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Time dilation when travelling

  1. Oct 2, 2004 #1
    i understand that time slows down as compared to a stationary perspective when an object is travelling at a certain velocity, but by how much does time slows down for that object? whats the formula behind this time difference? also, i understand that Einstein has derived this idea when he was comparing the different perspective of an observer with how he observed the speed of light approaching him ie. the result is that the observer will measure the same speed of light from whatever perspective he was measuring. so is it right for me to say that when one is travelling at the speed of light (just an analogy although i know it is impossible), his time should stop as compared to our perspective??
     
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  3. Oct 2, 2004 #2

    Doc Al

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    Imagine two frames moving with respect to each other with some speed v. (Two rocketships, for example.) Each frame will measure the clocks in the other frame to be running slow compared to their own clocks. If a clock measures a time interval of [itex]\Delta t_0[/itex], then the other frame will measure the time that passed according to their clocks to be [itex]\Delta t = \Delta t_0 / \sqrt{1 - v^2/c^2}[/itex].

    Of course, each frame views their own clocks as running normally.
     
  4. Oct 2, 2004 #3

    Fredrik

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    The time dilation factor is

    [tex]\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]

    Check out the thread "Am I thinking SR correctly?" for a lot more information about this.
     
    Last edited: Oct 2, 2004
  5. Oct 2, 2004 #4
    i dont understand. when u say 'a clock measures a time interval of...', which clock are u talking about?
     
  6. Oct 3, 2004 #5
    Either clock. Suppose I observe an oscillating pendulum whose centre of motion is at rest with respect to me. I measure the pendulum to execute one oscillation in an interval of time [itex]\Delta t[/itex]. If you move away from me at some fraction of the speed of light [itex]\beta[/itex], you will measure the period of the pendulum as [itex]\Delta t/\sqrt{1-\beta^2}[/itex]. The consequence is that intervals of time between events are measured as longer by observers in motion with respect to the events.
     
  7. Oct 3, 2004 #6
    thanks a lot i understand what u mean. but what i want to know precisely is that when using the equation for the example given by Doc Al, the velocity 'v' is the velocity of the frame of the moving clock that has a time interval of [itex]\Delta t_0[/itex], correct?
     
    Last edited: Oct 3, 2004
  8. Oct 3, 2004 #7
    Yes. In this case [itex]v[/itex] denotes the velocity of the clock wrt the observer. This is of course equal in magnitude to the velocity of the observer wrt the clock and opposite in sign. However, in this example the sign of the velocity is irrelevant as the equation involves only [itex]v^2[/itex].
     
  9. Oct 4, 2004 #8
    thank you very much jdstokes. because im quite new to this topic so that is y i posted a lot of questions. another thing i want to verify is when measuring [itex]\Delta t_0[/itex] and [itex]\Delta t[/itex], are we comparing both intervals from the same frame? if yes, what is that frame?
     
  10. Oct 4, 2004 #9

    Doc Al

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    different frames, different times

    The two time intervals are measured in different frames, moving with speed v with respect to each other. Let's call the two frames frame #1 and frame #2. Let [itex]\Delta t_0[/itex] be the time interval measured by a clock (call it clock A) in frame #1, the rest frame frame of clock A. (The "rest frame" is the frame in which the clock is at rest, that is, its speed is zero. Clock A is stationary in frame #1, but moves with speed v in frame #2.) [itex]\Delta t[/itex] is a measurement made from frame #2. According to frame #2, clock A is running slow; the actual time elapsed according to frame #2 is [itex]\Delta t[/itex], while clock A only reads [itex]\Delta t_0[/itex].
     
  11. Oct 5, 2004 #10
    Let [itex]\Delta t_0[/itex] be the time interval measured by a clock (call it clock A) in frame #1, the rest frame frame of clock A.

    when u are saying time interval, whats the time difference measured from?
     
  12. Oct 5, 2004 #11

    Doc Al

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    Between any two times measured by clock A there is an interval of time. For example, let [itex]t'_1 = 12:00[/itex] and [itex]t'_2 = 12:05[/itex] be two time measurements by clock A. Then [itex]\Delta t_0 =5[/itex] minutes is the time interval measured by clock A.

