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Time dilation questions

  1. Nov 14, 2008 #1
    Time dilation questions - Please help!

    In a standard configuration, which frame O' is moving along the x-axis of frame O with speed v, clock is synchronized to 0 when origin O' coincides origin O. Also, let us use normalized scale. The unit of time t is second; the unit of x-axis x is light-second; and thus the velocity v = x/t is the ratio of speed with the light speed, 0<=v<1, γ=1 / sqrt(1-v2)

    My question is, is it true that the famous time dilation only applies to an observer in frame O watching clocks at rest in frame O'? More specifically, when the clock rested at x'=0 in frame O' reaches the first tick at t'=1, the observer's clock in frame O has reached γ ticks. Also, the clock rested at x'=1 in frame O' ticks at the same rate as the clock rested at x'=0 in frame O'. But for an observer in frame O, the clock rested at x'=1 in frame O' will reach its first tick later than the clock rested at x'=0 in frame O'. This delay is vγ.

    Another question is, a uniformly moving clock in frame O' will cause different time dilation rates for an observer in frame O. Is it true? For example, a moving clock in frame O' with positive speed has an even higher time dilation rate. For example, if in frame O', a clock is traveling at the speed of light from origin O' at t'=0 (v'clock=1, normalized scale), when this clock is reaching x'=1 at t'=1, the observer's clock in frame O has ticked (1+v)γ ticks, which is a higher rate than γ. On the other hand, a moving clock in frame O' with negative speed has a lower time dilation rate and can even cause time contraction. For example, if in frame O', a clock is traveling at the speed of light but to the -x' direction, from origin O' at t'=0, when this clock is reaching x'=-1 at t'=1, the observer's clock in frame O has only ticked (1-v)γ ticks, appearing as the clock in frame O' is ticking faster.

    Could you please review my statement and tell me if it is correct or not? Thanks a lot!
     
    Last edited: Nov 14, 2008
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  3. Nov 14, 2008 #2

    JesseM

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    Time dilation is symmetrical, if that's what you're asking. An observer in frame O' watching clocks at rest in frame O will see them slowed down by the same time dilation factor. See this thread for an illustration.
    The clocks at x'=0 and x'=1 will reach there first ticks at different times in frame O because of the relativity of simultaneity--clocks which are synchronized in frame O' will be out-of-sync in frame O. They are both ticking at the same rate in frame O, but in frame O the two clocks showed different starting times at t=0, and their times continue to be out-of-sync by the same fixed amount at later times.

    As a general rule, if two clocks are synchronized and a distance of d apart from one another in their own rest frame, then in a frame where they are moving at speed v along the axis connecting them, they will be out-of-sync by the constant amount vd/c^2, with the trailing clock being ahead of the leading clock by this amount. So with d=1 for the clocks at x'=0 and x'=1, in the O frame they will be out-of-sync by v/c^2, meaning that when the clock at x'=0 reads a time of 1 second, the clock at x'=1 reads (1 second - v/c^2) in the O frame. Since the clock at x'=1 is slowed by a factor of gamma, it takes an additional gamma*v/c^2 seconds for this clock to reach a time of 1 second in the O frame.
    You can use the relativistic velocity addition formula to figure out the speed of this clock in frame O, and from there you can figure out the time dilation factor in frame O. But I don't understand what you mean by "normalized scale"--what units are you talking about when you say the speed is 1? Are you using a scale where c=1 (in which case it would be impossible for a clock to move at this speed) or some other set of units? Anyway, if the clock is traveling at velocity v1 in the +x' direction in frame O', and frame O' is moving at velocity v in the +x direction in frame O, then the clock is moving at velocity v2 = (v + v1)/(1 + v*v1/c^2) in the +x direction of frame O. So, its time dilation factor in frame O would be [tex]1/\sqrt{1 - v_2^2 /c^2}[/tex]. I don't think this is the same as [tex](v1 + v)/\sqrt{1 - v^2/c^2}[/tex] as you seem to suggest above.
    No clock is observed to experience "time contraction" (i.e. a faster rate of ticking than a clock at rest) in any inertial frame.
    A clock cannot travel at the speed of light in SR. And in the limit as a clock's speed approaches c, its time dilation factor approaches infinity (i.e. the clock approaches being completely frozen). Also, if a clock is traveling at v1 in the -x' direction in frame O' (i.e. negative velocity in the +x' direction of frame O', the phrase 'negative speed' is meaningless since speed is the magnitude of the velocity vector), then in frame O the velocity addition formula tells you that it must be moving at velocity v2 = (-v1 + v)/(1 - v1*v/c^2) in the +x direction of frame O (if v2 is negative, then this means the object is moving in the -x direction). The time dilation factor will still be [tex]1/\sqrt{1 - v_2^2/c^2}[/tex].
     
  4. Nov 14, 2008 #3

    Doc Al

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    No. Observers in any frame will observe time dilation for any clock moving with respect to that frame.
    OK.
    OK.
    OK. This is due to both time dilation and the relativity of simultaneity (clock desynchronization).

