10
$\begingroup$

Consider two black holes with masses $m_1,m_2$ and zero angular momenta merging to form a single one with the mass $m$ and the rotation parameter $a=J/m$. Hawking, in "Black Holes in General Relativity" Commun. math. Phys. 25 (1972), 152—166 proposed an inequality $$m^2+m\sqrt{m^2-a^2}>2(m_1^2+m_2^2)$$ for this process (in fact, for a more general one, see p. 14 of the paper). I learned about this bound ages ago from the Lightman-Press-Price-Teukolsky relativity problem book and had no doubt about it. But now I think that the proof given in this paper is total rubbish despite being published in a supposedly mathematical journal.

The inequality is derived from what is now called an area theorem which sates that the area of the event horizon never decreases. There is nothing wrong with the theorem itself except the way it is formulated makes it completely useless for obtaining an inequality of this sort. (And probably for any other meaningful conclusion.) The fishy point here is the assumption that the area of a black hole event horizon is given by the formula (in geometric units $c=G=1$) $$A=8\pi m(m+\sqrt{m^2-a^2}).$$ No doubt, this assumption is true for a Kerr black hole but there is a big problem. The event horizon as it is defined in the formulation of the area theorem depends on the (arbitrarily distant) future evolution of a black hole, so even it the thing looks exactly as a standard Kerr black hole now its event horizon may still well be very different from what one of a Kerr hole is supposed to be, with very different area. There is no formula for the actual area of this event horizon in terms of the mass and the angular momentum.

To see where the problem really lies it is convenient to consider a scattering of two black holes instead of their merger. This process has an inverse which is also perfectly physical even if not likely to ever happen in reality. (Because general relativity dynamics is, of course, time-symmetric.) Then exactly the same argument as in the paper when applied to both processes gives two inequalities which contradict each other.

Admittedly, from reading more recent physical literature I have the impression that the problem is more or less known. However, it is never mentioned explicitly. Apparently, physicists believe that the inequality is true anyway and do not care much about gaps in its proof. A mathematician like myself would rather like to see an actual proof though. Is such a proof already known or, at the very least, was the problem ever considered seriously? This is my question.

$\endgroup$
  • 2
    $\begingroup$ as far as I understand, the only generally valid principle is the increase of entropy upon a merger, which implies the surface area inequality $A(m)>A(m_1)+A(m_2)$; all other inequalities depend on additional assumptions for how the area $A$ of the event horizon depends on the mass and other parameters (angular momentum, charge). Some of this is discussed in arXiv:0909.4827 $\endgroup$ – Carlo Beenakker May 2 '19 at 13:37
5
$\begingroup$

A mathematical proof of Hawking's area theorem has been given by Chruściel, Delay, Galloway, and Howard, in Regularity of Horizons and The Area Theorem (2001). The proof identifies the conditions under which the area of sections of future event horizons in space–times satisfying the null energy condition is non–decreasing towards the future. The monotonicity is shown to hold without making requirements on the differentiability of event horizons.

The specific relation between the area of the event horizon and the angular momentum in the OP is model dependent, it does not have general validity.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you. This is an interesting work but it is related to a ``good'' part of the Hawking's parer (the one I have no issues with) about the area theorem. It does not address the question I am asking. $\endgroup$ – Alex Gavrilov May 3 '19 at 8:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.