## What’s the most-used textbook on statistical mechanics?

I was interested in working through a textbook on statistical mechanics, but as I’m no longer active in a physics department at a university, I was wondering what textbooks are commonly used these days. I have the book Fundamentals of Statistical and Thermal Physics by F. Reif, but this was published in 1965 and doesn’t seem to have had any later editions. It still seems to be well-regarded and used in several places, but I was wondering if there are any more recent books that are also used in current stat-mech courses.

If you take a course in statistical mechanics, please leave me a comment to say what book(s) you use (and what you think of them!). Thanks.

## Liénard-Wiechert potential for a charge moving on a hyperbolic trajectory

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problems 10.16.

We can now work out the Liénard-Wiechert potentials for a point charge moving on a hyperbolic trajectory. We’re trying to find

$\displaystyle V\left(\mathbf{r},t\right)=\frac{qc}{4\pi\epsilon_{0}}\frac{1}{\left(c\left|\mathbf{r}-\mathbf{w}\left(t_{r}\right)\right|-\left(\mathbf{r}-\mathbf{w}\left(t_{r}\right)\right)\cdot\mathbf{v}\right)} \ \ \ \ \ (1)$

The motion is in one dimension, given by

$\displaystyle \mathbf{w}\left(t\right)=\sqrt{b^{2}+c^{2}t^{2}}\hat{\mathbf{x}} \ \ \ \ \ (2)$

It’s worth noting here that ${w\ge b>0}$ for all times, and the velocity is

$\displaystyle \mathbf{v}\left(t\right)=\frac{d\mathbf{w}}{dt}=\frac{c^{2}t}{\sqrt{b^{2}+c^{2}t^{2}}}\hat{\mathbf{x}} \ \ \ \ \ (3)$

which is negative for ${t<0}$ (when the particle is moving in from the right) and positive for ${t>0}$ (when the particle is moving back out again).

We’ll consider only observation points ${\mathbf{r}}$ that lie on the ${x}$ axis to the right of the particle’s location, so the separation is

$\displaystyle d\left(t\right)=r-w\left(t\right)=x-\sqrt{b^{2}+c^{2}t^{2}}>0 \ \ \ \ \ (4)$

Using this notation, the potential is

$\displaystyle V\left(\mathbf{r},t\right)=\frac{qc}{4\pi\epsilon_{0}}\frac{1}{\left(cd\left(t_{r}\right)-\mathbf{d}\left(t_{r}\right)\cdot\mathbf{v}\right)} \ \ \ \ \ (5)$

The retarded time is given by

 $\displaystyle d\left(t_{r}\right)$ $\displaystyle =$ $\displaystyle c\left(t-t_{r}\right)\ \ \ \ \ (6)$ $\displaystyle x-\sqrt{b^{2}+c^{2}t_{r}^{2}}$ $\displaystyle =$ $\displaystyle c\left(t-t_{r}\right) \ \ \ \ \ (7)$

We can solve this by isolating the square root on one side and then squaring both sides, to get

$\displaystyle t_{r}=\frac{x^{2}-2xct+c^{2}t^{2}-b^{2}}{2c\left(ct-x\right)} \ \ \ \ \ (8)$

Note that for ${t=0}$, as ${x\rightarrow\infty}$, ${t_{r}\rightarrow-\infty}$ which makes sense, since the further out on the ${x}$ axis we place the observer, the further back in time we need to go to get a signal from the particle. We can substitute this back into 2 to get

$\displaystyle w\left(t_{r}\right)=\frac{1}{2}\sqrt{\frac{\left(x^{2}-2xct+c^{2}t^{2}+b^{2}\right)^{2}}{\left(ct-x\right)^{2}}} \ \ \ \ \ (9)$

Although the operand of the square root is a perfect square, we need to be careful when taking the square root to ensure we get the correct sign. Since ${w>0}$ we can look at ${t=0}$ as before, at which point

$\displaystyle w\left(t_{r}\right)=\frac{1}{2}\sqrt{\frac{\left(x^{2}+b^{2}\right)^{2}}{\left(-x\right)^{2}}}>0 \ \ \ \ \ (10)$

so we need to take the negative root of the operand to get ${w>0}$. That is

$\displaystyle w\left(t_{r}\right)=\frac{x^{2}-2xct+c^{2}t^{2}+b^{2}}{2\left(x-tc\right)} \ \ \ \ \ (11)$

We can now calculate

 $\displaystyle d\left(t_{r}\right)$ $\displaystyle =$ $\displaystyle x-w\left(t_{r}\right)\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{x^{2}-c^{2}t^{2}-b^{2}}{2\left(x-ct\right)} \ \ \ \ \ (13)$

after simplifying.

