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If a mass is accelerated by the application of a force, the mass/kinetic energy of the object approaches infinity as it approaches the speed of light.

Now let's consider the same mass falling toward a black hole from infinity. It should also approach the speed of light as it approaches the event horizon. What happens to the mass/ kinetic energy in this case as it approaches the speed of light?

Ashmit Dutta
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Humayoun Khan
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    "It should also approach the speed of light as it approaches the event horizon." Why should it? – hft Oct 30 '22 at 02:06
  • The “relativistic mass” (a sum of the rest mass and kinetic energy) is the conserved total energy of a free falling object. – safesphere Oct 30 '22 at 06:13
  • @safesphere Where does OP mention the Schwarzschild metric? – hft Oct 30 '22 at 07:38
  • @safesphere Then why are you bringing it up? – hft Oct 30 '22 at 15:05
  • @safesphere Oh, I see. OK, yeah, that does not answer my question. – hft Oct 30 '22 at 17:58
  • @safesphere You realize that my question was directed at OP, right? – hft Oct 31 '22 at 00:47
  • @safesphere You realize, too, that I am not looking for you to answer my question. Don't you? – hft Oct 31 '22 at 00:49
  • @safesphere You realize, too, that you are the one down here in the comments trying to answer questions that nobody asked you about. – hft Oct 31 '22 at 00:50
  • There is no general requirement that the speed of an object approaches the speed of light as the object approaches an event horizon. You can just pass on by the horizon without even knowing it. – hft Oct 31 '22 at 01:02
  • @hft I just added a chart of the actual speed of an object falling to the horizon from any distance (or equivalently with any initial speed). Notice that the speed of any free falling object always approaches the speed of light at the horizon: https://physics.stackexchange.com/questions/450032#524855 – safesphere Nov 01 '22 at 05:18

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Consider the situation from the point of view of an observer who is falling into the black hole along with the mass. The observer can be a little ahead, a little behind, or off to one side. The observer does not see the object accelerate to any spectacular velocity. If the observer is ahead of the object, then they see the object start from rest and then accelerate outward a little, relative to them. As the observer and the object pass the event horizon, nothing special happens to them. The observer's clock keeps ticking for the remaining time before they hit the singularity, and during that remaining time they can continue to observe the object. I think you can see from this description that it wouldn't make sense to expect the energy of the object to hit infinity just because it's at the moment of passing the horizon. If it did, then what would happen after that?

General relativity doesn't have global frames of reference, so it doesn't make sense to talk about the infalling mass's velocity or energy in the frame of reference of a distant observer. Popularizations often talk this way, but it's wrong.

There actually is no measure of mass-energy that is always conserved in all relativistic spacetimes. There is such a mass-energy in the case of a Schwarzschild black hole, but only because that spacetime is asymptotically flat, i.e., it describes an isolated object. That mass-energy is a property of the whole system and stays constant at all times, as measured by an observer who is far away compared to the infalling mass. That observer can measure the black hole's mass-energy by the gravitational forces it exerts on distant objects. It doesn't go to infinity. It stays constant. This is exactly what you would observe if an asteroid crashed into the sun. As measured by an observer a light-year away, the mass-energy of our solar system would not change at all during that process.

dkfhgsdlkh
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  • “General relativity doesn't have global frames of reference, so it doesn't make sense to talk about the infalling mass's velocity or energy *in the frame of reference of a distant observer” - directly contradicts - “That mass-energy is a property of the whole system and stays constant at all times,* *as measured by an observer who is far away* compared to the infalling mass” - This answer is wrong on too many points to list. It is a collection of quotes from the popular propaganda, an opinion with no mathematically justified answer. – safesphere Oct 30 '22 at 05:19