Part II · Special Relativity — Chapter 2.6

Electromagnetism Is Relativity

Maxwell's equations never needed fixing. They were relativistic before anyone knew what that meant. That is exactly why they produced a frame-independent $c$ and started the crisis in the first place.

Tools you'll need  — Chapter 0.7: divergence and curl in Cartesian components (§3.3, §4.2), the continuity equation (§6), and the two second-derivative identities $\nabla\times\nabla\phi=\vv 0$ and $\nabla\cdot(\nabla\times\vv A)=0$ (§7), together with §7.3 on why the vector potential exists and is not unique. Chapter 1.1 §3.3: the two moving charges that break the third law. Section 10 here is the resolution, so it is worth rereading the setup. Chapter 1.2 §8: the field Euler–Lagrange equation, and §8.1's table of actions. Section 9 here supplies the third entry. Chapter 1.4 §6: Noether's theorem for fields, and its grind box on improvement terms, which §10 uses without apology. Chapter 2.1 §2 (Maxwell produces $c$), §4 (the wave equation is not Galilean invariant) and the fork of §4.3. Chapter 2.4 §2.1 (the boost matrix $\Lambda^{\mu}{}_{\nu}$), §4 (raising and lowering), §5 (the tensor transformation law and the quotient theorem), §6 (the invariance theorem), §7.2 (the count of six), §8.1 (the Levi-Civita symbol). Chapter 2.5 §1.3 (four-velocity), §2 (four-momentum), §6.1 (the covariant force $\dd p^{\mu}/\dd\tau=f^{\mu}$ with $f\cdot u=0$).

1 · The four-current

We start with the simplest object in the theory. We are going to build it by requirement rather than by decree, which means we will state what we need of it and let that fix it completely.

Chapter 0.7 §6 derived, from nothing but "charge is neither created nor destroyed", the continuity equation

Here $\rho$ is charge per unit volume and $\vv J$ is current density. Now look at (2.6.1) with the eyes of Chapter 2.4. It is a sum of four terms: one time derivative of one function, and three space derivatives of three functions. That is exactly the shape of a four-divergence $\partial_{\mu}V^{\mu}$. The catch is the word if: it is that shape only if the four functions can be arranged into a four-vector.

So let us not guess. Let us demand. Suppose there exists a four-vector $j^{\mu}$ whose four-divergence reproduces (2.6.1), and ask what its components would have to be. Before we can compare anything we need the derivative operator in the right form. Recall from Chapter 2.4 §8 that $\partial_{\mu}=\pdv{}{x^{\mu}}$ carries a lower index, and that $x^{0}=ct$, so

Our goal now is to write out the four-divergence in ordinary three-dimensional language, so that we can set it side by side with the continuity equation. Write the unknown four-vector as $j^{\mu}=(j^{0},\vv j)$, split the sum over $\mu$ into its time part and its three space parts, and use the two derivatives we just wrote down:

Now compare term by term with (2.6.1). The spatial part matches immediately if $\vv j=\vv J$. The temporal part matches if $j^{0}/c=\rho$, that is, if $j^{0}=c\rho$. There is no freedom left anywhere. The two expressions agree as functions, for arbitrary $\rho$ and $\vv J$, only for that one choice. So the four-vector we demanded exists, and it is this one:

The factor of $c$ is not cosmetic. It is dimensional bookkeeping forced on us by $x^{0}=ct$, exactly as the $c$ in $x^{\mu}=(ct,\vv x)$ was.

Now let's look at what has happened to charge conservation. It arrived as a four-term partial differential equation whose behaviour under a change of frame was completely obscure. It now reads $\partial_{\mu}j^{\mu}=0$. That is a contraction of a $(0,1)$ index against a $(1,0)$ index, so it is a scalar by Chapter 2.4 §5.1, so it is the same statement in every frame by the invariance theorem of §6. Charge conservation is manifestly relativistic, and it takes one line to say so.

Grind box — why $\rho$ alone is not a scalar, and how the $\gamma$'s work out

A common first guess is that charge density should be an invariant, since charge is. It is not, and the reason is instructive.

Take $N$ charges $q$ sitting at rest in a box of volume $V_{0}$, so $\rho_{0}=Nq/V_{0}$ in their rest frame. View them from a frame in which the box moves at speed $v$ along $x$. The charge $Nq$ is unchanged, since that is the invariance assumption (ii). The volume is not unchanged. The box is contracted along $x$ by $\gamma$, so $V=V_{0}/\gamma$, and therefore

$$\rho \;=\; \frac{Nq}{V} \;=\; \gamma\,\rho_{0}, \qquad \vv J \;=\; \rho\vv v \;=\; \gamma\rho_{0}\vv v.$$

Now assemble the four-vector and compare with the four-velocity $u^{\mu}=\gamma(c,\vv v)$ of Chapter 2.5 §1.3:

$$j^{\mu} \;=\; \big(c\gamma\rho_{0},\ \gamma\rho_{0}\vv v\big) \;=\; \rho_{0}\,\gamma\big(c,\vv v\big) \;=\; \rho_{0}\,u^{\mu}.$$

So the four-current of a moving cloud of charge is the invariant rest-frame density times the four-velocity. That is manifestly a four-vector, being a scalar multiplying a $(1,0)$ tensor.

Notice how the single factor of $\gamma$ from length contraction is exactly the factor of $\gamma$ that $u^{\mu}$ already carries. It is the same $\gamma$, arriving twice for the same reason. Had charge density been invariant instead, $j^{\mu}$ would not have been a four-vector at all, and (2.6.4) would be false.

One caution. The relation $j^{\mu}=\rho_{0}u^{\mu}$ holds for a single species of charge all moving together. A copper wire has two species, a stationary lattice and a drifting electron gas, and you add their four-currents. That addition is the whole of §6, and the fact that the two species carry different factors of $\gamma$ is the whole of magnetism.

2 · The four-potential and the field tensor

Now for the six components that Chapter 2.4 promised. The plan of this section is short. We recall the two potentials that Maxwell's equations force into existence, we stack them into one four-component array, and we then find that $\vv E$ and $\vv B$ are what you get by differentiating that array.

2.1 · Recalling the potentials

Chapter 0.7 §7.3 established the first half. Since $\nabla\cdot(\nabla\times\vv A)=0$ identically, writing

makes $\nabla\cdot\vv B=0$ automatic. Feed that into Faraday's law:

The middle step there is the commutation of $\partial_{t}$ with the spatial derivatives inside the curl, which is Clairaut's theorem again, as in Chapter 2.1 §2.2.

Now use what that last line tells us. Chapter 0.7 §2.3 showed that a curl-free field on a simply connected region is a gradient. So $\vv E+\partial_{t}\vv A=-\nabla\phi$ for some scalar $\phi$. The minus sign is the standard convention, chosen so that $\phi$ reduces to the electrostatic potential in the static case. Rearranging that for $\vv E$, and carrying the curl formula for $\vv B$ along beside it, we have both fields written in terms of potentials:

Two of Maxwell's four equations have now been used up entirely. They are the reason the potentials exist, and once you work with $\phi$ and $\vv A$ they hold identically, with nothing left to check. Hold on to that thought. Section 3.4 makes it the punchline of the whole section.

2.2 · Assembling $A^{\mu}$

We now have four functions in hand: one scalar $\phi$ and the three components of $\vv A$. The pattern of §1 suggests trying $A^{\mu}=(\alpha\phi,\vv A)$ for some constant $\alpha$ that dimensions will fix. It also suggests checking afterwards that the result transforms as a four-vector, rather than assuming that it does.

Dimensions first. In SI, $\vv A$ has units of $\mathrm{T\,m}=\mathrm{V\,s\,m^{-1}}$ and $\phi$ has units of $\mathrm{V}$. So $\phi/c$ has the units of $\vv A$, and $\alpha=1/c$ is the only choice that makes the four components commensurable. It is the same $c$, appearing for the same reason, as in $x^{\mu}=(ct,\vv x)$ and $j^{\mu}=(c\rho,\vv J)$. With that settled, define

One warning about the letter. Chapters 2.2 to 2.5 used $\phi$ for rapidity, and from here to the end of the book it means the electric scalar potential. That is an unfortunate clash of symbols, and the whole subject lives with it. Rapidity does not appear again in this chapter, so nothing is ambiguous on this page. When Part V has to write both in one line, it writes the rapidity as $\varphi$.

Is $A^{\mu}$ a four-vector? Not by fiat. The check comes in §4. There we will find that in the Lorenz gauge $A^{\mu}$ satisfies $\Box A^{\mu}=\mu_{0}j^{\mu}$, where $\Box$ is a scalar operator and $j^{\mu}$ is a four-vector by §1. That forces $A^{\mu}$ to be a four-vector too, since otherwise the equation could not hold in every frame. Until we have that argument in hand, treat (2.6.8) as the definition of an array of four numbers and nothing more.

2.3 · The field tensor

Look at (2.6.7) again. Every entry is one derivative of one potential component, minus another derivative of another potential component. In four-dimensional language there is exactly one object with that shape:

It is antisymmetric, $F^{\nu\mu}=-F^{\mu\nu}$, and you can see that by inspection. Swapping $\mu$ and $\nu$ swaps the two terms. That is not a design choice anyone made. It is forced by the construction, and the callout at the end of §2 explains why the construction itself is forced.

The upper-index derivative is $\partial^{\mu}=\eta^{\mu\nu}\partial_{\nu}$, and since $\eta^{\mu\nu}=\mathrm{diag}(1,-1,-1,-1)$ this flips the sign of the spatial components:

That minus sign is where readers get hurt. Write it out once and keep it.

