### Lorentz Transform Summary and Levi-Cevita symbols: EM II Notes 2014_08_06

Summary of what's gone on before.  In the last two days, unbeknownst to me at the time, we first showed that the Lorentz/Fitzgerald contraction only happens in the direction of the velocity of a frame.  Coupling this with rotational symmetry, yesterday, it was shown how the vector version of the Lorentz transform could be derived.  Yesterday's treatment needs to be repeated from end to start to truly  count as a derivation though, I suppose.

Picking up where I left off yesterday, the second half of the vector Lorentz transform, the expression for time in the primed frame can be written as

$t^{\prime} = \gamma\left(t - \vec{v}\cdot\vec{r}\right)$

Once again, this shows that only the component of the position vector parallel to the velocity gets to play.  However, here you should notice that the velocity vector is not normalized, not divided by it's magnitude.  In other words, while only the component of the position vector parallel to the velocity gets to play, the entire magnitude of the velocity gets to play.  This is also noted in the notes where the expression for $\gamma$ in this case is

$\gamma = \left(1 - v^2\right)^{1/2}$

Finally, in the section 1.2 ending on page 9, it's observed that what will become known as the spacetime line element is invariant between frames

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

Consequently, we also just introduced hyperbolic geometry.

Index Notation
The first really interesting thing here is the totally-antisymmetric tensor.  I think this might be called the Levi-Civita symbol elsewhere.  I need to look it up.

Here's the definition:

$\epsilon_{ijk} = -\epsilon_{ikj} = -\epsilon_{jik} = -\epsilon_{kji}$.

Then, it only remains to define the value of $\epsilon_{123} = 1$ and we get that

$\epsilon_{123} = \epsilon_{231} = \epsilon_{312} = 1$, $\epsilon_{132} = \epsilon_{321} = \epsilon_{213} = -1$

With our new symbol, we can write down the vector cross product more concisely,

$\left(\vec{A} \times \vec{B}\right) = \epsilon_{ijk}A_jB_k$

Hunh, and now I have a question.

QUESTION:  Why is swizzling the indices between the following two versions of the same equation 'convenient'?

$\left[\vec{A}\times\left(\vec{B} \times \vec{C}\right)\right]_i = \epsilon_{ijk}\epsilon_{klm}A_jB_lC_m$

and the 'more convenient version'

$\left[\vec{A}\times\left(\vec{B} \times \vec{C}\right)\right]_i = \epsilon_{ijk}\epsilon_{lmk}A_jB_lC_m$

Memorize the following:

$\epsilon_{ijk}\epsilon_{lmk} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}$

QUESTION:  I'm guessing the second 'more convenient' version above was handy in deriving the Levi-Civita to delta function identity above?

### 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…