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When quantizing bosonic string theory by means of the path integral, one inverts the Faddeev-Popov determinant by going to Grassmann variables, yielding: $$ S_{\mathrm{ghosts}} = \frac{-i}{2\pi}\int\sqrt{\hat{g}}b_{\alpha\beta}\hat{\nabla^{\alpha}}c^{\beta}d^2\tau, $$ where the $-i/2\pi$ just comes from convention/Wick rotated or not. My first problem is the notion of the 'fudicial' metric $\hat{g}$. I find its role in the path integral procedure a bit confusing. What is its relation to the 'normal' metric $g$? Why is it introduced? Related to this confusion is that fact in my lecture notes it is said that the energy momentum tensor is given by: $$ T_{\alpha\beta} :=\frac{-1}{\sqrt{\hat{g}}}\frac{\delta S_g}{\delta\hat{g}^{\alpha\beta}} = \frac{i}{4\pi}\left( b_{\alpha\gamma}\hat{\nabla}_{\beta}c^{\gamma} + b_{\beta\gamma}\hat{\nabla}_{\alpha}c^{\gamma} - c^{\gamma}\nabla_{\gamma}b_{\alpha\beta} - g_{\alpha\beta}b_{\gamma\delta}\nabla^{\gamma}c^{\delta} \right), $$ I have trouble deriving this. Varying the $\sqrt{\hat{g}}$ in the action yields the last term I would say: $$ \delta\sqrt{\hat{g}} = -\frac{1}{2}\sqrt{\hat{g}}\hat{g}_{\alpha\beta}\delta \hat{g}^{\alpha\beta} $$ However, this term does not have a 'hat' on the covariant derivative, which I find strange. The first and second term follow easily when writing the action with all indices low (except for a factor of 1/2), but I really don't see where the third term comes from and it also misses a hat on the covariant derivative. It looks like there has been done a partial integration, but I don't see why. I guess I am missing the point of the fiducial metric here. Explanation greatly appreciated!

EDIT: In the discussion below I mentioned that $b_{\alpha\beta}$ is traceless: $b_{\alpha\beta}g^{\alpha\beta} = 0$, I forgot to place that here. It is a consequence of the path integral procedure.

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  • $\begingroup$ I've been meaning to do this for quite a long time. I've decided to work it out now. Will put up my answer once I'm done. $\endgroup$ – Prahar Jul 1 '13 at 21:24
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Here's part of my answer to the derivvation of the EM tensor for the ghost action. It does not match the expression you gave, but I may have made a mistake. CAn you check my work?

