### 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…

### Lost Phone

We were incredibly lucky to have both been in university settings when our kids were born.  When No. 1 arrived, we were both still grad students.  Not long after No. 2 arrived, (about 10 days to be exact), mom-person defended her dissertation and gained the appellation prependage Dr.

While there are lots of perks attendant to grad school, not the least of them phenomenal health insurance, that’s not the one that’s come to mind for me just now.  The one I’m most grateful for at the moment with respect to our kids was the opportunities for sheer independence.  Most days, we’d meet for lunch on the quad of whatever university we were hanging out at at the time, (physics research requires a bit of travel), to eat lunch.  During those lunches, the kids could crawl, toddle, or jog off into the distance.  There were no roads, and therefore no cars.  And, I realize now with a certain wistful bliss I had no knowledge of at the time, there were also very few people at hand that new what a baby…

### Lab Book 2014_07_10 More NaI Characterization

Summary: Much more plunking around with the NaI detector and sources today.  A Pb shield was built to eliminate cosmic ray muons as well as potassium 40 radiation from the concreted building.  The spectra are much cleaner, but still don't have the count rates or distinctive peaks that are expected.
New to the experiment?  Scroll to the bottom to see background and get caught up.
Lab Book Threshold for the QVT is currently set at -1.49 volts.  Remember to divide this by 100 to get the actual threshold voltage. A new spectrum recording the lines of all three sources, Cs 137, Co 60, and Sr 90, was started at approximately 10:55. Took data for about an hour.
Started the Cs 137 only spectrum at about 11:55 AM

Here’s the no-source background from yesterday
In comparison, here’s the 3 source spectrum from this morning.

The three source spectrum shows peak structure not exhibited by the background alone. I forgot to take scope pictures of the Cs137 run. I do however, have the printout, and…