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I Light Clock problem

  1. Mar 20, 2016 #1

    DAC

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    Hello PF.
    Re. the light clock on the train thought experiment. If the mirrors are one metre apart in both frames. And the speed of light is the same in both frames. Why isn't the time it takes light to travel one metre, the same in both frames?
    Thanks.
     
  2. jcsd
  3. Mar 20, 2016 #2

    Orodruin

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    It is.
     
  4. Mar 20, 2016 #3

    Ibix

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    They aren't one meter apart in both frames, assuming we're talking about one clock here. In one frame they are length contracted and moving, so the distance from where one mirror was to where the other will be is not one meter.
     
  5. Mar 20, 2016 #4

    Orodruin

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    In the usual light clock setup the light clock is perpendicular to the direction of motion. The mirrors are definitely one meter apart in both frames with this setup.

    The reason the light needs to travel more than one meter in the frame where the clock is moving is that the mirrors move.
     
  6. Mar 20, 2016 #5

    DAC

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    doesn't that mean time is the same in both frames?
     
  7. Mar 20, 2016 #6

    Orodruin

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    No. Even if the clock is perpendicular to the direction of motion, the light needs to travel longer in the frame where the mirrors are moving (it needs to travel additional distance to catch up with the mirrors). Therefore, the clock is dilated in the frame where it is moving.
     
  8. Mar 20, 2016 #7

    Ibix

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    True. Scratch the "length contracted" from what I wrote. The distance from one mirror to the other at any given instant is 1m. The distance from where one mirror was when the light hit it to where the other will be when it gets hit is more than 1m.
     
  9. Mar 20, 2016 #8

    FactChecker

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    Suppose you are an outside, stationary observer watching the light beam. You see the light travel along the hypotenuse of a right triangle. The triangle has one side of 1 meter. The other side will be the distance traveled while the light beam went from one mirror to the other. So the time you measure will be proportional to the length of the hypotenuse. It's a simple Pythagorean theorem calculation.
     
  10. Mar 20, 2016 #9

    Dale

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    Because the distance the light travels is not equal to the distance between the mirrors, in general.
     
  11. Mar 20, 2016 #10

    Janus

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    You have two light clocks, each with their mirrors 1 meter apart along a line perpendicular to the relative motion between the two clocks. You have an observer with each clock. At the moment the clocks pass each other, a light pulse leaves the bottom mirror of each clock. Each observer note how long it takes for his pulse to reach his upper clock. They both get the same answer. However when they consider the pulse for the other clock they note that it takes longer to reach its upper mirror than it did for their own pulse to reach their upper mirror.
    This is because both observers measure the speed of both pulses as being the same relative to themselves. Observer 1 measures his pulse moving at c and notes that it takes after it has traveled m meter vertically it has reached his mirror. The pulse of the other clock also moves 1 meter in the same time. However it is moving at an angle and after traveling for 1 meter has not yet reached the top mirror of the second clock. Thus it takes longer for pulse of clock 2 to reach the top mirror as measured by observer 1. Conversely, as far as observer 2 is concerned, it is clock 1's pulse that takes longer to reach its top mirror.

    So if you were to ask each observer how long it took for his pulse to reach his top mirror, they both will give the same answer, and if asked how long it took the othr clock's pulse to reach its top mirror, they will say longer than it took for their own pulse to reach their top mirror.
     
  12. Mar 20, 2016 #11
    Is
    Is there an easy way to show how you get from here to the v squared over c squared of Lorentz?
     
  13. Mar 20, 2016 #12

    Ibix

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    Say it takes the light clock a time ##t## to tick when it isn't moving, and the mirrors are a distance ##L## apart. Can you relate ##L## and ##t## using the speed of light?

    Now say the light clock is moving. It takes a time ##t'## to tick. How far do the mirrors move in this time? So how far did the light pulse have to travel? This distance must be equal to ##ct'##, by definition. Can you write down a relationship between ##L##, ##t'## and ##v## involving the speed of light?

    Using the last results of the last two paragraphs, eliminate ##L##. That should be it.
     
  14. Mar 20, 2016 #13

    stevendaryl

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    Pick a rest frame [itex]F[/itex], and let's suppose that you have a light clock moving in the x-direction at speed [itex]v[/itex], according to frame [itex]F[/itex], and it is oriented perpendicularly to its direction of motion, in the y-direction. Let a pulse of light go from one side of the clock, at [itex]y=0[/itex], to the other side, at [itex]y=L[/itex]. The position of the first mirror as a function of [itex]t[/itex] is given by [itex]x=vt, y=0[/itex]. The position of the second mirror is [itex]x=vt, y=L[/itex]. So if it takes [itex]T[/itex] seconds to travel from one mirror to the other, then the event of leaving the first mirror has coordinates

    [itex]x=0, y=0, t=0[/itex]

    and the event of arriving at the second mirror has coordinates

    [itex]x=vT, y=L, t=T[/itex]

    So the total distance traveled by the light is: [itex]D = \sqrt{\delta x^2 + \delta y^2} = \sqrt{v^2 T^2 + L^2}[/itex]. Since light travels at speed [itex]c[/itex], it must be that [itex]D = cT[/itex]. So we have:

    [itex]cT = \sqrt{v^2 T^2 + L^2}[/itex], or [itex]c^2 T^2 = v^2 T^2 + L^2[/itex]. Solving for [itex]T[/itex] in terms of [itex]L[/itex] and [itex]v[/itex] gives:

    [itex]T = L/\sqrt{1 - v^2/c^2}[/itex]
     
  15. Mar 20, 2016 #14

    Ibix

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    Nitpick: this is out by a factor of c.
     
  16. Mar 20, 2016 #15

    DAC

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    Is one metre the same in both frames?
     
  17. Mar 21, 2016 #16

    Orodruin

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    It is the distance light travels in 1/299792458 s regardless of the frame.

    That is not saying that an object which is one meter will be one meter in all frames though.
     
  18. Mar 21, 2016 #17

    DAC

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    Apart from length contraction are there any instances.
     
  19. Mar 21, 2016 #18

    Orodruin

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    Of what? Please separate the concepts of the length of an object being different in different frames from the fact that a meter (the unit!) is the same in all frames.
     
  20. Mar 21, 2016 #19

    DAC

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    Apart from length contraction does the length of an object differ in different frames?
     
  21. Mar 21, 2016 #20

    Orodruin

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    So what you are asking is essentially "apart from the fact that objects have different length in different frames, do objects have different lengths in different frames?" It is impossible to give a reasonable answer to this type of question.
     
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