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## PHY450HS1. Relativistic electrodynamics. Problem Set 5.

Posted by peeterjoot on April 7, 2011

# Problem 1. Sinusoidal current density on an infinite flat conducting sheet.

## Statement

An infinitely thin flat conducting surface lying in the $x-z$ plane carries a surface current density:

\begin{aligned}\boldsymbol{\kappa} = \mathbf{e}_3 \theta(t) \kappa_0 \sin\omega t\end{aligned} \hspace{\stretch{1}}(1.1)

Here $\mathbf{e}_3$ is a unit vector in the $z$ direction, $\kappa_0$ is the peak value of the current density, and $\theta(t)$ is the theta function: $\theta(t 0) = 1$.

\begin{enumerate}
\item Write down the equations determining the electromagnetic potentials. Specify which gauge you choose to work in.
\item Find the electromagnetic potentials outside the plane.
\item Find the electric and magnetic fields outside the plane.
\item Give a physical interpretation of the results of the previous section. Do they agree with your qualitative expectations?
\item Find the direction and magnitude of the energy flux outside the plane.
\item Consider a point at some distance from the plane. Sketch the intensity of the electromagnetic field near this point as a function of time. Explain physically.
\item Consider now a point near the plane. Are the electric and magnetic fields you found continuous across the conducting plane? Explain.
\end{enumerate}

## 1-2. Determining the electromagnetic potentials.

Augmenting the surface current density with a delta function we can form the current density for the system

\begin{aligned}\mathbf{J} = \delta(y) \boldsymbol{\kappa} = \mathbf{e}_3 \theta(t) \delta(y) \kappa_0 \sin\omega t.\end{aligned} \hspace{\stretch{1}}(1.2)

With only a current distribution specified use of the Coulomb gauge allows for setting the scalar potential on the surface equal to zero, so that we have

\begin{aligned}\square \mathbf{A} &= \frac{4 \pi \mathbf{J}}{c} \\ \mathbf{E} &= - \frac{1}{{c}} \frac{\partial {\mathbf{A}}}{\partial {t}} \\ \mathbf{B} &= \boldsymbol{\nabla} \times \mathbf{B}\end{aligned} \hspace{\stretch{1}}(1.3)

Utilizing our Green’s function

\begin{aligned}G(\mathbf{x}, t) = \frac{\delta(t - {\left\lvert{\mathbf{x}}\right\rvert}/c)}{4 \pi {\left\lvert{\mathbf{x}}\right\rvert}} = \delta^3(\mathbf{x})\delta(t),\end{aligned} \hspace{\stretch{1}}(1.6)

we can invert our vector potential equation, solving for $\mathbf{A}$

\begin{aligned}\mathbf{A}(\mathbf{x}, t) &= \int d^3 \mathbf{x}' dt' \square_{\mathbf{x}', t'} G(\mathbf{x} - \mathbf{x}', t - t') \mathbf{A}(\mathbf{x}', t') \\ &= \int d^3 \mathbf{x}' dt' G(\mathbf{x} - \mathbf{x}', t - t') \frac{4 \pi \mathbf{J}(\mathbf{x}', t')}{c} \\ &= \int d^3 \mathbf{x}' dt' \frac{\delta(t - t' - {\left\lvert{\mathbf{x} -\mathbf{x}'}\right\rvert}/c)}{4 \pi {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}\frac{4 \pi \mathbf{J}(\mathbf{x}', t')}{c} \\ &= \int d^3 \mathbf{x}' \frac{\mathbf{J}(\mathbf{x}', t - {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}/c}{c {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}} \\ &= \frac{1}{{c}} \int dx' dy' dz'\mathbf{e}_3 \theta(t - {\left\lvert{\mathbf{x} -\mathbf{x}'}\right\rvert}/c) \delta(y) \kappa_0 \sin(\omega(t - {\left\lvert{\mathbf{x} -\mathbf{x}'}\right\rvert}/c))\frac{1}{{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}} \\ &= \frac{\mathbf{e}_3 \kappa_0}{c} \int dx' dz'\theta(t - {\left\lvert{\mathbf{x} -(x', 0, z')}\right\rvert}/c) \sin(\omega(t - {\left\lvert{\mathbf{x} -(x', 0, z')}\right\rvert}/c))\frac{1}{{{\left\lvert{\mathbf{x} - (x', 0, z')}\right\rvert}}} \\ &= \frac{\mathbf{e}_3 \kappa_0}{c} \int dx' dz'\theta\left(t - \frac{1}{{c}} \sqrt{(x-x')^2 + y^2 + (z-z')^2}\right) \frac{\sin\left(\omega\left(t - \frac{1}{{c}} \sqrt{(x-x')^2 + y^2 + (z-z')^2}\right)\right)}{\sqrt{(x-x')^2 + y^2 + (z-z')^2}}\end{aligned}

Now a switch to polar coordinates makes sense. Let’s use

\begin{aligned}x' - x &= r \cos\alpha \\ z' - z &= r \sin\alpha \end{aligned} \hspace{\stretch{1}}(1.7)

This gives us

\begin{aligned}\mathbf{A}(\mathbf{x}, t) &= \frac{\mathbf{e}_3 \kappa_0}{c} \int_{r=0}^\infty \int_{\alpha=0}^{2\pi} r dr d\alpha\theta\left(t - \frac{1}{{c}} \sqrt{r^2 + y^2}\right) \frac{\sin\left(\omega\left(t - \frac{1}{{c}} \sqrt{r^2 + y^2 }\right)\right)}{\sqrt{r^2 + y^2}} \\ &= \frac{2 \pi \mathbf{e}_3 \kappa_0}{c} \int_{r=0}^\infty r dr \theta\left(t - \frac{1}{{c}} \sqrt{r^2 + y^2}\right) \frac{\sin\left(\omega\left(t - \frac{1}{{c}} \sqrt{r^2 + y^2 }\right)\right)}{\sqrt{r^2 + y^2}} \\ \end{aligned}

Since the theta function imposes a

\begin{aligned}t - \frac{1}{{c}} \sqrt{r^2 + y^2 } > 0\end{aligned} \hspace{\stretch{1}}(1.9)

constraint, equivalent to

\begin{aligned}c^2 t^2 > r^2 + y^2,\end{aligned} \hspace{\stretch{1}}(1.10)

we can reduce the upper range of the integral and drop the theta function explicitly

\begin{aligned}\mathbf{A}(\mathbf{x}, t) = \frac{2 \pi \mathbf{e}_3 \kappa_0}{c} \int_{r=0}^{\sqrt{c^2 t^2 - y^2}} r dr \frac{\sin\left(\omega\left(t - \frac{1}{{c}} \sqrt{r^2 + y^2 }\right)\right)}{\sqrt{r^2 + y^2}} \end{aligned} \hspace{\stretch{1}}(1.11)

