Energy Conversion from Mass to Gravitational Wave

How is mass converted to gravitational wave energy by inspiralling binary black holes?

Is the gravitational wave energy coming purely from the kinetic energy/gravitational potential of the two black holes? Or does the merged black hole emit mass beyond that?

Let's start by noting an alpha particle (aka a Helium-4 nucleus, aka 2 protons and 2 neutrons) has a mass of less than the sum of 2 free protons and 2 free neutrons. The reason is that there is some binding energy that is released when the alpha particle is formed. Using $$E=mc^2$$, some of the mass of the original particles is released as energy when the bound alpha state forms. In equations, $$\begin{equation} 2 m_p + 2m_n = m_{\alpha} + E_B \end{equation}$$ where $$m_p$$ is the proton mass, $$m_n$$ is the neutron mass, $$m_{\alpha}$$ is the mass of an alpha particle, and $$E_B$$ is the binding energy.

The same logic holds General Relativity. There are some technical points about how to define energy in GR, but for asymptotically spacetimes one can define the ADM mass, which is a measure of the energy of the entire spacetime and is conserved, and the Bondi mass, which can be used to measure the energy lost due to the emission of gravitational waves. The net result is that

$$\begin{equation} m_1 + m_2 = M_f + E_{\rm GW} \end{equation}$$ where $$m_1, m_2$$ are the initial masses of the component black holes of the binary, $$M_f$$ is the final mass of the remnant black hole, and $$E_{\rm GW}$$ is the energy emitted in gravitational waves.

$$E_{\rm GW}$$ can be calculated in post-Newtonian theory (at least the contribution from the inspiral), and depends on the initial masses (if we are dealing with black holes, it is a homogenous function of $$m_1$$ and $$m_2$$). (More accurate estimates can be obtained with numerical relativity). For GW150914 (the first binary black hole system to be detected with gravitational waves), the energy emitted was about $$3$$ solar masses. Gravitational wave events are quite extraordinary; even though they are very difficult to detect, for a brief shining moment near the merger they outshine the entire electromagnetic spectrum with gravitational-wave power.

• Do the inspiralling black holes release energy that's not in the form of GWs?
– gmz
Sep 28 '21 at 1:30
• @gmz Not unless they have accretion disks (or are otherwise surrounded by some kind of matter) that is ejected during the inspiral and merger. Sep 28 '21 at 2:28

There is no assurance for conversion of matter to gravitational waves. In one incident, gravitational waves were observed and a gamma ray burst was also observed after a delay of 1.7 seconds. It is expected that magnetic field of rotating bodies produced gravitational waves before actual collision, and gamma ray burst happened after 1.7 seconds when actual physical collision happened. So, production of gravitational waves should have reduced rotating-kinetic energy, and production of light rays from gamma rays burst should have reduced linear-kinetic energy. Only kinetic energies should have become gravitational waves and gamma ray burst.

I'm not sure mass being "converted" is the right way to put it. When the black holes merge the distribution of gravity changes so that it's stronger in some places, increasing the gravitational potential energy for mass in those areas.

How is mass converted to gravitational wave energy by inspiralling binary black holes?

A gravitational wave isn't actually carrying energy, but also energy isn't conserved in General Relativity. Everything we see or hear, even the gravitational waves, comes from events outside the event horizon. So it is immaterial whether they would form black holes or if they are things that are close to getting super compact, we see the interactions from events outside the event horizon. The point is there is some spacetime and some sources.

The spacetime is required to line up with the sources: the spacetime on one side of a source and on the other side have to connect up in a way determined by the source on the interface. If the sources were heading one way then they can destroy the spacetime in front of them and create new spacetime behind them (and they don't even have to destroy and create equal amounts) and the stuff behind them is like the stuff in front, except different based on how much source is there.

The spacetime around it has to connect up as well. So just like when you sail through a river the flow on the sides are impacted by you going through the middle. The sides saw some stronger curvature from when you were going by and some weaker curvature from when you weren't. This variation is now a variation that the spacetime around it has to follow.

That demand continues outwards and outwards. And it is the actual wave. It's just spacetime waving, there is no energy.

