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\begin{document}
\begin{multicols}{3}
Vector identities (Griffiths)
\begin{gather*}
\mathbf{A} \times (\mathbf{B} \times \mathbf{C}) = \mathbf{B}(\mathbf{A} \cdot
\mathbf{C}) - \mathbf{C}(\mathbf{A} \cdot \mathbf{B}) \\
\nabla(fg) = f\nabla g + g\nabla f \\
\begin{split}
\nabla(\mathbf{A} \cdot \mathbf{B}) &= \mathbf{A} \times (\nabla \times
\mathbf{B}) + \mathbf{B} \times (\nabla \times \mathbf{A}) \\
&+ (\mathbf{A} \cdot
\nabla)\mathbf{B} + (\mathbf{B} \cdot \nabla)\mathbf{A}
\end{split} \\
\nabla \cdot(f\mathbf{A}) = f(\nabla \cdot \mathbf{A}) + \mathbf{A} \cdot
\nabla f \\
\nabla \cdot(\mathbf{A} \times \mathbf{B}) = \mathbf{B} \cdot (\nabla \times
\mathbf{A}) - \mathbf{A} \cdot (\nabla \times \mathbf{B}) \\
\nabla \times (f\mathbf{A}) = f(\nabla \times \mathbf{A}) - \mathbf{A} \times
(\nabla f) \\
\begin{split}
\nabla \times(\mathbf{A} \times \mathbf{B}) &= (\mathbf{B} \cdot
\nabla)\mathbf{A} - (\mathbf{A} \cdot \nabla)\mathbf{B} \\
&+ \mathbf{A}(\nabla \cdot \mathbf{B}) - \mathbf{B}(\nabla \cdot \mathbf{A})
\end{split} \\
\nabla \times (\nabla \times \mathbf{A}) = \nabla(\nabla \cdot \mathbf{A}) -
\nabla^2\mathbf{A}
\end{gather*}
Vector identities (Jackson)
\begin{gather*}
(\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{c} \times \mathbf{d}) =
(\mathbf{a} \cdot \mathbf{c})(\mathbf{b}\cdot\mathbf{d}) -
(\mathbf{a}\cdot\mathbf{d})(\mathbf{b}\cdot\mathbf{c}) \\
\int_V \nabla \cdot \mathbf{A} \, d^3 x = \int_S \mathbf{A}
\cdot d\mathbf{a} \numberthis \label{eqn:divergence} \\
\int_V \nabla \psi \, d^3 x = \int_S \psi \, d\mathbf{a} \\
\int_V \nabla \times \mathbf{A} \, d^3 x = \int_S d\mathbf{a} \times \mathbf{A}
 \\
\int_V (\phi\nabla^2 \psi + \nabla\phi \cdot \nabla\psi) \, d^3 x = \int_S
(\phi\nabla\psi)\cdot d\mathbf{a} \numberthis \label{eqn:green1} \\
\int_V (\phi\nabla^2 \psi - \psi\nabla^2\phi) \, d^3 x = \int_S (\phi\nabla\psi
- \psi\nabla\phi) \cdot d\mathbf{a} \numberthis \label{eqn:green2} \\
\int_S (\nabla \times \mathbf{A}) \cdot d\mathbf{a} = \int_C\mathbf{A} \cdot
d\mathbf{l} \numberthis \label{eqn:stokes} \\
\int_S d\mathbf{a} \times \nabla\psi = \int_C \psi \, d\mathbf{l}
\end{gather*}
(\ref{eqn:divergence})~divergence theorem; (\ref{eqn:green1})~Green's first
identity; (\ref{eqn:green2})~Green's theorem; (\ref{eqn:stokes})~Stokes's
theorem \\
Spherical coordinates (Griffiths)
