# From where does the energy for gravitational waves come from?

As far as I understand, in the events detected by LIGO, about 4% of the total mass of merging binary black holes was converted to gravitational waves.

Where does this energy come from, i.e. what exactly gets converted into gravitational waves?

Is it simply the kinetic energy of the merging objects (velocities of these objects before merger are huge, up to 60% of c if I recall correctly), so does it mean emitting gravitational waves makes them orbit slower, but retain their original masses? Or do the compact objects really lose "real" mass, meaning they become lighter and in case of BHs their radius changes accordingly?

As an example, let's assume two BHs, both with 50 solar masses, orbiting each other far enough (say 1 light year) so that GWs nor kinetic energy has no significance to these initial mass measurements. During the merge, they should radiate about 5 solar masses in GWs. Would the resulting black hole have mass of 95 or 100 solar masses?

• Radiating gravitational waves makes an inspralling binary orbit closer and faster of course. Jun 10, 2017 at 19:25
• I edited my question slightly to reflect my intention. I do understand that GWs are the cause and essentially only mechanism that allows two BHs to eventually merge. I want to understand how this affects the resulting mass of the object. Jun 10, 2017 at 19:38
• It comes from the same source that the BHs' increased kinetic energy comes from as they fall towards each other: gravitational potential energy. Jun 11, 2017 at 2:09
• should it be "from whence..." Jun 11, 2017 at 13:49
• "Would the resulting black hole have mass of 95 or 100 solar masses?" it's a good question! Jun 11, 2017 at 13:51

Radiating gravitational waves makes an inspralling binary orbit closer and faster. (Rob Jefferies)

The source of the energy for both increased kinetic energy, and the gravitational radiation is the same: gravitational potential energy. (PM 2Ring)

Two black holes at a distance of 1 light year have a huge amount of potential energy, about 10^48 Joules of potential energy. As they spiral, a significant amount of that energy is radiated as gravitational waves

This is real mass lost. The mass of resulting black hole is smaller than the sum of the two merging black holes, though at no point does any black hole itself become smaller.

• Thanks for the answer, I want to understand what I am missing here: 1) neither BH loses any mass 2) Energy from GW comes from potential / kinetic energy 3) resulting BH is still smaller than sum of merging BHs; even though they both retain their original mass and still must have some (probably quite close to c!) velocity by the time they merge, so there must be still lots of kinetic energy, which should contribute to the mass of resulting BH (since net movement/momentum from orbiting objects is zero?). Jun 11, 2017 at 20:37
• How do you arrive at $10^{48}$ Joules? Jun 15, 2017 at 22:17
• I find the orbital energy conversion easy to grasp, but "This is real mass lost" loses me. Is this "matter"? If so, what is the process for "mass lost"? For now, it seems like a throw away line in this answer without any qualification, yet. Aug 14, 2018 at 6:33

As Rob correctly pointed out, the emission of gravitational waves reduces the orbital energy and result in an inspiral. This reduction in total energy also reduces the mass of the final BH, since $E=mc^2$. The bulk of the gravitational wave energy is emitted (and energy=mass lost) in the final chirp, when the separation approaches the Schwarzschild radius.

To quantify this, let's just make a simple energy budget calculation, starting from two equal-mass BHs of mass $M_\bullet$ orbiting each other at distance $d$ on a circular orbit. Then the orbital energy is $$E_{\mathrm{orbit}} = -\frac{GM^2_\bullet}{2d} = -M_\bullet c^2 \frac{R_s}{4d}$$ where $R_s=2GM/c^2$ the Schwarzschild radius of each BH and we have assumed that $d\gg R_s$ such that the orbit is Keplerian. The total initial energy is then given by the rest mass energies plus the orbital energy as $$E_{\mathrm{total}} = M_\bullet c^2 \left[2-\frac{R_s}{4d}\right].$$ After coalescence, a remnant of mass $M_{r}$ emerges. The energy deficit is the difference between the initial and final energies $$\delta E = M_\bullet c^2 \left[2-\frac{R_s}{4d}\right] - \frac{M_rc^2}{\sqrt{1-v^2/c^2}},$$ where $v$ is the speed of the remnant w.r.t. to the centre of mass of the progenitors. This energy has been lost by gravitational wave radiation. If this corresponds to a certain amount $\mu$ of rest mass, then from $\delta E = \mu c^2$ we find $$M_r = \sqrt{1-v^2/c^2}\left[2M_\bullet -\mu - M_\bullet\frac{R_s}{4d}\right].$$ Now for $v=0$ and $R_s\ll d$, the mass deficit $\delta m\equiv 2M_\bullet-M_r$ is identical to $\mu$: the radiated energy corresponds to the mass deficit; the final hole has 95$M_\odot$ if $M_\bullet=50M_\odot$ and $\mu=5M_\odot$. In particular, the gravitational wave energy cannot be taken merely from the orbital energy as suggested by another answer.

The mass deficit is even larger than the radiated energy if the remnant has undergone a considerable velocity kick, such that $v\neq0$ (caused by asymmetric gravitational wave radiation).

• "In particular, the gravitational wave energy cannot be taken merely from the orbital energy as suggested by another answer." - where else could it come from then? Atomic mass from the singularity? Aug 13, 2018 at 11:16
• @Todd As I said: from the rest mass energy ($mc^2$) of the holes. Aug 13, 2018 at 20:14
• Is that basically "matter"? It's hard to find a definition of "rest mass" online. Also, if it is "matter", is there a known process for how this occurs? or, is it more of an "effect" than a physical conversion of "matter" to gravitational wave energy? Aug 14, 2018 at 6:30