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\subsection{The $4$-potential}

We previously wrote
\[ \bf{E} = -\nabla \phi - \pfrac{\bf{A}}{t} ,\]
in terms of scalar and vector potentials. What
does this mean for $F_{\mu\nu}$?
\begin{align*}
    F_{i0}
    &= \frac{E_i}{c} \\
    &= v-\frac{1}{c} \pfrac{\phi}{x_i} -
    \frac{1}{c} \pfrac{A_i}{t} \\
    &= \partial_i \left( -\frac{\phi}{c} \right) -
    \partial_0 A_i \\
    F_{ij}
    &= \eps_{ijk} B_k \\
    &= \eps_{ijk} \eps_{klm} \partial_l A_m \\
    &= \partial_i A_j - \partial_j A_i
\end{align*}
We can interpret $F_{\mu\nu}$ as the antisymmetric
derivative
\[ \boxed{F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu
A_\mu} \]
of the covector
\[ A_\mu = \left( -\frac{\phi}{c}, \bf{A} \right) \]
The related $4$-vector is the \emph{4-potential}
\[ A^\mu = \left( \frac{\phi}{c}, \bf{A} \right) .\]
A gauge transformation
\[ \tilde{\bf{A}} = \bf{A} + \nabla \chi, \qquad
\tilde{\phi} = \phi - \pfrac{\chi}{t} \]
corresponds to replacing $A_\mu$ with
\[ \tilde{A}_\mu = A_\mu + \partial_\mu \chi \]
i.e. adding $\partial_\mu \chi = \left(
\frac{1}{c} \pfrac{\chi}{t}, \nabla \chi \right)$
to $A_\mu = \left( -\frac{\phi}{c}, \bf{A}
\right)$. This leaves $F_{\mu\nu}$ unaffected,
because
\[ \partial_\mu(\partial_\nu \chi) -
\partial_\nu(\partial_\mu \chi) = 0 .\]
The spacetime equivalent of the Coulomb gauge is
the \emph{Lorentz gauge} in which
\[ \partial_\mu A^\mu = 0 .\]
If $\partial_\mu A^\mu \neq 0$ then we can make a
gauge transformation such that $\partial_\mu
\tilde{A}^\mu = 0$ by choosing $\chi$ to be the
solution of the forced wave equation
\[ -\Box \chi = \partial_\mu A^\mu \]
(Note that $\partial_\mu \partial^\mu \chi =
\eta^{\mu\nu} \partial_\mu \partial_\nu \chi =
\Box \chi$).

