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Does a clock in free-fall slow due to gravitational time dilation?

  1. Dec 27, 2011 #1
    I imagine I have two clocks that are synchronised and are sitting a long way from a large massive body. I hold on to clock A and let clock B fall towards the mass. Let's assume that the falling clock B gets close to the massive body but somehow misses it, swings around and travels back to me.

    Would I expect clocks A and B to show the same time?

    One could argue that Clock B has been very close to the massive body so it should have experienced some gravitational time dilation. But on the other hand one could say that it has been in free fall at all times and therefore it has been in an inertial frame and not experienced any gravitational field. Thus it should show the same time as Clock A.

    Which argument is right?
  2. jcsd
  3. Dec 27, 2011 #2


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    No, not in general.


    No. There are no global inertial frames in GR. They exist only approximately locally. Free falling objects are affected by gravitational time dilation, just like other objects.
  4. Dec 28, 2011 #3


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    There is a free-fall trajectory (or frame, if you must, but the "frame" is only a frame sufficiently close to the trajectory) that maximizes "clock time", also known as proper time. However, that trajectory that maximizes "clock time" is not the one that falls towards the massive body, it's the one that goes _away_ from the massive body.

    Note that for short time intervals (say 1 second), there is not any free fall trajectory towards the massive body that winds up back at the start. The only trajectories that exist with this property for short time intervals are the ones that head away from the massive body.
  5. Dec 28, 2011 #4


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    To expand on this, there may also be a free-fall trajectory that drops down towards the massive body, swings around it, and then rises back up to the original radius--an elongated elliptical orbit with the original radius as its semi-major axis. If things were arranged right, object B could follow such a trajectory and object A could "hover" at the original radius until B returned to it. (I've assumed here that the original radius is large enough that the elliptical orbit does not precess measurably.) In this case, I would expect object B to experience *less* proper time than object A, even though B is in free fall and A is not. But that is because B's trajectory is not "close enough" to A's to fit within the "local" patch of spacetime in which the rule "free fall trajectories maximize proper time" applies. The only free-fall trajectories that *do* fit within that local patch are the ones, as pervect says, that rise away from the massive body and then fall back, such as A throwing up a ball and then catching it when it comes back down. The ball will experience more proper time than A does.
  6. Dec 30, 2011 #5


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    Simply arguing, I would say that clock B suffers a combined effect of SR (twin paradox) and GR (gravitational time dilatation), both of which slow clock B down compared to clock A.

    Too simpel?
  7. Dec 30, 2011 #6


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    Yes, this is basically why I said I would expect clock B to show less proper time than clock A when they meet again. But I'm not sure I would describe the "SR" effect as just like the twin paradox, because B is in free fall the whole time so the usual arguments about the "twin paradox" do not apply (B does not have to feel any acceleration in order to get back to meet up with A again; in fact, it's A who feels the acceleration in this scenario, not B, which may have been what prompted the OP's question).
  8. Dec 30, 2011 #7
    This hypothetical is like an observer being lowered and spending time near the horizon of a black hole and then returning to some distant observation point; also like a GPS satellite experiencing more accrued time than satellite earth stations.
  9. Dec 30, 2011 #8


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    Somewhat, yes, but in that case both observers feel acceleration (at least, that's what I take to be implied by your description of "an observer being lowered"). In the OP's case B is in free fall, and yet experiences less proper time, which is the opposite of the "naive" expectation from SR.

    In this case the GPS satellite, which experiences more proper time, is in free fall, and the earth station, which experiences less, is accelerated; this is the "naive" expectation so it doesn't seem like a paradox. I agree that's somewhat oversimplified, since the GPS satellite is also moving relative to the earth station, which creates an opposing "SR" effect; also, the GPS satellite and the earth station, technically, don't come back together to compare clock readings (but there are other ways of checking that time does "run faster" for the satellite).
  10. Dec 31, 2011 #9


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    Yes, understood.
    From B's free fall view, A accelerates away, deaccelerates then and meets B again. Would it be correct to show this in a Minkowski diagram with B's worldline remaining on the ct axis, while A is acceletating away and coming back. With the know acceleration of A's "trip" it should be possible to calculate the SR effect. Would you agree with that?

    I am aware, that one has to be careful in introducing gravity into a Minkowski space-time diagram. But in this special case one could possibly argue that gravity is equivalent to acceleration. And presumably acceleration can be shown in this diagramm.
    Last edited: Dec 31, 2011
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