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  What does SUSY have to do with the stability of black holes?

+ 3 like - 0 dislike

In this comment, Lumo says that

"Once SUSY is broken, it must be possible for all black holes to decay, to avoid black hole remnants"

I dont understand this remark, does it mean that if one has unbroken SUSY, stable black holes can exist but they are not black hole remnants? I just realize that I am somewhat confused about the difference between stable black hole remnants and stable (in particular microscopic) black holes too ...

asked Aug 26, 2014 in Theoretical Physics by Dilaton (6,240 points) [ no revision ]

1 Answer

+ 3 like - 0 dislike

Well, it has to do with the fact that some black holes within supersymmetric theories arises as BPS states of the theory, right? BPS states enjoy a lot of good properties. I am not sure if it applies to all supersymmetric black holes but let me give you an example which I have heard of.

In specific a Reissner-Nordström black hole which is a solution of the Einstein-Maxwell theory can be embedded in a $\mathcal{N}=2$ supegravity (not sure if it applies to any $d$) by adding some spinor-vector fields and taking the limit that they are 0. Now, this is indeed a supersymmetric solution since it possesses Killing spinors and by asking the variation $\delta_{\epsilon} \psi_{\mu \alpha}$ w.r.t. the supersymmetric parameter $\epsilon$ is zero you end up with half the supersymmetries being independent. Ok (things are a bit more messy but not dramatic), then only four Killing spinors are left, half of the total number, and these four transformations which do not act trivially correspond to four fermionic collective modes. Thus our black-hole is part of the $\mathcal{N}=2$ supergravity hypermultiplet which contains a supercharge $Z$ (seriously, think about it as a $\mathcal{N}=2$ susy), and this can be expressed as

$$Z = p-iq$$

where $p,q$ are the electric and magnetic charges of the theory. Now the Reissner-Nordström black hole mass is given by $M_{RN} = (q+p)^{1/2} = |Z|$. Thus we can say that this black hole solution is invariant under half the supersymmetries, it is a BPS state because of the BPS bound on its mass and lives in a short-multiplet thus it is a protected state and absolutely stable. 

I hope that this gives you a clear indication that BPS black holes are stable and if susy is broken they can lose this stability. You can find nice information in Ortin's book.

answered Jan 5, 2015 by conformal_gk (3,625 points) [ no revision ]
Thanks conformal_gk, I like this illustrative example and your nice explanations :-)

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