# How Does Mass Loss from Black Hole Mergers Affect Orbits?

• I
• tzimie
tzimie
TL;DR Summary
Question about the effects of the missing mass, lost due to the radiation of gravity waves
Hi

I have two black holes 1 Sun mass each, when they merge a significant percentage of the mass is lost and is radiated away, so the combined black hole remnant is only, say, 1.8M.

What is a behavior of the test body on an orbit far away from that binary? I assume that after some time (distance to the pair divided by c), test body would 'feel' that there is less mass inside and it would change the orbit? It is correct?

Can emitted gravitation waves be trapped on the last light orbit or fall back to the black hole?
That enourmous mass/energy loss - in what cells of metric tensor the radiated energy is hiding?

Thank you

tzimie said:
I assume that after some time (distance to the pair divided by c), test body would 'feel' that there is less mass inside and it would change the orbit? It is correct?
Yes. The test body will have its orbit change after it detects the passage of gravitational waves that are carrying away the emitted energy from the merger.

tzimie said:
Can emitted gravitation waves be trapped on the last light orbit or fall back to the black hole?
"Emitted" generally means "emitted to infinity", which then translates to "emitted to the rest of the universe". That is what the literature is describing when it talks about gravitational waves being emitted from events like a black hole merger.

It is of course possible in principle that there are gravitational waves that don't make it to infinity for whatever reason, but those aren't counted in the computations in the literature for things like the energy carried away by emission of gravitational waves.

tzimie said:
That enourmous mass/energy loss - in what cells of metric tensor the radiated energy is hiding?
Black holes are vacuum solutions, and so are gravitational waves. That means the energy contained in them cannot be localized. You can't point to a specific part of the metric at a specific point and say that that's where the energy is. You have to look at global quantities. Energy emitted by gravitational waves is usually evaluated by looking at the difference between the ADM mass and the Bondi mass, and subtracting out energy contained in non-gravitational radiation (like EM radiation or neutrinos).

PeroK
tzimie said:
what cells of metric tensor the radiated energy is hiding?
You can ask what's the difference between a flat spacetime metric and the metric of a spacetime that includes gravitational waves. This isn't the same as where the energy is (as Peter explained, that doesn't really make sense).

Unfortunately, the answer is that any component can be changed. If we simplify a bit by choosing the wave to be parallel to the z axis then the ##g_{iz}## and ##g_{zi}## components of the metric will be the same as in flat spacetime. It is always possible to arrange things so that the ##g_{it}## and ##g_{ti}## components are always zero, leaving you with only the ##g_{xx}##, ##g_{xy}##, ##g_{yx}## and ##g_{yy}## components differing from flat spacetime.

I'll repeat that this isn't an answer to your question, although it may be what you intended to ask. It's also a weak-field derivation, and I don't know if it holds close in. Finally, as anything talking about components of a tensor, it depends on your choice of coordinates and you can get different answers by picking a different coordinate system.

Thank you both. You helped me a lot.

Let me ask related question. Two one-solar mass neutron stars merge. Due to different initial conditions, different amount of mass in radiated away, so say remnants are 1.8Msun and 1.9Msun for 2 different scenarios

But wait, in both cases the number of neutrons are the same! Yes, neutrons are lighter to an observer in infinity when they are in gravitational well.

But! 1.9M remnant is heavier (and likely smaller), and gravity well is deeper, but neutrons in that remnant are, on average, heavier!

How is it possible?

You can say, ok, may be the number of neutrons is not conserved during violent merger. But let's keep GR but replace the Standard Model with an alternative physics, where there is only one fermion, "pokemon" instead of neutron, and "pokemonness" is strictly conserved, so the number of particles is guaranteed to be the same

What's radiated away from an inspiral binary systems isn't mass, its on the account of decreasing potential energy of the system.

tzimie said:
replace the Standard Model with an alternative physics
We can't do that since there is no such alternative theory. It's pointless to ask what a nonexistent theory predicts.

tzimie said:
Two one-solar mass neutron stars merge. Due to different initial conditions, different amount of mass in radiated away, so say remnants are 1.8Msun and 1.9Msun for 2 different scenarios

But wait, in both cases the number of neutrons are the same!
If the number of neutrons is the same, the final state will be the same. It's not possible to have two neutron stars with the same number of neutrons but different masses. For any given number of neutrons there is only one possible neutron star equilbrium.

