### 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

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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 differe…

### Division: Distributing the Work

Our unschooling math comes in bits and pieces.  The oldest kid here, seven year-old No. 1 loves math problems, so math moves along pretty fast for her.  Here’s how she arrived at the distributive property recently.  Tldr; it came about only because she needed it.
“Give me a math problem!” No. 1 asked Mom-person.

“OK, what’s 18 divided by 2?  But, you’re going to have to do it as you walk.  You and Dad need to head out.”

And so, No. 1 and I found ourselves headed out on our mini-adventure with a new math problem to discuss.

One looked at the ceiling of the library lost in thought as we walked.  She glanced down at her fingers for a moment.  “Is it six?”

“I don’t know, let’s see,” I hedged.  “What’s two times six?  Is it eighteen?”

One looked at me hopefully heading back into her mental math.

I needed to visit the restroom before we left, so I hurried her calculation along.  “What’s two times five?”

I got a grin, and another look indicating she was thinking about that one.

I flashed eac…