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Deriving the time dilation? equation of Lorentz transformation

  1. Sep 2, 2014 #1
    1. The problem statement, all variables and given/known data
    It is not an official question but a request for pointers. I am trying to derive the Lorentz time equation to understand the intuition behind it. My math skills are not very good so it might be an obvious question for you. Please see below for my attempt.
    The variables are x, t in one inertial reference frame and x', t' in another inertial reference frame. c is the speed of light, γ is the gamma factor, and u is the relative velocity of the primed IRF, from the perspective the unprimed one.
    The official equation (the end solution I want to get) is:
    t'=(t-(ux/c^2))/sqrt(1-(u/c)^2)

    2. Relevant equations
    I'm using the equations:
    γ=1/sqrt(1-(u/c)^2)
    x'=γ(x-ut)
    x=γ(x'+ut')

    3. The attempt at a solution
    So far I've been able to isolate t' from the equation x=γ(x'+ut') up to this state:
    t'=((x/γ)-x')/u
    I've tried substituting ct for x, substituting γ for 1/sqrt(1-(u/c)^2), and other substitutions that by now I've forgotten. It doesn't seem to come to the format I want in the official equation, mainly because I can't seem to isolate t in the RHS or get rid of the u in the denominator. I've been stuck on it for days :( I hope this makes sense!

    Any attempts to explain the intuition behind this equation/help me visualize it is welcome too, I've asked my instructor on it but somehow he seems shocked I would ask this question and told me to just try to derive it...

    Thanks for reading this over, hope everything actually makes sense!!
     
  2. jcsd
  3. Sep 2, 2014 #2
    A good way to "derive" this equation is to visualize it on paper by a series of diagrams.

    First consider a train moving at a speed v < c along the x direction. Inside the train is a set of mirrors on the ceiling and floor between which a photon is bouncing up and down in the frame of reference off the train. (Ignore absorption of photon etc.)
    In a given time t the photon will bounce from the bottom to the top back to the bottom, and travel a given distance d, and will travel at c.


    If there were a "stationary" observer outside the train watching the photon, it would take a given time t' for the photon to travel from the bottom to the top to the bottom again, and it would cover a given distance d', and it would still move at a speed c.

    Draw these two scenarios and use Pythagoras, and note that c is the same in both frames (invariance of the speed of light in any inertial frame) and you will "derive" your answer.
    After this try developing the result from your given equations.

    Here is a link to this sort of thought experiment, but I urge you to try it and visualize it yourself first before looking at the link.



    http://galileo.phys.virginia.edu/classes/252/srelwhat.html
     
  4. Sep 2, 2014 #3

    td21

    User Avatar
    Gold Member

    t'=((x/γ)-x')/u
    and substitute x'=γ(x-ut) in the above equation.
    After simplifying i think you can get t'=(t-(ux/c^2))/sqrt(1-(u/c)^2)
     
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