Solution

Introduce

$$\boldsymbol{\beta}=\frac{\mathbf v}{c},\qquad \gamma=\frac{1}{\sqrt{1-\beta^2}}.$$

The components of the electric and magnetic fields parallel and perpendicular to $\boldsymbol{\beta}$ will be denoted by the subscripts $\parallel$ and $\perp$, respectively.

The field transformation formulas are

$$\mathbf E'=\mathbf E_{\parallel}+\gamma\left(\mathbf E_{\perp}-[\boldsymbol{\beta}\times\mathbf B]\right),$$

$$\mathbf B'=\mathbf B_{\parallel}+\gamma\left(\mathbf B_{\perp}+[\boldsymbol{\beta}\times\mathbf E]\right).$$

a) Verification of the inverse transformation

After the first transformation with velocity $c\boldsymbol{\beta}$, the parallel components remain unchanged:

$$\mathbf E'_{\parallel}=\mathbf E_{\parallel},\qquad \mathbf B'_{\parallel}=\mathbf B_{\parallel}.$$

The transverse components are

$$\mathbf E'_{\perp}=\gamma\left(\mathbf E_{\perp}-[\boldsymbol{\beta}\times\mathbf B]\right),$$

$$\mathbf B'_{\perp}=\gamma\left(\mathbf B_{\perp}+[\boldsymbol{\beta}\times\mathbf E]\right).$$

Now perform the inverse transformation by replacing $\boldsymbol{\beta}$ with $-\boldsymbol{\beta}$.

For the transverse electric field,

$$\mathbf E''_{\perp}=\gamma\left(\mathbf E'_{\perp}+[\boldsymbol{\beta}\times\mathbf B']\right).$$

Since the parallel component of $\mathbf B'$ gives no contribution to the cross product,

$$[\boldsymbol{\beta}\times\mathbf B']=\gamma\left([\boldsymbol{\beta}\times\mathbf B_{\perp}]+[\boldsymbol{\beta}\times[\boldsymbol{\beta}\times\mathbf E]]\right).$$

Therefore,

$$\mathbf E''_{\perp}=\gamma^2\left(\mathbf E_{\perp}+[\boldsymbol{\beta}\times[\boldsymbol{\beta}\times\mathbf E]]\right).$$

Using

$$[\mathbf a\times[\mathbf a\times\mathbf b]]=\mathbf a(\mathbf a\cdot\mathbf b)-a^2\mathbf b,$$

and noting that only the transverse part of $\mathbf E$ contributes, we obtain

$$[\boldsymbol{\beta}\times[\boldsymbol{\beta}\times\mathbf E]]=-\beta^2\mathbf E_{\perp}.$$

Hence,

$$\mathbf E''_{\perp}=\gamma^2(1-\beta^2)\mathbf E_{\perp}.$$

Since

$$\gamma^2(1-\beta^2)=1,$$

we find

$$\mathbf E''_{\perp}=\mathbf E_{\perp}.$$

The parallel component is unchanged as well:

$$\mathbf E''_{\parallel}=\mathbf E_{\parallel}.$$

Thus,

$$\boxed{\mathbf E''=\mathbf E}.$$

The same calculation for the magnetic field gives

$$\boxed{\mathbf B''=\mathbf B}.$$

Therefore, two successive transformations with velocities $c\boldsymbol{\beta}$ and $-c\boldsymbol{\beta}$ return the fields to their original values.

