My training is in nuclear physics, so my intuition is trained by the liquid drop model, in which the nuclear fluid is incompressible. This leads to the wrong intuition when I try to apply it to the ring-down of a black-hole merger. A sphere has the minimum area for a fixed volume, and during the ring-down I expect the event horizon's shape to become less elongated and closer to a sphere. So if the ring-down proceeded with fixed volume, it would violate the Hawking area theorem (2nd law of black hole thermodynamics). Evidently the volume increases a lot over the course of the ring-down.

Is there a good way to understand why the volume increases so much? Maybe a semi-Newtonian argument? Is it true in Newtonian gravity that equipotentials grow a lot as a body or system undergoes gravitational collapse?

[EDIT] I got an opportunity to spring this question on a defenseless and unsuspecting relativist today. His off-the-cuff reaction was that the area of the event horizon is proportional to the square of the mass, so combining two black holes would clearly increase the total surface area by a lot, although this puts a bound on how much radiation you can have. This didn't seem to me to directly address the question of small perturbations around the final state, but it did suggest that there might be a semi-Newtonian explanation, since it's also true in Newtonian gravity that the area of an equipotential is proportional to the square of the mass. So I made some estimates in Newtonian gravity for an equipotential $\phi=\text{const.}$ of two equal point masses separated by a distance $2h$. Letting $\delta=(2h/\phi)^2$ and doing all calculations to $O(\delta)$, the long axis is proportional to $1+\delta$, while the short axes are proportional to $1-\delta/2$. Approximating this as an ellipsoid, the volume is constant to this order. Since area is minimized by a sphere at constant volume, it seems that this Newtonian approach gives the opposite of the correct relativistic result.

[EDIT] [Gupta et al.]1 calculate the evolution of the area during ring-down numerically for a symmetric merger between non-spinning black holes. There are some open qualitative questions about the behavior of the horizons. They compute the areas of the horizons as functions of time, and also the multipole moments of the horizon. For the mass quadrupole moment of the horizon, they use a parameter $m_2$ which is the quadrupole moment divided by the quadrupole moment of a Kerr black hole. This approaches 1 as the ring-down continues. One thing I notice from their graphs is that the area seems extremely insensitive to small perturbations around the Kerr state. At $t=30M$ (about $13M$ after merger), we have $m_2-1\approx0.1$, but the area differs from the Kerr value by $\sim -2\times 10^{-3}$.

  • 1
    $\begingroup$ The area is well-defined, but how would the volume be defined? For example, how would one define the volume of a Schwarzschild black hole? Normally we would define volume using the metric and integrating over the interior of the boundary within some Cauchy surface; but what Cauchy surface would we use inside a black hole? $\endgroup$ – Chiral Anomaly Apr 5 at 1:09
  • $\begingroup$ @ChiralAnomaly: Good point, thanks. I would be happy with any answer that would provide some insight into why the merger results in an increase of area. $\endgroup$ – Ben Crowell Apr 5 at 3:33
  • $\begingroup$ In GR, sheding "higher multipole moments" typically leads to increase of area. You can take the Kerr solution as an example of this. Reducing the dipole moment of Kerr at fixed energy (= monopole moment) leads to an increase in BH surface area. This is also why the Penrose process does not violate the area law. $\endgroup$ – mmeent Apr 5 at 9:39
  • $\begingroup$ Wouldn't the multipole have to be 2 or higher? I thought there was no such thing as a gravitational dipole. But yeah, I guess my question reduces to why it's true that reducing higher multipole moments increases area. The opposite is true for a liquid drop. $\endgroup$ – Ben Crowell Apr 5 at 12:22
  • $\begingroup$ There is no such thing as a gravitational dipole, but torsion in gravielectromagnetism has a dipole form. $\endgroup$ – Gareth Meredith Apr 7 at 0:25

Your Answer

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

Browse other questions tagged or ask your own question.