Field momentum
Next we would like to talk about the momentum in the electromagnetic field. Just as the field has energy, it will have a certain momentum per unit volume. Let us call that momentum density $\FLPg$. Of course, momentum has various possible directions, so that $\FLPg$ must be a vector. Let’s talk about one component at a time; first, we take the $x$-component. Since each component of momentum is conserved we should be able to write down a law that looks something like this: \begin{equation*} -\ddp{}{t} \begin{pmatrix} \text{momentum}\\ \text{of matter} \end{pmatrix} _x\!\!=\ddp{g_x}{t}+ \begin{pmatrix} \text{momentum}\\ \text{outflow} \end{pmatrix} _x. \end{equation*}  The left side is easy. The rate-of-change of the momentum of matter is just the force on it. For a particle, it is $\FLPF=q(\FLPE+\FLPv\times\FLPB)$; for a distribution of charges, the force per unit volume is $(\rho\FLPE+\FLPj\times\FLPB)$. The “momentum outflow” term, however, is strange. It cannot be the divergence of a vector because it is not a scalar; it is, rather, an $x$-component of some vector. Anyway, it should probably look something like \begin{equation*} \ddp{a}{x}+\ddp{b}{y}+\ddp{c}{z}, \end{equation*}  because the $x$-momentum could be flowing in any one of the three directions. In any case, whatever $a$, $b$, and $c$ are, the combination is supposed to equal the outflow of the $x$-momentum. Now the game would be to write $\rho\FLPE+\FLPj\times\FLPB$ in terms only of $\FLPE$ and $\FLPB$—eliminating $\rho$ and $\FLPj$ by using Maxwell’s equations—and then to juggle terms and make substitutions to get it into a form that looks like \begin{equation*} \ddp{g_x}{t}+\ddp{a}{x}+\ddp{b}{y}+\ddp{c}{z}. \end{equation*}  Then, by identifying terms, we would have expressions for $g_x$, $a$, $b$, and $c$. It’s a lot of work, and we are not going to do it. Instead, we are only going to find an expression for $\FLPg$, the momentum density—and by a different route. There is an important theorem in mechanics which is this: whenever there is a flow of energy in any circumstance at all (field energy or any other kind of energy), the energy flowing through a unit area per unit time, when multiplied by $1/c^2$, is equal to the momentum per unit volume in the space. In the special case of electrodynamics, this theorem gives the result that $\FLPg$ is $1/c^2$ times the Poynting vector: \begin{equation} \label{Eq:II:27:21} \FLPg=\frac{1}{c^2}\,\FLPS. \end{equation}  So the Poynting vector gives not only energy flow but, if you divide by $c^2$, also the momentum density. The same result would come out of the other analysis we suggested, but it is more interesting to notice this more general result. We will now give a number of interesting examples and arguments to convince you that the general theorem is true. First example: Suppose that we have a lot of particles in a box—let’s say $N$ per cubic meter—and that they are moving along with some velocity $\FLPv$. Now let’s consider an imaginary plane surface perpendicular to $\FLPv$. The energy flow through a unit area of this surface per second is equal to $Nv$, the number which flow through the surface per second, times the energy carried by each one. The energy in each particle is $m_0c^2/\sqrt{1-v^2/c^2}$. So the energy flow per second is \begin{equation*} Nv\,\frac{m_0c^2}{\sqrt{1-v^2/c^2}}. \end{equation*}  But the momentum of each particle is $m_0v/\sqrt{1-v^2/c^2}$, so the density of momentum is \begin{equation*} N\,\frac{m_0v}{\sqrt{1-v^2/c^2}}, \end{equation*}  which is just $1/c^2$ times the energy flow—as the theorem says. So the theorem is true for a bunch of particles. It is also true for light. When we studied light in Volume I, we saw that when the energy is absorbed from a light beam, a certain amount of momentum is delivered to the absorber. We have, in fact, shown in Chapter 34 of Vol. I that the momentum is $1/c$ times the energy absorbed [Eq. (34.24) of Vol. I]. If we let $\FLPU$ be the energy arriving at a unit area per second, then the momentum arriving at a unit area per second is $\FLPU/c$. But the momentum is travelling at the speed $c$, so its density in front of the absorber must be $\FLPU/c^2$. So again the theorem is right. Finally we will give an argument due to Einstein which demonstrates the same thing once more. Suppose that we have a railroad car on wheels (assumed frictionless) with a certain big mass $M$. At one end there is a device which will shoot out some particles or light (or anything, it doesn’t make any difference what it is), which are then stopped at the opposite end of the car. There was some energy originally at one end—say the energy $U$ indicated in Fig. 27–7(a)—and then later it is at the opposite end, as shown in Fig. 27–7(c). The energy $U$ has been displaced the distance $L$, the length of the car. Now the energy $U$ has the mass $U/c^2$, so if the car stayed still, the center of gravity of the car would be moved.  