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BH mass is a conserved quantity and cannot escape the horizon of a BH. Yet in mergers some percent (in GW150914 it was about 5 percent, or 3 solar masses) of the binary's mass escaped either before (one of the two BHs horizons) or after the merger started (one horizon merger, from the merged horizons), and turned into gravitational waves. The argument is that it is potential energy or maybe kinetic energy of the black holes that is radiated. It does not matter. The question is how did the BHs mass get reduced? Did some virtual negative mass particles fall into the BH, and if so it would have been gravitons? If so is it a quantum effects? If so has anyone worked out the linearized quantum gravity of that happening right outside the horizons, like they did for the very small Hawkings radiation, and isn't that a contradiction of the small effects expected for 30times solar mass BHs? So, if not that, how does the negative potential binding mutual gravitational energy of the two BHs steal mass from the binaries?

Please, no hand waving. I've looked everywhere and only see hand waving arguments about potential or other energy outside the BHs, but mass is an internal (inside the horizons) and conserved quantity of BHs.

Charge, another conserved quantity, and angular momentum, can be extracted (i.e., escape the EH). Angular momentum through the Penrose process, maybe charge too. Is it the same for mass? If so, where is it wel explained in some detail, and not simply 'it was calculated numerically'?

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The mass of a black hole is a surprisingly elusive quantity. If you write down the stress-energy tensor for a black hole you'll find it is zero everywhere except at the singularity where it is undefined. It is for this reason that the black hole is described as a vacuum solution.

When we talk about the mass of a black hole the quantity we are referring to is the ADM mass, and this is a quantity that is calculated from the spacetime geometry.

The point of all this is that if the geometry of some area of spacetime changes then its ADM mass will also change, and since two black holes merging certainly counts as the geometry changing we should not be surprised that the total mass changes. There is no reason to expect the sum of the masses of the initial black holes to be conserved, and indeed it is not.

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  • $\begingroup$ Thanks but though I know you are right, you can calculate that the gravitational wave carries energy. And you can estimate the M of the BH left behind, always less. I understand that more could be extracted through mergers, up to about 29 percent for equal BH masses, per the area theorem. But I also know that BHs ( just think spherical for now) acquire the M parameter through collapse of real mass and energy, and I'll bet that's conserved. So conserved is not the issue. It's how the mass absorbed given by M and determining spacetime geometry, could escape the event horizon. A GR calculation. $\endgroup$
    – Bob Bee
    Feb 21, 2016 at 23:25
  • $\begingroup$ I.e., a GR calculation that shows M being reduced as grav waves go out outside the horizon. $\endgroup$
    – Bob Bee
    Feb 21, 2016 at 23:54
  • $\begingroup$ -1 By definition, the ADM mass of a spacetime cannot change. It is associated to spatial infinity, and therefore it is either a single number for the entire spacetime, or its undefined. You may have been thinking of the Bondi energy. $\endgroup$
    – TimRias
    Oct 22, 2020 at 8:24
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BH mass is a conserved quantity and cannot escape the horizon of a BH.

Everything you interact with is outside the horizon. Absolutely everything. What you call the mass is a label you give to a certain kind of curvature that is found outside a black hole. What happened is the matter that fell into the black hole left the outside of it curved as it passed through. That is also how the spacetime outside the earth became curved and how the spacetime outside the sun got curved. The matter falling in leaves the spacetime outside curved. That's actually what matter does.

If you have a spherical shell of mass above a planet you have a solution of type $M+m$ above the shell and a solution of type $M$ between the shell and the surface of the planet. These are two different solutions, and the difference is of type $m$ and connecting together the two solutions is exactly what a mass of mass $m$ does. As the shell falls down the region that has type $M+m$ gets larger (it's the part above the shell and the shell falls down).

Yet in mergers some percent (in GW150914 it was about 5 percent, or 3 solar masses) of the binary's mass escaped either before (one of the two BHs horizons) or after the merger started (one horizon merger, from the merged horizons), and turned into gravitational waves.

That's not correct. Let's add another level of accuracy to that shell model. First we'll note that it isn't the mass in matter that connects the two type of solutions together. It is the energy density (and to a lesser extent the momentum density, the momentum flux, the pressure, and the stress, all five together make the Stress-Energy tensor). As the shell falls the energy increases but the type of solution outside stays the same. That's because more energy close in actually connects the two solutions together than is needed to connect them together farther out.

So you could extract energy from that falling shell and send it out. But then it has the same amount of matter, but less energy. And now it is closer too. So it actually connects different solutions together. So you have the planet, the shell layer and above that you have the energy you extracted and sent up. Outside all three you have type $M+m$ under the energy layer you have $m/2+M$ and under the shell you still have $M.$ So as that energy later expands upwards and eventually gets to you (it is so thin and weak now, from being stretched over a huge area that you barely notice it) then you become inside it and so you see the type $m/2+M$ solution. It looks like some "mass" is gone from the solution but every atom is still there.

