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Energy momentum tensor for RNS superstring

  1. Aug 12, 2012 #1
    1. The problem statement, all variables and given/known data

    The RNS action in flat gauge is given by
    [itex]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}[/itex]
    [itex]\mu[/itex] is a spacetime vector index, [itex]\alpha[/itex] a worldsheet vector intex and [itex]A,B[/itex] are Dirac spinor indices. Show that the energy-momentum tensor is given by
    [itex]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})[/itex]
    where Dirac indices are supressed.


    3. The attempt at a solution

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

    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.

    Best regards, physicus
     
  2. jcsd
  3. Aug 14, 2012 #2
    I don't know much about string theory, but are you sure the variation of gamma matrices vanishes? [itex] \frac{\delta \gamma^\rho}{\delta g^{\alpha \beta}}= 0[/itex] ?
     
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