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Two watches, one thrown in the air.

  1. Jun 11, 2005 #1

    CarlB

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    Standing on the North pole of the Earth, I have two watches. I synchronize the watches. I throw watch "A" into the air, for a height gain of 10 meters (I have a good arm). Watch "B", I keep in my hand. I catch watch "A" when it descends to its original height, the same height as "B".

    Obviously, the watches are no longer synchronized. Which shows the later time? By how much?

    Of course any high school student knows that the difference in time will be too small to distinguish in a practical experiment. What I'm asking for is the theoretical difference.

    Carl
     
    Last edited: Jun 12, 2005
  2. jcsd
  3. Jun 11, 2005 #2
    10 meters isnt high at all.
     
  4. Jun 11, 2005 #3

    HallsofIvy

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    I don't see why there would be any discernable difference. The speeds are no where near relativistic speeds.
     
  5. Jun 11, 2005 #4
    The larger a gravitational field (or the closer you get to one), the slower time runs. A clock in the ISS and on Earth differ by about 2E-9 seconds (Earth one is "slower"). So the magnitude of the difference you're talking about is ridiculous.

    My $0.02
     
  6. Jun 11, 2005 #5
    Don't know the answer to your question, but I never knew this could happen in real life.
     
  7. Jun 12, 2005 #6
    Ofcourse, this can happen for velocities approaching the value of 'c', the 'time-dilation' effect can take place.

    Time-dilation can take place if the watch A is in relative motion to the watch B, which is the case but the falling watch A will fall down reaching a speed of [itex]\sqrt 2gh[/itex] which is nowhere near c. If the watch A had fell with a velocity of 0.8c ....then time-dilation is observable.Anyways time-dilation takes place as:

    [itex]

    t= \frac{t_o}{\sqrt (1- \frac{v^2}{c^2})}
    [/itex]

    where denominator almost equals 1 in the case when watch A falls froma height of 10m and thus no observable time-dilation takes place.
     
    Last edited: Jun 12, 2005
  8. Jun 12, 2005 #7
    Hehe, yeah man, relativity is about real life, physics explains real life.

    Careful there, it could be for sufficient values of h :)
     
  9. Jun 12, 2005 #8
    He was asking a theoretical question and wanted a theoretical answer. Many people need to stop being rude by saying "there wont be any difference". A theoretical question is used to grasp the concept, even if it is unable to be performed. clearly he is trying to learn something here.
     
  10. Jun 12, 2005 #9
    Word. :smile:

    Can one use regular old special relativity to do this? For instance, the elapsed time on the thrown watch would be given by

    [tex]d \tau^2 = dt^2 - \frac{dx^2}{c^2}[/tex]

    where x and t are measured in the "stationary" ground frame. x and t are related by

    [tex]x = \sqrt{2gh} \ t - \frac{1}{2}g t^2[/tex]

    [tex]d \tau = \sqrt{1 - \frac{1}{c^2}(gt - \sqrt{2gh})^2} \ dt[/tex]

    Then the total elapsed time on the thrown watch would be

    [tex]\Delta \tau = \int d \tau = \int_0^{2\sqrt{\frac{2h}{g}}} \sqrt{1 - \frac{1}{c^2}(gt - \sqrt{2gh})^2} \ dt[/tex]

    [tex]\Delta \tau = \frac{c}{g} \int_{-\frac{\sqrt{2gh}}{c}}^{\frac{\sqrt{2gh}}{c}} \sqrt{1 - x^2} \ dx = 6 \times 10^7 \int_0^\epsilon \sqrt{1 - x^2} \ dx[/tex]

    where epsilon = Sqrt(200) / (3 x 10^8)

    [tex]\Delta \tau \approx 6 \times 10^7 \int_0^\epsilon 1 - \frac{x^2}{2} \ dx = 2 \sqrt{2} - 10^7 \epsilon^3[/tex]

    So if that's the way to do it, it looks like the difference is of order 10^-18 seconds.

    edit: forgot to mention I was using g=10. So the 2 Sqrt(2) should really be 2 Sqrt(20/g). The order of the magnitude of the difference should not be affected much, I think.
     
