0

I'm standing at a safe distance from a black hole and I'm in an Earthlike time frame of reference. Through my telescope I see a man starts to fall towards the black hole. I have looked at this question but it doesn't quite get to what I'm looking for.

I watch him accelerate towards the black hole until I know that he is getting close to $c$. But, the gravitational time dilation surrounding the black hole would be so great that I would actually see him slow down rather than speed up? And then just a short distance above the event horizon, gravitational time dilation would be so great that I would see him nearly stop, so that even the 14 billion year ago of the universe would not be long enough for him to actually reach the event horizon. Help me to understand this? How can he be falling at near $c$ at the same time that he is nearly stopped?

Of course, in his own frame, he just sees himself falling towards the black hole and never reaching it in the short time involved. He would be facing both kinetic and gravitational time dilation.

Help me to envision this from my frame.

Qmechanic
  • 201,751
foolishmuse
  • 4,551
  • What do you mean by "never reaching it in the short time involved"? In his frame, he quickly crosses the horizon. – PM 2Ring Aug 18 '20 at 17:58
  • Related, possible duplicate: https://physics.stackexchange.com/a/170506/123208 – PM 2Ring Aug 18 '20 at 18:01
  • @PM2Ring Remember that my question is only about my frame. But yes, in his frame he quickly crosses the horizon. Here's another conundrum for another day; if the universe ends before he reaches the horizon, would he actually cross it in his own frame? – foolishmuse Aug 18 '20 at 18:02
  • 1
    @PM2Ring I think the first graph you pointed at provides the answer. I would see him accelerate to .38c, and then rapidly slow to near 0 very close to the horizon. – foolishmuse Aug 18 '20 at 18:10
  • @foolishmuse Maybe also see here. – Charlie Aug 18 '20 at 18:39
  • You wouldn't actually "see" him stop. A photon whose wavelength is much greater than the Schwarzschild radius can't escape from the black hole. The photons from the guy falling in will become more and more red-shifted until there's a last photon that escapes, not long after he jumps in. – Peter Shor Aug 18 '20 at 22:19

1 Answers1

0

Your eyes don't directly perceive the outside world. They only perceive light that enters them through the pupil and strikes the retina. To reach your eyes, the light has to travel from wherever it was emitted.

By the definition of the event horizon, light that's emitted inside the event horizon can't reach any point outside the event horizon, so as long as your eyes are outside the event horizon, you'll never see it. Light that's emitted outside the event horizon, but very close to it, can eventually reach your eyes, but it takes a long time to escape the black hole's gravity. As the point of emission approaches the horizon, the time the light takes to escape approaches infinity. Therefore, it takes forever for all of the light that was released before crossing the horizon to reach your eyes. (Classically, anyway. Quantum mechanically, there are only finitely many photons, and you'll see the last one at some finite time.)

The fact that you see the infalling person "frozen" at the horizon is an optical illusion, like a mirage. Your brain assumes that light travels instantaneously in straight lines from its origin point to your eye, and when that approximation isn't accurate, it gets confused. In reality, falling through the event horizon happens very quickly.

benrg
  • 26,103
  • I think you are mistaken on this. In my frame time actually slows for the falling man (but not in his frame) So, in my frame his heart beat even slows down to near infinitely long. Its not just an optical illusion as you suggest. He actually ages slower in my frame. If he was to come back to earth after 1000 years, he would not have aged a day. – foolishmuse Aug 18 '20 at 20:32
  • @foolishmuse He can't come back because he crossed the event horizon. Someone else who accelerates very rapidly near the horizon could come back. That situation is like the twin paradox in special relativity, and doesn't need a black hole to work. There's no objective sense in which the man who crosses the horizon ages more slowly than you. – benrg Aug 18 '20 at 21:07
  • 2
    this is where we are disagreeing. I'm thinking that because time slows near the event horizon, he will never (in our frame) reach it. If he tried to reach it for 1000 years (in our frame but only 1 day in his frame) he could come back to earth and be just 1 day older. Time dilation is not an illusion. It is real. If you live on the sun, you actually age 66.4 seconds slower every year, compared to me at sea level. So if we both can read a book at 66.4 seconds per page, then after exactly 1 earth revolution around the sun, a man on the sun would have read 1 page less than a man on earth. – foolishmuse Aug 18 '20 at 21:20
  • 1
    @foolishmuse that may be true for a test particle, but it is not true for an extended object. A very naive heuristic argument to see this: adding the mass of the man falling into the black hole will actually increase the event horizon radius, even if only by a tiny amount. So the man needs to only reach a point that close to the black hole for the test particle approximation to break down, not all the way to the event horizon. That takes finite (but very long) time. Black holes also merge in finite time. – Prof. Legolasov Aug 19 '20 at 00:43
  • @Prof.Legolasov How long would that "finite (but very long) time" be? – foolishmuse Aug 19 '20 at 15:45