I am not sure I understand how this identity:

$$ \eta^{\alpha \beta} \left[ \partial_{\mu}\partial_{\beta} h_{\alpha \nu} - \frac{1}{2} \partial_{\mu} \partial_{\nu} h_{\alpha \beta} \right] = \partial_{\mu} \left[\eta^{\alpha \beta} \partial_{\beta} \left( h_{\alpha \nu} - \frac{1}{2} \eta_{\alpha \nu} \eta^{\sigma \tau} h_{\sigma \tau}\right) \right]$$

Of course the first term on the RHS makes complete sense, but I have no intuition for why the second term on the RHS is identical to the second term on the LHS and how these indices are working.

This has arisen when attempting to use the Lorenz gauge to simplify the components of the Ricci tensor.

Any insight would be appreciated.

  • 2
    $\begingroup$ $\eta^{\alpha\beta}\eta_{\alpha\nu}=\delta^\beta_\nu$ $\endgroup$
    – Phoenix87
    Apr 20 '20 at 23:35
  • $\begingroup$ Phonix87's identity was very useful. The right hand side was essentially two terms, expand out of the parenthesis then it's almost done. $\endgroup$ Apr 21 '20 at 1:47

Let's work back from RHS to LHS: Expand: $$ \eta^{\alpha\beta}\partial_\mu\partial_\beta h_{\alpha\nu}-\frac{1}{2}\eta^{\alpha\beta}\partial_\beta\partial_\mu\eta_{\alpha\nu}\eta^{\sigma\tau}h_{\sigma\tau} \tag{1} $$ You know can see clearly how the first term in the expansion is the same as the LHS so I won't use that any further but it's obviously still there. $$ =...-\frac{1}{2}\eta^{\alpha\beta}\partial_\beta\partial_\mu\eta_{\alpha\nu}\eta^{\alpha\beta}h_{\alpha\beta} \tag{2} $$ We're free to choose any indices for the last part as it's just the sum of h components (with correct signs from metric). I.e. $$ \eta^{\sigma\tau}h_{\sigma\tau}=\eta^{\alpha\beta}h_{\alpha\beta} \tag{3} $$ So carrying on: $$ =...-\frac{1}{2}\partial^\alpha\partial_\mu\eta_{\alpha\nu}\eta^{\alpha\beta}h_{\alpha\beta} \tag{4} $$ $$ =...-\frac{1}{2}\partial_\nu\partial_\mu\eta^{\alpha\beta}h_{\alpha\beta} \tag{5} $$ Hopefully this is clear enough to follow!

  • 1
    $\begingroup$ Very helpful thank you :) General relativity exam soon! $\endgroup$ Apr 21 '20 at 12:15

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.