    Of course, a different frame will measure different times ([itex]t_1[/itex] and [itex]t_2[/itex]) and a different time interval [itex]\Delta t[/itex].
     
  13. Oct 6, 2004 #12
    Thanks alot Doc Al, i understand now. To verify, i will give an example. Lets say we are going to conduct an experiment which is the same as the example given by you previously. I will start the experiment at 1pm and put a clock in the moving frame which is at first synchronised when at rest. I will then note down the time as the clock in the moving frame passes by me with velocity 'v' which is a significant fraction of 'c' (lets say i am able to see the clock). [itex]\Delta t_0[/itex] will then be the time interval of the difference in time between 1pm and the time i note down. Using [itex]\Delta t = \Delta t_0 / \sqrt{1 - v^2/c^2}[/itex], i will get the time interval which is at the rest frame where i am and that time interval is [itex]\Delta t[/itex]. [itex]\Delta t[/itex] will be an interval greater than [itex]\Delta t_0[/itex]. Am i correct?
     
  14. Oct 6, 2004 #13

    Doc Al

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    I think you've got it. But since I don't completely understand your example, I will replace it with an equivalent one of my own. (One must be very precise when setting up these special relativity problems!) Let's say I have a huge laboratory with synchronized clocks at both ends of the lab (call them clocks A and B). A fast rocket with its own clock zooms by, first passing my clock A, then B. Let's pretend I am able to see what that moving clock is reading as it zooms past.

    Let's also pretend that the rocket clock is set just right so that the moment it passes my clock A, both the rocket clock and clock A read 1 pm. Then I observe that the rocket clock reads a time [itex]\Delta t'[/itex] after 1pm exactly as it passes clock B, which reads a time of [itex]\Delta t[/itex] after 1pm. So the time interval measured in the rocket frame is [itex]\Delta t'[/itex], while the time interval according to my lab clocks is [itex]\Delta t[/itex]. Special relativity tells us that the two intervals are related by [itex]\Delta t = \Delta t' / \sqrt{1 - v^2/c^2}[/itex]. And, yes, [itex]\Delta t > \Delta t'[/itex], consistent with the time dilation maxim "moving clocks appear to run slow".
     
  15. Oct 7, 2004 #14
    I have understand how the equation works and what it means now thanks! But i will like to ask you a paradox. From my perspective ( lets say a stationary frame on a train station platform on Earth), the clock in the moving frame (lets say a clock in a train) will appear slow to me. However, from the perspective of the person in the moving frame he will see me as the one moving and not him, hence he will see my clock to be slow and not his. Therefore, whos clock is slow?
     
  16. Oct 8, 2004 #15

    selfAdjoint

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    The answer to that is Yes. Both clocks are "right", because "right time" can only be defined relative to the inertial frame.
     
  17. Oct 8, 2004 #16

    Doc Al

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    No paradox. Each observer views their own clocks operating normally and the other frame's clocks as being slow. (If it didn't work symmetrically, then there would be a problem as you would be able to tell which frame is "really" moving.)

    To become comfortable with this, and realize that there is no contradiction or paradox, you need to learn a bit more about relativity. To fully understand how space and time are affected by relative motion, one must consider several relativistic effects operating together:
    (1) time dilation
    (2) length contraction
    (3) the relativity of synchronization​
    Check out any elementary book on special relativity.
     
  18. Oct 8, 2004 #17
    If both clocks are 'right', doesnt that means both people in the stationary frame and the moving frame will see the same clock at one frame (lets say the stationary frame) to be 2 different times?
     
  19. Oct 8, 2004 #18

    Doc Al

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    No. To remove any ambiguity, imagine that someone standing right next to a clock (clock A) in the "stationary" frame fires a gun. If that clock reads time t, then everyone will agree that according to the stationary clock, the gun went off at time t. Even observers in the "moving" frame, if they could see clock A as it whizzed by, would agree that clock A read time=t when the gun went off. But the "moving" frame will observe the gun to fire at a different time according to its clocks (clock B, for example).
     
