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Time dilation for clock thrown up and caught back

  1. Jul 27, 2009 #1
    If you throw up a clock in a field of gravity and it falls back down, will it show the same time as a clock that stayed at the surface? In other words, will the gravitational time dilation of the stationary clock exactly equal the sum of the smaller gravitational time dilation of the other clock (higher up in the field) and the dilation caused by its speed?

    Just to make things perfectly clear: clock 1 stays at the surface, clock 2 is thrown up and falls back down, so its time passes more slowly because of its speed, but also more quickly because it is higher up in the field of gravity. The question is: do the two dilations balance each other out exactly?

    I know this may look like a homework question, but trust me, I'm a 35 year old airline pilot :)

    The reason I'm asking is that I want to understand gravitational time dilation in the expanding universe.

    If I understood correctly, you can use two different coordinate systems for the universe: the coordinate system that we would normally use based on special relativity (time for distant (and therefore moving) objects passes more slowly), or the god-like cosmological coordinate system in which the time coordinate at any location in the universe is chosen to be local time as experienced by a local observer who is moving together with the expansion of the universe.

    The first view just looks at the universe as a collection of things that happen to be flying apart, while the second view is closely tied to the particular structure of our expanding universe. Both are just a choice of coordinates, nothing more.

    The two systems give different, but consistent views of the universe. Just one example: In the first system, very distant objects will never reach our age because time grinds to a halt as they approach (but never exceed) the speed of light. In the second system, they are the same age as us but their light will never reach us because space between us is expanding faster than the speed of light (even the light that is trying to come towards us, is actually retreating away from us due to the expansion of the universe). Both views result in the same observation: we never see them reach our age.

    It all seems to make perfect sense, except if I consider a universe that shrinks back together (like people used to think before they found out the universe is actually accellerating, but never mind that). In the event of a Big Crunch, the first coordinate system results in a twin paradox, while the second obviously has no such problem.

    I know the Big Crunch will probably never happen, but I just want to figure out why the two models seem to be contradictory in such a theoretically possible case.

    If the two clocks from my first question (one thrown up and caught, the other stationary) show the same time, that would solve my paradox since any observer would consider himself to be at the center of the universe's great field of gravity. But I'm not very good at this kind of math (yet) and somebody else can probably show me a calculation in a minute or so, or even better, give an argument why the clocks (hopefully) must show the same time.
     
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  3. Jul 27, 2009 #2

    A.T.

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  4. Jul 27, 2009 #3
    no.

    nor will such a self powered clock that is launched without any gravity present run at the same rate as a "stationary clock"....for faster things time moves more slowly.... as does time in a high gravitational potential.

    So when when you skyjockies are airborne your clock runs a tad slower for both reasons and the GPS system you use for navigation must also be corrected for such relativistic time effects. http://en.wikipedia.org/wiki/Global_Positioning_System#Timekeeping
     
  5. Jul 27, 2009 #4

    Dale

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    The tossing clocks thread is a good one to go over for this question. The bottom line is that the tossed clock will record more time than the surface clock. In fact, the tossed clock will always record the maximum time possible between the "toss" and "catch" events. If you put a little rocket booster, or wings, or anything that makes it not purely free-falling, then the clock will record less time than a free-falling clock between the same two events.
     
  6. Jul 27, 2009 #5

    diazona

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    Interesting... never thought about it like that. I like that explanation.
     
  7. Jul 28, 2009 #6
    That is wrong. Time moves more slowly for objects with a lower gravitational potential (closer to a massive object). For example, time actually stops when someone approaches a black hole's event horizon (as seen from outside).

    So the two effects, speed and gravitational potential, have opposite effects on proper time. The tossed clock is higher up (further away from the earth) so it's running more quickly, but it's also moving which makes it run more slowly. I was kind of hoping the two would cancel each other out, but it appears they don't...

    Anyway, I'll have a good look at the tossing clocks thread and see if I can make any sense of it.
     
  8. Jul 28, 2009 #7

    A.T.

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    Yes, but at https://www.physicsforums.com/showthread.php?t=249722&p=1840160" it was sill not clear what time clock B (free fall, oscillating in a tunnel) measures. It would be interesting to compare that to clock D (free fall rest at the center) and clock E (free fall orbiting at surface level).
     
