From ##D3-Brane## to ##AdS_5-Schwarzschild##

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  • Thread starter ShayanJ
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I'm trying to read this paper. Of course there are lots of things about it that are beyond me but for now its only the calculations in the main body of the paper that I'm trying to, more or less, understand. So I'm trying to go through it and find out the things that I don't get.

For now, I have problem with the beginning of the section 2:
The metric of the near-extremal D3-brane is
## ds_{10}^2=H^{-\frac 1 2} (-h dt^2+d\vec x ^2)+H^{\frac 1 2} (\frac{dr^2}{h}+d\Omega_5^2) \ \ \ \ \ \ \ \left( H=1+(\frac{L}{r})^4 \ \ , \ \ h=1-(\frac{r_H}{r})^4 \right) \ \ \ \ \ (1)##
Here ##\vec x = (x, y, z) ## are the spatial coordinates along which the D3-brane is extended and ##dΩ^2_5## is the standard metric on the five-sphere ##S^5## with unit radius. The near-horizon limit consists of “dropping the 1” from H. Then the metric is ##AdS_5-Schwarzschild##,
##ds_5^2=(\frac{r}{L})^2 (-hdt^2+d\vec x^2)+(\frac{L}{r})^2 \frac{dr^2}{h} \ \ \ \ \ \ \ \ \ \ (2)##
times the metric for an ##S^5## of constant radius L.
I don't understand how he got equation (2) from equation (1) by just "dropping the 1 from H"! Because when I do that, I get:
## (\frac r L)^2 ( -h dt^2+ d \vec x ^2)+(\frac L r)^2 ( \frac{dr^2}{h}+d\Omega_5^2) ##
And this is really a mystery to me how that ## d\Omega_5^2 ## vanishes and reappears as an overall multiplicative factor! This really strange because in equation (1), the "radius" of the ##S^5## is ## H^{\frac 1 2} ## but in the near-horizon limit, it involves other coordinates too, actually their differentials which is non-sense because terms like ## dt^2 d\Omega_5^2 ## will appear in the metric. What is the author doing here?
Thanks
 

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  • #2
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Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 
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Ben Niehoff
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I think there might be a typo in (1) and it should be

$$ds^2 = H^{-1/2} (-h dt^2 + d \vec x^2) + H^{1/2} (\frac{dr^2}{h} + r^2 d\Omega^2)$$
Then you get Schwarzschild-AdS x S^5 as the near-horizon limit. You should see that the S^5 now gets a constant radius equal to the AdS radius.

In (2), Gubser also drops the S^5 part of the metric, because it will no longer play a role. (There are fancier situations where it does play a role...if you allow your fields to depend on the S^5 coordinates then things can quickly become a mess.)
 
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