# Varying the Polyakov Action

Beginning with the Polyakov action $$S=-\frac{1}{2}\frac{T}{c}\int d^2\xi \sqrt{-h}h^{ab}g_{\mu\nu}\partial_aX^\mu\partial_bX^\nu$$ and varying the action via $X^\mu\rightarrow X^\mu + \delta X^\mu$, the action becomes \begin{eqnarray} S+\delta S &=& -\frac{1}{2}\frac{T}{c}\int d^2\xi \sqrt{-h}h^{ab}\left[ (g_{\mu\nu}+\delta g_{\mu\nu})\partial_a(X^\mu + \delta X^\mu)\partial_B(X^\nu + \delta X^\nu)\right] \\ &=& -\frac{1}{2}\frac{T}{c}\int d^2\xi \sqrt{-h}h^{ab}g_{\mu\nu}\partial_aX^\mu\partial_bX^\nu \\ &+& -\frac{1}{2}\frac{T}{c}\int d^2\xi \sqrt{-h}h^{ab} \left[ 2g_{\mu\nu} \partial_aX^\mu\partial_b(\delta X^\nu) + \delta g_{\mu\nu}\partial_aX^\mu\partial_bX^\nu \right] + \mathcal{O}(\delta^2) \end{eqnarray} Where the terms on the $2^{nd}/3^{rd}$ line above are $0^{th}/1^{st}$ order in $\delta$. Writing $\delta g_{\mu\nu}=g_{\mu\nu}(X+\delta X)-g_{\mu\nu}(X) = \frac{\partial g_{\mu\nu}}{\partial X^\lambda}\delta X^\lambda$, the variation in the action becomes $$\delta S = -\frac{T}{c}\int d^2\xi \sqrt{-h}h^{ab}\left[g_{\mu\nu}\partial_aX^\mu\partial_b(\delta X^\nu) - \frac{1}{2}\frac{\partial g_{\mu\lambda}}{\partial X^\nu}\partial_aX^\mu\partial_b X^\lambda \delta X^\nu \right]$$ (I interchanged the indices $\nu\leftrightarrow\lambda$ in the second term). The first term above may be integrated by parts to have $\partial_b\left[ \sqrt{-h}h^{ab}g_{\mu\nu}\partial_aX^\mu \right]$ in the integrand.

I am not sure what to do with the second term above. The final equation of motion should look like $\Box X^\mu = \frac{1}{\sqrt{-h}}\partial_b\left[ \sqrt{-h}h^{ab}g_{\mu\nu}\partial_aX^\mu\right] = 0$. I think that the second term will add a connection that will account for the $1/\sqrt{-h}$, but am not sure how to arrive there. Any help would be greatly appreciated.

• The $1/\sqrt{-h}$ is just an overall factor that doesn't substantially change your target equation of motion. The extra term you are worried about is really there. People often work in a flat target space, in which case the derivative of the target space metric wrt $X^\nu$ is zero. There is also a second equation of motion that comes from varying wrt the ws metric $h_{ab}$. See for example (4.4) and (4.5) of arxiv.org/abs/hep-th/0605158. – user2309840 Mar 15 '17 at 16:05
• Thank you for the response, it helped clear up confusion on the second term containing $dg_{\mu\nu}/dX^\lambda$. You mention that the extra term is really there, but I still don't see where the $1/\sqrt{-h}$ is hiding. I agree that it will not change the equation of motion in this case, but if (for example) I wanted to include a mass term in the action, then this $1/\sqrt{-h}$ will become important for the equation of motion. – Bob Mar 18 '17 at 21:18