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Homework Help: Lorentz transformations

  1. May 13, 2007 #1
    Show that, with V = 4/5c, the Lorentz transformation of the equations, t^prime = y(V) (t-(v/c^2)x) and x^prime = y(V) (x-Vt). (where y(V) = the Lorentz factor).

    can be written as

    ct^prime = 5/3ct - 4/3x


    x^prime = 5/3x - 4/3ct

    Relevant equations
    y(V) = 1/(sqrt1-(V/c)^2)

    The attempt at a solution
    I have calculated y(V) = 5/3 (if V = 4/5c)
    and i can see how the left hand term in each equation becomes 5/3ct and 5/3x respectivley. But i cant figure where the 4/3 term comes from?? or how to derive it?
  2. jcsd
  3. May 13, 2007 #2
    The second term on the RHS in both equations contain a 'v', right?
  4. May 13, 2007 #3

    George Jones

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    [tex] \left( \frac{5}{3} \right) \left( \frac{4}{5} \right) = \frac{4}{3}.[/tex]
  5. May 15, 2007 #4
    Ahh... the penny drops. thank you
  6. May 15, 2007 #5
    A further Lorentz transformation problem.

    The question i have is.

    Use these Lorentz transformations ct'=5/3ct-4/3x and x'=5/3x-4/3ct. to determine the (ct', x') coordinates, in meters, that observer O' assigns to events e1 and e2.

    Relevent equations and information.

    from a previous question i have determined the coordinates of the events in the rest frame of observer O to be e1 = (0, 240)m, e2 = (60, 240)m.

    My attempt at answer

    Using the given Lorentz transformations i have found for event e1 as observed by O' is.

    e1 = (ct', x') = (5/3ct-4/3x, 5/3x-4/3ct) = (-180, 144)m
    e2 = (ct', x') = (5/3ct-4/3x, 5/3x-4/3ct) = (-144, 99) m

    However i feel uncomfortable with these answers, but cannot put my finger on why? Please can some one check my results?

    thank you
  7. May 15, 2007 #6
    Further to my last post, the reason i feel uncomfortable is that surely the x component of the coordinates should be the same for both events? I can check this via the Lorentz length contraction formula, which gives l = lo/y(V) =144?
  8. May 15, 2007 #7
    It seems to be errors in arithmetic to me.

    Surely, then, both observers must be referring to the same reference frame.
  9. May 15, 2007 #8
    yep your right i have just spotted my error. i think i have also calculated the transformed event coordinations incorrectly as well?
    Last edited: May 15, 2007
  10. May 15, 2007 #9
    Sorry about my previous post. I misread your statement, and gave a stupid reply (eyes can play tricks on you late at night!). I really cannot say whether the x coordinates of the two events in the O frame are the same or not, unless I have more information. But if you have managed to find you errors in spite of my comment, then well and good.
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