    Sure. Time dilation depends on the speed of the moving clock in the observing frame.
    This is a bit confusing (for example, clocks don't move at the speed of light), but since frame O will see that moving clock move faster than the clocks at rest in O', O will see that moving clock as being more strongly time dilated.
    No one will see anything "time contracted", whatever that might mean. Since, with respect to O, the moving clock is now moving slower than clocks at rest in O', O will see it less strongly time dilated. But both O and O' will see this moving clock as time dilated, as expected.
     
  5. Nov 14, 2008 #4
    Thank you Doc Al!

    I still have some concern regarding my second question: "Another question is, a uniformly moving clock in frame O' will cause different time dilation rates for an observer in frame O. Is it true?"

    I will prepare another two examples avoiding clocking traveling at the speed of light. I will reply with the examples as soon as possible. Hope you could comment on it again.

    Thank you!
     
  6. Nov 14, 2008 #5
    Again our measurement is done in a standard configuration, which frame O' is moving along the x-axis of frame O with speed v, clock is synchronized to 0 when origin O' coincides origin O. Also, let us use normalized scale. The unit of time t is second; the unit of x-axis x is light-second; and thus the velocity v=x/t is the ratio of speed with the light speed, 0<=v<1, γ=1/sqrt(1-v2)

    First example: In frame O', a clock is traveling at 0.75 times speed of light from origin O' at t'=0 (v'clock=0.75, normalized scale). When this clock reaches x'=0.75 at t'=1 in frame O', the observer in frame O sees the clock reaches x=(v+0.75)*γ at t=(1+0.75*v)*γ.

    Let v=0.66, thus γ=1.34. Therefore after one second in frame O', t=(1+0.75*0.66)*1.34=2.00 seconds has passed in frame O. So the time dilation factor is increased.

    Second example: In frame O', a clock is traveling at 0.75 times speed of light from origin O' at t'=0, towards -x' direction instead. When this clock reaches x'=-0.75 at t'=1 in frame O', the observer in frame O sees the clock reaches x=(v-0.75)*γ at t=(1-0.75*v)*γ.

    Let v=0.66, thus γ=1.34. Therefore after one second in frame O', t=(1-0.75*0.66)*1.34=0.67 seconds has passed in frame O. In this case, the observer in frame O thinks the clock in frame O' is running more quickly. Can I call this time contraction?
     
  7. Nov 14, 2008 #6

    JesseM

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    Yes, that's correct according to the Lorentz transformation:

    x = gamma * (x' + vt')
    t = gamma * (t' + vx'/c^2)
    Would actually be closer to 1.33, but OK.
    If you want three significant figures it should be 1.99 seconds, although for greater precision we might use 1.9899753 seconds.
    Yes, that's true. If we have marker at the origin of O' and another marker at x'=0.75 of O', then the time between the event of the clock passing the first marker and the event of the clock passing the second marker is 1 second in the O' frame but 1.9899753 seconds in the O frame. Both frames must agree on how much time elapsed on the moving clock between passing the two markers (because all frames agree on local events like what a clock reads as it passes next to some other object); since the O frame says it took twice as long to pass between the two markers, it must have been ticking twice as slow. To use your numbers, in the O' frame we can calculate that the clock must have been slowed by a factor of sqrt(1 - 0.75^2) = 0.66144, so if it read a time of T=0 as it passed the first marker it must have read a time of T=0.66144 as it passed the second marker. This means that in the O frame the clock took 1.9899753 seconds to go from reading T=0 to T=0.66144, which means it must have been ticking at a rate of 0.33238 ticks per second of time in the O frame.

    This is the same answer as you'd get if you used the velocity addition formula--if the clock is moving at 0.75c in frame O' and O' is moving at 0.66c relative to frame O, then the clock is moving at (0.75c + 0.66c)/(1 + 0.75*0.66) = 1.41c/1.495 = 0.943144c in frame O, and sqrt(1 - 0.943144^2) = 0.33238.
    More precisely it'd be t=0.672199
    No, you've made a mistake here, the clock is still running slow in frame O, just by a smaller amount. Again suppose in frame O' we have a marker at the origin, and another at position x'=-0.75. If the clock reads T=0 as it passes the first marker at the origin, it will read T=0.66144 as it passes the second marker. In frame O, 0.672199 seconds passed between the event of the clock passing next to the first marker and the event of it passing next to the second, so the clock was still running slightly slow, by a factor of 0.66144/0.672199 = 0.984 ticks per second of time in frame O.

    This is also what you get if you use the velocity addition formula to find the clock's speed in frame O--the formula tells you the clock's velocity in the +x direction must have been (0.66c - 0.75c)/(1 - 0.66*0.75) = -0.1782c (i.e. a speed of 0.1782c in the -x direction), so the time dilation factor would be sqrt(1 - 0.1782^2) = 0.984.
     