b) Applications of the transformation formulas

Problem 14.3.1

In the rest frame of the capacitor,

$$\mathbf B=0.$$

The capacitor moves parallel to its plates, whereas the electric field is perpendicular to them. Hence,

$$\mathbf E\perp\boldsymbol{\beta}.$$

Therefore,

$$\mathbf E'=\gamma\mathbf E,$$

and

$$\boxed{E'=\gamma E}.$$

Because of Lorentz contraction, the dimension of each plate along the direction of motion decreases by a factor $\gamma$. Since the charge of the plate is invariant, its surface charge density becomes

$$\boxed{\sigma'=\gamma\sigma}.$$

The magnetic field is

$$\mathbf B'=\gamma[\boldsymbol{\beta}\times\mathbf E].$$

Since $\mathbf E'=\gamma\mathbf E$,

$$\boxed{\mathbf B'=[\boldsymbol{\beta}\times\mathbf E']}.$$

In magnitude,

$$\boxed{B'=\beta E'=\gamma\beta E}.$$

Thus,

$$\boxed{\sigma'=\gamma\sigma,\qquad E'=\gamma E,\qquad B'=\beta E'}.$$

Problem 14.3.2

The capacitor now moves at an angle $\alpha$ to the planes of its plates.

Since the electric field is perpendicular to the plates,

$$E_{\parallel}=E\sin\alpha,$$

$$E_{\perp}=E\cos\alpha.$$

The parallel component is unchanged:

$$\boxed{E'_{\parallel}=E\sin\alpha}.$$

The transverse component is multiplied by $\gamma$:

$$\boxed{E'_{\perp}=\gamma E\cos\alpha}.$$

Therefore,

$$\boxed{E'=E\sqrt{\sin^2\alpha+\gamma^2\cos^2\alpha}}.$$

The magnetic field is

$$\mathbf B'=\gamma[\boldsymbol{\beta}\times\mathbf E],$$

so that

$$\boxed{B'=\gamma\beta E\cos\alpha}.$$

Since

$$E'_{\perp}=\gamma E\cos\alpha,$$

we have

$$\boxed{B'=\beta E'_{\perp}}.$$

Equivalently, in vector form,

$$\boxed{\mathbf B'=[\boldsymbol{\beta}\times\mathbf E']}.$$

If the transformed surface charge density is also required, the area of a plate transforms as

$$S'=S\sqrt{1-\beta^2\cos^2\alpha},$$

and therefore

$$\boxed{\sigma'=\frac{\sigma}{\sqrt{1-\beta^2\cos^2\alpha}}}.$$

Problem 14.3.3

Let $\rho$ be the charge per unit length of the wire in its rest frame.

The electric field of an infinite charged wire in Gaussian units is

$$E_r=\frac{2\rho}{r},$$

where $r$ is the distance from the wire.

There is no magnetic field in the rest frame:

$$\mathbf B=0.$$

Since the wire moves along its own direction, the electric field is perpendicular to $\boldsymbol{\beta}$. Hence,

$$E'_r=\gamma E_r.$$

Therefore,

$$\boxed{E'_r=\frac{2\gamma\rho}{r}}.$$

Lorentz contraction also gives the transformed linear charge density

$$\boxed{\rho'=\gamma\rho}.$$

The magnetic field is

$$\mathbf B'=\gamma[\boldsymbol{\beta}\times\mathbf E].$$

Thus,

$$\boxed{B'=\beta E'_r=\frac{2\gamma\beta\rho}{r}}.$$

Hence,

$$\boxed{\rho'=\gamma\rho,\qquad E'_r=\frac{2\gamma\rho}{r},\qquad B'=\frac{2\gamma\beta\rho}{r}}.$$

Problem 14.3.5

In the original frame, the straight conductor is electrically neutral.

Let the volume charge densities of the ions and electrons be

$$\rho_i=\rho,\qquad \rho_e=-\rho.$$

The ions are at rest, while the conduction electrons move with speed $\beta c$ relative to the conductor.

Consider a frame in which the conductor moves with speed

$$\beta_1c,$$

where

$$\beta_1=k\beta.$$

Define

$$\gamma_1=\frac{1}{\sqrt{1-\beta_1^2}}.$$

Ion charge density

The ions, initially at rest, move with speed $\beta_1c$ in the new frame. Their longitudinal separations are Lorentz-contracted, so

$$\boxed{\rho'_i=\gamma_1\rho}.$$

Electron charge density

First pass to the rest frame of the electrons.

