Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Time Dialation analysis

  1. Apr 3, 2008 #1
    see the attchments

    Referring to fig 1
    We have a clock that is used to set the time on a second clock. At point A is a laser. At B there is a photon detector. The laser at A fires a photon, shown as the red broken line. The photon makes its way from point A to point B.

    At A there is also an LED (the green circle). When the laser fires the LED emits a pulse of light. At B there is a second LED (the green circle). When the detector at B receives a photon from the laser at A the LED at B emits a pulse of light.

    There is an observer at the point Z and he has a stop watch. When he sees the first pulse of light he starts the stop watch when he sees a second pulse of light he stops the stop watch. The time difference on the stop watch is used to set another clock.

    The entire apparatus and the observer are in a space ship that can be stationary ie V = 0 or moving V > 0

    The length from A to B is L

    While the space ship is stationary a test is done. The laser at A is fired and the observer records the time for the first and second pulse of lights from the LEDs.

    The time would be t = L/c

    If the clock was rotated thru 180 degrees the same result would be obtained.

    Fig 2

    The space ship is now moving at a constant velocity V in the direction of the arrow shown. I note before you all go ballistic I have not shown the space ship at time zero and at some time in the future as it makes the picture to messy. This is the same for fig 3.

    At time zero the clock will be in position A and B. At some time in the future the clock will have moved with the space ship and be in the new position A hat and B hat.

    At time zero a pulse of light is emitted from the laser at point A. The LED at A also emits a pulse of light which the observer observes and starts his stop watch.

    At some time in the future the pulse of light emitted from the laser will strike the detector at B hat. And a pulse of light will be emitted from the LED at point B hat. The observer sees this pulse and stops his stop watch.

    Analysis

    The length from A to B is L
    Vt is the distance the clock travelled between the pulse leaving the laser at A and the pulse from the laser striking the detector at point B hat.
    The length from A to B hat is L + vt

    Time difference = (L + Vt)/ c c is the speed of light.

    I note two things
    1. As expected as the clock is moving the time difference is greater than the stationary clock.
    2. If the observer had been standing at position B hat he would have observed something very different to this result.


    Fig 3

    As in Fig 2 the space ship is moving at a constant velocity V in the direction of the arrow shown.

    The clock is now rotated thru 180 degrees compared with fig 2.

    At time zero the clock will be in position A and B. At some time in the future the clock will have moved with the space ship and be in the new position A hat and B hat.

    At time zero a pulse of light is emitted from the laser at point A. The LED at A also emits a pulse of light which the observer observes and starts his stop watch.

    At some time in the future the pulse of light emitted from the laser will strike the detector at B hat. And a pulse of light will be emitted from the LED at point B hat. The observer sees this pulse and stops his stop watch.

    Analysis

    The length from A to B is L
    Vt is the distance the clock travelled between the pulse leaving the laser at A and the pulse from the laser striking the detector at point B hat.

    The length from A to B hat is L - vt

    Time difference = (L - Vt)/ c c is the speed of light.

    I note two things
    1. As the clock is moving by theory the time difference must be greater than the stationary clock. The stationary clocks time difference is L/c (fig 1). In fig 3 the time difference is (L – Vt)/c which will always be less than L/c. As the observer uses this difference to set another clock the set clock will run faster than the stationary clock so a moving clock is running faster than a stationary clock.!
    2. If the observer had been standing at position B he would have obtained a different result.

    Is this analysis right?

    How is it that an observer at rest with a clock can get a different result from the clock simply by changing his position of observation? Time is a vector
    Surely this apparatus can be used as a test to identify if you are in a moving frame of reference? If so you can demonstrate you are in a moving frame of ref by an experiment internal to the frame of ref!
     