Here I got lazy and used Mathematica to help evaluate this integral, for an end result of

\begin{aligned}\mathbf{A}(\mathbf{x}, t) = \frac{2 \pi \kappa_0 \omega}{c^2} \mathbf{e}_3 (1 - \cos(\omega(t - {\left\lvert{y}\right\rvert}/c))).\end{aligned} \hspace{\stretch{1}}(1.12)

## 3. Find the electric and magnetic fields outside the plane.

Our electric field can be calculated by inspection

\begin{aligned}\mathbf{E} = -\frac{1}{{c}}\frac{\partial {\mathbf{A}}}{\partial {t}}= -\frac{2 \pi \kappa_0 \omega^2}{c^3} \mathbf{e}_3 \sin(\omega(t - {\left\lvert{y}\right\rvert}/c)).\end{aligned} \hspace{\stretch{1}}(1.13)

For the magnetic field we have

\begin{aligned}\mathbf{B} &= \boldsymbol{\nabla} \times \mathbf{A} \\ &= -\frac{2 \pi \kappa_0 \omega}{c^2} \mathbf{e}_3 \times \boldsymbol{\nabla} (1 -\cos(\omega(t - {\left\lvert{y}\right\rvert}/c)))) \\ &= \frac{2 \pi \kappa_0 \omega}{c^2} (-\sin(\omega(t - {\left\lvert{y}\right\rvert}/c))\mathbf{e}_3 \times \boldsymbol{\nabla} \omega(t - {\left\lvert{y}\right\rvert}/c) \\ &= \frac{2 \pi \kappa_0 \omega^2}{c^3} \sin(\omega(t - {\left\lvert{y}\right\rvert}/c))\mathbf{e}_3 \times \boldsymbol{\nabla} {\left\lvert{y}\right\rvert} \\ &= \frac{2 \pi \kappa_0 \omega^2}{c^3} \sin(\omega(t - {\left\lvert{y}\right\rvert}/c)) \mathbf{e}_3 \times \mathbf{e}_2,\end{aligned}

which gives us

\begin{aligned}\mathbf{B} = -\frac{2 \pi \kappa_0 \omega^2}{c^3}\mathbf{e}_1 \sin(\omega(t - {\left\lvert{y}\right\rvert}/c) \end{aligned} \hspace{\stretch{1}}(1.14)

## 4. Give a physical interpretation of the results of the previous section.

It was expected that the lack of boundary on the conducting sheet would make the potential away from the plane only depend on the $y$ components of the spatial distance, and this is precisely what we find performing the grunt work of the integration.

Given that we had a sinusoidal forcing function for our wave equation, it seems logical that we also find our non-homogeneous solution to the wave equation has sinusoidal dependence. We find that the sinusoidal current results in sinusoidal potentials and fields very much like one has in the electric circuits problem that we solve with phasors in engineering applications.

We find that the electric and magnetic fields are oriented parallel to the plane containing the surface current density, with the electric field in the direction of the current, and the magnetic field perpendicular to that, but having energy propagate outwards from the plane.

It’s kind of curious that despite introducing a step function in time for the current, that the end result appears to have no constraints on time $t > 0$, and that we have fields at all points in space and time from this current distribution, even the $t < 0$ case (ie: we have a non-causal system). I'd have expected a transient response to switching on the current. Perhaps this is because instantaneously inducing a current on all points in an infinite sheet is not physically realizable?

## 5. Find the direction and magnitude of the energy flux outside the plane.

Our energy flux, the Poynting vector, is

\begin{aligned}\mathbf{S} = \frac{c}{4\pi}\left( \frac{2 \pi \kappa_0 \omega^2}{c^3} \right)^2 \sin^2(\omega(t - {\left\lvert{y}\right\rvert}/c) \mathbf{e}_3 \times \mathbf{e}_1.\end{aligned} \hspace{\stretch{1}}(1.15)

This is

\begin{aligned}\mathbf{S} = \frac{ \pi \kappa_0^2 \omega^4 }{c^5} \sin^2(\omega(t - {\left\lvert{y}\right\rvert}/c) \mathbf{e}_2= \frac{ \pi \kappa_0^2 \omega^4 }{2 c^5} (1 - \cos( 2 \omega(t - {\left\lvert{y}\right\rvert}/c) ) ) \mathbf{e}_2.\end{aligned} \hspace{\stretch{1}}(1.16)

This energy flux is directed outwards along the $y$ axis, with magnitude oscillating around an average value of

\begin{aligned}{\left\lvert{\left\langle{{S}}\right\rangle}\right\rvert} = \frac{ \pi \kappa_0^2 \omega^4 }{2 c^5}.\end{aligned} \hspace{\stretch{1}}(1.17)

## 6. Sketch the intensity of the electromagnetic field far from the plane.

I’m assuming here that this question does not refer to the flux intensity $\left\langle{\mathbf{S}}\right\rangle$, since that is constant, and boring to sketch.

The time varying portion of either the electric or magnetic field is proportional to

\begin{aligned}\sin( \omega t - \omega {\left\lvert{y}\right\rvert}/c )\end{aligned} \hspace{\stretch{1}}(1.18)

We have a sinusoid as a function of time, of period $T = 2 \pi/\omega$ where the phase is adjusted at each position by the factor $\omega {\left\lvert{y}\right\rvert}/c$. Every increase of $\Delta y = 2 \pi c/\omega$ shifts the waveform back.

A sketch is attached.

## 7. Continuity across the plane?

It is sufficient to consider either the electric or magnetic field for the continuity question since the continuity is dictated by the sinusoidal term for both fields.

The point in time only changes the phase, so let’s consider the electric field at $t=0$, and an infinitesimal distance $y = \pm \epsilon c/\omega$. At either point we have

\begin{aligned}\mathbf{E}(0, \pm \epsilon c/\omega, 0, 0) = \frac{ 2 \pi \kappa_0 \omega^2 }{c^3} \mathbf{e}_3 \epsilon\end{aligned} \hspace{\stretch{1}}(1.19)

In the limit as $\epsilon \rightarrow 0$ the field strength matches on either side of the plane (and happens to equal zero for this $t= 0$ case).

