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I was interested in working through a textbook on statistical mechanics, but as I’m no longer active in a physics department at a university, I was wondering what textbooks are commonly used these days.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.

Fields due to a moving linear charge

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 10.19.

In his example 10.4, Griffiths works out the fields due to a point charge moving with constant velocity {\mathbf{v}}. They are

\displaystyle  \mathbf{E}\left(\mathbf{r},t\right)=\frac{q}{4\pi\epsilon_{0}}\frac{1-v^{2}/c^{2}}{\left(1-v^{2}\sin^{2}\theta/c^{2}\right)^{3/2}}\frac{\hat{\mathbf{R}}}{R^{2}} \ \ \ \ \ (1)

where

\displaystyle  \mathbf{R}\equiv\mathbf{r}-\mathbf{v}t \ \ \ \ \ (2)

is the vector from the particle’s present (not retarded) position to the observer (assuming the particle passes through the origin at {t=0}) and {\theta} is the angle between {\mathbf{R}} and {\mathbf{v}}. We can use this formula to rederive the equation for the electric field due to an infinite line charge with linear charge density {\lambda}. From electrostatics, we know the field is given by

\displaystyle  \mathbf{E}=\frac{\lambda}{2\pi\epsilon_{0}z} \ \ \ \ \ (3)

where {z} is the perpendicular distance from the line (wire). Let’s see if we can get the same result using the formula above.

The field due to a small segment of the wire of length {dx} at position is that due to a point charge {\lambda dx}. For an observation point at {\mathbf{r}}, the length of {\mathbf{R}} is

\displaystyle  R=\sqrt{z^{2}+x^{2}} \ \ \ \ \ (4)

and since the velocity is parallel to the wire, we have

\displaystyle  \sin\theta=\frac{z}{\sqrt{z^{2}+x^{2}}} \ \ \ \ \ (5)

Since {\mathbf{E}} is parallel to {\mathbf{R}}, by symmetry the components of {\mathbf{E}} parallel to the wire will cancel out, since there will be equal and opposite contributions from points {\pm x}. The perpendicular component is {\mathbf{E}\sin\theta} so the total field is

\displaystyle  \mathbf{E}=\frac{\lambda}{4\pi\epsilon_{0}}\left(1-\frac{v^{2}}{c^{2}}\right)\int_{-\infty}^{\infty}\frac{\sin\theta dx}{\left(1-v^{2}\sin^{2}\theta/c^{2}\right)^{3/2}}\frac{\hat{\mathbf{s}}}{\left(z^{2}+x^{2}\right)} \ \ \ \ \ (6)

where the {s} direction is radial. We can convert this to an integral over {\theta} by noting that

\displaystyle   \cos\theta d\theta \displaystyle  = \displaystyle  -\frac{xz}{\left(z^{2}+x^{2}\right)^{3/2}}dx\ \ \ \ \ (7)
\displaystyle  \frac{dx}{z^{2}+x^{2}} \displaystyle  = \displaystyle  -\frac{\sqrt{z^{2}+x^{2}}}{xz}\left(-\frac{xz}{\left(z^{2}+x^{2}\right)^{3/2}}dx\right)\ \ \ \ \ (8)
\displaystyle  \displaystyle  = \displaystyle  -\frac{\sqrt{z^{2}+x^{2}}}{xz}\cos\theta d\theta \ \ \ \ \ (9)

But

\displaystyle  \cos\theta=-\frac{x}{\sqrt{z^{2}+x^{2}}} \ \ \ \ \ (10)

so

\displaystyle   \frac{dx}{z^{2}+x^{2}} \displaystyle  = \displaystyle  \frac{d\theta}{z}\ \ \ \ \ (11)
\displaystyle  \frac{\lambda}{4\pi\epsilon_{0}}\left(1-\frac{v^{2}}{c^{2}}\right)\int_{-\infty}^{\infty}\frac{\sin\theta dx}{\left(1-v^{2}\sin^{2}\theta/c^{2}\right)^{3/2}}\frac{\hat{\mathbf{s}}}{\left(z^{2}+x^{2}\right)} \displaystyle  = \displaystyle  \frac{\lambda}{4\pi\epsilon_{0}z}\left(1-\frac{v^{2}}{c^{2}}\right)\int_{0}^{\pi}\frac{\sin\theta d\theta}{\left(1-v^{2}\sin^{2}\theta/c^{2}\right)^{3/2}}\hat{\mathbf{s}} \ \ \ \ \ (12)

