# Variation of the action using tensor algebra.

## Homework Statement

Hi, I have a problem calculating the variation of the action using tensor algebra because two derivative indices are causing some problem.

## Homework Equations

Generally you have the action $S = \int L(A^{\mu}, A^{\mu}_{\;,\nu}, x^{\mu})d^4x$
where:
$A ^{\mu}= A^{\mu}(x^{\nu})$
$A ^{\mu}_{\;,\nu} = \frac{\partial A^{\mu}}{\partial x^{\nu}}$
$x^{\nu} = (x^0, x^1, x^2 ,x^3)$
$d^{4}x = dx^0 dx^1dx^2dx^3$

Would I be correct in stating that the variation of the action is $\delta S = \int ( \frac{\partial L}{\partial A^{\mu} } \delta A^{\mu} + \frac{\partial L}{\partial A ^{\mu}_{\;,\nu} } \delta A ^{\mu}_{\;,\nu} ) d^{4} x$ ?

## The Attempt at a Solution

Say that our function L looks like this:
$L = A_{\mu, \nu}-A_{\nu, \mu}$
where $A_{\mu} = \eta _{\mu \nu} A^{\nu}$ and $\eta_{\mu \nu}$ is the Minkowski metric tensor.
How do I make sense of this in context of the variation $\delta S$ above? More specifically what should the derivative index of the variation $\delta A^{\mu}$ be? Because L is the $\nu$ derivative of $A_{\mu}$ minus $A_{\nu}$ derivated with respect to the $\mu$ derivative.
Specifically what I want to accomplish is to rewrite the right hand side of $\delta S$ as the sum of two parts that looks something like this
$\frac{\partial L}{\partial A_{\mu , \nu} } \delta A_{\mu, \nu}=\frac{\partial }{\partial x^{\nu}} (\frac{\partial L}{\partial A_{\mu,\nu}} \delta A_{\mu})-\delta A_{\mu} \frac{\partial }{\partial x^{\nu}}( \frac{\partial L}{\partial A_{\mu,\nu}})$ but for me to be able to do that I need a common derivative index in L which I don't have. I have two separate derivative indexes and I have not idea what to do with them. Thank you for your help.

I know the equations are physically nonsensical since I have removed all the clutter beside the actual problem, so this is mainly a mathematical question. I could not decide whether it should be in the Physics section or Math section. If a moderator think it should be moved somewhere else please feel free to do it there or tell me and I'll do it later.

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vanhees71
Gold Member
2019 Award
Your Lagrangian doesn't make sense, because it's a 2nd-rank tensor but should be a scalar (density). The action for the free em. field in Heaviside-Lorentz units with $c=1$ reads
$$\mathcal{L}=-\frac{1}{4} F_{\mu \nu} F^{\mu \nu},$$
where
$$F_{\mu \nu}=\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}.$$
Then you can use the antisymmetry of this tensor to write
$$\delta \mathcal{L}=-\frac{1}{2} F^{\mu \nu} \delta F_{\mu \nu}=-F^{\mu \nu} \partial_{\mu} \delta A_{\nu},$$
where I've also taken into account that in the Hamilton principle the space-time variables are not varied, i.e., that $\delta (\partial_{\mu} A_{\nu})=\partial_{\mu} \delta A_{\nu}$.

• 1 person
Thank you! It did not answer what I was looking for but it still helped in another way. Which is good since my question, as you pointed out was badly phrased. I will try and derive the result you posted as I cannot see the connection by just looking at it.

Regarding my original question. It does not have to be related to the Lagrangian. If you think of L as just some 2nd rank tensor, would it make more sense then or is there some other error which makes the expression illegal even if one just rewrites $\delta S \rightarrow \delta S^{\mu \nu}$?

Another question, my teacher did this on the whiteboard some lecture ago without motivation.
$\frac{\partial F_{\mu \nu}}{\partial A^{\mu}_{\;,\nu}} = \frac{\partial F_{\mu \nu}}{\partial A_{\mu, \nu}}$

Is this true generally for any function or only in the case where A is the 4-potential and F is Maxwell field tensor?

*edit*

I managed to solve the problem. Thank you for the help! But the question above still stands :)

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