    Last edited by a moderator: Apr 24, 2017
  9. Jul 28, 2009 #8
    From the very Wikipedia article you linked to: (emphasis mine)

    A number of sources of error exist due to relativistic effects. For example, the relativistic time slowing due to the speed of the satellite of about 1 part in 10^10, the gravitational time dilation that makes a satellite run about 5 parts in 10^10 faster than an Earth based clock, and the Sagnac effect due to rotation relative to receivers on Earth.
     
  10. Jul 28, 2009 #9
    That's exactly what I was really looking for (while only just realising it now): the comparison between a clock at the center of the earth, and a clock that is thrown up from the center and falls back again (without passing the surface, if that makes a difference).

    If they don't match, I'm at a loss to explain why a Big Crunch of the universe would not induce a twin paradox.
     
    Last edited by a moderator: Apr 24, 2017
  11. Jul 28, 2009 #10

    Dale

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    Well, that is not at all what you asked above. In your original question the metric was the exterior Schwarzschild metric, the tossed clock was travelling on a geodesic (free fall), and the surface clock was not. Since geodesics are always the locally longest path we know immediately that the tossed clock records the most time.

    Your revised question now deals with the interior Schwarzschild metric, and is comparing two different geodesics (both clocks are free falling). So you cannot immediately say which is shorter. There is no way around it other than to integrate the metric along the two geodesics.

    The above questions have little to do with any possible Big Crunch. The reason is that the metric is different. Your first question was about the exterior Schwarzschild metric, your second about the interior Schwarzschild metric, and this one is about the FLRW metric.
     
  12. Jul 28, 2009 #11
    I think I'm going to repost under a different title then, forgetting about tossed clocks and just asking the question what would happen in a big crunch...
     
  13. Jul 28, 2009 #12

    George Jones

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    This isn't always true in general relativity.
     
  14. Jul 28, 2009 #13

    A.T.

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    I'm bad at calculations. :frown: In the https://www.physicsforums.com/showthread.php?t=249722&p=1840160" kev didin't do clock B (free fall, oscillating in a tunnel) and said you might need a certain mass distribution to make clock B meet clock E (free fall orbiting at surface level) again. But B and D (free fall rest at the center) meet in any case. So you could assume interior Schwarzschild metric (which already assumes a certain mass distribution).

    The interior Schwarzschild metric, or at least the spatial part of it, is that of a hypersphere. It would make sense intuitively, if two geodesics on it had the same lengths between intersections, like on a normal sphere. I have conflicting statements on this. Maybe some day I will learn how to do these numeric integrations
    .
     
    Last edited by a moderator: Apr 24, 2017
  15. Jul 28, 2009 #14

    Dale

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    I am pretty sure that it is if we limit ourselves to timelike geodesics. Null geodesics, of course, all have 0 length, and spacelike geodesics are always the locally shortest path.

    Note that the word "locally" is very important here!

    EDIT: did you mean that there are some exotic geometries where a timelike geodesic is a locally minimal path or a "saddle point"? If so, I would be surprised, but I could understand.
     
    Last edited: Jul 28, 2009
  16. Jan 21, 2010 #15
    Am I the only person who, in appreciating alternate (UK/European) definitions of "toss" finds this dialogue a bit amusing? Is a tossed clock actually tossed, or is it a clock used to time the tossing?

    On a serious note, I'm with DaleSpam in being a bit confused by what you (George Jones) meant. It seems... contradictory unless Dale is on the right track.
     
  17. Jan 21, 2010 #16

    George Jones

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    I've always meant to get back to this thread. Okay, I'll try to type up some explicit calculations, and to talk about about what's happenintg.
     
  18. Jan 21, 2010 #17
    Thanks, I don't doubt that you have an interesting point to make. You have at least two of us on tenterhooks. ;)
     
  19. Jan 22, 2010 #18
    Make that 3 :wink:
     
  20. Jan 22, 2010 #19

    bcrowell

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    In both this thread and the earlier one on tossed clocks, I don't see anyone making what seems to me to be the simplest argument. By the equivalence principle, we can consider everything in the tossed clock's frame. Imagining ourselves in this frame, and adopting the coordinates that are natural there, there is zero gravitational field, and the metric is not only flat but also has the usual Minkowski form when written in our coordinates. In this frame, we observe a desk and a clock go zooming by. The desk exerts a force on the clock, causing it to stop and then come back. The whole thing is simply the standard twin paradox (for constant acceleration). The clock being manipulated by the desk reads a shorter elapsed time.

