Hamiltonian density of EM fields

Question

I'm learning about the field theory of electromagnetism. The Lagrangian density for an electromagnetic field can be taken to be $$ \mathcal{L} = -\frac{1}{4} F^{\mu\nu} F_{\mu\nu} + \mu_0 A^\mu J_\mu $$

such that the Euler-Lagrange equations reproduce the inhomogeneous Maxwell's equations \begin{align*} \partial_\nu F^{\mu\nu} = \partial_\nu \frac{\partial\mathcal{L}}{\partial(\partial_\nu A_\mu)} = \frac{\partial\mathcal{L}}{\partial A_\mu} = \mu_0 J^\mu \end{align*}

Is there a Hamiltonian formalism that gives the same result?

Answer 1

The metric convention is (-+++). Given the Lagrangian, the conjugate momenta are defined as \begin{align*} \pi^j = \frac{\partial \mathcal{L}}{\partial (\partial_0 A_j)} = F^{j0} = \frac{E_j}{c} \end{align*}

The Hamiltonian density is then \begin{align*} \mathcal{H} &= \pi^j \partial_0 A_j - \mathcal{L} \\ &= \pi^j (\partial_j A_0 - F_{j0}) + \frac{1}{4} (F^{jk} F_{jk} + F^{j0} F_{j0} + F^{0k} F_{0k}) - \mu_0 J^\mu A_\mu \\ &= \pi^j (\partial_j A_0 + \pi_j) + \frac{1}{4} (F^{jk} F_{jk} - \pi^j \pi_j - \pi^k \pi_k) - \mu_0 J^\mu A_\mu \\ &= \pi^j \partial_j A_0 + \frac{1}{4} F^{jk} F_{jk} + \frac{1}{2} \pi^j \pi_j - \mu_0 J^\mu A_\mu \\ &= \frac{1}{c^2} \vec{E} \cdot \nabla \varphi + \frac{1}{2} (|\vec{B}|^2 + \frac{1}{c^2} |\vec{E}|^2) - \mu_0 (\vec{J} \cdot \vec{A} - \rho \varphi) \end{align*}

The Hamilton equations of motion are then \begin{align*} \partial_0 \pi^j &= - \frac{\partial \mathcal{H}}{\partial A_j} - \partial_k \frac{\partial \mathcal{L}}{\partial (\partial_k A_j)} = - \mu_0 J^j - \partial_k F^{jk} \\ \partial_0 A^j &= \frac{\partial \mathcal{H}}{\partial \pi_j} = \partial^j A_0 + \pi^j \end{align*}

The second equation is just the definition of $\pi^j$ but the first equation is Faraday's law.