### Rotating Cross Product Inputs Rotates the Outputs, EMII Notes 2014_08_11

What's Gone Before and What Will Ensue:  Yesterday, he first step of an exercise regarding the rotation of cross products was worked out.  Today, the identity proved yesterday will be used to show that when the two vectors in a cross product are rotated by the same rotation matrix, the resulting vector of the cross product is rotated by the same rotation matrix.  In the end, yet another property will be proven using tensor index notation.

The identity from yesterday is:
$\epsilon_{ijk}W_{iq}W_{jl}W_{km} = det\left(W\right)\epsilon_{qlm}$

We'll also need the definition of the cross product in index notation
$\vec{A} \times \vec{B} = \epsilon_{ijk}A_jB_k$

and the rotation matrix transpose identity
$M^TM = 1$ also known as $M_{iq}M_{in} = \delta_{qn}$

We want to prove that if $\vec{A}$ and $\vec{B}$ are both rotated by the same rotation matrix, $M_{in}$, then so is the result of the cross product, $\vec{V}$

First, rotate the two input vectors

$\vec{A^{\prime}} \times \vec{B^{\prime}} = \epsilon_{ijk}M_{jl}A_l M_{km}B_m = V^{\prime}$

We'd like to show that $\vec{V}^{\prime} = M_{in}V_n$

Go ahead and substitute the desired result in for the right hand side now.  We'll see if we can make the two sides equal each other, completing the proof.

$\epsilon_{ijk}M_{jl}A_l M_{km}B_m = M_{in}V_n$

Now, rotate each side one more time using the same rotation matrix to get

$\epsilon_{ijk}M_{iq}M_{jl} M_{km} A_lB_m = M_{iq} M_{in}V_n$ .

Using the identities stated above, this gives
$\epsilon_{qlm} A_l B_m = \delta{qn}V_n$ or

$\epsilon_{qlm}A_l B_m = V_q$

The left hand side is the cross product of the unrotated input vectors and the right hand side is the result of that cross product.  That wraps up the proof.

These are kind of dry posts, so let's re-re-re-introduce the Picture of the Day!
This one is from Young Parisians In Lub!

### 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…