We have a discontinuity in the spatial derivative of either field near the plate, but not for the fields themselves. A plot illustrates this nicely

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.4\textheight]{phy450ps5P1Q7}
\caption{$\sin(t - {\left\lvert{y}\right\rvert})$}
\end{figure}

# Problem 2. Fields generated by an arbitrarily moving charge.

## Statement

Show that for a particle moving on a worldline parametrized by $(ct, \mathbf{x}_c(t))$, the retarded time $t_r$ with respect to an arbitrary space time point $(ct, \mathbf{x})$, defined in class as:

\begin{aligned}{\left\lvert{\mathbf{x} - \mathbf{x}_c(t_r)}\right\rvert} = c(t - t_r)\end{aligned} \hspace{\stretch{1}}(2.20)

obeys

\begin{aligned}\boldsymbol{\nabla} t_r = -\frac{\mathbf{x} - \mathbf{x}_c(t_r)}{c {\left\lvert{\mathbf{x} - \mathbf{x}_c(t_r)}\right\rvert} = c(t - t_r) - \mathbf{v}_c(t_r) \cot (\mathbf{x} - \mathbf{x}_c(t_r))}\end{aligned} \hspace{\stretch{1}}(2.21)

and

\begin{aligned}\frac{\partial {t_r}}{\partial {t}} = \frac{c {\left\lvert{\mathbf{x} - \mathbf{x}_c(t_r)}\right\rvert}}{c {\left\lvert{\mathbf{x} - \mathbf{x}_c(t_r)}\right\rvert} = c(t - t_r) - \mathbf{v}_c(t_r) \cot (\mathbf{x} - \mathbf{x}_c(t_r))}\end{aligned} \hspace{\stretch{1}}(2.22)

\begin{enumerate}
\item Then, use these to derive the expressions for $\mathbf{E}$ and $\mathbf{B}$ given in the book (and in the class notes).
\item Finally, re-derive the already familiar expressions for the EM fields of a particle moving with uniform velocity.
\end{enumerate}

## 0. Solution. Gradient and time derivatives of the retarded time function.

Let’s use notation something like our text [1], where the solution to this problem is outlined in section 63, and write

\begin{aligned}\mathbf{R}(t_r) &= \mathbf{x} - \mathbf{x}_c(t_r) \\ R &= {\left\lvert{\mathbf{R}}\right\rvert}\end{aligned} \hspace{\stretch{1}}(2.23)

where

\begin{aligned}\frac{\partial {\mathbf{R}}}{\partial {t_r}} = - \mathbf{v}_c.\end{aligned} \hspace{\stretch{1}}(2.25)

From $R^2 = \mathbf{R} \cdot \mathbf{R}$ we also have

\begin{aligned}2 R \frac{\partial {R}}{\partial {t_r}} = 2 \mathbf{R} \cdot \frac{\partial {\mathbf{R}}}{\partial {t_r}},\end{aligned} \hspace{\stretch{1}}(2.26)

so if we write

\begin{aligned}\hat{\mathbf{R}} = \frac{\mathbf{R}}{R},\end{aligned} \hspace{\stretch{1}}(2.27)

we have

\begin{aligned}R'(t_r) = -\hat{\mathbf{R}} \cdot \mathbf{v}_c.\end{aligned} \hspace{\stretch{1}}(2.28)

Proceeding in the manner of the text, we have

\begin{aligned}\frac{\partial {R}}{\partial {t}} = \frac{\partial {R}}{\partial {t_r}} \frac{\partial {t_r}}{\partial {t}} = -\hat{\mathbf{R}} \cdot \mathbf{v}_c \frac{\partial {t_r}}{\partial {t}}.\end{aligned} \hspace{\stretch{1}}(2.29)

From 2.20 we also have

\begin{aligned}R = {\left\lvert{\mathbf{x} - \mathbf{x}_c(t_r)}\right\rvert} = c(t - t_r),\end{aligned} \hspace{\stretch{1}}(2.30)

so

\begin{aligned}\frac{\partial {R}}{\partial {t}} = c\left(1 - \frac{\partial {t_r}}{\partial {t}}\right).\end{aligned} \hspace{\stretch{1}}(2.31)

This and 2.29 gives us

\begin{aligned}\boxed{\frac{\partial {t_r}}{\partial {t}} = \frac{1}{{ 1 -\hat{\mathbf{R}} \cdot \frac{\mathbf{v}_c}{c} }}}\end{aligned} \hspace{\stretch{1}}(2.32)

For the gradient we operate on the implicit equation 2.30 again. This gives us

\begin{aligned}\boldsymbol{\nabla} R = \boldsymbol{\nabla} (c t - c t_r) = - c \boldsymbol{\nabla} t_r.\end{aligned} \hspace{\stretch{1}}(2.33)

However, we can also use the spatial definition of $R = {\left\lvert{\mathbf{x} - \mathbf{x}_c(t')}\right\rvert}$. Note that this distance $R = R(t_r)$ is a function of space and time, since $t_r = t_r(\mathbf{x}, t)$ is implicitly a function of the spatial and time positions at which the retarded time is to be measured.

\begin{aligned}\boldsymbol{\nabla} R &=\boldsymbol{\nabla} \sqrt{(\mathbf{x} - \mathbf{x}_c(t_r))^2} \\ &=\frac{1}{{2R}} \boldsymbol{\nabla} (\mathbf{x} - \mathbf{x}_c(t_r))^2 \\ &=\frac{1}{{R}} (x^\beta - x_c^\beta) \mathbf{e}_\alpha \partial_\alpha (x^\beta - x_c^\beta(t_r)) \\ &=\frac{1}{{R}} (\mathbf{R})_\beta \mathbf{e}_\alpha ({\delta_\alpha}^\beta - \partial_\alpha x_c^\beta(t_r)) \\ \end{aligned}

We have only this bit $\partial_\alpha x_c^\beta(t_r)$ to expand, but that’s just going to require a chain rule expansion. This is easier to see in a more generic form

\begin{aligned}\frac{\partial {f(g)}}{\partial {x^\alpha}} = \frac{\partial {f}}{\partial {g}} \frac{\partial {g}}{\partial {x^\alpha}},\end{aligned} \hspace{\stretch{1}}(2.34)

so we have

\begin{aligned}\frac{\partial {x_c^\beta(t_r)}}{\partial {x^\alpha}} = \frac{\partial {x_c^\beta(t_r)}}{\partial {t_r}} \frac{\partial {t_r}}{\partial {x^\alpha}},\end{aligned} \hspace{\stretch{1}}(2.35)

which gets us close to where we want to be

\begin{aligned}\boldsymbol{\nabla} R&=\frac{1}{{R}} \left(\mathbf{R} - (\mathbf{R})_\beta \frac{\partial {x_c^\beta(t_r)}}{\partial {t_r}} \mathbf{e}_\alpha \frac{\partial {t_r}}{\partial {x^\alpha}} \right) \\ &=\frac{1}{{R}} \left(\mathbf{R} - \mathbf{R} \cdot \frac{\partial {\mathbf{x}_c^\beta(t_r)}}{\partial {t_r}} \boldsymbol{\nabla} t_r \right)\end{aligned}

