, radians, which I’ll explain in a moment. Let me first jot it down:

y = sin(2π(x–ct)/λ)

So what are we doing here? What’s going on? Well… First, we divide x–ct by the wavelength λ, so that’s the (x–ct)/λ in the argument of our sine function. So our ‘distance unit’ is no longer the meter but the wavelength of our wave, so we no longer measure in meter but in wavelengths. For example, if our argument x–ct was 20 m, and the wavelength of our wave is 4 m, we get (x–ct)/λ = 5 between the brackets. It’s just like comparing our length: ten years ago you were about half my size. Now you’re the same: one unit. 🙂 When we’re saying that, we’re using my length as the unit – and so that’s also your length unit now 🙂 – rather than meters or centimeters.

Now I need to explain the 2π factor, which is only slightly more difficult. Think about it: one wavelength corresponds to one full cycle, so that’s the full 360° of the circle below. In fact, we’ll express angles in radians, and the two animations below illustrate what a radian really is: an angle of 1 rad defines an arc whose length, as measured on the circle, is equal to the radius of that circle. […] Oh! Please look at the animations as two separate things: they illustrate the same idea, but they’re not synchronized, unfortunately! 🙂
Circle_radians

So… I hope it all makes sense now: if we add one wavelength to the argument of our wave function, we should get the same value, and so it’s equivalent to adding 2π to the argument of our sine function. Adding half a wavelength, or 35% of it, or a quarter, or two wavelengths, or e wavelengths, etc is equivalent to adding π, or 35%·2π ≈ 2.2, or 2π/4 = π/2, or 2·2π = 4π, or e·2π, etc to it. So… Well… Think about it: to go from the argument of our wavefunction expressed as a number of wavelengths − so that’s (x–ct)/λ – to the argument of our sine function, which is expressed in radians, we need to multiply by 2π.

[…] OK, Vincent. If it’s easier for you, you may want to think of the 1/λ and 2π factors in the argument of the sin(2π(x–ct)/λ) function as scaling factors: you’d use a scaling factor when you go from one measurement scale to another indeed. It’s like using vincents rather than meter. If one vincent corresponds to 1.8 m, then we need to re-scale all lengths by dividing them by 1.8 so as to express them in vincents. Vincent ten year ago was 0.9 m, so that’s half a vincent: 0.9/1.8 = 0.5. 🙂

[…] OK. […] Yes, you’re right: that’s rather stupid and makes nobody smile. Fine. You’re right: it’s time to move on to more complicated stuff. Now, read the following a couple of times. It’s my one and only message to you:

If there’s anything at all that you should remember from all of the nonsense I am writing about in this physics blog, it’s that any periodic phenomenon, any motion really, can be analyzed by assuming that it is the sum of the motions of all the different modes of what we’re looking at, combined with appropriate amplitudes and phases.

It really is a most amazing thing—it’s something very deep and very beautiful connecting all of physics with math.

We often refer to these modes as harmonics and, in one of my posts on the topic, I explained how the wavelengths of the harmonics of a classical guitar string – it’s just an example – depended on the length of the string only. Indeed, if we denote the various harmonics by their harmonic number n = 1, 2, 3,… n,… and the length of the string by L, we have λ₁ = 2L = (1/1)·2L, λ₂ = L = (1/2)·2L, λ₃ = (1/3)·2L,… λ_n = (1/n)·2L. So they look like this:

etcetera (1/8, 1/9,…,1/n,… 1/∞)

The diagram makes it look like it’s very obvious, but it’s an amazing fact: the material of the string, or its tension, doesn’t matter. It’s just the length: simple geometry is all that matters! As I mentioned in my post on music and physics, this realization led to a somewhat misplaced fascination with harmonic ratios, which the Greeks thought could explain everything. For example, the Pythagorean model of the orbits of the planets would also refer to these harmonic ratios, and it took intellectual giants like Galileo and Copernicus to finally convince the Pope that harmonic ratios are great, but that they cannot explain everything. 🙂 [Note: When I say that the material of the string, or its tension, doesn’t matter, I should correct myself: they do come into play when time becomes the variable. Also note that guitar strings are not the same length when strung on a guitar: the so-called bridge saddle is not in an exact right angle to the strings: this is a link to some close-up pictures of a bridge saddle on a guitar, just in case you don’t have a guitar at home to check.]

