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\markright{{\sf  Electrodinamics in covariant notation}}
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\section{11.9\quad Electrodynamics in covariant notation.}

\subsection{11.9.1}

\quad \thinspace \textbf{a. }The most elegant way of showing that Maxwell's
equations are invariant under Lorentz transformations is to rewrite them in
a ``manifestly covariant'' form. The procedure is pretty unique, based only
on the assumption that the electric charge is a Lorentz-invariant.

\smallskip\ 

\textbf{b}. The conservation of charge is expressed by the continuity
equation 
\begin{equation}
\frac{\partial \rho }{\partial t}=\mathbf{\nabla \cdot J}=0  \label{1}
\end{equation}
or by the equivalent statement that 
\[
q=\int \rho (\mathbf{x,\,}t)d^3\mathbf{x} 
\]
does not change in time, except by a flow of current through the surface
bounding the volume under consideration.

\smallskip\ 

\textbf{c. }Now $d^4x=dx^0dx^1dx^2dx^3$ is a Lorentz invariant because the
Lorentz transformation has Jacobian $J=\left\| \partial x^\alpha /\partial
x^{\prime \nu }\right\| =1.$ It follows that $\rho $ must transform like $%
dx^0$ (because $\int \rho d^3x$ and $\int dx^0d^3x$ are both invariant),
i.e. like the zeroth component of a contravariant vector.

\smallskip\ 

\textbf{d. }Even if we did not know the above conclusions, the simple
requirement that (\ref{1}) must hold in all Lorentz frames and the covariant
nature of $\partial /\partial x^\alpha $ are sufficient to conclude that the
four quantities 
\begin{equation}
J^\alpha =(c\rho ,\,\mathbf{J)}  \label{2}
\end{equation}
are contravariant components of a 4-vector.

\smallskip\ 

\textbf{e.} Therefore the current conservation law can be written in the
manifestly covariant form 
\begin{equation}
\partial _\alpha J^\alpha =0.  \label{1'}
\end{equation}

\newpage

\subsection{11.9.2}

\quad \quad \textbf{a. }Next we consider the potentials $\phi $ and $\mathbf{%
A.}$ In the Lorentz gauge 
\begin{equation}
\frac{1}{c}\frac{\partial \phi }{\partial t}+\mathbf{\nabla \cdot A}=0,
\label{3}
\end{equation}
they obey 
\begin{equation}
\mathbf{\Box A}=\frac{4\pi }{c}\mathbf{J\quad \quad \quad \quad \quad \quad
\Box }\phi =\frac{4\pi }{c}c\rho   \label{4}
\end{equation}
Since $(c\rho ,\,\mathbf{J)}$ is a four-vector and $\mathbf{\Box }$ is
invariant, it is clear that $A^{\alpha }=(\phi ,\,\mathbf{A)}$ is also a
contravariant four-vector and we can write 
\begin{equation}
\partial _{\alpha }A^{\alpha }=0\quad \quad \quad \quad (\text{Lorentz gauge)%
}  \label{3'}
\end{equation}
\begin{equation}
\mathbf{\Box }A^{\alpha }=0\quad \quad \quad \quad (\text{wave equation)}
\label{4'}
\end{equation}

\smallskip\ 

\textbf{b.} In a different gauge, 
\[
\phi ^{\prime }=\phi -\frac 1c\frac{\partial \Lambda }{\partial t}\,,\quad
\quad \quad \quad \mathbf{A}^{\prime }=\mathbf{A+\nabla }\Lambda \,. 
\]
Clearly this gives $A^{\prime \alpha }=A^\alpha -\partial ^\alpha \Lambda .$
(The contravariant derivative $\partial ^\alpha $ is given, of course, by $%
\partial ^\alpha =g^{\alpha \beta }\partial _\beta \,,$ or, explicitly, $%
\partial ^\alpha =((1/c)\partial /\partial t,\,-\mathbf{\nabla )).}$ Thus
the potentials form a 4-vector in any gauge, not just in the Lorentz gauge.
However, if $\mathbf{A}$ has been chosen to satisfy, say, the Coulomb gauge
condition $div\mathbf{A}=0\mathbf{\ }$in a certain Lorentz frame, it may not
satisfy it in other frames. We stress that it is the Coulomb condition that
fails to be Lorentz-invariant, not the potentials.

