# Relativistic EM Lagrangian and the derivation of equations of motion

As mentioned in my other post, I am attempting to learn from Gross'"Relativistic quantum mechanics and field theory", and I have a question concerning the manipulation of the antisymmetric 4x4 tensors involved.

There are two points where this does not make sense to me. Firstly, when proving that the correct equations of motion can be derived from the EM Lagrangian density:

$$L = -\frac{1}{4} F_{\mu \nu}F^{\mu \nu}-j_\mu A^{\mu}$$

Gross mentions that the equation can be simplified by expanding it the following way:

$$-\frac{1}{4} F_{\mu \nu}F^{\mu \nu} = -\frac{1}{4} (\partial_\mu A_\nu -\partial_\nu A_\mu)(\partial^\mu A^\nu -\partial^\nu A^\mu)$$ $$=-\frac{1}{2} g^{\mu \mu'}g^{\nu \nu'}(\partial_\mu A_\nu\partial_{\mu'} A_{\nu'} -\partial_\mu A_\nu\partial_{\nu'} A_{\mu'})$$

We then construct the Lagrangian equations of motion, which doesn't bother me. My question is (hopefully) much simpler. No matter how I treat the product shown above I cannot remove the multiple of 2. What 4-vector or metric tensor trickery is happening here?

The best I can do is to operate on all of the contravariant partials and A's with the metric tensor and get the following for the expansion: $$-\frac{1}{4} (\partial_\mu A_\nu -\partial_\nu A_\mu)(\partial^\mu A^\nu -\partial^\nu A^\mu)$$ $$= -\frac{1}{4} g^{\mu \mu'}g^{\nu \nu'}(\partial_\mu A_\nu\partial_{\mu'} A_{\nu'} -\partial_\mu A_\nu\partial_{\nu'} A_{\mu'}-\partial_\nu A_\mu\partial_{\mu'} A_{\nu'}+\partial_\nu A_\mu\partial_{\nu'} A_{\mu'})$$

I understand from here the argument of cancelling the factor of 2 is as simple as grouping these terms into two distinct factions, but I don't see how one can do so?

I can sort of see how, if $\mu = \mu'$ and $\nu = \nu'$ how the middle two terms could be reshuffled, ie: $$= -\frac{1}{4} g^{\mu \mu'}g^{\nu \nu'}(\partial_\mu A_\nu\partial_{\mu'} A_{\nu'} -2\partial_\mu A_\nu\partial_{\nu'} A_{\mu'}+\partial_\nu A_\mu\partial_{\nu'} A_{\mu'})$$

But even if I allow for that, the partials for the last two terms are completely different.

I expect that my misunderstanding here stems from being unfamiliar with 4-vectors. My background is Chemistry and I am just trying to understand some of these deeper concepts.

A similar occurrence happens two pages later, where Gross states that in making the relativistic lagrangian density, we can separate out the scalar potential terms (the time terms in the A 4-vector), by doing:

$$L = -\frac{1}{4} F_{\mu \nu}F^{\mu \nu}-j_\mu A^{\mu}$$ $$= -\frac{1}{2} \partial_\mu A^0(\partial^\mu A^0 -\partial^0 A^\mu)+\frac{1}{2} \partial_\mu A^i(\partial^\mu A^i -\bigtriangledown_i A^\mu) - \rho A^0 + j \cdot A$$ Where $\rho$ and $A^0$ are the time terms for the j and A 4-vectors, respectively.

Similarly, I don't know how the $-\frac{1}{4}$ term can be broken up here either. I know that this is something dead simple, but I'm stuck and I'd appreciate any help.

• – Frobenius Sep 10 '18 at 4:35

The answer to my question is that in the derivation: $$-\frac{1}{4} F_{\mu \nu}F^{\mu \nu} = -\frac{1}{4} (\partial_\mu A_\nu -\partial_\nu A_\mu)(\partial^\mu A^\nu -\partial^\nu A^\mu)$$

$$=-\frac{1}{2} g^{\mu \mu'}g^{\nu \nu'}(\partial_\mu A_\nu\partial_{\mu'} A_{\nu'} -\partial_\mu A_\nu\partial_{\nu'} A_{\mu'})$$

There is a step in between which is omitted, which is to expand as:

$$-\frac{1}{4}(\partial_\mu A_\nu \partial^\mu A^\nu-\partial_\mu A_\nu \partial^\nu A^\mu-\partial_\nu A_\mu\partial^\mu A^\nu+\partial_\nu A_\mu \partial^\nu A^\mu)$$

And then to consolidate the terms, which we are allowed to do BECAUSE the $\mu$ and $\nu$ are simply dummy variables. After expanding, all that is needed is to swap the dummy indices in the last two terms to....

$$-\frac{1}{4}(\partial_\mu A_\nu \partial^\mu A^\nu-\partial_\mu A_\nu \partial^\nu A^\mu-\partial_\mu A_\nu \partial^\nu A^\mu+\partial_\mu A_\nu \partial^\mu A^\nu)$$

You'll notice that this expression is identical to the last, except in the last two terms I have replaced $\mu$ everywhere with $\nu$ and vica versa. This is allowed because these are simply dummy indices which we are summing over, as bolbteppa helpfully pointed out in a comment.

At this point its obvious that the like terms can be summed and the factor of two can be brought out front to affect the change that had me stumped.

$$\frac{1}{4} \rightarrow \frac{1}{2}$$

Lastly, I'd like to point out that although this is an easy mistake, it stems from something non-trivial. I have no prior knowledge of the malleability of dummy indices. It was something that I had to look up when bolbteppa mentioned it. Without that knowledge, this series of simplifications looks impossible.

In your second equation the right hand side consists of four terms. In the next line only two terms remain because the four are two by two equal, giving the factor of two that you are looking for. By the way, for clarity you should get rid of the metric tensors, just write lower and upper indices.