## July 10, 2006

### Entropy of Extremal, Non-SUSY Blackholes

What with all the “fun” we’ve been having discussing LQG, I never did get around to posting about the recent paper of Emparan and Horowitz.

Better late than never…

One of the fundamental breakthroughs in quantum gravity was Strominger and Vafa’s tour de force calculation of the entropy of supersymmetric charged 5D blackhole. At weak coupling, the microstates can be completely accounted-for as the BPS states of the collection of D-branes with those charges.

In 4 dimensions, to get a nonzero horizon area, a supersymmetric blackhole must carry at least 4 charges1. And there, too, one finds perfect agreement between the microscopic D-brane and the macroscopic Bekenstein-Hawking entropy.

There have been many important refinements of the supersymmetric story (some of which we’ve discussed here before). And there are positive results in the nearly-supersymmetric case.

But it’s hard to escape the feeling that perhaps it is supersymmetry (in particular, the fact that the counting of BPS states is independent of the string coupling) that is behind the success of these calculations. Surely a weak-coupling D-brane description ought to be hopelessly poor in accounting for the micro-states of far-from-supersymmetric blackholes.

Well…

Emparan and Horowitz have a fascinating little paper, in which they study a far-from-supersymmetric (but still extremal) 2-charged blackhole in 4 dimensions.

Consider Type IIB string theory compactified on ${T}^{6}$. Write ${T}^{6}={T}^{2}×{T}^{2}×{T}^{2}$, and let ${a}_{\alpha },{b}_{\alpha },\phantom{\rule{thinmathspace}{0ex}}\alpha =1,2,3$ be a basis of 1-cycles on each of the ${T}^{2}$s. There are many, U-duality equivalent, ways to get a 4-charge, 1/8-BPS blackhole. One is to take 4 stacks of D3-branes wrapped, respectively around the 3-cycles

(1)
$\begin{array}{rl}{C}_{1}& ={a}_{1}×{a}_{2}×{a}_{3}\\ {C}_{2}& ={a}_{1}×{b}_{2}×{b}_{3}\\ {C}_{3}& ={b}_{1}×{a}_{2}×{b}_{3}\\ {C}_{4}& ={b}_{1}×{b}_{2}×{a}_{3}\end{array}$

Note that

• Each pair of cycles intersects along a curve.
• All 4 intersect at a point.
• The full configuration preserves 1/8 of the original 32 supercharges.

If, for simplicity, we take the tori to be square, ${\tau }_{\alpha }=i$, then the ADM mass of this configuration is $M=\frac{{V}_{3}\sum _{i=1}^{4}{N}_{i}}{g{l}_{s}^{4}}=\frac{\sum _{i=1}^{4}{N}_{i}}{\sqrt{{G}_{4}}}$ where the volumes of the ${C}_{i}$ are $\left(2\pi {\right)}^{3}{V}_{3}$, ${G}_{4}$ is the 4D Newton’s constant, and we’ve wrapped ${N}_{i}$ D3-branes around ${C}_{i}$.

For large ${N}_{i}$, the asymptotic degeneracy of states reproduces the Bekenstein-Hawking entropy ${S}_{\text{BH}}=\underset{{N}_{i}\gg 1}{\mathrm{lim}}d\left({N}_{i}\right)=2\pi \sqrt{{N}_{1}{N}_{2}{N}_{3}{N}_{4}}$

Emparan and Horowitz consider a closely related system. Let $k,l$ be a pair of positive coprime integers and let ${c}_{\alpha }^{±}=k{a}_{\alpha }±l{b}_{\alpha }$ be the circle which wraps $k$ times around the $a$-cycle and $±l$ times around the $b$-cycle of the $\alpha$th ${T}^{2}$. Instead of (1), consider the 3-cycles

(2)
$\begin{array}{rl}{C}_{1}& ={c}_{1}^{+}×{c}_{2}^{+}×{c}_{3}^{+}\\ {C}_{2}& ={c}_{1}^{+}×{c}_{2}^{-}×{c}_{3}^{-}\\ {C}_{3}& ={c}_{1}^{-}×{c}_{2}^{+}×{c}_{3}^{-}\\ {C}_{4}& ={c}_{1}^{-}×{c}_{2}^{-}×{c}_{3}^{+}\end{array}$

Wrapping $N$ D3-branes around each of the ${C}_{i}$ breaks all supersymmetries. And there are only two net D3-brane charges for this configuration: ${N}_{0}=4N{k}^{3}$ times the charge corresponding to wrapping ${a}_{1}×{a}_{2}×{a}_{3}$ and ${N}_{6}=4N{l}^{3}$ times the charge corresponding to wrapping ${b}_{1}×{b}_{2}×{b}_{3}$2.