Putting the pieces together we have only minor algebra left since we can now equate the two expansions of $\boldsymbol{\nabla} R$

\begin{aligned}- c \boldsymbol{\nabla} t_r = \hat{\mathbf{R}} - \hat{\mathbf{R}} \cdot \mathbf{v}_c(t_r) \boldsymbol{\nabla} t_r.\end{aligned} \hspace{\stretch{1}}(2.36)

This is given in the text, but these in between steps are left for us and for our homework assignments! From this point we can rearrange to find the desired result

\begin{aligned}\boxed{\boldsymbol{\nabla} t_r = -\frac{1}{{c}} \frac{\hat{\mathbf{R}} }{ 1 - \hat{\mathbf{R}} \cdot \frac{\mathbf{v}_c}{c} } = - \frac{\hat{\mathbf{R}}}{c} \frac{\partial {t_r}}{\partial {t}}}\end{aligned} \hspace{\stretch{1}}(2.37)

## 1. Solution. Computing the EM fields from the Lienard-Wiechert potentials.

Now we are ready to derive the values of $\mathbf{E}$ and $\mathbf{B}$ that arise from the Lienard-Wiechert potentials. We have for the electric field.

We’ll evaluate

\begin{aligned}\mathbf{E} &= -\frac{1}{{c}} \frac{\partial {\mathbf{A}}}{\partial {t}} - \boldsymbol{\nabla} \phi \\ \mathbf{B} &= \boldsymbol{\nabla} \times \mathbf{B}\end{aligned}

For the electric field we’ll use the chain rule on the vector potential

\begin{aligned}\frac{\partial {\mathbf{A}}}{\partial {t}} = \frac{\partial {t_r}}{\partial {t}} \frac{\partial {\mathbf{A}}}{\partial {t_r}}.\end{aligned} \hspace{\stretch{1}}(2.38)

Similarly for the gradient of the scalar potential we have

\begin{aligned}\boldsymbol{\nabla} \phi &=\mathbf{e}_\alpha \frac{\partial {\phi}}{\partial {x^\alpha}} \\ &=\mathbf{e}_\alpha \frac{\partial {\phi}}{\partial {t_r}} \frac{\partial {t_r}}{\partial {x^\alpha}} \\ &=\frac{\partial {\phi}}{\partial {t_r}} \boldsymbol{\nabla} t_r \\ &=- \frac{\hat{\mathbf{R}}}{c} \frac{\partial {t_r}}{\partial {t}}.\end{aligned}

Our electric field is thus

\begin{aligned}\mathbf{E} = - \frac{\partial {t_r}}{\partial {t}} \left( \frac{1}{{c}} \frac{\partial {\mathbf{A}}}{\partial {t_r}} - \frac{\hat{\mathbf{R}}}{c} \frac{\partial {\phi}}{\partial {t_r}} \right)\end{aligned} \hspace{\stretch{1}}(2.39)

For the magnetic field we have

\begin{aligned}\boldsymbol{\nabla} \times \mathbf{A}&=\mathbf{e}_\alpha \times \frac{\partial {\mathbf{A}}}{\partial {x^\alpha}} \\ &=\mathbf{e}_\alpha \times \frac{\partial {\mathbf{A}}}{\partial {t_r}} \frac{\partial {t_r}}{\partial {x^\alpha}}.\end{aligned}

The magnetic field will therefore be found by evaluating

\begin{aligned}\mathbf{B} = (\boldsymbol{\nabla} t_r) \times \frac{\partial {\mathbf{A}}}{\partial {t_r}} = - \frac{\partial {t_r}}{\partial {t}} \frac{\hat{\mathbf{R}}}{c} \times \frac{\partial {\mathbf{A}}}{\partial {t_r}} \end{aligned} \hspace{\stretch{1}}(2.40)

Let’s compare this to $\hat{\mathbf{R}} \times \mathbf{E}$

\begin{aligned}\hat{\mathbf{R}} \times \mathbf{E} &= \hat{\mathbf{R}} \times \left( - \frac{\partial {t_r}}{\partial {t}} \left( \frac{1}{{c}} \frac{\partial {\mathbf{A}}}{\partial {t_r}} - \frac{\hat{\mathbf{R}}}{c} \frac{\partial {\phi}}{\partial {t_r}} \right) \right) \\ &= \hat{\mathbf{R}} \times \left( - \frac{\partial {t_r}}{\partial {t}} \frac{1}{{c}} \frac{\partial {\mathbf{A}}}{\partial {t_r}} \right)\end{aligned}

This equals 2.40, verifying that we have

\begin{aligned}\mathbf{B} = \hat{\mathbf{R}} \times \mathbf{E},\end{aligned} \hspace{\stretch{1}}(2.41)

something that we can determine even without fully evaluating $\mathbf{E}$.

We are now left to evaluate the retarded time derivatives found in 2.39. Our potentials are

\begin{aligned}\phi(\mathbf{x}, t) &= \frac{e}{R(t_r)} \frac{\partial {t_r}}{\partial {t}} \\ \mathbf{A}(\mathbf{x}, t) &= \frac{e \mathbf{v}_c(t_r)}{c R(t_r)} \frac{\partial {t_r}}{\partial {t}}\end{aligned} \hspace{\stretch{1}}(2.42)

It’s clear that the quantity ${\partial {t_r}}/{\partial {t}}$ is going to show up all over the place, so let’s label it $\gamma_{t_r}$. This is justified by comparing to a particle’s boosted rest frame worldline

\begin{aligned}\begin{bmatrix}c t' \\ x'\end{bmatrix}=\gamma\begin{bmatrix}1 & -\beta \\ -\beta & 1\end{bmatrix}\begin{bmatrix}c t \\ 0\end{bmatrix}= \begin{bmatrix}\gamma c t \\ -\gamma \beta c t\end{bmatrix},\end{aligned} \hspace{\stretch{1}}(2.44)

where we have ${\partial {t'}}/{\partial {t}} = \gamma$, so for the remainder of this part of this problem we’ll write

\begin{aligned}\gamma_{t_r} \equiv \frac{\partial {t_r}}{\partial {t}} = \frac{1}{{ 1 -\hat{\mathbf{R}} \cdot \frac{\mathbf{v}_c}{c} }}.\end{aligned} \hspace{\stretch{1}}(2.45)

