Solutions "Special relativity and classical field theory - the theoretical minimum" (part II)

Solutions
Special Relativity and Classical Field Theory

Exercise list

Lecture 5

Exercise 5.1

Exercise 5.2

Lecture 6

Exercise 6.1

Exercise 6.2

Lecture 8

Exercise 8.1

Exercise 8.2

Exercise 8.3

Exercise 8.4

Lecture 11

Exercise 11.1

Exercise 11.2

Exercise 11.3

Exercise 11.4

Appendix

Exercise A.1

Exercise 5.1

Show that A^\nu A_\nu has the same meaning as A^\mu A_\mu.

Explicitly writing with the summation, as the index is dummy:

A^\nu A_\nu = \sum_{\nu = 0}^3 A^\nu A_\nu = \sum_{\mu = 0}^3 A^\mu A_\mu = A^\mu A_\mu

Exercise 5.2

Write an expression that undoes the effect of Eq. 5.20. In other words, how we “go backwards”?

Equation 5.20 is:

A_\mu = \eta_{\mu \nu} A^\nu

where:

\eta_{\mu\nu} = \begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}

Since the matrix is diagonal, its inverse is simply the matrix whose element of the diagonal are the reciprocal of the corresponding element, which for this matrix, since \left|a_{ii}\right| = 1, it is the matrix itself:

\eta^{\mu\nu} = \eta_{\mu\nu}

It is to go from a contravariant vector to a covariant vector using the matrix \eta_{\mu\nu}, and go from a covariant vector to a contravariant using \eta^{\mu\nu}:

\eta^{\mu \nu} A_\mu = \eta^{\mu \nu} \eta_{\mu \nu} A^\nu = A^\nu

Exercise 6.1

Given the transformation equation (Eq. 6.3) for the contravariant components of a 4-vector A^\nu, where L^\mu_\nu is a Lorentz transformation matrix, show that the Lorentz transformation for A’s covariant components is

\left(A^\prime \right)_\mu = {M_\mu}^\nu A_\nu

where

M = \eta L \eta

Eq. 6.3 is:

\left(A^\prime \right)^\mu = {L^\mu}_\nu A^\nu

The covariant components (A^\prime)_\mu in the primed frame are obtained by “lowering the index” of the contravariant components (A^\prime)^\alpha using the metric tensor \eta_{\mu\alpha}:

(A^\prime)_\mu = \eta_{\mu\alpha} (A^\prime)^\alpha

Let’s substitute the transformation for the contravariant components.

We know from Eq. 6.3 that (A^\prime)^\alpha = {L^\alpha}_\beta A^\beta. Substituting this into our equation:

(A^\prime)_\mu = \eta_{\mu\alpha} ({L^\alpha}_\beta A^\beta)

We can express the original contravariant vector in terms of its covariant version.

Our goal is to have an equation that transforms A_\nu to (A^\prime)_\mu. Right now we have an A^\beta.

We can “raise the index” on the original covariant vector A_\nu to get A^\beta using the inverse metric tensor \eta^{\beta\nu} (for the Minkowski metric, \eta^{-1} = \eta, so \eta^{\beta\nu} = \eta_{\beta\nu}):

A^\beta = \eta^{\beta\nu} A_\nu

We then substitute this back into our main equation:

(A^\prime)_\mu = \eta_{\mu\alpha} {L^\alpha}_\beta (\eta^{\beta\nu} A_\nu)

Finally we group the terms to identify the new transformation matrix. We can rearrange the parentheses since this is just a series of sums and products:

(A^\prime)_\mu = \left( \eta_{\mu\alpha} {L^\alpha}_\beta \eta^{\beta\nu} \right) A_\nu

This is exactly in the form we want: (A^\prime)_\mu = {M_\mu}^\nu A_\nu.

By comparing the two equations, we can identify the components of the matrix M:

{M_\mu}^\nu = \eta_{\mu\alpha} {L^\alpha}_\beta \eta^{\beta\nu}

This expression in index notation is precisely what the matrix equation M = \eta L \eta represents.

Now, let’s show that the matrix product \eta L \eta gives the exact matrix M that we identified.

Let’s define our matrices and set \gamma = \frac{1}{\sqrt{1-v^2}}:

\begin{aligned} L & = \begin{bmatrix} \gamma & -\gamma v & 0 & 0 \\ -\gamma v & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \\[10pt] \eta & = \begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \end{aligned}

The first step is to calculate \eta L:

\eta L = \begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} \gamma & -\gamma v & 0 & 0 \\ -\gamma v & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} = \begin{bmatrix} -\gamma & \gamma v & 0 & 0 \\ -\gamma v & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}

The second step is to calculate (\eta L) \eta:

(\eta L) \eta = \begin{bmatrix} -\gamma & \gamma v & 0 & 0 \\ -\gamma v & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} = \begin{bmatrix} \gamma & \gamma v & 0 & 0 \\ \gamma v & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}

The final matrix is:

\eta L \eta = \begin{bmatrix} \frac{1}{\sqrt{1-v^2}} & \frac{v}{\sqrt{1-v^2}} & 0 & 0 \\ \frac{v}{\sqrt{1-v^2}} & \frac{1}{\sqrt{1-v^2}} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} = M

which is precisely the matrix M, the original Lorentz transformation L with the sign of v reversed.

This completes the proof. We have shown that the covariant components A_\nu transform according to (A^\prime)_\mu = {M_\mu}^\nu A_\nu where the transformation matrix is given by M = \eta L \eta.

Exercise 6.2

Expression 6.28 was derived by identifying the index p with the z component of space, and then summing over n for the values (1,2,3). Why doesn’t the Expression 6.28 contains a v_z term?