2.4 · The six components, one at a time

The three $F^{0i}$. Put $\mu=0$, $\nu=i$ into (2.6.9):

the last step being (2.6.7) read backwards. So the first row of the matrix is $-\vv E/c$, and the first column is $+\vv E/c$ by antisymmetry.

The three $F^{ij}$. Both indices spatial, so both derivatives pick up the minus sign from (2.6.10):

Our goal now is to recognise that bracket as the curl, so that $\vv B$ appears. Chapter 0.7 §4.2 wrote the curl in components. In Levi-Civita notation, which is Chapter 2.4 §8.1 restricted to three indices, it reads $B^{k}=\epsilon^{klm}\partial_{l}A^{m}$. To get at the bracket we contract that with another epsilon and use the standard identity $\epsilon^{ijk}\epsilon^{klm}=\delta^{il}\delta^{jm}-\delta^{im}\delta^{jl}$:

The right-hand side there is precisely the bracket sitting inside (2.6.12), so we can substitute it and read off the purely spatial part of the field tensor:

Three entries, since the diagonal vanishes and the array is antisymmetric. Six in total. Assemble them:

Check one entry against (2.6.14) to be sure the epsilon signs landed where they should: $F^{12}=-\epsilon^{12k}B^{k}=-\epsilon^{123}B^{3}=-B_{z}$ ✓, and $F^{13}=-\epsilon^{132}B^{2}=+B_{y}$ ✓ since $\epsilon^{132}=-1$.

2.5 · Both indices down, and the signs that flip

We will need $F_{\mu\nu}$ constantly, and this is exactly the step at which sign errors are manufactured. Lower with two metrics, per Chapter 2.4 §4.1:

Since $\eta$ is diagonal, each entry simply picks up the factor $\eta_{\mu\mu}\eta_{\nu\nu}$, with no sum implied. For a $0i$ entry that factor is $(+1)(-1)=-1$. For an $ij$ entry it is $(-1)(-1)=+1$. Applying those two factors to the entries we found above gives the rule we will be using for the rest of the chapter:

Grind box — every entry of $F^{\mu\nu}$ from $F=\partial A-\partial A$, by hand

Nothing here is subtle. The point is to have watched all sixteen numbers appear once, so that the matrix above is a fact rather than a memory. Write $A^{\mu}=(\phi/c,A_{x},A_{y},A_{z})$ and $\partial^{\mu}=(\tfrac1c\partial_{t},-\partial_{x},-\partial_{y},-\partial_{z})$.

Diagonal. $F^{\mu\mu}=\partial^{\mu}A^{\mu}-\partial^{\mu}A^{\mu}=0$ for each fixed $\mu$ (no sum). Four zeros.

$F^{01}$.

$$F^{01} = \frac1c\pdv{A_{x}}{t} - \left(-\pdv{}{x}\right)\frac{\phi}{c} = \frac1c\left(\pdv{A_{x}}{t}+\pdv{\phi}{x}\right) = -\frac1c E_{x},$$

using $E_{x}=-\partial_{x}\phi-\partial_{t}A_{x}$. Identically $F^{02}=-E_{y}/c$ and $F^{03}=-E_{z}/c$.

$F^{12}$. Both spatial, so both derivatives carry the minus:

$$F^{12} = \left(-\pdv{}{x}\right)A_{y} - \left(-\pdv{}{y}\right)A_{x} = -\left(\pdv{A_{y}}{x}-\pdv{A_{x}}{y}\right) = -B_{z},$$

since $B_{z}=\partial_{x}A_{y}-\partial_{y}A_{x}$ from Chapter 0.7 §4.2. Cycling $x\to y\to z\to x$ gives $F^{23}=-B_{x}$ and $F^{31}=-B_{y}$, and the last of these is $F^{13}=+B_{y}$.

Lower half. By antisymmetry, $F^{10}=+E_{x}/c$, $F^{21}=+B_{z}$, and so on. Sixteen numbers, six of them independent, exactly as (2.6.15) says.

Sanity check on units. Every entry of $F^{\mu\nu}$ must have the same dimensions, or the object is not a tensor. $B$ is in tesla. And $E/c$ is $\mathrm{V\,m^{-1}}/(\mathrm{m\,s^{-1}}) = \mathrm{V\,s\,m^{-2}} = \mathrm{T}$ ✓. So the factor of $c$ in $E/c$ is not decoration. Without it the six components could not be six components of one thing.

3 · Four equations become two

This is the centrepiece of the chapter, and nothing in it will be waved through. We are going to write down two tensor equations and then expand every single component of both, so that you can see Maxwell's familiar four emerge one at a time.

3.1 · The inhomogeneous half

What can we build from $F^{\mu\nu}$, one derivative, and the current? The index structure almost writes the answer for us. Contracting the derivative into the first slot gives $\partial_{\mu}F^{\mu\nu}$, which has one free upper index. So does $j^{\nu}$. Two objects with matching index structure can be set equal, and there is only one way to do it, with a single dimensionful constant left undetermined:

We now show that (2.6.19) is Gauss's law together with the Ampère–Maxwell law, and that the constant is $\mu_{0}$. There are four components in it, and we will do all four.

Component $\nu=0$. The sum over $\mu$ runs over $0,1,2,3$, but $F^{00}=0$, so only the spatial terms survive:

reading $F^{i0}=+E^{i}/c$ off the first column of (2.6.15). The right-hand side is $\mu_{0}j^{0}=\mu_{0}c\rho$. Equate and multiply by $c$:

The last step there uses $\mu_{0}\epsilon_{0}=1/c^{2}$. That is not an extra input. It is Chapter 2.1's result $c=1/\sqrt{\mu_{0}\epsilon_{0}}$ rearranged.

That is Gauss's law. It also fixes the undetermined constant in (2.6.19) to be $\mu_{0}$, because any other choice would give the wrong Coulomb force.

Component $\nu=j$ (spatial). Now both the time term and the space terms contribute:

The first piece is already in ordinary language. The second piece is not, so let's handle it. We want to see a curl there, so swap the first two indices of the epsilon, which costs one sign, and compare with the component form of the curl:

Putting that back into (2.6.22), and setting the whole thing equal to $\mu_{0}j^{j}=\mu_{0}J^{j}$, the component equation reads $-\frac{1}{c^{2}}\partial_{t}E^{j}+(\nabla\times\vv B)^{j}=\mu_{0}J^{j}$. Collecting the three values of $j$ back into vectors and moving the time derivative to the right,

That is the Ampère–Maxwell law, displacement current and all. One tensor equation, four components, two of Maxwell's four laws.

Look at where the displacement term came from. Maxwell had to add it by hand, and Chapter 0.7's Problem 4 showed that charge conservation forces it. Here it is not added at all. It is the $\mu=0$ term of a sum that has to run over all four values of $\mu$, because the index is contracted and a contracted index sums over everything.

3.2 · The homogeneous half

Two of Maxwell's equations are left: $\nabla\cdot\vv B=0$ and Faraday's law. Neither contains a source, so the covariant version must have zero on the right. The object that works is the totally antisymmetrised derivative of $F$, written out as a cyclic sum:

Before expanding, notice two structural facts, because they save all the work.

It is totally antisymmetric in $\lambda\mu\nu$. Swap any two of the three indices and the three terms permute into one another with an overall sign. Check $\lambda\leftrightarrow\mu$ for yourself: the first term becomes $\partial_{\mu}F_{\lambda\nu}=-\partial_{\mu}F_{\nu\lambda}$, which is minus the second, and so on around. So the left-hand side is an antisymmetric rank-3 object.

Therefore there are only four independent components. By the count of Chapter 2.4 §7.2, generalised, a totally antisymmetric rank-3 array in four dimensions has $\binom{4}{3}=4$ independent entries. There is one entry for each way of choosing three distinct indices from $\{0,1,2,3\}$, namely $(1,2,3)$, $(0,1,2)$, $(0,2,3)$ and $(0,3,1)$. Any triple with a repeated index vanishes identically.

Four components, and four is exactly the number of scalar equations we need. $\nabla\cdot\vv B=0$ is one of them, and Faraday's law is the other three. The bookkeeping fits before we compute anything at all.

Triple $(\lambda\mu\nu)=(1,2,3)$. All three indices spatial. Using $F_{ij}=-\epsilon^{ijk}B^{k}$ from (2.6.17), so that $F_{23}=-B_{x}$, $F_{31}=-B_{y}$, $F_{12}=-B_{z}$:

No magnetic monopoles.

Triple $(\lambda\mu\nu)=(0,1,2)$. One time index. Now $F_{12}=-B_{z}$, $F_{20}=-F_{02}=-E_{y}/c$, $F_{01}=+E_{x}/c$, and $\partial_{0}=\tfrac1c\partial_{t}$:

Multiply by $-c$ and you have the $z$-component of Faraday's law, $(\nabla\times\vv E)_{z}=-\partial_{t}B_{z}$. The triples $(0,2,3)$ and $(0,3,1)$ give the $x$- and $y$-components by the identical computation with the labels cycled. The grind box does them explicitly, so that "by symmetry" is not doing any hidden work here.

