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Distance/rate = t = sqrt[2d/a] = Same Proper Time ?

  1. Apr 15, 2012 #1

    morrobay

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    If one clock is moving with constant velocity. The other clock moving with
    constant acceleration , both with respect to earth .
    And both clocks travel the same distance in the same time with respect to earth.
    distance/rate = [2d/a]1/2 Gamma is given for velocity clock.
    Would you expect the proper time for the accelerating clock to be greater or less than
    gamma t ?


    note: I have outlined a numerical problem with v=.6c and a = 414.m/sec^2
    that I can post in homework section if there is not an obvious answer.
     
    Last edited: Apr 15, 2012
  2. jcsd
  3. Apr 15, 2012 #2
    I would expect the proper time of the accelerating clock to be less that gamma t. The terminal velocity of the accelerating clock has to greater to catch up with the constant velocity clock. If you view the wordlines of the two clocks on a Minkowski diagram, the accelerating clock has the longer path through spacetime and therefore the least elapsed proper time.
     
  4. Apr 15, 2012 #3

    PeterDonis

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    Yes, this is obvious. If both clocks travel the same distance in the same time, then one can set up the problem so that they both travel between the same starting and ending events. Of all the worldlines between two fixed events, the one with the longest lapse of proper time is always the unaccelerated one.
     
  5. Apr 15, 2012 #4
    Except where gravity is involved. A hovering stationary observer in a gravitational field experiences proper acceleration and yet records more elapsed proper time than an orbiting inertial observer at the same altitude, between consecutive passing events.

    Even when we exclude gravity we have to be careful to distinguish between coordinate accelerations and proper acceleration. For an observer at rest with a rotating disc, a clock with inertial motion appears to be accelerating and moving in a curved path and yet it records more proper time. It is only when we consider proper acceleration in SR that that the accelerating observer records less elapsed proper time.
     
    Last edited: Apr 15, 2012
  6. Apr 15, 2012 #5

    PeterDonis

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    Both good points. I was assuming that the OP was referring to a situation where all motion is horizontal, so there is no variation in the strength of gravity, and motion "with respect to the earth" can be idealized as motion with respect to a fixed inertial frame. In that situation, coordinate acceleration and proper acceleration are the same.

    In the more complicated cases you mention, it is still true that an unaccelerated (meaning zero *proper* acceleration) worldline has longer elapsed proper time than all other "nearby" worldlines; but there can be accelerated worldlines that are not "nearby" that have less elapsed proper time than a given unaccelerated one between the same two events.

    For example, the orbiting inertial observer's worldline is not "nearby" with respect to the "hovering" observer's worldline--the orbiting observer's worldline passes around the Earth, so it can't be deformed continuously into the "hovering" observer's worldline or vice versa (the Earth is in the way). However, there is another unaccelerated worldline which *is* "nearby" to that of the hovering observer: the worldline of an observer who is moving upward through the first passing event with just the right velocity to rise up more and more slowly, finally come to a stop, and then fall back again to just pass through the second passing event. *That* observer experiences *more* proper time than the "hovering" observer does between the same two events.
     
  7. Apr 15, 2012 #6

    morrobay

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    Thanks, I thought the curve would have less proper time, now am looking for a numerical answer and will post problem in
    homework section that results in this integral: Integral to > 8.65*105 sec sqrt [1-1.904 *10-12 t2] dt
    From int t0 >t1 [1-v(t)2/c2]1/2dt
    From acc= 414m/s2
    d= 1.55*1014m
    In case someone can plug in values and solve quickly
    sqrt [1-ax2] is integral for evaluation
     
    Last edited: Apr 16, 2012
  8. Apr 16, 2012 #7

    morrobay

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    Good news. I just worked the problem with v=.3c and acc. = 208m/s2
    With t = 10 days :
    velocity clock = 9.53 days
    acceleration clock = 9.35 days

    From the equation from The Integrator , for [1-ax2]1/2
     
    Last edited: Apr 17, 2012
  9. Apr 19, 2012 #8
    AT. reminded me of an old thread where we concluded exactly that. https://www.physicsforums.com/showthread.php?p=1840160#post1840160

    I have just done a rerun of the calculations and come to a stronger conclusion. If I have two clocks in my hands and toss one up in the air, then the tossed clock always registers *more* elapsed proper time than the clock that remains in my hand, when the tossed clock returns.

    This seems to be true for any size or density of massive body, any distance from the massive body and for any height the tossed clock is launched to, as long as I am stationary with respect to the gravitational field and as long as the launch velocity is less than the escape velocity, (otherwise it won't come back :smile:).
     
    Last edited: Apr 20, 2012
  10. Apr 19, 2012 #9

    PeterDonis

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    You mean the tossed clock always registers more proper time than the clock that remains in your hand, correct?
     
  11. Apr 19, 2012 #10

    PAllen

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    Yes, this result is very general as long as there is only one significant mass. Synge had a colorful name for this problem: "ballistic suicide", which he introduced with his typical wit as follow: "Normally the aim of ballistics is to hit someone else. Here we examine in GR the problem if hitting yourself.".
     
  12. Apr 20, 2012 #11
    Oops, yes, that is what I meant. I have corrected the typo. Thanks.
     
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