Using primes to denote partial derivatives with respect to the retarded time $t_r$ we have

\begin{aligned}\phi' &= e \left( -\frac{R'}{R^2} \gamma_{t_r} + \frac{\gamma_{t_r}'}{R} \right) \\ \mathbf{A}' &= e \frac{\mathbf{v}_c}{c} \left( -\frac{R'}{R^2} \gamma_{t_r} + \frac{\gamma_{t_r}'}{R} \right)+ e \frac{\mathbf{a}_c}{c} \frac{\gamma_{t_r}}{R},\end{aligned} \hspace{\stretch{1}}(2.46)

so the electric field is

\begin{aligned}\mathbf{E} &= - \gamma_{t_r} \left( \frac{1}{{c}} \mathbf{A}' - \frac{\hat{\mathbf{R}}}{c} \phi' \right) \\ &= - \frac{e \gamma_{t_r}}{c} \left( \frac{\mathbf{v}_c}{c} \left( -\frac{R'}{R^2} \gamma_{t_r} + \frac{\gamma_{t_r}'}{R} \right)+ \frac{\mathbf{a}_c}{c} \frac{\gamma_{t_r}}{R}- \hat{\mathbf{R}} \left( -\frac{R'}{R^2} \gamma_{t_r} + \frac{\gamma_{t_r}'}{R} \right) \right) \\ &= - \frac{e \gamma_{t_r}}{c} \left( \frac{\mathbf{v}_c}{c} \left( \frac{c}{R^2} \gamma_{t_r} + \frac{\gamma_{t_r}'}{R} \right)+ \frac{\mathbf{a}_c}{c} \frac{\gamma_{t_r}}{R}- \hat{\mathbf{R}} \left( \frac{c}{R^2} \gamma_{t_r} + \frac{\gamma_{t_r}'}{R} \right) \right) \\ &= - \frac{e \gamma_{t_r}}{c R} \left( \gamma_{t_r} \left( \frac{\mathbf{a}_c}{c} +\frac{\mathbf{v}_c}{R} - \frac{\hat{\mathbf{R}} c}{R}\right)+ \gamma_{t_r}' \left( \frac{\mathbf{v}_c}{c} - \hat{\mathbf{R}} \right)\right).\end{aligned}

Here’s where things get slightly messy.

\begin{aligned}\gamma_{t_r}' &= \frac{\partial {}}{\partial {t_r}} \frac{1}{{1 - \frac{\mathbf{v}_c}{c} \cdot \hat{\mathbf{R}} }} \\ &= -\gamma_{t_r}^2 \frac{\partial {}}{\partial {t_r}} \left( 1 - \frac{\mathbf{v}_c}{c} \cdot \hat{\mathbf{R}} \right) \\ &= \gamma_{t_r}^2 \left( \frac{\mathbf{a}_c}{c} \cdot \hat{\mathbf{R}} + \frac{\mathbf{v}_c}{c} \cdot \hat{\mathbf{R}}' \right),\end{aligned}

and messier

\begin{aligned}\hat{\mathbf{R}}' &= \frac{\partial {}}{\partial {t_r}} \frac{ \mathbf{R} }{ R } \\ &= \frac{ \mathbf{R}' }{ R } - \frac{\mathbf{R} R'}{R^2} \\ &= -\frac{ \mathbf{v}_c }{ R } - \frac{\hat{\mathbf{R}} (-c)}{R} \\ &= \frac{1}{{R}} \left( -\mathbf{v}_c + c \hat{\mathbf{R}} \right),\end{aligned}

then a bit unmessier

\begin{aligned}\gamma_{t_r}'&= \gamma_{t_r}^2 \left( \frac{\mathbf{a}_c}{c} \cdot \hat{\mathbf{R}} + \frac{\mathbf{v}_c}{c} \cdot \hat{\mathbf{R}}' \right) \\ &= \gamma_{t_r}^2 \left( \frac{\mathbf{a}_c}{c} \cdot \hat{\mathbf{R}} + \frac{\mathbf{v}_c}{c R} \cdot (-\mathbf{v}_c + c \hat{\mathbf{R}}) \right) \\ &= \gamma_{t_r}^2 \left( \hat{\mathbf{R}} \cdot \left( \frac{\mathbf{a}_c}{c} + \frac{\mathbf{v}_c}{R} \right) - \frac{\mathbf{v}_c^2}{c R} \right).\end{aligned}

Now we are set to plug this back into our electric field expression and start grouping terms

\begin{aligned}\mathbf{E}&= - \frac{e \gamma_{t_r}^2}{c R} \left( \frac{\mathbf{a}_c}{c} +\frac{\mathbf{v}_c}{R} - \frac{\hat{\mathbf{R}} c}{R}+ \gamma_{t_r} \left( \hat{\mathbf{R}} \cdot \left( \frac{\mathbf{a}_c}{c} + \frac{\mathbf{v}_c}{R} \right) - \frac{\mathbf{v}_c^2}{c R} \right)\left( \frac{\mathbf{v}_c}{c} - \hat{\mathbf{R}} \right)\right) \\ &= - \frac{e \gamma_{t_r}^3}{c R} \left( \left(\frac{\mathbf{a}_c}{c} +\frac{\mathbf{v}_c}{R} - \frac{\hat{\mathbf{R}} c}{R}\right) \left(1 -\hat{\mathbf{R}} \cdot \frac{\mathbf{v}_c}{c} \right)+ \left( \hat{\mathbf{R}} \cdot \left( \frac{\mathbf{a}_c}{c} + \frac{\mathbf{v}_c}{R} \right) - \frac{\mathbf{v}_c^2}{c R} \right)\left( \frac{\mathbf{v}_c}{c} - \hat{\mathbf{R}} \right)\right) \\ &= - \frac{e \gamma_{t_r}^3}{c^2 R} \left( \mathbf{a}_c\left(1 -\hat{\mathbf{R}} \cdot \frac{\mathbf{v}_c}{c} \right)+\hat{\mathbf{R}} \cdot \mathbf{a}_c \left( \frac{\mathbf{v}_c}{c} - \hat{\mathbf{R}} \right)\right) \\ &\quad - \frac{e \gamma_{t_r}^3}{c R} \left( \left(\frac{\mathbf{v}_c}{R} - \frac{\hat{\mathbf{R}} c}{R}\right) \left(1 -\hat{\mathbf{R}} \cdot \frac{\mathbf{v}_c}{c} \right)+ \left( \hat{\mathbf{R}} \cdot \left( \frac{\mathbf{v}_c}{R} \right) - \frac{\mathbf{v}_c^2}{c R} \right)\left( \frac{\mathbf{v}_c}{c} - \hat{\mathbf{R}} \right)\right) \\ \end{aligned}

