- #1
SilasG
- 5
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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
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?
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
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?
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