Now we need to find ${\mathbf{v}\left(t_{r}\right)}$. Substituting 8 into 3

$\displaystyle \mathbf{v}\left(t_{r}\right)=-{\frac{c\left({b}^{2}-{x}^{2}+2\, xct-{c}^{2}{t}^{2}\right)}{2\left(ct-x\right)\sqrt{{b}^{2}+\,{\frac{\left({b}^{2}-{x}^{2}+2\, xct-{c}^{2}{t}^{2}\right)^{2}}{4\left(-x+ct\right)^{2}}}}}} \ \ \ \ \ (14)$

Simplifying the operand of the square root, we get

$\displaystyle \mathbf{v}\left(t_{r}\right)=-{\frac{c\left({b}^{2}-{x}^{2}+2\, xct-{c}^{2}{t}^{2}\right)}{2\left(ct-x\right)\sqrt{\frac{\left(x^{2}-2xct+c^{2}t^{2}+b^{2}\right)^{2}}{4\left(ct-x\right)^{2}}}}} \ \ \ \ \ (15)$

Again, we need to take the correct sign when taking the square root. For ${t=0}$ we get

$\displaystyle \mathbf{v}\left(t_{r}\right)=c\frac{b^{2}-x^{2}}{2x\sqrt{\frac{\left(x^{2}+b^{2}\right)^{2}}{4\left(-x\right)^{2}}}} \ \ \ \ \ (16)$

For large ${x}$, the signal comes from the particle in the distant past when it was moving to the left, so we should have ${v}$ negative in this case. This again requires taking the negative root, so we get

$\displaystyle \mathbf{v}\left(t_{r}\right)={\frac{c\left({b}^{2}-{x}^{2}+2\, xct-{c}^{2}{t}^{2}\right)}{x^{2}-2xct+c^{2}t^{2}+b^{2}}} \ \ \ \ \ (17)$

Putting it all together, we get

 $\displaystyle cd\left(t_{r}\right)-\mathbf{d}\left(t_{r}\right)\cdot\mathbf{v}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\,{\frac{\left(-{x}^{2}+{c}^{2}{t}^{2}+{b}^{2}\right)c}{-x+ct}}-\frac{1}{2}\,{\frac{\left(-{x}^{2}+{c}^{2}{t}^{2}+{b}^{2}\right)c\left({b}^{2}-\left(-x+ct\right)^{2}\right)}{\left(-x+ct\right)\left({b}^{2}+\left(-x+ct\right)^{2}\right)}}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle {\frac{\left(-x+ct\right)\left(-{x}^{2}+{c}^{2}{t}^{2}+{b}^{2}\right)c}{{x}^{2}-2\, xct+{c}^{2}{t}^{2}+{b}^{2}}} \ \ \ \ \ (19)$

We’ve used Maple to do some of the algebra. Therefore the potential is

 $\displaystyle V\left(\mathbf{r},t\right)$ $\displaystyle =$ $\displaystyle \frac{1}{4\pi\epsilon_{0}}\frac{q\left({x}^{2}-2\, xct+{c}^{2}{t}^{2}+{b}^{2}\right)}{\left(-x+ct\right)\left(-{x}^{2}+{c}^{2}{t}^{2}+{b}^{2}\right)}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{4\pi\epsilon_{0}}\frac{q\left(\left(x-ct\right)^{2}+{b}^{2}\right)}{\left(x-ct\right)\left({x}^{2}-{c}^{2}{t}^{2}-{b}^{2}\right)} \ \ \ \ \ (21)$

## Point charge in hyperbolic motion: visible and invisible points

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problems 10.15.

When we work out Liénard-Wiechert potentials for a moving point charge, we’ve been implicitly assuming that we can receive a signal from the charge from only one retarded time. That is, if the charge is moving on some trajectory ${\mathbf{w}\left(t\right)}$, there is only one point on that trajectory where a signal can be sent such that it will reach us at the time of observation. This assumption actually depends on special relativity with its postulate that nothing can travel faster than light. To see how this comes about, suppose there were, in fact, two different points ${P_{1}}$ and ${P_{2}}$ on the trajectory that could send signals that both arrived at the same time. Then if the distances to these two points are ${d_{1}}$ and ${d_{2}}$, the retarded times for these points are

 $\displaystyle t_{1}$ $\displaystyle =$ $\displaystyle t-\frac{d_{1}}{c}\ \ \ \ \ (1)$ $\displaystyle t_{2}$ $\displaystyle =$ $\displaystyle t-\frac{d_{2,}}{c} \ \ \ \ \ (2)$

or, in terms of the distances

$\displaystyle d_{1}-d_{2}=c\left(t_{2}-t_{1}\right) \ \ \ \ \ (3)$

If the particle gets closer to us between ${t_{1}}$ and ${t_{2}}$ then the difference ${d_{1}-d_{2}}$ represents the amount by which the distance to the particle has decreased. According to 3, the speed at which the particle must move to cover this distance is

$\displaystyle \frac{d_{1}-d_{2}}{t_{2}-t_{1}}=c \ \ \ \ \ (4)$

That is, the average speed in the radial direction has to be ${c}$. If the particle also has some transverse velocity on its trajectory, its total speed must be greater than ${c}$, which isn’t allowed. (If the velocity is entirely radial, then the particle would have to be moving at exactly ${c}$, which also isn’t allowed for any particle with rest mass.) Thus there can be at most one point where we receive a signal from a moving point charge.

