The relativity of magnetic and electric fields
When we said that the magnetic force on a charge was proportional to its velocity, you may have wondered: “What velocity? With respect to which reference frame?” It is, in fact, clear from the definition of $\FLPB$ given at the beginning of this chapter that what this vector is will depend on what we choose as a reference frame for our specification of the velocity of charges. But we have said nothing about which is the proper frame for specifying the magnetic field. It turns out that any inertial frame will do. We will also see that magnetism and electricity are not independent things—that they should always be taken together as one complete electromagnetic field. Although in the static case Maxwell’s equations separate into two distinct pairs, one pair for electricity and one pair for magnetism, with no apparent connection between the two fields, nevertheless, in nature itself there is a very intimate relationship between them that arises from the principle of relativity. Historically, the principle of relativity was discovered after Maxwell’s equations. It was, in fact, the study of electricity and magnetism which led ultimately to Einstein’s discovery of his principle of relativity. But let’s see what our knowledge of relativity would tell us about magnetic forces if we assume that the relativity principle is applicable—as it is—to electromagnetism.  Suppose we think about what happens when a negative charge moves with velocity $v_0$ parallel to a current-carrying wire, as in Fig. 13–10. We will try to understand what goes on in two reference frames: one fixed with respect to the wire, as in part (a) of the figure, and one fixed with respect to the particle, as in part (b). We will call the first frame $S$ and the second $S'$. In the $S$-frame, there is clearly a magnetic force on the particle. The force is directed toward the wire, so if the charge were moving freely we would see it curve in toward the wire. But in the $S'$-frame there can be no magnetic force on the particle, because its velocity is zero. Does it, therefore, stay where it is? Would we see different things happening in the two systems? The principle of relativity would say that in $S'$ we should also see the particle move closer to the wire. We must try to understand why that would happen. We return to our atomic description of a wire carrying a current. In a normal conductor, like copper, the electric currents come from the motion of some of the negative electrons—called the conduction electrons—while the positive nuclear charges and the remainder of the electrons stay fixed in the body of the material. We let the charge density of the conduction electrons be $\rho_-$ and their velocity in $S$ be $\FLPv$. The density of the charges at rest in $S$ is $\rho_+$, which must be equal to the negative of $\rho_-$, since we are considering an uncharged wire. There is thus no electric field outside the wire, and the force on the moving particle is just \begin{equation*} \FLPF=q\FLPv_0\times\FLPB. \end{equation*}  Using the result we found in Eq. (13.18) for the magnetic field at the distance $r$ from the axis of a wire, we conclude that the force on the particle is directed toward the wire and has the magnitude \begin{equation*} F=\frac{1}{4\pi\epsO c^2}\cdot\frac{2Iqv_0}{r}. \end{equation*}  Using Eqs. (13.3) and (13.5), the current $I$ can be written as $\rho_-vA$, where $A$ is the area of a cross section of the wire. Then \begin{equation} \label{Eq:II:13:20} F=\frac{1}{4\pi\epsO c^2}\cdot\frac{2q\rho_-Avv_0}{r}. \end{equation}  We could continue to treat the general case of arbitrary velocities for $v$ and $v_0$, but it will be just as good to look at the special case in which the velocity $v_0$ of the particle is the same as the velocity $v$ of the conduction electrons. So we write $v_0=v$, and Eq. (13.20) becomes \begin{equation} \label{Eq:II:13:21} F=\frac{q}{2\pi\epsO}\,\frac{\rho_-A}{r}\,\frac{v^2}{c^2}. \end{equation}  Now we turn our attention to what happens in $S'$, in which the particle is at rest and the wire is running past (toward the left in the figure) with the speed $v$. The positive charges moving with the wire will make some magnetic field $\FLPB'$ at the particle. But the particle is now at rest, so there is no magnetic force on it!  