Grind box — the other two Faraday components, written out

Triple $(0,2,3)$. The entries needed are $F_{23}=-B_{x}$, $F_{30}=-F_{03}=-E_{z}/c$, $F_{02}=+E_{y}/c$:

$$\begin{aligned} \partial_{0}F_{23}+\partial_{2}F_{30}+\partial_{3}F_{02} &= \frac1c\pdv{}{t}(-B_{x}) + \pdv{}{y}\left(-\frac{E_{z}}{c}\right) + \pdv{}{z}\left(\frac{E_{y}}{c}\right)\\[3pt] &= -\frac1c\left[\pdv{B_{x}}{t} + \left(\pdv{E_{z}}{y}-\pdv{E_{y}}{z}\right)\right] = -\frac1c\left[\pdv{B_{x}}{t}+(\nabla\times\vv E)_{x}\right]. \end{aligned}$$

Triple $(0,3,1)$. Entries $F_{31}=-B_{y}$, $F_{10}=-E_{x}/c$, $F_{03}=+E_{z}/c$:

$$\begin{aligned} \partial_{0}F_{31}+\partial_{3}F_{10}+\partial_{1}F_{03} &= \frac1c\pdv{}{t}(-B_{y}) + \pdv{}{z}\left(-\frac{E_{x}}{c}\right) + \pdv{}{x}\left(\frac{E_{z}}{c}\right)\\[3pt] &= -\frac1c\left[\pdv{B_{y}}{t} + \left(\pdv{E_{x}}{z}-\pdv{E_{z}}{x}\right)\right] = -\frac1c\left[\pdv{B_{y}}{t}+(\nabla\times\vv E)_{y}\right]. \end{aligned}$$

Setting each to zero gives the $x$- and $y$-components of Faraday's law. Together with (2.6.27) and (2.6.26) that is all four independent components of (2.6.25), and all four of Maxwell's homogeneous scalar equations. Nothing is left over and nothing is missing.

Why the cyclic pattern $(0,1,2),(0,2,3),(0,3,1)$ and not $(0,1,3)$? Because $(0,3,1)$ and $(0,1,3)$ are the same component up to a sign, and picking the cyclic representative keeps every equation's sign uniform. Choosing the other one is not an error. It gives the same equation multiplied by $-1$.

3.3 · The scoreboard

Covariant statement	Component	Traditional name
$\partial_{\mu}F^{\mu\nu}=\mu_{0}j^{\nu}$	$\nu=0$	Gauss's law
	$\nu=1,2,3$	Ampère–Maxwell
$\partial_{[\lambda}F_{\mu\nu]}=0$	$(1,2,3)$	$\nabla\cdot\vv B=0$
	$(0,ij)$	Faraday's law

Four equations, eight scalar equations in the traditional bookkeeping, compressed into two lines.

There is a second payoff, and it is the larger one. Both lines are equations between tensors of the same type, so the invariance theorem of Chapter 2.4 §6 applies to each of them. Each holds in every inertial frame the moment it holds in one. No chain-rule computation, no cross terms, none of the carnage of Chapter 2.1 §4.2. That is the whole purpose of the tensor apparatus, and this is the first place the book collects on it.

3.4 · Two of Maxwell's equations are not physics

Now for the structural point of the section. It is easy to prove and it deserves emphasis out of all proportion to that ease.

Our goal is to see what the homogeneous half says once the potentials are put back in. So substitute the definition (2.6.9) into (2.6.25), with all indices down so that $F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$:

Six terms. Let's pair them off. The first cancels the fourth, the second cancels the fifth, and the third cancels the sixth. Each pair differs only by the order of two partial derivatives, and those commute by Clairaut's theorem. So the sum is identically zero.

4 · Gauge invariance

4.1 · The freedom

The potentials are not unique, and Chapter 0.7 §7.3 already said why. Adding a gradient to $\vv A$ leaves $\nabla\times\vv A$ alone, because the curl of a gradient vanishes. In four-dimensional language the statement is cleaner, and it covers $\phi$ as well. Let $\chi(x)$ be any smooth scalar function of spacetime, and transform the four-potential by

The question is what that does to the fields, so our next step is to build the field tensor out of the new potential and see what survives. Feed $\tilde A^{\mu}$ straight into the definition (2.6.9):

That is Clairaut again, the third appearance in as many sections, and the last. The fields are untouched. Since the fields are what exert forces, no experiment can distinguish $A^{\mu}$ from $\tilde A^{\mu}$.

It is worth seeing what this looks like in the older notation, so let's unpack (2.6.29) into three-vector language. The one thing to watch is the minus sign in $\partial^{\mu}=(\tfrac1c\partial_{t},-\nabla)$:

You may have seen the familiar textbook form, which has $\vv A+\nabla\chi$ and $\phi-\partial_{t}\chi$. That is the same transformation with $\chi\to-\chi$. Nothing whatever depends on which sign convention you adopt.

4.2 · Choosing a gauge, and recovering 2.1's wave equation

Freedom is a nuisance if you want to solve for something, so we spend it. Impose the Lorenz condition

The equivalence there follows from $\partial_{\mu}A^{\mu}=\tfrac1c\partial_{t}(\phi/c)+\nabla\cdot\vv A$.

Notice that (2.6.32) is itself a scalar equation, being a contraction, so imposing it in one frame imposes it in all of them. That is not true of the other common choice, the Coulomb gauge $\nabla\cdot\vv A=0$. That one singles out a frame, which makes it useless for our purposes here. It is convenient elsewhere, and §10 uses it once, deliberately.

It is always attainable. Suppose you are handed potentials with $\partial_{\mu}A^{\mu}=f(x)\neq0$. Gauge-transform by some $\chi$ and compute the new divergence:

So the Lorenz condition holds for the new potentials precisely when $\Box\chi=-f$. That is an inhomogeneous wave equation for $\chi$ with a known source.

Now the payoff. Substitute the definition of $F$ into the field equation (2.6.19) and impose (2.6.32):

The middle step moved $\partial_{\mu}$ past $\partial^{\nu}$, which is legitimate because partial derivatives commute. That let the Lorenz condition kill the second term outright. So all of electromagnetism, in the Lorenz gauge, has come down to one line:

Four uncoupled wave equations, one per component, each with its own source. In vacuum ($j^{\mu}=0$) this is $\Box A^{\mu}=0$, that is,

That is the wave equation of Chapter 2.1 §2. It was derived there by taking the curl of Faraday's law and grinding through $\nabla\times(\nabla\times\vv E)=\nabla(\nabla\cdot\vv E)-\nabla^{2}\vv E$. Same equation, same $c$, obtained here in three lines because the bookkeeping was done first.

5 · $\vv E$ and $\vv B$ mix

If $\vv E$ and $\vv B$ really are the components of one tensor, then a boost must shuffle them into each other, and Chapter 2.4 tells us exactly how. For a $(2,0)$ tensor the law is

In matrix notation, with $F$ the array of (2.6.15), that reads $F'=\Lambda F\Lambda^{\mathsf T}$. There is one $\Lambda$ acting on the rows and one acting on the columns, which is what two upper indices buys you. To make it concrete we take the standard boost of Chapter 2.4 §2.1, with $S'$ moving at $+v$ along $x$:

Carrying out the two matrix multiplications, which the grind box does entry by entry, and then reading the primed fields off the result, gives the transformation rules in full.

Three things are worth noticing immediately.

The components along the boost do not change. That is the opposite of what happens to a four-vector, whose parallel component is the one that mixes with time. The reason is visible in (2.6.15). Here $E_{x}$ sits in the $01$ slot, so both of its indices lie in the block the boost acts on, and it acquires two factors of $\Lambda$ that undo each other. Meanwhile $B_{x}$ sits in the $23$ slot, which the boost does not touch at all.

The combination $\vv E+\vv v\times\vv B$ is already familiar. It is the Lorentz force per unit charge. Section 8 shows that is not a coincidence.

If $\vv B=\vv 0$ in one frame, $\vv B'\neq\vv 0$ in another. The one exception is when $\vv E$ happens to be parallel to $\vv v$. So a pure electric field is not a frame-independent notion at all. Magnetism is what an electric field looks like from a moving frame, and §6 makes that quantitative in the one case where you can check it against a laboratory measurement.

Grind box — the boost of $F^{\mu\nu}$, entry by entry

The transformation is $F'=\Lambda F\Lambda^{\mathsf T}$. Since $\Lambda$ is symmetric here, $\Lambda^{\mathsf T}=\Lambda$, and it acts nontrivially only on the $0$–$1$ block. Write $\Lambda$ as the $2\times2$ block $\begin{pmatrix}\gamma&-\gamma\beta\\-\gamma\beta&\gamma\end{pmatrix}$ on indices $(0,1)$ and the identity on $(2,3)$.

Entries with both indices in $\{0,1\}$. Only $F^{01}$ is nonzero there. The $2\times2$ block of $F$ is $\begin{pmatrix}0&-E_{x}/c\\ E_{x}/c&0\end{pmatrix}$, which is $(-E_{x}/c)$ times the antisymmetric symbol. For any $2\times2$ matrix $M$ and the antisymmetric $\varepsilon$ we have $M\varepsilon M^{\mathsf T}=\det(M)\,\varepsilon$, a two-dimensional special case of the determinant identity in Chapter 2.4 §8.1. Here $\det=\gamma^{2}-\gamma^{2}\beta^{2}=\gamma^{2}(1-\beta^{2})=1$. Hence

$$F'^{01}=F^{01} \qquad\Longrightarrow\qquad E'_{x}=E_{x}.$$

Entries with both indices in $\{2,3\}$. $\Lambda$ is the identity there, so $F'^{23}=F^{23}$, i.e. $B'_{x}=B_{x}$.

