# What is the mass of individual components in a gravitationally bound system?

When material of rest mass M falls from infinity onto a black hole accretion disk, it gets heated and then emits so much light that the energy radiated away can measure up to about 30% or so of M c^2. Let's say that ε is the fraction of rest mass energy radiated away.

My first question is, after this accreted material has crossed the event horizon, does the mass of the black hole (as an observer would measure by, say, examining keplerian orbits of nearby stars) grow by M or (1 - ε) M? I am quite certain that the answer is (1 - ε) M, but I am open to correction here.

My second question is, if the mass does indeed grow by (1 - ε)M, then if you wanted to compute, say, the gyroradius of electrons circling magnetic field lines in the accretion disk after it has cooled, would you use the regular electron mass or would you somehow include the mass defect?

The fine print: It would be helpful to agree on answers to the explicit questions before abstracting the answers. On the abstract side, I am comfortable with the idea that "the whole is not equal to the sum of the parts" when it comes to mass in bound systems. But that quote is often employed to compare the mass of the bound system with the masses of the unbound (free) components. I am trying to understand what the relation is between the masses of the components of the bound system to the total mass of the bound system, if an unambiguous relationship exists.

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## 1 Answer

1. The mass of the black hole only grows by $(1-\epsilon)M$, i.e. the mass that hasn't been radiated away yet. That's guaranteed by the mass conservation. However, one must be careful about dividing mass and energy to "individual places" in general relativity; in this case, it can be kind of done, but more detailed questions "where the mass/energy resides" could be meaningless. Only the total mass/energy is conserved in general relativity (in asymptotically flat and similar spaces).

2. The local physics of electrons moving in magnetic fields etc. is always the same. The electron mass is always the same constant. To describe what electron is doing in a situation like this, go to a freely falling frame, find out what the values of the electromagnetic fields are in this frame, and use exactly the same electron mass etc. as you would use in the absence of any black hole.

If you wanted to use a non-freely-falling frame (or coordinate system) to describe the behavior of an electron near the event horizon, you must be very careful to do it right. For example, the gravitational field near the event horizon makes the usual static coordinates extremely deformed relatively to the flat metric – a component of the metric tensor goes to zero or infinity near the event horizon – and there is a nonzero curvature etc. So I am sure that all people who think that general relativity is still essentially the same Newtonian mechanics – and in between the lines, you make it likely that you belong to this set – would almost certainly make the calculations incorrectly in a curved system. That's why I am urging you to go to a freely falling frame.

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That's useful, especially your answer to number 2. To make sure I've understood, consider this follow-up: Say an electron falls onto the accretion disk and the disk cools by emitting radiation with an energy of ϵ m_e c^2, adding (1 - ϵ) m_e c^2 to the measured mass of the hole/disk system. Then even though one should assign the electron its full rest mass energy of m_e c^2 in any calculation of its motion (using a freely-falling frame), this does not pose a violation of global energy conservation, because of the non-linearity of the Einstein field equations? –  kleingordon Jan 28 '12 at 6:44
Well, I wouldn't say that the reason why it's OK with the conservation law is the nonlinearity. The reason is the red shift, i.e. that the locally measured mass/energy in a gravitational field isn't the same thing as the contribution of this mass/energy to the total mass/energy as seen from infinity. Roughly speaking, the two quantities differ by the multiplicative constant $\sqrt{g_{00}}$, related to the gravitational potential, but studying motion of objects in general relativity is "harder" than just adding simple rescaling factors such as $\sqrt{g_{00}}$. –  Luboš Motl Jan 29 '12 at 6:43
In particular, all the components of the metric, $g_{\mu\nu}$, and not just $g_{00}$, matter for the reactions of electrons to themselves, external fields, etc. BTW when I said "redshift", I used the relationship between energy and frequency, $E=hf$: when an object "climbs out" of the gravitational field, its frequency is decreasing if it is a photon, and its kinetic (and total) energy is also decreasing as the gravity slows it down. This is really why the electron closer to the horizon is counted as "lighter" from the viewpoint of observers at infinity. –  Luboš Motl Jan 29 '12 at 6:47
Ah, that last point about redshift has clarified the matter for me very much. Thanks again –  kleingordon Jan 29 '12 at 22:07
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