What's the point in all that. Basically the gravitational "mass" of an object is not the sum of the masses of all the parts. And indeed it can be less than the sum of the masses of the parts. As in our example.

Even without gravity, the mass of a system can be wildly different than the sum of the masses of the parts. But the thing you called a mass in a gravitational system was probably an energy anyway.

So now the two black holes have a mass that can be less than the mass of the holes. Great. And energy could be extracted from that the same way. If the parts are moving we can slow them down while they are getting closer. Just like we did with the shell. Every atom of the objects might still be there. But we steal their energy and so the system they are a part of can have a smaller $M$ parameter.

The question is how did the BHs mass get reduced?

The system of two black holes has a mass parameter that is not the sum of the parameters of the parts. No real system has a mass parameter that is exactly the sum of the parameters of the parts. The system has some parts that get closer and that move. By slowing them down relative to each other you can reduce the parameter of the system. And that is what happens. Sure, they speed up. But they get slowed down compared to how much they would have sped up without the waves.

Did some virtual negative mass particles fall into the BH

No. And mass isn't the sum of the masses of the parts, so negative mass particles and a conserved additive mass isn't the way to think about anything.

So, if not that, how does the negative potential binding mutual gravitational energy of the two BHs steal mass from the binaries?

Without waves, the two objects orbiting would be a combined system wih it's own rate of rotation $J$ and size $M$ and amount of curvature. It's a type of curvature that could be sourced by fast rotating things that are closer in, or by slower rotating things that are farther out.

They start out as farther away things rotating slower. That's fine. When they move closer they could have moved in a way that produces he same $M$ and $J$ type curvature and that would requiring having some more rotation and some more kinetic energy. To have the exact same $M$ and $J$ they would have to have very specific additional energy and rotation.

But when they emit the waves they end up slower than that target speed. Do the solution outside of them is less. And it's just like the example of sending that layer of energy upwards. It's the solution in between that has a smaller parameter. As the waves go out we see the old $M$ and $J$ solution until the waves get to us, then we are underneath that layer and we see the smaller $M$ solution.

Please, no hand waving. I've looked everywhere and only see hand waving arguments about potential or other energy outside the BHs, but mass is an internal (inside the horizons) and conserved quantity of BHs.

Mass is not an internal quantity (it is a label assigned to spacetime regions outside horizons). And it isn't conserved either. Or even additive. If you insist it is when it isn't then you won't be able to learn correct physics. And energy isn't conserved in general relativity so I'm not sure what you are aiming for anyway.

angular momentum, can be extracted (i.e., escape the EH). Angular momentum through the Penrose process, maybe charge too. Is it the same for mass?

When you extract angular momentum by slowly down the parts then you also extract energy, which is what the parameter $M$ generally correlates with.

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  • $\begingroup$ The equations for the Kerr metric have the total energy (and of course it affects the Kerr metric) composed of a term with M and another term with J. You can reduce J w/o reducing M, as through the Penrose process. So, yes, E was reduced, and the gravitational field changed, but M is the same. That is the equivalent process for extracting energy from inside by reducing M, maybe through some perturbation outside the horizon? Your last sentence is surely wrong $\endgroup$
    – Bob Bee
    Feb 21, 2016 at 23:05
  • $\begingroup$ @BobBee What are you trying to say? Isn't the paper by Penrose and Floyd in Nature Physical Science 229, 177 (1971) called "Extraction of Rotational Energy from a Black Hole"? Sure you could increase it (throw more mass in). But a wave isn't going to do that. And just because you have a family of solutions parameterized by two parameters $M$ and $J$ doesn't mean that a particular process changes one while leaving the other constant. It slows the rotation by the mechanism of stealing energy from the parts. I'm sorry you have a completely baseless theory that $M$ doesn't change when it does. $\endgroup$
    – Timaeus
    Feb 21, 2016 at 23:21
  • $\begingroup$ So you are saying that to extract J you also reduce M? You are probably right, practically, since M was reduced in GW150914. $\endgroup$
    – Bob Bee
    Feb 21, 2016 at 23:32
  • $\begingroup$ @BobBee I'm saying the Penrose process was published as a way to extract energy and hence decrease $M.$ It is possible to decrease $J$ while leaving $M$ the same if you want (by throwing extra energy into the star). But I'm saying you have a fundamentally wrong idea about mass. The mass of a system is not the sum of the masses of the parts. And it isn't conserved either. It's not just GW150924, and it's not just every single simulation ever. It's that your idea is just totally and completely wrong. It's general chemistry level wrongness about what mass is and how it works. $\endgroup$
    – Timaeus
    Feb 21, 2016 at 23:38
  • $\begingroup$ @BobBee An atom has a smaller mass than the sum of the masses of the protons, neutrons and electrons that make it up. A proton or neutron has a hugely larger mass than the mass of the quarks that make it up. A static star has a stationary star has a smaller mass than the sum of the masses of its parts. A rotating star could have a larger mass than the sum of the parts. Buts its energy will be less than the enrgy of its parts. $\endgroup$
    – Timaeus
    Feb 21, 2016 at 23:42

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