Mixed entries. These carry one factor of the block and one identity. For $F'^{02}$:

$$F'^{02} = \Lambda^{0}{}_{\alpha}\Lambda^{2}{}_{\beta}F^{\alpha\beta} = \Lambda^{0}{}_{\alpha}F^{\alpha 2} = \gamma F^{02} - \gamma\beta F^{12}.$$

Substitute $F^{02}=-E_{y}/c$ and $F^{12}=-B_{z}$:

$$-\frac{E'_{y}}{c} = -\gamma\frac{E_{y}}{c} + \gamma\beta B_{z} \qquad\Longrightarrow\qquad E'_{y} = \gamma\big(E_{y} - \beta c B_{z}\big) = \gamma\big(E_{y}-vB_{z}\big).$$

For $F'^{03}$, with $F^{03}=-E_{z}/c$ and $F^{13}=+B_{y}$:

$$-\frac{E'_{z}}{c} = \gamma F^{03} - \gamma\beta F^{13} = -\gamma\frac{E_{z}}{c} - \gamma\beta B_{y} \;\Longrightarrow\; E'_{z} = \gamma\big(E_{z}+vB_{y}\big).$$

For $F'^{12}$, now the first index sits in the block as a row:

$$F'^{12} = \Lambda^{1}{}_{\alpha}F^{\alpha2} = -\gamma\beta F^{02} + \gamma F^{12} = \gamma\beta\frac{E_{y}}{c} - \gamma B_{z},$$

and since $F'^{12}=-B'_{z}$, $\;B'_{z}=\gamma\big(B_{z}-vE_{y}/c^{2}\big)$. Identically, $F'^{13}=-\gamma\beta F^{03}+\gamma F^{13}$ gives $B'_{y}=\gamma\big(B_{y}+vE_{z}/c^{2}\big)$.

Six components, six rules, no others. Every one of them was checked symbolically against $\Lambda F\Lambda^{\mathsf T}$ before this box was written.

A consistency check worth doing. Apply the rules twice, once with $v$ and once with $-v$, and you must get back what you started with. For $E_{y}$: $\gamma(E'_{y}+vB'_{z}) = \gamma^{2}\big[(E_{y}-vB_{z}) + v(B_{z}-vE_{y}/c^{2})\big] = \gamma^{2}E_{y}(1-v^{2}/c^{2}) = E_{y}$ ✓. The $\gamma^{2}(1-\beta^{2})=1$ that makes it work is the same identity that made $\det\Lambda=1$ above.

6 · The wire — magnetism as electrostatics in disguise

This is the argument that makes the chapter's title literal, and it can be checked against a current balance on a laboratory bench.

6.1 · The setup, in the lab

Take a long straight wire along the $x$-axis. We model it as Chapter 1.1's grind boxes model everything, with the crudest structure that still has the right physics. That means a rigid lattice of positive ions at rest, and a gas of conduction electrons drifting through it.

• Lattice: at rest in the lab, linear charge density $+\lambda_{0}$.
• Electrons: drifting with velocity $-v_{d}\hat{\vv x}$, linear charge density $-\lambda_{0}$ as measured in the lab.

The two densities are equal and opposite in the lab, so the wire is electrically neutral there. That is an experimental fact about wires rather than an assumption we are making. A current-carrying wire does not attract a stationary pith ball. With the densities set, the conventional current is

flowing in the $+\hat{\vv x}$ direction, because negative charge moving one way is positive current the other.

Now put a test charge $q\gt0$ at perpendicular distance $r$ from the wire, at position $r\hat{\vv y}$, moving parallel to the wire with velocity $u\hat{\vv x}$.

In the lab the wire is neutral, so $\vv E=\vv 0$ and there is no electric force whatever. That leaves the magnetic force to account for everything. The magnetic field of a long straight wire at distance $r$ is $\vv B=\dfrac{\mu_{0}I}{2\pi r}\,\hat{\vv x}\times\hat{\vv r}$. In our geometry $\hat{\vv r}=\hat{\vv y}$ and $\hat{\vv x}\times\hat{\vv y}=\hat{\vv z}$, so the field and the force on the test charge come out as

The force is attractive. It points from the charge toward the wire, since $\hat{\vv x}\times\hat{\vv z}=-\hat{\vv y}$. The force is purely magnetic and there is no ambiguity about that.

6.2 · The same situation, from the test charge's frame

Now board the test charge. Transform to the frame $S'$ moving at $+u\hat{\vv x}$. In $S'$ the test charge is at rest, so it cannot feel a magnetic force at all. The Lorentz force on a stationary charge has no magnetic term, whatever $\vv B'$ may be.

And yet the charge is certainly still accelerating toward the wire. Whether it hits the wire is not a matter of opinion. So some other force must be doing the work, and there is only one candidate.

Our next step is to compute the two charge densities in $S'$. This is the crux of the whole argument, so we go slowly.

The lattice. It is at rest in the lab with density $\lambda_{0}$, so its proper density, meaning the density in its own rest frame, is $\lambda_{0}$. In $S'$ it moves at speed $u$, so the lattice is length-contracted by $\gamma_{u}$ and the same charge occupies less length:

The electrons. These were already moving in the lab, so we must first back out their proper density. In the lab they move at $v_{d}$ and have density $-\lambda_{0}$. Running the logic of (2.6.41) backwards, their proper density is $-\lambda_{0}/\gamma_{d}$ with $\gamma_{d}=(1-v_{d}^{2}/c^{2})^{-1/2}$. To contract that proper density into $S'$ we need their speed in $S'$, which the velocity-addition law of Chapter 2.2 §5 supplies:

What the contraction actually needs is not that velocity but the Lorentz factor built from it, so substitute the last line into $\gamma=(1-v^{2}/c^{2})^{-1/2}$ and simplify. The grind box does the algebra. The result is remarkably tidy:

Now we have everything the electrons need. Their density in $S'$ is their proper density $-\lambda_{0}/\gamma_{d}$ multiplied by the contraction factor we just computed:

The two densities have been contracted by different factors. Both of them picked up $\gamma_{u}$. The electrons picked up an extra $(1+uv_{d}/c^{2})$ on top of that, because they were already moving before we boosted. Adding the two densities gives the net charge on the wire as $S'$ sees it:

The wire is not neutral in this frame. It carries a net negative charge per unit length. The test charge is positive, so it is attracted, and that is the direction we already know the force to have. To leading order in $uv_{d}/c^{2}$ we may set $\gamma_{u}\to1$, which gives $\lambda'\approx-\lambda_{0}uv_{d}/c^{2}$.

6.3 · The force, and the comparison

The field of an infinite line charge $\lambda'$ at perpendicular distance $r'$ is radial, with magnitude $E'=\lambda'/(2\pi\epsilon_{0}r')$. Distances perpendicular to the boost are unchanged, so $r'=r$. The test charge is stationary in $S'$, so the force on it is purely electric:

The last step there used $1/(\epsilon_{0}c^{2})=\mu_{0}$. Now set that beside the lab result (2.6.40), which is what we came here to compare it with:

That factor $\gamma_{u}$ is exactly the factor by which a transverse force is expected to change between the two frames. Here is why, in three lines.

Momentum transverse to the boost is unchanged, $p'_{y}=p_{y}$, because $p^{\mu}$ is a four-vector and the boost matrix of Chapter 2.4 §2.1 acts as the identity on the $y$ and $z$ rows. Time is not unchanged. For a particle instantaneously at rest in $S'$ we have $\dd x'=0$, so the time transformation gives $\dd t=\gamma_{u}(\dd t'+u\,\dd x'/c^{2})=\gamma_{u}\dd t'$. Dividing the first fact by the second,

which is (2.6.47) rearranged. The two calculations agree identically, not approximately.

6.4 · How small is $v_{d}/c$, and why you can still pick up a nail

Let's put numbers in, because they are startling. Copper has one conduction electron per atom, density $8960\ \mathrm{kg\,m^{-3}}$ and molar mass $63.55\ \mathrm{g\,mol^{-1}}$, so the number density of mobile electrons is

In a wire of cross-section $1\ \mathrm{mm^{2}}$ carrying $1\ \mathrm{A}$, the drift speed follows from $I=neAv_{d}$:

That is $0.074\ \mathrm{mm\,s^{-1}}$, slower than a growing fingernail. Expressed as a fraction of the speed of light, which is the form the physics above cares about, it is

So magnetism is an effect of relative order $10^{-13}$, and yet electromagnets lift cars. How can both of those be true?

The resolution is in (2.6.45). The surviving effect is not $\lambda_{0}(v_{d}/c)$ on its own. It is the difference between two enormous numbers, and the enormous number is $\lambda_{0}$ itself. In that same wire,

thirteen thousand coulombs of mobile charge per metre, exactly cancelled by thirteen thousand coulombs of lattice charge. A tiny fractional imbalance in a colossal cancellation is still a substantial charge. Worked example 1 does the arithmetic all the way to a force.

boost toward $q$'s frame

w = 0.000 c

drift $v_d/c$ (exaggerated)

0.500

lab frame

test-charge frame

maximise the drift

frame speed w = 0.000 c, gamma_w = 1.000 | test charge in this frame: u = 0.700 c

λ₊/λ₀ = +1.000 λ₋/λ₀ = −1.000 → net 0.000 (formula −gamma_w·w·v_d/c² = 0.000)

spacings: a₊ = 1.000 a, a₋ = 1.000 a, ratio a₋/a₊ = 1.000 (1+w v_d/c²)⁻¹ = 1.000

F_electric = 0.000 F₀, F_magnetic = 1.000 F₀, total = 1.000 F₀ | INVARIANT gamma·F⊥ = 1.400 F₀ (gamma_u = 1.400)

a real copper wire (1 mm², 1 A): v_d/c = 2.45e-13, λ₀ = 1.360e+4 C/m, net λ′ = -1.113e-12 C/m at u = 10⁵ m/s ⇒ F = 3.204e-19 N

Grind box — the composition identity $\gamma'_{-}=\gamma_{u}\gamma_{d}(1+uv_{d}/c^{2})$

