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
\begin{equation}
S = -\frac{T}{2}\int\sqrt{-h}h^{ab}g_{\mu\nu}\partial_aX^\mu\partial_bX^\nu\,d^2\xi
\end{equation}
the Lagrangian is
\begin{equation}
\mathcal{L}=-\frac{T}{2}\sqrt{-h}h^{ab}\partial_aX^\mu\partial_bX^\nu
\end{equation}
which after plugging into the Euler-Lagrange equation is
\begin{equation}
0-\frac{\partial}{\partial \xi^b}\left[-\frac{T}{2} \sqrt{-h}h^{ab}\partial_aX^\nu\right]=0
\end{equation}
which leads to (my result of)
\begin{equation}
\partial_b\left[ \sqrt{-h}h^{ab}\partial_a X^\nu\right] =0
\end{equation}
The equation of motion should be
\begin{equation}
\Box X^\nu = \frac{1}{\sqrt{-h}}\partial_b\left[ \sqrt{-h}h^{ab}\partial_a X^\nu\right] =0
\end{equation}
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.