It turns out that there are situations where a moving particle cannot be seen at all. For a simple example, suppose we have a particle moving along the ${x}$ axis with a trajectory

$\displaystyle \mathbf{w}\left(t\right)=\sqrt{b^{2}+c^{2}t^{2}}\hat{\mathbf{x}} \ \ \ \ \ (5)$

where ${-\infty.

It’s easiest to draw the trajectory on a spacetime diagram, with horizontal axis ${x}$ and vertical axis ${ct}$, as usual. In the following diagram, ${b=1}$:

The red curve (a hyperbola) is the trajectory, so we see that the particle approaches from the right until it reaches closest approach to the origin at ${x=1}$, then it moves away again. The grey lines are the asymptotes of the hyperbola. The yellow lines represent photons emitted by the particle at various points in its motion. As usual, they move upwards to the left and right parallel to the lines ${ct=\pm x}$. We can see from the diagram that no photons can ever reach points below the line ${ct=-x}$, so any observers in this region will be unaware of the particle’s existence (so the potentials are zero in this area).

For a stationary observer at location ${x}$, his world line is a vertical line travelling upwards, so he will first see the particle when he crosses the line ${ct=-x}$. Once the particle becomes visible, it will remain visible forever, since the particle is visible everywhere above the line ${ct=-x}$.

## Liénard-Wiechert potentials for a charge moving with constant velocity

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problems 10.14.

Griffiths shows in his example 10.3 that the Liénard-Wiechert potentials for a point charge ${q}$ moving at constant velocity ${\mathbf{v}}$ that passes through the origin at time ${t=0}$ are

 $\displaystyle V\left(\mathbf{r},t\right)$ $\displaystyle =$ $\displaystyle \frac{1}{4\pi\epsilon_{0}}\frac{qc}{\sqrt{\left(c^{2}t-\mathbf{r}\cdot\mathbf{v}\right)^{2}+\left(c^{2}-v^{2}\right)\left(r^{2}-c^{2}t^{2}\right)}}\ \ \ \ \ (1)$ $\displaystyle \mathbf{A}\left(\mathbf{r},t\right)$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{4\pi}\frac{qc\mathbf{v}}{\sqrt{\left(c^{2}t-\mathbf{r}\cdot\mathbf{v}\right)^{2}+\left(c^{2}-v^{2}\right)\left(r^{2}-c^{2}t^{2}\right)}} \ \ \ \ \ (2)$

These potentials can be expressed in a simpler form by defining the vector

$\displaystyle \mathbf{R}\equiv\mathbf{r}-\mathbf{v}t \ \ \ \ \ (3)$

We can eliminate ${\mathbf{r}}$ from 1 as follows.

 $\displaystyle \mathbf{R}\cdot\mathbf{v}$ $\displaystyle =$ $\displaystyle \mathbf{r}\cdot\mathbf{v}-v^{2}t\ \ \ \ \ (4)$ $\displaystyle \mathbf{r}\cdot\mathbf{v}$ $\displaystyle =$ $\displaystyle \mathbf{R}\cdot\mathbf{v}+v^{2}t\ \ \ \ \ (5)$ $\displaystyle R^{2}$ $\displaystyle =$ $\displaystyle r^{2}+v^{2}t^{2}-2\mathbf{r}\cdot\mathbf{v}t\ \ \ \ \ (6)$ $\displaystyle r^{2}$ $\displaystyle =$ $\displaystyle R^{2}-v^{2}t^{2}+2\mathbf{r}\cdot\mathbf{v}t\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle R^{2}+v^{2}t^{2}+2\mathbf{R}\cdot\mathbf{v}t\ \ \ \ \ (8)$ $\displaystyle \left(c^{2}t-\mathbf{r}\cdot\mathbf{v}\right)^{2}$ $\displaystyle =$ $\displaystyle \left(c^{2}t-\mathbf{R}\cdot\mathbf{v}-v^{2}t\right)^{2}\ \ \ \ \ (9)$ $\displaystyle \left(c^{2}-v^{2}\right)\left(r^{2}-c^{2}t^{2}\right)$ $\displaystyle =$ $\displaystyle \left(c^{2}-v^{2}\right)\left(R^{2}+v^{2}t^{2}+2\mathbf{R}\cdot\mathbf{v}t-c^{2}t^{2}\right) \ \ \ \ \ (10)$

Adding the last two RHSs together and cancelling terms, we get

$\displaystyle \left(c^{2}t-\mathbf{r}\cdot\mathbf{v}\right)^{2}+\left(c^{2}-v^{2}\right)\left(r^{2}-c^{2}t^{2}\right)=\left(\mathbf{R}\cdot\mathbf{v}\right)^{2}+\left(c^{2}-v^{2}\right)R^{2} \ \ \ \ \ (11)$

If ${\theta}$ is the angle between ${\mathbf{R}}$ and ${\mathbf{v}}$, then this becomes

