The gravity dual of $N$ D$p$-branes at zero temperature is

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


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

what is (tell me if I'm wrong) an extremal black $p$-brane.

When we consider that the boundary system is at temperature $T$, the dual metric then is

$$ ds^2= H^{-1/2}(-h(r)dt^2+dx_p^2) + H^{1/2}\left(\frac{dr^2}{h(r)} + r^2d\Omega_{8-p}^2\right) \qquad(*)$$


$$ h(r) = 1 + \left(\frac{r_0}{r}\right)^{7-p} $$

(for example, this is written here, in section 7.5, for $p=3$), but I don't know to what system corresponds this metric, except for $p=3$, which is an AdS black hole (for $r$ near the throat).

So the question is, what system has the metric $(*)$?


1 Answer 1


AdS black holes exist in various dimensions, $p=3$ is not the only choice. The parameter can take on values above or below $3$. One famous example is the three-dimensional BTZ black hole, and higher dimensional ones are also frequently used in the correspondence.

Furthermore, I think there is a misunderstanding on the concept of a "gravity dual". The metric you wrote down is not the the gravity dual of a $\mathrm{D}p$-brane at zero temperature, it is a solution of classical gravity that corresponds to the geometry of such a brane at low energies. In this sense, it is the D-brane in a particular limit. The word "gravity dual" refers to precisely this geometry, and it is the dual of a quantum field theory living on its boundary. The AdS/CFT correspondence is about formulating higher-dimensional gravity duals (D-Branes and their low energy limits) of quantum field theories in lower dimensions.

  • $\begingroup$ But for $p\neq 3$ the first metric is not AdS. If I understood correctly your answer, the metric (*) corresponds to a black hole in the spacetime of the first metric? $\endgroup$
    – dpravos
    Oct 1, 2014 at 10:59
  • $\begingroup$ Your statement is not true, AdS is not restricted to $p=3$. $\endgroup$ Oct 1, 2014 at 12:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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