Now, I already explained the need to express the argument of a wave function in radians – because we’re talking periodic functions and so we want to use sinusoidals − and how it’s just a matter of units really, and so how we can go from meter to wavelengths to radians. I also explained how we could do the same for seconds, i.e. for time. The key to converting distance units to time units, and vice versa, is the speed of the wave, or the phase velocity, which relates wavelength and frequency: c = λ·f. Now, as we have to express everything in radians anyway, we’ll usually substitute the wavelength and frequency by the wavenumber and the angular frequency so as to convert these quantities too to something expressed in radians. Let me quickly explain how it works:

The wavenumber k is equal to k = 2π/λ, so it’s some number expressed in radians per unit distance, i.e. radians per meter. In the example above, where λ was 4 m, we have k = 2π/(4 m) = π/2 radians per meter. To put it differently, if our wave travels one meter, its phase θ will change by π/2.
Likewise, the angular frequency is ω = 2π·f = 2π/T. Using the same example once more, so assuming a frequency of 5 Hz, i.e. a period of one fifth of a second, we have ω = 2π/[(1/5)·s] = 10π per second. So the phase of our wave will change with 10 times π in one second. Now that makes sense because, in one second, we have five cycles, and so that corresponds to 5 times 2π.

Note that our definition implies that λ = 2π/k, and that it’s also easy to figure out that our definition of ω, combined with the f = c/λ relation, implies that ω = 2π·c/λ and, hence, that c = ω·λ/(2π) = (ω·2π/k)/(2π) = ω/k. OK. Let’s move on.

Using the definitions and explanations above, it’s now easy to see that we can re-write our y = sin(2π(x–ct)/λ) as:

y = sin(2π(x–ct)/λ) = sin[2π(x–(ω/k)t)/(2π/k)] = sin[(x–(ω/k)t)·k)] = sin(kx–ωt)

Remember, however, that we were talking some wave that was traveling in the positive x-direction. For the negative x-direction, the equation becomes:

y = sin(2π(x+ct)/λ) = sin(kx+ωt)

OK. That should be clear enough. Let’s go back to our guitar string. We can go from λ to k by noting that λ = 2L and, hence, we get the following for all of the various modes:

k = k₁ = 2π·1/(2L) = π/L, k₂ = 2π·2/(2L) = 2k, k₃ = 2π·3/(2L) = 3k,,… k_n = 2π·3/(2L) = nk,…

That gives us our grand result, and that’s that we can write some very complicated waveform Ψ(x) as the sum of an infinite number of simple sinusoids, so we have:

Ψ(x) = a₁sin(kx) + a₂sin(2kx) + a₃sin(3kx) + … + a_nsin(nkx) + … = ∑ a_nsin(nkx)

The equation above assumes we’re looking at the oscillation at some fixed point in time. If we’d be looking at the oscillation at some fixed point in space, we’d write:

Φ(t) = a₁sin(ωt) + a₂sin(2ωt) + a₃sin(3ωt) + … + a_nsin(nωt) + … = ∑ a_nsin(nωt)

Of course, to represent some very complicated oscillation on our guitar string, we can and should combine some Ψ(x) as well as some Φ(t) function, but how do we do that, exactly? Well… We’ll obviously need both the sin(kx–ωt) as well as those sin(kx+ωt) functions, as I’ll explain in a moment. However, let me first make another small digression, so as to complete your knowledge of wave mechanics. 🙂

We look at a wave as something that’s traveling through space and time at the same time. In that regard, I told you that the speed of the wave is its so-called phase velocity, which we denoted as v_p = c and which, as I explained above, is equal to v_p = c = λ·f = (2π/k)·(ω/2π) = ω/k. The animation below (credit for it must go to Wikipedia—and sorry I forget to acknowledge the same source for the illustrations above) illustrates the principle: the speed of travel of the red dot is the phase velocity. But you can see that what’s going on here is somewhat more complicated: we have a series of wave packets traveling through space and time here, and so that’s where the concept of the so-called group velocity comes in: it’s the speed of travel of the green dot.

Now, look at the animation below. What’s going on here? The wave packet (or the group or the envelope of the wave—whatever you want to call it) moves to the right, but the phase goes to the left, as the peaks and troughs move leftward indeed. Huh? How is that possible? And where is this wave going? Left or right? Can we still associate some direction with the wave here? It looks like it’s traveling in both directions at the same time!

