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Relativistic Rapidity question

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  1. Dec 1, 2014 #1
    The question is as follows:
    We can define the rapidity, y, of a particle with respect to the x axis
    y≡tanh-1βx. Show that under a Lorentz transformation by rapidity yB
    y'=y-yB



    3. The attempt at a solution
    I started by working backwards (sorry if the LaTeX does not work
    $$ y'=y-y_B=\frac{1}{2}(\ln(\frac{1+\beta_x}{1-\beta_x})-\ln(\frac{1+\beta_B}{1-\beta_B}))=\frac{1}{2} \ln (\frac{1+\beta_x-\beta_B-\beta_x \beta_B}{1+\beta_B-\beta_x-\beta_B \beta_x})\\ $$
    But now I am stuck, what is the transform in this case?
     
    Last edited by a moderator: Dec 2, 2014
  2. jcsd
  3. Dec 2, 2014 #2

    Orodruin

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    How do velocities transform under Lorentz transformations? What does this tell you about the relation between ##\beta_B##, ##\beta_x## and the sought velocity ##\beta'##?

    You need to use two $ instead of one in order to get in and out of LaTeX mode. I fixed this in your original post, but it is a good thing to know.
     
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