Skip to main content

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,


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

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

Start with

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

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}\\

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

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


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


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

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


Popular posts from this blog

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…

Kids R Kapable

Just a little note to concerned ‘grownups’ everywhere.  If you look at a kid—and I mean really look—I don’t mean notice a person shorter than you, I mean make eye contact, notice their facial expression and observe their body language—If you look at a kid, don’t assume they need your help unless they’re obviously distressed, or ask for it.  You might think this is difficult call to make.  You might think, not having kids of your own, that you’re unable to make this determination.  You are.  You do in fact, already have the skills even if you’ve never been around kids  It’s a remarkably simple call to make, just use the exact same criteria you would for determining if an adult was in distress.  Because, guess what, kids and adults are in fact the same species of animal and communicate in the same way.  Honest.  If someone—adult or child—doesn’t need your help, feel free to say hello, give a wave, give a smile, but don’t—do not—try to force help on anyone that doesn’t want or need it.