Trying to understand the energy-momentum tensor from the Polyakov action I am reading the paper https://arxiv.org/abs/hep-th/0201124. The Polyakov action is defined as 
$$S_{str} =  \int d^2\zeta (-\frac{1}{2}\sqrt{-g}g^{ab}\partial_aX^\mu \partial_bX_\mu).$$
I am trying to understand the following expression for the symmetric energy-momentum tensor $T_{ab}$:
$$T_{ab} = \frac{2}{\sqrt{-g}} \frac{\delta S_{str}}{\delta g^{ab}}$$
$$= -\partial_aX^\mu \partial_bX_\mu + \frac{1}{2}g_{ab}g^{cd}\partial_cX^\mu \partial_dX_\mu$$
I have two questions. 


*

*Is the first equality just a definition? 

*Also, I understand how the first term in the second equality comes about but how does one get the term $\frac{1}{2}g_{ab}g^{cd}\partial_cX^\mu \partial_dX_\mu$?
 A: As alexarvanitakis has already said the answer to your first question is yes. When varying the action with respect to the metric you obtain the energy-momentum tensor, namely 
$\begin{equation}
T_{\alpha \beta} = - \frac{2}{T} \frac{1}{\sqrt{-g}} \frac{\delta S}{\delta g^{\alpha \beta}}
\end{equation}$
So, we have 
$\begin{equation}
\delta S = \int \frac{\delta S}{\delta g^{\alpha \beta}} \delta g^{\alpha \beta} = - \frac{T}{2} \int d^2 \zeta \sqrt{-g} ~ T_{\alpha \beta} ~ \delta g^{\alpha \beta}
\end{equation}$
And now let's perform the computations to find the terms. 
The variation is 
$\begin{equation}
\delta S = - \frac{T}{2} \int d^2 \zeta \left(\delta \sqrt{-g}~ g^{\alpha \beta} ~ \partial_{\alpha} X \cdot ~ \partial_{\beta} X + \sqrt{-g} ~ \delta g^{\alpha \beta} ~ \partial_{\alpha} X \cdot ~\partial_{\beta} X   \right)
\end{equation}$
We have to use the standard formulae for the metric variations (you can find these in G.R, string theory textbooks, etc)
$\begin{equation}
\begin{split}
\delta g &= - g ~ g_{\alpha \beta} ~ \delta g^{\alpha \beta} \\
\delta \sqrt{-g} &= - \frac{1}{2} \sqrt{-g} ~ g_{\alpha \beta} ~ \delta^{\alpha \beta}
\end{split}
\end{equation}$
for the above $\delta S$. This yields
$\begin{equation}
\delta S = -\frac{T}{2} \int d^2 \zeta ~ \sqrt{-g} ~ \delta g^{\alpha \beta} ~ \left( -\frac{1}{2} g_{\alpha \beta} g^{\gamma \delta} ~ \partial_{\gamma} X \cdot \partial_{\delta} X + \partial_{\alpha} X \cdot \partial_{\beta} X   \right) = 0
\end{equation}$
I hope this helps a bit. 
A: 1) Yes.
2) The second term comes from the variation of $\sqrt{\det g}$.
