Euler-Lagrange Equations: EM Field Term

[Gene Naden]

This problem is about one small step in the derivation of Maxwell's equations in free space from the field Lagrangian. The Lagrangian contains a term proportional to

$\partial \mu A_\nu \partial^\mu A^\nu - \partial \nu A\mu \partial ^\mu A^\nu$ where A is the four-vector potential.

The Euler-Lagrange equation requires a derivative $\partial \mu ( \frac{\partial \mathcal{L}}{\partial ( \partial \mu A \nu )})$

The problem is to get a term $\partial \mu ( \partial ^\mu A^\nu - \partial ^\nu A ^\mu)$

I want to apply the product rule of calculus to compute $\frac {\partial ( \partial_\alpha A_\beta \partial ^\alpha A^\beta)}{\partial ( \partial_\mu A_\nu )}$ and $\frac {\partial ( \partial_\beta A_\alpha \partial ^\alpha A^\beta)}{\partial ( \partial_\mu A_\nu )}$

This should bring in a factor of two.

But I am having difficulty managing the indices. The stack exchange says I can use the identity for $\frac{ \partial ( \partial_\alpha A _\beta )}{\partial ( \partial_\mu A_\nu)}$ in terms of the Kronecker $\delta$ but I am unsure of the corresponding identity for $\frac{ \partial ( \partial^\alpha A ^\beta )}{\partial ( \partial_\mu A_\nu)}$

 

[Orodruin]

but I am unsure of the corresponding identity for $\frac{ \partial ( \partial^\alpha A ^\beta )}{\partial ( \partial_\mu A_\nu)}$

Lower the indices in the numerator by extracting a metric from it, i.e., $\partial^\alpha A^\beta = \eta^{\alpha\gamma}\eta^{\beta\delta} \partial_\gamma A_\delta$. The metric components are constant and can be taken out of the derivative. You can then apply your previous relation with deltas, essentially leaving you with the metric tensors.

 

[Gene Naden]

Thanks, I get $\frac{\partial \left( \partial ^\alpha A^\beta \right)}{\partial \left( \partial _\mu A_\nu \right)} = \eta ^{\alpha^\mu} \eta ^{\beta \nu}$ and then it works out, just as you say.