    Attached Files:

  2. jcsd
  3. Apr 3, 2008 #2
    I am now doing the analysis from the perspective of an observer external to the space ship frame of ref
     
  4. Apr 3, 2008 #3
    I would like to make an amendment to this post I retract

    "How is it that an observer at rest with a clock can get a different result from the clock simply by changing his position of observation?"


    but I dont retract that time is or may be a vector
     
  5. Apr 3, 2008 #4

    JesseM

    User Avatar
    Science Advisor

    There's no point Z on your diagrams--did you mean to type point A, since they're next to each other on the keyboard?
    What do you mean "set another clock"? Where is this clock, and how is the time difference on the stopwatch used to set it?
    Stationary or moving relative to what? There is no absolute motion in relativity.
    In the ship's rest frame, presumably? Remember that the length would be different in a frame where the ship is moving due to length contraction.
    If I'm understanding the scenario correctly, wouldn't the time between the observer seeing each LED turn on be 2L/c? After all, he first sees the LED at A turn on, then it takes a time of L/c for the laser to get from A to B in his frame, then the LED at B turns on, then it takes an additional time interval of L/c for the light from the LED at B turning on to reach the eyes of the observer at A in his frame.
    Again, what is this clock that's separate from the stopwatch? What do you mean by "rotating" it?
    Again, you're ignoring the fact that in this frame, it also takes time for the light from the LED at B to return to A to make him stop his stopwatch. You're also ignoring the fact that because of time dilation, his stopwatch is running slow in this frame where the ship is moving at v, so the time in the frame between the pulse leaving A and him seeing the light from B will be greater than the time elapsed on his stopwatch by a factor of [tex]1/\sqrt{1 - v^2/c^2}[/tex]. And you're ignoring the fact that if the distance between A and B in the ship's rest frame is L, then the distance between them in this frame is only [tex]L' = L*\sqrt{1 - v^2/c^2}[/tex] due to length contraction. Finally, you haven't actually calculated the value of t, which is pretty easy to do: if the light starts at x=0 at t=0 and moves at c then its position as a function of time is x(t) = ct, while if B starts at x=L' and moves at v then its position as a function of time is x(t) = L' + vt, so the light will reach B when ct = L' + vt, and we can solve this for t to get t = L'/(c - v). We can do a similar analysis to show that the time for the light from the LED at B to get back to A is L'/(c + v), a smaller number since A is moving towards the position where B was when the LED turned on, while B was moving away from the position where A was when the pulse was emitted.

    So, the total time in this frame between the guy at A starting his stopwatch and stopping it would be L'/(c - v) + L'/(c + v), which is the same as:

    [L'*(c + v) + L'*(c - v)]/[(c - v)*(c + v)]

    which works out to:

    2cL' / (c^2 - v^2)

    which simplifies further to:

    2L' / c*(1 - v^2/c^2)

    Now, remember that L' = L * sqrt(1 - v^2/c^2)...so plugging that in, the time in this frame between the stopwatch starting and stopping is:

    2L / c*sqrt(1 - v^2/c^2)

    But as I said before, the stopwatch is slowed down in this frame due to time dilation, so it only ticks sqrt(1 - v^2/c^2) times this amount between starting and stopping, which comes out to 2L/c. You can see that this is exactly the amount of time we predicted in the frame where the ship was at rest, as it had better be if SR is correct that the laws of physics work the same way in all frames so there's no way to determine who is moving in any absolute sense.

    The analysis would be pretty much identical in the frame where the ship was moving in the opposite direction at speed v, except that in this frame it would take L'/(c + v) for the light to go from A to B, and L'/(c - v) for the light from the LED turning on at B to return to the eyes of the guy at A (the reverse order of the previous frame where the ship was moving forward at v).
    No, see above. Regardless of how the ship is moving, the observer will see a time of 2L/c elapse on his stopwatch between the time A turns on right next to him and the time he sees the light from B having turned on. On the other hand, an observer standing next to B will see the light from A turning on reach him at exactly the same time that this light causes B to turn on, but if each one factors out the signal delay between when events happened in his frame and when he sees them, they'll both conclude that the time delay in their frame between A turning on and B turning on is L/c.
     
  6. Apr 3, 2008 #5
    Back to the drawing board :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Time Dialation analysis
  1. Time dialation (Replies: 3)

  2. Time dialation (Replies: 12)

  3. Time dialation (Replies: 10)

Loading...