Using

\begin{aligned}\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = \mathbf{b} (\mathbf{a} \cdot \mathbf{c}) - \mathbf{c} (\mathbf{a} \cdot \mathbf{b})\end{aligned} \hspace{\stretch{1}}(2.48)

We can verify that

\begin{aligned}- \left( \mathbf{a}_c\left(1 -\hat{\mathbf{R}} \cdot \frac{\mathbf{v}_c}{c} \right)+\hat{\mathbf{R}} \cdot \mathbf{a}_c \left( \frac{\mathbf{v}_c}{c} - \hat{\mathbf{R}} \right)\right) &=-\mathbf{a}_c + \mathbf{a} \hat{\mathbf{R}} \cdot \frac{\mathbf{v}}{c} - \hat{\mathbf{R}} \cdot \mathbf{a}_c \frac{\mathbf{v}_c}{c} + \hat{\mathbf{R}} \cdot \mathbf{a}_c \hat{\mathbf{R}} \\ &= \hat{\mathbf{R}} \times \left( \left(\hat{\mathbf{R}} - \frac{\mathbf{v}_c}{c} \right) \times \mathbf{a}_c \right),\end{aligned}

which gets us closer to the desired end result

\begin{aligned}\mathbf{E}= \frac{e \gamma_{t_r}^3}{c^2 R} \hat{\mathbf{R}} \times \left( \left(\hat{\mathbf{R}} - \frac{\mathbf{v}_c}{c} \right) \times \mathbf{a}_c \right)- \frac{e \gamma_{t_r}^3}{c R^2} \left( \left(\mathbf{v}_c- \hat{\mathbf{R}} c\right) \left(1 -\hat{\mathbf{R}} \cdot \frac{\mathbf{v}_c}{c} \right)+ \left( \hat{\mathbf{R}} \cdot \mathbf{v}_c - \frac{\mathbf{v}_c^2}{c} \right)\left( \frac{\mathbf{v}_c}{c} - \hat{\mathbf{R}} \right)\right).\end{aligned} \hspace{\stretch{1}}(2.49)

It is also easy to show that the remaining bit reduces nicely, since all the dot product terms conveniently cancel

\begin{aligned}- \left( \left(\mathbf{v}_c- \hat{\mathbf{R}} c\right) \left(1 -\hat{\mathbf{R}} \cdot \frac{\mathbf{v}_c}{c} \right)+ \left( \hat{\mathbf{R}} \cdot \mathbf{v}_c - \frac{\mathbf{v}_c^2}{c} \right)\left( \frac{\mathbf{v}_c}{c} - \hat{\mathbf{R}} \right)\right) = c \left( 1 - \frac{\mathbf{v}_c^2}{c^2} \right) \left( \hat{\mathbf{R}} - \frac{\mathbf{v}}{c} \right)\end{aligned} \hspace{\stretch{1}}(2.50)

This completes the exercise, leaving us with

\begin{aligned}\boxed{\begin{aligned}\mathbf{E}&= \frac{e \gamma_{t_r}^3}{c^2 R} \hat{\mathbf{R}} \times \left( \left(\hat{\mathbf{R}} - \frac{\mathbf{v}_c}{c} \right) \times \mathbf{a}_c \right)+\frac{e \gamma_{t_r}^3}{R^2} \left( 1 - \frac{\mathbf{v}_c^2}{c^2} \right) \left( \hat{\mathbf{R}} - \frac{\mathbf{v}_c}{c} \right) \\ \mathbf{B} &= \hat{\mathbf{R}} \times \mathbf{E}.\end{aligned}}\end{aligned} \hspace{\stretch{1}}(2.51)

Looking back to 2.45 where $\gamma_{t_r}$ was defined, we see that this compares to (63.8-9) in the text.

## 2. Solution. EM fields from a uniformly moving source.

For a uniform source moving in space at constant velocity

\begin{aligned}\mathbf{x}_c(t) = \mathbf{v} t,\end{aligned} \hspace{\stretch{1}}(2.52)

our retarded time measured from the spacetime point $(ct, \mathbf{x})$ is defined implicitly by

\begin{aligned}R = {\left\lvert{\mathbf{x} - \mathbf{x}_c(t_r) }\right\rvert} = c (t - t_r).\end{aligned} \hspace{\stretch{1}}(2.53)

Squaring this we have

\begin{aligned}\mathbf{x}^2 + \mathbf{v}^2 t_r^2 - 2 t_r \mathbf{x} \cdot \mathbf{v} = c^2 t^2 + c^2 t_r^2 - 2 c t t_r,\end{aligned} \hspace{\stretch{1}}(2.54)

or

\begin{aligned}( c^2 -\mathbf{v}^2) t_r^2 + 2 t_r ( - c t + \mathbf{x} \cdot \mathbf{v} ) = \mathbf{x}^2 - c^2 t^2.\end{aligned} \hspace{\stretch{1}}(2.55)

Rearranging to complete the square we have

\begin{aligned}\left( \sqrt{c^2 - \mathbf{v}^2} t_r - \frac{ t c^2 - \mathbf{x} \cdot \mathbf{v} }{\sqrt{c^2 - \mathbf{v}^2}} \right)^2 &= \mathbf{x}^2 - c^2 t^2 +\frac{ (t c^2 - \mathbf{x} \cdot \mathbf{v})^2 }{c^2 - \mathbf{v}^2} \\ &= \frac{ (\mathbf{x}^2 - c^2 t^2)( c^2 - \mathbf{v}^2) + (t c^2 - \mathbf{x} \cdot \mathbf{v})^2}{ c^2 - \mathbf{v}^2} \\ &= \frac{ \mathbf{x}^2 c^2 - \mathbf{x}^2 \mathbf{v}^2 - {c^4 t^2} + c^2 t^2 \mathbf{v}^2 + {t^2 c^4} + (\mathbf{x} \cdot \mathbf{v})^2 - 2 t c^2 (\mathbf{x} \cdot \mathbf{v}) }{ c^2 - \mathbf{v}^2 } \\ &= \frac{ c^2 ( \mathbf{x}^2 + t^2 \mathbf{v}^2 -2 t (\mathbf{x} \cdot \mathbf{v})) - \mathbf{x}^2 \mathbf{v}^2 + (\mathbf{x} \cdot \mathbf{v})^2 }{ c^2 - \mathbf{v}^2 } \\ &= \frac{ c^2 ( \mathbf{x} - \mathbf{v} t)^2 - (\mathbf{x} \times \mathbf{v})^2 }{ c^2 - \mathbf{v}^2 } \\ \end{aligned}