Since their charge density in the original frame is $-\rho$, their proper charge density is

$$\rho_{e0}=-\frac{\rho}{\gamma},$$

where

$$\gamma=\frac{1}{\sqrt{1-\beta^2}}.$$

The relative velocity between the electron rest frame and the required frame is

$$\beta_2=\frac{\beta_1-\beta}{1-\beta_1\beta}.$$

The corresponding Lorentz factor satisfies

$$\gamma_2=\gamma\gamma_1(1-\beta\beta_1).$$

Hence,

$$\rho'_e=\gamma_2\rho_{e0}.$$

Therefore,

$$\boxed{\rho'_e=-\gamma_1(1-\beta\beta_1)\rho}.$$

The total charge density in the moving conductor is

$$\rho'_{\Sigma}=\rho'_i+\rho'_e.$$

Thus,

$$\boxed{\rho'_{\Sigma}=\gamma_1\beta\beta_1\rho}.$$

The conductor is therefore no longer electrically neutral in this frame.

Magnetic field

In the original frame,

$$\mathbf E=0.$$

The magnetic field of a straight conductor is perpendicular to the conductor and therefore perpendicular to $\boldsymbol{\beta}_1$.

The transformation law gives

$$\mathbf B_1=\gamma_1\mathbf B.$$

Hence,

$$\boxed{B_1=\gamma_1B}.$$

Electric field

Since $\mathbf E=0$,

$$\mathbf E_1=-\gamma_1[\boldsymbol{\beta}_1\times\mathbf B].$$

Using $\mathbf B_1=\gamma_1\mathbf B$, we obtain

$$\boxed{\mathbf E_1=-[\boldsymbol{\beta}_1\times\mathbf B_1]}.$$

Therefore, in magnitude,

$$\boxed{E_1=\beta_1B_1}.$$

The final results are

$$\boxed{\rho'_i=\gamma_1\rho},$$

$$\boxed{\rho'_e=-\gamma_1(1-\beta\beta_1)\rho},$$

$$\boxed{\rho'_{\Sigma}=\gamma_1\beta\beta_1\rho},$$

$$\boxed{B_1=\gamma_1B},$$

$$\boxed{E_1=\beta_1B_1}.$$

c) Problems 14.3.6 and 14.3.7

Problem 14.3.6

Suppose that in the original frame there is only a magnetic field:

$$\mathbf E=0.$$

The transformed electric field is

$$\mathbf E'=-\gamma[\boldsymbol{\beta}\times\mathbf B].$$

The transformed magnetic field is

$$\mathbf B'=\mathbf B_{\parallel}+\gamma\mathbf B_{\perp}.$$

Since

$$[\boldsymbol{\beta}\times\mathbf B_{\parallel}]=0,$$

we have

$$[\boldsymbol{\beta}\times\mathbf B']=\gamma[\boldsymbol{\beta}\times\mathbf B].$$

Therefore,

$$\boxed{\mathbf E'=-[\boldsymbol{\beta}\times\mathbf B']}.$$

At low velocities,

$$\beta\ll1,$$

so that

$$\gamma\approx1,\qquad \mathbf B'\approx\mathbf B.$$

Hence,

$$\boxed{\mathbf E'\approx-[\boldsymbol{\beta}\times\mathbf B]}.$$

Thus, a field that is purely magnetic in one inertial frame generally contains an electric component in another frame.

