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Time question

  1. Apr 5, 2009 #1
    I have a time question concerning relativity. I read that time slows down on a moving object from my perspective. This is so based on a lightbeam going up and down on the moving object. In my persepctive the light path is longer. In his perspective it is shorter. Thus, in my perspective his time slows down.

    The question is as follows:

    Why is a lightbeam going up and down the sole derterminor. If we use a lightbeam going from the front of the moving object to the back of the moving object to tell time, we will have the opposite effect. In my perspective the light path is shorter. In his perspective it is longer. We can thus say that in my perspective his time goes faster.

    This is my question. Any answers?
     
  2. jcsd
  3. Apr 5, 2009 #2

    Doc Al

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    How did you calculate the time for the round trip of the light beam? If you calculate it correctly, you'll find that the time you measure is always greater regardless of whether the light beam goes up and down or front to back.

    One reason for using a light beam that goes up and down is that it's much easier to calculate the travel time, since there's no length contraction to worry about.
     
  4. Apr 5, 2009 #3

    Mentz114

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    Hi Nathan,

    in the case of the longitudinal light-clock, you must take into account that from your frame the light takes different times for the in and out trips ( towards and away from you ). If you factor this in with the length contraction, you'll get the same time as the tranverse light-clock.

    [edit] Posted simultaneously with the above.
     
  5. Apr 5, 2009 #4

    A.T.

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    Empirical evidence shows that the same light beam moves at the same speed for all observers. Since the moving observer sees the light moving a shorter distance d, his clock has also to show less time passage t, in order for the light to have the same speed c as observed by you: c = d / t = const

    You cannot have it just go one way (front to back) to determine the time dilation in the moving system, because of relativity of simultaneity. Also don't forget the length contraction.
     
  6. Apr 5, 2009 #5
    In my question, I was timing only the path of the lightbeam from front to back to tell the time. We can use numerous front to back lightbeams to tell the time. In this case, the moving ship calculates longer paths from front to back, whereas from my perspective the back wall comes closer in for a shorter path from front to back.

    The answers given imply that you can't only time front to back. I don't see why not.

    Secondly, I don't see how timimng a round trip from front to back and back to front will come up with the same time difference as up and down. It would seem to me that a round trip from front to back and back to front would have effects that cancel each other out. The front to back is longer for him and shorter for me, and the back to front is shorter for him and longer for me.

    A third issue is as follows: How do we know that the lightsource does not effect the speed of light. Just like if I'm on a moving ship and throw a ball, we add both speeds together (or subtracts them for the opposite direction), perhaps we can add the speed of the flashlight to the speed of light that eminates from it. This would more easily explain why the ship's perception of the light speed and my perception of it is the same thing.
     
  7. Apr 5, 2009 #6

    Doc Al

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    Sure you can use a one-way trip, but it's more complicated since you must deal with length contraction and the relativity of simultaneity. Done correctly, you'll get the same answer. (A round trip is easier to analyze.)

    Why don't you figure it out carefully and see if the effects cancel?

    Why would that assumption (which is known to be false) more easily explain anything?
     
  8. Apr 5, 2009 #7
    Because you could also choose to measure the time from back to front. When you do this, you discover that the front-to-back time is shorter than the back-to-front, even though in the clock's rest frame the two times are equal. From this you deduce that there is disagreement between the two reference frames regarding simultaneity. In other words, that two tiny clocks, one on each mirror of the big clock, that are synchronized in the clock's rest frame, are not synchronized in the frame in which it moves. Thus when you time a one-way trip you are comparing the times read out on two different clocks (one in the front and one in the back) that are out of synch with each other. To find the rate of a single clock, the light flash must return to that clock.

    Imagine a double light clock, with an L-shaped mirror at the bottom left, and regular mirrors at the top and right, spaced equal distances away in its rest frame. There's an up-down moving photon and a left-right moving photon, respectively, in the two clocks that share the L-shaped mirror. Set it up so that both photons are at the L-shaped mirror simultaneously when they get there. The two clocks, of course, tick at the same rate.

    If the two photons meet up in the L-shaped mirror each time in the clock's rest frame, they do so in every frame, so the horizontal beam of the clock always ticks at the same rate as the vertical beam.

    They do not. The two clocks, horizontal and vertical, MUST tick at the same rate in every frame, because the two photons meet in every frame.