 $\displaystyle \left(\mathbf{R}\cdot\mathbf{v}\right)^{2}+\left(c^{2}-v^{2}\right)R^{2}$ $\displaystyle =$ $\displaystyle R^{2}\left(c^{2}-v^{2}\left(1-\cos^{2}\theta\right)\right)\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle R^{2}c^{2}\left(1-\frac{v^{2}}{c^{2}}\sin^{2}\theta\right) \ \ \ \ \ (13)$

Inserting this back into 1 we get

$\displaystyle V\left(\mathbf{r},t\right)=\frac{1}{4\pi\epsilon_{0}}\frac{q}{R\sqrt{\left(1-\frac{v^{2}}{c^{2}}\sin^{2}\theta\right)}} \ \ \ \ \ (14)$

Note that ${R}$ and ${\theta}$ are both functions of time since they vary as the charge moves. For non-relativistic speeds, ${v\ll c}$ and the formula reduces to the Coulomb potential from electrostatics:

$\displaystyle V=\frac{q}{4\pi\epsilon_{0}R} \ \ \ \ \ (15)$

## Liénard-Wiechert potentials for a moving point charge

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problems 10.13.

We now look at the retarded potentials for a moving point charge ${q}$. The potentials are

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

where

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

and

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

The charge density of a point charge is represented by a delta function in space, so if the charge’s trajectory is given by ${\mathbf{w}\left(t'\right)}$ then

$\displaystyle \rho\left(\mathbf{r}',t'\right)=q\delta^{3}\left(\mathbf{r}'-\mathbf{w}\left(t'\right)\right) \ \ \ \ \ (7)$

To work out ${V}$, we need the charge density at the retarded time ${t_{r}}$, which we can write as the integral over time of the charge density multiplied by another delta function:

$\displaystyle \rho\left(\mathbf{r}',t_{r}\right)=q\delta^{3}\left(\mathbf{r}'-\mathbf{w}\left(t'\right)\right)\int\delta\left(t'-t_{r}\right)dt' \ \ \ \ \ (8)$

We need to keep straight the different times we’re using here. The time ${t}$ is the observation time, ${t'}$ is the integration variable and ${t_{r}}$ is the retarded time, which is the time at which the signal that we are receiving at time ${t}$ left the moving charge, which is

$\displaystyle t_{r}=t-\frac{\left|\mathbf{r}-\mathbf{r}'\right|}{c} \ \ \ \ \ (9)$

The potential can now be written as an integral over both time and space:

 $\displaystyle V\left(\mathbf{r},t\right)$ $\displaystyle =$ $\displaystyle \frac{q}{4\pi\epsilon_{0}}\int d^{3}\mathbf{r}'\frac{\delta^{3}\left(\mathbf{r}'-\mathbf{w}\left(t'\right)\right)}{\left|\mathbf{r}-\mathbf{r}'\right|}\int dt'\delta\left(t'-t_{r}\right)\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{q}{4\pi\epsilon_{0}}\int d^{3}\mathbf{r}'\frac{\delta^{3}\left(\mathbf{r}'-\mathbf{w}\left(t'\right)\right)}{\left|\mathbf{r}-\mathbf{r}'\right|}\int dt'\delta\left(t'-\left(t-\frac{\left|\mathbf{r}-\mathbf{r}'\right|}{c}\right)\right) \ \ \ \ \ (11)$

We can do the spatial integration which sets ${\mathbf{r}'=\mathbf{w}\left(t'\right)}$

$\displaystyle \frac{4\pi\epsilon_{0}}{q}V\left(\mathbf{r},t\right)=\int dt'\frac{1}{\left|\mathbf{r}-\mathbf{w}\left(t'\right)\right|}\delta\left(t'-\left(t-\frac{\left|\mathbf{r}-\mathbf{w}\left(t'\right)\right|}{c}\right)\right) \ \ \ \ \ (12)$

The trick now is to transform the argument of the delta function so we can do the integral. To do this, we need to work out ${\delta\left(f\left(x\right)\right)}$ for some function ${f\left(x\right)}$. To work this out, we use the substitution

 $\displaystyle u$ $\displaystyle =$ $\displaystyle f\left(x\right)\ \ \ \ \ (13)$ $\displaystyle du$ $\displaystyle =$ $\displaystyle f'\left(x\right)dx \ \ \ \ \ (14)$

so we get

 $\displaystyle \int\delta\left(f\left(x\right)\right)dx$ $\displaystyle =$ $\displaystyle \int\frac{\delta\left(u\right)}{\left|f'\left(x\right)\right|}du\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\left|f'\left(x\left(0\right)\right)\right|} \ \ \ \ \ (16)$

where we need to solve for ${x}$ as a function of ${u}$ from 13 and then find ${x\left(u=0\right)}$.

