## Retarded potential of a wire loop

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 10.9.

One more of calculating the retarded potential. We have a loop of wire in the following shape. It extends along the ${x}$ axis from ${-b}$ to ${-a}$, then in a semicircular loop of radius ${a}$ clockwise around to ${x=+a}$, then along the ${x}$ axis from ${+a}$ to ${+b}$, then in a semicircular loop of radius ${b}$ back to ${x=-b}$. A linearly increasing current

$\displaystyle I\left(t\right)=kt \ \ \ \ \ (1)$

flows through the loop in the direction given above. Assuming the wire is electrically neutral, ${V=0}$ so our job is to find ${\mathbf{A}}$.

Calculating ${\mathbf{A}}$ in general is a complex task, so we’ll look only at the value of ${\mathbf{A}}$ at the origin. Consider first the inner loop of radius ${a}$. All points on this loop are at the same distance ${a}$ from the origin, so the retarded time is the same for all points on the loop. Since the current goes clockwise around the semicircle, the contribution to ${\mathbf{A}}$ is

 $\displaystyle \mathbf{A}_{a}$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{4\pi}\int_{\pi}^{0}\frac{k\left(t-\frac{a}{c}\right)}{a}\hat{\boldsymbol{\theta}}a\; d\theta\ \ \ \ \ (2)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}}{4\pi}k\left(t-\frac{a}{c}\right)\int_{0}^{\pi}\left(-\sin\theta\hat{\mathbf{x}}+\cos\theta\hat{\mathbf{y}}\right)d\theta\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{2\pi}k\left(t-\frac{a}{c}\right)\hat{\mathbf{x}} \ \ \ \ \ (4)$

We get a similar expression for the loop around the outer semicircle except this time the current flows counterclockwise so the sign is reversed:

$\displaystyle \mathbf{A}_{b}=-\frac{\mu_{0}}{2\pi}k\left(t-\frac{b}{c}\right)\hat{\mathbf{x}} \ \ \ \ \ (5)$

Adding these two together we get

 $\displaystyle \mathbf{A}_{ab}$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{2\pi}k\frac{b-a}{c}\hat{\mathbf{x}} \ \ \ \ \ (6)$

The contributions from each of the two horizontal segments are equal, so for these two segments we have

 $\displaystyle \mathbf{A}_{x}$ $\displaystyle =$ $\displaystyle 2\frac{\mu_{0}}{4\pi}k\hat{\mathbf{x}}\int_{a}^{b}\frac{t-\frac{x}{c}}{x}dx\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{2\pi}k\hat{\mathbf{x}}\left[t\ln\frac{b}{a}-\frac{b-a}{c}\right] \ \ \ \ \ (8)$

The total potential is then

 $\displaystyle \mathbf{A}\left(0,t\right)$ $\displaystyle =$ $\displaystyle \mathbf{A}_{ab}+\mathbf{A}_{x}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{2\pi}k\hat{\mathbf{x}}t\ln\frac{b}{a} \ \ \ \ \ (10)$

Because we have the potential at only a single point in space, we can’t calculate any of its derivatives, so we can’t calculate ${\mathbf{B}=\nabla\times\mathbf{A}}$. However we can calculate ${\mathbf{E}}$:

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle -\frac{\partial\mathbf{A}}{\partial t}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}}{2\pi}k\hat{\mathbf{x}}\ln\frac{b}{a} \ \ \ \ \ (12)$

The electric field is constant in time at the origin. An electrically neutral wire can produce an electric field since the changing current induces a changing magnetic field which in turn produces an electric field.

## Retarded potentials in an infinite straight wire

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 10.9.

Here are a couple of examples of calculating the retarded potential. In both examples, we have an infinite straight wire that carries a current. At time ${t=0}$ a current ${I\left(t\right)}$ is switched on, and we want to find the electric and magnetic fields after that time. We’ll assume the wire is electrically neutral so there is no free charge density and the scalar potential is thus ${V=0}$.

Example 1 The current is linearly increasing:

$\displaystyle I\left(t\right)=kt \ \ \ \ \ (1)$

For a linear current the vector potential is (assuming the current flows in the ${z}$ direction):

 $\displaystyle \mathbf{A}\left(\mathbf{r},t\right)$ $\displaystyle =$ $\displaystyle \hat{\mathbf{z}}\frac{\mu_{0}}{4\pi}\int\frac{I\left(\mathbf{r}',t_{r}\right)}{d}dz \ \ \ \ \ (2)$

where

$\displaystyle t_{r}\equiv t-\frac{d}{c} \ \ \ \ \ (3)$

and

 $\displaystyle d$ $\displaystyle =$ $\displaystyle \sqrt{r^{2}+z^{2}} \ \ \ \ \ (4)$

where we’re using cylindrical coordinates, so ${r}$ is the perpendicular distance from the wire and ${z}$ is the distance along the wire.

At time ${t}$ and distance ${r}$, only those parts of the wire where ${d will contribute to the potential, so the integral’s limits are ${z=\pm\sqrt{\left(ct\right)^{2}-r^{2}}}$ and we get

 $\displaystyle \mathbf{A}$ $\displaystyle =$ $\displaystyle \hat{\mathbf{z}}\frac{\mu_{0}k}{4\pi}\int_{-\sqrt{\left(ct\right)^{2}-r^{2}}}^{\sqrt{\left(ct\right)^{2}-r^{2}}}\frac{t-\sqrt{r^{2}+z^{2}}/c}{\sqrt{r^{2}+z^{2}}}dz\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hat{\mathbf{z}}\frac{\mu_{0}k}{4\pi c}2\int_{0}^{\sqrt{\left(ct\right)^{2}-r^{2}}}\left[\frac{ct}{\sqrt{r^{2}+z^{2}}}-1\right]dz\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hat{\mathbf{z}}\frac{\mu_{0}k}{2\pi c}\left[ct\ln\left(\frac{ct+\sqrt{\left(ct\right)^{2}-r^{2}}}{r}\right)-\sqrt{\left(ct\right)^{2}-r^{2}}\right] \ \ \ \ \ (7)$

From this we can find ${\mathbf{E}}$ and ${\mathbf{B}}$ using the equations

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle -\nabla V-\frac{\partial\mathbf{A}}{\partial t}\ \ \ \ \ (8)$ $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle \nabla\times\mathbf{A} \ \ \ \ \ (9)$

The derivatives are messy and best done with Maple. The results are

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}k}{2\pi}\ln\left(\frac{ct+\sqrt{\left(ct\right)^{2}-r^{2}}}{r}\right)\hat{\mathbf{z}}\ \ \ \ \ (10)$ $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}k\left(\sqrt{\left(ct\right)^{2}-r^{2}}ct+\left(ct\right)^{2}-r^{2}\right)}{2\pi rc\left(ct+\sqrt{\left(ct\right)^{2}-r^{2}}\right)}\hat{\boldsymbol{\theta}}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}k\sqrt{\left(ct\right)^{2}-r^{2}}}{2\pi rc}\hat{\boldsymbol{\theta}} \ \ \ \ \ (12)$