If there is any force on the particle, it must come from an electric field. It must be that the moving wire has produced an electric field. But it can do that only if it appears charged—it must be that a neutral wire with a current appears to be charged when set in motion. We must look into this. We must try to compute the charge density in the wire in $S'$ from what we know about it in $S$. One might, at first, think they are the same; but we know that lengths are changed between $S$ and $S'$ (see Chapter 15, Vol. I), so volumes will change also. Since the charge densities depend on the volume occupied by charges, the densities will change, too. Before we can decide about the charge densities in $S'$, we must know what happens to the electric charge of a bunch of electrons when the charges are moving. We know that the apparent mass of a particle changes by $1/\sqrt{1-v^2/c^2}$. Does its charge do something similar? No! Charges are always the same, moving or not. Otherwise we would not always observe that the total charge is conserved. Suppose that we take a block of material, say a conductor, which is initially uncharged. Now we heat it up. Because the electrons have a different mass than the protons, the velocities of the electrons and of the protons will change by different amounts. If the charge of a particle depended on the speed of the particle carrying it, in the heated block the charge of the electrons and protons would no longer balance. A block would become charged when heated. As we have seen earlier, a very small fractional change in the charge of all the electrons in a block would give rise to enormous electric fields. No such effect has ever been observed. Also, we can point out that the mean speed of the electrons in matter depends on its chemical composition. If the charge on an electron changed with speed, the net charge in a piece of material would be changed in a chemical reaction. Again, a straightforward calculation shows that even a very small dependence of charge on speed would give enormous fields from the simplest chemical reactions. No such effect is observed, and we conclude that the electric charge of a single particle is independent of its state of motion. So the charge $q$ on a particle is an invariant scalar quantity, independent of the frame of reference. That means that in any frame the charge density of a distribution of electrons is just proportional to the number of electrons per unit volume. We need only worry about the fact that the volume can change because of the relativistic contraction of distances. We now apply these ideas to our moving wire. If we take a length $L_0$ of the wire, in which there is a charge density $\rho_0$ of stationary charges, it will contain the total charge $Q=\rho_0L_0A_0$. If the same charges are observed in a different frame to be moving with velocity $v$, they will all be found in a piece of the material with the shorter length \begin{equation} \label{Eq:II:13:22} L=L_0\sqrt{1-v^2/c^2}, \end{equation}  but with the same area $A_0$ (since dimensions transverse to the motion are unchanged). See Fig. 13–11. If we call $\rho$ the density of charges in the frame in which they are moving, the total charge $Q$ will be $\rho LA_0$. This must also be equal to $\rho_0L_0A_0$, because charge is the same in any system, so that $\rho L=\rho_0L_0$ or, from (13.22), \begin{equation} \label{Eq:II:13:23} \rho=\frac{\rho_0}{\sqrt{1-v^2/c^2}}. \end{equation}  The charge density of a moving distribution of charges varies in the same way as the relativistic mass of a particle. We now use this general result for the positive charge density $\rho_+$ of our wire. These charges are at rest in frame $S$. In $S'$, however, where the wire moves with the speed $v$, the positive charge density becomes \begin{equation} \label{Eq:II:13:24} \rho_+'=\frac{\rho_+}{\sqrt{1-v^2/c^2}}. \end{equation}  The negative charges are at rest in $S'$. So they have their “rest density” $\rho_0$ in this frame. In Eq. (13.23) $\rho_0=\rho_-'$, because they have the density $\rho_-$ when the wire is at rest, i.e., in frame $S$, where the speed of the negative charges is $v$. For the conduction electrons, we then have that \begin{equation} \label{Eq:II:13:25} \rho_-=\frac{\rho_-'}{\sqrt{1-v^2/c^2}}, \end{equation}  or \begin{equation} \label{Eq:II:13:26} \rho_-'=\rho_-\sqrt{1-v^2/c^2}. \end{equation}  Now we can see why there are electric fields in $S'$—because in this frame the wire has the net charge density $\rho'$ given by \begin{equation*} \rho'=\rho_+'+\rho_-'. \end{equation*}  Using (13.24) and (13.26), we have \begin{equation*} \rho'=\frac{\rho_+}{\sqrt{1-v^2/c^2}}+\rho_-\sqrt{1-v^2/c^2}. \end{equation*}  Since the stationary wire is neutral, $\rho_-=-\rho_+$, and we have \begin{equation} \label{Eq:II:13:27} \rho'=\rho_+\,\frac{v^2/c^2}{\sqrt{1-v^2/c^2}}. \end{equation}  Our moving wire is positively charged and will produce an electric field $E'$ at the external stationary particle. We have already solved the electrostatic problem of a uniformly charged cylinder. The electric field at the distance $r$ from the axis of the cylinder is \begin{equation} \label{Eq:II:13:28} E'=\frac{\rho'A}{2\pi\epsO r}= \frac{\rho_+Av^2/c^2}{2\pi\epsO r\sqrt{1-v^2/c^2}}. \end{equation}  The force on the negatively charged particle is toward the wire. We have, at least, a force in the same direction from the two points of view; the electric force in $S'$ has the same direction as the magnetic force in $S$. The magnitude of the force in $S'$ is \begin{equation} \label{Eq:II:13:29} F'=\frac{q}{2\pi\epsO}\,\frac{\rho_+A}{r}\, \frac{v^2/c^2}{\sqrt{1-v^2/c^2}}. \end{equation}  Comparing this result for $F'$ with our result for $F$ in Eq. (13.21), we see that the magnitudes of the forces are almost identical from the two points of