For our problem, we have

 $\displaystyle f\left(t'\right)$ $\displaystyle =$ $\displaystyle t'-\left(t-\frac{\left|\mathbf{r}-\mathbf{w}\left(t'\right)\right|}{c}\right)\ \ \ \ \ (17)$ $\displaystyle \frac{df}{dt'}$ $\displaystyle =$ $\displaystyle 1+\frac{1}{c}\frac{d}{dt'}\left|\mathbf{r}-\mathbf{w}\left(t'\right)\right| \ \ \ \ \ (18)$

Let’s select our coordinate axes so that, at time ${t'}$, ${\mathbf{w}=-w\hat{\mathbf{x}}}$ and ${\frac{d\mathbf{w}}{dt'}=+\beta c\hat{\mathbf{x}}}$ where ${0<\beta<1}$. That is, the charge is on the negative ${x}$ axis and is moving in the ${+x}$ direction with a speed ${\beta c}$. Then we have

 $\displaystyle \frac{d}{dt'}\left|\mathbf{r}-\mathbf{w}\left(t'\right)\right|^{2}$ $\displaystyle =$ $\displaystyle 2\left|\mathbf{r}-\mathbf{w}\left(t'\right)\right|\frac{d}{dt'}\left|\mathbf{r}-\mathbf{w}\left(t'\right)\right|\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{d}{dt'}\left[\left(\mathbf{r}-\mathbf{w}\left(t'\right)\right)\cdot\left(\mathbf{r}-\mathbf{w}\left(t'\right)\right)\right]\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -2\left(\mathbf{r}-\mathbf{w}\left(t'\right)\right)\cdot\frac{d\mathbf{w}}{dt'}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -2\beta c\left(\mathbf{r}-\mathbf{w}\left(t'\right)\right)\cdot\hat{\mathbf{x}}\ \ \ \ \ (22)$ $\displaystyle \frac{d}{dt'}\left|\mathbf{r}-\mathbf{w}\left(t'\right)\right|$ $\displaystyle =$ $\displaystyle -\frac{\beta c\left(\mathbf{r}-\mathbf{w}\left(t'\right)\right)\cdot\hat{\mathbf{x}}}{\left|\mathbf{r}-\mathbf{w}\left(t'\right)\right|}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\left(\mathbf{r}-\mathbf{w}\left(t'\right)\right)\cdot\mathbf{v}}{\left|\mathbf{r}-\mathbf{w}\left(t'\right)\right|} \ \ \ \ \ (24)$

where in the last line ${\mathbf{v}\equiv\beta c\hat{\mathbf{x}}}$ is the velocity of the charge. Therefore

$\displaystyle \frac{df}{dt'}=1-\frac{\left(\mathbf{r}-\mathbf{w}\left(t'\right)\right)\cdot\mathbf{v}}{c\left|\mathbf{r}-\mathbf{w}\left(t'\right)\right|} \ \ \ \ \ (25)$

Returning to 12 we have ${f\left(t'\right)=0}$ when ${t'=t_{r}}$ so we can do the integral over the delta function to get

 $\displaystyle V\left(\mathbf{r},t\right)$ $\displaystyle =$ $\displaystyle \frac{q}{4\pi\epsilon_{0}}\int dt'\frac{1}{\left|\mathbf{r}-\mathbf{w}\left(t'\right)\right|}\delta\left(t'-\left(t-\frac{\left|\mathbf{r}-\mathbf{w}\left(t'\right)\right|}{c}\right)\right)\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{q}{4\pi\epsilon_{0}}\frac{1}{\left|\mathbf{r}-\mathbf{w}\left(t_{r}\right)\right|\left(1-\frac{\left(\mathbf{r}-\mathbf{w}\left(t_{r}\right)\right)\cdot\mathbf{v}}{c\left|\mathbf{r}-\mathbf{w}\left(t_{r}\right)\right|}\right)}\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{qc}{4\pi\epsilon_{0}}\frac{1}{\left(c\left|\mathbf{r}-\mathbf{w}\left(t_{r}\right)\right|-\left(\mathbf{r}-\mathbf{w}\left(t_{r}\right)\right)\cdot\mathbf{v}\right)} \ \ \ \ \ (28)$

The current density for a moving point charge is just

$\displaystyle \mathbf{J}=\rho\mathbf{v} \ \ \ \ \ (29)$

so the derivation of ${\mathbf{A}}$ from 2 follows exactly the same path and we get

$\displaystyle \mathbf{A}\left(\mathbf{r},t\right)=\frac{qc}{4\pi\epsilon_{0}}\frac{\mathbf{v}}{\left(c\left|\mathbf{r}-\mathbf{w}\left(t_{r}\right)\right|-\left(\mathbf{r}-\mathbf{w}\left(t_{r}\right)\right)\cdot\mathbf{v}\right)} \ \ \ \ \ (30)$

These are the Liénard-Wiechert potentials for a moving point charge.

