Jump to content

what is the entropy of a black hole ?


Recommended Posts

That was the value (1/4 is also a constant, btw :P). But for me the more interesting question is: What is the entropy of a black hole (or alternatively: Why would I expect it to be proportional to the volume and what is the volume)?

Link to comment
Share on other sites

Maybe I should have said it is proportional to a quarter of the area, the proportionality depends on the units you use.

 

The nieve way I view the blackhole entropy is a the property that "restores the laws of thermodynamics". As they have a temperature it is needed.

Link to comment
Share on other sites

Maybe I should have said it is proportional to a quarter of the area, the proportionality depends on the units you use.

I was not sure if I should comment on this or if at least I should do so in PM. But since it might also be of interest to others I'll do so in this post (despite being somewhat off-topic):

1) I do not see how that statement is different from your first one.

2) I really don't think a proportional constant depends on the units used! What you probably meant is that something like 10 m/s² can read 1000 cm/s², but that's just the same constant expressed in different units. In effect, I also doubt that there's any physics in the factor of 4 (which I expect to come from a squared R_s = 2M - see calculation below). It's just a factor that appears due to the canonical choice of setting physical constants to 1 (dimensionless). Not actually knowing the calculations I might be wrong on that, though.

 

The naive way I view the blackhole entropy is a the property that "restores the laws of thermodynamics". As they have a temperature it is needed.

I was initially thinking in the lines of John in which case the question of micro states is either trivial in classical physics (mass, charge and angular momentum completely fix the micro state) or super-complicated (when you consider qft on curved spacetimes; dunno what happens there but I know that some of our aqft people are still trying to understand what temperature could mean on relatively arbitrary spacetimes so it's probably not easy).

 

Introducing it because one introduced a temperature does sound somewhat reasonable. If (for some reason) I assume that Hawking Temperature is a temperature as in Thermodynamics (not sure to what extent that makes sense) then I could even reproduce the proportionality of S to R² with some (possibly oversimplified) easy calculation. I'll show it here, maybe someone would like to comment on it (let me emphasize here: I just pulled that out of my lowerback; it's not exactly backed up by serious literature).

 

 

(1) Let's call the mass of the BH it's energy. I am not sure to what extent that makes sense - I don't even know to what energy makes sense as a global property in curved spacetime. But it's perhaps a start.

(2) According to WP the Hawking Temperature is inversely proportional to the BH's mass [math]\stackrel{(1)}{\Rightarrow} T = \frac{\alpha}{E}[/math].

(3) Assume there is no pressure (whatever that would be) and that standard Thermo holds [math] \Rightarrow dE = TdS \Rightarrow dS = \frac{dE}{T} \stackrel{(2)}{=} \alpha E \, dE [/math]. Then, [math] S(E) - \underbrace{S(E=0)}_{=0} = \int_0^E \alpha E \, dE = \frac{\alpha}{2} E^2 = \tilde \alpha R^2[/math] where in the last step I used assumption (1) and the fact that the Schwarzschild radius is proportional to the BH's mass.

 

I would not be too surprised if the mainstream derivation would be similar to my attempt - hopefully with more clarity about pressure and relation of energy to mass. Even if that was the case and those two points were clear then there's still one important caveat to this that prevents the calculation from answering the question what the entropy of a BH is: Most if not all people on sfn (including me, possibly excluding you) will not really know what the temperature of a BH is (except when tricked by seeing an allegedly familiar word and thinking they therefore understood the concept that it stands for). And of course, even then the so-defined entropy could be just a result of tossing around equations without any real meaning.

 

EDIT: I indeed found that I could have saved some time asking Wikipedia first. Whatever, more fun trying it out myself.


Merged post follows:

Consecutive posts merged
Just a thought, I am sure I was told that

s=k ln w

Can someone who knows what the constants are please calculate w for me.

Say for a nice, simple, non-rotating black hole with the same mass as the sun.

As said previously, that would have been my initial approach (before Andrew mentioned an access via temperature), too. The basic answer to your question is probably: No, no one on sfn can do it in a manner that you will understand. This answer is based on the following section I found on Wikipedia (and -as every WP reader- blindly believe):

Although Hawking's calculations gave further thermodynamic evidence for black hole entropy, until 1995 no one was able to make a controlled calculation of black hole entropy based on statistical mechanics, which associates entropy with a large number of microstates. In fact, so called "no hair" theorems appeared to suggest that black holes could have only a single microstate. The situation changed in 1995 when Andrew Strominger and Cumrun Vafa calculated the right Bekenstein-Hawking entropy of a supersymmetric black hole in string theory, using methods based on D-branes[/b'].

From that quote I guess that there is no calculation that does not involve mathematics and physics beyond a level either of us will understand. I'll gladly be proven wrong on this, though. So any stringer or other guy familiar with such a calculation is gladly invited to try ;)

Edited by timo
Consecutive posts merged.
Link to comment
Share on other sites

From that quote I guess that there is no calculation that does not involve mathematics and physics beyond a level either of us will understand. I'll gladly be proven wrong on this, though. So any stringer or other guy familiar with such a calculation is gladly invited to try ;)

 

I think it would be an effort to understand all the details, and I expect beyond the average user of this forum.

 

Black holes have a temperature, the calculation of which involves semiclassical gravity and thus suffers the fate of the above comments. The deep origin of the temperature is the fact that we have no unique vacuum in semiclassical gravity. Different observers will not agree on empty or filled states.

 

This is almost identical to Unruh radiation as an observer would have to keep accelerating as not to fall into the black hole.

 

 

Anyway, any thermal system can have entropy defined via

 

 

[math]dE = T dS[/math].

 

So, it makes sense that black holes also have entropy.

 

Calculating this entropy directly form quantum/statistical physics is very involved.

 

In essence, by analogy with standard thermodynamics you get the proportionality to the area as timo has done. Well done :)

 

The important thing is that it is the area and not the volume. This is part of the holographic principle. It states roughly that "descriptions of the a volume of space are encoded in the area bounding the volume". This seems a general principle in quantum gravity and string theory.

 

For the black hole entropy this is slightly puzzling. The log of the number of states is proportional to the area of the horizon and not the volume. Basically everything about the black hole is encoded in the surface.


Merged post follows:

Consecutive posts merged

 

As for the constants of proportionality for the entropy, up to the usual collection of Planck's + Newton's constants and the speed of light (sometimes all or some of these are set to 1) we get 1/4.

 

[mp][/mp]

 

To devrimci_kürt .

 

Has any of this gone towards answering your question?

Edited by ajb
Consecutive posts merged.
Link to comment
Share on other sites

So, with all the constants of nature put in we get (from Timo's calculation for example)

 

[math]S = \left(\frac{k c^{3}}{\hbar G}\right)\frac{A}{4} = k log(w)[/math]

 

Or we can represent this as

 

[math] \frac{S}{k} = log(w) = \frac{A}{4 (l_{P})^{2}}[/math].

 

[math]l_{P}[/math] being the Planck length.

 

You could then interpret this as roughly one degree of freedom with finite number of states per Planck area. The horizon is essentially divided into Planck cells. String theory enters here as it provides candidates for these degrees of freedom. (But of course it is not quite that simple as they do not relate directly to the horizon area. )

 

Something you should also notice is that the entropy of an astrophysical black hole is going to be huge.

Edited by ajb
Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.