1
$\begingroup$

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.

Improve this question
$\endgroup$
1
  • $\begingroup$ The book does not say $\partial^2$. It must say $\nabla^2$. $\endgroup$ Commented Mar 28, 2016 at 15:40

1 Answer 1

Reset to default
1
$\begingroup$

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.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Can you elaborate on why the last sentence is valid? Why wouldn't you be able to use that if the field $X^\mu$ were not a scalar on the worldsheet? $\endgroup$ Commented Jun 22, 2020 at 18:24
  • $\begingroup$ When I apply that identity, I get: $\partial_a((-\gamma)^\frac{1}{2}\partial^aX^\mu) = (-\gamma)^\frac{1}{2}\nabla_a\left(\partial^a x^\mu\right)$. But that's not $\nabla^2$ as it should be. What am I missing? $\endgroup$ Commented Jun 22, 2020 at 18:29

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.