8
$\begingroup$

The low energy effective action for $N$ D$p$-branes in the string frame is

$$ S_\text{eff} = \frac{1}{16\pi G_{10}}\int d^{10}x \sqrt{-g}\ e^{-2\phi}(R+ 4(\partial\phi)^2+\cdots) $$

where $R$ is the Ricci curvature of $g$ and $\phi$ is the dilaton. There are another terms related to a antisymmetric tensor, but I'm not interested in them.

This gives us the metric:

$$ ds^2 = H^{-1/2} (-dt^2 + dx^2_p) + H^{1/2} (dr^2 + r^2d\Omega_{8-p}^2)$$

and the dilaton

$$ e^{-2\phi} = H^{\frac{p-3}{2}} $$

with

$$ H(r) = 1 + \left(\frac{R}{r}\right)^{7-p} .$$

If I want to obtain the Einstein metric I need to do a proper Weyl transformation, in that case

$$ g_{\mu\nu}\mapsto g^E_{\mu\nu} = e^{\phi/2}g_{\mu\nu} $$

My question is, in this tranformation do I have to put $H^{\frac{p-3}{8}}$ (the dilaton that I wrote above) or do I need to compute a new dilaton with the new action?

$\endgroup$
4
  • $\begingroup$ Even if you will compute a new dialton with the new action why would it be different than the previous one i mean you only add "a phase"(I think at least) $\endgroup$ Sep 14, 2014 at 16:32
  • $\begingroup$ Would you be so kind and tell me what the source of your expressions is? The answer might depend on the definition of $\phi$. $\endgroup$ Sep 14, 2014 at 23:40
  • $\begingroup$ @FredericBrünner Yes, it is the expression 39 in this PDF: members.ift.uam-csic.es/jfbarbon/Teaching_files/strbasics.pdf $\endgroup$
    – dpravos
    Sep 15, 2014 at 9:36
  • 1
    $\begingroup$ At the end of the page $53$, you see that the action in the Einstein frame, and the dilaton field is the same that in equation $(39)$. The metrics, in the Eistein frame, in $10$ dimensions is simply $ds_E^2 = e^{\phi/2} ds^2$ $\endgroup$
    – Trimok
    Sep 15, 2014 at 9:49

1 Answer 1

1
$\begingroup$

No, the expression for the dilaton it's the same, being a scalar. In a generic dimension D the relation is:

$$g_{\mu\nu}^E = e^{-\frac{4\phi}{D-2}} g_{\mu\nu}^S$$

Having used:

$$L_{Einstein} = \sqrt{-g_E} R_E$$

$$L_{String} = \sqrt{-g_S} R_S e^{-2\phi}$$

References:

Basic Concepts In String Theory, B-L-T, pag. 700

Gravity and Strings, Ortin, pag. 414

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.