# Hamiltonian density

Lagrangian density $\mathscr{L}(A, \partial^{\mu}A, x^{\mu})$ in absence of any source
$$
\begin{equation}
\begin{split}
\mathscr{L}
&=
-\frac{1}{4\mu_{0}}
F^{\mu\nu}F_{\mu\nu}
\\
&=
-\frac{1}{4\mu_{0}}
(
\partial^{\mu}A^{\nu}
-
\partial^{\nu}A^{\mu}
)
(
\partial_{\mu}A_{\nu}
-
\partial_{\nu}A_{\mu}
)
\\
&=
-\frac{1}{2\mu_{0}}
\eta^{\mu\mu^{\prime}}\eta^{\nu\nu^{\prime}}
(
\partial_{\mu}A_{\nu}
\partial_{\mu^{\prime}}A_{\nu^{\prime}}
-
\partial_{\mu}A_{\nu}
\partial_{\nu^{\prime}}A_{\mu^{\prime}}
)
\\
\end{split}
\end{equation}
$$
And the Euler-Lagrangian equation becomes
$$
\begin{split}
\frac{
\partial \mathscr{L}
}{
\partial (\partial^{\mu}A^{\mu})
}
&=
\frac{\partial
}{
\partial (\partial^{\mu}A^{\mu})
}
\left[
-\frac{1}{2\mu_{0}}
\eta_{\mu\mu^{\prime}}\eta_{\nu\nu^{\prime}}
(
\partial^{\mu}A^{\nu}
\partial^{\mu^{\prime}}A^{\nu^{\prime}}
-
\partial^{\mu}A^{\nu}
\partial^{\nu^{\prime}}A^{\mu^{\prime}}
)
\right]
\\
&=
-\frac{1}{2\mu_{0}}
\frac{\partial
}{
\partial (\partial^{\mu}A^{\mu})
}
\left[
\eta_{\mu\mu^{\prime}}\eta_{\nu\nu^{\prime}}
(
\partial^{\mu}A^{\nu}
\partial^{\mu^{\prime}}A^{\nu^{\prime}}
-
\partial^{\mu}A^{\nu}
\partial^{\nu^{\prime}}A^{\mu^{\prime}}
)
\right]
\\
&=
-\frac{1}{2\mu_{0}}
\frac{\partial
}{
\partial (\partial^{\mu}A^{\mu})
}
\left[
\eta_{\mu\mu^{\prime}}\eta_{\nu\nu^{\prime}}
(
\partial^{\mu^{\prime}}A^{\nu^{\prime}}
-
\partial^{\nu^{\prime}}A^{\mu^{\prime}}
)
\right]
\\
\end{split}
$$
Define the Legendre transform
$$
\pi^{0}
=
\frac{\partial \mathscr{L}}{\partial (\partial_{t} A_{0})}
=
\frac{\partial \mathscr{L}}{\partial \dot{A_{0}}}
=0
$$
$$
\pi^{i}
=
\frac{
\partial \mathscr{L}
}{
\partial (\partial_{t}A_{i})
}
=
\frac{\partial \mathscr{L}}{\partial \dot{A_{i}}}
=
-\frac{1}{\mu_{0}}F^{0i}
=
-\frac{1}{\mu_{0}}(-\frac{1}{c}E^{i})
=
\frac{1}{\mu_{0}c}E^{i}
$$

Therefore the Hamiltonian density $\mathscr{H}(A^{i}, \Pi^{\mu}, \partial^{i}\phi, x^{\mu})$ is
$$
\begin{split}
\mathscr{H}
&=\Pi^{\mu} \dot{\phi}_{\mu} - \mathscr{L}\\
&=\Pi^{i}\dot{A_{i}} - \mathscr{L}
\end{split}
$$
Solve the first term
$$
\begin{split}
\Pi^{i}\dot{A_{i}}
&=\frac{\partial\mathscr{L}}{\partial (\partial_{0}A_{0})}
(\partial_{t}A_{\nu})\\
&=(-\frac{1}{\mu_{0}}F^{0\nu})
(F_{0\nu}+\partial_{\nu}A_{0})\\
&=-\frac{1}{\mu_{0}}F^{0\nu}F_{0\nu}
-\frac{1}{\mu_{0}}F^{0\nu}\partial_{\nu}A_{0}\\
&=-\frac{1}{\mu_{0}}(\frac{1}{c}E^{\nu})(\frac{1}{c}E_{\nu})
-\frac{1}{\mu_{0}c}E^{i}(\vec{\nabla}\frac{\phi}{c})\\
\end{split}
$$
The second term has derived above
$$
\begin{split}
\mathscr{L}
&=-\frac{1}{4\mu_{0}}F^{\mu\nu}F_{\mu\nu}\\
&=\frac{1}{2}(B^{2}-E^{2})\\
\end{split}
$$
Then the Hamiltonain density (source free) becomes
$$
\mathscr{H}=\frac{1}{2}(E^{2}+B^{2})
$$