(Technically, the above is true for non-rotating neutron stars; if we allow rotation then two different neutron stars with the same number of neutrons but different rotation rates could have different masses. But I don't think the masses could differ by as much as they do in your example.)

Ibix
tzimie said:
How is it possible?
I don't think it is, except in the sense that that different impacts may eject different amounts of matter.

The energy of the systems pre-collision could, in principle, be different. (In practice this would usually be negligible.) But if the collision doesn't eject matter and the masses of the two stars are the same in each case the mass of the result will be the same. The extra energy would be radiated away.
tzimie said:
But let's keep GR but replace the Standard Model with an alternative physics,
If you invent your own physics, anything can happen. It's not a good way to study reality.

Ibix said:
If you invent your own physics, anything can happen. It's not a good way to study reality.
And asking us to explain it is at least as unlikely to be a good strategy.

PeterDonis said:
If the number of neutrons is the same, the final state will be the same. It's not possible to have two neutron stars with the same number of neutrons but different masses. For any given number of neutrons there is only one possible neutron star equilbrium.

(Technically, the above is true for non-rotating neutron stars; if we allow rotation then two different neutron stars with the same number of neutrons but different rotation rates could have different masses. But I don't think the masses could differ by as much as they do in your example.)

1. If the number of neutrons is the same, the final state will be the same
2. Effects of rotation are minor
3. Different merger scenarios result in different quadripole moment, so different amount if mass is radiated away

I beileve these 3 items are mutually inconsistent, at least one should be false. Do you agree?

berkeman
tzimie said:
1. If the number of neutrons is the same, the final state will be the same
2. Effects of rotation are minor
3. Different merger scenarios result in different quadripole moment, so different amount if mass is radiated away

I beileve these 3 items are mutually inconsistent, at least one should be false. Do you agree?
You stated them wrong. 1 is only true if rotation is exactly zero. 2 does not say the effects of rotation are zero; "minor" does not mean zero, so different rotation means different final mass (so different amount of energy radiated away). 3 is consistent with the correct 1 and 2 because different quadrupole moment means different rotation.

PeterDonis said:
You stated them wrong. 1 is only true if rotation is exactly zero. 2 does not say the effects of rotation are zero; "minor" does not mean zero, so different rotation means different final mass (so different amount of energy radiated away). 3 is consistent with the correct 1 and 2 because different quadrupole moment means different rotation.
So what you are saying is that the difference is explained by the rotation. I had almost agreed with you until I found a contradicting scenario

Imagine 2 non-rotating neutron star collision 'head-on', like billard balls, so the resulting neutron star has no rotation. Obviously, the amount of mass radiated away is less than a 'near miss' scenario, when neutron stars start their 'death dance' around each other emitting gravitation waves.

Hence non-rotating neutron star must be heavier than the rotating one, while I expected exactly an opposite.

tzimie said:
So what you are saying is that the difference is explained by the rotation.
The difference in rotation, yes.

tzimie said:
I had almost agreed with you until I found a contradicting scenario
No, you haven't. You just have not correctly analyzed your scenario. See below.

tzimie said:
Imagine 2 non-rotating neutron star collision 'head-on', like billard balls, so the resulting neutron star has no rotation.
Ok. This means the collision must be inelastic, so the resulting neutron star must gain a huge amount of heat (all of the original kinetic energy of the two colliding neutron stars) that must be radiated away. Of course it will be radiated away as electromagnetic waves, not gravitational waves (there will also be significant emission of neutrinos), but that makes no difference for what we are discussing here.

Also, for the final neutron star to reach its equilibrium configuration, even more energy will need to be radiated away, because the final neutron star will be more tightly bound gravitationally than the two original ones.

tzimie said:
Obviously, the amount of mass radiated away is less than a 'near miss' scenario, when neutron stars start their 'death dance' around each other emitting gravitation waves.
No, it's not obvious at all, it's wrong. In the kind of "near miss" scenario you are talking about, the final (rotating) neutron star does not have to gain any heat at all, in principle: all of the original kinetic energy of the two neutron stars (less whatever is radiated away by gravitational waves) can go into the rotational energy of the final neutron star (which is part of its externally measured mass). In practice the merger won't be in exactly the right configuration for that, but the heat gain will still be much smaller than the above case, and the total energy radiated away will be much less as well.

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