    Last edited: Jun 12, 2005
  11. Jun 12, 2005 #10

    Gokul43201

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    It looks like the time dilation from the variation of the gravitational field (in GR) might be slightly bigger ~ 10^-15 or thereabouts.
     
    Last edited: Jun 12, 2005
  12. Jun 12, 2005 #11

    HallsofIvy

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    He referred to throwing a watch up 30 meters- a completely "non-relativistic" situation- and said nothing about relativity. I said there would be no discernable difference. There is nothing rude about telling the truth.
     
  13. Jun 12, 2005 #12
    Cool. Is it still the thrown watch that has less time elapsed, though? Seems like I remember reading that the elapsed proper time is maximized in free fall, so I was wondering if in GR it was the thrown watch that read a later time.

    He asked for the difference in elapsed time on the two watches. What else could he be talking about?
     
  14. Jun 12, 2005 #13

    DaveC426913

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    Pah, you're rationalizing. You should know what a thought experiment is.

    It is not a "non-relativistic situation", since the watches are experiencing gravity - as in: changes in-. And the question is obviously about relativistic effects.

    The watch thrown in the air experiences less gravity briefly. It will fall out of sync. Pointing out that the AMOUNT that it is out-of-sync is vanishingly small that it might as well be zero does not shed any light on the poster's question.
     
  15. Jun 13, 2005 #14

    CarlB

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    Okay, here's another thrown watch problem, one that eliminates the gravitational interaction and may get you thinking about your answer to the first question.

    I'm in a spaceship somewhere free of significant gravitational effects. As before, I've got two perfect watches, A and B. The ship has its rocket engine running providing a forward acceleration of [tex]9.8 m/s^2[/tex].

    I throw watch A ahead of the ship at a speed of [tex]\sqrt{9.8 \times 10 \times 2}\;m/s[/tex] in a forward direction. Since the rocket is accelerating, I catch up with the thrown watch and catch it. Now which one shows the later time? By how much?

    Carl
     
  16. Jun 13, 2005 #15
    If you are just testing us to see if we can solve the problems you thought up, won't you at least tell us the right answer after we've given it a try?
     
  17. Jun 13, 2005 #16

    OlderDan

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    When are the clocks synchronized? Before you throw one out or as it leaves the spacecraft with its newly acquired velocity?
     
  18. Jun 13, 2005 #17

    CarlB

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    It doesn't matter. Under the special theory of relativity, acceleration itself (i.e. the act of throwing the watch) doesn't effect the passage of proper time. Only velocity (i.e. the integral of acceleration) effects the passage of proper time. The proper time discrepancy is proportional to the length of time you spend on the other reference frame, not on how you got there.

    If you don't like that answer, then choose either one. I'd love to hear your explanation for the difference.

    This is all under the assumption of perfect time keepers and all that. Most watches don't work so well when you put an infinite acceleration on them, even for a brief instant.

    Carl
     
  19. Jun 13, 2005 #18

    CarlB

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    You do see that the example of the watch thrown from the spaceship eliminates the need to consider the gravitational field of the earth, don't you? (Well, at least to first order, since the gravitational field of the earth isn't quite linear even over a distance of 10 meters.) It gets back to the principle that an acceleration is equivalent to a gravitational field.

    By the way, by transforming to the accelerated spaceship problem, you eliminate the need to include the mass and radius of the earth, and the gravitational constant G. That makes the problem a lot easier, I think. All that you need is the acceleration 9.8m/sec^2, the distance, 10m, and the speed of light 3x10^8 m/sec.

    Carl
     
  20. Jun 14, 2005 #19

    OlderDan

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    First let me say that I am not speaking with authority on this subject. But I'm going to respond anyway to present what prompted me to raise the question. I think it does matter, but that does not mean I will be able to do the calculation in either case.