We have that Eq. 6.28:

v_x \left( \frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x} \right) + v_y \left( \frac{\partial A_y}{\partial z} - \frac{\partial A_z}{\partial y} \right)

is derived from Eq. 6.27:

\dot{X}^n \left( \frac{\partial A_n}{\partial p} - \frac{\partial A_p}{\partial n} \right)

by setting p = z and then substituting x, y, z into the sum, we have that for n = z the term is:

v_z \left( \frac{\partial A_z}{\partial z} - \frac{\partial A_z}{\partial z} \right)

which is zero.

Exercise 8.1

Consider an electric charge at rest, with no additional electric of magnetic field present. In terms of the rest frame components, \left(E_x, E_y, E_z\right), what is the x component of the electric field for an observer moving in the negative x direction with velocity v? What are the y and z components? What are the corresponding components of the magnetic field?

We set some common definitions, because they will be used in all the exercises of the lecture.

The transformation of the electromagnetic field tensor from frame S to S^\prime is governed by the Lorentz transformation matrix, L. For a boost with velocity v along the x-axis, the matrix is:

L = \left[ \begin{array}{cc|cccc} & & 0 & 1 & 2 & 3 \\ & & ct & x & y & z \\ \hline 0 & ct & \gamma & -\gamma v & 0 & 0 \\ 1 & x & -\gamma v & \gamma & 0 & 0 \\ 2 & y & 0 & 0 & 1 & 0 \\3 & z & 0 & 0 & 0 & 1 \end{array} \right]

with \gamma = (1 - v^2)^{-1/2} (we consider c=1 as in the book).

To find the fields from the frame S in frame S^\prime we apply, component by component, the transformation:

\left(F^\prime\right)^{\mu\nu} = {L^\mu}_\sigma {L^\nu}_\tau F^{\sigma\tau} For a point charge q at rest at the origin in vacuum (these values will not be used, but written for reference), there is an electric field:

\begin{aligned} E_x & = \frac{q}{4\pi\epsilon_0} \frac{x}{(x^2 + y^2 + z^2)^{3/2}} \\ E_y & = \frac{q}{4\pi\epsilon_0} \frac{y}{(x^2 + y^2 + z^2)^{3/2}} \\ E_z & = \frac{q}{4\pi\epsilon_0} \frac{z}{(x^2 + y^2 + z^2)^{3/2}} \end{aligned}

The components of the electromagnetic field tensor, F^{\mu\nu}, are given by:

F^{\mu\nu} = \left[ \begin{array}{cc|cccc} & & 0 & 1 & 2 & 3 \\ & & ct & x & y & z \\ \hline 0 & ct & 0 & +E_x & +E_y & +E_z \\ 1 & x & -E_x & 0 & 0 & 0 \\ 2 & y & -E_y & 0 & 0 & 0 \\ 3 & z & -E_z & 0 & 0 & 0 \end{array} \right]

The electromagnetic tensor F^{\mu\nu} in this frame has six non-zero components:

\begin{aligned} & F^{0x} = -F^{x0} = E_x \\ & F^{0y} = -F^{y0} = E_y \\ & F^{0z} = -F^{z0} = E_z \end{aligned}

To simplify the summation all the ten zero components of F^{\mu\nu} will not be added to each formula, but leaving only the non-zeros. We will apply the transformation component by component, if we identify a zero term in a multiplication we will leave the other terms without explicitly written the value as the multiplication will lead to zero anyway.

For the x-component:

\begin{aligned} E_x^\prime & = \left(F^\prime \right)^{0x} = {L^0}_\sigma {L^x}_\tau F^{\sigma \tau} \\ & = {L^0}_0 {L^x}_x F^{0x} + {L^0}_0 {L^x}_y F^{0y} + {L^0}_0 {L^x}_z F^{0z} + {L^0}_x {L^x}_0 F^{x0} + {L^0}_y {L^x}_0 F^{y0} + {L^0}_z {L^x}_0 F^{z0} \\ & = {L^0}_0 {L^x}_x F^{0x} + {L^0}_0 (0) F^{0y} + {L^0}_0 (0) F^{0z} + {L^0}_x {L^x}_0 F^{x0} + (0) {L^x}_0 F^{y0} + (0) {L^x}_0 F^{z0} \\ & = {L^0}_0 {L^x}_x F^{0x} + {L^0}_x {L^x}_0 F^{x0} \\ & = \gamma \gamma E_x - (\gamma v) (\gamma v) E_x \\ & = \gamma^2 \left(1 - v^2\right)E_x = E_x \end{aligned}

For the y-component:

\begin{aligned} E_y^\prime & = \left(F^\prime \right)^{0y} = {L^0}_\sigma {L^y}_\tau F^{\sigma \tau} \\ & = {L^0}_0 {L^y}_x F^{0x} + {L^0}_0 {L^y}_y F^{0y} + {L^0}_0 {L^y}_z F^{0z} + {L^0}_x {L^y}_0 F^{x0} + {L^0}_y {L^y}_0 F^{y0} + {L^0}_z {L^y}_0 F^{z0} \\ & = {L^0}_0 (0) F^{0x} + {L^0}_0 {L^y}_y F^{0y} + {L^0}_0 (0) F^{0z} + {L^0}_x (0) F^{x0} + (0) {L^y}_0 F^{y0} + (0) {L^y}_0 F^{z0} \\ & = {L^0}_0 {L^y}_y F^{0y} \\ & = \gamma (1) E_y \\ & = \gamma E_y = \frac{1}{\sqrt{1-v^2}} E_y \end{aligned}

For the z-component:

\begin{aligned} E_z^\prime & = \left(F^\prime \right)^{0z} = {L^0}_\sigma {L^z}_\tau F^{\sigma \tau} \\ & = {L^0}_0 {L^z}_x F^{0x} + {L^0}_0 {L^z}_y F^{0y} + {L^0}_0 {L^z}_z F^{0z} + {L^0}_x {L^z}_0 F^{x0} + {L^0}_y {L^z}_0 F^{y0} + {L^0}_z {L^z}_0 F^{z0} \\ & = {L^0}_0 (0) F^{0x} + {L^0}_0 (0) F^{0y} + {L^0}_0 {L^z}_z F^{0z} + {L^0}_x (0) F^{x0} + (0) {L^z}_0 F^{y0} + (0) {L^z}_0 F^{z0} \\ & = {L^0}_0 {L^z}_z F^{0z} \\ & = \gamma (1) E_z \\ & = \gamma E_z = \frac{1}{\sqrt{1-v^2}} E_z \end{aligned}

We can compute the similar transformation for the magnetic field.

