Aberration of a Particle: Solving for Angle and Velocity in Relativistic Frames

[Ken Miller]

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.

 

[Orodruin]

In that case the y-component of velocity is the same as the y’-component of velocity; i.e. uy=u′y=u′sinθ′

This is incorrect. The y-component is frame dependent.

Do you know the 4-vector formalism? Otherwise you will have to use the Lorentz transformations.

 

[Ken Miller]

Oh...as soon as I saw your statement that the y-component is frame dependent, I realized that this is much easier than I thought. Anyway, I'm only vaguely familiar with 4-vectors, but I think I think I see how this can be done with Lorentz transformations. So, I think that part of where I was going wrong was in thinking that the $\gamma$ in the solution was connected with the x-component of velocity (simply because I saw it in the denominator). Instead it's connected with the y-component, right? That is, by using the Lorentz transformation on $\frac {\delta y} {\delta t},$ you get $$ u_y = \frac {u'_y \sqrt{1-\beta^2} } {1+\frac{\beta c u'_x} {c^2}},$$ I believe. And I believe that I am correct about the x-component in my Attempt at Solution (above). Then $$ \tan \theta = \frac {u_y} {u_x}, which yields the answer. Am I understanding now?
P.S. And thank you for your assistance. You folks are great!