3
$\begingroup$

It's widely belived that p-branes and Dp-branes are the same objects in string theory. P-branes are charged solutions of type IIB supergravity that have the topology of a black hole, one would expect to be able to construct the same black hole using a superposition of D-branes and to count the microstates to get a statistical entropy.

Maldacena showed in his thesis that one has to have D1-branes, D5-branes and some momentum in the $x^5$ direction in order to have a black whole with non zero area in order to be able to compute the Hawking-Bekenstein entropy.

In the count of microstates of the configuration the idea is to account for all the fields in the branes having in mind that the open strings can end in the D1 (1,1), D5 (5,5), and mixed D1-D5 (1,5), D5-D1 (5,1). It seems in the literature that one has to account for the bound state of the system in order to have a black hole, what is the explanation for this?.

If one sees the bound state of the system it's given by the Higgs branch, wich comes from giving the hypermultiples of the superpotential an expectation value:

$V = \frac{1}{(2\pi \alpha ')^2} \mid X_i \chi - \chi Y_i \mid^2 + \frac{g_1^2}{4} D_1^A D_1^A + \frac{g_5^2}{4V_4}D_5^2D_5^2 $

In the literature all the states that matter come from the massless excitations of the brane that i can get by setting the superpotential to zero, why I only care about the massless excitations of the branes?

$\endgroup$

1 Answer 1

0
$\begingroup$

It seems in the literature that one has to account for the bound state of the system in order to have a black hole, what is the explanation for this?.

BPS objects enjoy cancellations between gravitational and "electric" attraction, therefore in principle we can cluster a bunch of branes together, but still being able to remove a single brane without effort. This is not the scenario we are looking for to describe a black hole. We need a bound state of the branes, in which the branes cannot be freely moved apart. Microscopically we want open strings stretching between different branes, gluing them together.

In the literature all the states that matter come from the massless excitations of the brane that i can get by setting the superpotential to zero, why I only care about the massless excitations of the branes?

Since we are interested in a low energy dynamic, in which $g_s Q << 1$ (the opposite of the SUGRA limit, in which we have a macroscopic black hole), we can neglect open string excitations and Kaluza-Klein modes, of course with the exception of the units of momentum on the compact circle $S^1$ on which the D1 and the D5 are parallel. The greatest entropic contributions comes therefore from massless states. What we want to count is how many ways are there to distribute the momentum, and the maximum number of ways will arise in the configuration with the maximum number of massless world volume fields. The right choice is to give mass to scalars associated to the transverse motion of the branes, the Higgs branch, giving us a bound state.

$\endgroup$

Your Answer

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

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