The wave actually does travel in both directions at the same time. Well… Sort of. The point is actually quite subtle. When I started this post by writing that the pulses were ‘obviously’ traveling in the positive x-direction… Well… That’s actually not so obvious. What is it that is traveling really? Think about an oscillating guitar string: nothing travels left or right really. Each point on the string just moves up and down. Likewise, if our repeated pulse is some water wave, then the water just stays where it is: it just moves up and down. Likewise, if we shake up some rope, the rope is not going anywhere: we just started some motion that is traveling down the rope. In other words, the phase velocity is just a mathematical concept. The peaks and troughs that seem to be traveling are just mathematical points that are ‘traveling’ left or right.

What about the group velocity? Is that a mathematical notion too? It is. The wave packet is often referred to as the envelope of the wave curves, for obviously reasons: they’re enveloped indeed. Well… Sort of. 🙂 However, while both the phase and group velocity are velocities of mathematical constructs, it’s obvious that, if we’re looking at wave packets, the group velocity would be of more interest to us than the phase velocity. Think of those repeated pulses as real water waves, for example: while the water stays where it is (as mentioned, the water molecules just go up and down—more or less, at least), we’d surely be interested to know how fast these waves are ‘moving’, and that’s given by the group velocity, not the phase velocity. Still, having said that, the group velocity is as ‘unreal’ as the phase velocity: both are mathematical concepts. The only thing that’s ‘real’ is the up and down movement. Nothing travels in reality. Now, I shouldn’t digress too much here, but that’s why there’s no limit on the phase velocity: it can exceed the speed of light. In fact, in quantum mechanics, some real-life particle − like an electron, for instance – will be represented by a complex-valued wave function, and there’s no reason to put some limit on the phase velocity. In contrast, the group velocity will actually be the speed of the electron itself, and that speed can, obviously, approach the speed of light – in particle accelerators, for example – but it can never exceed it. [If you’re smart, and you are, you’ll wonder: what about photons? Well…The classical and quantum-mechanical view of an electromagnetic wave are surely not the same, but they do have a lot in common: both photons and electromagnetic radiation travel at the speed c. Photons can do so because their rest mass is zero. But I can’t go into any more detail here, otherwise this thing will become way too long.]

OK. Let me get back to the issue at hand. So I’ll now revert to the simpler situation we’re looking at here, and so that’s these harmonic waves, whose form is a simple sinusoidal indeed. The animation below (and, yes, it’s also from Wikipedia) is the one that’s relevant for this situation. You need to study it for a while to understand what’s going on. As you can see, the green wave travels to the right, the blue one travels to the left, and the red wave function is the sum of both.

Of course, after all that I wrote above, I should use quotation marks and write ‘travel’ instead of travel, so as to indicate there’s nothing traveling really, except for those mathematical points, but then no one does that, and so I won’t do it either. Just make sure you always think twice when reading stuff like this! Back to the lesson: what’s going on here?

As I explained, the argument of a wave traveling towards the negative x-direction will be x+ct. Conversely, the argument of a wave traveling in the positive x-direction will be x–ct. Now, our guitar string is going nowhere, obviously: it’s like the red wave function above. It’s a so-called standing wave. The red wave function has nodes, i.e. points where there is no motion—no displacement at all! Between the nodes, every point moves up and down sinusoidally, but the pattern of motion stays fixed in space. So that’s the kind of wave function we want, and the animation shows us how we can get it.

Indeed, there’s a funny thing with fixed strings: when a wave reaches the clamped end of a string, it will be reflected with a change in sign, as illustrated below: we’ve got that F(x+ct) wave coming in, and then it goes back indeed, but with the sign reversed.

The illustration above speaks for itself but, of course, once again I need to warn you about the use of sentences like ‘the wave reaches the end of the string’ and/or ‘the wave gets reflected back’. You know what it really means now: it’s some movement that travels through space. […] In any case, let’s get back to the lesson once more: how do we analyze that?

Easy: the red wave function is the sum of two waves: one traveling to the right, and one traveling to the left. We’ll call these component waves F and G respectively, so we have y = F(x, t) + G(x, t). Let’s go for it.

Let’s first assume the string is not held anywhere, so that we have an infinite string along which waves can travel in either direction. In fact, the most general functional form to capture the fact that a waveform can travel in any direction is to write the displacement y as the sum of two functions: one wave traveling one way (which we’ll denote by F, indeed), and the other wave (which, yes, we’ll denote by G) traveling the other way. From the illustration above, it’s obvious that the F wave is traveling towards the negative x-direction and, hence, its argument will be x+ct. Conversely, the G wave travels in the positive x-direction, so its argument is x–ct. So we write:

y = F(x, t) + G(x, t) = F(x+ct) + G(x–ct)

