2
$\begingroup$

The irreducible mass $ \rm m_{irr} $ of a black hole with charge $ \rm Q $ is (in natural units)

$$ \rm m_{irr} = \frac{r_{+}}{2} = \frac{M}{2} + \sqrt{\frac{M^2}{4} - \frac{Q^2}{4}} $$

which is less than the mass equivalent of the total energy $ \rm M $ (as measured from infinity), since $ \rm M $ contains not only the mass of the material but also the mass equivalent of the electromagnetic field energy.

Suppose we merge (neglecting gravitational waves) two charged black holes with the same mass and the opposite charge (so that the resulting net charge should be zero), would the resulting Schwarzschild black hole with $ \rm Q=0 $ have the total mass equivalent of $ \rm 2M $, or rather $ \rm 2 m_{irr} $?

In the reference Horizon Mass Theorem it says

enter image description here

so I would suspect the resulting mass equivalent to bei $ \rm 2 m_{irr} $, but I've also heard some arguments favoring $ \rm 2 M $

$\endgroup$
  • $\begingroup$ Related: physics.stackexchange.com/q/45448/2451. $m_{\rm irr}=r_+/2$. $\endgroup$ – Qmechanic Sep 7 '18 at 14:10
  • 1
    $\begingroup$ What do you mean by "neglecting gravitational waves" here? You only get to Schwarzschild at the end of this process because gravitational waves carry the non-spherically symmetric parts away from the merged black hole. $\endgroup$ – Brick Sep 7 '18 at 14:15
  • $\begingroup$ Therefore we assume a straight forward head on collision of two equal mass black holes, that should render the gravitational waves neglectable. This question focusses on the mass equivalent of the neutralized electromagnetic field energy. Gravitational waves are an additional effect which would only make the scenario more complicated than necessary, since we can also reduce the problem to a spherically symmetric and therefore guaranteed gravitational wave free scenario with an infalling shell of dust instead of two black holes. $\endgroup$ – Yukterez Sep 7 '18 at 14:20
  • $\begingroup$ @Qmechanic that is just an abbreviation, the question is, since $m_{irr}=M$ for Schwarzschild, would the new $M$ be the sum of the previous $M$s or the previous $m_{irr}$s. As I understand the Reissner Nordström metric it should be the latter, but on the other hand the neutral system of a proton and an electron is not the sum of their imaginary $m_{irr}$ but rather their $M$, that is what confuses me. $\endgroup$ – Yukterez Sep 7 '18 at 15:31
  • $\begingroup$ we assume a straight forward head on collision of two equal mass black holes, that should render the gravitational waves neglectable. Why would you assume that? From that geometry we could only conclude that the result would have $L=0$, but not the absence of gravitational radiation. For that matter, there would also be considerable EM radiation that would also carry away energy. $\endgroup$ – A.V.S. Sep 7 '18 at 16:21
1
$\begingroup$

This paper studies this question: https://arxiv.org/pdf/1311.6483.pdf

Note that the in-fall process for the configuration that you described results in both gravitational and electromagnetic radiation, both of which carry some energy away from the final black hole. The precise amount of each depends on the ratio $Q/M$. It looks like it's relatively small in all cases, so you'll end up with a black hole close to $2M$ in your notation. (The paper starts with each black hole having mass $M/2$ and so gets a final mass close to $M$ in their notation.)

EDIT

Here's some additional explanation following revision to the original question and the ensuing comments.

General relativity is a non-linear theory, so you cannot just take a linear combination of solutions like you can with, for example, the Maxwell equations. So let's consider to two limiting cases first:

  1. The black holes are "very close" together.
  2. The black holes are "very far" apart.

Quantifying "very far" and "very close" is beyond what I'll do here, but it should be some multiple of $M$.

When they are very close together, then the expectation should be that they are already approximately Schwarzschild with mass $2M$ (and no charge). The definition of "irreducible mass" provided by the OP should be applied to the binary (if at all) rather than separately to the individual black holes with their net charge. But since the net charge is 0 in this case presented, the irreducible mass is also $2M$. I see no sensible way to apply the definition of irreducible mass separately to the two black holes in this case for a variety of reasons:

  • The energy of a charge "lowered" into one of the black holes is path-dependent, so, at best, the expression would need to be more complicated.
  • The whole point of the definition is to separate the energy attributable to the EM field from everything else. In this case, the EM field is due to both black holes, so it's not immediately clear how such a definition could be made for one of the black holes. (Maybe for equal magnitude charge it's possible, but I think the nonlinearities may make it hard to separate outside of that case.)
  • To whatever extent you can compute separate contributions to the EM field from each BH, they will give partially cancelling contributions to the EM field. Since the irreducible mass is effectively trying to capture the energy in the EM field, whatever you did here would have to account for the fact that the EM field contains much less than the sum of the individual fields due to cancelling effects. Naively applying the definition is "more like" assuming that the field is doubled.

When they are very far apart, each experiences minimal curvature due to the other. In this case, you can approximately take the linear combination of solutions as initial data. For the same reason, it probably makes sense to talk about their individual "irreducible masses" (as defined by the OP). If you let them fall toward each other, however, a few things will happen:

  1. They will radiate gravitational radiation up to and past the time of merger, which will carry away the non-spherically-symmetric features of the spacetime structure.
  2. They will radiate electromagnetic energy, which will carry away the non-spherically-symmetric portions of the EM field.
  3. Some - as it turns out most - of the radiation will fall into the BHs and contribute to the mass of the steady-state, merged BH.

So in the "very far" case, you'll still end up with a merged black hole with mass that is approximately $2M$. As noted in the paper linked and in the comments, it is not necessarily obvious a priori how much of the energy would have been carried away, but the numerical results suggest that it's a small fraction.

Outside of these extreme cases, you'll need a computer to try to answer. You probably even need a computer to generate the initial data prior to the in-fall process.

Now I guess, perhaps, at the root of the question, is where does the "extra" mass come from in the cases that are farther apart, where "extra" means the difference between the $2M$ Schwarzschild mass of the merged BH and the sum of the "irreducible masses" pre-merger. The answer, I think, is that it comes from the EM field energy that is drawn into the final black hole.

$\endgroup$
  • $\begingroup$ Minor comment to the post (v1): In the future please link to abstract pages rather than pdf files. $\endgroup$ – Qmechanic Sep 7 '18 at 16:43
  • $\begingroup$ On page 2 it rather looks like 1M (in my notation): If you collide 2 Schwarzschild BHs you loose 29% and end up with 2M*(100-29)≈1.4M, and if Q is opposite Q=±M you lose 0.65% and end up with 2M*(100-65)≈0.7M. Divide 1.4M/0.7M=2, which means that two Schwarzschild BH with each having M=1 combined are twice as massive as two combined opposite charge BHs with also M=1. $\endgroup$ – Yukterez Sep 7 '18 at 16:52
  • $\begingroup$ Yeah, actually they change notation later in the paper. I think what you want is the result in Table 1? Top of third page, second column, you'll see the notation used for the numerical simulations. $\endgroup$ – Brick Sep 7 '18 at 16:54
  • $\begingroup$ Also notice that the part you mentioned is a bound only. $\endgroup$ – Brick Sep 7 '18 at 16:58
  • $\begingroup$ True, but it's nearer on M/2 than on M. $\endgroup$ – Yukterez Sep 7 '18 at 17:02

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.