# Tensor Notation for Triple Scalar Product Squared

1. Oct 1, 2013

### forestmine

1. The problem statement, all variables and given/known data

Hi all,

Here's the problem:

Prove, in tensor notation, that the triple scalar product of (A x B), (B x C), and (C x A), is equal to the square of the triple scalar product of A, B, and C.

2. Relevant equations

3. The attempt at a solution

I started by looking at the triple scalar product of A, B, and C. I know that I can write that as ε$_{ijk}$A$_{i}$B$_{j}$C$_{k}$. This is about as far as I got.

Initially, I did something different that got me further, but I think I was interpreting the problem incorrectly. What I did was set a = (AxB), b = (BxC), and c = (CxA), thinking that the square of the triple scalar product was referring to the triple scalar product of a, b, and c, but upon reading the problem again, I don't think that that's right.

That being said, I'm not sure where to go from here.

I know that the subject of tensors is a particularly difficult one to discuss over the web, especially without just spelling out the answer -- which I'm certainly not looking for. But that being said, I'm pretty lost, and could use a lot of help in the right direction.

Thanks so much!

2. Oct 1, 2013

### jfizzix

Knowing that $\vec{A}\cdot(\vec{B}\times\vec{C}) = \epsilon_{ijk}A_{i}B_{j}C_{k}$ is an excellent start.

The next thing I would try to figure out is how to express the cross product between two cross products $(\vec{A}\times \vec{B})\times(\vec{B}\times\vec{C})$

The following identity should be useful (note that the k index is repeated, and so summed over)
$\epsilon_{ijk}\epsilon_{k\ell m} = \delta_{i\ell}\delta_{jm}-\delta_{im}\delta_{j\ell}$
With this identity, you can show that
$\vec{A}\times (\vec{B}\times\vec{C}) =(\vec{A}\cdot\vec{C})\vec{B}-(\vec{A}\cdot\vec{B})\vec{C}$
among other vector identities.

3. Oct 1, 2013

### forestmine

Thanks for the response!

Ok, well, I know that (AxB)$_{i}$ = ε$_{ijk}$A$_{j}$B$_{k}$. And I can say that, (BxC)$_{i}$ = ε$_{ilm}$B$_{l}$C$_{m}$.

If I write that all as one term,
ε$_{ijk}$A$_{j}$B$_{k}$ε$_{ilm}$B$_{l}$C$_{m}$

then that equals,

(δ$_{jl}$δ$_{km}$ - δ$_{jm}$δ$_{kl}$)A$_{j}$B$_{k}$B$_{l}$C$_{m}$

and I know that I can do some further simplification of that, but I'm worried about simplifying it too much before incorporating yet the third cross product. Am I on the right track at least?

4. Oct 1, 2013

### jfizzix

Yes this is exactly on the right track. The formula you give would be for the dot product of the two cross products, i.e., $(\vec{A}\times\vec{B})\cdot(\vec{B}\times\vec{C})$. If each vector in the second cross product is expressed as a cross product of two other vectors, you should be able to get a final result.

5. Oct 1, 2013

### forestmine

Hm, ok, I'm not sure I follow here. If I computed the dot product of two cross products, isn't that wrong, in that I should first compute the cross product of (BxC) x (CxA), before taking the dot product that I did in the step before?

6. Oct 2, 2013

### vela

Staff Emeritus
The expression you started with, $(\varepsilon_{ijk}A_j B_k)(\varepsilon_{ilm}B_l C_m)$, is equal to $(\vec{A}\times\vec{B})_i (\vec{B}\times\vec{C})_i$, which is equal to $(\vec{A}\times\vec{B})\cdot(\vec{B}\times\vec{C})$. That's not what you want, right? So don't combine the two cross products that way.

Go back to your previous attempt where you said
\begin{align*}
\vec{a} &= \vec{A}\times\vec{B} \\
\vec{b} &= \vec{B}\times\vec{C} \\
\vec{c} &= \vec{C}\times\vec{A}
\end{align*} The triple scalar product of those three vectors is $\varepsilon_{ijk}a_i b_j c_k$. Now you want to substitute in expressions for $a_i$ in terms of the components of $\vec{A}$ and $\vec{B}$, and so on, and the simplify it.