Example 2 This time the current is an impulse at ${t=0}$ given by

$\displaystyle I\left(t\right)=q_{0}\delta\left(t\right) \ \ \ \ \ (13)$

The argument above still applies; only the integrand is different. We get

 $\displaystyle \mathbf{A}\left(\mathbf{r},t\right)$ $\displaystyle =$ $\displaystyle \hat{\mathbf{z}}\frac{\mu_{0}}{4\pi}\int\frac{I\left(\mathbf{r}',t_{r}\right)}{d}dz\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hat{\mathbf{z}}\frac{\mu_{0}q_{0}}{2\pi}\int_{0}^{\sqrt{\left(ct\right)^{2}-r^{2}}}\frac{\delta\left(t-\sqrt{r^{2}+z^{2}}/c\right)}{\sqrt{r^{2}+z^{2}}}dz \ \ \ \ \ (15)$

We can transform the integral using the substitution

 $\displaystyle u$ $\displaystyle =$ $\displaystyle t-\frac{\sqrt{r^{2}+z^{2}}}{c}\ \ \ \ \ (16)$ $\displaystyle \sqrt{r^{2}+z^{2}}$ $\displaystyle =$ $\displaystyle ct-cu\ \ \ \ \ (17)$ $\displaystyle z$ $\displaystyle =$ $\displaystyle \sqrt{\left(ct-cu\right)^{2}-r^{2}}\ \ \ \ \ (18)$ $\displaystyle dz$ $\displaystyle =$ $\displaystyle -\frac{c\left(ct-cu\right)}{\sqrt{\left(ct-cu\right)^{2}-r^{2}}}du \ \ \ \ \ (19)$

The limits transform as

 $\displaystyle z=0$ $\displaystyle \rightarrow$ $\displaystyle u=t-\frac{r}{c}\ \ \ \ \ (20)$ $\displaystyle z=\sqrt{\left(ct\right)^{2}-r^{2}}$ $\displaystyle \rightarrow$ $\displaystyle u=0 \ \ \ \ \ (21)$

The integral becomes

 $\displaystyle \mathbf{A}\left(\mathbf{r},t\right)$ $\displaystyle =$ $\displaystyle \hat{\mathbf{z}}\frac{\mu_{0}q_{0}}{2\pi}\int_{0}^{t-\frac{r}{c}}\frac{c\left(ct-cu\right)\delta\left(u\right)}{\left(ct-cu\right)\sqrt{\left(ct-cu\right)^{2}-r^{2}}}du\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \hat{\mathbf{z}}\frac{c\mu_{0}q_{0}}{2\pi\sqrt{\left(ct\right)^{2}-r^{2}}} \ \ \ \ \ (23)$

The corresponding fields are

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}q_{0}c^{3}t}{2\pi\left(c^{2}t^{2}-r^{2}\right)^{3/2}}\hat{\mathbf{z}}\ \ \ \ \ (24)$ $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}q_{0}cr}{2\pi\left(c^{2}t^{2}-r^{2}\right)^{3/2}}\hat{\boldsymbol{\theta}} \ \ \ \ \ (25)$

## Retarded potentials

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 10.8.

In electrostatics and magnetostatics, charge distributions and currents were all constant in time. When they vary, we need to take into account the finite speed of light in calculating potentials and fields. If we want the fields at some point ${P}$ then, if the charge or current changes at some point ${Q}$ a distance ${d}$ from ${P}$, an observer at ${P}$ won’t know about the change until the signal from ${Q}$ reaches him, which in vacuum takes a time ${d/c}$. To take account of this, the potentials at position ${\mathbf{r}}$ and time ${t}$ in a dynamic system are taken to be

 $\displaystyle V\left(\mathbf{r},t\right)$ $\displaystyle =$ $\displaystyle \frac{1}{4\pi\epsilon_{0}}\int\frac{\rho\left(\mathbf{r}',t_{r}\right)}{d}d^{3}\mathbf{r}'\ \ \ \ \ (1)$ $\displaystyle \mathbf{A}\left(\mathbf{r},t\right)$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{4\pi}\int\frac{\mathbf{J}\left(\mathbf{r}',t_{r}\right)}{d}d^{3}\mathbf{r}' \ \ \ \ \ (2)$

where

$\displaystyle t_{r}\equiv t-\frac{d}{c} \ \ \ \ \ (3)$

and

 $\displaystyle d$ $\displaystyle \equiv$ $\displaystyle \left|\mathbf{r}-\mathbf{r}'\right|\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\left(x-x'\right)^{2}+\left(y-y'\right)^{2}+\left(z-z'\right)^{2}}\ \ \ \ \ (5)$ $\displaystyle \hat{\mathbf{d}}$ $\displaystyle =$ $\displaystyle \frac{\mathbf{r}-\mathbf{r}'}{d} \ \ \ \ \ (6)$

That is, each potential is the sum over all locations where there is charge or current, and each location ${\mathbf{r}'}$ is sampled at the time ${t_{r}}$ in the past which is the time a light signal would have left ${\mathbf{r}'}$ to arrive at ${\mathbf{r}}$ at time ${t}$. These potentials are called retarded potentials, since they depend on the situation at various times in the past to get the fields at the present time.

Griffiths shows in his section 10.2.1 that these potentials (well ${V}$ anyway; the argument for ${\mathbf{A}}$ is similar) satisfy the wave equations in the Lorentz gauge

 $\displaystyle \square^{2}V$ $\displaystyle =$ $\displaystyle -\frac{\rho}{\epsilon_{0}}\ \ \ \ \ (7)$ $\displaystyle \square^{2}\mathbf{A}$ $\displaystyle =$ $\displaystyle -\mu_{0}\mathbf{J} \ \ \ \ \ (8)$

We also need to show that the potentials satisfy the Lorentz gauge condition

$\displaystyle \nabla\cdot\mathbf{A}=-\mu_{0}\epsilon_{0}\frac{\partial V}{\partial t} \ \ \ \ \ (9)$

Starting from 2 we need to find

$\displaystyle \nabla\cdot\mathbf{A}=\frac{\mu_{0}}{4\pi}\int\nabla\cdot\left(\frac{\mathbf{J}\left(\mathbf{r}',t_{r}\right)}{d}\right)d^{3}\mathbf{r}' \ \ \ \ \ (10)$

Note that the ${\nabla}$ is a derivative with respect to ${\mathbf{r}}$ (the observer’s position) and not ${\mathbf{r}'}$ (the source positions and variable of integration), and that both ${t_{r}}$ and ${d}$ depend on both ${\mathbf{r}}$ and ${\mathbf{r}'}$. We begin by writing