\begin{gather*}
d\mathbf{l} = \hat{\mathbf{r}} \, dr + r\hat\theta \, d\theta + r\sin\theta \,
\hat\phi \, d\phi \\
\begin{aligned}
\hat{\mathbf{x}} &= \sin \theta \cos \phi \, \hat{\mathbf{r}} + \cos \theta \cos
\phi \, \hat\theta - \sin \phi \, \hat\phi \\
\hat{\mathbf{y}} &= \sin \theta \sin \phi \, \hat{\mathbf{r}} + \cos \theta
\sin \phi \, \hat\theta + \cos \phi \, \hat\phi \\
\hat{\mathbf{z}} &= \cos \theta \, \hat{\mathbf{r}} - \sin \theta \, \hat\theta
\\
\hat{\mathbf{r}} &= \sin \theta \cos \phi \, \hat{\mathbf{x}} + \sin \theta
\sin \phi \, \hat{\mathbf{y}} + \cos \theta \, \hat{\mathbf{z}} \\
\hat\theta &= \cos \theta \cos \phi \hat{\mathbf{x}} + \cos \theta \sin \phi \,
\hat{\mathbf{y}} - \sin \theta \, \hat{\mathbf{z}} \\
\hat\phi &= -\sin \phi \, \hat{\mathbf{x}} + \cos \phi \, \hat{\mathbf{y}}
\end{aligned} \\
\nabla t = \partial_r t \, \hat{\mathbf{r}} + \frac{1}{r} \partial_\theta t \,
\hat\theta + \frac{1}{r\sin\theta} \partial_\phi t \, \hat\phi \\
\nabla \cdot \mathbf{v} = \frac{1}{r^2} \partial_r(r^2 v_r) +
\frac{1}{r\sin\theta}\partial_\theta (v_\theta \sin \theta) +
\frac{1}{r\sin\theta} \partial_\phi v_\phi \\
\begin{split}
\nabla \times \mathbf{v} &= \frac{1}{r\sin\theta}
\hat{\mathbf{r}}[\partial_\theta(v\phi \sin \theta) - \partial_\phi v_\theta]
\\
&+ \frac{1}{r}\hat\theta\bigg[\frac{1}{\sin\theta}\partial_\phi v_r 
- \partial_r(rv_\phi)\bigg] +
\frac{1}{r}\hat\phi[\partial_r(rv_\theta)-\partial_\theta v_r]
\end{split} \\
\nabla^2 t = \frac{1}{r^2} \partial_r(r^2 \partial_r t) + \frac{1}{r^2 \sin
\theta} \partial_\theta (\sin \theta \, \partial_\theta t) + \frac{1}{r^2
\sin^2 \theta} \partial_{\phi\phi}t
\end{gather*}
Cylindrical coordinates (Griffiths)
\begin{gather*}
d\mathbf{l} = \hat{\mathbf{s}} \, ds + s\hat\phi \, d\phi + \hat{\mathbf{z}} \,
dz \\
\begin{aligned}
\hat{\mathbf{x}} &= \cos \phi \, \hat{\mathbf{s}} - \sin \phi \, \hat\phi
&\quad
\hat{\mathbf{y}} &= \sin \phi \, \hat{\mathbf{s}} + \cos \phi \, \hat\phi \\
\hat{\mathbf{s}} &= \cos \phi \, \hat{\mathbf{x}} + \sin \phi \,
\hat{\mathbf{y}} &\quad
\hat\phi &= -\sin\phi \, \hat{\mathbf{x}} + \cos\phi \, \hat{\mathbf{y}}
\end{aligned} \\
\nabla t = \partial_s t \, \hat{\mathbf{s}} + \frac{1}{s} \partial_\phi t \,
\hat\phi + \partial_z t \, \hat{\mathbf{z}} \\
\nabla \cdot \mathbf{v} = \frac{1}{s} \partial_s (sv_s) + \frac{1}{s}