\subsection{Lorentz transformation of $\mathbf{E}$ and $\mathbf{B}$}

Since $F_{\mu\nu}$ is a $(0, 2)$ tensor, it
transforms as
\[ F'_{\mu\nu} = (\lambda^{-1})\indices{^\rho_\mu}
(\lambda^{-1})\indices{^\sigma_\nu} F_{\rho\sigma} \]
In matrix notation,
\[ F' = (\lambda^{-1})^\top F(\Lambda^{-1}) \]
For a Lorentz boots in standard configuration,
\[ \Lambda^{-1} =
\begin{pmatrix}
    \gamma & \gamma \beta & 0 & 0 \\
    \gamma\beta & \gamma & 0 & 0 \\
    0 & 0 & 1 & 0 \\
    0 & 0 & 0 & 1
\end{pmatrix}
\]
Hence (exercise):
\begin{align*}
    F'_{01}
    &= \gamma^2 (F_{01} + \beta^2 F_{10} +
    \beta(F_{00} + F_{11}) \\
    &= \gamma^2 (1 - \beta^2) F_{01} \\
    &= F_{01} \\
    F'_{02}
    &= \gamma(F_{02} + \beta F_{12}) \\
    F'_{03}
    &= \gamma(F_{03} + \beta F_{13}) \\
    F'_{12}
    &= \gamma(F_{12} + \beta F_{02}) \\
    F'_{13}
    &= \gamma(F_{13} + \beta F_{03}) \\
    F'_{23}
    &= F_{23}
\end{align*}
From which we deduce
\begin{equation*}
    \boxed{
    \begin{aligned}
        E_x'
        &= E_x
        &B_x'
        &= B_x \\
        E_y'
        &= \gamma(E_y - vB_z)
        &B_y'
        &= \gamma \left( B_y + \frac{v}{c^2} E_z
        \right) \\
        E_z'
        &= \gamma(E_z + vB_y) \qquad
        &B_z'
        &= \gamma \left( B_z - \frac{v}{c^2} E_y \right)
    \end{aligned}
    }
\end{equation*}
More generally if $\bf{v}$ is the velocity of $S'$
with respect to $S$,
\begin{equation*}
    \boxed{
    \begin{aligned}
        \bf{E}'_{\parallel}
        &= \bf{E}_{\parallel}
        &\bf{B}'_{\parallel}
        &= \bf{B}_{\parallel} \\
        \bf{E}'_{\perp}
        &= \gamma(\bf{E}_{\perp} + \bf{v} \times
        \bf{B}_{\perp}) \qquad
        &\bf{B}'_{\perp}
        &= \gamma \left( \bf{B}_{\perp} -
        \frac{1}{c^2} \bf{v} \times \bf{E}_{\perp}
        \right)
    \end{aligned}
    }
\end{equation*}
where we decompose $\bf{E}$ and $\bf{B}$ into
components parallel and perpendicular to $\bf{v}$:
\[ \bf{E} = \bf{E}_\parallel + \bf{E}_\perp \qquad
\bf{B} = \bf{B}_\parallel + \bf{B}_\perp \]
$F_{\mu\nu}$ can be double raised to make
\[ F^{\mu\nu} =
\begin{pmatrix}
    0 & E_x / c & E_y / c & E_z / c \\
    -E_x / c & 0 B_z & -B_y \\
    -E_y / c & -B_z & 0 & B_x \\
    -E_z / c & B_y & -B_x & 0
\end{pmatrix}
= -F^{\mu\nu} \]
Thus
\begin{align*}
    F^{00}
    &= 0 \\
    F^{0i}
    &= \frac{E_i}{c} \\
    F^{i0}
    &= -\frac{E_i}{c} \\
    F^{ij}
    &= \eps_{ijk} B_k
\end{align*}
We can find two Lorentz-invariant combinations of
$\bf{E}$ and $\bf{B}$ by forming scalars from
$F_{\mu\nu}$:
\begin{align*}
    F_{\mu\nu} F^{\mu\nu}
    &= F_{00} F^{00} + F_{0i} F^{0i} + F_{i0}
    F^{i0} + F_{ij} + F_{ij} \\
    &= 0 + \left( -\frac{E_i}{c} \right)
    \frac{E_i}{c} + \frac{E_i}{c} \left(
    -\frac{E_i}{c} \right) + \eps_{ijk} B_k
    \eps_{ijl} B_l \\
    &= -2 \frac{|\bf{E}|^2}{c^2} + 2\delta_{kl}
    B_k B_l \\
    &= -\frac{2}{c^2} \left(|\bf{E}|^2 - c^2
    |\bf{B}|^2\right) \\
    \eps_{\mu\nu\rho\sigma} F^{\mu\nu}
    F^{\rho\sigma}
    &= \eps_{0ijk} F^{0i} F^{jk} + \eps_{i0jk}
    F^{i0} F^{jk} + \eps_{ij0k} F^{ij} F^{0k} +
    \eps_{ijk0} F^{ij} F^{k0} \\
    &= \eps_{ijk} \frac{E_i}{c} \eps_{jkl} B_l -
    \eps_{ijk} \left( -\frac{E_i}{c} \right)
    \eps_{jkl} B_l + \eps_{ijk}\eps_{ijl} B_l
    \frac{E_k}{c} - \eps_{ijk} \eps_{ijl} B_l
    \left( -\frac{E_k}{c} \right) \\
    &= 4\eps_{ijk} \frac{E_i}{c} \eps_{jkl} B_l \\
    &= \frac{8}{c} \delta_{il} E_i B_l \\
    &= \frac{8}{c} \bf{E} \cdot \bf{B}
\end{align*}
where we used $\eps_{0ijk} = \eps_{ijk}$ and
permutations. Thus $|\bf{E}|^2 - c^2|\bf{B}|^2$ is
a scalar and $\bf{E} \cdot \bf{B}$ is a
pseudoscalar. For example, if $\bf{E} \perp
\bf{B}$ in one inertial frame then they are
perpendicular in any frame.

\begin{example*}[Boosted line charge]
    From \hyperlink{subsubsection.2.1.2}{Section
    2.1.2}, the electric field of a
    line charge $\lambda$ per unit length along
    the $x$-axis is
    \[ \bf{E} = \frac{\lambda}{2\pi\eps_0 (y^2 +
    z^2)} (0, y, z) \]
    while $\bf{B} = \bf{0}$ for static charge.
    Under a standard Lorentz boost,
    \begin{align*}
        \bf{E}'
        &= \frac{\gamma\lambda}{2\pi\eps_0(y^2 +
        z^2)}(0, y, z) \\
        &= \frac{\gamma\lambda}{2\pi\eps_0(y'^2 +
        z'^2}(0, y', z') \\
        \bf{B}'
        &= \frac{\gamma v}{c^2}
        \frac{\lambda}{2\pi\eps_0(y^2 + z^2)} (0,
        z, -y) \\
        &= \frac{\mu_0 \gamma \lambda v}{2\pi
        (y'^2 +z'^2)} (0, z', -y')
    \end{align*}
    $\bf{E}'$ can be understood as the electric
    field of a line charge $\lambda' =
    \gamma\lambda$ per unit length along the
    $x'$-axis. (Length contraction enhances the
    charge density). $\bf{B}'$ can be understood
    as the magnetic field due to a current $I' =
    -\lambda' v$ along the $x'$-axis. (The line
    charge moves with velocity $-v$ in $S'$).
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/ff37b234be6111ed.png}
    \end{center}
\end{example*}