$\displaystyle \nabla\cdot\left(\frac{\mathbf{J}}{d}\right)=\frac{1}{d}\nabla\cdot\mathbf{J}+\mathbf{J}\cdot\nabla\left(\frac{1}{d}\right) \ \ \ \ \ (11)$

We can also use the derivative with respect to ${\mathbf{r}'}$:

$\displaystyle \nabla'\cdot\left(\frac{\mathbf{J}}{d}\right)=\frac{1}{d}\nabla'\cdot\mathbf{J}+\mathbf{J}\cdot\nabla'\left(\frac{1}{d}\right) \ \ \ \ \ (12)$

We have (you can work this out by using 5 if you don’t believe me):

$\displaystyle \nabla\left(\frac{1}{d}\right)=-\frac{\hat{\mathbf{d}}}{d^{2}}=-\nabla'\left(\frac{1}{d}\right) \ \ \ \ \ (13)$

Therefore

 $\displaystyle \nabla'\cdot\left(\frac{\mathbf{J}}{d}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{d}\nabla'\cdot\mathbf{J}-\mathbf{J}\cdot\nabla\left(\frac{1}{d}\right)\ \ \ \ \ (14)$ $\displaystyle \mathbf{J}\cdot\nabla\left(\frac{1}{d}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{d}\nabla'\cdot\mathbf{J}-\nabla'\cdot\left(\frac{\mathbf{J}}{d}\right) \ \ \ \ \ (15)$

Inserting this into 11 we get

$\displaystyle \nabla\cdot\left(\frac{\mathbf{J}}{d}\right)=\frac{1}{d}\nabla\cdot\mathbf{J}+\frac{1}{d}\nabla'\cdot\mathbf{J}-\nabla'\cdot\left(\frac{\mathbf{J}}{d}\right) \ \ \ \ \ (16)$

Now we need to work out the two divergences ${\nabla\cdot\mathbf{J}}$ and ${\nabla\cdot\mathbf{J}'}$. To do this, we need to remember that ${\mathbf{J}=\mathbf{J}\left(\mathbf{r}',t_{r}\right)}$, so it depends on ${\mathbf{r}}$ only via ${t_{r}}$ but it depends on ${\mathbf{r}'}$ both explicitly through its first argument and implicitly through ${t_{r}}$. Using the chain rule, we get for the contribution from ${x}$

 $\displaystyle \left(\nabla\cdot\mathbf{J}\right)_{x}$ $\displaystyle =$ $\displaystyle \frac{\partial\mathbf{J}_{x}}{\partial t_{r}}\frac{\partial t_{r}}{\partial x}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{c}\dot{\mathbf{J}}_{x}\frac{\partial d}{\partial x}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{c}\dot{\mathbf{J}}_{x}\left(\nabla d\right)_{x} \ \ \ \ \ (19)$

where the dot over the ${\mathbf{J}}$ is a derivative with respect to ${t}$, which is the same as a derivative with respect to ${t_{r}}$ since ${t_{r}=t-d/c}$ and ${d}$ doesn’t depend on time.

The other two coordinates give similar results and we get

$\displaystyle \nabla\cdot\mathbf{J}=-\frac{1}{c}\dot{\mathbf{J}}\cdot\nabla d \ \ \ \ \ (20)$

For the other divergence, things are a bit trickier since ${\mathbf{J}}$ depends explicitly on ${\mathbf{r}'}$. Here we used the extended chain rule

$\displaystyle \frac{\partial g\left(x,f\left(x\right)\right)}{\partial x}=\frac{\partial g}{\partial x}+\frac{\partial g}{\partial f}\frac{\partial f}{\partial x} \ \ \ \ \ (21)$

Therefore

$\displaystyle \nabla'\cdot\mathbf{J}=\nabla'_{\mathbf{r}'}\cdot\mathbf{J}-\frac{1}{c}\dot{\mathbf{J}}\cdot\nabla'd \ \ \ \ \ (22)$

where we’ve used the specialized notation ${\nabla'_{\mathbf{r}'}}$ to indicate the divergence with respect to the explicit ${\mathbf{r}'}$ dependence in ${\mathbf{J}}$. From Maxwell’s fourth equation

$\displaystyle \nabla'_{\mathbf{r}'}\times\mathbf{B}=\mu_{0}\mathbf{J}+\mu_{0}\epsilon_{0}\frac{\partial\mathbf{E}}{\partial t} \ \ \ \ \ (23)$

where the fields and currents depend on ${\mathbf{r}'}$, we can take the explicit divergence to get

$\displaystyle \nabla'_{\mathbf{r}'}\cdot\left(\nabla'_{\mathbf{r}'}\times\mathbf{B}\right)=\mu_{0}\nabla'_{\mathbf{r}'}\cdot\mathbf{J}+\mu_{0}\epsilon_{0}\frac{\partial\nabla'_{\mathbf{r}'}\cdot\mathbf{E}}{\partial t} \ \ \ \ \ (24)$

The divergence of a curl is always zero, so we get

 $\displaystyle \nabla'_{\mathbf{r}'}\cdot\mathbf{J}$ $\displaystyle =$ $\displaystyle -\epsilon_{0}\frac{\partial\nabla'_{\mathbf{r}'}\cdot\mathbf{E}}{\partial t}\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\dot{\rho} \ \ \ \ \ (26)$

where we’ve used Maxwell’s first equation

$\displaystyle \nabla'_{\mathbf{r}'}\cdot\mathbf{E}=\frac{\rho}{\epsilon_{0}} \ \ \ \ \ (27)$

Plugging this into 22 we get

 $\displaystyle \nabla'\cdot\mathbf{J}$ $\displaystyle =$ $\displaystyle -\dot{\rho}-\frac{1}{c}\dot{\mathbf{J}}\cdot\nabla'd\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\dot{\rho}+\frac{1}{c}\dot{\mathbf{J}}\cdot\nabla d \ \ \ \ \ (29)$

since from 5

$\displaystyle \nabla'd=-\nabla d \ \ \ \ \ (30)$

Putting 20 and 29 into 16 we get

 $\displaystyle \nabla\cdot\left(\frac{\mathbf{J}}{d}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{d}\left[-\frac{1}{c}\dot{\mathbf{J}}\cdot\nabla d-\dot{\rho}+\frac{1}{c}\dot{\mathbf{J}}\cdot\nabla d\right]-\nabla'\cdot\left(\frac{\mathbf{J}}{d}\right)\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\dot{\rho}}{d}-\nabla'\cdot\left(\frac{\mathbf{J}}{d}\right) \ \ \ \ \ (32)$

Finally, from 2 we have

$\displaystyle \nabla\cdot\mathbf{A}\left(\mathbf{r},t\right)=\frac{\mu_{0}}{4\pi}\int\left[-\frac{\dot{\rho}}{d}-\nabla'\cdot\left(\frac{\mathbf{J}}{d}\right)\right]d^{3}\mathbf{r}' \ \ \ \ \ (33)$

Using the divergence theorem, the second term can be converted to a surface integral at infinity where (presumably) the current ${\mathbf{J}}$ is zero, so this term vanishes. Using 1 we then get

 $\displaystyle \nabla\cdot\mathbf{A}\left(\mathbf{r},t\right)$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}}{4\pi}\int\frac{\dot{\rho}}{d}d^{3}\mathbf{r}'\ \ \ \ \ (34)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\mu_{0}\epsilon_{0}\frac{\partial V}{\partial t} \ \ \ \ \ (35)$

which is the Lorentz gauge condition, as required.

