Transformation of Electromagnetic Fields

In this section, we shall investigate how electromagnetic fields transform when viewed in different inertial frames of reference. Our investigation is premised on two assumptions. First, Maxwell's equations (see Section 2.4.2) take equivalent forms in all inertial frames of reference. Of course, this is just a special case of the equivalence principle discussed in Section 3.2.1. Second, the electric charge of an elementary particle is the same in all inertial reference frames. Our second assumption is an experimentally verifiable fact.

Figure 3.16: Parallel plate capacitor.

Consider three inertial reference frames, $S_0$, $S$, and $S'$, that are all in standard configurations with respect to one another. (See Section 3.2.6.) Let frame $S$ move parallel to the $x$-axis at speed $v_0$ with respect to frame $S_0$. Let frame $S'$ move parallel to the $x$-axis at speed $v'$ with respect to frame $S_0$, and with speed $v$ with respect to frame $S$. See Figure 3.16. It follows from the relativistic transformation of velocity (see Section 3.2.9) that

$$v' = \frac{v_0+v}{1+v_0\,v/c^2}.$$ (3.249)

Let us define the Lorentz factors

$$\gamma_0 =\left(1-\frac{v_0^{\,2}}{c^2}\right)^{-1/2},$$ (3.250)

$$\gamma =\left(1-\frac{v^{2}}{c^2}\right)^{-1/2},$$ (3.251)

$$\gamma' =\left(1-\frac{v'^{\,2}}{c^2}\right)^{-1/2}.$$ (3.252)

The previous four equations yield

$$\gamma' = \left[1-\frac{1}{c^2}\left(\frac{v_0+v}{1+v_0\,v/c^2}\right)^2\right]^{-1/2}$$

$$= \left(1+\frac{v_0\,v}{c^2}\right)\left[\left(1+\frac{v_0\,v}{c^2}\right)^2 - \left(\frac{v_0}{c}+\frac{v}{c}\right)^2\right]^{-1/2}$$

$$= \left(1+\frac{v_0\,v}{c^2}\right)\left(1-\frac{v_0^{\,2}}{c^2}-\frac{v^2}{c^2}+\frac{v_0^{\,2}\,v^2}{c^4}\right)^{-1/2}$$

$$= \left(1+\frac{v_0\,v}{c^2}\right)\left(1-\frac{v_0^{\,2}}{c^2}\right)^{-1/2}\left(1-\frac{v^2}{c^2}\right)^{-1/2}$$

$$= \gamma_0\,\gamma \left(1+\frac{v_0\,v}{c^2}\right).$$ (3.253)

Suppose that frame $S_0$ contains a parallel plate capacitor that is at rest. Let the capacitor plates be parallel to the $x$-$z$ plane. Furthermore, let the lower (in $y$) plate have the uniform electric charge density $\sigma_0$, and let the upper plate have the uniform charge density $-\sigma_0$. See Figure 3.16. In frame $S$, the capacitor plates appear to move in the $-x$-direction with speed $v_0$. Thus, the lengths of the plates (in the $x$-direction) are contracted by a factor $\gamma_0$, whereas the widths of the plates (in the $z$-direction) are unchanged. (See Section 3.2.7.) Moreover, according to our second assumption, the net electric charges on the two capacitor plates in frame $S$ are the same as those in frame $S_0$. It follows that, in frame $S$, the charge densities on the two plates are $\pm\sigma$, where

$$\sigma =\gamma_0\,\sigma_0.$$ (3.254)

Analogous reasoning reveals that the charge densities on the two capacitor plates in frame $S'$ are $\pm\sigma'$, where

$$\sigma' = \gamma'\,\sigma_0.$$ (3.255)

All of the electric charges in frame $S_0$ are stationary, so the associated current density is zero. However, in frames $S$ and $S'$, the charges on the capacitor plates appear to move in the $-x$-direction with speeds $v_0$ and $v'$, respectively. Thus, in frame $S$, the current per unit width (in the $z$-direction) flowing on the lower (in $y$) capacitor plate takes the form ${\bf J}= J_x\,{\bf e}_x$, where

$$J_x= - \sigma\,v_0 = -\gamma_0\,v_0\,\sigma_0.$$ (3.256)

There is an equal and opposite current per unit width flowing on the upper plate. Likewise, in frame $S'$, the current per unit width flowing on the lower plate takes the form ${\bf J}'=J_x'\,{\bf e}_x$, where

$$J_{x}'= - \sigma'\,v' = -\gamma'\,v'\,\sigma_0.$$ (3.257)

Again, there is an equal and opposite current per unit width flowing on the upper plate.

