? on entrop of black holes

RasslinGod

My physics professor told us that string theory correctly predicts the entropy ofa black hoole. that leaves me wondering...how do u even measure what it's entropy is to even confirm a theoreticla calculation?

is S= k ln W even used at all?

nicksauce

A little reading can be done here...
http://en.wikipedia.org/wiki/Black_hole_thermodynamics

George Jones

As far I know, this relation has been used only for "exotic" black holes, i.e., black holes that have one or more of supersymmetry, Yang-Mills charge, and extra macroscopic spatial dimensions.

Also as far as I know, astrophysical black holes have none of these properties.

superpoincare

Recently people in strings have been doing non-supersymmetric as well.. of course these are not even close to astorphysical black holes...

Lawrence B. Crowell

There is the Hawking-Unruh effect, which tend to merge in a way for an accelerated frame near a black hole. To examine this it is best to transform to another set of coordinates to look at this. The Kruskal-Szekeres coordinates are better. These are given by

[tex]
u~=~\sqrt(r/2M - 1)e^{r/4M}cosh(t/2M),
[/tex]
[tex]
v~=~\sqrt(r/2M - 1)e^{r/4M}sinh(t/2M).
[/tex]

It is clear that

[tex]
(r/2M~-~1)e^{r/2M}~=~u^2~-~v^2
[/tex]
and

[tex]
tanh(t/4M)~=~v/u.
[/tex]

This has the advantage that there is no funny business at [itex]r~=~2M[/itex]. These coordinates provide a chart which covers the whole space. The [itex]u^2~-~v^2[/itex] equation shows that the inner and outer regions of the black hole are given by a branch cut connecting two sheets.

Consider [itex](r/2M~-~1)e^{r/2M}~=~x^2[/itex], and so [itex]x^2~=~u^2~-~v^2[/itex]. The u and v satisfy hyperbolic equations for a constant r and so

[tex]
u~=~sinh(gs),~ v~=~cosh(gs).
[/tex]

It is clear that for [itex]udu~-~vdv~=~e^{r/2M}/4M dr[/itex] that [itex]g~=~1/(r/2M~-~1)e^{t/2M}/4M[/itex] , and so the accleration diverges as [itex]r~\itex~2m[/itex] in order to hold a particle fixed near the event horizon.

The hyperbolic equations above are then identical in form to the ones which obtain for the Unruh effect. The spacetime trajectory in the u and v coordinates is then hyperbolic. Thus for an observer sitting on a frame fixed next to the event horizon of the black hole there would be a huge thermal bath of particles which would become very hot [itex]T~\rightarrow~\infty[/itex] as [itex]r~\rightarrow~2M[/itex]. The four velocities [itex]U_u[/itex] and [itex]U_v[/itex] will then satisfy

[tex]
U_u = 1/(4M(r/2M - 1))e^{t/2M}cosh(gs), U_v = 1/4M((r/2M - 1))e^{t/2M}sinh(gs),
[/tex]

with

[tex]
(U^u)^2~-~(U_v)^2~=~16M^2(1~-~r/2M)e^{-t/2M},
[/tex]

and so the temperature of the black hole is [itex]T~\sim~1/2\pi g[/itex] and

[tex]
T~=~\frac {e^{t/M}}{8\pi M}\sqrt{1~-~r/2M}).
[/tex]

Now for [itex]r~\rightarrow~\infty[/itex] and [itex]t~\rightarrow~\infty[/itex] this recovers the standard black hole temperature result of

[tex]
T~=~\frac{1}{8\pi M}.
[/tex]

Thus for an observer held close to the event horizon of a black the additional acceleration effectively heats up the black hole on that frame.

I think this might also indicate that for a distant observer there is a question as to what she would see of her compatriot held fixed near the black hole. This might be an interesting problem to examine.

Lawrence B. Crowell

I made a couple of Tex errors on the above discussion which I corrected

L. C.

Lawrence B. Crowell

These exotic black holes are BPS black holes, and pertain to issues of quantum gravity or dualities between D-branes and gauge charges. The astrophysical black hole is massive and these putative physics are in some infinitesimal region (or a small number of Planck lengths) from the event horizon. Hence for a stellar black hole these BPS things don't contribute much of anything.

Lawrence B. Crowell