## Lorentz gauge is always possible

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 10.7.

The Lorentz gauge is defined by setting

$\displaystyle \nabla\cdot\mathbf{A}=-\mu_{0}\epsilon_{0}\frac{\partial V}{\partial t} \ \ \ \ \ (1)$

but is it always possible to do this? We can show that it is using a similar technique to that for the Coulomb gauge. We want a function ${\lambda}$ which we can use to transform some arbitrary potentials ${\mathbf{A}'}$ and ${V}$ so that that ${\mathbf{A}}$ satisfies 1 as follows:

 $\displaystyle \mathbf{A}$ $\displaystyle =$ $\displaystyle \mathbf{A}'+\nabla\lambda\ \ \ \ \ (2)$ $\displaystyle V$ $\displaystyle =$ $\displaystyle V'-\frac{\partial\lambda}{\partial t} \ \ \ \ \ (3)$

Taking the divergence of both sides, we get

 $\displaystyle \nabla\cdot\mathbf{A}$ $\displaystyle =$ $\displaystyle \nabla\cdot\mathbf{A}'+\nabla^{2}\lambda\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\mu_{0}\epsilon_{0}\frac{\partial V}{\partial t}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\mu_{0}\epsilon_{0}\frac{\partial V'}{\partial t}+\mu_{0}\epsilon_{0}\frac{\partial^{2}\lambda}{\partial t^{2}} \ \ \ \ \ (6)$

Combining the first and last equations we get

 $\displaystyle \nabla^{2}\lambda-\mu_{0}\epsilon_{0}\frac{\partial^{2}\lambda}{\partial t^{2}}$ $\displaystyle =$ $\displaystyle -\mu_{0}\epsilon_{0}\frac{\partial V'}{\partial t}-\nabla\cdot\mathbf{A}'\ \ \ \ \ (7)$ $\displaystyle \square^{2}\lambda$ $\displaystyle =$ $\displaystyle -\mu_{0}\epsilon_{0}\frac{\partial V'}{\partial t}-\nabla\cdot\mathbf{A}' \ \ \ \ \ (8)$

That is, ${\lambda}$ is the solution of the wave equation with a driving term, which we can, in principle, always solve (although it may not be easy!). Therefore we can always find the function ${\lambda}$ to convert an arbitrary pair of potentials ${\mathbf{A}'}$ and ${V'}$ to the Lorentz gauge.

In general (not necessarily in the Lorentz gauge), we could always set ${V=0}$ by choosing

 $\displaystyle \frac{\partial\lambda}{\partial t}$ $\displaystyle =$ $\displaystyle V'\ \ \ \ \ (9)$ $\displaystyle \lambda$ $\displaystyle =$ $\displaystyle \int_{0}^{t}V'\left(\mathbf{r},t'\right)dt' \ \ \ \ \ (10)$

We can’t always choose ${\mathbf{A}=0}$ however, since ${\mathbf{B}=\nabla\times\mathbf{A}}$ and if the magnetic field is non-zero, then ${\mathbf{A}}$ can’t be zero everywhere.

## Coulomb and Lorentz gauges

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 10.6.

Two common gauges in electrodynamics are the Coulomb gauge and the Lorentz gauge. Each gauge amounts to specifying a value for ${\nabla\cdot\mathbf{A}}$. The Coulomb gauge sets ${\nabla\cdot\mathbf{A}=0}$, and is the gauge we used when introducing the magnetic vector potential. To see that it’s always possible to transform from an arbitrary gauge to the Coulomb gauge, we need to find a function ${\lambda}$ such that the transformation

 $\displaystyle \mathbf{A}$ $\displaystyle =$ $\displaystyle \mathbf{A}'+\nabla\lambda\ \ \ \ \ (1)$ $\displaystyle V$ $\displaystyle =$ $\displaystyle V'-\frac{\partial\lambda}{\partial t} \ \ \ \ \ (2)$

gives ${\nabla\cdot\mathbf{A}=0}$. To do this, we must have

 $\displaystyle \nabla\cdot\mathbf{A}$ $\displaystyle =$ $\displaystyle 0=\nabla\cdot\mathbf{A}'+\nabla^{2}\lambda\ \ \ \ \ (3)$ $\displaystyle \nabla^{2}\lambda$ $\displaystyle =$ $\displaystyle -\nabla\cdot\mathbf{A}' \ \ \ \ \ (4)$

The last line is just Poisson’s equation and for any “reasonable” original potential ${\mathbf{A}'}$ it is possible to solve it. In the Coulomb gauge, the potential forms of Maxwell’s equations:

 $\displaystyle \nabla^{2}V+\frac{\partial}{\partial t}\left(\nabla\cdot\mathbf{A}\right)$ $\displaystyle =$ $\displaystyle -\frac{\rho}{\epsilon_{0}}\ \ \ \ \ (5)$ $\displaystyle \nabla^{2}\mathbf{A}-\mu_{0}\epsilon_{0}\frac{\partial^{2}\mathbf{A}}{\partial t^{2}}-\nabla\left(\nabla\cdot\mathbf{A}+\mu_{0}\epsilon_{0}\frac{\partial V}{\partial t}\right)$ $\displaystyle =$ $\displaystyle -\mu_{0}\mathbf{J} \ \ \ \ \ (6)$

reduce to

 $\displaystyle \nabla^{2}V$ $\displaystyle =$ $\displaystyle -\frac{\rho}{\epsilon_{0}}\ \ \ \ \ (7)$ $\displaystyle \nabla^{2}\mathbf{A}-\mu_{0}\epsilon_{0}\frac{\partial^{2}\mathbf{A}}{\partial t^{2}}$ $\displaystyle =$ $\displaystyle -\mu_{0}\mathbf{J}+\mu_{0}\epsilon_{0}\nabla\left(\frac{\partial V}{\partial t}\right) \ \ \ \ \ (8)$

The scalar potential ${V}$ is therefore also a solution of Poisson’s equation, and once we have found it, we can, in principle, solve the second equation (which is a wave equation with a complicated driving term on the RHS) for the vector potential ${\mathbf{A}}$.

