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B Gravitational time dilation depth charge

  1. Jul 23, 2016 #1
    Hi. A student wishes to test gravitational time dilation near a black hole within her lifetime, so she travels to a location where a black hole is said to reside and parks her ship at distance where she is not affected by the hole's time dilation effects.

    She then fires a tethered capsule towards the BH with an explosive inside timed to detonate in exactly three hundred sixty-five days according to the clock in her ship. The capsule will not cross the event horizon, but will remain at a distance such that gravitational time dilation has an effect on it according to the clock back in the ship.

    After one year, the student reels the tether in. Will the unexploded capsule still be attached to it? Assume whatever convenient variables, such as the speed of the fired capsule and the time it took to get to the back hole, as well as the time to get it back in the ship to be ideal for such experiment.
     
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  3. Jul 23, 2016 #2

    Simon Bridge

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    Time dilation is relative: it depends on some other observer. Where is the other observer?
    If you mean that the student should be in flat space-time, then she cannot be anywhere near the black hole.
    Either way, the opening setup is internally inconsistent.

    That only makes sense if this timing is in the same reference frame as the clock in her ship.

    First you say that the capsule is fired towards the black hole, now you say it "remains at a distance"??
    The capsule is maybe lowered into the black hole's gravity?

    Is the capsule fired by a signal from it's own clock, initially synchronized with the ship, or is it fired by a signal from the ship's clock?
    But it does not seem to matter for the question whether the ship is in curved space-time or not. What you care about is the difference in the clocks.
     
    Last edited: Jul 23, 2016
  4. Jul 23, 2016 #3
    Hi, Simon. Apologies for not being clear. Yes, the ship is at a distance from the BH in flat-spacetime and the capsule is lowered into it. Then, after a year, it's reeled back inside the ship. The timer is of course synchronized with the ship's clock. What I care about is whether the explosive goes off or not.
     
  5. Jul 23, 2016 #4

    Dale

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    It will not have detonated
     
  6. Jul 23, 2016 #5
    Yes! That is what I thought, but for the timer inside the capsule, time is running normally all the way to the horizon, so at what time according to my clock does the bomb goes off?
     
  7. Jul 23, 2016 #6

    Dale

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    Time is normal inside the capsule, there is just less of it.

    As far as when it goes off, that depends on the motion of the capsule. Just specify the worldline and then integrate the spacetime interval over it.
     
  8. Jul 23, 2016 #7
    One thing I don't understand is when does it happen. The capsule is ejected/fired/whatever and it's moving with uniform acceleration until it's stopped by the length of the tether which stops it from going further. During this time, both the clock on the ship and the timer in the capsule are in agreement, correct? When do they become desynchronized?
     
  9. Jul 23, 2016 #8

    Nugatory

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    The desynchronization starts the moment that the capsule is lower in the gravity well than the ship. You could imagine the clock in the capsule sends a signal (at the speed of light, just to keep thing simple) up the cable every time it counts off another second. Before we start lowering the capsule, we on the ship will receive one signal every second according to our clock. As soon as we start lowering it, the signals will become more widely spaced so we no longer receive one signal for every second that our clock counts - the capsule clock is running slow compared to our clock so the clocks must fall out of sync.
     
  10. Jul 23, 2016 #9
    I see. So if I wait those 365 days, then pull the capsule out of the BH, it will still be there attached to the cable, meaning the capsule experienced less time according to my clock. So based on my time limit of 365 days, there in no way for the capsule not to return, meaning when I reel the tether, I will always get the capsule back because of the time difference?
     
  11. Jul 23, 2016 #10

    PeterDonis

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    This is impossible as you state it. If a black hole is present, spacetime is not flat, period. What you could say is that the ship is far enough from the BH that the effects of spacetime curvature are negligible--so, for example, the rocket thrust required for it to hold station at a fixed distance from the hole is too small to observe.

    Does the capsule cross the horizon or not? I think you intend not, but I'm not sure because of some things you say in later posts. If it doesn't, then saying the capsule is lowered "into" the hole is misleading; that would imply that it crosses the horizon. If the capsule does cross the horizon, then you can't reel it back in or receive any signals from it after it does, so asking "when" it detonates relative to you is meaningless.