So… Well… We know that the string is actually not infinite, but that it’s fixed to two points. Hence, y is equal to zero there: y = 0. Now let’s choose the origin of our x-axis at the fixed end so as to simplify the analysis. Hence, where y is zero, x is also zero. Now, at x = 0, our general solution above for the infinite string becomes y = F(ct) + G(−ct) = 0, for all values of t. Of course, that means G(−ct) must be equal to –F(ct). Now, that equality is there for all values of t. So it’s there for all values of ct and −ct. In short, that equality is valid for whatever value of the argument of G and –F. As Feynman puts it: “G of anything must be –F of minus that same thing.” Now, the ‘anything’ in G is its argument: x – ct, so ‘minus that same thing’ is –(x–ct) = −x+ct. Therefore, our equation becomes:

y = F(x+ct) − F(−x+ct)

So that’s what’s depicted in the diagram above: the F(x+ct) wave ‘vanishes’ behind the wall as the − F(−x+ct) wave comes out of it. Now, of course, so as to make sure our guitar string doesn’t stop its vibration after being plucked, we need to ensure F is a periodic function, like a sin(kx+ωt) function. 🙂 Why? Well… If this F and G function would simply disappear and ‘serve’ only once, so to speak, then we only have one oscillation and that’s it! So the waves need to continue and so that’s why it needs to be periodic.

OK. Can we just take sin(kx+ωt) and −sin(−kx+ωt) and add both? It makes sense, doesn’t it? Indeed, −sinα = sin(−α) and, therefore, −sin(−kx+ωt) = sin(kx−ωt). Hence, y = F(x+ct) − F(−x+ct) would be equal to:

y = sin(kx+ωt) + sin(kx–ωt) = sin(2π(x+ct)/λ) + sin(2π(x−ct)/λ)

Done! Let’s use specific values for k and ω now. For the first harmonic, we know that k = 2π/2L = π/L. What about ω? Hmm… That depends on the wave velocity and, therefore, that actually does depend on the material and/or the tension of the string! The only thing we can say is that ω = c·k, so ω = c·2π/λ = c·π/L. So we get:

sin(kx+ωt) = sin(π·x/L + π·c·t/L) = sin[(π/L)·(x+ct)]

But this is our F function only. The whole oscillation is y = F(x+ct) − F(−x+ct), and − F(−x+ct) is equal to:

–sin[(π/L)·(−x+ct)] = –sin(−π·x/L+π·c·t/L) = −sin(−kx+ωt) = sin(kx–ωt) = sin[(π/L)·(x–ct)]

So, yes, we should add both functions to get:

y = sin[π(x+ct)/L] + sin[π(x−ct)/L]

Now, we can, of course, apply our trigonometric formulas for the addition of angles, which say that sin(α+β) = sinαcosβ + sinβcosα and sin(α–β) = sinαcosβ – sinβcosα. Hence, y = sin(kx+ωt) + sin(kx–ωt) is equal to sin(kx)cos(ωt) + sin(ωt)cos(kx) + sin(kx)cos(ωt) – sin(ωt)cos(kx) = 2sin(kx)cos(ωt). Now, that’s a very interesting result, so let’s give it some more prominence by writing it in boldface:

y = sin(kx+ωt) + sin(kx–ωt) = 2sin(kx)cos(ωt) = 2sin(π·x/L)cos(π·c·t/L)

The sin(π·x/L) factor gives us the nodes in space. Indeed, sin(π·x/L) = 0 if x is equal to 0 or L (values of x outside of the [0, L] interval are obviously not relevant here). Now, the other factor cos(π·c·t/L) can be re-written cos(2π·c·t/λ) = cos(2π·f·t) = cos(2π·t/T), with T the period T = 1/f = λ/c, so the amplitude reaches a maximum (+1 or −1 or, including the factor 2, +2 or −2) if 2π·t/T is equal to a multiple of π, so that’s if t = n·T/2 with n = 0, 1, 2, etc. In our example above, for f = 5 Hz, that means the amplitude reaches a maximum (+2 or −2) every tenth of a second.

The analysis for the other modes is as easy, and I’ll leave it you, Vincent, as an exercise, to work it all out and send me the y = 2·sin[something]·cos[something else] formula (with the ‘something’ and ‘something else’ written in terms of L and c, of course) for the higher harmonics. 🙂

[…] You’ll say: what’s the point, daddy? Well… Look at that animation again: isn’t it great we can analyze any standing wave, or any harmonic indeed, as the sum of two component waves with the same wavelength and frequency but ‘traveling’ in opposite directions?