The Lorentz gauge sets

$\displaystyle \nabla\cdot\mathbf{A}=-\mu_{0}\epsilon_{0}\frac{\partial V}{\partial t} \ \ \ \ \ (9)$

which transforms 5 and 6 to

 $\displaystyle \nabla^{2}V-\mu_{0}\epsilon_{0}\frac{\partial^{2}V}{\partial t^{2}}$ $\displaystyle =$ $\displaystyle -\frac{\rho}{\epsilon_{0}}\ \ \ \ \ (10)$ $\displaystyle \nabla^{2}\mathbf{A}-\mu_{0}\epsilon_{0}\frac{\partial^{2}\mathbf{A}}{\partial t^{2}}$ $\displaystyle =$ $\displaystyle -\mu_{0}\mathbf{J} \ \ \ \ \ (11)$

In terms of the d’Alembertian operator

$\displaystyle \square^{2}\equiv\nabla^{2}-\mu_{0}\epsilon_{0}\frac{\partial^{2}}{\partial t^{2}} \ \ \ \ \ (12)$

we can write these two equations as

 $\displaystyle \square^{2}V$ $\displaystyle =$ $\displaystyle -\frac{\rho}{\epsilon_{0}}\ \ \ \ \ (13)$ $\displaystyle \square^{2}\mathbf{A}$ $\displaystyle =$ $\displaystyle -\mu_{0}\mathbf{J} \ \ \ \ \ (14)$

so that both potentials now become solutions of the wave equation with a driving term, but now ${V}$ and ${\mathbf{A}}$ are decoupled.

Example 1 For the potentials

$\displaystyle V=0 \ \ \ \ \ (15)$

$\displaystyle \mathbf{A}=\begin{cases} \frac{\mu_{0}k}{4c}\left(ct-\left|x\right|\right)^{2}\hat{\mathbf{z}} & \mbox{for }\left|x\right|ct \end{cases} \ \ \ \ \ (16)$

we have

$\displaystyle \nabla\cdot\mathbf{A}=0 \ \ \ \ \ (17)$

and because ${V=0}$,

$\displaystyle \nabla\cdot\mathbf{A}=-\mu_{0}\epsilon_{0}\frac{\partial V}{\partial t} \ \ \ \ \ (18)$

These potentials satisfy the conditions for both the Coulomb and Lorentz gauges.

Example 2 For the potentials

 $\displaystyle V\left(\mathbf{r},t\right)$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (19)$ $\displaystyle \mathbf{A}\left(\mathbf{r},t\right)$ $\displaystyle =$ $\displaystyle -\frac{1}{4\pi\epsilon_{0}}\frac{qt}{r^{2}}\hat{\mathbf{r}} \ \ \ \ \ (20)$

we have

$\displaystyle \nabla\cdot\mathbf{A}=-\frac{qt}{\epsilon_{0}}\delta_{3}(\mathbf{r}) \ \ \ \ \ (21)$

so it uses neither the Coulomb nor Lorentz gauge.

Example 3 For the potentials

 $\displaystyle V$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (22)$ $\displaystyle \mathbf{A}$ $\displaystyle =$ $\displaystyle A_{0}\sin\left(kx-\omega t\right)\hat{\mathbf{y}} \ \ \ \ \ (23)$

we again have ${\nabla\cdot\mathbf{A}=0}$ so it satisfies both the Coulomb and Lorentz gauges.

## Gauge transformations in electrodynamics

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 10.5.

The potentials ${V}$ and ${\mathbf{A}}$ that give rise to a particular configuration of fields ${\mathbf{E}}$ and ${\mathbf{B}}$ are not unique. For example, for a point charge at the origin, the usual potentials would be

 $\displaystyle V$ $\displaystyle =$ $\displaystyle \frac{q}{4\pi\epsilon_{0}r}\ \ \ \ \ (1)$ $\displaystyle \mathbf{A}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (2)$

but as we’ve seen, the potentials

 $\displaystyle V\left(\mathbf{r},t\right)$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (3)$ $\displaystyle \mathbf{A}\left(\mathbf{r},t\right)$ $\displaystyle =$ $\displaystyle -\frac{1}{4\pi\epsilon_{0}}\frac{qt}{r^{2}}\hat{\mathbf{r}} \ \ \ \ \ (4)$

give rise to the same fields. The question arises: what is the most general set of potentials that give a particular configuration of fields? That is, what changes can we make to ${V}$ and ${\mathbf{A}}$ without changing ${\mathbf{E}}$ and ${\mathbf{B}}$?

The fields can be calculated from the potentials via the equations

 $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle \nabla\times\mathbf{A}\ \ \ \ \ (5)$ $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle -\nabla V-\frac{\partial\mathbf{A}}{\partial t} \ \ \ \ \ (6)$

We can’t multiply ${V}$ and ${\mathbf{A}}$ by anything and leave the fields unchanged, but we might be able to add something. Suppose we add a vector function ${\boldsymbol{\alpha}\left(\mathbf{r},t\right)}$ to ${\mathbf{A}}$ and a scalar function ${\beta\left(\mathbf{r},t\right)}$ to ${V}$:

 $\displaystyle \mathbf{A}'$ $\displaystyle =$ $\displaystyle \mathbf{A}+\boldsymbol{\alpha}\ \ \ \ \ (7)$ $\displaystyle V'$ $\displaystyle =$ $\displaystyle V+\beta \ \ \ \ \ (8)$

To get the same magnetic field, we must have

 $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle \nabla\times\mathbf{A}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \nabla\times\mathbf{A}'\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \nabla\times\left(\mathbf{A}+\boldsymbol{\alpha}\right)\ \ \ \ \ (11)$ $\displaystyle \nabla\times\boldsymbol{\alpha}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (12)$

Since its curl is zero, we can write ${\boldsymbol{\alpha}}$ as the gradient of a scalar field:

$\displaystyle \boldsymbol{\alpha}\left(\mathbf{r},t\right)=\nabla\lambda'\left(\mathbf{r},t\right) \ \ \ \ \ (13)$

We also have to get the same electric field ${\mathbf{E}}$ so

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle -\nabla V-\frac{\partial\mathbf{A}}{\partial t}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\nabla\left(V+\beta\right)-\frac{\partial\left(\mathbf{A}+\boldsymbol{\alpha}\right)}{\partial t}\ \ \ \ \ (15)$ $\displaystyle \nabla\beta+\frac{\partial\boldsymbol{\alpha}}{\partial t}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (16)$

Combining these results, we get

$\displaystyle \nabla\left(\beta+\frac{\partial\lambda'}{\partial t}\right)=0 \ \ \ \ \ (17)$

The term in parentheses must not depend on position (since its gradient is identically zero everywhere), but it might depend on time. Define this term to be ${k\left(t\right)}$; then

$\displaystyle \beta=k\left(t\right)-\frac{\partial\lambda'}{\partial t} \ \ \ \ \ (18)$

If we define

$\displaystyle \lambda\equiv\lambda'-\int_{0}^{t}k\left(t'\right)dt' \ \ \ \ \ (19)$

then

$\displaystyle \beta=-\frac{\partial\lambda}{\partial t} \ \ \ \ \ (20)$

Since ${\lambda}$ and ${\lambda'}$ differ only by a function of time (and not of space), their gradients are equal, so we can use ${\lambda}$ in place of ${\lambda'}$ in 13 without changing ${\boldsymbol{\alpha}}$. Therefore we can change the potentials as follows, without changing the fields they produce:

 $\displaystyle \mathbf{A}'$ $\displaystyle =$ $\displaystyle \mathbf{A}+\nabla\lambda\ \ \ \ \ (21)$ $\displaystyle V'$ $\displaystyle =$ $\displaystyle V-\frac{\partial\lambda}{\partial t} \ \ \ \ \ (22)$

where ${\lambda=\lambda\left(\mathbf{r},t\right)}$ is an arbitrary scalar field. Such a change to the potentials is called a gauge transformation.

