Extremal Single-charge Small Black holes Aninda Sinha DAMTP, Cambridge University, UK hep-th/0601183 (to appear in CQG) with Nemani Suryanarayana(Perimeter),

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Presentation transcript:

Extremal Single-charge Small Black holes Aninda Sinha DAMTP, Cambridge University, UK hep-th/ (to appear in CQG) with Nemani Suryanarayana(Perimeter), hep-th/06????? in progress with Nemani Suryanarayana. Eurostrings 2006

2 Outline Motivation Entropy function formalism [Wald, Sen, OSV, Dabholkar, de Wit] Single-charge small black holes in M-theory Solution with M-theory R 4 term [Tseytlin, Green, Deser, Howe] Features of solution and discussion

3 Motivation Certain charged extremal black holes in string theory S BH =0 using lowest order lagrangian while S stat  0. These are called small black holes. Ignored higher derivative terms. Area/4 law is modified and is now given by Wald’s formula. Example: A single long Heterotic string wrapping S 1 momentum n winding w, only left moving excitations to make a BPS state. The number of such states for large n and w is » exp (4  p(nw)). The statistical entropy of this system is S stat =4  p(nw) Using Wald’s formula we can associate an entropy S BH to this. Is S stat =S BH ? [Dabhokar] What happens when n or w is zero? Can we associate non-zero entropy with single-charge black holes?

4 Entropy function—Sen’s derivation Following Sen, definition of extremal black holes in higher derivative theory: Black hole with near horizon geometry AdS 2 £ S D. Start with general gauge and coordinate invariant theory with metric g , Abelian gauge fields A  i and neutral scalar fields  s. Assume background fields to respect the SO(2,1)£ SO(D+1) symmetry of AdS 2 £ S D.

5 General form of solution(4d)

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7 An example: Heterotic F-string with winding and momentum

8 Setting n or w to zero makes the string coupling blow up.

9 Note that till now we have not used any specific form for the higher derivative correction and this derivation is general. It is also applicable to the type II F- string wrapping a circle and having momentum. For the heterotic string the leading  ’ correction begins with the well known Gauss-Bonnet type S R 2 interaction while for type II it begins with S R 4. Sen found that just using the S R 2 term and nothing else, the entire result could be exactly reproduced in 4 and 5 dimensions. Setting either n or w to zero seems to make the entropy result vanish. However this conclusion is too naïve as in this case string coupling is large and string loop corrections will play a crucial role. QUESTION: What happens when you set n or w to zero? Hard question to tackle in type II theories in general as higher derivative terms involving fluxes are not known except where topological string arguments can be used. Study a related question in M-theory. Consider a large number of D0-particles and ask what happens to the entropy of a black-hole formed from such a collection.

10 Single charge small black holes Large number q of D0-particles in type IIA. KK modes of the 11-d graviton. Lowest order solution, dilaton blows up, null singularity— zero horizon area. Use strong coupling limit, lift to M theory. Assume near horizon geometry to be AdS 2 x S 8 and see if solution exists using the entropy function formalism.

11 Now incorporate higher derivative corrections to EH terms

12 Steps in the calculation

13 So have seen possibility of getting partitioning of integer as the macroscopic entropy. What gives rise to this degeneracy? We can form multigraviton states with fixed total KK momentum. Their degeneracy is given by partitioning of q, given by the Hardy- Ramanujan formula in the large charge limit, Thus expect resulting black hole to have entropy given by Have to demonstrate that leading order result (using R 4 ) is non- vanishing.

14 The M-theory R 4 term In order to proceed we need to know more details about the M-theory R 4 term. This is inherited from the well known 1-loop type IIA R 4 term. Field redefinition allows us to write this as [Tsyetlin]

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16 Features of solution and discussion Satisfies v 1 =y 2. Sets two of the three equations to be equal to one another and hence we need to solve only two equations for two unknowns. Presumably can be done algebraically. v 1 =y 2 implies probe D0-brane experiences vanishing force and D M R ABCD =0 (note that this is not required in the 11-d lift and is only required in 10-d geometry). The leading entropy is still proportional to area. Our n.h. geometry AdS 2 £ S 8 is a new solution not seen in SUGRA. The 11-d lift is locally AdS 3 £ S 8 with a 1-d null boundary. Ratio EH/R 4 =-1/3. Ratio in 2 charge case for Sen EH/R 2 =-1. Is this rationality a coincidence?

17 Is the solution supersymmetric? We are counting all multiparticle states in order to get the entropy. For the two charge system, suppose n=1 and w is large. Then A more precise calculation yields leading term to be To see this need to consider perturbative string loop corrections as well as non-perturbative effects in the macroscopic calculation. We find that for the single charge case, it is the multiparticle configuration that dominates the counting.

18 Recently Kallosh and Linde have provided a quantum information way of looking at our result. They propose that our result is connected to the quantum stretching/non-normalizability of a qubit wave-function. Thank you for listening

19 Features of the solution and discussion

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