Griffiths gives a heuristic argument as to why the extra term ${-\left(\mathbf{r}-\mathbf{w}\left(t_{r}\right)\right)\cdot\mathbf{v}}$ turns up in the denominator. The effect arises because of the perception of size of a moving object. If we see a metre stick coming directly at us with a speed ${v}$, we will perceive it to be slightly longer than it actually is, since the light from the far end of the stick left the stick when it was further away from us than the light from the near end. Although this argument does give the right answer and the argument doesn’t depend ultimately on the size of the object approaching, I find the argument unsatisfying when applied to a point object, since I’d still expect that the effect should disappear in that case. The argument above, using delta functions, is a lot more abstract than the moving metre stick argument, but at least it shows rigorously how the effect arises.

Example We have a point charge ${q}$ moving in a circle of radius ${a}$ in the ${xy}$ plane at constant angular speed ${\omega}$ so that its position is given by

$\displaystyle \mathbf{w}\left(t\right)=a\hat{\mathbf{x}}\cos\omega t+a\hat{\mathbf{y}}\sin\omega t \ \ \ \ \ (31)$

The velocity is

$\displaystyle \mathbf{v}\left(t\right)=\frac{d\mathbf{w}}{dt}=-a\omega\hat{\mathbf{x}}\sin\omega t+a\omega\hat{\mathbf{y}}\cos\omega t \ \ \ \ \ (32)$

For an observation point ${\mathbf{r}=z\hat{\mathbf{z}}}$ the retarded time is

$\displaystyle t_{r}=t-\frac{\left|\mathbf{r}-\mathbf{w}\left(t'\right)\right|}{c}=t-\frac{\sqrt{z^{2}+a^{2}}}{c} \ \ \ \ \ (33)$

This is independent of the charge’s position, since it’s always at the same distance from a point on the ${z}$ axis. Also by direct calculation

$\displaystyle \left(\mathbf{r}-\mathbf{w}\left(t_{r}\right)\right)\cdot\mathbf{v}=0 \ \ \ \ \ (34)$

so the potentials are

 $\displaystyle V\left(\mathbf{r},t\right)$ $\displaystyle =$ $\displaystyle \frac{qc}{4\pi\epsilon_{0}}\frac{1}{\left(c\left|\mathbf{r}-\mathbf{w}\left(t_{r}\right)\right|-\left(\mathbf{r}-\mathbf{w}\left(t_{r}\right)\right)\cdot\mathbf{v}\right)}\ \ \ \ \ (35)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{q}{4\pi\epsilon_{0}\left|\mathbf{r}-\mathbf{w}\left(t_{r}\right)\right|}\ \ \ \ \ (36)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{q}{4\pi\epsilon_{0}\sqrt{z^{2}+a^{2}}}\ \ \ \ \ (37)$ $\displaystyle \mathbf{A}\left(\mathbf{r},t\right)$ $\displaystyle =$ $\displaystyle \frac{qa\omega}{4\pi\epsilon_{0}\sqrt{z^{2}+a^{2}}}\left(-\hat{\mathbf{x}}\sin\omega t_{r}+\hat{\mathbf{y}}\cos\omega t_{r}\right) \ \ \ \ \ (38)$

## Jefimenko’s equation for time-dependent magnetic field

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problems 10.12.

Using the retarded potentials, we can find a time-dependent expression for the magnetic field ${\mathbf{B}}$ to complement the equation we got earlier for the electric field. The potentials are

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

where

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

and

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

The magnetic field is

 $\displaystyle \mathbf{B}$ $\displaystyle =$ $\displaystyle \nabla\times\mathbf{A}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{4\pi}\int\left[\frac{\left(\nabla\times\mathbf{J}\left(\mathbf{r}',t_{r}\right)\right)}{d}-\mathbf{J}\left(\mathbf{r}',t_{r}\right)\times\nabla\left(\frac{1}{d}\right)\right]d^{3}\mathbf{r}' \ \ \ \ \ (8)$

where we’ve used the identity

$\displaystyle \nabla\times\left(f\mathbf{A}\right)=f\left(\nabla\times\mathbf{A}\right)-\mathbf{A}\times\nabla f \ \ \ \ \ (9)$

We have

$\displaystyle \nabla\left(\frac{1}{d}\right)=-\frac{\hat{\mathbf{d}}}{d^{2}} \ \ \ \ \ (10)$

To get ${\nabla\times\mathbf{J}}$, it’s easiest to look at a single component of the curl, say the ${x}$ component:

$\displaystyle \left(\nabla\times\mathbf{J}\right)_{x}=\frac{\partial J_{z}}{\partial y}-\frac{\partial J_{y}}{\partial z} \ \ \ \ \ (11)$

Using the chain rule

 $\displaystyle \frac{\partial J_{z}}{\partial y}$ $\displaystyle =$ $\displaystyle \frac{\partial J_{z}}{\partial t_{r}}\frac{\partial t_{r}}{\partial y}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{c}\frac{\partial J_{z}}{\partial t}\frac{\partial d}{\partial y}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{c}\dot{J}_{z}\left(\nabla d\right)_{y} \ \ \ \ \ (14)$

