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Energy-momentum tensor of Bosonic Ghost Action in String Theory

Physics Asked by Funzies on June 13, 2021

When quantizing bosonic string theory by means of the path integral, one inverts the Faddeev-Popov determinant by going to Grassmann variables, yielding:
$$
S_{mathrm{ghosts}} = frac{-i}{2pi}intsqrt{hat{g}}b_{alphabeta}hat{nabla^{alpha}}c^{beta}d^2tau,
$$
where the $-i/2pi$ 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:
$$
T_{alphabeta} :=frac{-1}{sqrt{hat{g}}}frac{delta S_g}{deltahat{g}^{alphabeta}} = frac{i}{4pi}left( b_{alphagamma}hat{nabla}_{beta}c^{gamma} + b_{betagamma}hat{nabla}_{alpha}c^{gamma} – c^{gamma}nabla_{gamma}b_{alphabeta} – g_{alphabeta}b_{gammadelta}nabla^{gamma}c^{delta} right),
$$
I have trouble deriving this. Varying the $sqrt{hat{g}}$ in the action yields the last term I would say:
$$
deltasqrt{hat{g}} = -frac{1}{2}sqrt{hat{g}}hat{g}_{alphabeta}delta hat{g}^{alphabeta}
$$
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 $b_{alphabeta}$ is traceless: $b_{alphabeta}g^{alphabeta} = 0$, I forgot to place that here. It is a consequence of the path integral procedure.

One Answer

Here's part of my answer to the derivation 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}{2pi} int d^2 sigma sqrt{g} g^{alphamu} b_{alphabeta} 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}{2pi} int d^2 sigma left( delta sqrt{g} right) g^{alphamu} b_{alphabeta} nabla_mu c^beta &~~~~~~~~~~~~~~~~~- frac{i}{2pi} int d^2 sigma sqrt{g} left( delta g^{alphamu} right) b_{alphabeta} nabla_mu c^beta &~~~~~~~~~~~~~~~~~- frac{i}{2pi} int d^2 sigma sqrt{g} g^{alphamu} b_{alphabeta} delta left( nabla_mu c^beta right) &= - frac{i}{4pi} int d^2 sigma sqrt{g} left[ b_{alphamu} nabla_beta c^mu + b_{betamu} nabla_alpha c^mu - g_{alphabeta} b_{rhosigma} nabla^rho c^sigma right] delta g^{alphabeta} &~~~~~~~~~~~~~~~~~ - frac{i}{2pi} int d^2 sigma sqrt{g} g^{alphamu} b_{alphabeta} c^lambda delta Gamma^beta_{mulambda} end{split} end{equation} We now use begin{equation} begin{split} delta Gamma^beta_{mulambda} = frac{1}{2} g^{betarho} left[ nabla_lambda delta g_{rho mu} + nabla_mu delta g_{rho lambda} - nabla_rho delta g_{mulambda}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}{2pi} int d^2 sigma sqrt{g} g^{alphamu} b_{alphabeta} c^lambda delta Gamma^beta_{mulambda} &= - frac{i}{4pi} int d^2 sigma sqrt{g} b^{murho} c^lambda left[ nabla_lambda delta g_{rho mu} + nabla_mu delta g_{rho lambda} - nabla_rho delta g_{mulambda}right] &= - frac{i}{4pi} int d^2 sigma sqrt{g} b^{murho} c^lambda nabla_lambda delta g_{rho mu} &= - frac{i}{4pi} int d^2 sigma sqrt{g} nabla_lambda left( b_{alphabeta} c^lambda right) delta g^{alphabeta} end{split} end{equation} We then have begin{equation} begin{split} delta S_{gh} &= - frac{i}{4pi} int d^2 sigma sqrt{g} left[ b_{alphamu} nabla_beta c^mu + b_{betamu} nabla_alpha c^mu - g_{alphabeta} b_{rhosigma} nabla^rho c^sigma + nabla_lambda left( b_{alphabeta} c^lambda right) right] delta g^{alphabeta} end{split} end{equation}

Correct answer by Prahar Mitra on June 13, 2021

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