Let a particle move at velocity $w$ along $x$ in frame $S$, and let $S'$ move at $+u$. The velocity in $S'$ is $w'=(w-u)/(1-uw/c^{2})$. Compute $1-w'^{2}/c^{2}$:

$$\begin{aligned} 1 - \frac{w'^{2}}{c^{2}} &= 1 - \frac{(w-u)^{2}}{c^{2}\left(1-\frac{uw}{c^{2}}\right)^{2}} = \frac{c^{2}\left(1-\frac{uw}{c^{2}}\right)^{2} - (w-u)^{2}}{c^{2}\left(1-\frac{uw}{c^{2}}\right)^{2}}. \end{aligned}$$

Expand the numerator, writing $a=u/c$, $b=w/c$ and factoring out $c^{2}$:

$$(1-ab)^{2} - (b-a)^{2} = 1 - 2ab + a^{2}b^{2} - b^{2} + 2ab - a^{2} = (1-a^{2})(1-b^{2}).$$

The cross terms cancel exactly, and that is the only step in this box with any content. Putting that factorised numerator back and taking the reciprocal square root,

$$1-\frac{w'^{2}}{c^{2}} = \frac{(1-u^{2}/c^{2})(1-w^{2}/c^{2})}{\left(1-uw/c^{2}\right)^{2}} \qquad\Longrightarrow\qquad \gamma_{w'} = \gamma_{u}\gamma_{w}\left(1-\frac{uw}{c^{2}}\right).$$

For the electrons, $w=-v_{d}$, and the minus sign in the bracket becomes a plus: $\gamma'_{-}=\gamma_{u}\gamma_{d}(1+uv_{d}/c^{2})$, which is (2.6.43). ✓

Bonus. Read the same identity in reverse and it says that $\gamma_{w'}=\gamma_{u}\gamma_{w}(1-uw/c^{2})$ is the $0$-component of the boosted four-velocity. That is to say, it is $u'^{0}=\Lambda^{0}{}_{\mu}u^{\mu}$ written out. Nothing new was needed here. Velocity addition is the transformation of $u^{\mu}$, as Chapter 2.5 remarked.

And the same identity again. Notice that the spacings drawn in the figure obey $a'_{-}/a'_{+}=(1+wv_{d}/c^{2})^{-1}$, which is this identity divided by itself. The two rows of charges compress by different amounts because they enter the boost with different $\gamma$'s, and for no other reason.

7 · The invariants

If $\vv E$ and $\vv B$ separately are observer-dependent, what do all observers agree about? Chapter 2.4 §5.1 answers that in general terms. What everybody agrees about is whatever you can build by contracting every index away. From a single antisymmetric rank-2 tensor there are exactly two such quantities, and this section constructs both of them.

7.1 · The first invariant

The simplest thing you can do with two copies of $F$ is contract every index of one against every index of the other, which leaves nothing free:

To see what that is in terms of $\vv E$ and $\vv B$, split the double sum by index type. The diagonal contributes nothing, since the diagonal of $F$ is zero. The $0i$ and $i0$ entries contribute twice over, once each way round, and by (2.6.17) we have $F_{0i}F^{0i}=(E^{i}/c)(-E^{i}/c)=-(E^{i})^{2}/c^{2}$. So those terms give

That accounts for the electric entries. Now for the purely spatial ones. Lowering both indices leaves them alone, so $F_{ij}=F^{ij}=-\epsilon^{ijk}B^{k}$, and their contribution is a sum over two epsilons:

using the contraction identity $\epsilon^{ijk}\epsilon^{ijl}=2\delta^{kl}$ (sum on $i,j$), which is the previous identity $\epsilon^{ijk}\epsilon^{klm}=\delta^{il}\delta^{jm}-\delta^{im}\delta^{jl}$ with one more index contracted. Adding:

7.2 · The second invariant

The other contraction available uses the Levi-Civita symbol of Chapter 2.4 §8.1:

Total antisymmetry means the only surviving terms are those in which $\{\mu\nu\}$ and $\{\rho\sigma\}$ are complementary pairs. There are three such pair-splittings, namely $\{01|23\}$, $\{02|13\}$ and $\{03|12\}$. Each of the three accounts for eight of the $24$ nonzero permutations: two orderings inside the first pair, two inside the second, and two choices of which pair comes first.

Flipping any one of those three things costs a sign in $\epsilon$ and a sign in one factor of $F$, so the two signs cancel and all eight terms are equal. That reduces the whole sum to three representative terms with a factor of eight in front:

using $\epsilon_{0123}=+1$, $\epsilon_{0213}=-1$ (one transposition) and $\epsilon_{0312}=+1$ (two transpositions), together with the entries of (2.6.15). All three terms come out positive:

7.3 · What the invariants forbid

We now have two numbers that everyone agrees on. They are strong constraints, and it is worth being exact about what each of them rules out.

A plane light wave stays a plane light wave. Chapter 2.1 §2 found that in an electromagnetic wave $\vv E\perp\vv B$ and $E=cB$. Then both invariants vanish:

Both are zero in every frame, so $E'=cB'$ and $\vv E'\perp\vv B'$ for every observer, no matter how fast. You may Doppler-shift a light wave to any frequency you like and change its amplitude by any factor, and Chapter 2.5 §7.2 computed exactly how. What you can never do is make it into something other than a light wave.

In particular, you cannot bring light to rest. A stationary field configuration would have to have $\vv E$ and $\vv B$ independent of each other, and there is no frame in which the ratio $E/cB$ is anything but $1$.

When can you kill the magnetic field? Suppose there exists a frame with $\vv B'=\vv 0$. Then in that frame the invariants read $-E'^{2}/c^{2}$ and $0$. Since invariants are invariants, that requires

Those two conditions must hold in every frame, and in particular in yours. So the condition is checkable without leaving home. The fields must be perpendicular, and the electric one must dominate. These conditions are also sufficient, and Worked example 2 constructs the frame explicitly in the mirror-image case.

When can you kill the electric field? Run the identical argument with the roles swapped and the conditions come out as

Worked example 2 does this case in full and finds the frame velocity, which turns out to be the drift velocity every mass spectrometer relies on.

And if $\vv E\cdot\vv B\neq0$, neither field can be removed in any frame, ever. The two fields are then irreducibly both present, and the best you can do is find a frame where they are parallel.

That table has a shape, and it is worth seeing it. Take $\vv E=(E_{\parallel},E_{\perp},0)$ and $\vv B=(B_{\parallel},0,B_{\perp})$. That is a configuration on which a boost along $\hat{\vv x}$ closes in two dimensions, so the picture stays flat, and the four rows of the table become four regions of one plane.

rapidity $\phi$

φ = +0.000 β = tanh φ = +0.000000

$E_{\perp}$

0.80

$B_{\perp}$

0.35

$E_{\parallel}$

0.30

$B_{\parallel}$

0.50

a light wave

the wire of §6

crossed fields

$\vv E\cdot\vv B\neq0$

reset

E′ = ( 0.300000, 0.800000, 0.000000) B′ = ( 0.500000, 0.000000, 0.350000)

F_μν F^μν by contraction of F′: -0.7150000000 2(B′²−E′²): -0.7150000000 Δ 1.1e-16

ε F F by the 24-term sum: +1.2000000000 8 E′·B′: +1.2000000000 Δ 0.0e+0

type, from the invariants alone: GENERIC an x-boost at φ* = +0.469135 (β* = 0.437500) drives B⊥ to zero, but B∥ = 0.50 survives it.

over the whole range |φ| ≤ 3: E∥ moves 9.9e-15, B∥ moves 0.0e+0, max(|E′_z|,|B′_y|) = 0.0e+0, B⊥²−E⊥² drifts 4.1e-14

drag the rapidity: the marker moves along its curve and never off it.

8 · The Lorentz force, covariantly

Chapter 2.5 §6.1 established the shape any relativistic force law must have:

What can $f^{\mu}$ be for a charge in an electromagnetic field? Three requirements narrow it almost to nothing. It must be a four-vector, it must be linear in the charge, and it must be built from the only field object we have. With $u^{\mu}$ the particle's four-velocity, there is essentially one candidate:

Chapter 2.4's grind box on the quotient theorem already flagged this construction. Requiring $F^{\mu\nu}u_{\nu}$ to be a four-vector for every four-velocity is what certifies $F^{\mu\nu}$ as a tensor in the first place. Our job now is to verify that (2.6.64) is the force law you already know.

8.1 · The spatial components

Lower the index on the four-velocity $u^{\mu}=\gamma(c,\vv v)$ of Chapter 2.5 §1.3, which flips the sign of the spatial part:

Take $\mu=i$ and split the sum over $\nu$ into its time and space parts:

The left-hand side is $\dd p^{i}/\dd\tau=\gamma\,\dd p^{i}/\dd t$, since $\dd t/\dd\tau=\gamma$ (Chapter 2.5 §1.2). Cancel the common $\gamma$:

That is the Lorentz force law, the one Chapter 1.1 §3.3 had to quote without justification. Here it arrives as the spatial part of a four-vector equation. Notice also that $\vv p=\gamma m\vv v$ here rather than $m\vv v$, so the relativistic correction rides along for free.

8.2 · The time component is the power

Take $\mu=0$. The $\nu=0$ term dies because $F^{00}=0$:

And the left-hand side is $\dd p^{0}/\dd\tau=\gamma\,\dd(E/c)/\dd t$, where $E$ is now the particle's energy of Chapter 2.5 §3 (an unfortunate clash of symbols that the whole subject lives with). Cancelling $\gamma/c$:

The magnetic field does no work. That familiar fact is not a separate observation here. It is a consequence of where $\vv B$ sits in the matrix. The $F^{0j}$ entries are purely electric, so only $\vv E$ can appear in the energy equation at all.

And (2.6.69) is Chapter 2.5's work–energy theorem $\dd E/\dd t=\vv F\cdot\vv v$ with $\vv F$ the Lorentz force, since $(\vv v\times\vv B)\cdot\vv v=0$ automatically.

