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I Time dilation in relativistic Doppler effect

  1. May 23, 2016 #1
    Hello,
    It kind of bothers me that the derivation for the Lorentz transformation relies on two dimensions of space. (Here I am referring to the standard derivation where one person is using a vertical light clock in a trolley traveling horizontally at speed v, and an observer outside is watching). My problem is that this derivation relies on the Pythagorean theorem which is an equation only meaningful in (at least) 2 spatial dimensions. So, how can the Lorentz transformation (and the resulting formula for time dilation) be used in one-dimensional derivations (e.g., the 1D relativistic Doppler effect)?
     
  2. jcsd
  3. May 23, 2016 #2
    I think the idea applies to N dimensions, but 2 is as much as I can cope with.
     
  4. May 23, 2016 #3

    Orodruin

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    What you are referring to is just a heuristic derivation. It is perfectly possible to derive it directly from the assumption that the light speed is invariant.
     
  5. May 24, 2016 #4

    Nugatory

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    Those aren't 1D derivations. They just look like they are because we've chosen the axes so that ##y'=y## and ##z'=z## so we don't have to do anything with them.
     
  6. May 24, 2016 #5

    Dale

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    Easy. You do the derivation of the transform in the full 1+3 D spacetime. By doing so you find that the formula for two of the space dimensions is trivial. So then you work problems in 1+1 D for simplicity.

    The Doppler effect isn't inherently 1+1 D.
     
  7. May 24, 2016 #6

    stevendaryl

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    If you start with the assumption that the coordinates are linearly related:

    [itex]x' = A x + B t[/itex]
    [itex]t' = D x + E t[/itex]

    and you insist that:
    1. Light has speed [itex]c[/itex] in both coordinate systems (so [itex]x= \pm ct \Rightarrow x' = \pm c t'[/itex])
    2. The spatial origin of one coordinate system moves at speed [itex]\pm v[/itex] relative to the other coordinate system (so [itex]x' = 0 \Rightarrow x = vt[/itex] and [itex]x=0 \Rightarrow x' = -vt'[/itex])
    Then this implies the transformation:

    [itex]x' = A (x - vt)[/itex]
    [itex]t' = A (t - \frac{vx}{c^2})[/itex]

    with the inverse:
    [itex]x = \frac{\gamma^2}{A} (x' + vt')[/itex]
    [itex]t = \frac{\gamma^2}{A} (t' + \frac{vx'}{c^2})[/itex]

    (where [itex]\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}[/itex])

    Any choice other than [itex]A = \gamma[/itex] would spoil the symmetry between forward and inverse transforms. A little more of an argument is needed to say what's wrong with that.
     
  8. May 24, 2016 #7
    Thank you very much to all. I knew there had to be a way of deriving gamma in 1+1 D but I was not sure how.
     
  9. May 25, 2016 #8
    Does not seem to be a very satisfying proof.

    First it starts with a presupposition that coordinates are linearly related (treating this as a postulate?).
    Second it never invokes the second of the SR postulates that the laws of physics are the same in all inertial frames of reference.
     
  10. May 25, 2016 #9

    Orodruin

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    As was mentioned, this follows directly from the assumption of homogeneity, although the argumentation is more involved. The argument is presented in, eg, Rindler if I recall correctly.

    Furthermore, it is exactly the same assumption that is made in deriving the Galilei transformation. The only difference lies in the second postulate. In the Galilean case that t=t' and in the Lorentz case that the speed of light is invariant.
     
  11. May 25, 2016 #10

    Orodruin

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    The principle of relativity is used in finding the linear form ...
     
  12. May 25, 2016 #11
    Exactly.
    I would think that a proper proof would start with the principal of relativity, and if the linear relationship follows from that then fine. Presupposing the linear relationship without demonstrating how you get there from the principal of relativity seems like skipping a step in the proof.
     
  13. May 25, 2016 #12

    stevendaryl

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    You can look up dozens of proofs online. I was just sketching the derivation.

    People have ridiculous (in my opinion) expectations for Physics Forums posts.
     
  14. May 25, 2016 #13
    Agreed, all of this can be found online but the OPs original issue was specifically that Lorentz transformation derivations he was seeing depend on assumptions outside of the postulates of SR, such as the need for 2 dimensions for the benefit Pythagorean theorem. Seemed like we just provided another derivation with different non-SR postulates.
     
  15. May 25, 2016 #14

    stevendaryl

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    Well, I was just pointing out that the 2 spatial dimensions is not an essential assumption.
     
  16. May 25, 2016 #15

    Mister T

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    You can use a light clock where the light beam moves parallel to the motion. In that case you are using only one dimension of space.

    The derivations you usually see are the ones that are the easiest to follow. The analysis of the traditional light clock is easier to follow because it's more straight forward.
     
  17. May 25, 2016 #16

    Orodruin

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    In this approach, the light clock parallel to the direction of motion is usually used to derive the length contraction. You can only get either length contraction or time dilation from this consideration, not both.
     
  18. May 25, 2016 #17

    robphy

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  19. May 25, 2016 #18
    Just to clarify, my original issue was that 2 spatial dimensions are used in the derivation, not that well-established facts (such as the Pythagorean theorem) are used in the derivation.

    I'm a little confused by this statement. Once you have time dilation, can you not use the fact that speed of light is constant to immediately derive length contraction?
     
  20. May 25, 2016 #19

    stevendaryl

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    Here's the issue: We pick one coordinate system, and call it the "stationary" frame. We pick another coordinate system that is moving relative to the "stationary" frame, and call that the "moving" frame. One way to derive the Lorentz transformations is by insisting that the observer in the "moving" frame uses "light-based measurements". The way you set up light-based measurements is this:
    1. Use a light clock to measure time.
    2. Measure distances between objects that are at rest in your frame by using radar: You bounce a light signal off the distant object, and time the round-trip. Divide the time by two to get the distance in light-seconds.
    3. Synchronize distant clocks that are at rest in your frame by using light synchronization: Pick one clock to be the standard. Note the time on that clock, and send a signal to the distant clock. Set the distant clock to the time on the standard clock, plus the travel time (computed using the distance between clocks).
    This procedure gives you a coordinate system for the moving frame, which you can relate to the coordinate system in the stationary frame. But it doesn't quite give you the Lorentz transformations. That's because there is one unknown parameter, namely the rate of the moving light clock as measured by the stationary observer. So that gives you an unknown time dilation factor. Since you use light clocks to measure distances, this unknown time dilation factor gives rise to an unknown length contraction factor.

    I'm going to skip the tedious details, but it turns out to be the following:

    The moving light clock runs slower by a factor [itex]F[/itex] as measured in the stationary frame.
    A moving measuring rod (oriented in the direction of motion) that has length 1 meter in its own rest frame has length [itex]\frac{1-\frac{v^2}{c^2}}{F}[/itex] as measured in the "stationary" frame.

    Light-based measurements alone don't allow you to compute [itex]F[/itex]; it could be anything. However, if you use light-clocks that are oriented perpendicular to the direction of motion, and insist there is no length contraction in that direction, then you can uniquely compute [itex]F[/itex]:

    [itex]F = \sqrt{1-\frac{v^2}{c^2}}[/itex]

    Which also gives the length contraction [itex]\frac{1-\frac{v^2}{c^2}}{F} = \sqrt{1-\frac{v^2}{c^2}}[/itex]
     
  21. May 25, 2016 #20

    Mister T

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    Well, you have to use constancy of light speed to derive time dilation. But if you then try to derive length contraction using the same apparatus (the unconventional light clock with the light moving parallel to the direction of the clock's motion) your reasoning will be circular. At least I think that's right.
     
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