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