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The polyakov action is defined as

\begin{equation} S=-\frac{T}{2}\int d\sigma d\tau \sqrt{-h}h^{\alpha \beta} \partial_{\alpha}X^{\mu}\partial_{\beta}X^{\nu}\eta_{\mu \nu} \end{equation}

by varing this action with respect to metric yield: \begin{equation} T_{\alpha \beta}= \partial_{\alpha} X \cdot \partial_{\beta}X -\frac{1}{2}h_{\alpha \beta}h^{\gamma \delta}\partial_{\gamma}X \cdot \partial_{\delta}X \end{equation}

and by variation the action with respect to $X$ yield: \begin{equation} \partial_{\alpha}(\sqrt{-h}h^{\alpha \beta}\partial_{\beta}X^{\mu})=0. \end{equation}

It's mentioned in several literature that the equation of motion with respect to $h$ imply the energy-momentum tensor is zero, how this can be shown?

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  • $\begingroup$ You can find it on page 23 in David Tong lectures on String Theory damtp.cam.ac.uk/user/tong/string.html $\endgroup$
    – Navid
    Nov 1, 2023 at 4:01
  • $\begingroup$ It can be shown explicitly by checking each indices, nonetheless, It's not imply from the equation of motion, a more appropriate derivation would be expanding the equation of motion that the functional form is equal or similar to the stress tensor, yet I am not able to show this. $\endgroup$
    – wong tom
    Nov 2, 2023 at 14:59

1 Answer 1

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Varying the action with respect to $h^{\alpha \beta}$ yields \begin{equation} T_{\alpha \beta} = \partial_\alpha X \cdot \partial_\beta X - \frac{1}{2} h_{\alpha \beta} h^{\gamma \delta} \partial_\gamma X \cdot \partial_\delta X=0. \end{equation} Why did you include the "$=0$" in the equation from varying with respect to the $X^\mu$ fields but not in the one from varying with respect to $h^{\alpha \beta}$? The equations of motion are always found by setting to zero the variation of the action with respect to each of the fields, that includes the metric.

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