Example Earlier, we saw the unusual potentials 3 and 4 for a point charge at the origin. We can transform them using the gauge function

$\displaystyle \lambda=-\frac{qt}{4\pi\epsilon_{0}r} \ \ \ \ \ (23)$

We get

 $\displaystyle \nabla\lambda$ $\displaystyle =$ $\displaystyle \frac{qt}{4\pi\epsilon_{0}r^{2}}\hat{\mathbf{r}}\ \ \ \ \ (24)$ $\displaystyle \frac{\partial\lambda}{\partial t}$ $\displaystyle =$ $\displaystyle -\frac{q}{4\pi\epsilon_{0}r} \ \ \ \ \ (25)$

so the new potentials are

 $\displaystyle V$ $\displaystyle =$ $\displaystyle \frac{q}{4\pi\epsilon_{0}r}\ \ \ \ \ (26)$ $\displaystyle \mathbf{A}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (27)$

which is the more usual set of potentials 1 and 2 for a point charge.

## Potentials for an electromagnetic wave

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 10.4.

We can define an electromagnetic wave in terms of electric and magnetic potentials as follows. Using rectangular coordinates, let

 $\displaystyle V$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (1)$ $\displaystyle \mathbf{A}$ $\displaystyle =$ $\displaystyle A_{0}\sin\left(kx-\omega t\right)\hat{\mathbf{y}} \ \ \ \ \ (2)$

These potentials give rise to the fields

 $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle \nabla\times\mathbf{A}=A_{0}k\cos\left(kx-\omega t\right)\hat{\mathbf{z}}\ \ \ \ \ (3)$ $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle -\nabla V-\frac{\partial\mathbf{A}}{\partial t}=A_{0}\omega\cos\left(kx-\omega t\right)\hat{\mathbf{y}} \ \ \ \ \ (4)$

We can check that these fields satisfy Maxwell’s equations in vacuum. First,

$\displaystyle \nabla\cdot\mathbf{E}=\nabla\cdot\mathbf{B}=0 \ \ \ \ \ (5)$

For the curls, we have

 $\displaystyle \nabla\times\mathbf{E}$ $\displaystyle =$ $\displaystyle -A_{0}k\omega\sin\left(kx-\omega t\right)\hat{\mathbf{z}}=-\frac{\partial\mathbf{B}}{\partial t}\ \ \ \ \ (6)$ $\displaystyle \nabla\times\mathbf{B}$ $\displaystyle =$ $\displaystyle A_{0}k^{2}\sin\left(kx-\omega t\right)\hat{\mathbf{y}} \ \ \ \ \ (7)$

The second equation should be equal to ${\mu_{0}\epsilon_{0}\frac{\partial\mathbf{E}}{\partial t}=\frac{1}{c^{2}}\frac{\partial\mathbf{E}}{\partial t}}$ so

 $\displaystyle A_{0}k^{2}\sin\left(kx-\omega t\right)\hat{\mathbf{y}}$ $\displaystyle =$ $\displaystyle \frac{1}{c^{2}}\frac{\partial\mathbf{E}}{\partial t}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{c^{2}}A_{0}\omega^{2}\sin\left(kx-\omega t\right)\hat{\mathbf{y}} \ \ \ \ \ (9)$

which can be true only if

$\displaystyle k^{2}=\frac{\omega^{2}}{c^{2}} \ \ \ \ \ (10)$

which is the usual relation between wave number ${k}$ and angular frequency ${\omega}$.

## Potentials for a point charge

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 10.3.

As a simple (though unusual) example of specifying a system through electric and magnetic potentials suppose we have

 $\displaystyle V\left(\mathbf{r},t\right)$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (1)$ $\displaystyle \mathbf{A}\left(\mathbf{r},t\right)$ $\displaystyle =$ $\displaystyle -\frac{1}{4\pi\epsilon_{0}}\frac{qt}{r^{2}}\hat{\mathbf{r}} \ \ \ \ \ (2)$

These potentials give rise to the fields

 $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle \nabla\times\mathbf{A}=0\ \ \ \ \ (3)$ $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle -\nabla V-\frac{\partial\mathbf{A}}{\partial t}=\frac{1}{4\pi\epsilon_{0}}\frac{q}{r^{2}}\hat{\mathbf{r}} \ \ \ \ \ (4)$

The expression for ${\mathbf{E}}$ is just that of a point charge ${q}$ at the origin, while a zero magnetic field indicates that there is no current. If we want to be pedantic, we can also get these results from the potentials. For the charge density, we had

$\displaystyle \nabla^{2}V+\frac{\partial}{\partial t}\left(\nabla\cdot\mathbf{A}\right)=-\frac{\rho}{\epsilon_{0}} \ \ \ \ \ (5)$

Using the formula

$\displaystyle \nabla\cdot\left(\frac{1}{r^{2}}\hat{\mathbf{r}}\right)=4\pi\delta_{3}(\mathbf{r}) \ \ \ \ \ (6)$

we get

 $\displaystyle \nabla\cdot\mathbf{A}$ $\displaystyle =$ $\displaystyle -\frac{qt}{\epsilon_{0}}\delta_{3}(\mathbf{r})\ \ \ \ \ (7)$ $\displaystyle \rho$ $\displaystyle =$ $\displaystyle q\delta_{3}(\mathbf{r}) \ \ \ \ \ (8)$

For the current, we can use the equation

$\displaystyle \nabla\times\left(\nabla\times\mathbf{A}\right)=\mu_{0}\mathbf{J}-\mu_{0}\epsilon_{0}\nabla\frac{\partial V}{\partial t}-\mu_{0}\epsilon_{0}\frac{\partial^{2}\mathbf{A}}{\partial t^{2}} \ \ \ \ \ (9)$

and we see that all the terms not involving ${\mathbf{J}}$ are zero, so ${\mathbf{J}=0}$ as well. Thus this is a bizarre way of writing potentials for a point charge. This illustrates that the potentials giving rise to a particular charge and current distribution are not unique.

