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:

### Cool Math Tricks: Deriving the Divergence, (Del or Nabla) into New (Cylindrical) Coordinate Systems

The following is a pretty lengthy procedure, but converting the divergence, (nabla, del) operator between coordinate systems comes up pretty often. While there are tables for converting between common coordinate systems, there seem to be fewer explanations of the procedure for deriving the conversion, so here goes!

What do we actually want?

To convert the Cartesian nabla

to the nabla for another coordinate system, say… cylindrical coordinates.

What we’ll need:

1. The Cartesian Nabla:

2. A set of equations relating the Cartesian coordinates to cylindrical coordinates:

3. A set of equations relating the Cartesian basis vectors to the basis vectors of the new coordinate system:

How to do it:

Use the chain rule for differentiation to convert the derivatives with respect to the Cartesian variables to derivatives with respect to the cylindrical variables.

The chain rule can be used to convert a differential operator in terms of one variable into a series of differential operators in terms of othe…

### The Valentine's Day Magnetic Monopole

There's an assymetry to the form of the two Maxwell's equations shown in picture 1.  While the divergence of the electric field is proportional to the electric charge density at a given point, the divergence of the magnetic field is equal to zero.  This is typically explained in the following way.  While we know that electrons, the fundamental electric charge carriers exist, evidence seems to indicate that magnetic monopoles, the particles that would carry magnetic 'charge', either don't exist, or, the energies required to create them are so high that they are exceedingly rare.  That doesn't stop us from looking for them though!

Keeping with the theme of Fairbank[1] and his academic progeny over the semester break, today's post is about the discovery of a magnetic monopole candidate event by one of the Fairbank's graduate students, Blas Cabrera[2].  Cabrera was utilizing a loop type of magnetic monopole detector.  Its operation is in concept very simpl…

### Unschooling Math Jams: Squaring Numbers in their own Base

Some of the most fun I have working on math with seven year-old No. 1 is discovering new things about math myself.  Last week, we discovered that square of any number in its own base is 100!  Pretty cool!  As usual we figured it out by talking rather than by writing things down, and as usual it was sheer happenstance that we figured it out at all.  Here’s how it went.

I've really been looking forward to working through multiplication ala binary numbers with seven year-old No. 1.  She kind of beat me to the punch though: in the last few weeks she's been learning her multiplication tables in base 10 on her own.  This became apparent when five year-old No. 2 decided he wanted to do some 'schoolwork' a few days back.

"I can sing that song... about the letters? all by myself now!"  2 meant the alphabet song.  His attitude towards academics is the ultimate in not retaining unnecessary facts, not even the name of the song :)

After 2 had worked his way through the so…