8.3 · The constraint is automatic, and why

Chapter 2.5 §6.1 proved that any four-force must satisfy $f\cdot u=0$, because $u\cdot u=c^{2}$ is constant. That is a strong condition, and it is not obvious that an arbitrary guess for $f^{\mu}$ would satisfy it. Check (2.6.64):

Why zero? Because $F^{\mu\nu}$ is antisymmetric under $\mu\leftrightarrow\nu$ while $u_{\mu}u_{\nu}$ is symmetric, and a symmetric object fully contracted with an antisymmetric one vanishes.

Here is that argument once, in case it has not been made explicit before. Relabel the dummy indices $\mu\leftrightarrow\nu$ in the sum, which changes nothing, and then use the two symmetry properties to put the labels back the way they were:

9 · The action

Chapter 1.2 §8.1 listed eight actions and promised each would be constructed in its own chapter. Line three said electromagnetic field, Chapter 2.6. Here it is, with the source term and the SI constants restored:

Two terms, one line, and it contains every result of §3. The claim we have to verify is that feeding (2.6.72) to the field Euler–Lagrange equation of Chapter 1.2 §8 returns $\partial_{\mu}F^{\mu\nu}=\mu_{0}j^{\nu}$.

9.1 · Varying it

The dynamical variable here is $A_{\nu}$, which is four fields, one for each value of $\nu$. So the Euler–Lagrange equation carries a free index:

The second term is immediate. Only $-j_{\mu}A^{\mu}=-j^{\nu}A_{\nu}$ contains $A_{\nu}$ undifferentiated, so $\partial\mathcal{L}/\partial A_{\nu}=-j^{\nu}$. The first term is where the work is, and the grind box does it slowly. The answer is

We now have both pieces the Euler–Lagrange equation asks for, so put them into (2.6.73) and see what field equation comes out:

That is (2.6.19). Gauss's law and Ampère–Maxwell, both of them, out of one scalar. The homogeneous pair needs no varying at all. It holds identically the moment you write $F$ in terms of $A$, which was the point of §3.4.

Grind box — differentiating $F_{\alpha\beta}F^{\alpha\beta}$ with respect to $\partial_{\mu}A_{\nu}$

This is pure index gymnastics and it is worth doing once with every step visible, because the same manipulation reappears in Chapters 5.2, 6.3 and 6.4 with more indices.

Step 1. The chain rule on a product. Write $\mathcal{L}_{F}=-\tfrac{1}{4\mu_{0}}F_{\alpha\beta}F^{\alpha\beta}$ and note $F_{\alpha\beta}F^{\alpha\beta}=\eta^{\alpha\gamma}\eta^{\beta\delta}F_{\alpha\beta}F_{\gamma\delta}$, so it is quadratic in the $F_{\alpha\beta}$ with constant coefficients. Hence

$$\pdv{}{(\partial_{\mu}A_{\nu})}\big(F_{\alpha\beta}F^{\alpha\beta}\big) \;=\; 2F^{\alpha\beta}\,\pdv{F_{\alpha\beta}}{(\partial_{\mu}A_{\nu})}.$$

Step 2. The inner derivative. Since $F_{\alpha\beta}=\partial_{\alpha}A_{\beta}-\partial_{\beta}A_{\alpha}$ and the sixteen quantities $\partial_{\mu}A_{\nu}$ are treated as independent variables,

$$\pdv{(\partial_{\alpha}A_{\beta})}{(\partial_{\mu}A_{\nu})} \;=\; \delta^{\mu}{}_{\alpha}\,\delta^{\nu}{}_{\beta} \qquad\Longrightarrow\qquad \pdv{F_{\alpha\beta}}{(\partial_{\mu}A_{\nu})} \;=\; \delta^{\mu}{}_{\alpha}\delta^{\nu}{}_{\beta} - \delta^{\mu}{}_{\beta}\delta^{\nu}{}_{\alpha}.$$

The two deltas in each product are what turn "differentiate with respect to a particular component" into "select that component". The two terms are the two places $\partial A$ appears in $F$.

Step 3. Contract. Now feed Step 2 back into Step 1. The deltas rename indices:

$$2F^{\alpha\beta}\big(\delta^{\mu}{}_{\alpha}\delta^{\nu}{}_{\beta} - \delta^{\mu}{}_{\beta}\delta^{\nu}{}_{\alpha}\big) \;=\; 2\big(F^{\mu\nu} - F^{\nu\mu}\big) \;=\; 4F^{\mu\nu},$$

where the last step used antisymmetry. There is the factor of $4$ that the $\tfrac14$ in (2.6.72) was put there to cancel. Carrying the prefactor along,

$$\pdv{\mathcal{L}_{F}}{(\partial_{\mu}A_{\nu})} \;=\; -\frac{1}{4\mu_{0}}\cdot 4F^{\mu\nu} \;=\; -\frac{1}{\mu_{0}}F^{\mu\nu},$$

which is (2.6.74). The source term contains no derivatives of $A$ and contributes nothing here. ✓

Why the sign of the whole Lagrangian is what it is. Expand (2.6.72) into three-vector language using (2.6.56):

$$-\frac{1}{4\mu_{0}}F_{\mu\nu}F^{\mu\nu} \;=\; -\frac{1}{2\mu_{0}}\left(B^{2}-\frac{E^{2}}{c^{2}}\right) \;=\; \frac{\epsilon_{0}E^{2}}{2} - \frac{B^{2}}{2\mu_{0}}.$$

That is kinetic minus potential. The electric term carries the time derivatives of $\vv A$ and plays the role of $T$, while the magnetic term carries the spatial derivatives and plays the role of $V$. The overall sign of (2.6.72) is fixed by demanding that $T$ come out positive, which is Chapter 1.2 §5's discussion transplanted verbatim to a field. Get it wrong and the field has negative kinetic energy, which Chapter 5.2 shows is a catastrophe rather than a sign convention.

9.2 · Why this Lagrangian and not another

The variation above is a verification rather than a derivation. We wrote down (2.6.72) and checked it. So where did it come from? The honest answer is a short list of demands.

1. Lorentz invariance. $\mathcal{L}$ must be a scalar, or the action is not the same number for every observer and the theory has a preferred frame. That immediately restricts us to contractions: $F_{\mu\nu}F^{\mu\nu}$, $\epsilon FF$, $A_{\mu}A^{\mu}$, $j_{\mu}A^{\mu}$, and products and derivatives of these.
2. Gauge invariance. $\mathcal{L}$ must be unchanged, up to a four-divergence, under (2.6.29). This kills $A_{\mu}A^{\mu}$ outright, since that term is not gauge invariant. It would have been a photon mass term. The photon is massless because a mass term is not gauge invariant. The same demand permits $j_{\mu}A^{\mu}$, and only because charge is conserved, as §4.1 showed.
3. Second-order equations of motion. $\mathcal{L}$ should contain at most first derivatives of $A$, so that Euler–Lagrange returns something second order and the initial-value problem is the usual one.
4. Parity, if you want it. $\epsilon_{\mu\nu\rho\sigma}F^{\mu\nu}F^{\rho\sigma}$ passes tests 1–3 but is a pseudoscalar, and moreover is a total derivative, so it does not affect the classical field equations at all.

What survives at lowest order is $F_{\mu\nu}F^{\mu\nu}$ and $j_{\mu}A^{\mu}$, with two constants in front. One of those constants is fixed by matching Coulomb's law, and the other is absorbed into the normalisation of $A$. The form is very nearly forced.

10 · Field momentum, and Newton's third law repaired

Chapter 1.1 §3.3 put two moving charges alone in the universe and showed that the total force on them does not vanish. Momentum appeared from nowhere at a computable rate.

That chapter offered two escapes. Either momentum conservation is false, or momentum is stored somewhere that is not a particle. It took the second escape, named the electromagnetic field as the only candidate, and left an IOU. This section pays it, and the payment is exact.

10.1 · The energy–momentum tensor, from Noether

Chapter 1.4 proved that every continuous symmetry yields a conserved current. The symmetry we have not yet used is the most basic one there is. The laws are the same here as there, and now as then, which is invariance under spacetime translations $x^{\mu}\to x^{\mu}+a^{\mu}$. Since $a^{\mu}$ has four components, we get four conserved currents, one per direction, and those four assemble into a rank-2 object.

Run Chapter 1.4 §6's machinery. Under a translation the field changes by $\delta A_{\lambda}=-a^{\nu}\partial_{\nu}A_{\lambda}$ and the Lagrangian density by $\delta\mathcal{L}=-a^{\nu}\partial_{\nu}\mathcal{L}=\partial_{\mu}\big(-a^{\nu}\delta^{\mu}{}_{\nu}\mathcal{L}\big)$ That last expression is a four-divergence, which is exactly the loosened invariance condition of Chapter 1.4 §1.2. So the symmetry qualifies, and the Noether current for the $\nu$-th translation is

using (2.6.74) and setting $j^{\mu}=0$ (we are asking what the free field carries). It satisfies $\partial_{\mu}T^{\mu\nu}_{\text{can}}=0$ by Noether's theorem.

And it is wrong in two ways. It is not symmetric in $\mu\nu$. It is also not gauge invariant, because it contains $A_{\lambda}$ bare, so different gauges would assign different energy densities to the same physical field.