Taking roots (and keeping the negative so that we have $t_r = t - {\left\lvert{\mathbf{x}}\right\rvert}/c$ for the $\mathbf{v} = 0$ case, we have

\begin{aligned}\sqrt{1 - \frac{\mathbf{v}^2}{c^2}} c t_r =\frac{1}{{\sqrt{1 - \frac{\mathbf{v}^2}{c^2}}}} \left(c t - \mathbf{x} \cdot \frac{\mathbf{v}}{c} - \sqrt{ \left( \mathbf{x} - \mathbf{v} t \right)^2 - \left(\mathbf{x} \times \frac{\mathbf{v}}{c}\right)^2 }\right),\end{aligned} \hspace{\stretch{1}}(2.56)

or with $\boldsymbol{\beta} = \mathbf{v}/c$, this is

\begin{aligned}c t_r = \frac{1}{{1 - \boldsymbol{\beta}^2}} \left( c t - \mathbf{x} \cdot \boldsymbol{\beta} - \sqrt{ \left( \mathbf{x} - \mathbf{v} t \right)^2 - \left(\mathbf{x} \times \boldsymbol{\beta}\right)^2 } \right).\end{aligned} \hspace{\stretch{1}}(2.57)

What’s our retarded distance $R = c t - c t_r$? We get

\begin{aligned}R = \frac{\boldsymbol{\beta} \cdot (\mathbf{x} - \mathbf{v} t) + \sqrt{ (\mathbf{x} - \mathbf{v} t)^2 - (\mathbf{x} \times \boldsymbol{\beta})^2 }}{ 1 - \boldsymbol{\beta}^2 }.\end{aligned} \hspace{\stretch{1}}(2.58)

For the vector distance we get (with $\boldsymbol{\beta} \cdot (\mathbf{x} \wedge \boldsymbol{\beta}) = (\boldsymbol{\beta} \cdot \mathbf{x}) \boldsymbol{\beta} - \mathbf{x} \boldsymbol{\beta}^2$)

\begin{aligned}\mathbf{R} = \frac{\mathbf{x} -\mathbf{v} t + \boldsymbol{\beta} \cdot (\mathbf{x} \wedge \boldsymbol{\beta}) + \boldsymbol{\beta} \sqrt{ (\mathbf{x} - \mathbf{v} t)^2 - (\mathbf{x} \times \boldsymbol{\beta})^2 }}{ 1 - \boldsymbol{\beta}^2 }.\end{aligned} \hspace{\stretch{1}}(2.59)

For the unit vector $\hat{\mathbf{R}} = \mathbf{R}/R$ we have

\begin{aligned}\hat{\mathbf{R}} = \frac{\mathbf{x} - \mathbf{v} t + \boldsymbol{\beta} \cdot (\mathbf{x} \wedge \boldsymbol{\beta}) + \boldsymbol{\beta} \sqrt{ (\mathbf{x} - \mathbf{v} t)^2 - (\mathbf{x} \times \boldsymbol{\beta})^2 }}{ \boldsymbol{\beta} \cdot (\mathbf{x} - \mathbf{v} t) + \sqrt{ (\mathbf{x} - \mathbf{v} t)^2 - (\mathbf{x} \times \boldsymbol{\beta})^2 } }.\end{aligned} \hspace{\stretch{1}}(2.60)

The acceleration term in the electric field is zero, so we are left with just

\begin{aligned}\mathbf{E}= \frac{e \gamma_{t_r}^3}{R^2} \left( 1 - \frac{\mathbf{v}_c^2}{c^2} \right) \left( \hat{\mathbf{R}} - \frac{\mathbf{v}_c}{c} \right).\end{aligned} \hspace{\stretch{1}}(2.61)

Leading to $\gamma_{t_r}$, we have

\begin{aligned}\hat{\mathbf{R}} \cdot \boldsymbol{\beta} = \frac{\boldsymbol{\beta} \cdot (\mathbf{x} - \mathbf{v} t + R^{*} \boldsymbol{\beta})}{\boldsymbol{\beta} \cdot (\mathbf{x} - \mathbf{v} t) + R^{*}},\end{aligned} \hspace{\stretch{1}}(2.62)

where, following section 38 of the text we write

\begin{aligned}R^{*} = \sqrt{(\mathbf{x} - \mathbf{v} t)^2 - (\mathbf{x} \times \boldsymbol{\beta})^2 }\end{aligned} \hspace{\stretch{1}}(2.63)

This gives us

\begin{aligned}\gamma_{t_r} = \frac{\boldsymbol{\beta} \cdot (\mathbf{x} - \mathbf{v} t) + R^{*}}{R^{*}(1 - \boldsymbol{\beta}^2)}.\end{aligned} \hspace{\stretch{1}}(2.64)

Observe that this equals one when $\boldsymbol{\beta} = 0$ as expected.

We can also compute

\begin{aligned}\hat{\mathbf{R}} - \boldsymbol{\beta} &= \frac{\mathbf{x} + \boldsymbol{\beta} \cdot (\mathbf{x} \wedge \boldsymbol{\beta}) - \mathbf{v} t + \boldsymbol{\beta} \sqrt{ (\mathbf{x} - \mathbf{v} t)^2 - (\mathbf{x} \times \boldsymbol{\beta})^2 }}{ \boldsymbol{\beta} \cdot (\mathbf{x} - \mathbf{v} t) + \sqrt{ (\mathbf{x} - \mathbf{v} t)^2 - (\mathbf{x} \times \boldsymbol{\beta})^2 } } - \boldsymbol{\beta} \\ &=\frac{(\mathbf{x} - \mathbf{v} t)(1 -\boldsymbol{\beta}^2)}{\boldsymbol{\beta} \cdot (\mathbf{x} - \mathbf{v} t) + \sqrt{ (\mathbf{x} - \mathbf{v} t)^2 - (\mathbf{x} \times \boldsymbol{\beta})^2 } }.\end{aligned}

