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Does Time Dilation Really Happen?

  1. May 11, 2014 #1
    I will just describe a common thought experiment to do with special relativity and a question I have about it.
    Note this question only applies to the realms of SPECIAL relativity.

    Imagine a scenario in which there are 2 spaceships A and B. Each spaceship has its own clock. Spaceship B moves a uniform speed, v, past A. When A and B are exactly adjacent the clocks are synchronized to read the same times.

    Now considering the reference frame in which A is stationary. B will move away from A and A will thus observe B's clock going slow due to time dilation.
    My question is what does this time dilation apply to? Does this not just apply to the perception of Bs clock in As reference frame? Surely to an observer in B looking at clock B no dilation would have occurred.
    So technically an observer in A looking at clock A would read the same time as an observer in B looking at clock B at any point in time.
    So is time dilation just something that an observer observes when observing a clock moving relative to it. In which case would it not be correct to say that in actual fact both clocks read the same time, it's just that by someone in A looking at B's clock an incorrect perception is obtained? In which case time dilation does not really occur and is just a misperception.

    Surely this is true in the sense that a stationary individual and someone moving at a constant velocity will observe the same rate of clock ticking on their own watch...?

    Moreover the fact that if we were to be in a reference frame with B at rest B would observe A's clock going slow leads to a kind of paradox in that each spaceships sees the others clock going slow. I can make sense of this apparent paradox by applying my above logic in that the spaceships are only perceiving the others clocks to be slow, where in actuality both clocks keep the same time for observers looking at their own clock.

    If I'm talking rubbish can you please explain why and what you think I am misunderstanding? And also how to explain the apparent paradox in that each spaceship sees the other clock going slow.
  2. jcsd
  3. May 11, 2014 #2


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    For better or worse that's what special relativity predicts. There have been many observations which show this to be so. One well known example, muons from cosmic rays (going at close to the speed of light) live much longer (using clocks on earth) than they are expected to (when moving at slower speed).
  4. May 11, 2014 #3


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    This is the crucial point. What does "at any point in time" really mean? It is implicit that A and B both look at their own clocks "at the same time as each other", but what does this really mean? In relativity A and B have their own definitions of "at the same time", which disagree with each other. There is no definition that everyone agrees with. This phenomenon is called "relativity of simultaneity". You can search for that phrase for more information.

    However if B turns round and comes back to A again, and A & B compare clocks, their clocks really do show different times to each other. The symmetry was broken because it was B that turned round, not A. This phenomenon is called the "twins paradox". You can search for that phrase for more information.
  5. May 11, 2014 #4
    So if you are in frame B and measure a time between two events would that tell you absolutely nothing about what A reads on As clock. All one can deduce is the time B observes on As clock by applying the time dilation formula?
    Is that correct?
  6. May 11, 2014 #5


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    Yes. The SEQUENCE of causally related events will be the same for both observers, the time between them is frame variant and the "start time" of the first is frame variant.

    Also, some of your terminology leads to confusion. There is no such thing as a "stationary individual" but there are frames of reference in which one individual is stationary and another is moving.
  7. May 11, 2014 #6


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    I think the best way to understand scenarios like yours is through spacetime diagrams. Here's one that depicts your scenario from the moment that spaceship B (depicted in red) passes spaceship A (in blue) and they synchronize their clocks. Spaceship B is traveling at 0.6c. The one-second ticks of their clocks are shown as dots. Since A is stationary in this frame, his clock ticks at the same rate as the Coordinate Time. Since B is moving at 0.6c, we can calculate his Time Dilation as gamma which equals 1.25. You can see that his dots are spaced 1.25 times the Coordinate Time:


    Not just Time Dilation but Time Dilation plus the time it takes for the images of the clock's ticking to traverse the ever increasing distance from B's clock to A. This is known as Relativistic Doppler and it has a different formula than Time Dilation. For a clock moving away at 0.6c it evaluates to 0.5. That is, A sees B's clock ticking at one half the rate of his own and B sees A's clock ticking also at one half the rate of his own. We can add the images of each clock ticking traveling at the speed of light between the two clocks:


    Both A and B have the same perception of the other ones clock. They each perceive their own clock to be ticking normally and the other ones clock to be ticking at half the rate of their own. Neither one of them can perceive any Time Dilation, either in their own clock or in the other ones clock.

