Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Why does the azimuthal angle, $\phi$, remain unchanged between reference frames in special relativity?

Edit: Here is a link from "Radiative Processes in Astrophysics": http://i937.photobucket.com/albums/ad216/Kyanise/Capture-5.jpg.

I think this comes from the aberration formula, showing dependence only on the polar angle, $\theta$. The aberration formula is:

$\tan \theta = \frac{u_\perp}{u_{||}} =\frac{ u' \sin \theta '}{\gamma (u' \cos \theta ' + v)}$

where ' indicates a property of the moving frame.

But I'm not even sure why this is true. I know it comes from the transforms of the velocities, but why should $\tan \theta = \frac{u_\perp}{u_{||}}$ anyways? I mean $v$ can be in any direction, right? How come there is no dependence on $\phi$?

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

Actually this happens because we chooses our coordinates in that way, if you will review how aberration formula was derived you will notice that they was choosing the observer velocity to be perpendicular to XZ plane, this is done solely to simplify calculations, you may test that by your hand.

For more intuitive approach for that, is if you are familiar with the basics of Electrodynamics (the source of special relativity historically) you may know that $\vec{H},\vec{E}$ are always in the same plane, and if the observer will move in a perpendicular orientation to that plane, this will cause those vectors to rotate (the angel between them will change) but they will remain in the same plane.

share|improve this answer
    
Ok right this was just confusing me since if we know that there is always a coordinate system in which we can boost in the $x$ direction, then it was a bit strange that we would need to derive a general equation for when $v$ is not in the $x$ direction –  Atreyu Dec 2 '12 at 1:08
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.