# Calculate energy from vacuum electromagnetic-field action

I'm reading Rubakov's "Classical Theory of Gauge Fields", and I'm having a little bit of trouble with problem 7, p 15:

Using an expression of the type $E = \int d^{3} x \frac{\delta L}{\delta \dot{\varphi} (x)} \dot{\varphi} (x) - L$, find the energy of an electromagnetic field with action $S = -\frac{1}{4} \int d^{4}x F_{\mu \nu} F_{\mu \nu}$ and dynamical coordinates $A_{\mu} (x,t)$. Show that when the field equations $\partial_{\mu} F_{\mu \nu} = 0$ are satisfied, the energy thus found agrees with $E =\frac{1}{2} \int d^{3}x (\vec{E}^2 + \vec{H}^2)$.

I used the expression $E = \int d^{3}x \frac{\delta L}{\delta (\partial_{0} A_{\nu})} \partial_{0}A_{\nu} - L$, where $L = \int d^{3}x (-\frac{1}{4} F_{\mu \nu} F_{\mu \nu})$. First though,

$\frac{\delta }{\delta (\partial_{0} A_{\nu})} (\int d^{3} x (\partial_{\alpha} A_{\beta} - \partial_{\beta} A_{\alpha})^2) = (?) 2F_{\alpha \beta} (\delta_{0 \alpha} \delta_{\nu \beta} - \delta_{0 \beta} \delta_{\nu \alpha}) = 4F_{0 \nu}$.

Substituting, I got

$E = \int d^{3}x (-F_{0 \nu} \partial_{0}A_{\nu} + \frac{1}{4} F_{\mu \nu} F_{\mu \nu})$,

which is where I'm now stuck; I'm suspicious of the first term. Any hints/ help would be much appreciated.

• It appears that there are some major problems with your use of index notation. Are you aware of the precise workings of the Einstein summation convention (which you are violating; you're not using raised indices where they are necessary)? – Danu Apr 21 '14 at 8:06
• I have a little familiarity with upstairs/downstairs notation, but upstairs indices are rarely used in the portion of Rubakov's book I've read so far, and I've been following his convention, but you're correct, I ought to translate to safer notation – L. Ron Hubbard Apr 21 '14 at 11:30

First, and most important of all: as said in the comments, you are not using the Einstein summation convention! This way you will have mistakes at your computation! Solve it.

Solving that problem you may find an expression like

$$E = \int d^3 x \left[ F^{\sigma 0}\partial^0 A_\sigma + \frac{1}{4} F^{\mu \nu} F_{\mu \nu} \right]$$

As you can see, it's very similar to what you have found

Now comes the "trick". We can modify the integral adding surface terms. Always will be possible to get a surface with radius going to infinity and to assume that the fields goes to zero as it approaches infinity. What I'm saying is that

$$\int d^3 x \partial_\sigma (F^{\sigma 0} A^0) = 0$$

If you subtract the expression for the Energy and this surface term,

$$E = \int d^3 x \left[ F^{\sigma 0}\partial^0 A_\sigma - \partial_\sigma (F^{\sigma 0} A^0) + \frac{1}{4} F^{\mu \nu} F_{\mu \nu} \right]$$

Of course, $\partial_\sigma (F^{\sigma 0} A^0) = \partial_\sigma F^{\sigma 0} A^0 + F^{\sigma 0} \partial_\sigma A^0$, and using that $\partial_\sigma F^{\sigma 0}=0$ we get $\partial_\sigma (F^{\sigma 0} A^0) = F^{\sigma 0} \partial_\sigma A^0$.

So,

$$E = \int d^3 x \left[ F^{\sigma 0}\partial^0 A_\sigma - F^{\sigma 0} \partial_\sigma A^0 + \frac{1}{4} F^{\mu \nu} F_{\mu \nu} \right]$$

Where we can identify the electromagnetic field tensor,

$$E = \int d^3 x \left[ F^{\sigma 0} F^0_{\;\;\sigma} + \frac{1}{4} F^{\mu \nu} F_{\mu \nu} \right]$$

Now it's just a matter of substitution to find the exactly expression you want

Adding a surface term is a thing more comprehensive when you are looking to the Stress-Energy-Momentum tensor.

From that point of view what you get as conserved current via Noether first theorem is

$$\Theta^{\mu \nu} = F^{\sigma \mu} \partial^\nu A_\sigma + \eta^{\mu\nu} \frac{1}{4} F^{\alpha \beta} F_{\alpha \beta}$$

Satisfying

$$\partial_\mu \Theta^{\mu \nu} = 0$$

You can note that $E = \int d^3x \Theta^{00}$, then turning back to your problem

As we can see, $\Theta^{\mu \nu}$ isn't a symmetric quantity. We can add to it a surface term, which will not contribute when computing the integrals $d^3 x$:

$$T^{\mu \nu} = \Theta^{\mu \nu} - \partial_\sigma (F^{\sigma \mu} A^\nu)$$

It's important to note that $T^{\mu \nu}$ is also conserved,

$$\partial_\mu T^{\mu \nu} = \partial_\mu \Theta^{\mu \nu} - \partial_\mu \partial_\sigma (F^{\sigma \mu} A^\nu)$$ $$\partial_\mu T^{\mu \nu} = \partial_\mu \Theta^{\mu \nu} = 0$$

This last expression follows from the fact that $\partial_\mu \partial_\sigma$ is symmetric in $\mu$ and $\sigma$ while $F^{\sigma \mu}$ is antisymmetric.

One canonical reference you may like to see is Jackson, Chap 12 (Dynamics of Relativistic Particles and Electromagnetic Fields). In the 1999 edition you can look at equation 12.104, where it's exhibit the stress-energy-tensor. And equation 12.111 - 12.113 where he defines the symmetric stress-energy-momentum tensor.

I'm using a different convention because I like to use $T^{\mu \nu}$ for the symmetric case. There's no problem on it.

• Nice answer, very prompt, thank you very much. I found the information about the stress-energy-momentum tensor and reference very useful too. – L. Ron Hubbard Apr 21 '14 at 12:11