\partial_\phi v_\phi + \partial_z v_z \\
\begin{split}
\nabla \times \mathbf{v} &= \hat{\mathbf{s}}\bigg[\frac{1}{s}\partial_\phi v_z -
\partial_z v_\phi\bigg] + \hat\phi[\partial_z v_s - \partial_s v_z] \\
&+ \frac{1}{s}\hat{\mathbf{z}}[\partial_s(sv_\phi)-\partial_\phi v_s]
\end{split} \\
\nabla^2 t = \frac{1}{s}\partial_s(s\partial_s t) +
\frac{1}{s^2}\partial_{\phi\phi}t + \partial_{zz}t
\end{gather*}
Useful variation \\
$\displaystyle \delta(ds/d\theta) = \frac{(dx^i/d\theta)(d\delta x_i/d\theta)}
{ds/d\theta}$ (54.2)\\
Primed frame moves with velocity $v\hat{\mathbf{x}}$ relative to unprimed
frame. \\
$\begin{pmatrix} ct' \\ x' \\ y' \\ z' \end{pmatrix} = \begin{pmatrix} \gamma &
-\beta\gamma & 0 & 0 \\ -\beta\gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0
& 0 & 1 \end{pmatrix}\begin{pmatrix} ct \\ x \\ y \\ z \end{pmatrix}$ \\
Transformation of the field (HW3, problem~2.1)
\begin{alignat*}{3}
E_x' &= E_x &\quad E_y' &= \gamma(E_y - \beta B_z) &\quad
E_z' &= \gamma(E_z+\beta B_y) \\
B_x' &= B_x &\quad B_y' &= \gamma(B_y + \beta E_z) &\quad
B_z' &= \gamma(B_z - \beta E_y)
\end{alignat*}
Basic objects of relativistic electrodynamics \\
$A^i = (\phi, \mathbf{A}) \quad A_i = (\phi, -\mathbf{A}) \quad F_{ij} =
\partial_i A_j - \partial_j A_i$ \\
$\mathbf{E} - \nabla \phi - \partial\mathbf{A}/\partial t \quad \mathbf{B} =
\nabla \times \mathbf{A}$
\[
F_{ij} = \begin{pmatrix} 0 & E_x & E_y & E_z \\
-E_x & 0 & -B_z & B_y \\
-E_y & B_z & 0 & -B_x \\
-E_z & -B_y & B_x & 0 \end{pmatrix}
\]
\[
F^{ij} = \begin{pmatrix} 0 & -E_x & -E_y & -E_z \\
E_x & 0 & -B_z & B_y \\
E_y & B_z & 0 & -B_x \\
E_z & -B_y & B_x & 0 \end{pmatrix}
\]
$F_{\alpha\beta} = -\epsilon_{\alpha\beta\gamma}B_\gamma$ (p.~80) \quad
$\epsilon^{0123} = +1 \quad \epsilon_{0123} = -1$ \\
$F^{ij}F_{kl} \propto B^2 - E^2 \quad F^{ij}F^{kl}\epsilon_{ijkl} \propto
\mathbf{E} \cdot \mathbf{B}$ (p.~85) \\
$F^{ij}F^{kl}\epsilon_{ijkl}$ is a boundary term (4-divergence) \\
Four-current (p.~96)
\begin{gather*}
e \int u^i A_i \, ds = \frac{1}{c} \int A_i j^i \, d^4 x \\
j^i(x^j) = \sum_A \left[ ce_A \int u^i_A \delta^4(x^j - x^j_A(\tau)) \, ds
\right]
\end{gather*}
Four-current for EM field (p.~101): $j^i = (c\rho, \mathbf{j})$ \\
Action for scalar field (\ref{eqn:scalar}) and EM field (\ref{eqn:vector})
\begin{gather}
S = -mc \int ds + \int \phi(x^i) \, ds \label{eqn:scalar} \\
