# Extremal Kerr Black holes, and their associated temperature.

1. Nov 17, 2008

### Harenil

So I was calculating the temperature for a extremal Kerr black hole. G2*M2 = a2, where a = J/m. Using T = kappa / 2 /Pi... I'm assuming this is the correct approach. But my question is...

The surface gravity of an extremal Kerr black hole appears to be zero, leading to a zero temperature. Well, this baffles me, but seems to point to the non existence of extremal black holes in the physical universe. In my GR class we discussed the possibility of creating a naked singularity by hurling additional momentum into a black hole, and decided that it was simply impossible, but we never stated whether or not an extremal black hole could be formed. It seems that as a2 approaches G2*M2, the surface gravity, and thus the temperature drops. This seems counter intuitive to me, and points to a likely possibility that astrophysical Kerr black holes have a near zero surface temperature. Anyone care to shed some light/insight on the subject?

Daniel

2. Nov 17, 2008

### xantox

According to the third law of black hole mechanics it should be indeed impossible to accrete a black hole to the critical ratio for which the temperature is zero in finite time. See Bardeen, Carter, Hawking, "The four laws of black hole mechanics", Commun. math. Phys. 31, 161-170 (1973).

3. Nov 17, 2008

### Harenil

Thank you very much, I will have to check that out.

So I am correct that an extremal black hole will possess zero temperature and entropy?

4. Nov 18, 2008

### stevebd1

Technically, if a Kerr black hole attained a/M=1 then it would no longer be a black hole. The inner and outer event horizons would converge at M (gravitational radius), leaving a naked singularity in time-like space.

5. Nov 18, 2008

### Harenil

I was under the impression that you could not have a naked singularity unless a/M was greater than 1. And if a/M = 1, then there was a single event horizon, but an event horizon none the less.

6. Nov 18, 2008

### stevebd1

At a=M the horizon would be degenerate due to the Killing surface gravity becoming zero at the Killing horizon. If there is no surface gravity (κ) then there is technically no temperature. It seems the subject of extremal Kerr black holes brings up the subject of higher dimensions. In one source that mentions multiple dimensions, in d=4 the degenerate horizon still has an area but in d=5, the area becomes zero and a maximal Kerr solution is considered a naked singularity.

source (beneath equation 37)-
http://relativity.livingreviews.org/Articles/lrr-2008-6/articlese4.html

Another paper that might be of interest-
'The Entropy Function for the extremal Kerr-(anti-)de Sitter Black Holes'
http://arxiv.org/PS_cache/arxiv/pdf/0804/0804.3811v2.pdf

Another source states-
'Introductory Lectures on Black Hole Thermodynamics' by Ted Jacobson
http://www.physics.umd.edu/grt/taj/776b/lectures.pdf
page 14

So it looks like in d=4, the extremal Kerr black hole has zero temperature but non-zero entropy.

Last edited: Nov 19, 2008
7. Nov 18, 2008

### Harenil

Quite fascinating, thank you very much.

8. Nov 18, 2008

### xantox

The temperature of an extremal black hole is generally believed to be zero, there are however recent views suggesting that an extremal black hole must have a non-thermal spectrum and that its temperature cannot be defined.

Concerning the entropy of an extremal black hole, string theory calculations enforce the Bekenstein-Hawking entropy of the horizon area. However, this is also debated. Semiclassical calculations indicate zero entropy despite a nonzero horizon area, such as Hawking and Horowitz, "Entropy, area, and black hole pairs", Phys. Rev. D 51, 4302-4314 (1995).

Last edited: Nov 18, 2008