- #1
Ken Miller
- 26
- 1
Homework Statement
A particle has speed u‘ in the S’ frame, its track making an angle ## \theta’ ## with the x’ axis. The particle is viewed by an observer in frame S, the two frames having a relative speed parameter ## \beta ##.
(a) Show that the angle ## \theta ## made by the track of the particle with the x-axis is given by
$$ \tan \theta = \frac {u’ \sin\theta’} { \gamma \{u’ \cos\theta’ + \beta c\} } $$
(b) Show that this reduces to the standard aberration formula $$ \tan \theta = \frac {\sin \theta’ \sqrt{1-\beta^2} } {\cos \theta’ + \beta} $$ if the “particle” is a light pulse, so that ## u’ = c##.
Homework Equations
Given in problem statement
The Attempt at a Solution
Solving (b) is easy. But (a) is a mystery to me. I think I see how the authors get there, but it seems too simple. I can get there if I argue as follows:
I believe the authors mean that the relative movement of the two frames is along the common x-x’ axis, though they aren’t explicit. In that case the y-component of velocity is the same as the y’-component of velocity; i.e. ## u_y = u’_y = u’ \sin \theta’.## Also the x-component of velocity is some combination of the relative velocity of the frames (## \beta c##) and the x’-component of velocity (## u’ \cos \theta’##); here is seems clear that the authors just added ## u’ \cos \theta’## and ## \beta c## and expanded by multiplying by ##\gamma##. But I am puzzled as to why that is the right thing to do. After all, why shouldn’t the two velocities be added using the relativistic velocity-addition formula: $$ u_x = \frac {u’_x \cos \theta’ + \beta c} {1+ \frac {u’ \cos \theta’ \beta c} {c^2} }? $$ Or am I misunderstanding completely what they have done to get (a)?
I also considered trying to derive (a) in the same way the authors derived (b). But they did that assuming a light wave and making use of the fact that that ## \lambda f = \lambda’ f’ = c.## But in (a), we have particle motion, not wave propagation (we haven’t gotten to quantum mechanics yet), so it seems to me that I cannot treat this as a propagating wave.