# Effective mass of a black hole?

Suppose a black hole forms from a given mass of particles such as the core of a star going supernova. The black hole formed will have an effective mass due to the curvature of space time induced. Such a mass is presumably the inferred mass deduced from the effect of the black hole on the motion of nearby bodies. How does this effective mass compare with the original mass of particles all of which ended up "in the black hole"? Is the effective mass less greater or the same as the formation mass?

## 1 Answer

Answer: slightly less mass.

In general relativity, mass/energy of any system is conserved from the point of view of a distant observer (I forget the name of this theorem). We use this mass when we talk about how heavy a black hole is because we are (very) far away.

If you start with particles that are stationary, the total mass of the system is slightly less than the mass of the particles (because of gravitational binding energy). When they fall toward each-other to make a hole, additional (a few % typically) energy is released as gravitational waves (and light rays, neutrinos, etc). Due to conservation of energy, this means less mass for the final hole. In fact, asymmetric radiation of gravitational waves also can create a recoil effect that ejects the hole formed (although this case is for two black holes merging):

http://www.youtube.com/watch?v=dNXWvJoN_sw

Note that mass/energy never leaves the hole itself, the expelled stuff originates from spacetime regions before/outside of the hole.

• Thanks for the reply. Isn't the binding energy of an object the size of the Schwarzschild radius comparable the the rest energy of the unbound mass? A naive Newtonian calculation would suggest it is 3/5 of the unbound rest mass energy.... – user34870 Nov 29 '13 at 17:17
• In relativity binding energy of a black hole doesn't make sense since it's impossible to "unbind" it! However, for non-black holes that barely are large enough to meet the minimum stability requirement of 4/3 the Schwarzschild radius, your statement is correct, however the Newtonian answer will be inaccurate. – Kevin Kostlan Dec 4 '13 at 16:09
• Yes absolutely right that binding energy doesn't make sense for a black hole conceptually since it obviously can't be unbound (at least not without quantum effects). However the Newtonian calculation does indicate that the binding energy of a body somewhat larger than the Schwarzschild radius must be approximately as I quoted. Thus as a black hole forms this energy will likely be radiated somehow either by electromagnetic radiation or through gravitational waves and it will be a significant fraction of the rest energy of the original mass making up the forming black hole. – user34870 Dec 5 '13 at 17:46
• Additionally GR makes conservation of energy problematic because of gravitational waves which are part of space-time rather than the energy mass tensor. Thus this question I asked is complicated. I asked it because I read a recent paper explaining dark energy as the result of (solitonic) gravitational waves radiated by the initial massive black holes formed in the early universe. – user34870 Dec 5 '13 at 17:51
• Caveat: in spherical symmetry, the ADM mass is EXACTLY the mass of the final black hole (so long as you don't have any matter expelled from the hole), which follows directly from Birchoff's theorem. – Jerry Schirmer Feb 26 '14 at 21:50