Problem 14.3.7

Now suppose that in the original frame there is only an electric field:

$$\mathbf B=0.$$

Then

$$\mathbf E'=\mathbf E_{\parallel}+\gamma\mathbf E_{\perp},$$

and

$$\mathbf B'=\gamma[\boldsymbol{\beta}\times\mathbf E].$$

Since the parallel component of $\mathbf E'$ does not contribute to the cross product,

$$[\boldsymbol{\beta}\times\mathbf E']=\gamma[\boldsymbol{\beta}\times\mathbf E].$$

Therefore,

$$\boxed{\mathbf B'=[\boldsymbol{\beta}\times\mathbf E']}.$$

For

$$\beta\ll1,$$

we have

$$\gamma\approx1,\qquad \mathbf E'\approx\mathbf E.$$

Hence,

$$\boxed{\mathbf B'\approx[\boldsymbol{\beta}\times\mathbf E]}.$$

Thus, a field that is purely electric in one frame generally contains a magnetic component in another frame.

d) The limit $\beta\to1$

Consider the transverse components

$$\mathbf E'_{\perp}=\gamma\left(\mathbf E_{\perp}-[\boldsymbol{\beta}\times\mathbf B]\right),$$

$$\mathbf B'_{\perp}=\gamma\left(\mathbf B_{\perp}+[\boldsymbol{\beta}\times\mathbf E]\right).$$

Let

$$\mathbf n=\frac{\boldsymbol{\beta}}{\beta}$$

be a unit vector along the direction of motion.

As $\beta\to1$, define

$$\mathbf A=\mathbf E_{\perp}-[\mathbf n\times\mathbf B_{\perp}].$$

Then the leading transverse electric field is

$$\mathbf E'_{\perp}\sim\gamma\mathbf A.$$

Now,

$$[\mathbf n\times\mathbf A]=[\mathbf n\times\mathbf E_{\perp}]-[\mathbf n\times[\mathbf n\times\mathbf B_{\perp}]].$$

Since $\mathbf B_{\perp}\perp\mathbf n$,

$$[\mathbf n\times[\mathbf n\times\mathbf B_{\perp}]]=-\mathbf B_{\perp}.$$

Therefore,

$$[\mathbf n\times\mathbf A]=\mathbf B_{\perp}+[\mathbf n\times\mathbf E_{\perp}].$$

Hence the leading transverse magnetic field is

$$\mathbf B'_{\perp}\sim\gamma[\mathbf n\times\mathbf A].$$

Thus,

$$\mathbf B'_{\perp}\sim[\mathbf n\times\mathbf E'_{\perp}].$$

A cross product is perpendicular to the vector from which it is constructed. Therefore, in the ultrarelativistic limit,

$$\boxed{\mathbf E'\perp\mathbf B'}.$$

The same conclusion follows from the Lorentz invariant

$$\boxed{\mathbf E'\cdot\mathbf B'=\mathbf E\cdot\mathbf B}.$$

The scalar product remains finite, whereas in the generic ultrarelativistic case

$$E'\sim\gamma,\qquad B'\sim\gamma.$$

Thus,

$$E'B'\sim\gamma^2\to\infty.$$

If $\theta'$ is the angle between the transformed fields,

$$\cos\theta'=\frac{\mathbf E'\cdot\mathbf B'}{E'B'}.$$

Therefore,

$$\cos\theta'\to0,$$

and hence

$$\boxed{\theta'\to\frac{\pi}{2}}.$$

This statement refers to the non-degenerate case in which the leading transverse terms do not cancel.

Final results

The Lorentz transformations of the fields are mutually inverse:

$$\boxed{\mathbf E''=\mathbf E,\qquad \mathbf B''=\mathbf B}.$$

For a purely magnetic field,

$$\boxed{\mathbf E'=-[\boldsymbol{\beta}\times\mathbf B']}.$$

For a purely electric field,

$$\boxed{\mathbf B'=[\boldsymbol{\beta}\times\mathbf E']}.$$

At low velocities,

$$\boxed{\mathbf E'\approx-[\boldsymbol{\beta}\times\mathbf B]},$$

$$\boxed{\mathbf B'\approx[\boldsymbol{\beta}\times\mathbf E]}.$$

In the generic ultrarelativistic limit,

$$\boxed{\beta\to1\quad\Longrightarrow\quad\mathbf E'\perp\mathbf B'}.$$