S = -mc \int ds - \frac{e}{c} \int u^i A_i \, ds - \frac{1}{16\pi c} \int
F^{ij}F_{ij} \, d^4 x \label{eqn:vector}
\end{gather}
Equation of motion for charged particle: $mc \frac{du_i}{ds} =
\frac{e}{c}F_{ij} u_j$ \\
$j=0$ gives: $\frac{d\mathcal{E}}{dt} = e\mathbf{E} \cdot \mathbf{v}$ \\
$j=1,2,3$ gives: $\frac{d\mathbf{p}}{dt} = e\left(\mathbf{E} +
\frac{\mathbf{v}}{c} \times \mathbf{B}\right)$ \\
Useful variation \\
$\delta(F^{ij} F_{ij}) = F^{ij} \delta F_{ij} + F_{ij} \delta F^{ij} = 2 F^{ij}
\delta F_{ij} = 2F^{ij} \delta(\partial_i A_j - \partial_j A_i) = 2F^{ij}
\partial_i \delta A_j - 2F^{ij} \partial_j \delta A_i = 4F^{ij}\partial_i
\delta A_j = 4\partial_i (F^{ij}\delta A_j) - 4(\delta A_j)\partial_i F^{ij}$
\\
Bianchi identity (p.~88): $\epsilon^{ijkl} \partial_j F_{kl} = 0$ \\
Implications: $\nabla \cdot \mathbf{B} = 0$ and $\nabla \times \mathbf{E} =
-\frac{1}{c} \frac{\partial \mathbf{B}}{\partial t}$ \\
Equation of motion for EM field (p.~107):
$\partial_i F^{ij} = \frac{4\pi}{c} j^j$ \\
Implications: $\nabla \cdot \mathbf{E}  = 4\pi\rho$ and $\nabla \times
\mathbf{B} = \frac{4\pi}{c} \mathbf{j} +
\frac{1}{c}\frac{\partial\mathbf{E}}{\partial t}$ \\
Gauge invariance: $\phi' \leftarrow \phi + \frac{1}{c} \frac{\partial
\chi}{\partial t}; \mathbf{A}' \leftarrow \mathbf{A} - \nabla \chi$
\textit{i.e.} $A^{i\prime} \leftarrow A^i + \partial^i \chi$ ($\chi$ is a
Lorentz scalar field) \\
Properties of Levi--Civita symbol (HW3)
\begin{gather*}
\begin{aligned}
\epsilon_{\alpha\beta\gamma} \epsilon_{\mu\nu\lambda} &=
\delta^{\alpha\beta\gamma}_{\mu\nu\lambda} \\
\epsilon_{\alpha\beta\gamma}\epsilon_{\mu\nu\gamma} &=
\delta_{\alpha\mu}\delta_{\beta\nu} - \delta_{\alpha\nu} \delta_{\beta\mu} \\
\epsilon_{\alpha\beta\gamma} \epsilon_{\mu\beta\gamma} &= 2\delta_{\alpha\mu}
\end{aligned} \\
\epsilon_{\alpha\beta\gamma} \epsilon_{\mu\nu\lambda} A_{\alpha\mu}A_{\beta\nu}
A_{\gamma\lambda} = 6\det A
\end{gather*}
Useful relations for relativistic mechanics
\begin{gather}
\mathbf{v} = \frac{\mathbf{p}c^2}{\mathcal{E}} \label{eqn:vfromp} \\
\mathbf{p} = \frac{\mathcal{E} \mathbf{v}}{c^2} \label{eqn:pfromv}
\end{gather}
Motion in uniform E-field: Integrate equations of motion to obtain energy and
momentum, then use (\ref{eqn:vfromp}) to get velocity. \\
Motion in uniform B-field: Note that energy is constant. Use (\ref{eqn:pfromv})
to write momentum in terms of velocity, then integrate equations of motion to
