What is the Mistake in Deriving the EM Hamiltonian?

[RedX]

For some reason I can't derive the Hamiltonian from the Lagrangian for the E&M field. Here's what I have (using +--- metric):

<br /> \begin{equation*}<br /> \begin{split}<br /> \mathcal L=\frac{-1}{4}F_{ \mu \nu}F^{ \mu \nu}<br /> <br /> \\<br /> <br /> \Pi^\mu=\frac{\delta \mathcal L}{\delta \dot{A_\mu}}=-F^{0 \mu}<br /> <br /> \\<br /> <br /> \mathcal H=\Pi^\mu \dot{A}_\mu -\mathcal L=-F^{0 \mu}\dot{A}_\mu +\frac{1}{4}F_{ \mu \nu}F^{ \mu \nu}<br /> =-F^{0 \mu}\dot{A}_\mu+\frac{1}{4}(2F_{0i}F^{0i}+F_{ij}F^{ij})<br /> \end{split}<br /> \end{equation*}<br />

But F_0i=Ei, and F_ij=-B_k, so this is equal to:

<br /> \mathcal H=-F^{0 \mu}\dot{A}_\mu+\frac{1}{2}(-E_{i}^2+B_{i}^2)


The Hamiltonian however should be one half the sum of the squares of the electric and magnetic fields. But I can't figure out what I did wrong. I almost have it, as the first term almost adds to the 2nd term to give that, but not quite.

Also, I'm not quite sure when using the (+---) metric whether the canonical momenta is:

<br /> \Pi^\mu=\frac{\delta \mathcal L}{\partial^0 A_\mu}<br />
or

<br /> \Pi^\mu=\frac{\delta \mathcal L}{\partial_0 A_\mu}

I don't think it matters in the derivation of the Hamiltonian, but which one do you use in the canonical commutation relations for example?

 

[fzero]

OK, I missed a term the first time around. We can write

<br /> -F^{0 \mu}\dot{A}_\mu = -F^{0 i}\partial_0{A}_i = - F^{0i}( F_{0i} + \partial_i A_0 ) = E_i^2 + E_i (\partial_i A_0)<br />

Then

<br /> \mathcal H=-F^{0 \mu}\dot{A}_\mu+\frac{1}{2}(-E_{i}^2+B_{i}^2) =\frac{1}{2}(E_{i}^2+B_{i}^2) + E_i (\partial_i A_0).<br />

The last term can be written as

\partial_i ( A_0 E_i) - A_0 ( \partial_i E_i) .

The total derivative will vanish in the volume integral, while the second term vanishes by Gauss' law in the absence of sources.I believe that there's a cleaner way of doing things by treating the choice of gauge as either a 1st or 2nd class constraint, but I don't remember the details.

 

[dextercioby]

Also, I'm not quite sure when using the (+---) metric whether the canonical momenta is:

<br /> \Pi^\mu=\frac{\delta \mathcal L}{\partial^0 A_\mu}<br />
or

<br /> \Pi^\mu=\frac{\delta \mathcal L}{\partial_0 A_\mu}

It doesn't matter whether the 0 is upstairs or downstairs, if you're using the (+---) metric. It would have mattered however if you were using the (-+++) metric, of course. However, for consistency and esthetical reasons, the 0 should be downstairs when it's the time derivative of the vector potential. So the second formula in my quote is the right one to use.

And the canonical hamiltonian for a constrained system is always computed on the hypersurface of primary constraints. In your case, it's the hypersurface \Pi_{0} = \Pi^{0} = 0.

To find the hamiltonian in terms of the classical fields E and B, first compute wrt the potentials and the canonical momenta.