## Wednesday, October 29, 2014

Here's today's special relativistic EM question.  Can the Thomas precession be shown to be a special case of the perihelion advance of relativistic elliptical orbits?  Any ideas?  Here's what's going on:

We've been deriving the special relativistic  orbit of a charged particles around another fixed charged particle.  At the end of the day, you wind up with a perihelion advance which is a fancy way to say that major axis of the elliptical orbit won't stay put.  It swivels around, (orbits), the charged particle as well.  The advance angle of the major axis winds up being\\

$\delta\phi = 2\pi\left[\left(1 - \dfrac{\kappa^2}{l^2}\right)^{-1/2} - 1\right]$

Which is very, very, similar to the Thomas angle for the spin precession, or gyroscopic precession along a circular orbit at special relativistic speeds:\\

$\delta\phi = 2\pi\left[cosh\left(w\right) - 1\right]$

$= 2\pi\left[\left(1 - \dfrac{v^2}{c^2}\right)^{-1/2} - 1\right]$

In the expression for the perihelion advance, $\kappa = eQ$ where $e$ is the charge of the orbiting particle and $Q$ is the charge of the fixed particle.  $l$ is the angular momentum and is written as $l = mr^2\left(\dfrac{d\phi}{d\tau}\right)$

Just a little more massaging of the above before I'm done for the day.  For a circular orbit, $\dfrac{d\phi}{d\tau} = \omega$ is constant and we can write $l = m\omega r^2$.  If we plug this in as to the Perihelion advance equation we wind up with

$\dfrac{\kappa^2}{l^2} = \dfrac{Q^2 e^2}{m^2v^2r^2}$ where $v = \omega r$

Here's where I get into trouble for playing fast and loose with things that migh actually be rapidities instead of velocities like $v$ above.  However, if you carry the simplificatins a bit further out, I believe you wind up with something that looks like an units of potential energy over momentum squared.

$\dfrac{Q^2 e^2}{m^2v^2r^2} = \dfrac{F}{m} \dfrac{1}{mv^2} \sim \dfrac{a}{E}$

Notes du jour: