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\subsubsection*{Raising and Lowering}

The metric tensor provides an isomorphism between
vectors $(V^\mu)$ and covectors $(V_\mu)$:
\[ V^\mu = \eta^{\mu\nu} V_\nu, \qquad V_\mu =
\eta_{\mu\nu} V^\nu \]
These operations are known as \emph{raising and
lowering indices}.

\myskip
Raising or lowering simply changes the sign of the
$0$ (time) component, while leaving the spatial
components unchanged. So raising or lowering a
Roman index changes nothing. For example: for the
$4$-velocity:
\[ U^\mu = \gamma_{\bf{u}} = \gamma_{\bf{u}}(c,
\bf{u}), \qquad U_\mu = \gamma_{\bf{u}} (-c,
\bf{u}) \]
or the derivative:
\[ \partial_\mu f = \left( \frac{1}{c}
\pfrac{f}{t}, \nabla f \right), \qquad
\partial^\mu f = \left( -\frac{1}{c} \pfrac{f}{t},
\nabla f \right) \]
The inner product of two $4$-vectors can be
written variously as
\begin{align*}
    V \cdot W
    &= \eta_{\mu\nu} V^\mu W^\nu \\
    &= V^\mu W_\mu \\
    &= V_\mu W^\mu \\
    &= \eta^{\mu\nu} V_\mu W_\nu
\end{align*}
Raising and lowering can be applied to any tensor
index, for example
\begin{align*}
    T\indices{_\mu^\nu}
    &= \eta\indices{_\mu_\rho}
    T\indices{^\rho^\nu} \\
    T\indices{^\mu_\nu}
    &= \eta\indices{_\nu_\rho} T\indices{^\mu^\rho}
\end{align*}
Exercise: show that $\eta\indices{^\mu_\nu} =
\delta\indices{^\mu_\nu}$.

\myskip
A tensor may be \emph{symmetric} or
\emph{antisymmetric} on a pair of its indices if
they are both ``down'' or both ``up''. Exercise:
Show that this property is preserved if
\emph{both} indices are raised or lowered.

\subsection{Charge conservation}

The equation of charge conservation,
\[ \pfrac{\rho}{t} + \nabla \cdot \bf{J} = 0 \]
can be written in the form of a spacetime
divergence,
\[ \boxed{\partial_\mu J^\mu = 0} \tag{1} \]
by defining the \emph{charge-curent density}
$4$-vector (or $4$-current density)
\[ \boxed{J^\mu = (\rho c, \bf{J})} \]
Equation (1) is a scalar (Lorentz invariant)
equation, i.e. valid in any inertial frame.

\myskip
Consider a Lorentz boost in standard
configuration. Suppose, in frame $S$, we have
uniform distribution of charge particles at rest,
with charge $q$ and (proper) number density
(number per unit volume) $n$. Then $\rho = nq$ and
$\bf{J} = \bf{0}$, giving
\[ J^\mu = (nqc, \bf{0}) \qquad \text{ in $S$} \]
In frame $S'$, the particles have $3$-velocity
$(-v, 0, 0)$. Their number density increases to
$\gamma n$ because of length contraction in the
$x$-direction. So $\rho' = \gamma nq$ and $\bf{J}'
= \gamma nq(-v, 0, 0)$, giving
\[ J'^\mu = (\gamma nqc, -\gamma nqv, 0, 0) \qquad
\text{ in $S'$} \]
These are consistent with the Lorentz
Transformation of a $4$-vector.

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Now suppose that, in $S$, we have a wire in which
particles of charge $+q$ and number density $n_+$
move along the wire with $3$-velocity $(u, 0, 0)$
and particles of charge $-q$ and number density
$n_- = n_+$ move with $3$-velocity $(-u, 0, 0)$.
The net charge density is $\rho = n_+ q - n_- q =
0$ and the net current density is
\begin{align*}
    \bf{J}
    &= n_+ q(u, 0, 0) - n_- q(-u, 0, 0) \\
    &= (2n_+ qu, 0, 0)
\end{align*}
The wire is uncharged but carries a current. So
$\bf{E} = \bf{0}$ and $\bf{B} \neq \bf{0}$.
\begin{center}
    \includegraphics[width=0.4\linewidth]
    {images/5f0941d4bccb11ed.png}
\end{center}
In $S'$, the $+$ and $-$ particles have different
$x$-velocities
\[ \pfrac{\pm - v}{1 \mp \frac{uv}{c^2}} \]
and therefore different charge densities because
of length contraction. The wire has a net charge
and therefore $\bf{E}' = \bf{0}$.