    One way in which this description is simpler is that you don't have to worry about whether geodesics are minima or maxima (which actually depends on issues like +--- versus -+++ metrics), whether they're global or local extrema, etc. The metric is purely Minkowskian when written in our coordinates, geodesics are linear equations written in our coordinates, and we just recycle the usual result of the twin paradox.

    Viewed in this frame, the example also doesn't require reasoning about two competing effects, one from gravitational time dilation and one from time dilation due to motion.
     
  21. Jan 22, 2010 #20

    George Jones

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    In the quote below, clocks E and B are freely falling and thus are both in local inertial frames, and clock C (sitting on a desk) is accelerated. Doesn't your argument above lead to T_C < T_E and T_C < T_B? But this isn't true, as, actually, T_C > T_E and T_C > T_B. This is the whole point: this type of reasoning breaks down.
     
  22. Jan 22, 2010 #21

    bcrowell

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    No. My argument is based on the equivalence principle, which is local. A and C occupy the same local piece of spacetime. You can't use the equivalence principle in a nonlocal way to compare with B, D, and E. (I assume that E is going to make a complete orbit and then be compared with C again when it gets back -- otherwise it's just like A.)

    So I would agree that my argument is not the way to go if you also want to analyze B, D, and E, but it still seems to me like the most direct way to compare A and C.
     
  23. Jan 22, 2010 #22

    George Jones

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    The metric is not flat, because, even with respect to a local inertial coordinate system, the second derivatives of the components of the metric are non-zero, and, consequently, the curvature tensor in non-zero at all points.
    For a clock that actually is tossed physically, the situation involving A and C is just as non-local as the other situations. In particular, all the situations are non-local because definite integration (over a non-infinitesimal interval for which a single inertial coordinate system is not possible) is required to calculate times in all the situations.

    The difference is that the situations involving C and E and C and B involve conjugate points, while the situation involving C and A doesn't involve conjugate points. I hope to say more about this in later posts.
     
  24. Jan 22, 2010 #23

    atyy

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    So may I guess that your objection to time-like geodesics being "locally longest" is that "length" is intrinsically non-local. Locally, a geodesic has zero acceleration, which only translates into the longest distance between two points if there is one time-like geodesic connecting those two points.
     
  25. Jan 22, 2010 #24
    OK, now that makes sense in the context of George's answer.
     
  26. Jan 22, 2010 #25

    bcrowell

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    Hmm...I don't think so. You can start from a Minkowski metric and apply a coordinate transformation into the frame of an observer with constant proper acceleration, which gives a metric [itex]ds^2=(1+ax)^2dt^2-dx^2[/itex]. The curvature tensor is zero at all points, because curvature is intrinsic, so you can't get a nonvanishing curvature by doing a coordinate transformation on a Minkowski spacetime. Its time-time component has a second derivative that's nonzero, but that isn't sufficient to prove that it has a nonvanishing curvature. You get some Christoffel symbols that are nonzero, but the Riemann tensor vanishes.

    No, I don't think that's right. If you were going to be that strict about interpreting the locality of the e.p., then it would be meaningless, because it would never apply to anything. It's true that this is a subtle point. The reason the e.p. is not a totally well defined mathematical law is that it's very hard to provide a perfectly rigorous definition of how local is local. But I'm convinced that A and C are local enough to qualify, for the following reason. The e.p. does not prevent you from measuring accelerations. That is, the guy in the elevator can't tell whether his elevator is accelerating with acceleration [itex]a[/itex] or whether it's in a gravitational field [itex]a[/itex], but he *can* tell what the numerical value of [itex]a[/itex] is. A typical experiment he would do to determine that would be to release a rock above the floor and time how long it takes to reach the floor. That's a measurement of a certain world-line which is finite, but not so extended that it falls outside the kind of local measurement that the e.p. allows. If you compare it with the observation of the clocks A and C, it should be clear that those observations are no less local than the rock in the elevator; they're simply whole parabolas rather than half parabolas, but a factor of 2 doesn't make the e.p. suddenly not apply. The kind of measurement that the e.p. doesn't apply to is something like measuring tidal forces, e.g., determining whether the rock's acceleration is different near the floor than near the ceiling. Basically you can determine [itex]a[/itex], but you can't determine [itex]a[/itex]'s derivatives.

    The reason that the e.p. can't be applied to compare, say, C sitting at the surface and E orbiting the planet is that E's world-line extends so far through space that it clearly demonstrates a tidal effect: its acceleration vector differs measurably from point to point along its trajectory. By measuring it, you're measuring an acceleration's derivatives.
     
    Last edited: Jan 22, 2010
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