Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

The classical Lagrangian for the electromagnetic field is

$$\mathcal{L} = -\frac{1}{4\mu_0} F^{\mu \nu} F_{\mu \nu} + J^\mu A_\mu.$$

Is there also a Hamiltonian? If so, how to derive it? I know how to write down the Hamiltonian from the Lagrangian where derivatives are taken only with respect to time, but I can't see the obvious way to generalize this.

share|improve this question
More on singular Legendre transformations: physics.stackexchange.com/q/30192/2451 , physics.stackexchange.com/q/47847/2451 and links therein. –  Qmechanic Dec 15 '13 at 10:25
add comment

1 Answer 1

Yes. There is a standard way to generalize to field theory.

Let a theory of $n\geq 1$ fields $\phi^i$ with a Lagrangian density $\mathcal L = \mathcal L(\phi^i, \partial_\mu\phi^i)$ be given. Here we use that standard abuse of notation in which $\phi^i$ denotes the vector whose components are the fields; $\phi^i = (\phi^1, \dots, \phi^n)$.

To obtain the corresponding hamiltonian density, one first defines the following canonical momentum corresponding to the field $\phi^i$: \begin{align} \pi_i(x) = \frac{\partial \mathcal L}{\partial\dot \phi^i}(\phi^i(x), \partial_\mu\phi^i(x)), \qquad \dot\phi^i := \partial_t\phi^i \tag{1} \end{align} Then, the Hamiltian density is \begin{align} \mathcal H = \pi_i\dot\phi^i - \mathcal L \end{align} where a sum over $i$ is implied. Note that as in classical mechanics, on the right hand side of this expression, $\dot \phi^i$ should be replaced with its expression in terms of $\pi_i,\phi^i$ so that the Hamiltonian is a function of $(\pi_i, \phi^i)$ only, namely \begin{align} \mathcal H(\pi_i, \phi^i) = \pi_j \,\dot\phi^j(\pi_i,\phi^i) - \mathcal L(\phi^i, \dot\phi(\pi_i,\phi^i)). \end{align} Again we have abused notation slightly here in writing $\dot\phi^i$ as a function of $\pi_i$ and $\phi^i$. What we mean is the expression for $\dot\phi^i$ is obtained by solving the definition $(1)$ of the canonical momentum for $\dot\phi^i$ in terms of $\pi_i$ and $\phi^i$.

In your case, the fields are $A^\mu$ with corresponding momenta $\pi_\mu$.

share|improve this answer
It's worth mentioning that you need to add Lagrange multipliers to handle the constraints, e.g., $\Pi_0 = \frac{\delta L}{\delta \dot A_0} = 0$ and so on. Here's a random link –  lionelbrits Dec 14 '13 at 19:38
@lionelbrits Yes. Thanks for that. –  joshphysics Dec 14 '13 at 20:31
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.