The integral can be evaluated using Maple, and we get

\displaystyle   \int_{0}^{\pi}\frac{\sin\theta d\theta}{\left(1-v^{2}\sin^{2}\theta/c^{2}\right)^{3/2}}\hat{\mathbf{s}} \displaystyle  = \displaystyle  \left.\frac{-\cos\theta}{\left(1-\frac{v^{2}}{c^{2}}\right)\sqrt{1-\frac{v^{2}}{c^{2}}\sin^{2}\theta}}\hat{\mathbf{s}}\right|_{0}^{\pi}\ \ \ \ \ (13)
\displaystyle  \displaystyle  = \displaystyle  \frac{2}{1-\frac{v^{2}}{c^{2}}}\hat{\mathbf{s}} \ \ \ \ \ (14)

so we get back the correct field

\displaystyle  \mathbf{E}=\frac{\lambda}{2\pi\epsilon_{0}z}\hat{\mathbf{s}} \ \ \ \ \ (15)

The magnetic field of a point charge is given by Griffiths as

\displaystyle  \mathbf{B}\left(\mathbf{r},t\right)=\frac{1}{c^{2}}\mathbf{v}\times\mathbf{E}\left(\mathbf{r},t\right) \ \ \ \ \ (16)

Since {\mathbf{v}} is a constant, the total magnetic field can be found from the same integral as above. Its direction is given by {\hat{\mathbf{x}}\times\hat{\mathbf{s}}=\hat{\boldsymbol{\phi}}} which circles the wire in a direction given by the usual right-hand rule. Since {\lambda\mathbf{v}=\mathbf{I}} (the current), we get

\displaystyle   \mathbf{B} \displaystyle  = \displaystyle  \frac{I}{2\pi\epsilon_{0}c^{2}z}\hat{\boldsymbol{\phi}}\ \ \ \ \ (17)
\displaystyle  \displaystyle  = \displaystyle  \frac{\mu_{0}I}{2\pi z}\hat{\boldsymbol{\phi}} \ \ \ \ \ (18)

which agrees with the magnetostatic formula using Ampère’s law.

Fields of a point charge moving in one dimension

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problem 10.18.

Here’s a simple example of calculating the fields due to a moving point charge. Suppose we have a charge constrained to move on the {x} axis in the {+x} direction, so that {\mathbf{v}=v\hat{\mathbf{x}}}.

The fields are

\displaystyle   \mathbf{E}\left(\mathbf{r},t\right) \displaystyle  = \displaystyle  \frac{q}{4\pi\epsilon_{0}\left(\boldsymbol{\mathfrak{r}}\cdot\mathbf{u}\right)^{3}}\left[\left(c^{2}-v^{2}\right)\mathbf{u}+\boldsymbol{\mathfrak{r}}\times\left(\mathbf{u\times}\mathbf{a}\right)\right]\ \ \ \ \ (1)
\displaystyle  \mathbf{B}\left(\mathbf{r},t\right) \displaystyle  = \displaystyle  \frac{1}{c}\hat{\boldsymbol{\mathfrak{r}}}\times\mathbf{E}\left(\mathbf{r},t\right) \ \ \ \ \ (2)

where

\displaystyle   \boldsymbol{\mathfrak{r}} \displaystyle  \equiv \displaystyle  \mathbf{r}-\mathbf{w}\left(t_{r}\right)\ \ \ \ \ (3)
\displaystyle  \mathbf{u} \displaystyle  \equiv \displaystyle  c\hat{\boldsymbol{\mathfrak{r}}}-\mathbf{v} \ \ \ \ \ (4)

and {\mathbf{w}\left(t_{r}\right)} is the particle’s position at the retarded time. If the observer is to the right of the particle then

\displaystyle   \boldsymbol{\mathfrak{r}} \displaystyle  = \displaystyle  +\mathfrak{r}\hat{\mathbf{x}}\ \ \ \ \ (5)
\displaystyle  \mathbf{u} \displaystyle  = \displaystyle  \left(c-v\right)\hat{\mathbf{x}} \ \ \ \ \ (6)

