# The classical explanation for the electron’s mass and radius

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

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

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

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

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

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

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

What does this say? It says that the momentum of the field – i.e. the electromagnetic momentum, integrated over all of space – is proportional to the velocity v of our charge. That makes sense: when v = 0, we’ll have an electrostatic field all over space and, hence, some inertia, but it’s only when we try to move our charge that Newton’s Law comes into play: then we’ll need some force to overcome that inertia. It all works through the Poynting formula: S = E×B0. If nothing’s moving, then B = 0, and so we’ll have some E and, therefore, we’ll have field energy alright, but the energy flow will be zero. But when we move the charge, we’re moving the field, and so then B ≠ 0 and so it’s through B that the E in our S equation start kicking in. Does that make sense? Think about it: it’s good to try to visualize things in your mind. 🙂

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

• e2 = qe2/(4πε0), with qthe electron charge (without the minus sign) and ε0 our ubiquitous electric constant. [Note that, unlike Feynman, I prefer to not write e in italics, so as to not confuse it with Euler’s number ≈ 2.71828 etc. However, I know I am not always consistent in my notation. We don’t need Euler’s number in this post, so e or is always an expression for the electron charge, not Euler’s number. Stupid remark, perhaps, but I don’t want you to be confused.]
• a is the radius of our charge—see we got away from the idea of a point charge? 🙂
• c2 is just c2, i.e. our weird constant (the square of the speed of light) which seems to connect everything to everything. Indeed, think about stuff like this: S/g = c= 1/(ε0μ0).

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

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

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

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

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

That’s good, because if we equate mand melec and switch melec and a in our formula for melec, we get:

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

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

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

Let me enlarge the formula:

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

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

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

U = mc2 = (melec + mPoincaré)c2

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

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

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

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

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

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

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

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

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

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

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

In short, Feynman’s approach here is disappointing. Naive even, but then… Well… Who knows? Perhaps he didn’t do this Lecture himself. Perhaps it’s just an assistant or so. In fact, I should wonder why there’s still physicists wasting time on this! I should also note that naively comparing that a radius with the classical electron radius also makes little or no sense. Unlike what you’d expect, the classical electron radius re and the Thomson scattering cross-section σare not related like you might think they are, i.e. like σ= π·re2 or σ= π·(re/2)2 or σre2 or σ= π·(2·re)2 or whatever circular surface calculation rule that might make sense here. No. The Thomson scattering cross-section is equal to:

σ= (8π/3)·re2 = (2π/3)·(2·re)2 = (2/3)·π·(2·re)2 ≈ 66.5×10−30 m= 66.5 (fm)2

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

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

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

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

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

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

Post scriptum on the math versus the physics

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

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

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

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

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

So the field (E) goes as for r ≤ R and as 1/rfor r ≥ R. So, for r ≥ R, the integral will have (1/r2)2 = 1/rin it. Now, you know that the integral of some function is the surface under the graph of that function. Look at the 1/r4 function below: it blows up between 1 and 0. That’s where the problem is: there needs to be some kind of cut-off, because that integral will effectively blow up when the radius of our little sphere of charge gets ‘too small’. So that makes it clear why it doesn’t make sense to use this formula to try to calculate the energy of a point charge. It just doesn’t make sense to do that.

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

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

Uelect  = α/, with α = e2/2

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

me = αe/re

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

Ee = αe/re

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

α = eP2 and, therefore, Ee = eP2/re

Now, you can check for yourself: it’s just a matter of re-expressing everything in standard SI units, and relating eP2 to e2, and it should all work: you should get the Uelect  = (1/2)·e2/expression. So… Well… At least this takes some of the magic out the fine-structure constant. It’s still a wonderful thing, but so you see that the fundamental relationship between (a) the energy (and, hence, the mass), (b) the radius and (c) the charge of an electron is not something God-given. What’s God-given are Maxwell’s equations, and so the Ee = αe/r= eP2/re is just one of the many wonderful things that you can get out of  them🙂

# Field energy and field momentum

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

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

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

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

1. The heat flow vector h is the heat flux density vector, from which we get the heat flux through an area through the (hn)·da = hn·da product.
2. The energy flow is the energy flux density vector, from which we get the energy flux through the (Sn)·da = Sn·da product.

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

#### Local energy conservation

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

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

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

1. E− p2c2 = m02c4
2. pc = E·(v/c) ⇔ p = v·E/c= m·v

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

g is the momentum of the field per unit volume.

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

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

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

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

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

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

Let’s look at all of the three terms:

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

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

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

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

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

g = S/c2

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

OK. Two good examples. However, it’s still frustrating to not see how we get the g = S/c2 in the specific context of the electromagnetic force, so let’s do a dimensional analysis at least. In my previous post, I showed that the dimension of S must be J/(m2·s), so [S/c2] = [J/(m2·s)]/(m2/s2) = [N·m/(m2·s)]·(s2/m2) = [N·s/m3]. Now, we know that the unit of mass is 1 kg = N/(m/s2). That’s just the force law: a force of 1 newton will give a mass of 1 kg an acceleration of 1 m/s per second, so 1 N = 1 kg·(m/s2). So the [N·s/m3] dimension is equal to [kg·(m/s2)·s/m3] = [(kg·(m/s)/m3] = [(kg·(m/s)]/m3, which is the dimension of momentum (p = mv) per unit volume, indeed. So, yes, the dimensional analysis works out, and it’s also in line with the p = v·E/c2 = m·v equation, but… Oh… We did a dimensional analysis already, where we also showed that [g] = [S/c2] = (J·s/m4). Well… In any case… It’s a bit frustrating to not see the detail here, but let us note the the Grand Result once again:

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

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

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

1. [c·g = (m/s)·(N·s)/m= N/m= J/m3.
2. [S/c] = (s/m)·(N·m)/(s·m2) = N/m= J/m3.

Isn’t that nice? Both sides of the equation now have a dimension like ‘the force per unit area’, or ‘the energy per unit volume’. To get that, we just re-scaled g and S, by c and 1/c respectively. As far as I am concerned, this shows an underlying unity we probably tend to mask with our ‘related but different’ energy and momentum concepts. It’s like E and B: I just love it we can write them together in our Poynting formula = ε0c2E×B. In fact, let me show something else here, which you should think about. You know that c= 1/(ε0μ0), so we can write also as SE×B0. That’s nice, but what’s nice too is the following:

1. S/c = c·= ε0cE×B = E×B/μ0c
2. S/g = c= 1/(ε0μ0)

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

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

#### Einstein’s car

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

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

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

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

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

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

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

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

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