For the x component:

\begin{aligned} B_x^\prime & = (F^\prime)^{yz} = {L^y}_\sigma {L^z}_\tau F^{\sigma\tau} \\ & = {L^y}_0 {L^z}_x F^{0x} + {L^y}_0 {L^z}_y F^{0y} + {L^y}_0 {L^z}_z F^{0z} + {L^y}_x {L^z}_0 F^{x0} + {L^y}_y {L^z}_0 F^{y0} + {L^y}_z {L^z}_0 F^{z0} \\ & = (0) {L^z}_x F^{0x} + (0) {L^z}_y F^{0y} + (0) {L^z}_z F^{0z} + (0) {L^z}_0 F^{x0} + {L^y}_y (0) F^{y0} + (0) {L^z}_0 F^{z0} \\ & = 0 \end{aligned}

For the y component:

\begin{aligned} B_y^\prime & = (F^\prime)^{zx} = {L^z}_\sigma {L^x}_\tau F^{\sigma\tau} \\ & = {L^z}_0 {L^x}_x F^{0x} + {L^z}_0 {L^x}_y F^{0y} + {L^z}_0 {L^x}_z F^{0z} + {L^z}_x {L^x}_0 F^{x0} + {L^z}_y {L^x}_0 F^{y0} + {L^z}_z {L^x}_0 F^{z0} \\ & = (0) {L^x}_x F^{0x} + (0) {L^x}_y F^{0y} + (0) {L^x}_z F^{0z} + (0) {L^x}_0 F^{x0} + (0) {L^x}_0 F^{y0} + {L^z}_z {L^x}_0 F^{z0} \\ & = {L^z}_z {L^x}_0 F^{z0} \\ & = (1) (-\gamma v) (-E_z) \\ & = \gamma v E_z = \frac{v}{\sqrt{1-v^2}} E_z \end{aligned}

For the z component:

\begin{aligned} B_z^\prime & = (F^\prime)^{xy} = {L^x}_\sigma {L^y}_\tau F^{\sigma\tau} \\ & = {L^x}_0 {L^y}_x F^{0x} + {L^x}_0 {L^y}_y F^{0y} + {L^x}_0 {L^y}_z F^{0z} + {L^x}_x {L^y}_0 F^{x0} + {L^x}_y {L^y}_0 F^{y0} + {L^x}_z {L^y}_0 F^{z0} \\ & = {L^x}_0 (0) F^{0x} + {L^x}_0 {L^y}_y F^{0y} + {L^x}_0 (0) F^{0z} + {L^x}_x (0) F^{x0} + (0) (0) F^{y0} + (0) (0) F^{z0} \\ & = {L^x}_0 {L^y}_y F^{0y} \\ & = (-\gamma v) (1) E_y \\ & = -\gamma v E_y = -\frac{v}{\sqrt{1-v^2}} E_y \end{aligned}

The fields in the moving frame (the primed frame) for this specific scenario are:

\begin{aligned} \mathbf E^\prime & = \begin{pmatrix} E_x, & \frac{1}{\sqrt{1 - v^2}} E_y, & \frac{1}{\sqrt{1-v^2}} E_z \end{pmatrix} \\ \mathbf B^\prime & = \begin{pmatrix} 0, & \frac{v}{\sqrt{1-v^2}} E_z, & - \frac{v}{\sqrt{1-v^2}} E_y\end{pmatrix} \end{aligned}

Exercise 8.2

Art is sitting in the station as the train passes by. In terms of Lenny’s field components, what is the x component of E observed by Art? What are the y and z components? What are the components of the magnetic field seen by Art?

The components of the electromagnetic field tensor in Lenny reference frame, F^{\mu\nu}, are given by:

F^{\mu\nu} = \left[ \begin{array}{cc|cccc} & & 0 & 1 & 2 & 3 \\ & & ct & x & y & z \\ \hline 0 & ct & 0 & +E_x & +E_y & +E_z \\ 1 & x & -E_x & 0 & +B_z & -B_y \\ 2 & y & -E_y & -B_z & 0 & +B_x \\ 3 & z & -E_z & +B_y & -B_x & 0 \end{array} \right] In this case there are no null component in the tensor, except the diagonal, since it is antisymmetric. We will apply the transformation component by component. We will apply the transformation component by component, if we identify a zero term in a multiplication we will leave the other terms without explicitly written the value as the multiplication will lead to zero anyway.

For the x-component:

\begin{aligned} E_x^\prime = & \left(F^\prime \right)^{0x} = {L^0}_\sigma {L^x}_\tau F^{\sigma \tau} \\ = & {L^0}_0 {L^x}_x F^{0x} + {L^0}_0 {L^x}_y F^{0y} + {L^0}_0 {L^x}_z F^{0z} + {L^0}_x {L^x}_0 F^{x0} + {L^0}_x {L^x}_y F^{xy} + {L^0}_x {L^x}_z F^{xz} \\ & + {L^0}_y {L^x}_0 F^{y0} + {L^0}_y {L^x}_x F^{yx} + {L^0}_y {L^x}_z F^{yz} + {L^0}_z {L^x}_0 F^{z0} + {L^0}_z {L^x}_x F^{zx} + {L^0}_z {L^x}_y F^{zy} \\ = & {L^0}_0 {L^x}_x F^{0x} + {L^0}_0 (0) F^{0y} + {L^0}_0 (0) F^{0z} + {L^0}_x {L^x}_0 F^{x0} + {L^0}_x (0) F^{xy} + {L^0}_x (0) F^{xz} \\ & + (0) {L^x}_0 F^{y0} + (0) {L^x}_x F^{yx} + (0) {L^x}_z F^{yz} + (0) {L^x}_0 F^{z0} + (0) {L^x}_x F^{zx} + (0) {L^x}_y F^{zy} \\ = & {L^0}_0 {L^x}_x F^{0x} + {L^0}_x {L^x}_0 F^{x0} \\ = & \gamma \gamma E_x - (\gamma v) (\gamma v) E_x \\ = & \gamma^2 \left(1 - v^2\right)E_x = E_x \end{aligned}