\smallskip\ 

\textbf{c. }Proceeding now to the fields $\mathbf{E}$ and $\mathbf{B,}$ it
is clear that, being given by the derivatives of a vector, they must form a
tensor. In fact if we define the antisymmetric tensor 
\begin{equation}
F^{\alpha \beta }=\partial ^\alpha A^\beta -\partial ^\beta A^\alpha
\label{5}
\end{equation}
we see by explicit calculation and comparison with $\mathbf{B}=\mathbf{%
\nabla \times A,}$ $\mathbf{E}=-(1/c)\partial \mathbf{A/}\partial t-\mathbf{%
\nabla }\phi $ that 
\begin{equation}
F^{\alpha \beta }=\left( 
\begin{tabular}{llll}
$\,\,0$ & $-E_x$ & $-E_y$ & $-E_z$ \\ 
$E_x$ & $\quad 0$ & $-B_z$ & $\,\,\,\,B_y$ \\ 
$E_y$ & $\,\,\,\,B_z$ & $\quad 0$ & $-B_x$ \\ 
$E_z$ & $-B_y$ & $\,\,\,\,B_x$ & $\quad 0$%
\end{tabular}
\right) .  \label{6}
\end{equation}

\newpage 

\subsection{11.9.3}

\quad \quad \textbf{a. }The covariant form of the field-strength tensor $F$
is 
\begin{equation}
F_{\alpha \beta }=g_{\alpha \gamma }F^{\gamma \delta }g_{\delta \beta
}=\left( 
\begin{tabular}{llll}
$\quad 0$ & $\,\,\,\,E_x$ & $\,\,\,\,E_y$ & $\,\,\,\,E_z$ \\ 
$-E_x$ & $\quad 0$ & $-B_z$ & $\,\,\,\,B_y$ \\ 
$-E_y$ & $\,\,\,\,B_z$ & $\quad 0$ & $-B_x$ \\ 
$-E_z$ & $-B_y$ & $\,\,\,\,B_x$ & $\quad 0$%
\end{tabular}
\right) .  \label{6'}
\end{equation}

\smallskip

\textbf{b.} We can also write this, if we wish, as $E_i=F_{0i}\,,$ $%
-B_i=(1/2)\epsilon _{ijk}F^{jk}$ (latin indices run over spatial
directions). This is consistent with the fact that $\mathbf{E}$ is a radial
vector in 3-space, while $\mathbf{B}$ is an axial vector, odd under parity
and related in the standard way to the components of an antisymmetric tensor
in 3-space.

\smallskip

\textbf{c.} One may be disturbed by the fact that $\mathbf{E}$ and $\mathbf{B%
}$ appear in different ways in $F^{\alpha \beta },$ while Maxwell's
equations have a special symmetry under duality transformations $\mathbf{E}%
\rightarrow \mathbf{B,B}\rightarrow \mathbf{-E.}$ But the duality is
restored and given a further interpretation by introducing the dual field
strength tensor 
\begin{equation}
\frak{F}^{\alpha \beta }=\frac 12\epsilon ^{\alpha \beta \gamma \delta
}F_{\gamma \delta }=\left( 
\begin{tabular}{llll}
$\,\,0$ & $-B_x$ & $-B_y$ & $-B_z$ \\ 
$B_x$ & $\quad 0$ & $\,\,\,\,E_z$ & $-E_y$ \\ 
$B_y$ & $-E_z$ & $\quad 0$ & $\,\,\,\,E_x$ \\ 
$B_z$ & $\,\,\,\,E_y$ & $-E_x$ & $\quad 0$%
\end{tabular}
\right) .  \label{6''}
\end{equation}

\smallskip

\textbf{d.} The connection between an antisymmetric tensor and its dual
through the Ricci tensor $\epsilon ^{\alpha \beta \gamma \delta }$ is the
analogue of the relation between an axial vector and an antisymmetric tensor
in 3-space, such as $-B_i=(1/2)\epsilon _{ijk}F^{jk}.$

\smallskip

\textbf{e. }The Ricci symbol $\epsilon ^{\alpha \beta \gamma \delta }$ is
actually a pseudotensor (tensor analogue of axial vector) since its
components do not change sign under inversion. That makes $\frak{F}^{\alpha
\beta }$ a pseudotensor too. No problem.

