### EMII Notes: Special Relativity and Gamma via Initial Conditions

I'm working on a new writing project as I prepare for Electricity and Magnetism II this semester.  I'll be reviewing the class lecture notes each day leading up to the start of the semester.  After each half hour review, I intend to write up a summary of the notes.  (Yes, I took a reflective writing workshop this summer, and I'm trying it out.)

Pages covered today:  7 & 8

We consider light as a spherical wavefornt

$x^2 + y^2 + z^2 - c^2t^2 = 0$

$x^{\prime 2} + y^{\prime 2} + z^{\prime 2} - c^2t^{\prime 2} = 0$

We also look at the usual situation with two frames, $S$ and $S^{\prime}$ that are moving parallel to each other where $S^{\prime}$ has positive velocity $v$ in the x direction with respect to $S$.

We have the usual postulate that c, the speed of light is constan in all frames.

We need a way to transform the distance and time coordinates in the two frames.  Because we want to preserve invariance with respect to translation, (conservation of momentum), and invariance with respectc to translations in space-time, (conservation of energy), we're stuck with a linear transformation.  so

$x^{\prime} = Ax + Bt$

$t^{\prime} = Cx + Dt$

Question:  Is the above an example of a bilinear transform?

We won't really bother with C and D I don't think.

At time t = 0, we set the origins of both frames at the same position and get

$x^{\prime} = 0 => x = vt$

so

$0 = Avt + Bt$
$B = -Av$

Now, since we know where we're headed, go ahead and rename $A$ to be $\gamma$.

This all gives:

$x^{\prime} = \gamma x - \gamma v t$
$x^{\prime} = \gamma\left( x - v t\right)$

For the $S$ frame at $x = 0$, we have $x^{\prime} = -vt^\{prime}$

We can use this reversed velocity to do a nice little trick giving

$x = \gamma\left( x^{\prime} - v t^{\prime}\right)$

without doing any math to solve for $x$ in terms of $x^{\prime}$

Since the speed of light is constant in every frame.  That's our postulate.  we get

$x = ct <==> x^{\prime} = ct^{\prime}$

Plugging this into the $AB$ transform equation above we get

$ct^{\prime} = \gamma x - \gamma v t$
$ct^{\prime} = \gamma\left(ct - vt\right)$

and the following string of algebra from solving the two equations above for gamma and then multiplying them together.

$\gamma = \dfrac{ct}{t^{\prime}\left(c+v\right)}$
$\gamma = \dfrac{ct^{\prime}}{t\left(c - v\right)}$

multiply the two equations for gamma to get.  (Why?  Because I know the answer I'm headed for and this seems simplest).

$\gamma^2 = \dfrac{c^2 t t^{\prime}}{t t^{\prime} \left(c + v\right)\left(c - v\right)}$
$= \dfrac{c^2}{c^2 - v^2}$
$= \dfrac{1}{1-\dfrac{v^2}{c^2}}$
$\gamma = \dfrac{1}{\sqrt{1 - \dfrac{v^2}{c^2}}}$

Question:
Since we never used the $BD$ transform equation, is there more information that can be gleaned from it?

### 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…