Since the motion is constrained to the {x} axis, any velocity and acceleration must be parallel, so {\mathbf{u}\times\mathbf{a}=0} and we have

\displaystyle   \mathbf{E} \displaystyle  = \displaystyle  \frac{q}{4\pi\epsilon_{0}\left(\boldsymbol{\mathfrak{r}}\cdot\mathbf{u}\right)^{3}}\left(c^{2}-v^{2}\right)\mathbf{u}\ \ \ \ \ (7)
\displaystyle  \displaystyle  = \displaystyle  \frac{q\left(c^{2}-v^{2}\right)}{4\pi\epsilon_{0}\left(c-v\right)^{3}}\left(c-v\right)\hat{\mathbf{x}}\ \ \ \ \ (8)
\displaystyle  \displaystyle  = \displaystyle  \frac{q\left(c+v\right)}{4\pi\epsilon_{0}\left(c-v\right)}\hat{\mathbf{x}} \ \ \ \ \ (9)

Because {\hat{\boldsymbol{\mathfrak{r}}}} is parallel to {\mathbf{E}}, {\mathbf{B}=0} from 2.

If the observer is to the left of the charge, then

\displaystyle   \boldsymbol{\mathfrak{r}} \displaystyle  = \displaystyle  -\mathfrak{r}\hat{\mathbf{x}}\ \ \ \ \ (10)
\displaystyle  \mathbf{u} \displaystyle  = \displaystyle  -\left(c+v\right)\hat{\mathbf{x}}\ \ \ \ \ (11)
\displaystyle  \mathbf{E} \displaystyle  = \displaystyle  \frac{q}{4\pi\epsilon_{0}\left(\boldsymbol{\mathfrak{r}}\cdot\mathbf{u}\right)^{3}}\left(c^{2}-v^{2}\right)\mathbf{u}\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  -\frac{q\left(c^{2}-v^{2}\right)}{4\pi\epsilon_{0}\left(c+v\right)^{3}}\left(c+v\right)\hat{\mathbf{x}}\ \ \ \ \ (13)
\displaystyle  \displaystyle  = \displaystyle  -\frac{q\left(c-v\right)}{4\pi\epsilon_{0}\left(c+v\right)}\hat{\mathbf{x}}\ \ \ \ \ (14)
\displaystyle  \mathbf{B} \displaystyle  = \displaystyle  0 \ \ \ \ \ (15)

Fields of a moving point charge

References: Griffiths, David J. (2007), Introduction to Electrodynamics, 3rd Edition; Pearson Education – Problems 10.17.

We can use the Liénard-Wiechert potentials for a moving point charge to calculate the fields produced by the charge. 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)}\ \ \ \ \ (1)
\displaystyle  \mathbf{A}\left(\mathbf{r},t\right) \displaystyle  = \displaystyle  \frac{q}{4\pi\epsilon_{0}c}\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)} \ \ \ \ \ (2)

where {\mathbf{w}\left(t_{r}\right)} is the position of the particle at the retarded time {t_{r}} and {\mathbf{v}=d\mathbf{w}/dt_{r}} is the velocity at the same time.

I’ll use the shorthand variable

\displaystyle  \boldsymbol{\mathfrak{r}}\equiv\mathbf{r}-\mathbf{w}\left(t_{r}\right) \ \ \ \ \ (3)

in what follows (since I don’t know of any Latex symbol to match the script-r used by Griffiths). Thus the potentials are written as

\displaystyle   V\left(\mathbf{r},t\right) \displaystyle  = \displaystyle  \frac{qc}{4\pi\epsilon_{0}}\frac{1}{\left(c\mathfrak{r}-\boldsymbol{\mathfrak{r}}\cdot\mathbf{v}\right)}\ \ \ \ \ (4)
\displaystyle  \mathbf{A}\left(\mathbf{r},t\right) \displaystyle  = \displaystyle  \frac{q}{4\pi\epsilon_{0}c}\frac{\mathbf{v}}{\left(c\mathfrak{r}-\boldsymbol{\mathfrak{r}}\cdot\mathbf{v}\right)} \ \ \ \ \ (5)

