distorted event horizons and temperature of black holes

verdigris

Black hole event horizons are circular in shape, in isolation.But if
one black hole approaches another black hole the event horizons of
both holes must distort away from circular.Indeed the closer the holes
get together the closer the distorted part of the event horizons
between the holes must get to the centres of the black holes (because
the gravity of one black hole can assist particles to escape from the
other - if this were not true then we would have to say that one black
hole has no gravity and therefore no mass -clearly ridiculous!). Some
of the event horizon will be further away
from the centre of the black holes because the gravity of one hole is
helping the gravity of the other to keep particles from leaving it.
So we would expect that overall the Hawking Radiation emitted for two
Black holes approaching one another is the same as that for two
distant "independent" black holes.However if we had one central black
hole surrounded by thousands of others - all holes of equal mass - the
event horizon of the central black hole would shift throughout its
circumference towards its centre.Now,
a black hole with a smaller scharzschild radius usually is a black
hole with a smaller mass and a hotter temperature.So the central black
hole would have a hotter temperature than expected.Is my logic here
correct?

carlip-nospam@physics.ucdavis.edu

verdigris <verywellsaid@yahoo.co.uk> wrote:
> Black hole event horizons are circular in shape, in isolation.But if
> one black hole approaches another black hole the event horizons of
> both holes must distort away from circular.

Right. In general, any matter outside the horizon can distort the
shape. Visser calls such a distorted black hole a "dirty black
hole."

[...]
> However if we had one central black
> hole surrounded by thousands of others - all holes of equal mass - the
> event horizon of the central black hole would shift throughout its
> circumference towards its centre. Now,
> a black hole with a smaller scharzschild radius usually is a black
> hole with a smaller mass and a hotter temperature.So the central black
> hole would have a hotter temperature than expected.

Actually, you end up with a cooler black hole. You can find the details
in Visser, "Dirty blackholes: Thermodynamics and horizon structure,"
Phys.Rev. D46 (1992) 2445, http://arxiv.org/abs/hep-th/9203057. He works
out in detail, for instance, a black hole surrounded by a thin shell of
matter.

The piece you're missing, I think, is that the Hawking temperature of a
black hole is the temperature as measured at infinity. (The temperature
diverges at the horizon.) Photons emitted by a "dirty black hole" still
have to get to infinity, and are red-shifted in the process by the field
of the matter surrounding the black hole.

It might be possible to ask about the temperature near the black hole --
inside the shell of matter distorting it -- to see how that is affected.
This is a little tricky, though: you need to figure out what you mean by
the "same position" in two different spacetimes. Perhaps you could look
at the effect of slowly decreasing the radius of a surrounding shell of
matter; this should be possible using section V.B of Visser's paper.

Steve Carlip

carlip-nospam@physics.ucdavis.edu

carlip-nospam@physics.ucdavis.edu wrote:
> verdigris <verywellsaid@yahoo.co.uk> wrote:
>> Black hole event horizons are circular in shape, in isolation.But if
>> one black hole approaches another black hole the event horizons of
>> both holes must distort away from circular.

> Right. In general, any matter outside the horizon can distort the
> shape. Visser calls such a distorted black hole a "dirty black
> hole."

> [...]
>> However if we had one central black
>> hole surrounded by thousands of others - all holes of equal mass - the
>> event horizon of the central black hole would shift throughout its
>> circumference towards its centre.

I had replied earlier, but missed a key point.

Start with Newtonian gravity. Consider a uniform shell of mass. What
is the net gravitational force on an object inside the shell?

The answer is that it's zero. This is obviously true at the exact
center. It's less obvious elsewhere, but moving an object from the
center towards the shell brings part of the shell closer, increasing
attraction in that direction, while increasing the total mass in the
opposite direction; it turns out that the two effects cancel. (You
may be familiar with this is a different setting -- the electric
field inside a uniformly charged shell is also zero.)

It turns out that by Birkhoff's theorem, the same is true in general
relativity. So if you have surrounded your central black hole by a
uniform shell of other black holes, the net effect on it is zero.
(Since your "shell" is not quite continuous, this won't be exactly
true, but it's a good approximation.)

The upshot is that the temperature of the black hole as measured
inside the shell will be unchanged, while the temperature outside
the shell will be decreased by the added red shift due to the mass
of the shell.

Steve Carlip