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If a particle with rest mass falls from r = infinity to r = Rs of a black hole it is supposed to reach a velocity of c. But where does all that energy (infinite) come from to bring the rest mass to a velocity c? Even the Big Bang doesn’t have infinite energy.

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Well my guess is that the particle won't come from the $\infty$... – PML Feb 22 '13 at 13:55

No observer will see the object cross the event horizon at the speed of light.

In GR you need to be careful to specify what you mean by statements like:

it is supposed to reach a velocity of c

For example, you and I sitting on the Earth watching the object falling will see the object slow as it approaches the event horizon, and in fact we would never see it reach the event horizon let alone cross it. For us the object takes an infinite time to reach the event horizon.

If you were falling alongside the object you're not moving at the speed of light because of course in your rest frame you're not moving at all.

The nearest you get is in what a shell observer sees. A shell observer is someone hovering at a fixed distance from the event horizon (presumably using a powerful rocket) and watching the object fall past. If the shell observer is at a radius $r$ then they will measure the velocity of the passing object to be:

$$ \frac{dr_{shell}}{dt_{shell}} = -c\sqrt{\frac{r_s}{r}} $$

where $r_s$ is the radius of the event horizon. So if the shell observer is at the event horizon, i.e. $r = r_s$, they would see the object pass at $c$. But you can't have a shell observer at $r = r_s$ because that would require an infinitely powerful rocket. The more powerful a rocket you have the closer to $r_s$ you can et and the closer to $c$ you see the object pass you, but you can never see the object reach $c$. In fact as soon as the object has passed you it starts to slow and it never reaches the event horizon just as observed by you and I on the Earth.

So in no observer's frame does the object acquire infinite energy.

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A particle falling into a black hole doesn't get accelerated via "propulsion" (i.e. no energy is required to accelerate the particle, and the particle never "feels" any acceleration forces). It simply follows a geodesic along the (curved) spacetime in the vicinity of the black hole.

Near a black hole, or at the cosmological horizon of the universe, the curvature is so great that spacetime itself "travels" faster than light (relative to us), which causes particles in the vicinity to travel along with it. This is very much permitted in General Relativity.

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A rest mass particle with velocity has added kinetic energy. The energy has to come from somewhere. It sounds like you’re saying the geodesic supplies the energy. The geodesic can’t supply infinite energy – nothing can. The conservation of energy is a more fundamental theory than general relativity. – Greg M Feb 22 '13 at 16:11
As John Rennie points out, an outside observer will never actually see the infalling particle move faster than light, so no conservation laws are violated. – Dmitry Brant Feb 22 '13 at 16:23
@GregM A particle falling in the gravitational field of the earth turns potential energy into kinetic energy, no? The geodesic is a different formulation of the same concept, in the limit of flat space it is the same. – anna v Feb 22 '13 at 16:31
If a rest mass particle travels at the speed of light in any reference frame (observable or not) it must have infinite energy. The only way out of this conflict is to say that it doesn’t have a velocity of c which is the same as saying that the escape velocity is less than c at Rs. – Greg M Feb 22 '13 at 16:34
"observable or not" << There's the flaw in your reasoning. – Dmitry Brant Feb 22 '13 at 16:52

I apologize for the round-about approach, but I will borrow an equation from I promise you, I've seen this expression before for accretion matter, and there are other ways to arrive at it. Anyway, the following is an expression for potential energy around a black hole. It has assumptions, but it's sufficient for us here.

$$ V(r) = \frac{mc^{2}}{2} \left[ - \frac{r_{s}}{r} + \frac{a^{2}}{r^{2}} - \frac{r_{s} a^{2}}{r^{3}} \right] $$

In this formulation, $a$ is proportional to angular momentum. We're interested in a particle headed for the singularity dead-on so $a=0$ and we find only one term left in $V(r)$. We're interested in $r=r_s$.

$$V(r) = - \frac{m c^2 }{2}$$

This is a very important result, and if I understand correctly, it's the kinetic energy a particle will have getting close to the event horizon from the reference of $r=\infty$. This is how black holes can make really really powerful x-ray sources. In the turbulent chaos of in-falling matter, collisions can be happening with energy up to this amount. Now, x-rays from the interactions will be gravitationally red-shifted, so how the balance falls out in the end may be non-trivial. Obviously the above energy is just a maximum and things have to interact above the horizon for the products of their collisions to make it back out to $\infty$.

I'm also sure there's some extremely deep reason as to why the energy near the event horizon is half the rest mass energy, but I'm officially a layman on the subject so I'll hold off on speculation.

aside: Dmitry's answer is the correct perspective, so consider mine to be building off his. I wanted to add real numbers. The most important things to know is the geodesic concept, the light-cone orientation at the horizon, and that things travel faster than $c$ in GR depending on your geometry.

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