  20. Oct 9, 2004 #19
    But why should the person in the moving frame reads the same time 't'? Since from the perspective of the person in the moving frame, the stationary frame should be the one moving and not his, then he should read time 't' to be slower isnt it?
     
  21. Oct 9, 2004 #20

    Doc Al

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    I'm not clear as to what you are talking about. Each frame's clocks read their own time. And each frame will observe the other's clocks to go slow. Do you have a concrete example in mind?

    As I said in an earlier post, in order to analyze a particular situation you need to consider all three SR effects.
     
  22. Oct 10, 2004 #21
    A good example will be the 'Twin Paradox'. In this case whos time will be slowing down?
     
  23. Oct 10, 2004 #22

    Doc Al

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    I would say that the "twin paradox" is not a good example, since one twin accelerates and thus does not remain in a single inertial frame. The twin that accelerates ages less.
     
  24. Oct 11, 2004 #23
    What im not sure about is when you said 'And each frame will observe the other's clocks to go slow'. Are you saying that both clocks will observed to be slow by each person in each of the frame? But the equation [itex]\Delta t = \Delta t_0 / \sqrt{1 - v^2/c^2}[/itex] will only mean one of the time to be slow so im quite confused. As about the 3 SR effects, i have only know the first 2 so far ie. time dilation and length contraction and i think to go deep in these 3 fields will take quite some time. Let me give an example to show you what i dont understand.

    Lets say im on a train platform and i observed a train pass by me with speed 0.8c (assuming im able to see the train). Theres a clock behind me and also a clock in the train. Both of us agree that it is noon at that time. When the train pass by me, i see a time of 12.15pm in the train's clock. The clock behind me will show a time that is longer than 12.15 and the time interval of my frame will correspond to the answer given by the above equation and adding it to 12.00 we have 12.25pm. But if theres a person in the train and he sees the clock behind me as the train pass by, what time will he sees? Will the time he sees on the clock behind me be 12.25pm or a time slower than his? I come at this question because if the person behind me uses the equation to solve the time interval from his frame he will get an answer that is slower than 12.25pm.
     
  25. Oct 11, 2004 #24

    russ_watters

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    Since any observer can arbitrarily call himself stationary, the "v" in the equation always applies to the other guy (his own "v" is 0).
     
  26. Oct 11, 2004 #25

    Doc Al

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    Here's what I mean: Two frames, A and B, each with their own clocks. Time dilation says: (1) Frame A will make measurements of the behavior of the B clocks and will deduce that they are slow compared to his A clocks, and (2) Frame B will make measurements of the behavior of the A clocks and will deduce that they are slow compared to his B clocks. Note that this is a specific statement about the behavior of a single moving clock; it is not a statement about time intervals deduced from the readings of multiple moving clocks. Each frame is equally permitted to use the time dilation formula to describe the behavior of the other frame's clocks compared to his own.
    I repeat, there is no way to understand SR without considering all three SR effects working together. The one you left out--the desynchronization of moving clocks--is key!

    OK, so far there's a clock on the platform and a clock in the front of the train.
    On which clock? The one in the front of the train? (That one's long since past. How do you see it?) Or another one in the back of the train?
    Careful. If you are saying this: When the front of the train passed you, that clock at the front of the train showed 12:00. Then when the rear of the train passed you, that same clock at the front of the train showed 12:15, then you are correct that the clock behind you must show a longer time (say 12:25). It is a simple application of the time dilation formula for a moving clock: the clock at the front of the train.

    But this is according to you, on the ground. The train observers will disagree! The train observers say that when the back of the train passes you, their train clocks will read a time greater than 12:25. You, on the ground, however, see the train clocks (at front and rear of the train) as be out of synchronization -- to you they don't read the same times at all! (That's the desynchronization effect.)
    If he's looking at your clock, he'd better see the same thing you do! :smile: Maybe you are asking: What will he see on his clock at the back of the train when the back of the train passes you? In that case, his clock at the rear of the train will read greater than 12:25.
    And that makes sense, since from the view of the train, your clock is a single moving clock, which the train observer sees as going slow.
     
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