Energy momentum tensor for RNS superstring

[physicus]

Homework Statement

The RNS action in flat gauge is given by
$S=-\frac{1}{8\pi}\int d\sigma d\tau\,\frac{2}{\alpha'}\partial_\alpha X^{\mu}\partial^\alpha X_\mu+2i\bar{\psi}_A^\mu\gamma_{AB}^\alpha\partial_\alpha\psi_{\mu B}$
$\mu$ is a spacetime vector index, $\alpha$ a worldsheet vector intex and $A,B$ are Dirac spinor indices. Show that the energy-momentum tensor is given by
$T_{\alpha\beta}=\partial_\alpha X^\mu\partial_\beta X_\mu+\frac{1}{4}\bar{\psi}^\mu\gamma_\alpha \partial_\beta \psi_\mu+\frac{1}{4}\bar{\psi}^\mu\gamma_\beta \partial_\alpha \psi_\mu-(\text{trace})$
where Dirac indices are supressed.

The Attempt at a Solution

I really have serious trouble solving this. The energy momentum tensor is defined as
$T_{\alpha\beta}=\frac{4\pi}{\sqrt{-h}}\frac{\delta S}{\delta h^{\alpha\beta}}$
where $h_{\alpha\beta}$ is the worldsheet metric and $h$ its determinant.

My first problem is, that the expression above is already gauge fixed. In order to take the variation of the action with respect to the metric I have to reintroduce it in the action which yields
$S=-\frac{1}{8\pi}\int d\sigma d\tau\,\sqrt{-h}\left[\frac{2}{\alpha'}h^{\alpha\beta}\partial_\alpha X^{\mu}\partial_\beta X_\mu+2ih^{\alpha\beta}\bar{\psi}^\mu \gamma_{\alpha} \partial_{\beta} \psi_{\mu B}\right]$
Is that correct? I have read that also the metric $h_{\alpha\beta}$ has a superpartner, do I have to introduce the superpartner instead of the metric itself in the fermionic part of the action?

Although I am not sure if my form of the action with the metric reintroduced is correct I tried to do the calculation:
$\frac{\delta S}{\delta h^{\alpha\beta}}$
$= -\frac{1}{8\pi}\int d\sigma d\tau \left[\frac{1}{2\sqrt{-h}}\frac{-\delta h}{\delta h^{\alpha\beta}}h^{\epsilon\rho}( \partial_{\epsilon} X^{\mu}\partial_\rho X_\mu+2i\bar{\psi}^\mu\gamma_{\epsilon} \partial_{\rho} \psi_{\mu B})\right]-\frac{1}{8\pi}\sqrt{-h}(\partial_\alpha X^{\mu}\partial_\beta X_\mu+2i\bar{\psi}^\mu\gamma_{\alpha} \partial_{\beta} \psi_{\mu B})$
$= -\frac{1}{8\pi}\left[-\frac{1}{2}\sqrt{-h}h_{\alpha\beta}h^{\epsilon\rho}( \partial_{\epsilon} X^{\mu} \partial_{\rho} X_\mu+2i\bar{\psi}^\mu\gamma_{\epsilon} \partial_{\rho} \psi_{\mu B})+\sqrt{-h}(\partial_\alpha X^{\mu}\partial_\beta X_\mu+2i\bar{\psi}^\mu\gamma_{\alpha} \partial_{\beta} \psi_{\mu B})\right]$
$\Rightarrow T_{\alpha\beta}=\frac{4\pi}{\sqrt{-h}}\frac{\delta S}{\delta h^{\alpha\beta}} = \frac{1}{4}h_{\alpha\beta}h^{\epsilon\rho}( {\partial}_{\epsilon} X^{\mu} \partial_{\rho} X_\mu+2i\bar{\psi}^\mu\gamma_{\epsilon} \partial_{\rho} \psi_{\mu B})-\frac{1}{2}(\partial_\alpha X^{\mu}\partial_\beta X_\mu+2i\bar{\psi}^\mu\gamma_{\alpha} \partial_{\beta} \psi_{\mu B})$

Unfortunately, that does not look like the given solution. Does anybody know what I did wrong or which other approach I could choose? I am very thenkful for any help. Also, I am grateful for any reference to literature.

physicus

 

[clamtrox]

I don't know much about string theory, but are you sure the variation of gamma matrices vanishes? $\frac{\delta \gamma^\rho}{\delta g^{\alpha \beta}}= 0$ ?