The integral form of the Maxwell equation (2.484) is

$$\oint_S {\bf E} \cdot d{\bf S} = \frac{1}{\epsilon_0} \int_V \rho \,dV,$$ (3.258)

where $S$ is some surface enclosing a volume $V$, and where use has been made of the divergence theorem. (See Section A.20.) As described in Sections 2.1.12 and 2.1.13, if the previous equation is applied to a Gaussian pill-box in frame $S$ that encloses one or other of the capacitor plates, and co-moves with the plates, then it is easily demonstrated that the electric field in the region between the plates is uniform, taking the form ${\bf E} = E_{y}\,{\bf e}_y$, where

$$E_y = \frac{\sigma}{\epsilon_0} = \gamma_0\,\frac{\sigma_0}{\epsilon_0}.$$ (3.259)

Note, incidentally, that there is nothing in Equation (3.258) that precludes volume $V$ from being a moving volume. Analogous reasoning reveals that the electric field between the capacitor plates in frame $S'$ is uniform, taking the value ${\bf E}'=E_{y}'\,{\bf e}_y$, where

$$E_{y}' = \frac{\sigma'}{\epsilon_0} = \gamma'\,\frac{\sigma_0}{\epsilon_0}.$$ (3.260)

The integral form of the Maxwell equation (2.487) is

$$\oint_C {\bf B}\cdot d{\bf r} = \int_S\left(\mu_0\,{\bf j} + \epsilon_0\,\mu_0\,\frac{\partial {\bf E}}{\partial t}\right)\cdot d{\bf S},$$ (3.261)

where $S$ is a surface bounded by a loop $C$, and where use has been made of the curl theorem. (See Section A.22.) However, the electric field between the capacitor plates is constant in time in all three of our reference frames, so the previous equation simplifies to give

$$\oint_C {\bf B}\cdot d{\bf r} = \mu_0\int_S{\bf j} \cdot d{\bf S},$$ (3.262)

If the previous equation is applied to an Ampèrian loop in the $y$-$z$ plane of frame $S$ that straddles one or other of the capacitor plates, and co-moves with the plates, then it is easily demonstrated that the magnetic field in the region between the plates is uniform, taking the form ${\bf B} = B_z\,{\bf e}_z$, where

$$B_z = \mu_0\,J_x = -\gamma_0\,v_0\,\mu_0\,\sigma_0.$$ (3.263)

Again, there is nothing in Equation (3.262) that prohibits surface $S$ from being a moving surface. Likewise, in frame $S'$, the magnetic field between the plates is uniform, taking the form ${\bf B} = B_z\,{\bf e}_z$, where

$$B_{z}' = \mu_0\,J_{x}' = -\gamma'\,v'\,\mu_0\,\sigma_0.$$ (3.264)

According to Equations (3.253), (3.259), (3.260), and (3.263),

$$E_{y}' =\gamma_0\,\gamma\left(1+\frac{v_0\,v}{c^2}\right)\frac{\sigma_0}{\epsilon_0}$$

$$= \gamma\left(\gamma_0\,\frac{\sigma_0}{\epsilon_0}+ v\,\gamma_0\,v_0\,\mu_0\,\sigma_0\right)$$

$$= \gamma\,(E_y-v\,B_z),$$ (3.265)

where use has been made of $c=1/\!\sqrt{\epsilon_0\,\mu_0}$. Likewise, Equation (3.249), (3.253), (3.259), (3.263), and (3.264) yield

$$B_{z}' =- \gamma_0\,\gamma\left(1+\frac{v_0\,v}{c^2}\right)\left( \frac{v_0+v}{1+v_0\,v/c^2}\right)\mu_0\,\sigma_0$$

$$=\gamma\left(-\gamma_0\,v_0\,\mu_0\,\sigma_0 -\frac{v}{c^2}\,\gamma_0\,\frac{\sigma_0}{\epsilon_0}\right)$$

$$= \gamma\left(B_z - \frac{v}{c^2}\,E_y\right).$$ (3.266)

If we repeat the previous exercise with capacitor plates that are parallel to the $x$-$y$ plane, instead of the $x$-$z$ plane, then it is easily demonstrated that

$$E_{z}' = \gamma\,(E_z+v\,B_y),$$ (3.267)

$$B_{y}' = \gamma\left(B_y + \frac{v}{c^2}\,E_z\right).$$ (3.268)

Figure 3.17: Parallel plate capacitor.

