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Time dilation when falling into a black hole

  1. Aug 2, 2012 #1
    Let's say I start out a few thousand kilometers from a black hole, and I begin to move toward the black hole due to it's gravitational pull.

    What type of time dilation would I experience as I fell into the black hole before the event horizon, and after the event horizon? By the time I die inside of the black hole, approximately how many years on earth would have gone by (no years, a year, a few hundred years...)?

    Thanks.
     
  2. jcsd
  3. Aug 2, 2012 #2
    Well, remember that you never experience time dilation. For you, time is always normal, and it's everyone else's clocks that are running at a different rate.

    An observer on earth will never see you cross the event horizon. As you get closer and closer to it, light from you gets more and more redshifted, and you appear to move slower and slower. Eventually, you'll just redshift out if the visible spectrum, and the astronomers on earth will have seen you 'frozen' at the event horizon. So, no amount of time can pass on earth until you hit the center, as that will never happen according to them.

    As I mentioned in the beginning, time is always normal for you. So, if you jump into a black hole, you won't notice anything peculiar. However, you'll either be torn apart by tidal forces, or you'll hit the singularity and die after a very short period of time.
     
  4. Aug 2, 2012 #3
    Okay, thanks. So I understand that time will always be normal for me, and that people on earth won't be able to see me. But, what about the time on earth relative to the person falling into the black hole's time? By the time the person died, (very) roughly how much time on earth would have passed?

    I have heard, since spacetime inside the black hole is infinitely warped, that an infinite or near infinite amount of time will pass on earth before I die. But, I'm not sure if this is correct.
     
  5. Aug 2, 2012 #4

    phinds

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    If people on earth could SEE you die, it would take approximately forever, but in actual fact you would die in quite a small amount of earth time and physicists would pronounce you dead. Amateurs with good telescopes would be shouting, NO ... he's still outside the event horizon and quite alive.
     
  6. Aug 2, 2012 #5
    Haha, thanks, good way to put it. Could you distinguish between your first two points a bit? I don't understand why physicists would pronounce you dead even though you are still technically alive inside of the event horizon due to the warped spacetime. Or are you? And if you are not technically alive, then why would you die quickly in earth time in this infinitely warped time?
     
  7. Aug 2, 2012 #6

    phinds

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    You would NOT be technically alive. You would be technically and actually dead. The fact that there are some gravitationally red-shifted photons that say otherwise is on the order of an optical illusion. You need to read up on the relativity of simultaneity and how people in different frames of reference see things differently.
     
  8. Aug 2, 2012 #7

    A.T.

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    Nope. You would die in quite a small amount of your own time. It would take forever of earth time.
     
  9. Aug 2, 2012 #8

    A.T.

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    Relativistic effects have nothing to do with "optical illusions" or just seeing things differently. They are what's left after you factor out the signal delay and optical effects.
     
  10. Aug 2, 2012 #9

    Nugatory

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    If the cause of death is being ripped apart by tidal forces (which doesn't necessarily have to happen at all) before you've fallen through the event horizon... That will happen in a finite and possibly a quite short amount of earth time.

    If we're going to declare the moment of death to be when you actually are observed to pass through the event horizon... yes, that will take forever in earth time.
     
  11. Aug 2, 2012 #10

    phinds

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    DOH ! Once again I have proven how stupid/careless I sometimes am.
     
  12. Aug 2, 2012 #11
    Suppose that there is a large mass of material that is compacting. The center is a black hole that is expanding. Suppose that you, a small inhomogeneity, are falling into the black hole.
    Also assume that this mass of material is spinning. So the black hole that forms is also spinning. You are falling toward the equator. However, centrifugal forces are slowing your approach to the black hole.
    Wouldn't you see an ellipsoidal black hole that is expanding near the equator?
    Couldn't the black hole expand to envelope you before you are torn apart or crushed?

    The original Kerr calculations where done for a static black hole with no spin. In this model, the only expansion is spherically symmetrical. There are no inhomogeneities. However, a space ship falling into a black hole is a small inhomogeneity. I am not sure the calculations for a spherically symmetric black hole apply to a black hole system with both inhomogeneity and anisotropy.
     
  13. Aug 2, 2012 #12
    So, wait, I'm confused now. Would it take forever/a long time in earth time for you to technically die?
     
  14. Aug 2, 2012 #13

    phinds

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    Sorry for mis-stating previously.

    In YOUR time, you would die rather quickly, and by the time you hit the singularity (not long after dying) the earth would have aged an ENORMOUS amount.
     
  15. Aug 2, 2012 #14
    When you run into a wall with 500 MPH, the amount of time people wait before cleaning your remains from the wall is quite irrelevant.
     