For the y-component:

\begin{aligned} E_y^\prime & = \left(F^\prime \right)^{0y} = {L^0}_\sigma {L^y}_\tau F^{\sigma \tau} \\ = & {L^0}_0 {L^y}_x F^{0x} + {L^0}_0 {L^y}_y F^{0y} + {L^0}_0 {L^y}_z F^{0z} + {L^0}_x {L^y}_0 F^{x0} + {L^0}_x {L^y}_y F^{xy} + {L^0}_x {L^y}_z F^{xz} \\ & + {L^0}_y {L^y}_0 F^{y0} + {L^0}_y {L^y}_x F^{yx} + {L^0}_y {L^y}_z F^{yz} + {L^0}_z {L^y}_0 F^{z0} + {L^0}_z {L^y}_x F^{zx} + {L^0}_z {L^y}_y F^{zy} \\ = & {L^0}_0 (0) F^{0x} + {L^0}_0 {L^y}_y F^{0y} + {L^0}_0 (0)F^{0z} + {L^0}_x (0) F^{x0} + {L^0}_x {L^y}_y F^{xy} + {L^0}_x (0) F^{xz} \\ & + (0) {L^y}_0 F^{y0} + (0){L^y}_x F^{yx} + (0) {L^y}_z F^{yz} + (0) {L^y}_0 F^{z0} + (0) {L^y}_x F^{zx} + (0) {L^y}_y F^{zy} \\ = & \gamma (1) E_y + (-\gamma v) (1) B_z \\ = & \gamma E_y -\gamma v B_z \\ = & \frac{1}{\sqrt{1-v^2}} E_y -\frac{v}{\sqrt{1-v^2}} B_z \end{aligned}

For the z-component:

\begin{aligned} E_z^\prime & = \left(F^\prime \right)^{0z} = {L^0}_\sigma {L^z}_\tau F^{\sigma \tau} \\ & = {L^0}_0 {L^z}_x F^{0x} + {L^0}_0 {L^z}_y F^{0y} + {L^0}_0 {L^z}_z F^{0z} + {L^0}_x {L^z}_0 F^{x0} + {L^0}_x {L^z}_y F^{xy} + {L^0}_x {L^z}_z F^{xz} \\ & + {L^0}_y {L^z}_0 F^{y0} + {L^0}_y {L^z}_x F^{yx} + {L^0}_y {L^z}_z F^{yz} + {L^0}_z {L^z}_0 F^{z0} + {L^0}_z {L^z}_x F^{zx} + {L^0}_z {L^z}_y F^{zy} \\ = & {L^0}_0 (0) F^{0x} + {L^0}_0 (0) F^{0y} + {L^0}_0 {L^z}_z F^{0z} + {L^0}_x (0) F^{x0} + {L^0}_x (0) F^{xy} + {L^0}_x {L^z}_z F^{xz} \\ & + (0){L^z}_0 F^{y0} + (0) {L^z}_x F^{yx} + (0){L^z}_z F^{yz} + (0) {L^z}_0 F^{z0} + (0) {L^z}_x F^{zx} + (0) {L^z}_y F^{zy} \\ = & {L^0}_0 {L^z}_z F^{0z} + {L^0}_x {L^z}_z F^{xz} \\ = & (\gamma) (1) (E_z) + (-\gamma v) (1) (-B_y) \\ = & \gamma E_z + \gamma v B_y \\ = & \frac{1}{\sqrt{1-v^2}} E_z + \frac{v}{\sqrt{1-v^2}} B_y \end{aligned}

We can compute the similar transformation for the magnetic field.

For the x component:

\begin{aligned} B_x^\prime & = (F^\prime)^{yz} = {L^y}_\sigma {L^z}_\tau F^{\sigma\tau} \\ & = {L^y}_0 {L^z}_x F^{0x} + {L^y}_0 {L^z}_y F^{0y} + {L^y}_0 {L^z}_z F^{0z} + {L^y}_x {L^z}_0 F^{x0} + {L^y}_x {L^z}_y F^{xy} + {L^y}_x {L^z}_z F^{xz} \\ & + {L^y}_y {L^z}_0 F^{y0} + {L^y}_y {L^z}_x F^{yx} + {L^y}_y {L^z}_z F^{yz} + {L^y}_z {L^z}_0 F^{z0} + {L^y}_z {L^z}_x F^{zx} + {L^y}_z {L^z}_y F^{zy} \\ = & (0){L^z}_x F^{0x} + (0){L^z}_y F^{0y} + (0){L^z}_z F^{0z} + (0){L^z}_0 F^{x0} + (0){L^z}_y F^{xy} + (0){L^z}_z F^{xz} \\ & + {L^y}_y (0) F^{y0} + {L^y}_y (0) F^{yx} + {L^y}_y {L^z}_z F^{yz} + (0){L^z}_0 F^{z0} + (0){L^z}_x F^{zx} + (0){L^z}_y F^{zy} \\ = & {L^y}_y {L^z}_z F^{yz} \\ = & (1) (1) (B_x) \\ = & B_x \end{aligned}

