Question

5. What fraction of rest mass energy is converted from potential energy to kinetic energy when a particle comes from infinity to the event horizon of a black hole?

Answer 1

Let d be the distance (in some coordinate system) from a particle to the event horizon of a black hole. For a particle starting at distance d>0 above the event horizon to rise away to a very large (infinite distance) would require an energy E. As the initial distance d shrinks to zero, E rises without limit (to ?). At d=0, the particle is at the event horizon. Alright, reverse this sequence.....the particle falls back toward the event horizon from a large distance. It would seem that the limit of kinetic energy E acquired by the particle would be infinite as d shrunk to zero. Since this is clearly not possible, the particle should not be able to cross the event horizon (in either direction). The issue remains even if the acquired kinetic energy is radiated away or converted to mass, since the potential well is still infinitely large.

When an object of the mass m (small in comparison) falls into a black hole from infinity, the object gains a certain speed and therefore kinetic energy. Therefore it would seem that the mass of the black hole after consuming this object would increase more than just by m. In fact, in a classical (not applicable) way of thinking, the energy released in a fall to a singularity would be infinite, but surely it is not infinite in General Relativity. Can someone clarify what is the total gravitational energy released by a mass m falling from infinity to the singularity of a black hole?

A different interpretation of this may be the Frozen Star where the object never actually crosses the event horizon in the frame of a remote observer. In this case the speed actually is reduced to zero at the event horizon, but what happens with the kinetic energy? What is the extra mass added by the object in this case, as easily measured by a remote observer based on the change in his speed and the size of his distant orbit around the black hole

Consider the non-relativistic problem of a particle falling into a potential well and releasing all its energy in there. The quantity which is conserved during the infall is the total energy E=T+V+EinternalE=T+V+Einternal, where TT is kinetic energy, VV is the potential energy, and EinternalEinternal is some internal "chemical" energy of the particle.

Now we assume that the particle starts at rest at infinity where the potential is zero so that E=0+0+Einternal=EinternalE=0+0+Einternal=Einternal. As the particle starts falling to the potential well, nothing happens to internal energy, VV becomes negative and since energy is conserved, that must be countered with a positive TT. When we arrive all the way in the potential well, no matter what happens with the kinetic and potential energy, we always have E=EinternalE=Einternal and when the energy release in the well comes about, it is exactly E=EinternalE=Einternal which is released.

You can think similarly about the black hole along with the realization that Einternal=m0c2Einternal=m0c2, where m0m0 is the object's rests mass. In other words, if a particle of (rest) mass m0m0 falls into a stationary black hole starting at rest at infinity, the black hole will receive exactly m0c2m0c2 in terms of energy, no matter what happens to the kinematic or potential parts of the energy.