0
$\begingroup$

I am doing some tensor product calculations, which involve terms like this:

$F^{\mu\lambda}\partial^\nu A_\lambda$

I am trying to write it as a total differential of some quantity. To achieve this I hope to use equation of motion for free field $F^{\mu\lambda}$: $$\partial_\lambda F^{\mu\lambda}=0$$

Are following transformations correct:

$F^{\mu\lambda}\partial^\nu A_\lambda = F^{\mu\lambda}\partial_\alpha A_\lambda g^{\alpha\nu} = F^{\mu\lambda}\partial_\lambda\delta^\lambda_\alpha A_\lambda g^{\alpha\nu}=F^{\mu\lambda}\partial_\lambda A_\alpha g^{\alpha\nu}=F^{\mu\lambda}\partial_\lambda A^\nu=\partial_\lambda(F^{\mu\lambda}A^\nu)$

The main question is: Can I use Kronecker delta after second equal sign to change $\partial_\alpha$ with $\partial_\lambda$? I am worried because $\lambda$ is already used in that term as a summation index.

$\endgroup$
  • 2
    $\begingroup$ You can’t have four $\lambda$s in a contracted tensor. $\endgroup$ – G. Smith Jun 16 at 23:20
  • $\begingroup$ Use $\rho$ instead of $\lambda$: $F^{\mu\lambda}\partial_\rho \delta^\rho_\alpha A_\lambda g^{\alpha\nu}$ $\endgroup$ – Avantgarde Jun 17 at 1:12
  • $\begingroup$ @Avantgarde I want to match one of the index of $F^{\mu\lambda}$, to use equation of motion afterwards. In that sense there is no difference between $\rho$ and $\alpha$. $\endgroup$ – abendrot Jun 17 at 6:44
  • $\begingroup$ But that is another problem. The question was about renaming the indices, so from @G.Smith answer it appears it is not possible to match one of the index which is already used in a tensor. If one really wants to change the index, it should be a completely new one, like $\rho$. $\endgroup$ – abendrot Jun 17 at 7:08
  • $\begingroup$ @abendrot Yes. Repeated indices appear twice in an expression. Not more times than that. $\endgroup$ – Avantgarde Jun 17 at 7:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.