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:

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

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.

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.

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  :

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

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.

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:

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

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.