I'm trying to go from the low-energy effective action in the string-frame to the corresponding action in the Einstein frame. The action in the string frame has the form

$$S = \frac{1}{(2\pi)^7 l_s^8}\int d^{10}x \sqrt{-\gamma}\left[e^{-2\Phi}(R[\gamma_{\mu\nu}] + 4(\nabla\Phi)^2) - \frac{1}{2}\left|F_{p+2}\right|^2\right]$$

Here $\gamma_{\mu\nu}$ is the string-frame metric, and $R[\gamma_{\mu\nu}]$ is the usual Einstein curvature for this metric, which appears in the first (dilaton corrected) Einstein-Hilbert like term in the action.

The process is two fold:

  1. Introduce a field $\phi$ for fluctuations $e^{\Phi} = g_s e^{\phi}$ ($\Phi$ in this language is the dilaton).

  2. Weyl rescale the metric $g_{\mu\nu} = e^{-\phi/2}\gamma_{\mu\nu}$.

The second step ought to yield the following relationship between the scalar curvatures:

$$R[\gamma_{\mu\nu}] = e^{-\phi/2}\left[R[g_{\mu\nu}] - \frac{9}{2}\nabla^2\phi - \frac{9}{2}(\nabla\phi)^2\right]$$

So my problem right now is to cleanly derive this curvature relationship, which is really a GR problem. One way to go about it is to start from the Christoffel symbols for $g$ and show that they are equal to the Christoffel symbols for $\gamma$ plus $\phi$-dependent correction terms, then go the Riemann curvature tensor, and then finally to the Ricci tensor, in the usual canonical way.

However, I am hoping there's a faster way to to do this (vielbeins maybe? I can't think of a direct application) because the algebra in this standard procedure is quite tedious.

The general problem has a simple statement: how do you relate the curvature of a Weyl rescaled metric to its original curvature?

  • $\begingroup$ How did you get that equation in the first place? $\endgroup$ Mar 31 '15 at 14:32
  • $\begingroup$ [1] R.M. Wald, General Relativity (1984), Appendix D. [2] S. Carroll, Spacetime and Geometry (2003), Appendix G. [3] A. Zee, Einstein Gravity in a Nutshell (2013), p. 353. $\endgroup$
    – Ryan Unger
    Mar 31 '15 at 14:34

Let $\Omega^2=\exp\phi/2$ so that $\gamma=\Omega^2 g$. Then we have the standard formula $$R_\gamma=\Omega^{-2}\big(R_g-2(n-1)\Delta\ln\Omega-(n-2)(n-1)[\nabla\ln\Omega]^2\big)$$ This is proven in any number of General Relativity texts, but the one in R.M. Wald, General Relativity (1984), is particularly easy to follow and is proved in the form shown here. From above we get $\Omega=\exp\phi/4$ so that $\ln\Omega=\phi/4$. Furthermore, we have $n=10$ so that the coefficients of the last two terms are $2\cdot 9$ and $8\cdot 9$. From the $\ln\Omega$s we get factors of $1/4$ and $1/4^2$ on the second and third terms, respectively. Thus we have $9/2$ in front of those terms. Putting it all together, we obtain $$R_\gamma=\exp(-\phi/2)\big(R_g-\tfrac{9}{2}\Delta\phi-\tfrac{9}{2}(\nabla\phi)^2\big)$$ as was to be shown.


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