### Showing Anti-Symmetries with Levi-Civita: EMII notes 2014_08_11

Summary of what's gone on before  The use of index notation to indicate a transpose was explained and shown with a concrete example.  Today, work on old homeworks begin.  The big issue of the day was figuring out an elegant way of showing that a contracted product was antisymmetric. The insights about commuting terms in index based products and how the Levi-Civita symbol works were worth the effort, but it was in fact a lot of effort!  There's a heap of broken attempts to get the short answer at the bottom of the post.

The first part of the rotated cross product problem is to show that

$W_{il}W_{jm}W_{kn}\epsilon_{lmn} = det\left(W\right)\epsilon_{lmn}$

We're to do this by first showing that the left hand side is antisymmetric with respect to the $i$, $j$, and $k$ indices and therefore proportional to $\epsilon_{lmn}$ and then by showing that for a concreted example, $i=1$, $j=2$, $k=3$, the left hand side is equal to $det\left(W\right)$

There's a long way and a short was to show the antyisymmetry of the left hand side.

The Short Way

Swapping the first two indices of any two terms in the product and then commuting the terms is indistinguishable from swapping the corresponding summation, (second), indices in the two terms.  Heres an example:

Swap the first two indices and commute

$W_{il}W_{jm}W_{kn} \rightarrow W_{jl}W_{im}W_{kn} = W_{im}W_{jl}W_{kn}$

Swap the summation indices

$W_{il}W_{jm}W_{kn} \rightarrow W_{im}W_{jl}W_{kn}$

Swapping the summation indices however, swaps the two indices in the Levi-Civita symbol which changes the sign of the symbol.  This means that, interchanging the first indices of any two terms in the product introduces a negative sign and the product as a whole is anti-symmetric under the swap of $i$, $j$, or $k$.

The Long Way

The second indices in each term are limited by the non-zero values of the Levi-Civita symbol that is contacted with the first three terms of the product.  We can fairly easily write out the six available combinations,

$W_{il}W_{jm}W_{kn}\epsilon_{lmn}$

$= W_{i1}W_{j2}W_{k3} + W_{i3}W_{j1}W_{k2} + W_{i2}W_{j3}W_{k1} - \\W_{i3}W_{j2}W_{k1} - W_{i2}W_{j1}W_{k3} - W_{i1}W_{j3}W_{k2}$

Number the above terms 1 through 6.  It is easy to show that by swapping $i$ and $j$ we get the following swaps, $1\leftrightarrow 6$,  $2\leftrightarrow 5$, and $3\leftrightarrow 4$.

When the first term becomes the sixth term, there is a corresponding sign change.  The same thing happens for terms 2 and 5 as well as 3 and 4.

{\bf Showing the Concrete determinant}

The second part of the problem is to show that for the index choices, $i = 1$, $j = 2$, $k = 3$, the l.h.s. is equal to $det\left(W\right)$.  This is just plug and chug work

$W_{il}W_{jm}W_{kn} \rightarrow W_{1l}W_{2m}W_{3n}$

There are six terms in the contracted sum

$= W_{11}W_{22}W_{33} + W_{13}W_{21}W_{32} + W_{12}W_{23}W_{31} -\\ W_{13}W_{22}W_{31} - W_{12}W_{21}W_{33} - W_{11}W_{23}W_{32}$

Now, work out the determinant of $W_{ij}$

$det\left(W\right) = \begin{vmatrix} W_{11} & W_{12} & W_{13}\\ W_{21} & W_{22} & W_{23}\\ W_{31} & W_{32} & W_{33}\\ \end{vmatrix}$

$=W_{11}W_{22}W_{33} - W_{11}W_{23}W_{32} - W_{12}W_{21}W_{33} + \\W_{12}W_{23}W_{31} + W_{13}W_{21}W_{32} - W_{13}W_{22}W_{31}$

The results are the same.

The Many Missteps to the Short Way

These are the attempts at the short way that didn't quite pan out just so I can see where I was misled.

To show the l.h.s. is antisymmetric, we need to show that by swapping any two of the $i$, $j$, and $k$ indices, we wind up with a quantity fo the same magnitude, but different components.  Let's start with the expression

$W_{il}W_{jm}W_{kn}\epsilon_{lmn}$,

and swap two of the indices, say $i$, and $j$.  This gives

$W_{jl}W_{im}W_{kn}\epsilon_{lmn}$.

Now, if we switch the l and m indices, the swap in the indices of the Levi-Civita symbol will produce a negative sign. Then, keeping in mind that in index notation, multiplication always commutes, we can write the above down as:

$W_{jl}W_{im}W_{kn}\epsilon_{lmn} = -W_{jm}W_{il}W_{kn}\epsilon_{lmn} = -W_{il}W_{jm}W_{kn}\epsilon_{mln}$.

New try

In all cases, the second index is restricted by the non-zero values of the Levi-Civita tensor.  If I switch the first indices on any two terms in the expression, then I've just effectively switched the second indices.

The swap of any two fo the first two indices is the same thing as swapping the second two indices for those two terms.  However, swapping the second two indices will introduce a negative sign from the Levi-Civita symbol.  This proves that swapping any two first indices is the same operation as introducing a negative sign.  An exmple may serve to make this more clear

$W_{il}W_{jm}W_{kn} \leftrightarrow W_{jl}W_{im}W_{kn}$

### 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…

### Lab Book 2014_07_10 More NaI Characterization

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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…

### 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…