  8. Nov 15, 2008 #7
    Thank you very much Jesse for your answer in such great detail.

    Now I agree with you in the second example, that the moving clock in frame O' is running slightly slow. However, I do have another question if we think in a different way - we simply replace the moving clock with a moving baseball:

    Second example: In frame O', a baseball is traveling at 0.75 times speed of light from origin O' at t'=0, towards -x' direction instead. When this baseball reaches x'=-0.75 at t'=1 in frame O', the observer in frame O sees the baseball reaches x=(v-0.75)*γ at t=(1-0.75*v)*γ. Let v=0.66, thus γ=1.33. Therefore, when x'=-0.75, t'=1, x=-0.12, t=0.67.

    You can still put two markers on x'=0 and x'=0.75 on frame O'. Note that the two markers are at rest in frame O', and are moving together with frame O' respective to frame O. And the observer in frame O even sees the distance between the two markers shorter by γ times, which is 0.75/1.33=0.56. Also, at t=0.67, the observer in frame O checks his own clock, it reads 0.67. At the same time, the observer also checks the clock resting at origin O' in frame O', it reads 0.67/γ=0.67/1.33=0.50.

    Could you please confirm everything is still correct above?

    Now, someone in frame O' can clearly claim, the time span between the baseball passing the first marker and the baseball passing the second marker is 1 second. Or in other words, baseball moving process takes 1 second.

    But the observer in frame O is confused. According to his own clock, the baseball moving process takes 0.67 second. According to the clock in frame O', the baseball moving process takes 0.50 second. Which is correct? Is the time dilated (0.67 > 0.50) or contracted (0.67 < 1)?
     
  9. Nov 15, 2008 #8

    JesseM

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    OK, that works.
    You mean x'=-0.75, right?
    Yup, that's about right (the exact time on the clock would be 0.505 seconds), but note that the clock resting at at x'=-0.75 in O' does *not* read 0.505 at that same moment in frame O, because of the relativity of simultaneity. As I mentioned in my first post on this thread:
    In this case, d=0.75 in the rest frame of the two clocks, and they are both moving at 0.66c in the O frame, so in the O frame the trailing clock (the one at x'=-0.75c) is ahead of the leading clock (the one at x'=0) by 0.75*0.66 = 0.495 seconds. This means that when the clock at x'=0 reads 0.505 seconds, the clock at x'=-0.75 reads 0.505 + 0.495 = 1 second, according to the frame O definition of simultaneity.
    Yes.
    No, in order to time a moving process, you must compare the time on a local clock at the starting point with the time on a local clock at the ending point. Frame O sees that when the baseball was next to the clock at x'=0, the clock at x'=0 read 0 seconds (while the clock at x'=-0.75 already read 0.495 seconds at that moment in frame O), and that when the baseball was next to the clock at x'=-0.75, the clock at x'=-0.75 read 1 second (while the clock at x'=0 read only 0.505 seconds at that moment in frame O). So, frame O can see that frame O' will measure the time to be 1 second, not 0.505 seconds.
     
  10. Nov 15, 2008 #9
    Thank you very much for your time and patience, Jesse. Now comes to the last question regarding this scenario:

    Due to time dilation, the observer in frame O will think most processes take more time to finish in frame O than in frame O'. For example, a tick of a resting clock in frame O' takes 1 second in frame O' but 1.33 second in frame O.

    But in this case, the observer in frame O can see and deduce frame O' will measure the baseball moving process to be 1 second. But the observer in frame O measures the process in his own frame, it is 0.67 second. So this time, this process in takes less time to finish in frame O than in frame O'.

    So the observer feels time dilation rate is not reliable. It varies with different processes. How can it be?
     
    Last edited: Nov 15, 2008
  11. Nov 16, 2008 #10

    JesseM

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    Not "most processes", only processes along the worldlines of objects which have a lower speed in frame O' than in frame O will take longer to finish in frame O. As a counterexample, a clock at rest in frame O (which naturally has a greater speed in frame O') takes 1.33 seconds between ticks in frame O' but 1 second between ticks in frame O. Strictly speaking, the time dilation equation as it's normally written is only intended to compare the time between events along the worldline of an object at rest in one frame with the time between the same events in a frame where that object is moving at speed v; if the time between the events in the frame where the object is at rest is T, then the time T' between the events in the second frame where the object is moving at speed v will be given by [tex]T' = \frac{T}{\sqrt{1 - v^2/c^2}}[/tex]. If you wanted to talk about the time in each frame between events on the wordline of an object that was not at rest in either frame, this equation wouldn't be the correct one to use (although each frame could use it to determine how much the time was expanded in their own frame relative to the object's own rest frame, and then compare their answers).
     
    Last edited: Nov 16, 2008
  12. Nov 16, 2008 #11
    I got your point Jesse. Thank you so much! Tomorrow I am going to post another thread about length contraction - an experiment that will cause a gradient colored moving rod. I really hope you could take a look, because I find myself easier to understand your responds.
     
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