da_willem said:
Look at the example of e.g. a neutron falling into a black hole. You say it is rest mass it is gradually loosing while moving closer to the black hole. This cannot be correct, for the mass of a neutron is a constant of nature independent of it's speed or position in a gravitational field (just look it up, it's the same everywhere). So something else must be converted into radiation...
That's not what I'm saying at all. I'm saying that you can't add the mass of the neutron to the mass of the black hole to get the total mass.
More to the point, neither can you add the energy of the neutron to the energy of the black hole to get the total energy of the system (except for the special case where the neutron is at infinity). The reason you can't do this is because of the gravitational binding energy has to be included.
Couldn't it be the potential energy which is converted into kinetic energy when the particle approaches the black hole which is converted into radiation? So when the particle bumps into other masses it slows down and release energy in the form of radiation.
Well as I mentioned before in GR you have to be a bit careful about assuming that a potential energy even exists, but in this case I think it does. (I'm open to corrections on this point - while I think it probably exisits, I can't write down an expression for it). For more on where I'm coming from see
http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html
If we assume that the potential energy exists, you can say that the potential energy, which starts out as zero, becomes negative as the neutron approaches the black hole, and that the sum of the potential energy and the kinetic energy remains constant, at least until the neutron bumps into stuff on the way down, at which point some of the kinetic energy is converted to radiation
Using the nonrelativistic potential U=-GMm/r and by using the Swarschild radius of a black hole r=2GM/c^2 we see that the particle can loose a maximum of .5mc^2 of potential energy by moving from infinity to the event horizon of the black hole.
I'm fairly sure that the Newtonian potential is not going to give the correct answer for a relativistic problem like this.
So if you count this potential energy as mass the statement quoted above can be correct. The total relativistic mass of a particle at infinity would be mc^2+.5mc^2=1.5mc^yielding a 'mass' of 1,5m. So the particle can loose 2/3 of it's mass in the process of falling in the black hole. The actual number is about 10% because the particle still has some kinetic energy (wich is added to the mass of the black hole) when it falls into the black hole.
Even if you use the Newtonian potenital, -GmM/r is zero at r=infinity, which is what I previously assumed.
It's worthwhile to try and clear up a few points of terminology. I used the overloaded term "mass of the system", which caused some unfortunate confusion. I really meant the energy of the system. The energy of the system is what an observer at infinity can measure, and find to be constant.
Using this more accurate terminology gives me an idea - it might actually be helpful to consider the issue of dropping a photon of energy E into a black hole instead of dropping a neutron.
IF you have an observer close to the black hole, the energy E of the photon just before it crosses the event horizon is going to appear to be larger, it will appear blueshifted, due to gravitational blue-shift. But that's not the observer we are interested in. We are interested in the observer at infinity. If we let that same photon climb out to infinity again, and measured it's energy, the photon would be red-shifted again, and we would get the same v alue E that we started out with. So, for the observer at infinity, E is the energy of the photon, it doesn't matter that it got blueshifted going in.
Something roughly similar happens when we drop a neutron in. If you look at the neutron from the viewpoint of someone close to the black hole, you can argue that it's total energy must be E>= m_0*c^2, where m_0 is the invariant rest mass of the neutron.
But if we convert the neutron to a photon of equivalent total energy (i.e. a photon of energy m_0*c^2 + kinetic energy terms), we see that the observer at infinity will assign a different, lower, energy to it. And he's the observer that we are interested in when computing the total system energy.