# Proof of energy-momentum tensor is zero in Polyakov action

The polyakov action is defined as

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

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

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

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?

• You can find it on page 23 in David Tong lectures on String Theory damtp.cam.ac.uk/user/tong/string.html Commented Nov 1, 2023 at 4:01
• 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. Commented Nov 2, 2023 at 14:59

Varying the action with respect to $$h^{\alpha \beta}$$ yields $$$$T_{\alpha \beta} = \partial_\alpha X \cdot \partial_\beta X - \frac{1}{2} h_{\alpha \beta} h^{\gamma \delta} \partial_\gamma X \cdot \partial_\delta X=0.$$$$ Why did you include the "$$=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.