# Tensor Notation for Triple Scalar Product Squared

## Homework Statement

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.

## 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!

jfizzix
Gold Member

## Homework Statement

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.

## 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!

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.

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?

jfizzix
Gold Member
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.

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.

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?

vela
Staff Emeritus
Homework Helper