# From String Frame to Einstein Frame for 10D supergravity

This question is related to but not answered in the post String frame and Einstein frame for a Dp-brane, so it should be treated as a separate question.

Beginning with the gravity action

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

in the string frame, I want to derive the action in the Einstein frame, which is

$$S = \frac{1}{(2\pi)^7 l_s^8 g_s^2}\int d^{10}x \sqrt{-g}\left[R - 4(\nabla\phi)^2 - \frac{1}{2}g_s^2 e^{(3-p)\phi/2}\left|F_{p+2}\right|^2\right]$$

where $e^{\Phi} = g_s e^{\phi}$, $g_{\mu\nu} = e^{-\phi/2}\gamma_{\mu\nu}$, and $|F_{p}|^2 = \frac{1}{p!}F_{\mu_1\mu_2\ldots\mu_p}F^{\mu_1\mu_2\ldots\mu_p}$.

I understand that

$$R_\gamma = e^{-\phi/2}\left[R_g - \frac{9}{2}\nabla^2\phi - \frac{9}{2}(\nabla\phi)^2\right]$$

(Note: the above expression for the Ricci scalar has been derived here: Curvature of Weyl-rescaled metric from curvature of original metric). The interpretation is that the derivative terms (gradient squared, and Laplacian) on the right hand side have been computed using the $g$ metric, and hence are "already" in Einstein frame form.

Now, I also understand that

$$\sqrt{-\gamma} = e^{5\phi/2}\sqrt{-g}$$

$$|F_{p+2}|^2_{\mbox{string frame}} = e^{-(p+2)\phi/2} |F_{p+2}|^2_{\mbox{Einstein frame}}$$

(for the particular normalization stated above) and

$$(\nabla\phi)^2_{\mbox{string frame}} = e^{-\phi/2}(\nabla\phi)^2_{\mbox{Einstein frame}}$$

but substituting all this into the first expression for the action still leaves behind the Laplacian term $\nabla^2\phi$, which does not appear in the (correct) expression for the string frame action.

What am I missing here?

• Integrate by parts? – Prahar Apr 5 '15 at 17:08
• The action after these substitutions is $$S = \frac{1}{(2\pi)^7 l_s^8 g_s^2}\int d^{10}x\sqrt{-g}\left[R_g - \frac{9}{2}\nabla^2\phi - \frac{1}{2}|\nabla\phi|^2 - \frac{1}{2}e^{(3-p)\phi/2}g_s^2 |F_{p+2}|^2\right]$$ You're probably right about integration by parts, but I don't see how it kills the laplacian unless I have some wrong coefficients... – leastaction Apr 5 '15 at 17:39
• Note that the Laplacian term is just a boundary term. However, I got the same result ^^, minus the Laplacian, which I discarded. – Ryan Unger Apr 6 '15 at 3:59