\newpage

\subsection{11.9.4}

\quad \quad \textbf{a. }Next\textbf{\ }we write Maxwell's equations in
manifestly covariant form. The correspondence is 
\begin{eqnarray}
\mathbf{\nabla \cdot E\quad \quad \quad \quad } &=&4\pi \rho \quad \quad
\quad \quad \Longrightarrow \partial _\alpha F^{\alpha \beta }=\frac{4\pi }%
cJ^\beta \quad \quad \quad \quad \quad  \label{7} \\
\mathbf{\nabla \times B}-\frac 1c\frac{\partial \mathbf{E}}{\partial t} &=&%
\frac{4\pi }c\mathbf{J}  \nonumber
\end{eqnarray}

\begin{eqnarray}
\mathbf{\nabla \cdot B\quad \quad \quad \quad } &=&0\quad \quad \quad \quad
\Longrightarrow \partial _\alpha \frak{F}^{\alpha \beta }=0\quad \quad \quad 
\text{or}  \label{8} \\
\mathbf{\nabla \times E}+\frac 1c\frac{\partial \mathbf{B}}{\partial t}
&=&0\quad \quad \quad \quad \quad \quad \partial _\alpha F_{\beta \gamma
}+\partial _\beta F_{\gamma \alpha }+\partial _\gamma F_{\alpha \beta }=0 
\nonumber
\end{eqnarray}

\smallskip

\textbf{b.} The last equation looks unwieldy, until one realizes that only 4
components of the formidable-looking third rank tensor are essentially
distinct: 123, 230, 301 and 012.

\smallskip

\textbf{c.} Finally, we require that the Lorentz-Newton force equation 
\begin{equation}
m\frac{d\mathbf{v}}{dt}=q\left( \mathbf{E+}\frac{\mathbf{v}}c\times \mathbf{B%
}\right)  \label{9}
\end{equation}
be made into a Lorentz invariant equation. This actually requires a
modification of the left-hand side, since Newton's law is Galileo-covariant
and not Lorentz-covariant.

\smallskip

\textbf{d.} The modification, as suggested by Einstein, is to write $md%
\mathbf{v/}dt$ as $d\mathbf{p}/dt$ and then replace the Newtonian mass $m$
by $m\gamma $, or in other words the momentum should be $m\gamma \mathbf{v}$
and not simply $m\mathbf{v.}$ The Lorentz-Einstein force equation is then 
\begin{equation}
\frac{d\mathbf{p}}{dt}\equiv \frac{d(m\gamma \mathbf{v)}}{dt}=q\left( 
\mathbf{E+}\frac{\mathbf{v}}c\times \mathbf{B}\right) .  \label{10}
\end{equation}

\smallskip

\textbf{e.} The covariance of this is seen by introducing the covariant
four-velocity $u^\alpha =cdx^\alpha /d\tau $ (so that $p^\alpha =mu^\alpha
). $ With $dt=\gamma d\tau $ and $u^\alpha =(\gamma c,\gamma \mathbf{v),}$ (%
\ref{10}) becomes 
\begin{equation}
\frac{dp^\alpha }{d\tau }\equiv m\frac{du^\alpha }{d\tau }=\frac qcF^{\alpha
\beta }u_\beta  \label{11}
\end{equation}

\smallskip

\textbf{f.} The zeroth component of (\ref{11}) gives 
\begin{equation}
\frac{d(m\gamma c^2)}{dt}=q\mathbf{v\cdot E}  \label{12}
\end{equation}
which is the rate of energy change.

\smallskip

\end{document}
