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I was reading Polchinski book and there he tries to make a gauge choice to fix the redundacies in the Polyakov action.

In order to make this choice, he fixes $$\tau=X^{+}\tag{1.3.8a}$$ and then he says $f=\gamma_{\sigma\sigma}\left(-\det\gamma\right)^{-1/2}$ transforms as $$f'd\sigma'=fd\sigma \tag{1.3.9},$$ under reparameterizations of $\sigma$ where $\tau$ is left fixed. I would like to show this last equation, but I couldn't. Can you help me?

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  • $\begingroup$ Hint: Write down the transformation law for a metric tensor $\gamma_{ab}$ under a coordinate transformation $(\tau,\sigma)\to (\tau,\sigma^{\prime})$. $\endgroup$
    – Qmechanic
    Commented Feb 21, 2017 at 21:20
  • $\begingroup$ I make that , I got this transformation under the supposition $\sigma'\left(\sigma\right)$. But I would like to know if It's unnecessary this supposition. $\endgroup$
    – 7919
    Commented Feb 21, 2017 at 21:33
  • $\begingroup$ $\gamma'_{\sigma'\sigma'}=\frac{\partial\sigma^{a}}{\partial\sigma'}\frac{\partial\sigma^{b}}{\partial\sigma'}\gamma_{ab}$ . In order to eliminate terms as $\gamma_{\tau\sigma}$ or $\gamma_{\tau\tau}$ . I used the assumption $\sigma'(\sigma)$ $\endgroup$
    – 7919
    Commented Feb 21, 2017 at 21:45
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    $\begingroup$ $\uparrow$ You should. $\endgroup$
    – Qmechanic
    Commented Feb 22, 2017 at 14:14

1 Answer 1

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Note that the determinant of the metric behave as a $()_{\sigma\sigma}$ (two lower $\sigma$ indices) under $\sigma$-reparametrizations with $\tau$ held fixed.

$$ \det(\gamma_{ab})=\gamma_{\sigma\sigma}\gamma_{\tau\tau}-\gamma_{\sigma\tau}\gamma_{\sigma\tau} $$ You can see that from the counting how many $\sigma$ indices there are in each term of the determinant. The square root of it, $\sqrt{\det(\gamma_{ab})}$ will behave as $()_{\sigma}$, just one index. Finally, taking the inverse this will behave as $()^{\sigma}$, just one upper index. Now, $f=\gamma_{\sigma\sigma}(\det(\gamma_{ab}))^{-1/2}$ will behave as $()_{\sigma}$ since $2-1=1$. Knowing that $d\sigma$ behave as $()^{\sigma}$ under $\sigma$-reparametrization you get:

$$ fd\sigma=f'd\sigma' $$

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