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Proper Time vs Time

  1. Nov 5, 2013 #1
    I'm having a hard time figuring out in problems which is proper time and which is just time, and I think it follows from my misunderstanding of time dilation. I've read my book and looked over my notes but I can't seem to figure it out.

    Here's an example: A person in frame S' moves with a speed of .95c along the positive x axis relative to a person in frame S. If the time interval between two events is 20 s according to a clock in S' what would the person in frame S measure for this time interval?

    So what I'm thinking is (because I already know the answer) tp=20s [itex]\gamma[/itex]=3.2 and therefore frame S measures 64s.

    This is easy but what if the question was backwards and frame S measured the time between events to be 64s and it wanted to know what would the person in frame S' measure. tp would be 64 and [itex]\gamma[/itex] would be 3.2 right? This means frame S measures 204s which HAS to be wrong because we already know it was 20s.

    What am I doing wrong? Please and Thank you!
     
  2. jcsd
  3. Nov 5, 2013 #2

    Nugatory

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    Leave "proper time" alone for a moment - it's equal to the undilated time for some observers under some circumstances, but it's not what you're asking about here.

    Actually, both results are right. Both observers correctly view themselves as at rest, with the other one moving and having a slowed clock. This seems paradoxical to you because you've overlooked the relativity of simultaneity. The two events "S' clock reads 20" and "S clock reads 64" are simultaneous in frame S, so S correctly concludes that the S' clock is slow by a factor of 3.2. However, these two events are not simultaneous in the S' frame; in that frame the events "S' clock reads 20" and "S clock reads 6.25" are simultaneous so the S' observer will correctly conclude that the S clock is the one that running slow by a factor of 3.2.

    (The above is assuming that both clocks were set to zero as the two observers were passing each at the same point in space. Change this assumption and the math gets more complicated without adding anything new).
     
  4. Nov 5, 2013 #3

    stevendaryl

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    The relationship between frame S and frame S' is not as simple as: "Clocks in frame S' run slow." If you have two events, [itex]e_1[/itex] and [itex]e_2[/itex], and let

    [itex]\delta x [/itex] = distance between the events in frame S.
    [itex]\delta x'[/itex] = distance between the events in frame S'.
    [itex]\delta t[/itex] = time between the events in frame S.
    [itex]\delta t'[/itex] = time between the events in frame S'.

    Then the relationship between these quantities (for one-dimensional motion in the x-direction) is:

    [itex]\delta x' = \gamma (\delta x - v \delta t)[/itex]
    [itex]\delta t' = \gamma (\delta t - \frac{v}{c^2} \delta x)[/itex]

    with the inverse equations:

    [itex]\delta x = \gamma (\delta x' + v \delta t')[/itex]
    [itex]\delta t = \gamma (\delta t' + \frac{v}{c^2} \delta x')[/itex]

    So in the particular case where
    [itex]\delta x = 0[/itex], (so the events are at the same location in frame S), we have:

    [itex]\delta x' = - \gamma v \delta t[/itex]
    [itex]\delta t' = \gamma \delta t[/itex]

    So the time between the events is longer in frame S' than in frame S. On the other hand, if [itex]\delta x' = 0[/itex], (so the events are at the same location in frame S'), we have:

    [itex]\delta x = \gamma v \delta t'[/itex]
    [itex]\delta t = \gamma \delta t'[/itex]

    So the time between the events is longer in frame S than in frame S'. You can't compute [itex]\delta t'[/itex] in terms of [itex]\delta t[/itex] or vice-verse without knowing something about [itex]\delta x[/itex] or [itex]\delta x'[/itex].
     
  5. Nov 5, 2013 #4

    Doc Al

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    The key is "a clock", meaning that the time interval is measured on a single moving clock in S'.

    OK. Note that the two events (that you are measuring the interval between) occur at different locations in S. Thus S must use multiple synchronized clocks to measure that interval.

    The time dilation "rule of thumb" is that moving clocks run slow. The time dilation formula doesn't apply to time intervals in general, but to intervals measured on a single clock. (You, of course, can apply the full Lorentz transformation for any intervals.)

    According to S', the clocks in S are not synchronized and they do not each show an elapsed time of 64s. (In fact, according to S', those clocks only show an elapsed time of about 6.25 seconds during the interval in question.)

    Edit: Nugatory beat me to it!
     
  6. Nov 5, 2013 #5
    How do you know S clock reads 6.25?
     
  7. Nov 5, 2013 #6

    Nugatory

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    I used the same 3.2 gamma factor that you calculated :smile:
    And I could do that because I already knew that the situation would be symmetrical: In the S' frame the S clock is dilated by the gamma factor calculated from the relative speed; in the S frame the S' clock is dilated by the gamma factor calculated from the same relative speed.

    However, you can double-check these results by using the Lorentz transforms from which the time dilations are derived. I've done this calculation often enough to know that it will come out right, but the first few hundred times that I had to solve one of these problems I did it the hard way, using the Lorentz transforms to verify that the symmetry really was there and relativity of simultaneity would lead to consistent results.
     
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