    I know almost nothing about general relativity, or how to properly treat accelerating reference frames from a special relativity viewpoint, but I do recall the notion that gravity and acceleration are more or less equivalent in general relativity, so if one affects the way a clock runs, so should the other. I was not thinking in terms of an infinite acceleration when I asked the question. If infinite acceleration avoids the issue, then so be it. Let's just say the clock is locally synchronized when it gets up to speed and leaves the spaceship, however that is accomplished.

    Beyond that, I'd also like to see a solution to the modified problem. I do understand the motivtion for eliminating the gravitional effects and focusing on the acceleration alone. I suppose one could do some sort of integral over quasi-inertial reference frames, but I think there is more involved than just time dilation. As in the case of resolving all those special relativity paradoxes like the twins, I expect those synchronization issues will come into play.
     
  21. Jun 15, 2005 #20

    CarlB

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    Okay, here's the answer.

    Dear Dan,

    > I do recall the notion that gravity and acceleration
    > are more or less equivalent in general relativity,
    > so if one affects the way a clock runs, so should the other.

    This is true, and it is why we can work the problem out for the accelerated case, and the answer we get will be identical to the situation on the earth.

    Rather than work with accelerated frames of reference, don't you think it would be easier if you used an inertial reference frame? There is one that is available, and that is the one attached to the thrown watch.

    In that reference frame, the equation for the passage of proper time (using only one spatial dimension for simplicity) is

    [tex]ds^2 = dt^2 - dx^2/c^2[/tex]

    The position of the thrown watch in the (accelerated) reference frame of the held watch is given by

    [tex]x_t = V_0\, t -4.9 t^2[/tex]

    where "4.9" is half the gravitational acceleration of the earth. Therefore, the position of the held (accelerated) watch, in the frame of reference of the thrown (inertial) watch is the negative of this:

    [tex]x_h = -V_0\,t +4.9 t^2[/tex]

    This gives the velocity of the held watch as:

    [tex]\frac{d x_h}{dt} = -V_0 + 9.8 t[/tex]

    The duration of the flight is given by [tex]2 V_0 / 9.8 = V_0/4.9 = \tau[/tex] seconds, and this is the proper time as experienced by the thrown watch.

    The proper time of the accelerated (held) watch is given by integrating [tex]ds[/tex] over this time period. That is,

    [tex]\Delta s = \int_0^\tau ds = \int_0^\tau \sqrt{1 - (\frac{dx_h}{dt})^2/c^2}dt < \tau[/tex]

    where the final inequality is due to the fact that the integrand is less than one.

    So there is no need to know anything about general relativity, other than, as you mentioned, an accelerated reference frame is equivalent to a gravitational field, from the point of view of a small observer. Just choose the right (inertial) reference frame and special relativity will crank out the answer.

    The fact that proper time depends on velocity and not on acceleration is encoded in the fact that

    [tex]ds = \sqrt{1 - (v/c)^2} dt [/tex]

    The above equation has no acceleration in it, only a velocity.

    Now the thing to note is that it is the accelerated (i.e. held) watch that is the one which is slowed down (i.e. the travelling twin has [tex]\Delta s < \tau[/tex]). When you compare the watches upon catching the thrown one, the one you catch will show the later time.

    This is contrary to the intuitive expectation that the thrown watch is the one that corresponds to the travelling twin, which is why I posed the problem.

    By the way, if you don't like using the inertial reference frame attached to the watch, you can do the problem in any other inertial reference frame of your choosing, and you will similarly avoid needing to write out a solution to Einstein's field equations. It's just that the equation for the proper time experienced by the thrown watch is particularly simple in the inertial frame of that watch.

    With this example, I'll bet that you are now eager to solve some more problems in general relativity using special relativity mathematics.

    Carl
     
    Last edited: Jun 15, 2005
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