For the y component:

\begin{aligned} B_y^\prime & = (F^\prime)^{zx} = {L^z}_\sigma {L^x}_\tau F^{\sigma\tau} \\ & = {L^z}_0 {L^x}_x F^{0x} + {L^z}_0 {L^x}_y F^{0y} + {L^z}_0 {L^x}_z F^{0z} + {L^z}_x {L^x}_0 F^{x0} + {L^z}_x {L^x}_y F^{xy} + {L^z}_x {L^x}_z F^{xz} \\ & + {L^z}_y {L^x}_0 F^{y0} + {L^z}_y {L^x}_x F^{yx} + {L^z}_y {L^x}_z F^{yz} + {L^z}_z {L^x}_0 F^{z0} + {L^z}_z {L^x}_x F^{zx} + {L^z}_z {L^x}_y F^{zy} \\ = & (0) {L^x}_x F^{0x} + (0){L^x}_y F^{0y} + (0){L^x}_z F^{0z} + (0){L^x}_0 F^{x0} + (0){L^x}_y F^{xy} + (0){L^x}_z F^{xz} \\ & + (0){L^x}_0 F^{y0} + (0){L^x}_x F^{yx} + (0){L^x}_z F^{yz} + {L^z}_z {L^x}_0 F^{z0} + {L^z}_z {L^x}_x F^{zx} + {L^z}_z (0) F^{zy} \\ = & {L^z}_z {L^x}_0 F^{z0} + {L^z}_z {L^x}_x F^{zx} \\ = & (1) (-\gamma v) (-E_z) + (1) (\gamma) (B_y) \\ = & \gamma v E_z + \gamma B_y \\ = & \frac{1}{\sqrt{1-v^2}} B_y + \frac{v}{\sqrt{1-v^2}} E_z \end{aligned}

For the z component:

\begin{aligned} B_z^\prime & = (F^\prime)^{xy} = {L^x}_\sigma {L^y}_\tau F^{\sigma\tau} \\ & = {L^x}_0 {L^y}_x F^{0x} + {L^x}_0 {L^y}_y F^{0y} + {L^x}_0 {L^y}_z F^{0z} + {L^x}_x {L^y}_0 F^{x0} + {L^x}_x {L^y}_y F^{xy} + {L^x}_x {L^y}_z F^{xz} \\ & + {L^x}_y {L^y}_0 F^{y0} + {L^x}_y {L^y}_x F^{yx} + {L^x}_y {L^y}_z F^{yz} + {L^x}_z {L^y}_0 F^{z0} + {L^x}_z {L^y}_x F^{zx} + {L^x}_z {L^y}_y F^{zy} \\ = & {L^x}_0 (0) F^{0x} + {L^x}_0 {L^y}_y F^{0y} + {L^x}_0 (0)F^{0z} + {L^x}_x (0) F^{x0} + {L^x}_x {L^y}_y F^{xy} + {L^x}_x (0) F^{xz} \\ & + (0) {L^y}_0 F^{y0} + (0){L^y}_x F^{yx} + (0) {L^y}_z F^{yz} + (0) {L^y}_0 F^{z0} + (0) {L^y}_x F^{zx} + (0) {L^y}_y F^{zy} \\ = & {L^x}_0 {L^y}_y F^{0y} + {L^x}_x {L^y}_y F^{xy} \\ = & (-\gamma v) (1) (E_y) + (\gamma) (1) (B_z) \\ = & \gamma B_z - \gamma v E_y \\ = & \frac{1}{\sqrt{1-v^2}} B_z - \frac{v}{\sqrt{1-v^2}} E_y \end{aligned}

The fields in the moving frame (the primed frame) for this specific scenario are:

\begin{aligned} \mathbf E^\prime & = \begin{pmatrix} E_x, & \frac{1}{\sqrt{1 - v^2}} E_y - \frac{v}{\sqrt{1 - v^2}}B_z, & \frac{1}{\sqrt{1-v^2}} E_z + \frac{v}{\sqrt{1-v^2}} B_y \end{pmatrix} \\ \mathbf B^\prime & = \begin{pmatrix} B_x, & \frac{1}{\sqrt{1-v^2}} B_y + \frac{v}{\sqrt{1-v^2}} E_z, & \frac{1}{\sqrt{1-v^2}} B_z - \frac{v}{\sqrt{1-v^2}} E_y\end{pmatrix} \end{aligned}

Exercise 8.3

For Einstein’s example, work out all components of the electric and magnetic fields in the electron’s rest frame.

In the Einstein example, the components of the electromagnetic field tensor, F^{\mu\nu}, are given by:

F^{\mu\nu} = \left[ \begin{array}{cc|cccc} & & 0 & 1 & 2 & 3 \\ & & ct & x & y & z \\ \hline 0 & ct & 0 & 0 & 0 & 0 \\ 1 & x & 0 & 0 & +B_z & 0 \\ 2& y & 0 & -B_z & 0 & 0 \\ 3 & z & 0 & 0 & 0 & 0 \end{array} \right]

The electromagnetic tensor F^{\mu\nu} in this frame has only two non-zero components:

F^{xy} =- F^{yx} = B_z

To simplify the summation all the fourteen zero components of F^{\mu\nu} will not be added to each formula, but leaving only the two non-zeros. We will apply the transformation component by component.

For the x-component:

\begin{aligned} E_x^\prime & = \left(F^\prime \right)^{0x} = {L^0}_\sigma {L^x}_\tau F^{\sigma \tau} \\ & = {L^0}_x {L^x}_y F^{xy} + {L^0}_y {L^x}_x F^{xy}\\ & = (-\gamma \beta) (0) F^{x y} + (0) (\gamma) F^{y x} = 0 \end{aligned}

For the y-component:

\begin{aligned} E_y^\prime & = \left(F^\prime \right)^{0y} = {L^0}_\sigma {L^y}_\tau F^{\sigma \tau} \\ & = {L^0}_x {L^y}_y F^{x y} + {L^0}_y {L^y}_x F^{y x}\\ & = (-\gamma \beta) (1) F^{x y} + (0) (0) F^{y x} \\ & = -\beta\gamma F^{x y} = -\frac{v}{\sqrt{1-v^2}} B^z \end{aligned}

For the z-component:

\begin{aligned} E_z^\prime & = \left(F^\prime \right)^{0z} = {L^0}_\sigma {L^z}_\tau F^{\sigma \tau} \\ & = {L^0}_x {L^z}_y F^{x y} + {L^0}_y {L^z}_x F^{y x}\\ & = (-\gamma \beta) (0) F^{x y} + (0)(0) F^{y x} = 0 \end{aligned}

We can compute the similar transformation for the magnetic field.

