Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Lorentz transformations ( synchronising reference frames?)

  1. Apr 12, 2010 #1
    Lorentz transformations ("synchronising" reference frames?)

    1. The problem statement, all variables and given/known data

    A particle moves from (x,y,z,t) = (0 m,0 m,0 m,0 s) to (1 m,1 m,0 m,10 ns).
    1. i. What is the speed of the particle in this reference frame?
    2. ii. What is the speed of the particle in a reference frame moving along the x axis at 0.7c?

    2. Relevant equations

    The usual Lorentz transformation equations:
    • β ≡ V/c
    • γ ≡ 1/sqrt(1 − β2)
    • x′ = γ(x − tβc)
    • t′ = γ(t − xβ/c)
    • x = γ(x′ + t′βc)
    • t = γ(t′ + x′β/c)
    • c ≈ 3×108 ms−1

    3. The attempt at a solution

    I understand that the answer to part (i) is just 1.414×108 ms−1, that's just simple trigonometry. However I do not understand how to do part (ii).

    My lecturer gives the following answer to the problem:
    But I don't understand the answer. What does he mean by "synchronise clocks" and how did he calculate those answers? (unfortunately my lecturer is away for another 2 weeks so I can't ask him!)

    Thank you
  2. jcsd
  3. Apr 12, 2010 #2
    Re: Lorentz transformations ("synchronising" reference frames?)

    By synchronizing clocks and rulers, he meant that in both frames, the particle passes through the origin at time 0.

    What he then did was transform the second event (The end of the particle's path in the original frame) to the moving frame.

    y'=y, so 1=1
    x'=γ(x − tβc) = 1.4(1-2.1) = -1.54 m
    t'=γ(t - xβ/c) = 10^-8 *(1.4(10-2.3)) = 10.7 ns

    So the velocity is just the path divided by the time.

    Another way to approach the problem would be the velocity addition formula.

    [tex]v_x ' =\frac{v_x - V}{1-\frac{v_x V}{c^2}}[/tex]
    [tex]v_y' = \gamma{v_y'}[/tex]
    Where [tex]V[/tex] is the velocity of the primed frame relative to the unprimed frame.

    Which immediately gives:
    [tex]\beta_x ' = -0.34[/tex]

    [tex]\beta_y ' =0.23[/tex]

    [tex]\beta ' = 0.57[/tex]
    Last edited: Apr 12, 2010
  4. Apr 12, 2010 #3
    Re: Lorentz transformations ("synchronising" reference frames?)

    Thank you, RoyalCat - makes sense now! Those velocity addition formulae are new to me...
  5. Apr 12, 2010 #4
    Re: Lorentz transformations ("synchronising" reference frames?)

    No problem. ^^

    You'll probably come across them in the next chapter or two.

    A quick rundown of how they're developed:

    x′ = γ(x − tβc)
    t′ = γ(t − xβ/c)

    dx′ = γ(dx − (dt)βc)
    dt′ = γ(dt − (dx)β/c)

    The x velocity in the primed frame is, by definition:

    [tex]v_x' = \frac{dx'}{dt'}[/tex]

    Plugging in those expressions:

    [tex]v_x' = \frac{\gamma(dx-dt\beta c)}{\gamma(dt-dx\frac{\beta}{c})}[/tex]

    Some algebra massage (Dividing denominator and numerator by dt, and recalling that [tex]v_x=\frac{dx}{dt}[/tex] by definition):

    [tex]v_x'=\frac{v_x-\beta c}{1-\frac{v_x\beta}{c}}[/tex]

    Which recalling that [tex]\beta\equiv \frac{V}{c}[/tex] turns into:

    [tex]v_x'=\frac{v_x-V}{1-\frac{v_x V}{c^2}}[/tex]

    If you're not familiar with differentials, or if you haven't studied the subject as a class, there's another way to approach the derivation, using time dilation and length contraction, or some some other lie. :)

    Since we're dealing with constant velocities throughout, we could have just as well taken a non-infinitesimal difference [tex](\Delta)[/tex] and come upon the same result.
  6. Apr 12, 2010 #5
    Re: Lorentz transformations ("synchronising" reference frames?)

    Superbly explained, thank you muchly! You are awesome! :)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook