# ? on entrop of black holes

1. Mar 14, 2008

### 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?

2. Mar 15, 2008

### nicksauce

3. Mar 19, 2008

### George Jones

Staff Emeritus
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.

4. Mar 19, 2008

### superpoincare

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

5. Mar 19, 2008

### Lawrence B. Crowell

Hawking and Unruh effect

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

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

It is clear that

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

$$tanh(t/4M)~=~v/u.$$

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

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

$$u~=~sinh(gs),~ v~=~cosh(gs).$$

It is clear that for $udu~-~vdv~=~e^{r/2M}/4M dr$ that $g~=~1/(r/2M~-~1)e^{t/2M}/4M$ , and so the accleration diverges as $r~\itex~2m$ 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 $T~\rightarrow~\infty$ as $r~\rightarrow~2M$. The four velocities $U_u$ and $U_v$ will then satisfy

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

with

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

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

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

Now for $r~\rightarrow~\infty$ and $t~\rightarrow~\infty$ this recovers the standard black hole temperature result of

$$T~=~\frac{1}{8\pi M}.$$

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.

6. Mar 20, 2008

### Lawrence B. Crowell

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

L. C.

7. Mar 20, 2008

### 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