We start with the action \begin{equation} \begin{split} S_{gh} &= - \frac{i}{2\pi} \int d^2 \sigma \sqrt{g} g^{\alpha\mu} b_{\alpha\beta} \nabla_\mu c^\beta \\ \end{split} \end{equation} Let us now vary the action w.r.t. metric. We get \begin{equation} \begin{split} \delta S_{gh} &= - \frac{i}{2\pi} \int d^2 \sigma \left( \delta \sqrt{g} \right) g^{\alpha\mu} b_{\alpha\beta} \nabla_\mu c^\beta \\ &~~~~~~~~~~~~~~~~~- \frac{i}{2\pi} \int d^2 \sigma \sqrt{g} \left( \delta g^{\alpha\mu} \right) b_{\alpha\beta} \nabla_\mu c^\beta \\ &~~~~~~~~~~~~~~~~~- \frac{i}{2\pi} \int d^2 \sigma \sqrt{g} g^{\alpha\mu} b_{\alpha\beta} \delta \left( \nabla_\mu c^\beta \right) \\ &= - \frac{i}{4\pi} \int d^2 \sigma \sqrt{g} \left[ b_{\alpha\mu} \nabla_\beta c^\mu + b_{\beta\mu} \nabla_\alpha c^\mu - g_{\alpha\beta} b_{\rho\sigma} \nabla^\rho c^\sigma \right] \delta g^{\alpha\beta} \\ &~~~~~~~~~~~~~~~~~ - \frac{i}{2\pi} \int d^2 \sigma \sqrt{g} g^{\alpha\mu} b_{\alpha\beta} c^\lambda \delta \Gamma^\beta_{\mu\lambda} \\ \end{split} \end{equation} We now use \begin{equation} \begin{split} \delta \Gamma^\beta_{\mu\lambda} = \frac{1}{2} g^{\beta\rho} \left[ \nabla_\lambda \delta g_{\rho \mu} + \nabla_\mu \delta g_{\rho \lambda} - \nabla_\rho \delta g_{\mu\lambda}\right] \end{split} \end{equation} Note that in particular, it is a tensor. The last term then becomes \begin{equation} \begin{split} I &= - \frac{i}{2\pi} \int d^2 \sigma \sqrt{g} g^{\alpha\mu} b_{\alpha\beta} c^\lambda \delta \Gamma^\beta_{\mu\lambda} \\ &= - \frac{i}{4\pi} \int d^2 \sigma \sqrt{g} b^{\mu\rho} c^\lambda \left[ \nabla_\lambda \delta g_{\rho \mu} + \nabla_\mu \delta g_{\rho \lambda} - \nabla_\rho \delta g_{\mu\lambda}\right]\\ &= - \frac{i}{4\pi} \int d^2 \sigma \sqrt{g} b^{\mu\rho} c^\lambda \nabla_\lambda \delta g_{\rho \mu} \\ &= - \frac{i}{4\pi} \int d^2 \sigma \sqrt{g} \nabla_\lambda \left( b_{\alpha\beta} c^\lambda \right) \delta g^{\alpha\beta} \end{split} \end{equation} We then have \begin{equation} \begin{split} \delta S_{gh} &= - \frac{i}{4\pi} \int d^2 \sigma \sqrt{g} \left[ b_{\alpha\mu} \nabla_\beta c^\mu + b_{\beta\mu} \nabla_\alpha c^\mu - g_{\alpha\beta} b_{\rho\sigma} \nabla^\rho c^\sigma + \nabla_\lambda \left( b_{\alpha\beta} c^\lambda \right) \right] \delta g^{\alpha\beta} \end{split} \end{equation}

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  • $\begingroup$ I am pretty sure that my expression for the energy-momentum tensor is correct, I have found it on page 50 of Lust and Theisen's Lectures on String Theory. I think the first two terms in your expression are correct, but the third and fourth are not.. $\endgroup$ – Funzies Jul 2 '13 at 20:32
  • $\begingroup$ THe third term is also correct? I thought it matches what you have. I am worried about the last term though. $\endgroup$ – Prahar Jul 2 '13 at 20:56
  • $\begingroup$ maybe we can use the fact that $\beta_{\alpha\beta}$ is traceless? $\endgroup$ – Funzies Jul 4 '13 at 16:56
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    $\begingroup$ @Erik There is a sign error in the last integration by parts, so you get in fact: $- \nabla_\lambda \left( b_{\alpha\beta} c^\lambda \right) \delta g^{\alpha\beta}$. Assuming a flat frame give $- c^\lambda \partial_\lambda ( b_{\alpha\beta}) \delta g^{\alpha\beta} - \partial_\lambda c^\lambda ( b_{\alpha\beta} \delta g^{\alpha\beta})$. But $( b_{\alpha\beta} \delta g^{\alpha\beta}) = \delta(b_ {\alpha\beta} g^{\alpha\beta}) = 0$, because $b$ is traceless. And you can reintroduce the Christoffel symbols and you find $- c^\lambda\nabla_\lambda ( b_{\alpha\beta}) \delta g^{\alpha\beta}$ $\endgroup$ – Trimok Jul 5 '13 at 18:05
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    $\begingroup$ No. Note that $g^{\alpha\rho} g_{\rho\beta} = \delta^\alpha_\beta$. Thus $\delta \left( g^{\alpha\rho} g_{\rho\beta} \right) = 0 \implies \delta g^{\alpha\rho} g_{\rho\beta} + g^{\alpha\rho} \delta g_{\rho\beta} = 0$. Thus $\delta g^{\alpha\beta} = - g^{\alpha\rho} g^{\beta\sigma} \delta g_{\rho\sigma} $. So there is an extra minus sign. $\endgroup$ – Prahar Aug 10 '13 at 15:43

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