    It is an observed fact of nature.

    No, that does absolutely nothing to explain why onboard the ship the light ray travels at c, whereas on the Earth the same light ray travels at c. Not c+v_flashlight.
     
  9. Apr 5, 2009 #8
    I have problems with the light clock derivation for a number of reasons. It's not a particularly intuitive framework when you start to look closely at it, it's more intuitive if you are happy to gloss over things.

    The OP touches on something interesting.

    Has anyone worked out what happens with two synchronised light clocks which are set in motion?

    If the clocks are in motion relative to A, A will see one clock with two even phases (mirror to mirror) and one clock with uneven phases.

    If those clocks are at rest relative to B, B will see both clocks with two even phases.

    Below are woefully inaccurate diagrams of what A will see, Transverse clock first, then Longitudinal.


    B------.Tick----->|
    B<-----Tock------|

    ___
    ^
    |
    |
    |
    Tick
    |
    |
    |
    |
    B
    ___
    |
    |
    Tock
    |
    V
    B

    I'd be interested to hear how this oddity is usually explained (ie why there is agreement on ticks and tocks being equal for the Transverse clock, but not for the Longitudinal clock).

    I have an idea of how it should be worked out, but I'd be interested to see if I am on the right track.

    cheers,
    neopolitan
     
  10. Apr 5, 2009 #9

    Doc Al

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    As long as the rest lengths of your two light clocks are the same, the round trips "tocks" are the same.

    The one way "ticks" occur at different points along the direction of motion, thus would not be simultaneous in all frames. No reason for them to be equal in all frames.
     
  11. Apr 5, 2009 #10

    JesseM

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    I would say it's just the relativity of simultaneity. Imagine replacing the two mirrors with two digital watches which are synchronized in their own rest frame. If we're in a frame where the watches are moving, then if the axis between the watches is perpendicular to their direction of motion, they will still be synchronized in this frame; if the axis between them is parallel to their direction of motion, they will be out-of-sync in this frame. So, for example, suppose in their own rest frame each watch lights up once every 2 seconds, and there is a 1 second offset between the two (so if the watches are 1 light-second apart, their lighting up corresponds to when a photon bouncing between mirrors at the same positions would hit each mirror). In terms of the time-coordinate of their rest frame it might look like this:

    t=0 seconds: watch A reads 0 seconds, lights up
    t=1 second: watch B reads 1 second, lights up
    t=2 seconds: watch A reads 2 seconds, lights up
    t=3 seconds: watch B reads 3 seconds, lights up
    t=4 seconds: watch A reads 4 seconds, lights up
    t=5 seconds: watch B reads 5 seconds, lights up

    ...and so forth. Now in a frame where they are moving at 0.6c in a direction perpendicular to their direction of motion, their ticks will be dilated by a factor of 1.25 but they will remain in sync, so in this frame's coordinates it would look like:

    t=0 seconds: watch A reads 0 seconds, lights up
    t=1.25 seconds: watch B reads 1 second, lights up
    t=2.5 seconds: watch A reads 2 seconds, lights up
    t=3.75 seconds: watch B reads 3 seconds, lights up
    t=5 seconds: watch A reads 4 seconds, lights up
    t=6.25 seconds: watch B reads 5 seconds, lights up

    But in a frame where they are moving at 0.6c parallel to their direction of motion, they will be out-of-sync by vL/c^2 where L is the distance between them in their rest frame, which is 1 light-second, so B's time is behind A's time by 0.6 seconds. So you have to modify the times that B reads 1, 3 and 5 seconds above; for example, although it's true that A reads 2 seconds at t=1.25 seconds, B will only read 2 - 0.6 = 1.4 seconds at that moment, so it will take an additional 0.75 seconds of coordinate time to tick 0.6 seconds forward to a reading of 2 seconds (since its rate of ticking is dilated by 1.25, and 0.6*1.25=0.75), meaning B will not read 2 seconds until a coordinate time of t=1.25+0.75=2 seconds in this frame. The modified numbers will be:

    t=0 seconds: watch A reads 0 seconds, lights up
    t=2 seconds: watch B reads 1 second, lights up
    t=2.5 seconds: watch A reads 2 seconds, lights up
    t=4.5 seconds: watch B reads 3 seconds, lights up
    t=5 seconds: watch A reads 4 seconds, lights up
    t=7 seconds: watch B reads 5 seconds, lights up

    So you can see that because of the relativity of simultaneity, in this frame the time from A to B lighting up is longer than the time from B to A lighting up.
     