In order to determine the transformation rule for $E_x$, consider the situation shown in Figure 3.17. Here, a parallel plate capacitor is stationary in frame $S$, and is aligned such that its plates are parallel to the $y$-$z$ plane. Frame $S'$ is in a standard configuration, and moves with velocity ${\bf v} = v\,{\bf e}_x$, with respect to frame $S$. The cross-sectional areas of the plates are the same in reference frames $S$ and $S'$. (See Section 3.2.7.) Furthermore, the electric charges on the plates are identical in both frames. Hence, we deduce that the electric charge densities on the plates are the same in the two reference frames. In other words,

$$\sigma'=\sigma.$$ (3.269)

Now, the parallel distance between the two plates is contracted by a factor $\gamma$ in frame $S'$, compared to frame $S$. However, the electric field generated between the capacitor plates only depends on the charge density residing on the plates, and is independent of the inter-plate spacing. In fact, the electric fields in frames $S$ and $S'$ are ${\bf E}= E_x\,{\bf e}_x$ and ${\bf E}' = E_x'\,{\bf e}_x$, respectively, where

$$E_x = \frac{\sigma}{\epsilon_0},$$ (3.270)

$$E_x' = \frac{\sigma'}{\epsilon_0}.$$ (3.271)

Hence, we deduce from the previous three equations that

$$E_{x}' = E_x.$$ (3.272)

In order to determine the transformation for $B_x$, consider a long, thin, solenoid whose axis runs parallel to the $x$-direction. Let the solenoid have $N$ turns per unit length, and carry a current $I$, in frame $S$, and let the solenoid have $N'$ turns per unit length, and carry a current $I'$, in frame $S'$. Frame $S'$ is in a standard configuration, and moves with velocity ${\bf v} = v\,{\bf e}_x$, with respect to frame $S$. Because of relativistic length contraction (see Section 3.2.4),

$$N' = \gamma\,N.$$ (3.273)

On the other hand, because current is electric charge per unit time, and electric charge is invariant between different inertial frames, whereas time is dilated in frame $S'$, relative to frame $S$, we have

$$I' = \frac{I}{\gamma}.$$ (3.274)

According to Section 2.2.11, the magnetic fields generated inside the solenoid in frames $S$ and $S'$ are ${\bf B} = B_x\,{\rm e}_x$ and ${\bf B}' = B_x'\,{\bf e}_x$, respectively, where

$$B_x = \mu_0\,N\,I,$$ (3.275)

$$B_{x}' = \mu_0\,N'\,I'.$$ (3.276)

The previous four equations imply that

$$B_{x}'=B_x.$$ (3.277)

Thus, we can now state the complete set of transformation laws for the components of the electric and magnetic field between an inertial reference frame $S$, and a second inertial reference frame, $S'$, that is in a standard configuration, and moves at velocity ${\bf v} = v\,{\bf e}_x$, with respect to the first. The transformation laws are as follows:

$$E_{x}' = E_x,$$ (3.278)

$$E_{y}' = \gamma\,(E_y-v\,B_z),$$ (3.279)

$$E_{z}' = \gamma\,(E_z+v\,B_y),$$ (3.280)

and

$$B_{x}' = B_x,$$ (3.281)

$$B_{y}' = \gamma\left(B_y + \frac{v}{c^2}\,E_z\right),$$ (3.282)

$$B_{z}' = \gamma\left(B_z - \frac{v}{c^2}\,E_y\right).$$ (3.283)

Here, $\gamma=(1-v^2/c^2)^{-1/2}$.

It is easily demonstrated from the transformation rules (3.278)–(3.283) that

$${\bf E}'\cdot{\bf B}' = {\bf E}\cdot {\bf B},$$ (3.284)

$$E'^{\,2}-c^2\,B'^{\,2} = E^{\,2}-c^2\,B^2.$$ (3.285)

Thus, if electric and magnetic fields in one inertial frame of reference are in the configuration of an electromagnetic wave traveling through a vacuum—in other words, if ${\bf E}\cdot{\bf B}=0$ and $E=c\,B$ (see Section 2.4.4)—then they are in this configuration in all inertial frames of reference.