Therefore

 $\displaystyle \left(\nabla\times\mathbf{J}\right)_{x}$ $\displaystyle =$ $\displaystyle -\frac{1}{c}\left(\dot{J}_{z}\left(\nabla d\right)_{y}-\dot{J}_{y}\left(\nabla d\right)_{z}\right)\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{c}\left(\dot{\mathbf{J}}\times\nabla d\right)_{x} \ \ \ \ \ (16)$

The other two components work out the same way so we have

 $\displaystyle \nabla\times\mathbf{J}$ $\displaystyle =$ $\displaystyle \frac{1}{c}\dot{\mathbf{J}}\times\nabla d\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{c}\dot{\mathbf{J}}\times\hat{\mathbf{d}} \ \ \ \ \ (18)$

Putting this back into 8 we get

$\displaystyle \mathbf{B}\left(\mathbf{r},t\right)=\frac{\mu_{0}}{4\pi}\int\left[\frac{\dot{\mathbf{J}}\left(\mathbf{r}',t_{r}\right)\times\hat{\mathbf{d}}}{cd}+\mathbf{J}\left(\mathbf{r}',t_{r}\right)\times\frac{\hat{\mathbf{d}}}{d^{2}}\right]d^{3}\mathbf{r}' \ \ \ \ \ (19)$

This is Jefimenko’s equation for the magnetic field. For steady currents, ${\dot{\mathbf{J}}=0}$ and the dependence on ${t_{r}}$ disappears, so we’re left with the Biot-Savart law:

$\displaystyle \mathbf{B}\left(\mathbf{r}\right)=\frac{\mu_{0}}{4\pi}\int\mathbf{J}\left(\mathbf{r}'\right)\times\frac{\hat{\mathbf{d}}}{d^{2}}d^{3}\mathbf{r}' \ \ \ \ \ (20)$

In the special case where the current changes slowly enough that we can approximate it by a first-order Taylor series:

 $\displaystyle \mathbf{J}\left(\mathbf{r},t_{r}\right)$ $\displaystyle =$ $\displaystyle \mathbf{J}\left(\mathbf{r},t\right)+\left(t_{r}-t\right)\dot{\mathbf{J}}\left(\mathbf{r},t\right)\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \mathbf{J}\left(\mathbf{r},t\right)-\frac{d}{c}\dot{\mathbf{J}}\left(\mathbf{r},t\right)\ \ \ \ \ (22)$ $\displaystyle \frac{\partial\mathbf{J}}{\partial t_{r}}$ $\displaystyle =$ $\displaystyle \dot{\mathbf{J}}\left(\mathbf{r},t\right) \ \ \ \ \ (23)$

we get from 19

 $\displaystyle \mathbf{B}\left(\mathbf{r},t\right)$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{4\pi}\int\left[\frac{\dot{\mathbf{J}}\left(\mathbf{r},t\right)}{cd}+\frac{1}{d^{2}}\left(\mathbf{J}\left(\mathbf{r},t\right)-\frac{d}{c}\dot{\mathbf{J}}\left(\mathbf{r},t\right)\right)\right]\times\hat{\mathbf{d}}d^{3}\mathbf{r}'\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\mu_{0}}{4\pi}\int\frac{\mathbf{J}\left(\mathbf{r},t\right)\times\hat{\mathbf{d}}}{d^{2}}d^{3}\mathbf{r}' \ \ \ \ \ (25)$

That is, for slowly varying currents, to first order, Jefimenko’s equation gives the Biot-Savart law where all currents are evaluated at the current time. We’re essentially assuming that the travel time for signals from all locations of the current to the observation point is zero. This was the quasistatic approximation.

## Jefimenko’s equation for time-dependent electric field

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 10.11.

Using the retarded potentials, we can find a time-dependent expression for the electric field ${\mathbf{E}}$. The potentials are

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

where

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

and

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

The expression for ${\mathbf{E}}$ is given by

$\displaystyle \mathbf{E}=-\nabla V-\frac{\partial\mathbf{A}}{\partial t} \ \ \ \ \ (7)$

The derivatives are complicated by the fact that the integrands in 1 and 2 depend on ${\mathbf{r}}$ both via ${t_{r}}$ and ${d}$. We get (note that ${\nabla}$ indicates derivatives with respect to components of ${\mathbf{r}}$ only, not ${\mathbf{r}'}$):

 $\displaystyle \nabla V$ $\displaystyle =$ $\displaystyle \frac{1}{4\pi\epsilon_{0}}\int\nabla\left(\frac{\rho\left(\mathbf{r}',t_{r}\right)}{d}\right)d^{3}\mathbf{r}'\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{4\pi\epsilon_{0}}\int\rho\nabla\left(\frac{1}{d}\right)+\frac{1}{d}\nabla\rho d^{3}\mathbf{r}' \ \ \ \ \ (9)$

