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.

  1. Is the first equality just a definition?

  2. Also, I understand how the first term in the second equality comes about but how does one get the term $\frac{1}{2}g_{ab}g^{cd}\partial_cX^\mu \partial_dX_\mu$?


2 Answers 2


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.


1) Yes. 2) The second term comes from the variation of $\sqrt{\det g}$.

  • 1
    $\begingroup$ Can you please elaborate on how to vary the $\sqrt{det\ g}$ term. $\endgroup$
    – Matrix23
    Mar 6, 2020 at 15:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.