Yes, Vincent. I can hear you sigh: “Daddy, I really do not see why I should be interested in this.”

Well… Your call… What can I say? Maybe one day you will. In fact, if you’re going to go for engineering studies, you’ll have to. 🙂

To conclude this post, I’ll insert one more illustration. Now that you know what modes are, you can start thinking about those more complicated Ψ and Φ functions. The illustration below shows how the first and second mode of our guitar string combine to give us some composite wave traveling up and down the very same string.

Think about it. We have one physical phenomenon here: at every point in time, the string is somewhere, but where exactly, depends on the mathematical shape of its components. If this doesn’t illustrate the beauty of Nature, the fact that, behind every simple physical phenomenon − most of which are some sort of oscillation indeed − we have some marvelous mathematical structure, then… Well… Then I don’t know how to explain why I am absolutely fascinated by this stuff.

Addendum 1: On actual waves

My examples of waves above were all examples of so-called transverse waves, i.e. oscillations at a right angle to the direction of the wave. The other type of wave is longitudinal. I mentioned sound waves above, but they are essentially longitudinal. So there the displacement of the medium is in the same direction of the wave, as illustrated below.

Real-life waves, like water waves, may be neither of the two. The illustration below shows how water molecules actually move as a wave passes. They move in little circles, with a systemic phase shift from circle to circle.

Why is this so? I’ll let Feynman answer, as he also provided the illustration above:

“Although the water at a given place is alternately trough or hill, it cannot simply be moving up and down, by the conservation of water. That is, if it goes down, where is the water going to go? The water is essentially incompressible. The speed of compression of waves—that is, sound in the water—is much, much higher, and we are not considering that now. Since water is incompressible on this scale, as a hill comes down the water must move away from the region. What actually happens is that particles of water near the surface move approximately in circles. When smooth swells are coming, a person floating in a tire can look at a nearby object and see it going in a circle. So it is a mixture of longitudinal and transverse, to add to the confusion. At greater depths in the water the motions are smaller circles until, reasonably far down, there is nothing left of the motion.”

So… There you go… 🙂

Addendum 2: On non-periodic waves, i.e. pulses

A waveform is not necessarily periodic. The pulse we looked at could, perhaps, not repeat itself. It is not possible, then, to describe its wavelength. However, it’s still a wave and, hence, its functional form would still be some y = F(x−ct) or y = F(x+ct) form, depending on its direction of travel.

The example below also comes out of Feynman’s Lectures: electromagnetic radiation is caused by some accelerating electric charge – an electron, usually, because its mass is small and, hence, it’s much easier to move than a proton 🙂 – and then the electric field travels out in space. So the two diagrams below show (i) the acceleration (a) as a function of time (t) and (ii) the electric field strength (E) as a function of the distance (r). [To be fully precise, I should add he ignores the 1/r variation, but that’s a fine point which doesn’t matter much here.]

He basically uses this illustration to explain why we can use a y = G(t–x/c) functional form to describe a wave. The point is: he actually talks about one pulse only here. So the F(x±ct) or G(t±x/c) or sin(kx±ωt) form has nothing to do with whether or not we’re looking at a periodic or non-periodic waveform. The gist of the matter is that we’ve got something moving through space, and it doesn’t matter whether it’s periodic or not: the periodicity or non-periodicity, of a wave has nothing to do with the x±ct, t±x/c or kx±ωt shape of the argument of our wave function. The functional form of our argument is just the result of what I said about traveling along with our wave.

So what is it about periodicity then? Well… If periodicity kicks it, you’ll talk sinusoidal functions, and so the circle will be needed once more. 🙂

Now, I mentioned we cannot associate any particular wavelength with such non-periodic wave. Having said that, it’s still possible to analyze this pulse as a sum of sinusoids through a mathematical procedure which is referred to as the Fourier transform. If you’re going for engineer, you’ll need to learn how to master this technique. As for now, however, you can just have a look at the Wikipedia article on it. 🙂

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

The four-force vector fμ

The force and the tensor

The invariance of physics and the use of vector equations

Electrodynamics in relativistic notation

Four-vectors and invariance under Lorentz transformations

The electromagnetic tensor

The Lorentz transformation of the electric and magnetic fields

The impedance concept

Resistors

Capacitors (condensers)

Inductors

Summary of conclusions

Addendum: Why is V = − Ɛ?

On Maxwell’s equations and the properties of empty space

On quantum oscillations, Planck’s constant, and Planck units

The fine-structure constant

Addendum: How to think about space and time?

Power = Ɛ·I

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

The four-force vector f_μ