Our long and messy expression for the field is therefore

\begin{aligned}\mathbf{E} &=e \gamma_{t_r}^3 \frac{1}{{R^2}} (1 - \boldsymbol{\beta}^2)(\hat{\mathbf{R}} - \boldsymbol{\beta}) \\ &=e \left( \frac{\boldsymbol{\beta} \cdot (\mathbf{x} - \mathbf{v} t) + R^{*}}{R^{*}(1 - \boldsymbol{\beta}^2)}\right)^3\frac{(1 - \boldsymbol{\beta}^2)^2 }{ (\boldsymbol{\beta} \cdot (\mathbf{x} - \mathbf{v} t) + R^{*})^2 } (1 -\boldsymbol{\beta}^2)\frac{(\mathbf{x} - \mathbf{v} t)(1 -\boldsymbol{\beta}^2)}{\boldsymbol{\beta} \cdot (\mathbf{x} - \mathbf{v} t) + R^{*}} \\ \end{aligned}

This gives us our final result

\begin{aligned}\mathbf{E} =e \frac{1}{{(R^{*})^3}}(1 -\boldsymbol{\beta}^2)(\mathbf{x} - \mathbf{v} t)\end{aligned} \hspace{\stretch{1}}(2.65)

As a small test we observe that we get the expected result

\begin{aligned}\mathbf{E} = e \frac{\mathbf{x}}{{\left\lvert{\mathbf{x}}\right\rvert}^3}\end{aligned} \hspace{\stretch{1}}(2.66)

for the $\boldsymbol{\beta} = 0$ case.

When $\mathbf{v} = V \mathbf{e}_1$ this also recovers equation (38.6) from the text as desired, and if we switch to primed coordinates

\begin{aligned}x' &= \gamma( x - v t) \\ y' &= y \\ z' &= z \\ (1 - \beta^2) {r'}^2 &= (x - v t)^2 + (y^2 + z^2)(1 - \beta^2),\end{aligned} \hspace{\stretch{1}}(2.67)

we recover the field equation derived twice before in previous problem sets

\begin{aligned}\mathbf{E} = \frac{e}{(r')^3} ( x', \gamma y', \gamma z')\end{aligned} \hspace{\stretch{1}}(2.71)

## 2. Solution. EM fields from a uniformly moving source along x axis.

Initially I had errors in the vector treatment above, so tried with the simpler case using uniform velocity $v$ along the $x$ axis instead. Comparison of the two showed where my errors were in the vector algebra, and that’s now also fixed up.

Performing all the algebra to solve for $t_r$ in

\begin{aligned}{\left\lvert{\mathbf{x} - v t_r \mathbf{e}_1}\right\rvert} = c(t - t_r),\end{aligned} \hspace{\stretch{1}}(2.72)

I get

\begin{aligned}c t_r = \frac{c t - x \beta - \sqrt{ (x- v t)^2 + (y^2 + z^2)(1-\beta^2) } }{ 1 - \beta^2 } = - \gamma (\beta x' + r' )\end{aligned} \hspace{\stretch{1}}(2.73)

This matches the vector expression from 2.57 with the special case of $\mathbf{v} = v \mathbf{e}_1$ so we at least started off on the right foot.

For the retarded distance $R = ct - c t_r$ we get

\begin{aligned}R = \frac{ \beta( x - v t) + \sqrt{ (x- v t)^2 + (y^2 + z^2)(1-\beta^2) } }{ 1 - \beta^2 } = \gamma( \beta x' + r' )\end{aligned} \hspace{\stretch{1}}(2.74)

This also matches 2.58, so things still seem okay with the vector approach. What’s our vector retarded distance

\begin{aligned}\mathbf{R} &= \mathbf{x} - \beta c t_r \mathbf{e}_1 \\ &= (x - \beta c t_r, y, z) \\ &= \left( \frac{ x - v t + \beta \sqrt{ (x- v t)^2 + (y^2 + z^2)(1-\beta^2) } }{ 1 - \beta^2 }, y, z \right) \\ &= \left( \gamma (x' + \beta r'), y', z' \right)\end{aligned}

So

\begin{aligned}\hat{\mathbf{R}} &= \frac{1}{{ \gamma (\beta x' + r') }} \left( \gamma(x' + \beta r'), y', z' \right) \\ &= \frac{1}{{ \beta x' + r' }} \left( x' + \beta r', \frac{y'}{\gamma}, \frac{z'}{\gamma} \right)\end{aligned}

\begin{aligned}\hat{\mathbf{R}} -\boldsymbol{\beta}&= \frac{1}{{ \gamma (\beta x' + r') }} \left( \gamma(x' + \beta r'), y', z' \right) - (\beta, 0, 0) \\ &= \frac{1}{{ \beta x' + r' }} \left( x'(1- \beta^2), \frac{y'}{\gamma}, \frac{z'}{\gamma} \right) \\ &= \frac{1}{{\gamma (\beta x' + r')}} ( x - v t, y, z)\end{aligned}

For ${\partial {t_r}}/{\partial {t}}$, using $\hat{\mathbf{R}}$ calculated above, or from 2.73 calculating directly I get

\begin{aligned}\frac{\partial {t_r}}{\partial {t}} = \frac{r' + \beta x'}{r'(1 - \beta^2)} = \frac{\gamma( r' + \beta x') }{R^{*}},\end{aligned} \hspace{\stretch{1}}(2.75)

where, as in section 38 of the text, we write

\begin{aligned}R^{*} = \sqrt{ (x - v t)^2 + (y^2 + z^2)(1-\beta^2) }.\end{aligned} \hspace{\stretch{1}}(2.76)

Putting all the pieces together I get

\begin{aligned}\mathbf{E} &= e (1 -\beta^2) \frac{(x - v t, y, z)}{{\gamma( \beta x' + r'})} \left( \frac{{\gamma( r' + \beta x')}}{R^{*}} \right)^3 \frac{1}{{{ \gamma^2 (\beta x' + r')^2 } }} \\ \end{aligned}

so we have

\begin{aligned}\mathbf{E} = e \frac{1 -\beta^2}{(R^{*})^3} (x - v t, y, z) \end{aligned} \hspace{\stretch{1}}(2.77)

This matches equation (38.6) in the text.

# Problem 3.

FIXME: TODO.

Only the first question was graded (I lost 1.5 marks). I got my units wrong when I integrated 1.11, and my $\omega/c$ factor in the result is circled. Should have also done a dimensional analysis check. There was also a remark that the integral is zero if $t < y/c$, which introduces a $\theta$ function. I think that I incorrectly dropped my $\theta$ function, and should have retained.