    This is impossible. The only way to keep the two clocks synchronized would be for them both to be (a) at rest relative to each other, and (b) in the region of spacetime where the effects of spacetime curvature were negligible. This is certainly not the case for the capsule.
     
  12. Jul 23, 2016 #11

    Dale

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    Let me give you another way to think of it.

    Consider a piece of paper with two points, A and B, connected by two paths. One path is a straight line and the other path is a semicircle. The straight line is shorter than the semicircle. Distance is normal along the semicircle, meaning that the Pythagorean theorem determines distances just as with any curve. There is just more ordinary distance along the semicircle than there is along the straight line.

    Now ask "where is that extra distance located"? Does it make sense to say that the extra distance is at the beginning or the end or middle? No. The path is longer, but it isn't like the additional length is located in a particular part of the semicircle.
     
  13. Jul 23, 2016 #12
    Hi, Peter. You're quoting my reply to Simon. In the OP, I said ''parks her ship at distance where she is not affected by the hole's time dilation effects.'' Isn't that region good enough to be called flat spacetime?

    Read my OP. You are quoting what I said to Simon in reply to his question ''The capsule is maybe lowered into the black hole's gravity?'' to which I replied ''the capsule is lowered into it.'' Meaning the capsule is lowered into the black holes gravity. I never intended for the capsule to cross the EH.

    I don't even think it's possible to keep clocks synchronized in a two-story building, but my point was that the timer inside the capsule departed with its clock synchronized with the one on the ship.
     
  14. Jul 23, 2016 #13
    My main concern was to lower the tethered bomb-capsule into a strong gravitational field and test whether there was any chance of it going off before I pulled the cable back. Because of the time dilation effect, it seem I will always get it back unexploded.:smile:
     
  15. Jul 23, 2016 #14

    Simon Bridge

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    Of course there is a chance for the capsule to detonate before it is pulled back.
    Just wait longer... until the clock with the bomb has ticked off 31536000 of it's seconds.
    Exactly how long that is by the ship-clock depends on what exactly happened when the capsule was lowered. You can work that out by figuring out what path that is (say, it is winched down at a constant rate, or maybe free-falls till it reaches a specific distance - whatever) and then integrate over the space-time interval for that trajectory.

    Note:
    the ship does not have to be in flat spacetime for the point of your question to work, and you do not need a black hole (though it kinda helps since you can easily get very large time dilation effects, avoiding the issue of the bomb blowing up as it is being winched in).
     
    Last edited: Jul 23, 2016
  16. Jul 24, 2016 #15
    Hey, that's true. What if I send the capsule flying off at close to c for a year then winch it back at the same speed. Would I still get the same result?
     
  17. Jul 25, 2016 #16

    Simon Bridge

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    You do the same thing: integrate over the space-time integral for the trajectory.
    Do you mean, here, to change from a gravitational time dilation to more of a twin's paradox situation?
     
  18. Jul 25, 2016 #17
    Yes. I want to find out in which scenario I get the bomb to explode before it's retrieved. Obviously I could wait until it goes off like you said, but I want to constrain the experiment to the time allotted (365 days) otherwise what is the point. BTW, I don't know how to integrate integrals or what that even means in any rigorous way, but thanks for even suggesting that I could.
     
  19. Jul 25, 2016 #18

    PeterDonis

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    Then the only option you have is to set up a scenario where your elapsed time is less than that of the capsule, not more. For example, you could put yourself deep in the gravity well of something like a black hole, and send the capsule out into deep space, far away from the hole, in such a way that if nothing else happened, it would return to you in 365 days by your clock. Then the capsule would explode before it returned.
     
  20. Jul 25, 2016 #19
    Yes, indeed. I'm having a tough time visualizing the time frame of the capsule going out of the gravity well, though. If I go by what Nugatory said, then the pulses traveling along the cable as the capsule makes it way out of the well should return to me faster?
     
  21. Jul 25, 2016 #20

    PeterDonis

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    I'm not sure why. Just take the original scenario where the capsule was deep in the gravity well and you were way outside it, and reverse that. Now you are the capsule and the capsule is you, so to speak. Obviously your "time frames" will be switched.

    Yes. Which means that the 365th pulse from the capsule (if we suppose there is one pulse per day--or the 365*86400th pulse if there is one pulse per second), emitted just before it explodes, will reach you before your own clock has shown 365 days elapsed.
     
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