# 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? Apr 5, 2015 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... Apr 5, 2015 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. Apr 6, 2015 at 3:59

The previous answers are correct, in general you can ignore the laplacian terms since they are total derivatives (equations of motion are not affected by total derivatives). Perhaps your confusion is due to the fact that you should be using the covariant laplacian as in this wiki page: $$\Delta \phi = \frac{1}{\sqrt{-det \;g}}\partial _ \mu \left( \sqrt{-det \;g} g^{\mu\nu} \partial _ \nu \phi \right)$$ Much like for the divergence in this question, that part of the action gives: $$\int d^{10} x \sqrt{-det \;g}\frac{1}{\sqrt{-det \;g}}\partial _ \mu \left( \sqrt{-det \;g} g^{\mu\nu} \partial _ \nu \phi \right)=\int d^{10} x \partial _ \mu \left( \sqrt{-det \;g} g^{\mu\nu} \partial _ \nu \phi \right)$$ which is clearly a total derivative.
Importantly, things are not as simple when you go from Einstein frame to string frame: you end up with a term $$e^{-2 \phi} \Delta \phi$$ which is NOT a total derivative. In that case, you need to use that $$e^{-2 \phi} \Delta \phi = -\frac12 \Delta e^{-2 \phi} + 2 e^{-2 \phi} \partial _ \mu \phi \partial ^\mu \phi$$ the first term on the right IS a total derivative and can be killed, but the last one contributes to give the right normalization for the dilaton in the string frame (a factor of $$+4$$).
• thanks, i see, so the question was why should $$\sqrt{-det g} g^{\mu\nu} \partial_ \nu \phi$$ vanish at infinity? you mean to have some kind of linear dilaton background or something like that? Aug 19, 2023 at 8:29