    In Special Relativity, when you say "at any point in time", you mean "at any Coordinate Time" according to the particular reference frame you are describing the scenario in. As you can see, in A's rest frame, when A's clock is at 5 seconds the Coordinate Time is also at 5 seconds and B's clock is at 4 seconds.

    There's no paradox. We can transform from the rest frame of A to the rest frame of B using the Lorentz Transformation process:


    Now it's spaceship A's clock that is Time Dilated while spaceship B's clock ticks at the same rate as the Coordinate Time but both continue to perceive the other ones clock ticking at one half the rate of their own. It's important to realize that both these diagrams contain exactly the same information. Neither one is better or more important or more real than the other. You might think that A's rest frame is better for A and B's is better for B, but that is not even true. Both depict equally what each one perceives. In fact, we can also transform to a frame in which both spaceships are moving at the same speed in opposite directions, this speed being 0.3333c:


    Again, you can see that both spaceships perceive the other ones clock to be ticking at one half the rate of their own, even though, in this frame they are Time Dilated to the same extent (1.06) and ticking at the same rate.

    I hope this helps you understand what Time Dilation is and how it is different from Relativistic Doppler and how it is the later that determines perception and not the former.

    Any questions?
  8. May 12, 2014 #7


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    As I pointed out in the previous post, it doesn't matter which frame you use to depict a scenario, if you just use your clock to measure the time interval between remote events according to when you see them, you will get an answer that is not related to any particular frame, not even the one in which you are at rest. For example let's say B wants to measure the time between A's clock displaying 2 and A's clock displaying 4. B's clock will display 4 and 8 between the times that he sees those times on A's clock so he will not get the correct time according to his own rest frame. If we look again at this diagram for B's rest frame we will see that when A's clock is at 2, the Coordinate Time is 2.5 and when A's clock is at 4, the Coordinate Time is 5, so the correct answer is the difference or 2.5 seconds:


    Yes, but in order to apply the time dilation formula, B needs to know A's speed. How can he know that? Well, as long as he knows that both of them have been inertial, that is, not accelerating (not changing speed or direction) and as long as he can see A's clock, then he can measure the Doppler between them and from that, B can calculate A's speed relative to himself.

    So in this scenario, we have already determined that the Doppler Factor is 0.5. He can calculate the relative speed using the following formula (Doppler Factor is DF):

    speed = (1 - DF2)/(1 + DF2) = (1 - 0.52)/(1 + 0.52) = (1 - 0.25)/(1 + 0.25) = (0.75)/(1.25) = 0.6

    Next he calculates gamma = 1/√(1-speed2) = 1/√(1-0.62) = 1/√(1-0.36) = 1/√(0.64) = 1/0.8 = 1.25

    Finally, he takes the time interval according to A's clock (4-2) = 2 and multiplies that by gamma which is 2(1.25) = 2.5 seconds.

    But what if he couldn't see the time on A's clock? Let's just suppose that something happened at A's time of 2 and something else happened at A's time of 4. B can still measure the time interval according to his rest frame by continually sending out radar signals noting the times they were sent and the times the echoes were received according to his own clock and then noting the pairs of times corresponding to reflections off the two events of interest. Then after doing some calculation, he can determine the correct time interval.

    Here is a spacetime diagram for the rest frame of A showing just the radar signals and their reflections that happen to hit the two events of interest:


    For the first radar, B has logged a sent time of 1 second and a received time of 4 seconds. He averages these two signals to get 2.5 seconds and sets that as the Coordinate Time of the first event. Similarly, for the second radar, B has logged a sent time of 2 seconds and a received time of 8 seconds for an average of 5 seconds and sets that as the Coordinate Time of the second event. Taking the difference, he establishes that the time interval between those two events is 2.5 seconds according to his rest frame.

    The reason this works is because he is applying Einstein's second postulate for his rest frame when he averages the sent and received times. He is assuming that the radar signal took the same time to get to the target as the reflection took to get back to him even though it isn't true in the frame we were viewing this in.

    But it is true in his own rest frame as you can see here:


    Does this all make sense? Any questions?

    Attached Files:

    Last edited: May 12, 2014
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