With many field theories you can take two solutions and add them to get another solution so you can talk about the energy and of one and the energy of the other and the energy of the interaction. But for General Relativity there is no such additivitiy. The spacetime is just forced to line up with what it is in the past in a particular way, the vacuum way. And it has to keep doing that until it meets another source that allows it to align up in a different way.

Now if something is rotating, it can lose that rotation from waves emitted. The idea is that for an extended body with waves coming from each part, the wave it is making is from the other parts in now the spacetime against which it has to make a difference. When the motion is such as to make those waves bigger then you make more of the same curvature and this requires a loss.

If you want to think of working with the wave to make it stronger that corresponds to cases where you lose energy to the field. And working with the wave to make it weaker corresponds to cases where you gain energy from the field. But it is just a correspondence.

The reality is the wave would normally make more of itself nearby and the source nearby could instead interface to make the wave stronger or weaker beyond the interface.

Now compact objects head towards states closer and closer to one of the static solutions. A static solution is a curvature that creates more of itself. The deviations from that don't make more of itself. Mathematically you could have waves coming in or going out. But physically the waves are going away from the newly forming combined compact object.

So if there is a rotating because the parts are orbiting each other then a wave heading out just goes out, but one going in beads over to the other body, where it is now heading outward. Since the wave has an amount of time between peaks and troughs and peaks again (called a period) and it also takes some time to get from one part of the orbiting system to another you can imagine some periods of waves pas back and forth in a coordinated wave that make them stronger each time, and other periods don't. The others that are matched to get stronger each time and slip by sometimes will build up and slip away and once heading away they don't come back (the universe is large). So that is where the emission comes from. Now the object changes too because sources react to curvature as well as creating it, so the whole process can go faster as it gets started.

Again you could imagine some periods that would encourage it to go faster at a shorter period, and those are the ones that can do more.

If you want to see something going away and becoming the wave. Think about the difference between the current objects and the new combined static one it is turning into, that deviation is what is being destroyed. And the wave travelling out is what it is turning into.

The static combined object it turns into have an unchanging quadrupole moment. So that is, in a sense, the thing that is turning into the waves. For instance a spherically symmetric shell of collapsing matter produces no waves because it didn't have a nonzero quadrupole to begin with, so none of that kinetic energy turns into waves. It needs a twist since the waves themselves have a twist.

if a gravitational wave does not carry energy, how can it move a mirror on earth. Assuming that a photon hits the mirror at the right time, the photon will gain some energy.

Firstly, energy is not conserved in general relativity. Secondly, motion is not obvious. I could stand between you and your friend. And be 5m from you and 5m from your friend and move towards your friend (and away from you) while creating space between me and you. So I end up 7m from you and 4m from your friend. And still on a line between you two. And the 4m next to your friend might look like it did before, and the first 5m next to you might look the same as before. But I destroyed 1m between me and your friend and created 2m between you and me. That happens all the time. Stars do that when they collapse, they move father down from the high altitudes than they get closer to the center.

• Ok, I guess I am missing a bit of background in general relativity. However, if a gravitational wave does not carry energy, how can it transfer energy to photons on earth (eg by moving a mirror). Would it be correct that this moving mirror could increase the energy of a reflected photon by extracting energy from the gravitational wave? Feb 15 '16 at 0:02
• @ScienceAndPhilosophy Firstly, energy is not conserved in general relativity. Secondly, motion is not obvious. I could be 5m from you and 5m from your friend and move towards your friend while creating space between me and you. So I end up 7m from you and 4m from your friend. The 4m might look like it did before, the first 5m to you might look the same as before. But I destroyed 1m between me and your friend and created 2m between you and me. That happens all the time. Stars do that when they collapse, they move father down from the high altitudes than they get closer to the center. Feb 15 '16 at 0:09
• This is completely wrong. There is energy content in gravitational waves. This can be measured in the difference between the Bondi and ADM masses. Sep 22 '16 at 18:30
• And energy is globally conserved in asymptotically flat spacetimes, even if there is no covariantly defined local gravitational energy density. Sep 22 '16 at 18:31
• And finally, in the LIGO experiment, the mirror moves even after the wave has passed, because you are transferring energy to it. This should also be apparent in the perturbative case, where you do get back-reaction of the waves on emission and absorption, which shows that the waves have energy-momentum content, relative to the flat background. Sep 22 '16 at 18:33