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\markright{{\sf E+M Field Lagrangian    }}
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\section{12.8\quad EM Field Lagrangian.}

\subsection{12.8.1}

\quad \thinspace \textbf{a. }The Lagrangian formalism becomes truly useful,
at least as a conceptual tool, when applied to a field.

\smallskip\ 

\textbf{b}. Historically, the formalism arose in the context of fluid
dynamics and continuum elastic theory. In these theories one is led to
replace the discrete indices of the $i$-th particle by a continuum label
denoting the position of the particle in the medium. Furthermore, the
interactions between particles are described in terms of derivatives of the
displacements with respect to the continuum labels. In this way the ``field
variables'' depend on time and space in an analogous way and space and time
derivatives appear on equal footing.

\smallskip\ 

\textbf{c. }In the electromagnetic theory, the potentials $A^{\alpha }$ play
naturally the role of field variables (displacements of the ``ether''). It
is best to skip over the history of the subject and go directly to the fully
covariant formulation, which has great conceptual simplicity. Let us say
that we just approach the problem mathematically and try to construct an
action integral $\mathcal{A},$ such that $\delta \mathcal{A}=0$ implies the
Maxwell equations.

\smallskip\ 

\textbf{d. }The action integral is written in terms of the Lagrangian
density $\frak{L}$ as 
\begin{equation}
\mathcal{A}=\int d^{4}x\frak{L\,.}  \label{1}
\end{equation}
Clearly, if $\mathcal{A}$ is an invariant (scalar), so is $\frak{L}$. The
form (\ref{1}) is suggested by $\mathcal{A}=\sum_{i}\int L_{i}\left(
t\right) dt$ for many particles.

\smallskip\ 

\textbf{e. }The search for a proper $\frak{L}$ is restricted to the
invariants of the theory. In the e.m. case, only $F_{\alpha \beta }F^{\alpha
\beta }=2\left( B^2-E^2\right) $ is invariant under all Lorentz
transformations, proper and improper.

\newpage 

\subsection{12.8.2}

\quad \quad \textbf{a. }We conclude that the free-field Lagrangian density
must be a function of $F_{\alpha \beta }F^{\alpha \beta }.$ The simplest
function, just a multiple, is found to work (see \textbf{c. }below).

\smallskip\ 

\textbf{b.} The interaction part of the Lagrangian must give correctly both
the sources in Maxwell's equations and the Lorentz force in the equations of
motion for the particles. Therefore (see 12.1) it must have the form 
\[
-\sum_{i}\frac{e}{c}\frac{dx_{i}^{\alpha }}{dt}A_{\alpha }(\mathbf{x}%
_{i}(t),t).
\]
Introducing the current density $4$-vector $J^{\alpha }\mathbf{(}\mathsf{x}%
\mathbf{)}$ by the equations
\begin{eqnarray*}
J^{0}\mathbf{(}\mathsf{x}\mathbf{)} &=&\sum_{i}ec\,\delta (\mathbf{x-x}%
_{i}(t)) \\
\mathbf{J(}\mathsf{x}\mathbf{)} &=&\sum_{i}e\mathbf{v}_{i}(t)\,\delta (%
\mathbf{x-x}_{i}(t))\,
\end{eqnarray*}
where $\mathbf{v}_{i}\left( t\right) =d\mathbf{x}_{i}/dt,$ we see that 
\[
-e\int \sum_{i}\left[ \phi \left( \mathbf{x}_{i},t\right) -\frac{\mathbf{v}%
_{i}}{c}\cdot \mathbf{A}\left( \mathbf{x}_{i},t\right) \right] \,dt=-\frac{1%
}{c}\int d^{4}x\,J^{\alpha }\mathbf{(}\mathsf{x}\mathbf{)}A_{\alpha }(%
\mathsf{x})\,.
\]
On the l.h.s. of this equation it is understood that $\mathbf{x}_{i}=\mathbf{%
x}_{i}(t)$ and $\mathbf{v}_{i}=\mathbf{v}_{i}\left( t\right) ,$ while on the
r.h.s. $\mathsf{x}=(ct,\mathbf{x)}$ and the explicit particle paths $\mathbf{%
x}_{i}(t)$ have formally disappeared (they are hidden in $J^{\alpha }\mathbf{%
(}\mathsf{x}\mathbf{)}$).

\smallskip \ 

\textbf{c. }The correct Lagrangian density (fixing the undetermined factor
for the free-field part) is 
\begin{equation}
\frak{L}=-\frac 1{16\pi }F_{\alpha \beta }F^{\alpha \beta }-\frac 1cJ_\alpha
A^\alpha  \label{2}
\end{equation}
where, of course, $F^{\alpha \beta }=\partial ^\alpha A^\beta -\partial
^\beta A^\alpha $.

\smallskip\ 

\textbf{d. }It is a straightforward exercise to show that $\delta \mathcal{A}%
=0,$ with $\mathcal{A}$ given by (\ref{1}) and $\frak{L}$ given by (\ref{2}%
), leads to the Euler-Lagrange equations. 
\begin{equation}
\partial ^{\beta }\frac{\partial \frak{L}}{\partial \left( \partial ^{\beta
}A^{\alpha }\right) }-\frac{\partial \frak{L}}{\partial A^{\alpha }}=0
\label{3}
\end{equation}
and that these are equivalent to the inhomogeneous Maxwell equations $%
\partial ^{\beta }F_{\alpha \beta }=\left( 4\pi /c\right) J_{\alpha },$
since 
\[
\frac{\partial \frak{L}}{\partial \left( \partial ^{\beta }A^{\alpha
}\right) }=\frac{1}{4\pi }F_{\alpha \beta }\quad \quad \quad \text{and\quad
\quad \quad }\frac{\partial \frak{L}}{\partial A^{\alpha }}=-\frac{1}{c}%
J_{\alpha }\,.
\]

\newpage 

\subsection{12.8.3}

\quad \quad \textbf{a. }The Lagrangian formulation is in terms of the
potentials $A^\alpha ,$ rather than the fields. It is nevertheless
gauge-invariant. By inspection of (\ref{2}), only the interaction part $%
\left( 1/c\right) J_\alpha A^\alpha $ appears to violate gauge invariance.
But a gauge transformation $A^\alpha \rightarrow A^\alpha +\partial ^\alpha
\Lambda $ adds a term proportional to $J_\alpha \partial ^\alpha \Lambda .$
With an integration by parts, this reduces to $\Lambda \partial ^\alpha
J_\alpha \,,$ which vanishes by current conservation. The connection between
gauge invariance and charge (current) conservation is an important point
that will not be pursued.

\smallskip\ 

\textbf{b. }Of course, since the fields are expressed in terms of
potentials, the homogeneous Maxwell equations are automatically satisfied.

\smallskip\ 

\textbf{c. }Modified or generalized or new field theories are most easily
constructed by postulating appropriate Lagrangians. For instance, the effect
of a photon ``mass'' $\mu $ would be to add a term $\left( \mu ^2/8\pi
\right) A^\alpha A_\alpha $ to $\frak{L.}$ No evidence for such an effect
exists.

\smallskip\ 

\textbf{d. }Not the least advantage of Lagrangians, which makes them a
favorite of theorists, is that the various fields and their interactions
each add a separate term to the total $\frak{L}.$

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