Einstein didn’t like the idea that the center of gravity of an object could be moved by fooling around only on the inside, so he assumed that it is impossible to move the center of gravity by doing anything inside. But if that is the case, when we moved the energy $U$ from one end to the other, the whole car must have recoiled some distance $x$, as shown in part (c) of the figure. You can see, in fact, that the total mass of the car, times $x$, must equal the mass of the energy moved, $U/c^2$ times $L$ (assuming that $U/c^2$ is much less than $M$): \begin{equation} \label{Eq:II:27:22} Mx=\frac{U}{c^2}\,L. \end{equation}  Let’s now look at the special case of the energy being carried by a light flash. (The argument would work as well for particles, but we will follow Einstein, who was interested in the problem of light.) What causes the car to be moved? Einstein argued as follows: When the light is emitted there must be a recoil, some unknown recoil with momentum $p$. It is this recoil which makes the car roll backward. The recoil velocity $v$ of the car will be this momentum divided by the mass of the car: \begin{equation*} v=\frac{p}{M}. \end{equation*}  The car moves with this velocity until the light energy $U$ gets to the opposite end. Then, when it hits, it gives back its momentum and stops the car. If $x$ is small, then the time the car moves is nearly equal to $L/c$; so we have that \begin{equation*} x=vt=v\,\frac{L}{c}=\frac{p}{M}\,\frac{L}{c}. \end{equation*}  Putting this $x$ in Eq. (27.22), we get that \begin{equation*} p=\frac{U}{c}. \end{equation*}  Again we have the relation of energy and momentum for light, from which the argument above shows the momentum density is \begin{equation} \label{Eq:II:27:23} \FLPg=\frac{\FLPU}{c^2}. \end{equation}   You may well wonder: What is so important about the center-of-gravity theorem? Maybe it is wrong. Perhaps, but then we would also lose the conservation of angular momentum. Suppose that our boxcar is moving along a track at some speed $v$ and that we shoot some light energy from the top to the bottom of the car—say, from $A$ to $B$ in Fig. 27–8. Now we look at the angular momentum of the system about the point $P$. Before the energy $U$ leaves $A$, it has the mass $m=U/c^2$ and the velocity $v$, so it has the angular momentum $mvr_A$. When it arrives at $B$, it has the same mass and, if the linear momentum of the whole boxcar is not to change, it must still have the velocity $v$. Its angular momentum about $P$ is then $mvr_B$. The angular momentum will be changed unless the right recoil momentum was given to the car when the light was emitted—that is, unless the light carries the momentum $U/c$. It turns out that the angular momentum conservation and the theorem of center-of-gravity are closely related in the relativity theory. So the conservation of angular momentum would also be destroyed if our theorem were not true. At any rate, it does turn out to be a true general law, and in the case of electrodynamics we can use it to get the momentum in the field. We will mention two further examples of momentum in the electromagnetic field. We pointed out in Section 26–2 the failure of the law of action and reaction when two charged particles were moving on orthogonal trajectories. The forces on the two particles don’t balance out, so the action and reaction are not equal; therefore the net momentum of the matter must be changing. It is not conserved. But the momentum in the field is also changing in such a situation. If you work out the amount of momentum given by the Poynting vector, it is not constant. However, the change of the particle momenta is just made up by the field momentum, so the total momentum of particles plus field is conserved. Finally, another example is the situation with the magnet and the charge, shown in Fig. 27–6. We were unhappy to find that energy was flowing around in circles, but now, since we know that energy flow and momentum are proportional, we know also that there is momentum circulating in the space. But a circulating momentum means that there is angular momentum. So there is angular momentum in the field. Do you remember the paradox we described in Section 17–4 about a solenoid and some charges mounted on a disc? It seemed that when the current turned off, the whole disc should start to turn. The puzzle was: Where did the angular momentum come from? The answer is that if you have a magnetic field and some charges, there will be some angular momentum in the field. It must have been put there when the field was built up. When the field is turned off, the angular momentum is given back. So the disc in the paradox would start rotating. This mystic circulating flow of energy, which at first seemed so ridiculous, is absolutely necessary. There is really a momentum flow. It is needed to maintain the conservation of angular momentum in the whole world.