view. In fact, \begin{equation} \label{Eq:II:13:30} F'=\frac{F}{\sqrt{1-v^2/c^2}}, \end{equation}  so for the small velocities we have been considering, the two forces are equal. We can say that for low velocities, at least, we understand that magnetism and electricity are just “two ways of looking at the same thing.” But things are even better than that. If we take into account the fact that forces also transform when we go from one system to the other, we find that the two ways of looking at what happens do indeed give the same physical result for any velocity. One way of seeing this is to ask a question like: What transverse momentum will the particle have after the force has acted for a little while? We know from Chapter 16 of Vol. I that the transverse momentum of a particle should be the same in both the $S$- and $S'$-frames. Calling the transverse coordinate $y$, we want to compare $\Delta p_y$ and $\Delta p_y'$. Using the relativistically correct equation of motion, $\FLPF=d\FLPp/dt$, we expect that after the time $\Delta t$ our particle will have a transverse momentum $\Delta p_y$ in the $S$-system given by \begin{equation} \label{Eq:II:13:31} \Delta p_y=F\,\Delta t. \end{equation}  In the $S'$-system, the transverse momentum will be \begin{equation} \label{Eq:II:13:32} \Delta p_y'=F'\,\Delta t'. \end{equation}  We must, of course, compare $\Delta p_y$ and $\Delta p_y'$ for corresponding time intervals $\Delta t$ and $\Delta t'$. We have seen in Chapter 15 of Vol. I that the time intervals referred to a moving particle appear to be longer than those in the rest system of the particle. Since our particle is initially at rest in $S'$, we expect, for small $\Delta t$, that \begin{equation} \label{Eq:II:13:33} \Delta t=\frac{\Delta t'}{\sqrt{1-v^2/c^2}}, \end{equation}  and everything comes out O.K. From (13.31) and (13.32), \begin{equation*} \frac{\Delta p_y'}{\Delta p_y}=\frac{F'\,\Delta t'}{F\,\Delta t}, \end{equation*}  which is just $=1$ if we combine (13.30) and (13.33). We have found that we get the same physical result whether we analyze the motion of a particle moving along a wire in a coordinate system at rest with respect to the wire, or in a system at rest with respect to the particle. In the first instance, the force was purely “magnetic,” in the second, it was purely “electric.” The two points of view are illustrated in Fig. 13–12 (although there is still a magnetic field $B'$ in the second frame, it produces no forces on the stationary particle).  If we had chosen still another coordinate system, we would have found a different mixture of $\FLPE$ and $\FLPB$ fields. Electric and magnetic forces are part of one physical phenomenon—the electromagnetic interactions of particles. The separation of this interaction into electric and magnetic parts depends very much on the reference frame chosen for the description. But a complete electromagnetic description is invariant; electricity and magnetism taken together are consistent with Einstein’s relativity. Since electric and magnetic fields appear in different mixtures if we change our frame of reference, we must be careful about how we look at the fields $\FLPE$ and $\FLPB$. For instance, if we think of “lines” of $\FLPE$ or $\FLPB$, we must not attach too much reality to them. The lines may disappear if we try to observe them from a different coordinate system. For example, in system $S'$ there are electric field lines, which we do not find “moving past us with velocity $v$ in system $S$.” In system $S$ there are no electric field lines at all! Therefore it makes no sense to say something like: When I move a magnet, it takes its field with it, so the lines of $\FLPB$ are also moved. There is no way to make sense, in general, out of the idea of “the speed of a moving field line.” The fields are our way of describing what goes on at a point in space. In particular, $\FLPE$ and $\FLPB$ tell us about the forces that will act on a moving particle. The question “What is the force on a charge from a moving magnetic field?” doesn’t mean anything precise. The force is given by the values of $\FLPE$ and $\FLPB$ at the charge, and the formula (13.1) is not to be altered if the source of $\FLPE$ or $\FLPB$ is moving (it is the values of $\FLPE$ and $\FLPB$ that will be altered by the motion). Our mathematical description deals only with the fields as a function of $x$, $y$, $z$, and $t$ with respect to some inertial frame. We will later be speaking of “a wave of electric and magnetic fields travelling through space,” as, for instance, a light wave. But that is like speaking of a wave travelling on a string. We don’t then mean that some part of the string is moving in the direction of the wave, we mean that the displacement of the string appears first at one place and later at another. Similarly, in an electromagnetic wave, the wave travels; but the magnitude of the fields change. So in the future when we—or someone else—speaks of a “moving” field, you should think of it as just a handy, short way of describing a changing field in some circumstances.