I'm stuck at a step in the derivation of the equations of motions of a string using the Polyakov action.

In Polchinski's textbook in String Theory , Page 14 ; Equation ( 1.2.25 ) ,

Varying the Polyakov action , One gets :

$\delta S\sim \int \partial _{a}(\sqrt{-det\mid \gamma\mid}\partial ^{a}X^{\mu})\delta X_{\mu} + $ Surface Terms

So $\partial _{a}(\sqrt{-\mid \gamma\mid}\partial ^{a}X^{\mu})=(\sqrt{-\mid \gamma\mid})\partial^{2}X^{\mu} + \partial _{a}(\sqrt{-\mid \gamma\mid})\partial^{a}X^{\mu}=\sqrt{-\mid\gamma\mid}\gamma^{cd}\partial_{a}\gamma_{cd}\partial^{a}X^{\mu}+(\sqrt{-\mid \gamma\mid})\partial^{2}X^{\mu} = 0$

if the surface terms vanish

Now , The Term which is proportional to the derivative of $\gamma_{cd}$ is zero in the conformal gauge. But the book did not show this. The result of the above derivative is

$(\sqrt{-\mid \gamma\mid})\partial^{2}X^{\mu}$

I don't know how to show this without using the conformal gauge.

  • $\begingroup$ The book does not say $\partial^2$. It must say $\nabla^2$. $\endgroup$ – Prahar Mar 28 '16 at 15:40

To see that the term is zero one can try to figure out that $\partial _{a}(\sqrt{-\mid \gamma\mid}\partial ^{a}X^{\mu})$ is equivalent to the covariant derivative $\nabla_a(\sqrt{-\mid \gamma\mid}\partial ^{a}X^{\mu})$, which would make the term with the metric derivative vanish ($\nabla g=0$).

A more convenient way is just to use the identity $$\nabla_a A^a=\frac{1}{|\gamma|^{1/2}}\partial_a (|\gamma|^{1/2} A^a),$$ (e.g. see here for proof).

It is possible to use here $A^a=\partial^a X^\mu$ since the field $X^\mu$ is just a scalar on the worldsheet.


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