This might be a little bit of a technical question, so bear with me. Ok, so from string theory we know that the action for a relativistic string is found from the worldsheet when we embed the string in spacetime and so we get $$ S_{Nambu-Goto}=T \int d^2 \sigma \sqrt{h_{\alpha \beta}}$$ where $h_{\alpha \beta} = \eta_{\mu \nu} \partial_{\alpha} X^{\mu} \partial_{\beta} X^{\nu},$ the derivatives taken with respect to the the worldsheet coordinates $\sigma^{\alpha}=(\tau, \sigma)$.

That is all nice and well, and then we want to get rid of this ugly square root so we introduce the auxiliary variable $\gamma_{\alpha \beta}$ and write $$S_{Polyakov}= \frac{T}{2} \int d^2 \sigma \sqrt {\gamma} \gamma^{\alpha \beta} h_{\alpha \beta}$$ Upon varying the action with respect to $\gamma_{\mu \nu}$ to find the EOM, you get $\gamma_{\alpha \beta}=h_{\alpha \beta},$ so this auxiliary variable is just the induced metric, and everything simplifies nicely into the Nambu-Goto action when substituted.

My question is, during the derivation of the EOM, you had to vary the action ofcourse, which gave you (which is just the chain rule) $$\delta S_P =\bigg[ \frac{\partial \sqrt{\gamma}}{\partial \gamma_{\mu \nu}} (\gamma^{\alpha \beta} h_{\alpha \beta} ) + \sqrt {\gamma} \Big( \frac{\partial \gamma^{\alpha \beta}}{\partial \gamma_{\mu \nu}} h_{\alpha \beta} \Big) \bigg] \delta \gamma_{\mu \nu} $$ When computing $\frac{\partial \sqrt {\gamma}}{\partial \gamma_{\mu \nu}}$, I got this: $$ \frac{\partial \sqrt {\gamma}}{\partial \gamma_{\mu \nu}} = \frac{1}{2 \sqrt{\gamma}} \frac{\partial (det \gamma)}{\partial \gamma_{\mu \nu}}$$ Then I use Jacobi's formula which says that $$\frac{\partial (det \gamma)}{\partial \gamma_{\mu \nu}} = \gamma \cdot tr \Big( \gamma^{\alpha \beta} \frac{\partial \gamma_{\alpha \beta}}{\partial \gamma_{\mu \nu}} \Big) = \gamma \cdot tr (\gamma^{\alpha \beta} \delta_{\alpha}^\mu \delta_{\beta}^\nu) = \gamma \cdot tr(\gamma^{\mu \nu})$$ This is the part that stumped me. The book says that the answer to this should be $$\delta \gamma = \gamma \gamma^{\nu \mu} \delta \gamma_{\mu \nu}$$ However, I am getting this result above, i.e. $$\delta \gamma = \gamma tr (\gamma^{\mu \nu}) \delta \gamma_{\mu \nu}$$ How is $tr(\gamma^{\mu \nu}) = \gamma^{\nu \mu}$ or am I just missing a stupid mistake.

Btw, I am using the fact that $\delta f(x) = \frac{df(x)}{dx}\delta x$ as you saw to find the variation. Sorry if this might be a dumb question, I was just learning GR when I came across an appendix with this and got interested!


1 Answer 1


As you say, $\gamma_{\mu \nu}$ is a metric. Then, $\gamma_{\mu \nu}=\gamma_{ \nu \mu}$ and it makes no difference which one you write down. This already deals with one half of your discrepancy.

Now for the trace: In the einstein summation notation, a the trace of $\gamma_{\mu \nu}$ is $\gamma_{\mu}^{\; \mu}$. The looking at your $$\frac{\partial (det \gamma)}{\partial \gamma_{\mu \nu}} = \gamma \cdot tr \Big( \gamma^{\alpha \beta} \frac{\partial \gamma_{\alpha \beta}}{\partial \gamma_{\mu \nu}} \Big)$$ The $tr \Big( \gamma^{\alpha \beta} \frac{\partial \gamma_{\alpha \beta}}{\partial \gamma_{\mu \nu}} \Big)$ is really $\Big( \gamma^{\alpha \beta} \frac{\partial \gamma_{\alpha \beta}}{\partial \gamma_{\mu}^{\; \mu}} \Big)$. Now in $$\frac{\partial (det \gamma)}{\partial \gamma_{\mu \nu}} = \gamma \cdot \Big( \gamma^{\alpha \beta} \frac{\partial \gamma_{\alpha \beta}}{\partial \gamma_{\mu}^{\; \mu}} \Big)$$ there is clearly something wrong as the indices don't match on either side. The correct expression is $$\frac{\partial (det \gamma)}{\partial \gamma_{\mu \nu}} = \gamma \cdot \Big( \gamma^{\alpha \beta} \frac{\partial \gamma_{\alpha \beta}}{\partial \gamma_{\mu \nu}} \Big)$$ Think about what you should be taking the trace of in your formula. It is only the $\gamma^{\alpha \beta}\partial \gamma_{\alpha \beta}$ part; but as this is scalar, taking the trace would be meaningless.

  • $\begingroup$ Ok cool I understand the symmetry of the metric part, thanks. However, I'm still confused as to why we only take the trace of $\gamma^{\alpha \beta} \partial \gamma_{\alpha \beta}$ since I thought the whole expression with the partial derivative is in the trace. $\endgroup$ Jul 4, 2016 at 1:44
  • $\begingroup$ remember that $\partial (det\gamma )/\partial \gamma_{\mu \nu}$ is not one equation but $d^2$ equations for $d$ spacetime dimensions. Consider writing out the individual equations $\partial (det\gamma )/\partial \gamma_{0 0}$, $\partial (det\gamma )/\partial \gamma_{0 1}$,.. and working some of those out individually. That should make it clear. If not, you might want to revisit Einstein summation convention. $\endgroup$
    – Gerben
    Jul 4, 2016 at 1:48
  • $\begingroup$ Oooooh I got it now, that whole thing is just a scalar! Thanks a lot man! $\endgroup$ Jul 4, 2016 at 1:54

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