Using the chain rule

 $\displaystyle \nabla\rho$ $\displaystyle =$ $\displaystyle \frac{\partial\rho}{\partial t_{r}}\nabla t_{r}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{c}\frac{\partial\rho}{\partial t}\nabla d\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\dot{\rho}}{c}\nabla d \ \ \ \ \ (12)$

since ${\frac{\partial\rho}{\partial t_{r}}=\frac{\partial\rho}{\partial t}}$ because of 3. By direct calculation we have

 $\displaystyle \nabla d$ $\displaystyle =$ $\displaystyle \hat{\mathbf{d}}\ \ \ \ \ (13)$ $\displaystyle \nabla\left(\frac{1}{d}\right)$ $\displaystyle =$ $\displaystyle -\frac{\hat{\mathbf{d}}}{d^{2}} \ \ \ \ \ (14)$

Plugging everything into 9 we get

$\displaystyle \nabla V\left(\mathbf{r},t\right)=-\frac{1}{4\pi\epsilon_{0}}\int\left(\frac{\rho}{d^{2}}+\frac{\dot{\rho}}{cd}\right)\hat{\mathbf{d}}d^{3}\mathbf{r}' \ \ \ \ \ (15)$

The second term in 7 is just

$\displaystyle \frac{\partial\mathbf{A}}{\partial t}=\frac{\mu_{0}}{4\pi}\int\frac{\dot{\mathbf{J}}\left(\mathbf{r}',t_{r}\right)}{d}d^{3}\mathbf{r}' \ \ \ \ \ (16)$

so the time-dependent field is (using ${\mu_{0}\epsilon_{0}=1/c^{2}}$):

 $\displaystyle \mathbf{E}\left(\mathbf{r},t\right)$ $\displaystyle =$ $\displaystyle \frac{1}{4\pi\epsilon_{0}}\int\left(\frac{\rho}{d^{2}}+\frac{\dot{\rho}}{cd}\right)\hat{\mathbf{d}}d^{3}\mathbf{r}'-\frac{\mu_{0}}{4\pi}\int\frac{\dot{\mathbf{J}}\left(\mathbf{r}',t_{r}\right)}{d}d^{3}\mathbf{r}'\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{4\pi\epsilon_{0}}\int\left(\frac{\rho\left(\mathbf{r}',t_{r}\right)}{d^{2}}\hat{\mathbf{d}}+\frac{\dot{\rho}\left(\mathbf{r}',t_{r}\right)}{cd}\hat{\mathbf{d}}-\frac{\dot{\mathbf{J}}\left(\mathbf{r}',t_{r}\right)}{c^{2}d}\right)d^{3}\mathbf{r}' \ \ \ \ \ (18)$

This is Jefimenko’s equation for the electric field. In the static case, all time derivatives are zero and there is no dependence on ${t_{r}}$ so we get

$\displaystyle \mathbf{E}\left(\mathbf{r}\right)=\frac{1}{4\pi\epsilon_{0}}\int\frac{\rho\left(\mathbf{r}\right)}{d^{2}}\hat{\mathbf{d}}d^{3}\mathbf{r}' \ \ \ \ \ (19)$

which is just Coulomb’s law from electrostatics.

A special case is that of constant current but varying charge. In that case

$\displaystyle \rho\left(\mathbf{r},t\right)=\dot{\rho}\left(\mathbf{r},0\right)t+\rho\left(\mathbf{r},0\right) \ \ \ \ \ (20)$

where

$\displaystyle \dot{\rho}\left(\mathbf{r},0\right)=-\nabla\cdot\mathbf{J}\left(\mathbf{r}\right) \ \ \ \ \ (21)$

In this case, the integrand of 18 is

 $\displaystyle \frac{\rho\left(\mathbf{r}',t_{r}\right)}{d^{2}}\hat{\mathbf{d}}+\frac{\dot{\rho}\left(\mathbf{r}',t_{r}\right)}{cd}\hat{\mathbf{d}}-\frac{\dot{\mathbf{J}}\left(\mathbf{r}',t_{r}\right)}{c^{2}d}$ $\displaystyle =$ $\displaystyle \frac{\dot{\rho}\left(\mathbf{r}',0\right)\left(t-\frac{d}{c}\right)+\rho\left(\mathbf{r}',0\right)}{d^{2}}\hat{\mathbf{d}}+\frac{\dot{\rho}\left(\mathbf{r}',0\right)}{cd}\hat{\mathbf{d}}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\dot{\rho}\left(\mathbf{r}',0\right)t+\rho\left(\mathbf{r}',0\right)}{d^{2}}\hat{\mathbf{d}}\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\rho\left(\mathbf{r}',t\right)}{d}\hat{\mathbf{d}} \ \ \ \ \ (24)$

so the field is

$\displaystyle \mathbf{E}\left(\mathbf{r},t\right)=\frac{1}{4\pi\epsilon_{0}}\int\frac{\rho\left(\mathbf{r},t\right)}{d^{2}}\hat{\mathbf{d}}d^{3}\mathbf{r}' \ \ \ \ \ (25)$

That is, Coulomb’s law is valid with the charge density evaluated at the current (non-retarded) time.

## Retarded potential of a wire loop

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 10.10.

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.