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I Why y=y' in the Lorentz transformation?

  1. Oct 13, 2016 #1
    By y I mean any perpendicular axis to the axis in which there is movement.
     
  2. jcsd
  3. Oct 13, 2016 #2

    vanhees71

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    By definition, a boost in ##x## direction doesn't change components of four-vectors in ##y## and ##z## direction. For an introduction to special relativity, see my SRT manuscript (unfinished, but the basic concepts on Lorentz transformations are all in):

    http://th.physik.uni-frankfurt.de/~hees/pf-faq/srt.pdf
     
  4. Oct 13, 2016 #3
    Thanks for the response!
    By definition?
    I can Imagine length contraction in the y axis without it violating any definitions.
    The link doesn't work for me.
     
  5. Oct 13, 2016 #4

    vanhees71

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  6. Oct 13, 2016 #5

    Ibix

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    Can you really? If I have a 1m rule held perpendicular to my motion and so do you, and we are on a collision course - what happens when we collide? Whose rod is shorter? Why?
     
  7. Oct 13, 2016 #6

    PeroK

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    Can you come up with a simple thought experiment that shows that there cannot be length contraction in the ##y## direction when the relative motion is in the ##x## direction?
     
  8. Oct 13, 2016 #7
  9. Oct 13, 2016 #8
    I was about to say:
    "I can imagine that in my reference frame your ruler will be shorter and in your reference frame my ruler will be shorter."
    but then I realized that if I place a chalk on to top and bottom of each ruler the argument by symmetry makes perfect sense, because there should be only one result. Is this a valid thought experiment?
     
  10. Oct 13, 2016 #9

    PeroK

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    Yes, if the two rulers (with the same rest length) move towards each other "vertically", then they must, by symmetry, be of equal height as they pass each other. Each is a valid measurement of length for the other, because the two ends meet simultaneously (in both frames).
     
  11. Oct 13, 2016 #10

    Here’s one example of a paradox if length contraction occurred along axes other than the direction of motion:

    Consider these initial conditions
    1. A sphere with proper diameter 1 meter with position (x, y, z) = (-8, 0, 0) and velocity (vx, vy, vz) = (.866c, 0, 0) with respect to you the observer.
    2. A vertical plate with a 1.1 meter diameter hole centered on the origin at rest with respect to you.

    Using Relativistic Physics, the x-dimension if the sphere is 50% contracted but it still passes through the hole in the plate without interference.

    Now switch reference frames to one in which the sphere is at rest. The vertical plate is moving at velocity (vx, vy, vz) = (-.866c, 0, 0) toward the sphere. Standard Relativistic Physics says the sphere is a back to being a real sphere and plate is now half its thickness but the hole is still a 1.1 meter circle and the sphere still passes through the hole without interference.

    What if length contraction occurred in all 3 dimensions?
    Consider again the initial conditions. The sphere is contracted in all dimensions and again passes through the plate without interference.
    Now switch reference frames to one in which the sphere is at rest. The vertical plate is again moving at velocity (vx, vy, vz) = (-.866c, 0, 0) toward the sphere. But the plate is contracted in all dimensions requiring the hole to be contracted too. The sphere cannot make it through the hole. Paradox.
     
  12. Oct 15, 2016 #11

    Andrew Mason

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    y = y' follows from the Lorentz transformation for time, i.e. ##t' = \gamma (t - vx/c^2)##. An observer in the stationary frame measures a distance y in the direction perpendicular to the direction of motion (of an observer in a rocket ship) by sending a light signal from the origin (0,0) to a mirror located at position (0,y) and measuring the time, t, that it takes to return to the point (0,0). It takes t = 2y/c. But the observer in the moving rocket frame sees that same light signal travel a longer path over a time t' = 2d/c, where d is the hypotenuse of a right angle triangle whose base is v(t'/2) and whose height is y'. This is because, in the moving frame, the light is emitted at time t'=0 at a point on the x' axis (-vt'/2,0), travels to a point (0,y') and returns to a point (vt'/2,0) at a time t'.

    So let's compare y and y':

    (1) ##y = ct/2##
    (2) ##y' = \sqrt{d^2-(vt'/2)^2} = \sqrt{(ct'/2)^2 - (vt'/2)^2)} = (ct'/2)(\sqrt{1-v^2/c^2}) = (ct'/2)/\gamma##

    Since x = 0, ##t' = \gamma t##. So substituting this for t' in (2) results in:

    (3) ##y' = c(\gamma t)/2\gamma = ct/2 = y##

    AM
     
    Last edited: Oct 16, 2016
  13. Oct 24, 2016 #12
    Andrew Mason's example is pretty good.
     
  14. Oct 24, 2016 #13
    Think of the Lorentz transformation with only one dimension of space, x.

    Now, consider a photon moving in the x and y directions. What transformation would allow the photon to still posses a speed of c? y' = y

    This is equivalent to Andrew Mason's example.
     
  15. Oct 24, 2016 #14
    Most teachers jump the gun by assuming that y' = y and use it to derive time dilation, working backwards, which ends up confusing students more ironically.


    image017.gif
     
  16. Oct 24, 2016 #15

    PeroK

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    Although, assuming the Lorentz Transformation not only jumps the gun but jumps half way to the finishing line!
     
  17. Oct 24, 2016 #16
    haha, but you see, teachers don't want to derive the LT the proper way, so they just skip it and use a light clock.

    the LT itself is not a jump but based on Einstein's postulates.
     
  18. Oct 24, 2016 #17

    robphy

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    Well.... it's not an unreasonable assumption since it is already there in the Galilean transformation and there has been no issues of length contraction in the Galilean case. It's only after deducing length contraction along the direction of relative motion in special relativity that one goes back and questions the assumption in the transverse direction.
     
  19. Oct 24, 2016 #18
    hmmm, I'm not so sure. there is an issue in the case of the light clock, which is time dilation being a result of the height remaining the same.

    conversely, if we assumed that there was no time dilation then there would be a height contraction of the light clock.
     
  20. Oct 24, 2016 #19

    robphy

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    The issue of time-dilation arises because one is invoking the principle of relativity with the light clock....
    that the round-trip of the light signal in that light clock marks one tick for that clock, as it would for a clock at rest,
    but that elapsed time measured by "the clock at rest" is different than that of "the clock at rest".

    While one may be so unhappy with this time-dilation result that one tries to use height-contraction to avoid it,
    the arguments above show that no such height-contraction exists (as in the Galilean case, which we already presumed)... since to do so violates the principle of relativity. [I guess one could have also questioned the principle of relativity... and be led to consider other implications... ultimately being tested by experiment.]

    So, my point is that consideration of transverse contraction is an afterthought... since transverse-noncontraction is already practically presumed in the Galilean case.

    edit: I guess one could have started from scratch altogether [not starting with Galilean physics]... then one has many more assumptions to question.
     
  21. Oct 24, 2016 #20
    Although there is no height contraction in the GT, there is no length contraction either.
    Ultimately, it has been shown that there should be no height contraction, though I don't agree that that is a reasonable assumption to make from the start.
    The simplest assumptions would be Einstein's two postulates.
     
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