  16. Aug 2, 2012 #15
    No worries. So, would it be correct to say that if you fell towards and into a black hole, you would outlive earth itself, and maybe the human race?

    And, I assume this is because as you fall towards the event horizon, time slows and slows relative to earth time, then as you are almost touching the event horizon time practically stops?
     
  17. Aug 2, 2012 #16

    PAllen

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    It depends on who is correlating with earth - which simultaneity is used. Barring tidal shredding, a point particle sees signals continuing to arrive from earth between passing the horizon and reaching the singularity. They can add signal delay, and arrive at a specific earth time for when they reached the singularity.

    As for an earth observer, the only strictly physical statements you can make are what light and signals they will receive. It is philosophy whether you call this an illusion or not. Specifically, simultaneity is not a physical observable and there is nothing inherently real about SC simultaneity for the outside observer. They could choose a different simultaneity convention (a foliation into spacelike surfaces parameterized by proper time along their world line), and also conclude a fixed time when singularity was reached. From which they would be inclined to think of a trapped light interpretation.
     
  18. Aug 2, 2012 #17
    According to the observers on earth, you'll never pass the event horizon. This is because the time dilation factor goes to infinity as you approach the Schwarzschild radius. For a black hole, the SR is the event horizon.
     
  19. Aug 2, 2012 #18

    PAllen

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    Observer's on earth will never see you reach the event horizon. How this is interpreted, is a choice, not physics. See my prior post.

    Analogy: If you are accelerating in a rocket at 1 g forever, you are not required to believe that time on earth slowed to a stop corresponding to your Rindler horizon. You can choose a different simultaneity convention and conclude no such thing. However, it is undisputed that no signals from earth past a certain time will reach you as long as you continue accelerating. But you can stop accelerating and then see what you missed. Similarly, you can 'stop accelerating away from the SC horizon' - i.e. fall into it - and then see all the history of infaller's you missed.
     
  20. Aug 2, 2012 #19

    PeterDonis

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    I think you mean the original Schwarzschild calculations, or perhaps the original Oppenheimer-Snyder calculations of a spherically symmetric collapse of matter to form a black hole. The Kerr solution is for a spinning black hole.

    We don't have an *analytical* solution that includes inhomogeneity and anisotropy. But there have been plenty of numerical solutions done of such cases. Basically what happens is that any anisotropy/inhomogeneity in the BH's horizon, due to something like a spaceship falling in, quickly gets radiated away as gravitational waves, and the BH settles down to a new spherically symmetric state with a slightly larger mass. None of this affects what other people have been saying about the proper time elapsed for an infalling observer, or the relationship of that to the time elapsed on Earth.
     
  21. Aug 2, 2012 #20

    pervect

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    You need to define a little bit more about how you measure time dilation. What comes to my mind is observing the frequency of a laser beam directed toward you from far away, "at infinity"m, to speak loosely.

    The laser beam is a constant frequency as measured at its source, far away from the black hole, but as you fall into it, the frequency you directly measure will change, due to doppler shift. We can imagine that the laser beam also is amplitude modulated with some timestamp information, , so we know what time any light signal was sent.

    As you fall into the black hole, you'll see the beam redshift. You'll attribute this redshift to tidal forces, as there won't be any other sort of forces affecting your trajectory. You will see that as the laser beam redshifts, and that as it redshifts the encoded "timestamp" information slows down at exactly the same rate as the carrier frequency of the beam itself is reduced.

    This is what you will directly measure, in terms of doppler shift.

    I calculated once that from a fall from infinity, the redshift factor at the event horizon would be 2:1 at the event horizon. I haven't done any more detailed calculations for the exact scenario you describe.

    There isn't any good answer to " By the time I die inside of the black hole, approximately how many years on earth would have gone by ", because simultaneity is relative. What you can answer in principle is the timstamp of the last signal you'll ever recieve just before you reach the singularity. You can also answer, perhaps more easily, what the reading of a clock you carry with you will be just before you reach the singularity as well - i.e. the proper time it takes you to reach the singularity.

    Unfortunately, while the exterior geometry of the black hole is well understood, the interior geometry of a black hole is much less well understood. It hasn't been measured for reasons which I hope are obvious, and there is good reason to believe that the usual Schwarzschild geometry is not stable - and there are other puzzling issues that arise in particular with rotating black holes. And you'll be unlikely to find any other sort of black hole in reality, almost any black hole you'll find will have some angular momentum.

    So while we can answer the above questions (what's the proper time, and the timestamp of the last signal you ever see, before you reach the singularity) for a Schwarzschild blacak hole, we can't be too terribly confident what the actual answer would be with a real black hole.

    http://casa.colorado.edu/~ajsh/schw.shtml
    might be a useful online reference, the author of the webpage has written a number of peere reviewed papers on black holes.
     
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