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\subsection{Maxwell's equations}

How do we write Maxwell's equations
\begin{align*}
    \nabla \cdot \bf{E}
    &= \frac{\rho}{\eps_0}
    \tag{M1} \\
    \nabla \cdot \bf{B}
    &= 0
    \tag{M2} \\
    \nabla \times \bf{E}
    &= -\pfrac{\bf{B}}{t}
    \tag{M3} \\
    \nabla \times \bf{B}
    &= \mu_0 \left( \bf{J} + \eps_0
    \pfrac{\bf{E}}{t} \right)
    \tag{M4}
\end{align*}
in a covariant form? Using the relations:
\[ F^{0i} = -F^{i0} = \frac{E_i}{c}, \quad F^{ij}
= \eps_{ijk} B_k, \quad J^0 = \rho c, \quad J^i =
J_i, \quad c^2 = \frac{1}{\mu_0\eps_0} \]
(M1) becomes
\[ c \partial_i ^{0i} = \frac{J^0}{\eps_0 c}
\implies \partial_i F^{0i} = \mu_0 J^0 \]
and (M4) becomes
\[ \eps_{ijk} \partial_j B_k = \mu_0 J_0 +
\frac{1}{c} \partial_0 E_i \implies \partial_0
F^{i0} + \partial_j F^{ij} = \mu_0 J^i \]
These are the components of the 4-vector equation
\[ \boxed{\partial_\nu F^{\mu\nu} = \mu_0 J^\mu} \]
Now
\[ F_{ij} = \eps_{ijk} B_k \implies B_i = \half
\eps_{ijk} F_{jk} \]
So (M2) becomes
\begin{align*}
    \partial_i B_i
    &= \half \eps_{ijk} \partial_i F_{jk} \\
    &= \partial_1 F_{23} + \partial_2 F_{31} +
    \partial_3 F_{12} \\
    &= 0
    \tag{1}
\end{align*}
equivalent to
\[ \partial_i F_{jk} + \partial_j F_{ki} +
\partial_k F_{ij} = 0 \tag{3} \]
(trivially satisfied if any of $i, j, k$ coincide
because $F$ is antisymmetric).

\myskip
(M3) becomes
\[ \eps_{ijk} \partial_j E_k = -c \partial_0 B_i
\implies c \eps_{ijk} \partial_j F_{k0} +
\frac{c}{2} \eps_{ijk} \partial_0 F_{jk} = 0
\tag{3} \]
Multiply by $\eps_{ilm}$ and divide by $c$:
\[ \partial_l F_{m0} - \partial_m F_{l0} +
\partial_0 F_{lm} = 0 .\]
Using antisymmetry and relabelling:
\[ \partial_i F_{j0} + \partial_j F_{0i} +
\partial_0 F_{ij} = 0 \tag{4} \]
Combine (2) and (4) into the covariant form
\[ \boxed{\partial_\mu F_{\nu\rho} + \partial_\nu
F_{\rho\mu} + \partial_\rho F_{\mu\nu} = 0} \tag{5} \]
Alternatively, we can write this as a 4-vector
equation
\[ \boxed{\eps^{\mu\nu\rho\sigma} \partial_\nu
F_{\rho\sigma} = 0} \tag{6} \]
The $0$-component of (6) is
\[ \eps^{0ijk} \partial_i F_{jk} = -\eps_{ijk}
\partial_i F_{jk} = 0 \]
equivalent to (1). The $i$-component of (6) is
\begin{align*}
    \eps^{i0jk} \partial_0 F_{jk} + \eps^{ij0k}
    \partial_j F_{0k} + \eps^{ijk0} \partial_j
    F_{k0}
    &= \eps_{ijk} \partial_0 F_{jk} - \eps_{ijk}
    \partial_j F_{0k} + \eps_{ijk} \partial_j
    F_{k0} \\
    &= \eps_{ijk} \partial_0 F_{jk} + 2\eps_{ijk}
    \partial_j F_{k0} \\
    &= 0
\end{align*}
equivalent to (3) $\times \frac{2}{c}$. In
summary, Maxwell's equations can be written in
covariant form as
\[ \boxed{\partial_\nu F^{\mu\nu} = \mu_0 J^\mu}
\tag{7} \]
and either
\[ \boxed{\partial_\mu F_{\nu\rho} + \partial_\nu
F_{\rho\mu} + \partial_\rho F_{\mu\nu} = 0} \tag{8} \]
or
\[ \boxed{\eps^{\mu\nu\rho\sigma} \partial_\nu
F_{\rho\sigma}} \tag{9} \]
The equation of motion of a charged particle is
\[ \dfrac{P^\mu}{\tau} = f^\mu \quad \text{with}
\quad P^\mu = mU^\mu, f_\mu = qF_{\mu\nu} U^\nu \]
Charge conservation follows easily from (7):
\[ \frac{1}{\mu_0} \partial_\mu \partial_\nu
F^{\mu\nu} = \partial_\mu J^\mu = 0 ,\]
since $\partial_\mu \partial_\nu$ is symmetric
while $F^{\mu\nu}$ is antisymmetric. If we write
$F_{\mu\nu}$ in terms of the 4-potentials as
\[ F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu
A_\mu ,\]
then (8) and (9) are automatically satisfied:
\[ \partial_\mu \partial_\nu A_\rho - \partial_\mu
\partial_\rho A_\nu + \partial_\nu \partial_\rho
A_\mu - \partial_\nu \partial_\mu A_\rho +
\partial_\rho \partial_\mu A_\nu - \partial_\rho
\partial_\nu A_\mu = 0 ,\]
\[ \eps^{\mu\nu\rho\sigma} \partial_\nu
\partial_\rho A_\sigma - \eps^{\mu\nu\rho\sigma}
\partial_\nu \partial_\sigma A_\rho = 0 ,\]
while (7) becomes
\[ \partial_\nu (\partial^\mu A^\nu) -
\partial_\nu (\partial^\nu A^\mu) = \mu_0 J^\mu \]
In the Lorentz gauge, the first term vanishes:
\[ \partial_\nu (\partial^\mu A^\nu) =
\partial^\mu (\partial_\nu A^\nu) = 0 ,\]
leaving
\[ \boxed{-\Box A^\mu = \mu_0 J^\mu} \tag{10} \]
This is an inhomogeneous wave equation for the
4-potential, with a source $\alpha$ charge-current
density:
\begin{itemize}
    \item In the absence of charges and currents,
        (10) describes free EM waves.
    \item In a time-independent situation, (10)
        reduces to
        \[ -\nabla^2 \Phi = \frac{\rho}{\eps_0},
        \qquad -\nabla^2 \bf{A} = \mu_0 \bf{J} \]
        as we found in electrostatics and
        magnetostatics. (Lorentz gauge becomes
        Coulomb gauge when there is no
        time-dependence).
    \item More generally, (10) describes how EM
        waves are generated by time-dependent
        charge and current distributions. (See
        Part II Electrodynamics).
\end{itemize}

\subsubsection*{Non-examinable application}

Motion in a uniform magnetic field
\begin{align*}
    \dfrac{\bf{p}}{t}
    &= q \bf{u} \times \bf{B} \\
    &= \frac{q}{\gamma m} \bf{p} \times \bf{B}
\end{align*}
($\bf{p} = m \gamma \bf{u}$). Circular gyromotion
in plane perpendicular to $\bf{B}$ with frequency
$\frac{qB}{\gamma m}$.