Both defects are fixed by the freedom that Chapter 1.4's grind box identified. Adding $\partial_{\lambda}\Sigma^{\lambda\mu\nu}$, with $\Sigma$ antisymmetric in its first two indices, changes no conserved charge. So we are free to choose a $\Sigma$ that cleans both problems up at once. Take

Its divergence is, using $\partial_{\lambda}F^{\mu\lambda}=-\partial_{\lambda}F^{\lambda\mu}=-\mu_{0}j^{\mu}=0$ in the source-free region,

Add it to (2.6.76) and the two $\partial A$ terms combine into an $F$:

One last cosmetic step puts the indices in their conventional places. Since $F^{\nu}{}_{\lambda}=-F_{\lambda}{}^{\nu}$, the minus sign in front can be absorbed, and we are left with the object we were after:

Now it is manifestly gauge invariant, since only $F$ appears in it. It is also symmetric, and that is worth checking rather than believing. The first term is $\eta_{\lambda\rho}F^{\mu\lambda}F^{\rho\nu}$. Swap $\mu\leftrightarrow\nu$ and relabel the dummies $\lambda\leftrightarrow\rho$, and you get it back, with the two minus signs from antisymmetry cancelling each other.

That symmetry is not cosmetic. Chapter 3.6 puts $T^{\mu\nu}$ on the right-hand side of Einstein's equation, and the left-hand side of that equation is symmetric by construction.

10.2 · What its components are

$T^{00}$. Only $\lambda$ spatial contributes to $F^{0\lambda}F_{\lambda}{}^{0}$, and lowering the index costs a minus:

using $F^{i0}=-F^{0i}$ and $F^{0i}=-E^{i}/c$. With $\eta^{00}=1$ and (2.6.56) for the second term,

The last step used $1/(\mu_{0}c^{2})=\epsilon_{0}$. That is the energy density of the electromagnetic field. You have probably met that expression as a stated fact. Here it is derived, as the conserved Noether charge density of time translation, which is what the word "energy" has meant since Chapter 1.4 §3.1.

$T^{0i}$. Now $\eta^{0i}=0$, so only the first term survives:

Only $\lambda=j$ contributes there, since $F^{00}=0$. The metric $\eta_{jk}$ supplies one minus sign and the antisymmetry $F^{ji}=-F^{ij}$ supplies the other, so the two cancel. Putting that into the definition of $T^{\mu\nu}$,

$\vv S$ is the Poynting vector. Now recall what $T^{0\mu}/c$ is by construction. It is the density of $p^{\mu}$, with the $0$-index labelling the conserved density and the $\mu$ labelling which component of momentum we mean. Reading the spatial $\mu$ off, the momentum density of the field is

Which is precisely the expression Chapter 1.1 promised, arriving here as a component of a Noether current rather than as an assertion.

10.3 · The conservation law, with sources

Now restore the charges, because it is the exchange between field and matter that we care about. Take the divergence of (2.6.80):

Term (1) is the field equation itself, $\partial_{\mu}F^{\mu\lambda}=\mu_{0}j^{\lambda}$, so it is $\mu_{0}j^{\lambda}F_{\lambda}{}^{\nu}$. Terms (2) and (3) cancel identically, by the Bianchi identity, and the grind box does that cancellation in full. Dividing through by $\mu_{0}$,

The right-hand side is not zero, and it should not be. The field is exchanging energy and momentum with the charges. Let's see what that exchange term actually is. For $\nu=i$, using $j_{0}=c\rho$ and $j_{j}=-J^{j}$,

That is the Lorentz force per unit volume, which is (2.6.67) written for a continuous distribution. Meanwhile the left-hand side, with $\partial_{0}=\tfrac1c\partial_{t}$ and $T^{0i}=cg^{i}$, is $\partial_{t}g^{i}+\partial_{j}T^{ji}$. Putting the two sides together,

Here $\mathsf{T}$ is the spatial block $T^{ij}$, called the Maxwell stress tensor.

We want a statement about totals rather than densities, so integrate that over all space and take the three terms one at a time. The middle term becomes a surface integral at infinity by the divergence theorem. For a system of charges confined to a finite region the fields fall off at least as fast as $1/r^{2}$, so $T^{ij}\sim1/r^{4}$ while the area grows only as $r^{2}$, and the surface term vanishes. The last term integrates to the total force on the matter, which is $\dd\vv P_{\text{mech}}/\dd t$. What is left is

The $\nu=0$ component of (2.6.87) gives, by the identical route, $\partial_{t}u+\nabla\cdot\vv S=-\vv J\cdot\vv E$. That is Poynting's theorem. It says the field energy in a region falls by the flux of $\vv S$ out of it, plus the work done on the charges inside. The two statements are one four-vector equation.

Grind box — why terms (2) and (3) cancel

Write term (2) with cleaner dummy names, $\mu\to\alpha$ and $\lambda\to\beta$:

$$(2) \;=\; F^{\alpha\beta}\,\partial_{\alpha}F_{\beta}{}^{\nu}.$$

Because $F^{\alpha\beta}$ is antisymmetric, only the antisymmetric part of whatever it multiplies survives. So we may replace the second factor by half of it minus its $\alpha\leftrightarrow\beta$ swap:

$$(2) \;=\; \tfrac12 F^{\alpha\beta}\big(\partial_{\alpha}F_{\beta}{}^{\nu} - \partial_{\beta}F_{\alpha}{}^{\nu}\big).$$

Now bring in the Bianchi identity (2.6.25), with the third index raised. Raising is legitimate here because we raise it on every term with the same $\eta$:

$$\partial_{\alpha}F_{\beta}{}^{\nu} + \partial_{\beta}F^{\nu}{}_{\alpha} + \partial^{\nu}F_{\alpha\beta} \;=\; 0.$$

Use $F^{\nu}{}_{\alpha}=-F_{\alpha}{}^{\nu}$ on the middle term and rearrange:

$$\partial_{\alpha}F_{\beta}{}^{\nu} - \partial_{\beta}F_{\alpha}{}^{\nu} \;=\; -\,\partial^{\nu}F_{\alpha\beta}.$$

Substitute:

$$(2) \;=\; -\tfrac12 F^{\alpha\beta}\,\partial^{\nu}F_{\alpha\beta} \;=\; -\,(3). \qquad\blacksquare$$

Note what did the work there. It was the homogeneous half of Maxwell, the half §3.4 called bookkeeping. It is not decorative after all. Without it $T^{\mu\nu}$ would not be conserved, and energy and momentum would not balance.

10.4 · Back to Chapter 1.1

Here is the configuration again, unchanged. Charge $q_{1}$ sits at the origin moving with $\vv v_{1}=v\hat{\vv x}$. Charge $q_{2}$ sits at $\vv r_{2}=d\hat{\vv x}$, directly ahead of it, moving with $\vv v_{2}=v\hat{\vv y}$. Both are in uniform motion at the instant considered.

What 1.1 found. Charge 1 produces no magnetic field directly ahead of itself, so charge 2 feels no magnetic force. Charge 2 does produce one at charge 1's location, so charge 1 does feel a magnetic force. The magnetic residue was

What 1.1 was entitled to drop, and we are not. That chapter said the electric forces "are equal and opposite along the line joining them, so they cancel". That is true to zeroth order in $v/c$ and false at order $v^{2}/c^{2}$, which is the very order the magnetic residue lives at.

Chapter 1.1 was making a qualitative point, so the omission cost it nothing. We are about to make a quantitative one, so we need the correction. It is three lines with §5's transformation rules.

Charge 1's field at charge 2. Boost to charge 1's rest frame $S'$. The field point $(t=0,\ x=d)$ has $x'=\gamma(d-v\cdot0)=\gamma d$, so in $S'$ the charge sits at a Coulomb distance $\gamma d$ and the field there is $q_{1}/(4\pi\epsilon_{0}\gamma^{2}d^{2})$, pointing along $\hat{\vv x}$. That direction is parallel to the boost, and by §5 the parallel component $E_{\parallel}$ is unchanged on transforming back. So

Charge 2's field at charge 1. Boost to charge 2's rest frame, which moves along $\hat{\vv y}$. The displacement from charge 2 to the origin, $-d\hat{\vv x}$, is perpendicular to that boost, so it is unchanged and the Coulomb distance is $d$. The field there is $-q_{2}/(4\pi\epsilon_{0}d^{2})\,\hat{\vv x}$. That direction is also perpendicular to the boost, so by §5 the field gets multiplied by $\gamma$ on transforming to the lab:

The two electric forces are $q_{2}\vv E_{1}(\vv r_{2})$ and $q_{1}\vv E_{2}(\vv r_{1})$, and they no longer cancel. One is weakened by $\gamma^{-2}$ and the other is strengthened by $\gamma$, because charge 1 is moving along the line joining them while charge 2 is moving across it. Adding them, and expanding to order $\beta^{2}$ with $\gamma^{-2}=1-\beta^{2}$ and $\gamma\approx1+\tfrac12\beta^{2}$, gives the electric residue:

The last step used $\beta^{2}/\epsilon_{0}=v^{2}/(c^{2}\epsilon_{0})=\mu_{0}v^{2}$. So the third law fails in the $\hat{\vv x}$ direction too, and by a larger amount than it fails in $\hat{\vv y}$. Adding the two residues, the total rate at which mechanical momentum is being created is

10.5 · And where it went

Now we compute the field momentum, to see whether it is disappearing at the same rate. Two simplifications are available at the order we need. Each charge's magnetic field is $\vv B_{a}=\vv v_{a}\times\vv E_{a}/c^{2}$, which §5 gives exactly, by boosting a rest frame where $\vv B=\vv 0$. And each $\vv E_{a}$ may be taken as Coulomb, since $\vv E\times\vv B$ is already first order in $v/c$ while corrections to $\vv E$ are second. With the fields written as sums of the two charges' contributions, (2.6.90) expands to

The two self terms $\vv E_{a}\times\vv B_{a}$ depend only on $\vv v_{a}$, which is constant. So however large they are, and they are in fact infinite, as the caution below explains, they contribute nothing to $\dd\vv P/\dd t$. That leaves the cross terms. Expanding each of those with $\vv a\times(\vv b\times\vv c)=\vv b(\vv a\cdot\vv c)-\vv c(\vv a\cdot\vv b)$,

Two integrals are needed to finish this, and both are elementary. The grind box evaluates them. Writing $\vv d=\vv r_{2}-\vv r_{1}$, $d=\abs{\vv d}$ and $\hat{\vv n}=\vv d/d$, they are

Now substitute those two results back. Write $\vv V\equiv\vv v_{1}+\vv v_{2}$. The two tensor terms combine into $M^{ij}V^{j}$, using the symmetry of $M^{ij}\equiv\int E_{1}^{i}E_{2}^{j}\,\dd^{3}x$, and the prefactors collapse with $1/(\epsilon_{0}c^{2})=\mu_{0}$:

That is the general answer. For the configuration at hand, at the instant $t=0$, we have $\hat{\vv n}=\hat{\vv x}$, $\vv V=v(\hat{\vv x}+\hat{\vv y})$ and $\hat{\vv n}\cdot\vv V=v$, so

There is momentum in the empty space between the charges.

