Relativity: Calculating Separation r' between Emission and Reception

  • Thread starter Thread starter Void123
  • Start date Start date
  • Tags Tags
    Relativity
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 2K views
Void123
Messages
138
Reaction score
0

Homework Statement



A light pulse is emitted at a position [tex]x_{A}[/tex] (horizontally) and is received at position [tex]x_{B} = x_{A} + r[/tex]. Considering that [tex]v = \beta c[/tex] for a moving reference frame, I must calculate the separation [tex]r'[/tex] between the point of emission and reception.

Homework Equations



[tex]x_{B} - x_{A} = \gamma(x_{B}' - x_{A}')[/tex]



The Attempt at a Solution



I used the above equation, solving for [tex]x_{B}' - x_{A}'[/tex], but the answer provided by the book gives me an answer that differs from mine by a few operation signs. But I cannot think how they got it.
 
Physics news on Phys.org
This is precisely what I have written above:

[tex]l_{moving} = \frac{l_{rest}}{\gamma}[/tex]
 
Okay, for the time:

[tex]t_{2} = t_{1}'\sqrt{\frac{1 - \beta}{1 + \beta}}[/tex]

Essentially, this is the doppler effect.

Can I rewrite the times in terms of [tex]l[/tex] and [tex]l'[/tex] to get the answer I want?