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Suppose I'm on a non rotating planet. I have two identical, perfect watches. I synchronize them. Then I throw one of them into the air and catch it. Does the one I throw into the air gain or lose time with respect to the one I was holding?

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I believe this is precisely answered by the previous question (and answers therein): physics.stackexchange.com/questions/8314/… The wording is a little different but otherwise it seems like a duplicate to me. –  Marek Apr 9 '11 at 20:47
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@Carl: you have a point there but still, if the answers are essentially the same it means the questions are too, in my opinion. That's because answers usually tend to be a little longer than a single digit. Also, if some question has an answer as short as one sentence (as yours does) I don't think it's a very good question. By the way, from the close button box: "This question covers exactly the same ground as earlier questions on this topic; its answers may be merged with another identical question." I think this is satisfied. –  Marek Apr 9 '11 at 22:00
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@Carl: I agree with Marek. This question is not necessary. Especially if you thought of it after reading the other question. –  MBN Apr 9 '11 at 22:22
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I agree with @Marek, too. Sorry, @Carl. Moreover, I don't think that an Internet user who doesn't think about these matters physically will find it via Google, anyway. There are many ways (wordings) to formulate the same problem. A female extraterrestrial alien is in love but she's older than her sweetheart, and is offered to be repeatedly shot by a cannon by the Jovian Hegaxon Department of Defense. Will she accept? I write it to Google and it doesn't give me your question (later, it will give, when my comment is added to the index haha). –  Luboš Motl Apr 10 '11 at 5:15
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You also write: "In addition, acceleration and gravitational fields are only theoretically identical. One can imagine gravitational theories where they are different. Consequently the physical situation is not identical." - Nope. You are completely wrong. First of all, gravitation appeared in both question: the two questions are identical. Moreover, the effect of gravitation and acceleration is identical in practice, not just theory. The fact that you don't believe general relativity has nothing to do with it: GR has been proved by science while your-like alternatives have been refuted. –  Luboš Motl Apr 10 '11 at 5:21

2 Answers 2

up vote 3 down vote accepted

Lawrence has the details, but this question can actually be determined exactly without assuming anything about mass distribution, etc. A geodesic (which is what the watch in free-fall follows) in GR has maximal proper time along it. The "stationary" watch, which is actually accelerated, is following some other path and so must experience a shorter proper time.

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And so the held or "stationary" watch loses time, and the thrown watch gains time. –  Carl Brannen Apr 12 '11 at 21:35
    
Hmmm. But the watch is not in free-fall while it is being thrown and caught. So it seems to me this answer is not sufficient, we would have to give a reason why that is negligible. (We are dealing with very small time differences in the first place.) –  Retarded Potential Mar 28 '13 at 17:39

What astounds me is there is considerable quibbling over the nature of the question, but nobody answers it! This is comparatively simple to address. Let us consider the Schwarzschild metric in a weak gravity field $$ ds^2~=~-(1~-~2\phi/c^2)dt^2~+~dr^2~+~r^2d\Omega^2 $$ for $\phi~=~GM/r$ the Newtonian gravity potential. The unit velocity is then $$ 1~=~-(1~-~2\phi/c^2)u_t^2~+~u_r^2~-~\dots $$ where we can consider the motion in the radial direction for simplicity. The derivative of this with respect to the proper time $s$ is then $$ 0~=~-(1~-~2\phi/c^2)u_ta_t~+~u_ra_r. $$ If the gravity potential is zero the solutions are $t~=~g^{-1}sinh(gs)$ $r~=~g^{-1}cosh(gs)$, for $g$ the acceleration parameter. Here $g$ counters the gravitation of the Earth. If the gravity potential is turned on we can the write the time solution solution as $t~=~g^{-1}sinh(gs~+~\gamma)$, which we input into the third equation $$ 0~=~-(1~-~2\phi/c^2)g~cosh(gs)sinh(gs)~+~g~cosh(gs~+~\gamma) sinh(gs~+~\gamma) $$ $$ =~(1~-~2\phi/c^2)\frac{g}{2}sinh(2gs~+~2\gamma) ~+~sinh(2gs) $$ If we consider weak fields, small accelerations and small proper time $s$ we have $$ 0~\simeq~-(1~-~2\phi/c^2)(g^2s~+~g\gamma)~+~g^2s, $$ where $\gamma~\simeq~-2\phi gs/c^2$.

The coordinate time is reduced with the turning on of the acceleration. This implies that the watch on the accelerated frame will mark off a shorter interval of time than the watch which is placed on a geodesic motion in the local gravity field with acceleration $g$. This is a gravitational version of the twin paradox. The twin which travels outwards and back is on an accelerated frame, which is a path in spacetime that is non-extremal, or maximal. As a result the proper time marked off is shorter.

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Schwarzchild metric seems like overkill -- if we are talking about a normal person throwing a normal watch, we could just use an accelerated frame of reference in flat space. Also the comment I made to genneth's answer about throw/catch time applies also -- this is qualitatively different from the twin paradox in that respsect. –  Retarded Potential Mar 28 '13 at 17:18

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