get velocity. \\
Coulomb gauge: $\phi = 0; \nabla \cdot \mathbf{A} = 0; -\nabla^2 \mathbf{A} +
\frac{1}{c^2} \frac{\partial^2 \mathbf{A}}{\partial t^2} = \mathbf{0}$ \\
Lorentz gauge: $\partial_i A^i = 0; \partial_i \partial^i A_j = \square A_j =
0$ \\
Fourier series
\begin{gather*}
f(x) = \frac{1}{L} \sum_{n=-\infty}^\infty \tilde f(k_n) e^{ik_n x} \\
\tilde f(k_n) = \int_{-L/2}^{L/2} f(x) e^{-ik_n x} \, dx \\
k_n = \frac{2\pi n}{L}
\end{gather*}
Fourier transform
\begin{gather*}
f(x) = \frac{1}{2\pi} \int_{-\infty}^\infty \tilde f(k) e^{ikx} \, dk \\
\tilde f(k) = \int_{-\infty}^\infty f(x) e^{-ikx} \, dx \\
\int_{-\infty}^\infty e^{i(k-k')x} \, dx = 2\pi \delta(k-k')
\end{gather*}
Need $\tilde f(k) = \tilde f^*(-k)$ in order for $f(x)$ to be real. \\
Solution of wave equation for EM field in Coulomb gauge (p.~121)
\[\mathbf{A}(\mathbf{x}, t) = \iiint \vec\beta^*(-\mathbf{k})
e^{i(\mathbf{k}\cdot\mathbf{x} + \omega t)} + \vec\beta(\mathbf{k})
e^{i(\mathbf{k} \cdot \mathbf{x} - \omega t)} \frac{d^3 k}{(2\pi)^3}\]
Monochromatic plane wave \\
$\vec\beta(\mathbf{k}) = \vec\beta \delta^3(\mathbf{k}-\mathbf{p}) (2\pi)^3$
which yields\\
$\mathbf{A}(\mathbf{x}, t) = 2\vec\beta
\cos(\mathbf{p}\cdot\mathbf{x} - \omega t)$ (p.~122) \\
$\mathbf{k} \cdot (\mathbf{E} \times \mathbf{B}) > 0$ (p.~123) \\
$\omega = c\|\mathbf{k}\|$ (p.~124) \\
Poynting vector and EM energy density (p.~126)
\begin{gather*}
\mathbf{S} = \frac{c}{4\pi} \mathbf{E} \times \mathbf{B} \\
\mathcal{E}_{\text{em}} = \frac{1}{8\pi}(E^2 + B^2) \\
\frac{\partial}{\partial t} \left[\frac{1}{8\pi}(E^2 + B^2)\right] =
-\mathbf{j} \cdot \mathbf{E} - \nabla \cdot \mathbf{S} \\
\frac{d}{dt} \int_V \frac{E^2+B^2}{8\pi} \, d^3 x = -\int_V \mathbf{j} \cdot
\mathbf{E} \, d^3 x - \int_S \mathbf{S} \cdot d\mathbf{a}
\end{gather*}
Laplace Green function (GF1) \\
$G(\mathbf{x}) = \frac{1}{4\pi\|\mathbf{x}\|}; \nabla^2 G(\mathbf{x}) =
-\delta^3(\mathbf{x})$ \\
d'Alembert Green function (GF5) \\
$G(x^i) = -\frac{\delta(ct-r)}{4\pi r}; \square G(x^i) = -\delta^4(x^i)$ \\
Energy of plane wave (p.~128): $\mathbf{S} = \hat{\mathbf{k}}c
\mathcal{E}_{\text{em}}$ \\
EM momentum density (p.~129): $\vec{\mathcal{P}}_{\text{em}} =
\frac{\mathbf{S}}{c^2}$ \\
Inhomogeneous wave equation in Lorenz gauge (p.~133): $\square A^i =
\frac{4\pi}{c} j^i$ \\
Retarded four-potentials in Lorenz gauge (142.1)
