# What kind of systems of black holes satisfy the laws of black hole thermodynamics?

+ 4 like - 0 dislike
1757 views

I've come across black holes thermodynamics multiple times recently (both at this site and elsewhere) and some things started bugging me.

For one thing, first law bothers me a little. It is a reflection of the law of conservation of energy. This is fine when the space-time is stationary (as in Kerr solution) and is consistent with system being in a thermal equilibrium (so that thermodynamics apply at all). But what about more general systems of black holes?

What are the assumptions on the system of black holes for it to be in thermal equilibrium so that laws of BH thermodynamics can apply?

Note: the reason I am asking is that I heard that the laws should also be correct for system of multiple BH (so that their total event horizon area is increasing, for example). But I cannot wrap my head about how might the system of BH be in thermal equilibrium. I mean, they would be moving around, generating gravitational waves that carry energy away (violating the first law) all the time. Right?

This post imported from StackExchange Physics at 2014-04-01 16:46 (UCT), posted by SE-user Marek

+ 6 like - 0 dislike

The first law is not violated if stated properly:

$$dM = \frac{\kappa}{2 \pi}dA + \Omega dJ + \Phi dQ$$

(reference: wikipedia)

where $\kappa,\Omega,J,\Phi,Q$ are the surface gravity,angular velocity, angular momentum, electric potential and charge of the black hole, respectively. Compare this to the usual expression for the first law:

$$dE = TdS + PdV + \mu dN$$

One can (heuristically) make the identifications $T = \frac{\kappa}{2\pi}$, $S = A/4$, $\mu = \Phi$ and $N = Q$. The first two of these are well-established. Hawking showed that the temperature of the black hole is proportional to its surface gravity ($T \propto \kappa$) and Bekenstein showed that its entropy should be proportional to its area ($S=A/4$). The third and fourth equalities ($\mu = \Phi$ and $N = Q$) can be understood if we think of the black hole as an aggregate of N particle with unit charge. Adding another charged particle to this ensemble of $N$ particles, with total charge $Q$, will cost an amount of work given by $\Phi dQ$.

For the case of a single black hole, one can use the framework of dynamical horizons developed by Ashtekar, Badri Krishnan, Sean Hayward and others [refs 1, 2]. Turns out the laws of black hole entropy can be extended to completely dynamical black holes with a well-defined expressions for the first and second laws in terms of fluxes through the dynamical horizon.

The definition of a dynamical horizon is in terms of the expansion of the inward pointing, null normal vector field on the 2+1 d boundary of a 3+1 d region. I can't think of the detailed expressions of the top of my head, but you can find them in the above reference.

I can't give a concrete answer for the case of multiple black holes, but I would think you could extend the dynamical horizon framework to that case - probably not without some serious effort though.

In any case, there will be no violation of energy conservation. The sum of the energy and momentum carried off by gravitational waves (and detected by an observer at infinity) and the change in the black hole's energy and momentum (again w.r.t such an asymptotic observer) will remain a constant.

Hope that helps !

Edit: Corrected proportionality constants for $T$ and $S$. Thanks @Jeff!

This post imported from StackExchange Physics at 2014-04-01 16:46 (UCT), posted by SE-user user346
answered Dec 7, 2010 by (1,985 points)
Also, the definition of an IH/DH isn't dependent on a 3+1 slicing. It can be defined intrinsically on a null/spacelike hypersurface. And heh, I know a reference: arxiv.org/abs/1009.0934

This post imported from StackExchange Physics at 2014-04-01 16:46 (UCT), posted by SE-user Jerry Schirmer
Congratulations on finishing up @Jerry :) The slice-independence of IH/DHs is a very important fact, indeed. Thanks for pointing it out!

This post imported from StackExchange Physics at 2014-04-01 16:46 (UCT), posted by SE-user user346

This post imported from StackExchange Physics at 2014-04-01 16:46 (UCT), posted by SE-user Jerry Schirmer
@user346: Marek may have accepted the answer that the gravitational waves are reducing the "omega" in the BH thermodynamics, but it is completely false. The potential energy of the two black holes is increasingly negative, and this is what balances the outgoing radiation. The black hole theromodynamics is for the hole itself.

This post imported from StackExchange Physics at 2014-04-01 16:46 (UCT), posted by SE-user Ron Maimon
@user346: -1 this answer is totally wrong! The radiated energy is mostly reducing the potential energy of the orbiting black holes, and you have not included that energy contribution, and pretended that the total energy of the multi-black-hole system is just the sum of the energies of its constituent black holes, without their gravitational potential energy.

This post imported from StackExchange Physics at 2014-04-01 16:46 (UCT), posted by SE-user Ron Maimon
You might want to edit your answer so that the numerical factors are correct. $T= \kappa/2 \pi$, $S=A/4$.
 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOver$\varnothing$lowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). Please complete the anti-spam verification