Given the potentials, we can work out the fields using

\displaystyle   \mathbf{E} \displaystyle  = \displaystyle  -\nabla V-\frac{\partial\mathbf{A}}{\partial t}\ \ \ \ \ (6)
\displaystyle  \mathbf{B} \displaystyle  = \displaystyle  \nabla\times\mathbf{A} \ \ \ \ \ (7)

Because of the convoluted dependence of the various quantities on the observer’s location {\mathbf{r}} and time {t}, these derivatives get quite involved. Griffiths works out {\nabla V} in his section 10.3.2, with the result

\displaystyle  \nabla V=\frac{qc}{4\pi\epsilon_{0}\left(c\mathfrak{r}-\boldsymbol{\mathfrak{r}}\cdot\mathbf{v}\right)^{3}}\left[\left(c\mathfrak{r}-\boldsymbol{\mathfrak{r}}\cdot\mathbf{v}\right)\mathbf{v}-\left(c^{2}-v^{2}+\boldsymbol{\mathfrak{r}}\cdot\mathbf{a}\right)\boldsymbol{\mathfrak{r}}\right] \ \ \ \ \ (8)

so we’ll deal with {\frac{\partial\mathbf{A}}{\partial t}} here.

First, it’s useful to work out {\dot{t}_{r}\equiv dt_{r}/dt}. The retarded time {t_{r}} is defined implicitly by the equation

\displaystyle  \mathfrak{r}=\left|\mathbf{r}-\mathbf{w}\left(t_{r}\right)\right|=c\left(t-t_{r}\right) \ \ \ \ \ (9)

We start with

\displaystyle   \frac{\partial}{\partial t}\sqrt{\boldsymbol{\mathfrak{r}}\cdot\boldsymbol{\mathfrak{r}}} \displaystyle  = \displaystyle  \frac{1}{2\mathfrak{r}}\frac{\partial}{\partial t}\left(\boldsymbol{\mathfrak{r}}\cdot\boldsymbol{\mathfrak{r}}\right)\ \ \ \ \ (10)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{2\mathfrak{r}}\frac{\partial}{\partial t}\left(r^{2}-2\mathbf{r}\cdot\mathbf{w}+w^{2}\right)\ \ \ \ \ (11)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{2\mathfrak{r}}\left(-2\mathbf{r}\cdot\frac{d\mathbf{w}}{dt_{r}}\dot{t}_{r}+2\mathbf{w}\cdot\frac{d\mathbf{w}}{dt_{r}}\dot{t}_{r}\right)\ \ \ \ \ (12)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{\mathfrak{r}}\dot{t}_{r}\mathbf{v}\cdot\left(\mathbf{w}-\mathbf{r}\right)\ \ \ \ \ (13)
\displaystyle  \displaystyle  = \displaystyle  -\frac{1}{\mathfrak{r}}\dot{t}_{r}\mathbf{v}\cdot\boldsymbol{\mathfrak{r}} \ \ \ \ \ (14)

From the RHS of 9 we get

\displaystyle  c\frac{d}{dt}\left(t-t_{r}\right)=c\left(1-\dot{t}_{r}\right) \ \ \ \ \ (15)

Combining these two equations we get

\displaystyle   c\left(1-\dot{t}_{r}\right) \displaystyle  = \displaystyle  -\frac{1}{\mathfrak{r}}\dot{t}_{r}\mathbf{v}\cdot\boldsymbol{\mathfrak{r}}\ \ \ \ \ (16)
\displaystyle  \dot{t}_{r} \displaystyle  = \displaystyle  \frac{\mathfrak{r}c}{\mathfrak{r}c-\boldsymbol{\mathfrak{r}}\cdot\mathbf{v}}\ \ \ \ \ (17)
\displaystyle  \displaystyle  = \displaystyle  \frac{\mathfrak{r}c}{\boldsymbol{\mathfrak{r}}\cdot\mathbf{u}} \ \ \ \ \ (18)

where

\displaystyle  \mathbf{u}\equiv c\hat{\boldsymbol{\mathfrak{r}}}-\mathbf{v} \ \ \ \ \ (19)