\[
A^i(\mathbf{x}, t) = \frac{1}{c} \int \frac{j^i(\mathbf{x}', t - \|\mathbf{x} -
\mathbf{x}'\|/c)}{\|\mathbf{x} - \mathbf{x}'\|} \, d^3 x'
\]
Four-current for point particle (p.~142)
\begin{align*}
\rho(\mathbf{x}, t) &= e\delta^3(\mathbf{x} - \mathbf{x}_c(t)) \\
\mathbf{j}(\mathbf{x}, t) &= e\dot{\mathbf{x}}_c(t) \delta^3(\mathbf{x} -
\mathbf{x}_c(t))
\end{align*}
Li\'enard--Wiechert potentials (p.~146)
\begin{align*}
A^0(\mathbf{x}, t) &= \frac{ec}{cr_r - \mathbf{v}_r \cdot \mathbf{r}_r} \\
\mathbf{A}(\mathbf{x}, t) &= \frac{e\mathbf{v}_r}{cr_r - \mathbf{v}_r \cdot
\mathbf{r}_r}
\end{align*}
where $\mathbf{r}_r = \mathbf{x} - \mathbf{x}_c(t_r)$; $\mathbf{v}_r =
\dot{\mathbf{x}}_c(t_r)$; and $t_r$ is the retarded time and satisfies
$c(t-t_r) = \|\mathbf{x} - \mathbf{x}_c(t_r)\|$. \\
E and B fields of point charge (p.~148)
\begin{align*}
\mathbf{u} &= c\hat{\mathbf{r}}_r - \mathbf{v}_r \\
\mathbf{E} &= e\frac{r_r}{(\mathbf{r}_r \cdot \mathbf{u})^3} [(c^2 -
v_r^2)\mathbf{u} + \mathbf{r}_r \times (\mathbf{u} \times
\ddot{\mathbf{x}}_c(t_r))] \\
\mathbf{B} &= \hat{\mathbf{r}}_r \times \mathbf{E}
\end{align*}
Approximate fields of localized charge configuration ($x' \ll r, x' \ll
\lambda$) (p.~158)
\begin{align*}
A^0(\mathbf{r}, t) &\approx \frac{q}{r} + \frac{\hat{\mathbf{r}}}{r^2} \cdot
\mathbf{d}(t_0) + \frac{1}{c} \frac{\hat{\mathbf{r}}}{r} \cdot
\dot{\mathbf{d}}(t_0) \\
\mathbf{A}(\mathbf{r}, t) &\approx \frac{1}{c} \frac{1}{r}
\dot{\mathbf{d}}(t_0)
\end{align*}
where $t_0 = t - r/c$, and $\mathbf{d}$ is the dipole moment. \\
E and B fields for pure dipole (p.~163), Poynting vector (p.~164), and total
power radiated (p.~165)
\begin{align*}
\mathbf{E} &= \frac{1}{c^2 r} (\ddot{\mathbf{d}}(t_0) \times \hat{\mathbf{r}})
\times \hat{\mathbf{r}} \\
\mathbf{B} &= \hat{\mathbf{r}} \times \mathbf{E} \\
\mathbf{S} &= \hat{\mathbf{r}} \sin^2 \theta
\frac{\|\ddot{\mathbf{d}}(t_0)\|^2}{4\pi c^3 r^2} \\
P_{\text{tot}} &= \frac{2}{3c^2} \|\ddot{\mathbf{d}}(t_0)\|^2
\end{align*}
Power of oscillating dipole, $d = qa \cos \omega t$ (p.~165) \\
$\langle P_{\text{tot}} \rangle = \frac{e^2 a^2}{3c^3} \omega^4$ \\
Electromagnetic stress-energy tensor
\begin{gather*}
T^{km} = -\frac{1}{4\pi} F^{kj} F^m{}_j + \frac{\eta^{km}}{16\pi} F^{ij} F_{ij}
\\
\partial_k T^{km} = 0 \quad \text{(172.1)} \\
T^{km} = T^{mk} \quad \text{(p.~172)} \\