The lengths of the curves, ${c}_{\alpha }^{±}$ are a factor $\left({k}^{2}+{l}^{2}{\right)}^{1/2}$ times longer than the previous case. So the volumes of the ${C}_{i}$ are $\left(2\pi {\right)}^{3}\left({k}^{2}+{l}^{2}{\right)}^{3/2}{V}_{3}$ and the ADM mass of this configuration is $M=\frac{4N\left({k}^{2}+{l}^{2}{\right)}^{3/2}{V}_{3}}{g{l}_{s}^{4}}=\frac{4N\left({k}^{2}+{l}^{2}{\right)}^{3/2}}{\sqrt{{G}_{4}}}$

A weak-coupling calculation of the asymptotic degeneracy of states for this system gives $S=16\pi {k}^{3}{l}^{3}{N}^{2}$

Relative to the supersymmetric case (with ${N}_{i}=N$), the entropy is a factor $\left(2kl{\right)}^{3}$ larger, reflecting the fact that the ${C}_{i}$ in (2) intersect in $\left(2kl{\right)}^{3}$ points instead of 1 point.

These formulæ precisely reproduce the mass and entropy of the extremal dyonic blackholes. Indeed, lifted to M-theory, the Type-IIA description becomes a dyonic Kaluza-Klein blackhole ($×{T}^{6}$). (For a recent discussion, see here.) The mass is $M=\frac{{\left({Q}^{2/3}+{P}^{2/3}\right)}^{3/2}}{2{G}_{4}}$ where the electric and magnetic charges are, respectively $Q=\frac{2{G}_{4}{N}_{0}}{R},\phantom{\rule{1em}{0ex}}P=\frac{{N}_{6}R}{4}$ $R$ is the asymptotic radius of the Kaluza-Klein circle, ${N}_{6}$ is the Euler class of the ${S}^{1}$ bundle over the 2-sphere at spatial infinity, and ${N}_{0}$ is the (quantized) momentum around the circle. The entropy $S=\frac{A}{4{G}_{4}}=\frac{2\pi PQ}{{G}_{4}}=\pi {N}_{0}{N}_{6}$ is independent of $R$. The $R\to \infty$ limit is the 5 dimensional extremal Myers-Perry blackhole. In the $R\to 0$ limit, one obtains (a T-dual version of) the weakly-coupled D-brane description above.

The independence of the entropy of the radius, $R$, translates into independence of the string coupling, $g$, which, presumably, is the reason for the success of the weakly-coupled D-brane calculation despite the absence of any BPS property to protect the counting of microstates.

1 OK, there are the small blackholes which carry only two charges. Classically, they have zero horizon area, but quantum effects give them a finite horizon.

2 The notation is suggested by the fact that, in a T-dual description, these are the numbers of D0- and D6-branes.

Posted by distler at July 10, 2006 4:36 PM

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## 1 Comment & 0 Trackbacks

### Re: Entropy of Extremal, Non-SUSY Blackholes

I cannot understand a word of what you are saying on this post. Aside from accents this usually isnt a problem for me with people from Texas.

Accordingly I am going to shamelessly hijack it on to the topic of intelligent design/creationism and given that you take an interest in this do you know the context of the following quote attributed to Arno Penzias which suspiciously can only be found on ID/creationist websites:

“Astronomy leads us to a unique event, a universe which was created out of nothing and delicately balanced to provide exactly the conditions required to support life. In the absence of an absurdly-improbable accident, the observations of modern science seem to suggest an underlying, one might say, supernatural plan.”

Posted by: Atiyah on July 10, 2006 7:42 PM | Permalink | Reply to this

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