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Tensor Notation for Triple Scalar Product Squared

  1. Oct 1, 2013 #1
    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[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!
  2. jcsd
  3. Oct 1, 2013 #2


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    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.
  4. Oct 1, 2013 #3
    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,

    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?
  5. Oct 1, 2013 #4


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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.
  6. Oct 1, 2013 #5
    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?
  7. Oct 2, 2013 #6


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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
    \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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