Returning to 5 we now have

\displaystyle   \frac{4\pi c\epsilon_{0}}{q}\frac{\partial\mathbf{A}}{\partial t} \displaystyle  = \displaystyle  \frac{\partial\mathbf{v}}{\partial t}\frac{1}{\left(c\mathfrak{r}-\boldsymbol{\mathfrak{r}}\cdot\mathbf{v}\right)}-\mathbf{v}\frac{1}{\left(c\mathfrak{r}-\boldsymbol{\mathfrak{r}}\cdot\mathbf{v}\right)^{2}}\frac{\partial}{\partial t}\left(c\mathfrak{r}-\boldsymbol{\mathfrak{r}}\cdot\mathbf{v}\right)\ \ \ \ \ (20)
\displaystyle  \displaystyle  = \displaystyle  \frac{\partial\mathbf{v}}{\partial t_{r}}\dot{t_{r}}\frac{1}{\left(c\mathfrak{r}-\boldsymbol{\mathfrak{r}}\cdot\mathbf{v}\right)}-\mathbf{v}\frac{1}{\left(c\mathfrak{r}-\boldsymbol{\mathfrak{r}}\cdot\mathbf{v}\right)^{2}}c\frac{\partial\mathfrak{r}}{\partial t}-\frac{\partial\boldsymbol{\mathfrak{r}}}{\partial t}\cdot\mathbf{v}-\boldsymbol{\mathfrak{r}}\cdot\frac{\partial\mathbf{v}}{\partial t_{r}}\dot{t}_{r} \ \ \ \ \ (21)

The derivative

\displaystyle  \frac{\partial\mathbf{v}}{\partial t_{r}}\equiv\mathbf{a}\left(t_{r}\right) \ \ \ \ \ (22)

is the acceleration of the particle at the retarded time. The other two derivatives are, from 9

\displaystyle   \frac{\partial\mathfrak{r}}{\partial t} \displaystyle  = \displaystyle  c\left(1-\dot{t}_{r}\right)\ \ \ \ \ (23)
\displaystyle  \frac{\partial\boldsymbol{\mathfrak{r}}}{\partial t} \displaystyle  = \displaystyle  -\frac{d\mathbf{w}}{dt_{r}}\dot{t}_{r}=-\mathbf{v}\dot{t}_{r} \ \ \ \ \ (24)

Therefore

\displaystyle   \frac{4\pi\epsilon_{0}c}{q}\frac{\partial\mathbf{A}}{\partial t} \displaystyle  = \displaystyle  \dot{t}_{r}\left[\frac{\mathbf{a}}{\left(c\mathfrak{r}-\boldsymbol{\mathfrak{r}}\cdot\mathbf{v}\right)}-\frac{\mathbf{v}}{\left(c\mathfrak{r}-\boldsymbol{\mathfrak{r}}\cdot\mathbf{v}\right)^{2}}\left(-c^{2}+v^{2}-\boldsymbol{\mathfrak{r}}\cdot\mathbf{a}\right)\right]-\frac{c^{2}\mathbf{v}}{\left(c\mathfrak{r}-\boldsymbol{\mathfrak{r}}\cdot\mathbf{v}\right)^{2}}\ \ \ \ \ (25)
\displaystyle  \displaystyle  = \displaystyle  \frac{\mathfrak{r}c}{\mathfrak{r}c-\boldsymbol{\mathfrak{r}}\cdot\mathbf{v}}\left[\frac{\mathbf{a}}{\left(c\mathfrak{r}-\boldsymbol{\mathfrak{r}}\cdot\mathbf{v}\right)}-\frac{\mathbf{v}}{\left(c\mathfrak{r}-\boldsymbol{\mathfrak{r}}\cdot\mathbf{v}\right)^{2}}\left(-c^{2}+v^{2}-\boldsymbol{\mathfrak{r}}\cdot\mathbf{a}\right)\right]-\frac{c^{2}\mathbf{v}}{\left(c\mathfrak{r}-\boldsymbol{\mathfrak{r}}\cdot\mathbf{v}\right)^{2}}\ \ \ \ \ (26)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{\left(c\mathfrak{r}-\boldsymbol{\mathfrak{r}}\cdot\mathbf{v}\right)^{3}}\left[\mathfrak{r}c\left(\mathbf{a}\left(c\mathfrak{r}-\boldsymbol{\mathfrak{r}}\cdot\mathbf{v}\right)+\mathbf{v}\left(c^{2}-v^{2}+\boldsymbol{\mathfrak{r}}\cdot\mathbf{a}\right)\right)-c^{2}\mathbf{v}\left(c\mathfrak{r}-\boldsymbol{\mathfrak{r}}\cdot\mathbf{v}\right)\right]\ \ \ \ \ (27)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{\left(c\mathfrak{r}-\boldsymbol{\mathfrak{r}}\cdot\mathbf{v}\right)^{3}}\left[\left(\mathfrak{r}c\mathbf{a}-c^{2}\mathbf{v}\right)\left(c\mathfrak{r}-\boldsymbol{\mathfrak{r}}\cdot\mathbf{v}\right)+\mathfrak{r}c\mathbf{v}\left(c^{2}-v^{2}+\boldsymbol{\mathfrak{r}}\cdot\mathbf{a}\right)\right]\ \ \ \ \ (28)
\displaystyle  \frac{\partial\mathbf{A}}{\partial t} \displaystyle  = \displaystyle  \frac{qc}{4\pi\epsilon_{0}\left(c\mathfrak{r}-\boldsymbol{\mathfrak{r}}\cdot\mathbf{v}\right)^{3}}\left[\left(\frac{\mathfrak{r}\mathbf{a}}{c}-\mathbf{v}\right)\left(c\mathfrak{r}-\boldsymbol{\mathfrak{r}}\cdot\mathbf{v}\right)+\frac{\mathfrak{r}\mathbf{v}}{c}\left(c^{2}-v^{2}+\boldsymbol{\mathfrak{r}}\cdot\mathbf{a}\right)\right] \ \ \ \ \ (29)

