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Relative time

  1. Jan 1, 2012 #1
    When 2 objects A,B are moving wrt to each other (lets say @0.86c)then from frame A if local time is 10 years then time at B is 5 years. What does this mean?is it any event happening now in A(10 years) is simultaneous with the events happened in B when its clock ticked 5 years?
  2. jcsd
  3. Jan 1, 2012 #2

    You can't specifiy just "local time" because in the frame of the other object, a similar thing is happening. The rate at which times flows depends on the individual observer.

    If you start with clocks together, separate them, and bring them back together, they will generally differ (the path with greater deviation from an inertial path will accumulate less time). If two clocks compare readings, move apart, and then later reunite and compare readings again, if one clock moved inertially between meetings while the other accelerated at some point, the one that accelerated will have elapsed less time.

    In special relativity greater speed means less elapsed time; In general relativity (with gravity) we find that greater gravitational potential also slows time.
    Last edited: Jan 1, 2012
  4. Jan 1, 2012 #3
    why not local time?i m talking considering A as my frame and called its time local. Consider this 2 objects A and B meet at a point and synchronized to read 0. Now A and B continue to move as they did earlier(let relative velocity be 0.86c for convenience). Now if i am in A and 10 years passed for me then only 5 years has passed for B according to me isn't it?i want to know what this means?is it any event happening now in A(10 years) is simultaneous with the events happened in B when its clock ticked 5 years?
  5. Jan 1, 2012 #4


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    You do understand that when you say 2 objects are moving with respect to each other at 0.86c, you don't mean they are both moving at the same time at that rate, do you? Don't you in your mind think about one frame at a time--from A's frame, B is moving, and from B's frame, A is moving? You don't think that both are moving at the same 0.86c which would add up to a total speed of 1.72c, do you? So when you are thinking about A's frame, B is the one that is experiencing speed which means B is the one that is experiencing time dilation. Conversely, when you are thinking about B's frame, A is the one that is experiencing speed which means A is the one that is experiencing time dilation.

    Time dilation simply means that moving clocks are ticking at a slower rate than stationary clocks according to whichever frame you are considering using the Einstein's formula:

    τ = t√(1-β2) = t/γ

    where τ is the time on a moving clock and t is the time on a stationary clock.

    So, yes, in frame A, when 10 years ticks by, 5 years will tick by for B since he is moving at 0.86c in A's frame.
    When you're talking about comparing the time coordinate of an event in two frames of reference, you need to use the Lorentz Transform which is:

    t' = γ(t-βx)

    So clearly, these are not the same formulas because the LT involves the location of the event as well as the time. But if you are careful to always calculate the location of B as being x = β*t, then:

    t' = γ(t-β2t) = γt(1-β2)

    And since γ = 1/√(1-β2) then

    t' = t(1-β2)/√(1-β2) = t(√(1-β2)) = t/γ
  6. Jan 2, 2012 #5
    Snip3r, the answer to your question is "Yes." That is, A's clock can be reading 10 years while the clock on B's world line (that is simultaneous with A in A's frame) is reading 5 years.

    See if you can comprehend this space-time diagram below, because it depicts this situation.

    Red and blue guys are moving in opposite directions at the same speed with respect to a rest frame, and carry clocks that give us readings for t1 and t2. I have labeled times t1 and t2 along the X4 axis in each of the red and blue frames. The time, t1, reads exactly the same in the red frame and the blue frame. The t2 times are also exactly the same.

    The thing that provides the enlightenment for this problem is the understanding of the 4-dimensional universe. Observers move along their world lines (their own X4 dimension) at the speed of light, c. The red guy and the blue guy live in different 3-dimensional worlds, because their 3-D worlds cut across cross-sections of the 4-dimensional universe at different angles (X1 and X4 axes are always rotated such that the 45-degree angle lines, corresponding photons (green lines in sketch), always bisect the angle between X1 and X4 for any frame--that's why the speed of light is c in every frame).

    You can see that red's clock reading time, t1, is simultaneous with blue's clock reading t2 in the blue 3-D world shown below. But, when red is at his space-time position at t2, the 3-D world he is living in cuts across a cross-section of the 4-D universe in a direction that includes the blue guy when he (blue) is at blue's t1 position. Thus, red and blue literallly live in two different 3-D worlds which are different cross-sections of the 4-dimensional universe.
    Last edited: Jan 2, 2012
  7. Jan 3, 2012 #6
    thanks. i understood this from your STD if A' is an event in A, B' in B and A' is simultaneous with B' in A then it doesn't mean B' is simultaneous with A' in B. however i find it hard to visualize may be its because it is hard to imagine 4-D. is there a way by which you could draw an analogy between 4-D and 3-D, 3-D worlds and 2-D worlds and explain the scenario in only 3-dimensions?
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