  12. Apr 5, 2009 #11
    Help me out here to see how the back front back clock will be the same as the up down clock.

    For demostrative purposes say the ship is going at the speed of light or close to it, if that is not possible.

    If "up" takes one second, up and down will take two seconds for the guy on the ship.
    For me down below, I will see two diagnals each the leangth of (or close to) the square root of two. So it will take about 2.8 seconds.

    From back to front and back, on the other hand, will take much longer. Just to get to the front is immpossible if both are going at the speed of light. If the ship goes a bit less than the speed of light, it will still take much time to get there, and then it has to get back. As such, how do we come upon the same time difference as up and down?
     
  13. Apr 5, 2009 #12

    A.T.

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    Length contraction. Compute it if you don't believe it.
     
  14. Apr 5, 2009 #13

    robphy

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    Here's my calculation:
    http://arxiv.org/abs/physics/0505134
    with animations at
    http://physics.syr.edu/courses/modules/LIGHTCONE/LightClock/ [Broken]
    (My avatar is taken from those animations.)

    As JesseM says, the "oddity" is an indication of the relativity of simultaneity... which is made more visual if one considers not just two mirrors... but a circular array of mirrors.
     
    Last edited by a moderator: May 4, 2017
  15. Apr 5, 2009 #14

    Doc Al

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    This implies that the speed of the ship is v = 0.7c.

    Figure it out using these facts and assuming the same speed as in your example:
    1) The observed length of the clock is shortened by a factor of 1.4. Thus its length is not 1 light-second, but 1/1.4 light-seconds. (This is length contraction.)
    2) When the light moves forward, it must cover a distance equal to the length of the clock plus the distance that the front of the clock has moved in that time. Call that time T1.
    3) When the light moves backward, it must cover a distance equal to the length of the clock minus the distance that the back of the clock has moved in that time. Call that time T2.

    If you do the calculation, you'll find that the round trip time is T1 + T2 = 2.8 seconds, exactly the same as for the vertical light clock.
     
  16. Apr 5, 2009 #15
    Thanks for the replies.

    They are pretty much in accordance with what I thought, although noone really touched on what I see as the crux. It is a simultaneity issue for sure.

    I had B in motion with respect to A, but stationary with respect to the synchronised clocks.

    If there was one more synchronised clock (orientation unimportant) which was half the length of the other two and at rest with respect to B, B would assume/calculate that whenever the "half clock" ticks, the photons in other clocks would be bouncing off one mirror or the other.

    Would he be right or wrong, according to A?

    If I understood Jesse correctly, then according to A, B would be right in the case of the Tranverse (long) lock and wrong in the case of the Longitudinal (long) clock.

    Because we are talking about photons, B can only know that a photon has hit a distant mirror (call it a "tock mirror") when the photon that hit the tock mirror hits the "tick" mirror where he is nominally located. It's difficult to split up the two way journey into two meaningful legs for B who can only be in one place at a time.

    However, say we had a "centiclock", one hundredth the length of a standard light clock, attached to the "tock mirror" of the Longitudinal light clock. With each tick of the centiclock, a photon is transmitted towards B, until the photon of the Longitudinal light clock hits the "tock mirror". The action of both the centiclock and the Longitudinal clock would be initiated by a photon emitted from the centre of the Longitudinal clock's length (ie the photons from the centiclock would start being emitted simultaneously with the first tick of the Longitudinal clock, according to B).

    B would have these facts to hand: photon arrived from initiator/clock photon left "tick mirror", photons started arriving from the centiclock and photons stopped arriving from centiclock/clock photon arrived back from the "tock mirror".

    As far as I can tell, B should get 50 photons from the centiclock in half the time it takes for the clock photon to complete one full circuit. Assume then that those photons pass by B and reach A. Like B, A will also see 50 photons, in half the time it takes the clock photon in the longitudinal clock to complete a full circuit.

    Or do they?

    If my description is unclear, please let me know.

    cheers,

    neopolitan
     
  17. Apr 6, 2009 #16

    atyy

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    Cool! I've long wondered what your avatar was.
     
    Last edited by a moderator: May 4, 2017
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