T^{00} = \frac{1}{8\pi} (E^2 + B^2) = \mathcal{E}_{\text{em}} \\
T^{\alpha 0} = T^{0\alpha} = \frac{\mathbf{S}}{c} \\
\begin{split}
T^{\alpha\beta} &= -\frac{1}{4\pi} \bigg[E_\alpha E_\beta + B_\alpha B_\beta \\
&\qquad- \frac{1}{2} \delta_{\alpha\beta}(E^2 + B^2)\bigg]
\quad \text{(p.~176)}
\end{split} \\
\sigma_{\alpha\beta} = -T^{\alpha\beta}
\end{gather*}
$\sigma$ is Maxwell stress tensor. $T^{\alpha 0}$ is $c$ times momentum
density. $T^{0\alpha}$ is $1/c$ times energy flux density. $T^{\alpha\beta}$ is
momentum flux.\\
Total electromagnetic force on volume (p.~178)
\begin{align*}
F^\beta_V &= \int_S n^\alpha T^{\alpha\beta} da \\
F^\beta_V &= f^\beta_V + \frac{1}{c^2} \frac{d}{dt} \int_V \mathbf{S} \, d^3 x
\end{align*}
$n$ is inward unit normal. $f^\beta_V$ is total force on charges and currents,
\[
f^\beta_V = \int_V \rho\mathbf{E} + \frac{\mathbf{j}}{c} \times \mathbf{B} \,
d^3 x
\]
$\frac{1}{c^2} \int_V \mathbf{S} \, d^3 x$ is total electromagnetic momentum.
\\
Classical electron radius (p.~187): $r_e = \frac{e}{m_e c^2} \approx 10^{-13}$
cm \\
Radiation reaction (p.~198): $m\mathbf{a} = \mathbf{F} + \frac{2e^2}{3c^3}
\dot{\mathbf{a}}$ \\
Electromagnetic duality:
\begin{align*}
\mathbf{E} &\to \mathbf{B} \\
\mathbf{B} &\to -\mathbf{E} \\
\mathbf{d} &\to \frac{\vec\mu}{c} \\
\vec\mu &\to -c\mathbf{d}
\end{align*}
$\mathbf{d}$ is electric dipole moment, $\vec\mu$ is magnetic dipole moment \\
Potential of uniformly moving charge ($\mathbf{x}' = (vt, 0, 0)$, HW2)
\[
\phi = \frac{\gamma e}{\sqrt{\gamma^2(x-vt)^2 + y^2 + z^2}}
\]
$A_x = \beta \phi \quad A_y = 0 \quad A_z = 0$ \\
Field of uniformly moving charge (HW3)
\begin{align*}
E_x &= \frac{\gamma e(x-vt)}{(\gamma^2(x-vt)^2 + y^2 + z^2)^{3/2}} &\quad& B_x =
0 \\
E_y &= \frac{\gamma ey}{(\gamma^2(x-vt)^2 + y^2 + z^2)^{3/2}} &\quad& B_y =
-\beta E_z \\
E_z &= \frac{\gamma ez}{(\gamma^2(x-vt)^2 + y^2 + z^2)^{3/2}} &\quad& B_z =
\beta E_y
\end{align*}
\textit{i.e.}, $\mathbf{B} = \vec\beta \times \mathbf{E}$ \\
Boundary conditions for E and B fields
\begin{align*}
E^+_\perp - E^-_\perp = 4\pi \sigma &\quad& B^+_\perp - B^-_\perp = 0 \\
\mathbf{E}^+_\parallel - \mathbf{E}^-_\parallel = \mathbf{0} &\quad& 
\mathbf{B}^+_\parallel - \mathbf{B}^-_\parallel = \frac{4\pi}{c} \vec\kappa
\times \hat{\mathbf{n}}
\end{align*}
$\sigma$ is surface charge density, $\vec\kappa$ is surface current density
\end{multicols}
\end{document}
