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\begin{document}

The field can be expressed as \( A^{u}(x^{i},t) \)

\( S=\int d^{4}xL \) where L is the lagrangian density. Action should be a lorentz scalar
if this equation is to apply in all reference frames. Since \( d^{4}x \) is lorentz
invariant, this means \( L \) should be lorentz invariant.

\begin{enumerate}

\item Easiest L: \( A^{u}A_{u} \) problem: changes with gauge transformations. L should
be gauge invariant. \( A^{i'}=A^{i}+\nabla ^{i}x \) and \( \phi '=\phi -\frac{1}{c}\frac{\partial x}{\partial t} \) gives same electric and magnetic field.
More simple expressed, this is: \( A'^{u}=A^{u}+\partial ^{u}x \).

\item Next L: use field strength tensor \( L=\frac{-1}{16\pi }F^{uv}F_{uv} \) so \( \delta L=\frac{-1}{8\pi }F_{uv}\delta F^{uv} \) so \( \delta S=-\frac{1}{8\pi }\int d^{4}x(\partial ^{u}F_{uv}\delta A^{v}-\partial ^{v}F_{uv,}\delta A^{u}) \) \( =-\frac{1}{4\pi }\int d^{4}x(\partial ^{u}F_{uv}\delta A^{v}) \) so \( 0=-\frac{1}{4\pi }\partial ^{u}F_{uv} \) \( =-\frac{1}{4\pi }(\partial ^{u}\partial _{u}A^{v}-\partial _{v}\partial ^{u}A_{u}) \). The lorentz
gauge gives: \( \partial ^{u}A_{u}=0 \) so \( \partial _{u}\partial ^{u}A_{v}=0 \). This is just the wave equation.

\item With a charge or current, \( J^{u} \), assume \( L_{int}=\frac{1}{c}J_{u}A^{u} \). Then \( S=\int d^{4}x(\frac{-1}{16\pi }F_{uv}F^{uv}+\frac{1}{c}J_{u}A^{u}) \) so \( \delta S \) \( =\frac{1}{4\pi }\int d^{4}x(\partial ^{u}F_{uv}\delta A^{v}+\frac{J_{v}}{c}) \) which gives:
\( \partial _{u}F^{uv}=-\frac{4\pi }{c}J^{v} \). Expanding into space and time, you get: \( \frac{1}{c}\frac{\partial }{\partial t}F^{0\nu }+\frac{\partial }{\partial x^{i}}F^{i\nu }=\frac{-4\pi }{c}J^{v} \) . This equation splits
into: \( \nabla ^{i}E_{i}=4\pi \rho  \) and \( \epsilon _{ijk}\nabla ^{j}B^{k}=\frac{4\pi }{c}J_{i}+\frac{1}{c}\frac{\partial E_{i}}{\partial t} \) . The other 2 maxwell's equations are automatically
satisfied using the dual tensor.

\end{enumerate}

\end{document}