For the x component:

\begin{aligned} B_x^\prime & = (F^\prime)^{y z} = {L^y}_\sigma {L^z}_\tau F^{\sigma\tau} \\ & = {L^y}_x {L^z}_y F^{x y} + {L^y}_y {L^z}_x F^{y x} \\ & =(0)(0)F^{x y} + (1)(0)F^{y x} = 0 \end{aligned}

For the y component:

\begin{aligned} B_y^\prime & = (F^\prime)^{z x} = {L^z}_\sigma {L^x}_\tau F^{\sigma\tau} \\ & = {L^z}_x {L^x}_y F^{x y} + {L^z}_y {L^x}_x F^{y x} \\ & = (0)(0)F^{x y} + (0)(\gamma)F^{y x} = 0 \end{aligned}

For the z component:

\begin{aligned} B_z^\prime & = (F^\prime)^{x y} = {L^x}_\sigma {L^y}_\tau F^{\sigma\tau} \\ & = {L^x}_x {L^y}_y F^{x y} + {L^x}_y {L^y}_x F^{y x} \\ & = (\gamma)(1)F^{x y} + (0)(0)F^{y x} \\ &= \gamma F^{x y} = \gamma B_z = \frac{1}{\sqrt{1 - v^2}} B_z \end{aligned}

The fields in the moving frame (the primed frame) for this specific scenario are:

\begin{aligned} \mathbf E^\prime & = \begin{pmatrix} 0, & -\frac{v}{\sqrt{1 - v^2}} B_z, & 0 \end{pmatrix} \\ \mathbf B^\prime & = \begin{pmatrix} 0, & 0, & \frac{1}{\sqrt{1 - v^2}} B_z \end{pmatrix} \end{aligned}

A final point to consider is the force experienced by the electron in its own rest frame (the primed frame). The Lorentz force is given by the equation:

F = q(\mathbf E + \mathbf v \times \mathbf B)

In the electron’s rest frame, its velocity is, by definition, zero. Consequently, the magnetic component of the Lorentz force vanishes. This means that even though an observer in this frame measures a non-zero magnetic field \mathbf B^\prime, it exerts no force on the stationary electron. The only force acting on the electron is the electric force.

Therefore, in its rest frame, the electron is accelerated solely by the electric field component that arises from the relativistic transformation of the original magnetic field.

A discussion of Einstein’s example is available here.

Exercise 8.4

Use the second group of Maxwell’s equations from Table 8.1 with the two vector identities from Section 8.2.1 to derive the continuity equation.

The second group of Maxwell’s equations from table 8.1 are:

\begin{aligned} & \nabla \cdot \mathbf E = \rho \\ & \nabla \times \mathbf B - \frac{\partial}{\partial t} \mathbf E = \mathbf j \end{aligned}

And the two vector identities from section 8.2.1 are:

\begin{aligned} & \nabla \cdot \left(\nabla \times \mathbf A\right) = 0 \\ & \nabla \times \left(\nabla S\right) = 0 \end{aligned}

The continuity equation is:

\frac{\partial \rho}{\partial t} + \nabla \cdot \mathbf j = 0

Taking the divergence of the second Maxwell’s equation:

\nabla \cdot (\nabla \times \mathbf B) - \nabla \cdot \left(\frac{\partial}{\partial t} \mathbf E \right) = \nabla \cdot \mathbf j

The first term is zero using the vector identity:

- \nabla \cdot\left(\frac{\partial}{\partial t} \mathbf E\right) = \nabla \cdot \mathbf j

Exchanging the derivative orders and using the first of the Maxwell’s equations the left hand side:

- \frac{\partial}{\partial t} \left(\nabla \cdot \mathbf E\right) = - \frac{\partial \rho}{\partial t}

Substituting gives the continuity equation:

\begin{aligned} & - \frac{\partial \rho}{\partial t} = \nabla \cdot \mathbf j \\ & \frac{\partial \rho}{\partial t} + \nabla \cdot \mathbf j = 0 \end{aligned}

Exercise 11.1

Show that T^{0n} is the Poynting vector.

The Poynting vector is:

\mathbf S = \mathbf E \times \mathbf B = \begin{vmatrix} \mathbf i & \mathbf j & \mathbf k \\ E_x & E_y & E_z \\ B_x & B_y & B_z \end{vmatrix}

whose components are:

\begin{aligned} & S_x = E_yB_z - E_zB_y \\ & S_y = E_zB_x - E_xB_z \\ & S_z = E_xB_y - E_yB_x \end{aligned}

Let’s list the electromagnetic tensor will all its variant as it will be used in this and the next exercise:

\begin{aligned} & F_{\mu\nu} = \begin{bmatrix} 0 & -E_x & -E_y & -E_z \\ +E_x & 0 & +B_z & -B_y \\ +E_y & -B_z & 0 & +B_x \\ +E_z & +B_y & -B_x & 0 \end{bmatrix} \\[25pt] & F^{\mu\nu} = \begin{bmatrix} 0 & +E_x & +E_y & +E_z \\ -E_x & 0 & +B_z & -B_y \\ -E_y & -B_z & 0 & +B_x \\ -E_z & +B_y & -B_x & 0 \end{bmatrix}\\[25pt] & {F^\mu}_\nu = \begin{bmatrix} 0 & +E_x & +E_y & +E_z \\ +E_x & 0 & +B_z & -B_y \\ +E_y & -B_z & 0 & +B_x \\ +E_z & +B_y & -B_x & 0 \end{bmatrix}\\[25pt] & {F_\mu}^\nu = \begin{bmatrix} 0 & -E_x & -E_y & -E_z \\ -E_x & 0 & +B_z & -B_y \\ -E_y & -B_z & 0 & +B_x \\ -E_z & +B_y & -B_x & 0 \end{bmatrix} \end{aligned} The energy-momentum tensor is:

T^{\mu\nu} = F^{\mu\sigma}{F^\nu}_\sigma - \frac{1}{4} \eta^{\mu\nu}F^{\sigma\tau}F_{\sigma\tau}

Which for energy flux is:

T^{0n} = F^{0\sigma}{F^n}_\sigma - \frac{1}{4} \eta^{0n}F^{\sigma\tau}F_{\sigma\tau} = F^{0\sigma}{F^n}_\sigma

as \eta^{0n} = 0.