What we actually want is the rate at which that momentum changes, so differentiate. The separation vector evolves as $\vv d(t)=(d-vt)\hat{\vv x}+vt\,\hat{\vv y}$, since charge 1 moves in $+x$ and charge 2 moves in $+y$. At $t=0$ that gives $\dot d=-v$ and $\dot{\hat{\vv n}}=(v/d)\hat{\vv y}$. Grinding through the product rule, as the grind box does,

Now set that beside (2.6.95), which is the rate at which the particles were gaining momentum from nowhere. The two match term by term, including the awkward $\tfrac32$ that Chapter 1.1 never saw. So the books balance exactly:

Grind box — the two field integrals of (2.6.98)

Both are done with the same two moves. Integrate by parts, then use Gauss's law $\nabla\cdot\vv E_{a}=\rho_{a}/\epsilon_{0}$ with $\rho_{a}=q_{a}\delta^{3}(\vv x-\vv r_{a})$. Surface terms vanish because the fields fall off as $1/r^{2}$ and the integrands as $1/r^{3}$ or faster.

The scalar one. Write $\vv E_{1}=-\nabla\phi_{1}$ and integrate by parts:

$$\int\!\vv E_{1}\!\cdot\!\vv E_{2}\,\dd^{3}x = -\!\int\!\nabla\phi_{1}\cdot\vv E_{2}\,\dd^{3}x = \int\!\phi_{1}\,\big(\nabla\!\cdot\!\vv E_{2}\big)\,\dd^{3}x = \frac{q_{2}}{\epsilon_{0}}\,\phi_{1}(\vv r_{2}) = \frac{q_{1}q_{2}}{4\pi\epsilon_{0}^{2}d}.$$

Multiplied by $\epsilon_{0}$, that is the familiar interaction energy $q_{1}q_{2}/4\pi\epsilon_{0}d$, which is a useful check that the normalisation is right.

The tensor one. Call it $M^{ij}(\vv d)$. Three facts pin it down.

(i) It is symmetric. Invert through the midpoint, $\vv x\to\vv r_{1}+\vv r_{2}-\vv x$. Under that map $\vv x-\vv r_{1}\mapsto-(\vv x-\vv r_{2})$, so $\vv E_{1}\mapsto-(q_{1}/q_{2})\vv E_{2}$ and $\vv E_{2}\mapsto-(q_{2}/q_{1})\vv E_{1}$, and the Jacobian is $1$. Hence $M^{ij}=M^{ji}$.

(ii) Its form. The only vectors in the problem are $\hat{\vv n}$ and the coordinate axes, so by rotational symmetry about $\hat{\vv n}$ and (i),

$$M^{ij} \;=\; \frac{A}{d}\,\delta^{ij} \;+\; \frac{B}{d}\,n^{i}n^{j},$$

with $A$ and $B$ constants. The $1/d$ is there because $M$ has the dimensions of the scalar integral, which scales as $1/d$.

(iii) Two equations. The trace is the scalar integral just computed:

$$3A + B \;=\; \frac{q_{1}q_{2}}{4\pi\epsilon_{0}^{2}}.$$

For the second, differentiate with respect to $\vv d$ and contract. Since $\vv E_{2}$ depends on $\vv x-\vv r_{2}$ and $\vv r_{2}=\vv r_{1}+\vv d$, we have $\partial E_{2}^{j}/\partial d^{k}=-\partial_{k}E_{2}^{j}$, so

$$\pdv{M^{ij}}{d^{j}} = -\!\int\! E_{1}^{i}\,\partial_{j}E_{2}^{j}\,\dd^{3}x = -\frac{q_{2}}{\epsilon_{0}}E_{1}^{i}(\vv r_{2}) = -\frac{q_{1}q_{2}}{4\pi\epsilon_{0}^{2}}\frac{n^{i}}{d^{2}}.$$

Evaluate the same derivative on the parametrised form, using $\partial_{j}(1/d)=-n^{j}/d^{2}$ and $\partial_{j}\big(d^{i}d^{j}/d^{3}\big)=n^{i}/d^{2}$:

$$\pdv{}{d^{j}}\left[\frac{A\delta^{ij}+Bn^{i}n^{j}}{d}\right] = -\frac{A\,n^{i}}{d^{2}} + \frac{B\,n^{i}}{d^{2}} = \frac{(B-A)\,n^{i}}{d^{2}}.$$

Comparing, $B-A=-q_{1}q_{2}/4\pi\epsilon_{0}^{2}$. Solve the two equations: $A=q_{1}q_{2}/8\pi\epsilon_{0}^{2}$ and $B=-A$. Hence

$$M^{ij} \;=\; \frac{q_{1}q_{2}}{8\pi\epsilon_{0}^{2}d}\big(\delta^{ij}-n^{i}n^{j}\big),$$

which is (2.6.98). Note the structure. Here $\delta-nn$ is the projector onto directions perpendicular to the line joining the charges, so $M^{ij}n^{j}=0$, and the longitudinal part of the integral vanishes identically. This was also checked by direct numerical integration before being written down.

Grind box — differentiating (2.6.99)

Write $C=\mu_{0}q_{1}q_{2}/8\pi$ and $\vv V=v(\hat{\vv x}+\hat{\vv y})$, a constant. Then

$$\vv P_{\text{field}} \;=\; C\left[\frac{\vv V}{d} + \frac{\hat{\vv n}\,(\hat{\vv n}\cdot\vv V)}{d}\right].$$

The kinematics at $t=0$. $\vv d(t)=\vv r_{2}-\vv r_{1}=(d-vt)\hat{\vv x}+vt\,\hat{\vv y}$, so $\dot{\vv d}=v(\hat{\vv y}-\hat{\vv x})$ and

$$\dot d = \hat{\vv n}\cdot\dot{\vv d} = \hat{\vv x}\cdot v(\hat{\vv y}-\hat{\vv x}) = -v, \qquad \dot{\hat{\vv n}} = \frac{\dot{\vv d}}{d} - \frac{\vv d\,\dot d}{d^{2}} = \frac{v(\hat{\vv y}-\hat{\vv x})}{d} + \frac{v\hat{\vv x}}{d} = \frac{v}{d}\hat{\vv y}.$$

Note $\dot{\hat{\vv n}}\perp\hat{\vv n}$, as it must be for a unit vector.

First term.

$$\dv{}{t}\frac{\vv V}{d} = -\frac{\dot d}{d^{2}}\vv V = \frac{v}{d^{2}}\,v(\hat{\vv x}+\hat{\vv y}) = \frac{v^{2}}{d^{2}}\big(\hat{\vv x}+\hat{\vv y}\big).$$

Second term. Three pieces by the product rule:

$$\begin{aligned} \dv{}{t}\left[\frac{\hat{\vv n}(\hat{\vv n}\cdot\vv V)}{d}\right] &= -\frac{\dot d}{d^{2}}\hat{\vv n}(\hat{\vv n}\cdot\vv V) + \frac{\dot{\hat{\vv n}}(\hat{\vv n}\cdot\vv V)}{d} + \frac{\hat{\vv n}(\dot{\hat{\vv n}}\cdot\vv V)}{d}\\[4pt] &= \frac{v}{d^{2}}\,\hat{\vv x}\,(v) \;+\; \frac{(v/d)\hat{\vv y}\,(v)}{d} \;+\; \frac{\hat{\vv x}\,\big((v/d)\hat{\vv y}\cdot v(\hat{\vv x}+\hat{\vv y})\big)}{d}\\[4pt] &= \frac{v^{2}}{d^{2}}\hat{\vv x} + \frac{v^{2}}{d^{2}}\hat{\vv y} + \frac{v^{2}}{d^{2}}\hat{\vv x} \;=\; \frac{v^{2}}{d^{2}}\big(2\hat{\vv x}+\hat{\vv y}\big). \end{aligned}$$

Total.

$$\dv{\vv P_{\text{field}}}{t} = C\frac{v^{2}}{d^{2}}\Big[(\hat{\vv x}+\hat{\vv y}) + (2\hat{\vv x}+\hat{\vv y})\Big] = \frac{\mu_{0}q_{1}q_{2}v^{2}}{8\pi d^{2}}\big(3\hat{\vv x}+2\hat{\vv y}\big),$$

which is (2.6.101) after pulling out a factor of $2$: $\tfrac{1}{8\pi}(3\hat{\vv x}+2\hat{\vv y}) = \tfrac{1}{4\pi}\big(\tfrac32\hat{\vv x}+\hat{\vv y}\big)$. ✓ Every term of this box, and of (2.6.94), was verified symbolically. The sum (2.6.102) is zero identically rather than numerically.

11 · Worked examples

12 · Your turn