Friday, 17 December 2010

Can we (theoretically) spin the black hole so strong that it will be broken apart by centrifugal force?


Can we (theoretically) spin the black hole so strong so it will be broken apart by centrifugal force?




For a Kerr-Newman (rotating, charged, isolated) black hole of mass MM, angular momentum JJ, and charge QQ, the surface area of the event horizon is given by
A=8Mleft[M2+(M2a2Q2)1/2Q2/2right]text,A=8Mleft[M2+(M2a2Q2)1/2Q2/2right]text,
where a=J/Ma=J/M. An extremal black hole occurs when M2=a2+Q2M2=a2+Q2. Beyond that, if the black hole is even more overspun or overcharged, is an "overextremal" Kerr-Newman spacetime, which wouldn't really be a black hole at all, but rather a naked singularity.



Thus, I interpret your question as asking whether or not a black hole can be be spun up to the extremal limit and beyond, so as to destroy the event horizon. It's very probable that it can't be done.



Wald proved in 1974 that as one flings matter into a black hole to try to increase its angular momentum, the nearer to an extremal black hole it is, the harder it is to continue this process: a fast-spinning black hole will repel matter that would take it beyond the extremal limit. There are other schemes, and though I'm not aware of any completely general proof within classical general relativity, the continual failure of schemes like this is well-motivated by the connection between black hole dynamics and thermodynamics.



For example, the Hawking temperature of the black hole is TtextH=kappa/2piTtextH=kappa/2pi, where
kappa=fracsqrtM2a2Q22Mleft(M+sqrtM2a2Q2right)Q2kappa=fracsqrtM2a2Q22Mleft(M+sqrtM2a2Q2right)Q2
is the black hole's surface gravity. Thus, even reaching the extremal limit is thermodynamically equivalent to cooling a system to absolute zero.

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