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Tensor Notation for Triple Scalar Product Squared

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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.

Homework Equations





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[itex]_{ijk}[/itex]A[itex]_{i}[/itex]B[itex]_{j}[/itex]C[itex]_{k}[/itex]. 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!
 

Answers and Replies

  • #2
jfizzix
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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.

Homework Equations





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[itex]_{ijk}[/itex]A[itex]_{i}[/itex]B[itex]_{j}[/itex]C[itex]_{k}[/itex]. 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 [itex]\vec{A}\cdot(\vec{B}\times\vec{C}) = \epsilon_{ijk}A_{i}B_{j}C_{k}[/itex] is an excellent start.

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

The following identity should be useful (note that the k index is repeated, and so summed over)
[itex]\epsilon_{ijk}\epsilon_{k\ell m} = \delta_{i\ell}\delta_{jm}-\delta_{im}\delta_{j\ell}[/itex]
With this identity, you can show that
[itex]\vec{A}\times (\vec{B}\times\vec{C}) =(\vec{A}\cdot\vec{C})\vec{B}-(\vec{A}\cdot\vec{B})\vec{C}[/itex]
among other vector identities.
 
  • #3
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Thanks for the response!

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

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

then that equals,

(δ[itex]_{jl}[/itex]δ[itex]_{km}[/itex] - δ[itex]_{jm}[/itex]δ[itex]_{kl}[/itex])A[itex]_{j}[/itex]B[itex]_{k}[/itex]B[itex]_{l}[/itex]C[itex]_{m}[/itex]

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
jfizzix
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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., [itex](\vec{A}\times\vec{B})\cdot(\vec{B}\times\vec{C})[/itex]. 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
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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., [itex](\vec{A}\times\vec{B})\cdot(\vec{B}\times\vec{C})[/itex]. 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?
 
  • #6
vela
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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.
 

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