# Equation of motion from Polyakov action

If I write (my understanding of) how to derive the equations of motion from the Polyakov action, I come up missing a term.

Beginning with the basic Polyakov action
$$S = -\frac{T}{2}\int\sqrt{-h}h^{ab}g_{\mu\nu}\partial_aX^\mu\partial_bX^\nu\,d^2\xi$$ the Lagrangian is $$\mathcal{L}=-\frac{T}{2}\sqrt{-h}h^{ab}\partial_aX^\mu\partial_bX^\nu$$ which after plugging into the Euler-Lagrange equation is $$0-\frac{\partial}{\partial \xi^b}\left[-\frac{T}{2} \sqrt{-h}h^{ab}\partial_aX^\nu\right]=0$$ which leads to (my result of) $$\partial_b\left[ \sqrt{-h}h^{ab}\partial_a X^\nu\right] =0$$ The equation of motion should be $$\Box X^\nu = \frac{1}{\sqrt{-h}}\partial_b\left[ \sqrt{-h}h^{ab}\partial_a X^\nu\right] =0$$ My question is that I do not see where the $1/\sqrt{-h}$ comes from. (This result is ok for the Polyakov action on its own, but I want to be able to add a mass term to the action which will make the $1/\sqrt{-h}$ matter then.)

I think that there is something fundamental that I am missing about deriving the equation of motion. I understand that the operator $\nabla_\mu X^\nu=\partial_\mu X^\nu +\Gamma^\nu_{\mu\lambda}X^\lambda$, but I am bothered by deriving the equations of motion when the Lagrangian density is $L(X^\nu,\partial_aX^\nu)$ instead of $L(X^\nu,\nabla_aX^\nu)$. If I derive the Euler-Lagrange equation of motion from varying an action containing $L(X^\nu,\partial_aX^\nu)$: \begin{eqnarray} \delta S &=& \int \delta L (X^\nu,\partial_a X^\nu)\sqrt{-g}d^nX \\ &=& \int \left[ \frac{\delta L}{\delta(\partial_aX^\nu)}\delta(\partial_aX^\nu) + \delta X^\nu\nabla_a\frac{\delta L}{\delta(\partial_aX^\nu)}\right]\sqrt{-g}d^nX + \int \left[ \frac{\delta L}{\delta X^\nu}-\nabla_a\frac{\delta L}{\delta (\partial_a X^\nu)}\right]\delta X^\nu\sqrt{-g}d^nX \end{eqnarray} can I still send the surface term (the first term above) to zero? Thanks for any insight.

• Your result and the "right" result differ just by an overall factor. Just multiply both sides of equation $\Box X^{\mu}=0$ by $\sqrt{-h}$ and you are done. – Andrey Feldman Feb 11 '17 at 6:13
• I can see that. The problem is that the $1/\sqrt{-h}$ term should come from solving the Euler-Lagrange equations, but I am finding that it does not (and I want to know what I am missing here). In the minimal theory (with no matter action) it won't make a difference, but I imagine that this will be important if there is a matter term added to the action. – Bob Feb 11 '17 at 7:09
• The point is that both equations are equivalent. You obtain one or the other by multiplying or dividing by $\sqrt{-h}$ – Iván Mauricio Burbano Jul 2 at 3:35

$$\frac{\delta S}{\delta h^{ab}}=0$$ You can use the distributive property of the variational operator and make use of the identity $$\frac{1}{\sqrt{-h}}\frac{\delta\sqrt{-h}}{\delta h^{ab}}=-\frac{1}{2}h_{ab}$$