Retarded Potentials - variable change to actual position.

[JesseC]

Homework Statement

A charged particle is moving at velocity \vec{v}=c\vec{\beta} along the z-axis. We're working in cylindrical co-ordinates. Here's a picture:

[PLAIN]http://img696.imageshack.us/img696/9789/retardedpotential.png

The problem is to get the Lienard-wiechert potential, which is normally analysed at the retarded time, in terms of the vector \vec{R_p} which points from the actual position of the particle. The point P is where we're analysing the field.

Homework Equations

L-W potential at retarded time:
V=\frac{1}{4 \pi \epsilon_0}\frac{q}{R_r(1-\vec{\beta}\cdot\hat{R_r})}

The Attempt at a Solution

I'm following through a solution to this problem, and out of the blue comes this statement. "Perpendicular components are equal such that:"

|\vec{R_r} \times \vec{\beta}|^2=|\vec{R_p} \times \vec{\beta}|^2

Now it isn't immediately obvious to me why this is true, can anyone shed some light on this?

 

[fzero]

It's true because, from the diagram,

\vec{R}_r = \vec{R}_p + \vec{\beta}.