Combining this with 8 and inserting into 6 gives the result quoted in Griffiths as equation 10.65:

\displaystyle  \mathbf{E}\left(\mathbf{r},t\right)=\frac{q}{4\pi\epsilon_{0}\left(\boldsymbol{\mathfrak{r}}\cdot\mathbf{u}\right)^{3}}\left[\left(c^{2}-v^{2}\right)\mathbf{u}+\boldsymbol{\mathfrak{r}}\times\left(\mathbf{u\times}\mathbf{a}\right)\right] \ \ \ \ \ (30)

For reference, the magnetic field is

\displaystyle   \mathbf{B}\left(\mathbf{r},t\right) \displaystyle  = \displaystyle  \nabla\times\mathbf{A}\ \ \ \ \ (31)
\displaystyle  \displaystyle  = \displaystyle  \frac{1}{c}\hat{\boldsymbol{\mathfrak{r}}}\times\mathbf{E}\left(\mathbf{r},t\right) \ \ \ \ \ (32)

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<t<+\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}:

Griffiths 10.15a

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) \displaystyle  = \displaystyle  \frac{\mu_{0}qc}{4\pi}\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)
\displaystyle  \displaystyle  = \displaystyle  \frac{\mathbf{v}}{c^{2}}V\left(\mathbf{r},t\right) \ \ \ \ \ (31)

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 \ \ \ \ \ (32)

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 \ \ \ \ \ (33)

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} \ \ \ \ \ (34)

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 \ \ \ \ \ (35)

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)}\ \ \ \ \ (36)
\displaystyle  \displaystyle  = \displaystyle  \frac{q}{4\pi\epsilon_{0}\left|\mathbf{r}-\mathbf{w}\left(t_{r}\right)\right|}\ \ \ \ \ (37)
\displaystyle  \displaystyle  = \displaystyle  \frac{q}{4\pi\epsilon_{0}\sqrt{z^{2}+a^{2}}}\ \ \ \ \ (38)
\displaystyle  \mathbf{A}\left(\mathbf{r},t\right) \displaystyle  = \displaystyle  \frac{qa\omega}{4\pi\epsilon_{0}c^{2}\sqrt{z^{2}+a^{2}}}\left(-\hat{\mathbf{x}}\sin\omega t_{r}+\hat{\mathbf{y}}\cos\omega t_{r}\right) \ \ \ \ \ (39)

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.

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