## Energy flow in time-dependent fields

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 10.2.

Suppose we have time-dependent electric and magnetic potentials given by

$\displaystyle V=0 \ \ \ \ \ (1)$

$\displaystyle \mathbf{A}=\begin{cases} \frac{\mu_{0}k}{4c}\left(ct-\left|x\right|\right)^{2}\hat{\mathbf{z}} & \mbox{for }\left|x\right|ct \end{cases} \ \ \ \ \ (2)$

where ${k}$ is a constant and ${c}$ is the speed of light.

We can get the fields by using our earlier formulas

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle -\nabla V-\frac{\partial\mathbf{A}}{\partial t}\ \ \ \ \ (3)$ $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle \nabla\times\mathbf{A} \ \ \ \ \ (4)$

We get, for ${\left|x\right|

 $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}k}{2}\left(ct-\left|x\right|\right)\ \ \ \ \ (5)$ $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle -\frac{\mu_{0}k}{4c}\frac{\partial}{\partial x}\left(ct-\left|x\right|\right)^{2}\hat{\mathbf{y}}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \pm\frac{\mu_{0}k}{2c}\left(ct-\left|x\right|\right)\hat{\mathbf{y}} \ \ \ \ \ (7)$

where in the last equation the + applies to ${x>0}$ and the ${-}$ to ${x<0}$. For ${\left|x\right|>ct}$, all fields are zero.

Griffiths shows in his example 10.1 that these fields satisfy Maxwell’s equations and arise from a surface current flowing in the ${x=0}$ plane given by

$\displaystyle \mathbf{K}=kt\hat{\mathbf{z}} \ \ \ \ \ (8)$

for ${t\ge0}$. That is, the current starts at ${t=0}$ and its effects travel outwards in the ${x}$ direction at the speed ${c}$. What we’ll do here is calculate the energy flow due to this current. We’ll consider a rectangular box of width ${w}$ in the ${y}$ direction and length ${\ell}$ in the ${z}$ direction extending from ${x=d}$ to ${x=d+h}$. At time ${t_{1}=d/c}$ the energy is zero inside the box since the fields from the current have only just reached the lower face of the box at ${x=d}$. At time ${t_{2}=\left(d+h\right)/c}$, the fields have reached the outer face of the box so the entire box is filled with fields. At this time the energy within the box can be calculated from the energy density

 $\displaystyle U$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\epsilon_{0}\int E^{2}d^{3}\mathbf{r}+\frac{1}{\mu_{0}}\int B^{2}d^{3}\mathbf{r}\right]\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{w\ell}{2}\left[\epsilon_{0}\left(\frac{\mu_{0}k}{2}\right)^{2}\int_{d}^{d+h}\left(d+h-x\right)^{2}dx+\frac{1}{\mu_{0}}\left(\frac{\mu_{0}k}{2c}\right)^{2}\int_{d}^{d+h}\left(d+h-x\right)^{2}dx\right]\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{w\ell\mu_{0}k^{2}}{4c^{2}}\int_{d}^{d+h}\left(d+h-x\right)^{2}dx\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{w\ell\mu_{0}k^{2}}{4c^{2}}\int_{0}^{h}v^{2}dv\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{w\ell\mu_{0}k^{2}h^{3}}{12c^{2}} \ \ \ \ \ (13)$

where in the second line we used ${\epsilon_{0}\mu_{0}=1/c^{2}}$ and in the third line we used the substitution ${v=d+h-x}$ in the integral.

Now we can calculate the flow of energy into the box as a function of time using the Poynting vector which gives the rate per unit area at which energy crosses a surface. We have for the surface ${x=d}$

 $\displaystyle \mathbf{S}$ $\displaystyle =$ $\displaystyle \frac{1}{\mu_{0}}\mathbf{E}\times\mathbf{B}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\mu_{0}}\frac{\mu_{0}^{2}k^{2}}{4c}\left(ct-d\right)^{2} \ \ \ \ \ (15)$

The total rate at which energy flows into the box via this face is thus

$\displaystyle \frac{dU}{dt}=\frac{w\ell\mu_{0}k^{2}}{4c}\left(ct-d\right)^{2} \ \ \ \ \ (16)$

and the total energy that flows into the box between times ${t_{1}=d/c}$ and ${t_{2}=\left(d+h\right)/c}$ is

 $\displaystyle U$ $\displaystyle =$ $\displaystyle \frac{w\ell\mu_{0}k^{2}}{4c}\int_{d/c}^{\left(d+h\right)/c}\left(ct-d\right)^{2}dt\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{w\ell\mu_{0}k^{2}}{4c^{2}}\int_{0}^{h}v^{2}dv\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{w\ell\mu_{0}k^{2}h^{3}}{12c^{2}} \ \ \ \ \ (19)$

which of course agrees with our earlier calculation of the energy in the box. [We used the substitution ${v=ct-d}$ in the first line.]

## Maxwell’s equations in terms of potentials

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 10.1.

 $\displaystyle \nabla\cdot\mathbf{E}$ $\displaystyle =$ $\displaystyle \frac{\rho}{\epsilon_{0}}\ \ \ \ \ (1)$ $\displaystyle \nabla\cdot\mathbf{B}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (2)$ $\displaystyle \nabla\times\mathbf{E}$ $\displaystyle =$ $\displaystyle -\frac{\partial\mathbf{B}}{\partial t}\ \ \ \ \ (3)$ $\displaystyle \nabla\times\mathbf{B}$ $\displaystyle =$ $\displaystyle \mu_{0}\mathbf{J}+\mu_{0}\epsilon_{0}\frac{\partial\mathbf{E}}{\partial t} \ \ \ \ \ (4)$

The are completely general, in that they describe the interactions between electric and magnetic fields and the charges and currents that generate them, in both the static (fixed charge and current densities) and dynamic (time-varying) cases. In the static case, the two time derivatives are zero, which means that ${\nabla\times\mathbf{E}=0}$. Since any vector field whose curl is zero can be represented as the gradient of a scalar field, we could write ${\mathbf{E}=-\nabla V}$ where ${V}$ is the potential. The presence of the term ${-\frac{\partial\mathbf{B}}{\partial t}}$, however, means that we can’t, in general, write ${\mathbf{E}}$ as the gradient of a scalar field in the dynamic case.