This gives:

T^{0n} = F^{00}{F^n}_0 + F^{01}{F^n}_1 + F^{02}{F^n}_2 + F^{03}{F^n}_3

We can identify the components in the above matrix:

F^{\mu\nu} = \begin{bmatrix} \boxed{0} & \boxed{+E_x} & \boxed{+E_y} & \boxed{+E_z} \\ -E_x & 0 & +B_z & -B_y \\ -E_y & -B_z & 0 & +B_x \\ -E_z & +B_y & -B_x & 0 \end{bmatrix}

and substituting:

T^{0n} = 0{F^n}_0 + E_x{F^n}_1 + E_y{F^n}_2 + E_z{F^n}_3

Finally, writing the components for n=1,2,3:

\begin{aligned} T^{01} & = E_x{F^1}_1 + E_y{F^1}_2 + E_z{F^1}_3 = E_x(0) + E_yB_z + E_z(-By) \\ & = E_yB_z - E_zB_y = S_x\\ T^{02} & = E_x{F^2}_1 + E_y{F^2}_2 + E_z{F^2}_3 = E_x(-B_z) + E_y(0) + E_zB_x\\ & = E_zB_x - E_xB_z = S_y\\ T^{03} & = E_x{F^3}_1 + E_y{F^3}_2 + E_z{F^3}_3 = E_xB_y + E_y(-B_x) + E_z(0) \\ & = E_xB_y - E_yB_x= S_z \end{aligned}

Exercise 11.2

Calculate T^{11} and T^{12} in term of the field components \left(E_x, E_y, E_z\right) and \left(B_x, B_y, B_z\right).

The component T^{11} is:

T^{11} = F^{1\sigma}{F^1}_\sigma - \frac{1}{4} \eta^{11}F^{\sigma\tau}F_{\sigma\tau} = F^{1\sigma}{F^1}_\sigma - \frac{1}{4}F^{\sigma\tau}F_{\sigma\tau}

as \eta^{11} = 1.

Computing F^{1\sigma}{F^1}_\sigma:

\begin{aligned} F^{1\sigma}{F^1}_\sigma =\ & F^{10}{F^1}_0 + F^{11}{F^1}_1 + F^{12}{F^1}_2 + F^{13}{F^1}_3 \\ & (-E_x)(E_x) + (0)(0) + (-B_z)(-B_z) + (B_y)(B_y) \\ & -E_x^2 + 0 + B_z^2 + B_y^2 \\ & -E_x^2 + B_y^2 + B_z^2 \end{aligned}

Computing F^{\sigma\tau}F_{\sigma\tau}:

\begin{aligned} F^{\sigma\tau}F_{\sigma\tau} &= F^{01}F_{01} + F^{02}F_{02} + F^{03}F_{03} \\ & + F^{10}F_{10} + F^{12}F_{12} + F^{13}F_{13} \\ & + F^{20}F_{20} + F^{21}F_{21} + F^{23}F_{23} \\ & + F^{30}F_{30} + F^{31}F_{31} + F^{32}F_{32} \\ &= (E_x)(-E_x) + (E_y)(-E_y) + (E_z)(-E_z) \\ & + (-E_x)(E_x) + (-B_z)(-B_z) + (B_y)(B_y) \\ & + (-E_y)(E_y) + (B_z)(B_z) + (-B_x)(-B_x) \\ & + (-E_z)(E_z) + (-B_y)(-B_y) + (B_x)(B_x) \\ &= -E_x^2 - E_y^2 - E_z^2 \\ & - E_x^2 + B_z^2 + B_y^2 \\ & - E_y^2 + B_z^2 + B_x^2 \\ & - E_z^2 + B_y^2 + B_x^2 \\ &= 2\left(B_x^2 + B_y^2 + B_z^2 - E_x^2 - E_y^2 - E_z^2\right)\\ &= 2\left(B^2 - E^2\right) \end{aligned}

Combining:

\begin{aligned} T^{11} &= -E_x^2 + B_y^2 + B_z^2 - \frac{1}{4} \left[2(B^2 - E^2)\right] \\ &= -E_x^2 + B_y^2 + B_z^2 - \frac{1}{2}(B_x^2 + B_y^2 + B_z^2 - E_x^2 - E_y^2 - E_z^2) \\ &= -E_x^2 + B_y^2 + B_z^2 - \frac{1}{2}B_x^2 - \frac{1}{2}B_y^2 - \frac{1}{2}B_z^2 + \frac{1}{2}E_x^2 + \frac{1}{2}E_y^2 + \frac{1}{2}E_z^2 \\ &= \left(-E_x^2 + \frac{1}{2}E_x^2\right) - \frac{1}{2}B_x^2 + \left(B_y^2 - \frac{1}{2}B_y^2\right) + \left(B_z^2 - \frac{1}{2}B_z^2\right) + \frac{1}{2}E_y^2 + \frac{1}{2}E_z^2 \\ &= -\frac{1}{2}E_x^2 - \frac{1}{2}B_x^2 + \frac{1}{2}B_y^2 + \frac{1}{2}B_z^2 + \frac{1}{2}E_y^2 + \frac{1}{2}E_z^2 \\ &= -E_x^2 - B_x^2 + \frac{1}{2}(E_x^2 + E_y^2 + E_z^2) + \frac{1}{2}(B_x^2 + B_y^2 + B_z^2) \\ &= -E_x^2 - B_x^2 + \frac{1}{2}(E^2 + B^2) \end{aligned}

The component T^{12} is:

T^{12} = F^{1\sigma}{F^2}_\sigma - \frac{1}{4} \eta^{12}F^{\sigma\tau}F_{\sigma\tau} = F^{1\sigma}{F^2}_\sigma as \eta^{12} = 0.