\myskip
So a magnetic field in one frame can produce an
electric field in another (and vice versa).

\subsection{Electromagnetic tensor}

The relativistic equation of motion for a massive
particle
\[ \dfrac{P^\mu}{\tau} = f^\mu \tag{2} \]
where $\tau$ is the proper time and $f^\mu$ is the
$4$-force. Since $P \cdot P = -m^2 c^2$ is
constant, $f^\mu$ must satisfy $f \cdot P = 0$. In
detail,
\begin{align*}
    \dfrac{}{\tau} (\eta\indices{_\mu_\nu} P^\mu
    P^\nu)
    &= \eta_{\mu\nu} \left(
    \dfrac{P^\mu}{\tau}P^\nu + P^\mu
    \dfrac{P^\nu}{\tau} \right) \\
    &= \eta_{\mu\nu} (f^\mu P^\nu + P^\mu f^\nu)
    \\
    &= 2\eta_{\mu\nu} f^\mu P^\nu \\
    &= 2f_\mu P^\mu \\
    &= 0
\end{align*}
How can we express the Lorentz force as a
$4$-vector (or covector)? (All known charged
particles are massive). It should be $\propto q$
and linear in $U^\mu$. Assume
\[ f_\mu = qF_{\mu\nu} U^\nu \tag{3} \]
for some $(0, 2)$ tensor $F_{\mu\nu}$. Then
\[ f\cdot P = 0 \iff qmF_{\mu\nu} U^\mu U^\nu = 0 \]
and is satisfied for all $U$ if $F_{\mu\nu}$ is
\emph{antisymmetric}. The spatial components of
(3) are
\begin{align*}
    f_i
    &= q (F_{i0} U^0 + F_{ij} U^j) \\
    &= \gamma_{\bf{u}} q(F_{i0} c + F_{ij} u_j)
\end{align*}
For this to agree with $\gamma_{\bf{u}} q(E_i +
\eps_{ijk} u_j B_k)$ (the factor of
$\gamma_{\bf{u}}$ comes from converting
$\dfrac{}{t}$ into $\dfrac{}{\tau}$) we require
\[ F_{i0} = \frac{E_i}{c}, \qquad F_{ij} =
\eps_{ijk} B_k .\]
By antisymmetry,
\[ F_{00} = 0, \qquad F_{0i} = -\frac{E_i}{c} \]
Thus,
\[ \boxed{F_{\mu\nu} = 
\begin{pmatrix}
    0 & -\frac{E_x}{c} & -\frac{E_y}{c} & -\frac{E_z}{c} \\
    \frac{E_x}{c} & 0 & B_z & -B_y \\
    \frac{E_y}{c} & -B_z & 0 & B_x \\
    \frac{E_z}{c} & B_y & -B_x & 0
\end{pmatrix}
} \]
is the \emph{electromagnetic (field) tensor}.

\myskip
Neither $\bf{E}$ nor $\bf{B}$ is the spatial part
of a $4$-vector. However, for a particle with
$4$-velocity $U$, $F_{\mu\nu} U^\nu$ is a
covector that equals $(0, \bf{E})$ in the
particle's rest frame.

\myskip
By construction, the spatial components of (2)
give the relativistic equation of motion
\[ \dfrac{\bf{p}}{\tau} = \gamma_{\bf{u}} q(\bf{E}
+ \bf{u} \times \bf{B}) \]
where $\bf{p} = \gamma_{\bf{u}} m\bf{u}$ is the
relativistic $3$-momentum. Equivalently,
\[ \dfrac{\bf{p}}{\tau} = q(\bf{E} + \bf{u} \times
\bf{B}) = \bf{F} \]
which is familiar. The time component of (2) is
\begin{align*}
    \dfrac{P^0}{\tau}
    &= f^0 \\
    &= -f_0 \\
    &= -qF_{0i} U^i \\
    &= q \frac{E_i}{c} \gamma_{\bf{u}} u_i
\end{align*}
equivalent to
\[ \dfrac{E}{\tau} = q \gamma_{\bf{u}} \bf{E}
\cdot \bf{u} \]
or
\[ \dfrac{E}{t} = q\bf{E} \cdot \bf{u} = \bf{F}
\cdot \bf{u} \]
which is also familiar.