Because ${\nabla\cdot\mathbf{B}=0}$ we can write ${\mathbf{B}}$ as the curl of a vector field

$\displaystyle \mathbf{B}=\nabla\times\mathbf{A} \ \ \ \ \ (5)$

This is still true in the dynamic case. Substituting this into 3 we get

 $\displaystyle \nabla\times\mathbf{E}$ $\displaystyle =$ $\displaystyle -\frac{\partial\nabla\times\mathbf{A}}{\partial t}\ \ \ \ \ (6)$ $\displaystyle \nabla\times\left(\mathbf{E}+\frac{\partial\mathbf{A}}{\partial t}\right)$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (7)$

The LHS is the curl of a vector field that combines the electric field and the magnetic vector potential, and since it is zero even in the dynamic case, we can write this vector field as the gradient of a scalar field ${V}$. [We should be able to use the same symbol ${V}$ for the potential in the dynamic case, since if ${\frac{\partial\mathbf{A}}{\partial t}=0}$ we regain the static equations ${\nabla\times\mathbf{E}=0}$ and ${\mathbf{E}=-\nabla V}$.] That is, we can write

 $\displaystyle \mathbf{E}+\frac{\partial\mathbf{A}}{\partial t}$ $\displaystyle =$ $\displaystyle -\nabla V\ \ \ \ \ (8)$ $\displaystyle \mathbf{E}$ $\displaystyle =$ $\displaystyle -\nabla V-\frac{\partial\mathbf{A}}{\partial t} \ \ \ \ \ (9)$

Equations 5 and 9 effectively replace 2 and 3. Using 5 and 9 we can eliminate ${\mathbf{E}}$ and ${\mathbf{B}}$ from 1 and 4:

 $\displaystyle \nabla\cdot\mathbf{E}$ $\displaystyle =$ $\displaystyle -\nabla^{2}V-\frac{\partial}{\partial t}\left(\nabla\cdot\mathbf{A}\right)\ \ \ \ \ (10)$ $\displaystyle \nabla^{2}V+\frac{\partial}{\partial t}\left(\nabla\cdot\mathbf{A}\right)$ $\displaystyle =$ $\displaystyle -\frac{\rho}{\epsilon_{0}}\ \ \ \ \ (11)$ $\displaystyle \nabla\times\mathbf{B}$ $\displaystyle =$ $\displaystyle \nabla\times\left(\nabla\times\mathbf{A}\right)\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \mu_{0}\mathbf{J}+\mu_{0}\epsilon_{0}\frac{\partial\mathbf{E}}{\partial t}\ \ \ \ \ (13)$ $\displaystyle \nabla\times\left(\nabla\times\mathbf{A}\right)$ $\displaystyle =$ $\displaystyle \mu_{0}\mathbf{J}-\mu_{0}\epsilon_{0}\nabla\frac{\partial V}{\partial t}-\mu_{0}\epsilon_{0}\frac{\partial^{2}\mathbf{A}}{\partial t^{2}} \ \ \ \ \ (14)$

We can rearrange the last equation using the identity

$\displaystyle \nabla\times\left(\nabla\times\mathbf{A}\right)=\nabla\left(\nabla\cdot\mathbf{A}\right)-\nabla^{2}\mathbf{A} \ \ \ \ \ (15)$

so we get

 $\displaystyle \nabla\left(\nabla\cdot\mathbf{A}\right)-\nabla^{2}\mathbf{A}$ $\displaystyle =$ $\displaystyle \mu_{0}\mathbf{J}-\mu_{0}\epsilon_{0}\nabla\frac{\partial V}{\partial t}-\mu_{0}\epsilon_{0}\frac{\partial^{2}\mathbf{A}}{\partial t^{2}}\ \ \ \ \ (16)$ $\displaystyle \nabla^{2}\mathbf{A}-\mu_{0}\epsilon_{0}\frac{\partial^{2}\mathbf{A}}{\partial t^{2}}-\nabla\left(\nabla\cdot\mathbf{A}+\mu_{0}\epsilon_{0}\frac{\partial V}{\partial t}\right)$ $\displaystyle =$ $\displaystyle -\mu_{0}\mathbf{J} \ \ \ \ \ (17)$

To summarize:

$\displaystyle \boxed{\nabla^{2}V+\frac{\partial}{\partial t}\left(\nabla\cdot\mathbf{A}\right)=-\frac{\rho}{\epsilon_{0}}} \ \ \ \ \ (18)$

$\displaystyle \boxed{\nabla^{2}\mathbf{A}-\mu_{0}\epsilon_{0}\frac{\partial^{2}\mathbf{A}}{\partial t^{2}}-\nabla\left(\nabla\cdot\mathbf{A}+\mu_{0}\epsilon_{0}\frac{\partial V}{\partial t}\right)=-\mu_{0}\mathbf{J}} \ \ \ \ \ (19)$

These two equations comprise 4 equations (one from 18 and one for each vector component in 19) for four functions (${V}$ and ${\mathbf{A}}$), and their solution allows us to calculate both ${\mathbf{E}}$ and ${\mathbf{B}}$ by means of 9 and 5, so they form a complete replacement for the original set of 4 Maxwell equations that we started with. This reduces the number of equations to be solved from 6 (for the 3 components of each of ${\mathbf{E}}$ and ${\mathbf{B}}$) to 4.

We can write these equations in a more compact form using the d’Alembertian operator

$\displaystyle \square^{2}\equiv\nabla^{2}-\mu_{0}\epsilon_{0}\frac{\partial^{2}}{\partial t^{2}} \ \ \ \ \ (20)$

and defining the function

$\displaystyle L\equiv\nabla\cdot\mathbf{A}+\mu_{0}\epsilon_{0}\frac{\partial V}{\partial t} \ \ \ \ \ (21)$

We get

 $\displaystyle \square^{2}V$ $\displaystyle =$ $\displaystyle \nabla^{2}V-\mu_{0}\epsilon_{0}\frac{\partial^{2}V}{\partial t^{2}}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \nabla^{2}V-\frac{\partial L}{\partial t}+\frac{\partial}{\partial t}\left(\nabla\cdot\mathbf{A}\right)\ \ \ \ \ (23)$ $\displaystyle \square^{2}V+\frac{\partial L}{\partial t}$ $\displaystyle =$ $\displaystyle \nabla^{2}V+\frac{\partial}{\partial t}\left(\nabla\cdot\mathbf{A}\right)\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\rho}{\epsilon_{0}} \ \ \ \ \ (25)$

using 18 in the last line. Also,

 $\displaystyle \square^{2}\mathbf{A}$ $\displaystyle =$ $\displaystyle \nabla^{2}\mathbf{A}-\mu_{0}\epsilon_{0}\frac{\partial^{2}\mathbf{A}}{\partial t^{2}}\ \ \ \ \ (26)$ $\displaystyle \square^{2}\mathbf{A}-\nabla L$ $\displaystyle =$ $\displaystyle -\mu_{0}\mathbf{J} \ \ \ \ \ (27)$

using 19. In summary

$\displaystyle \boxed{\square^{2}V+\frac{\partial L}{\partial t}=-\frac{\rho}{\epsilon_{0}}} \ \ \ \ \ (28)$

$\displaystyle \boxed{\square^{2}\mathbf{A}-\nabla L=-\mu_{0}\mathbf{J}} \ \ \ \ \ (29)$