Expanding:

\begin{aligned} T^{12} & = F^{10}{F^2}_0 + F^{11}{F^2}_1 + F^{12}{F^2}_2 + F^{13}{F^2}_3 \\ & = (-E_x)(E_y) + (0)(B_z) + (-B_z)(0) + (B_y)(-B_x) \\ & = -E_xE_y -B_xB_y \end{aligned}

In general, the spatial stress tensor can be written as:

T^{mn} = -E_mE_n - B_mB_n + \frac{1}{2}\delta_{mn}\left(E^2 + B^2\right)

Exercise 11.3

As an exercise, you can calculate the force on a solar sail a million square meters in area, at. Earth distance from the Sun. The result is tiny, about 8 newtons or roughly 2 pounds. On the other hand, if the same sail absorbed (rather than reflected) the sunlight that fell on it, the power absorbed would be about a million kilowatts.

When sunlight is absorbed by a surface, the radiation pressure (and thus the force) is given by F = \frac{P}{c}, where P is the total power of the incident light and c is the speed of light.

If the light is reflected, the momentum change is twice as large, so the force is twice that of absorption: F = 2 \times P.

We will use the solar constant (1361 \text{ W/m}^2) for the irradiance in space.

Let’s calculate with these assumptions for a 10^6 \text{ m}^2 area the total incident power:

P = 1361 \text{ W/m}^2 \times 10^6 \text{ m}^2 = 1.361 \times 10^9 \text{ W}

The force with perfect reflection (using c \approx 3 \times 10^8 \text{ m/s}):

F = \frac{2 \times P}{c} = \frac{2 \times 1.361 \times 10^9 \text{ W}}{3 \times 10^8 \text{ m/s}} \approx 9.07 \text{ N}

Applying a 90\% efficiency:

  • 9.07 \text{ N} \times 0.90 \approx 8.16 \text{ N}

Rounding to a single significant figure we arrive to 8 \text{ N}.

If all this power would be absorbed it would amount to:

1.361 \times 10^9 \text{ W} \approx 10^6 \text{ KW}

Exercise 11.4

Convert “I refuse to join any club that would have me as a member.” to a mathematical formula.

\mathcal C = \{C \mid \text{me} \notin C \}

Exercise A.1

Derive Eq. A.13, based on Eq. 9.18. Hint: The derivation follow the same logic as the derivation of Eq. 9.22 in Section 9.2.5.

Equation A.13 is:

E = \frac{\dot \phi}{2\pi r}

Equation 9.18 is:

EMF = -\frac{d\Phi}{dt}

Equation 9.22 is:

B(r) = \frac{j}{2\pi rc^2}

Let’s consider a ring surrounding a thin string-like solenoid with a charged electron slides along the ring. A changing magnetic field creates an electric field:

\nabla \times \mathbf E = -\frac{\partial \mathbf B}{\partial t}

The increasing magnetic flux induces an electric field that exerts a force on the electron. If the flux through the string is \phi(t), we can apply Equation 9.18:

EMS = -\frac{\mathrm d \phi}{\mathrm dt} = -\dot \phi

We apply Stokes’ theorem (here) to the area within the ring (which is a disc with the normal aligned with the solenoid. The left side is the integral of \nabla \times \mathbf E which, applying Stokes’s theorem, is the line integral of \mathbf E over the circle:

\iint_{S} (\nabla \times \mathbf{E}) \cdot \mathbf n \, \mathrm{d}S = \oint_C \mathbf{E} \cdot \mathrm{d}s Here, the line integral of the electric field \mathbf E around a closed path C is equal to the negative rate of change of the magnetic flux through the surface enclosed by the path.

For a long, thin solenoid, the magnetic field is contained within the solenoid, and the induced electric field outside of it has circular symmetry.

The electric field lines are circles centered on the solenoid’s axis. We consider a circular path C of radius r concentric with the solenoid.

At every point on this circular path, the electric field E has the same magnitude E(r) and is tangent to the path. Therefore, \mathbf E and the line element are parallel.

By Lenz’s Law, the induced electric field \mathbf E oppose the change in flux. Adopting the standard mathematical convention of a counter-clockwise path for \mathrm ds, the resulting field \mathbf E must be clockwise. Therefore, \mathbf E and \mathrm ds are in opposite directions at every point on the path:

\oint_C \mathbf{E} \cdot \mathrm{d}\mathbf{s} = \oint_C (-E(r)) ds = -E(r) \oint_C ds

The integral of \mathrm ds around the circle is simply its circumference, 2\pi r:

\oint_C \mathbf{E} \cdot \mathrm{d}\mathbf{s} = -E(r) (2\pi r)

From Faraday’s Law, the EMF is equal to the negative rate of change of flux. Since the rate of change is \mathrm d\phi / \mathrm dt, the EMF is -\dot \phi.

We apply the integral form of Faraday’s Law, which is equivalent to Equation 9.18:

EMF = \oint_C \mathbf{E} \cdot \mathrm{d}\mathbf{s} = -\frac{d\Phi_B}{dt}

Substituting this and the result of the line integral back into Faraday’s Law:

-E(r) (2\pi r) = -\dot{\phi}

Finally, we solve for the magnitude of the electric field, E(r), which we can simply call E:

E = \frac{\dot{\phi}}{2\pi r}

This is the desired Equation A.13.

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