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Prove Vector Quadruple Product with Levi-Civita/Index Notation

  1. Sep 6, 2014 #1
    I'm asked to prove the following using Levi-Civita/index notation:
    [itex]
    (\mathbf{a \times b} )\mathbf{\times} (\mathbf{c}\times \mathbf{d}) = [\mathbf{a,\ b, \ d}] \mathbf c - [\mathbf{a,\ b, \ c}] \mathbf d \
    [/itex]

    I'm able to prove it using triple product identities, but I'm completely stuck with the index notation. I was previously able to prove Lagrange's Identity with index notation, but applying similar concepts I just get stuck on the first step with the quadruple product.

    Using the same first step of proving Lagrange's identity, I transformed [itex]
    (\mathbf{a \times b} )[/itex] into [itex]\varepsilon_{ijk} a^j b^k[/itex] and [itex]
    (\mathbf{c \times d} )[/itex] into [itex]\varepsilon_{imn} c^m d^n[/itex] but then I'm just left with [itex](\varepsilon_{ijk} a^j b^k) \mathbf{\times} (\varepsilon_{imn} c^m d^n)[/itex] which is seemingly unhelpful.

    I also tried letting AxB = W and CxD = Z and transforming WxZ to index notation. Then I tried to 'un-nest' the original cross products in index notation, but it quickly ended up in a place where I couldn't understand what the different indexes represented.

    Any help would be appreciated. Thanks.
     
  2. jcsd
  3. Sep 6, 2014 #2
    I am not entirely sure how you defined ##[\mathbf{a},\mathbf{b},\mathbf{c}]##. Probably as a determinant? Either way, you must have ##[\mathbf{a},\mathbf{b}, \mathbf{c}] = \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})##. That might be helpful.

    There is also an identity that relates ##\epsilon_{ijk}\epsilon_{lmk}## to Kronecker deltas. It might also prove useful.
     
  4. Sep 7, 2014 #3
    Yes, [a,b,c]=a⋅(b×c). I'm sure it will be helpful, but only after I can get past the initial steps.

    I'm aware of the Kronecker delta identity you refer to, I used it to prove the scalar quadruple product/Lagrange's Identity as part of the same assignment. I don't doubt that that will be involved as well, but again, only after I can get past this initial step.

    I don't know how to combine [itex](\varepsilon_{ijk} a^j b^k) \mathbf{\times} (\varepsilon_{imn} c^m d^n)[/itex] without the index notation blowing up on me to a point where I don't understand it. It could just be that I'm approaching it from the wrong direction in the one step I did make in re-writing the two inner cross products in index notation.
     
  5. Sep 7, 2014 #4
    Thought I had a breakthrough after referencing another forum post. Was able to work through this:

    [itex]\begin{align*}
    [(A\times B)\times(C\times D)]_{i} &= \varepsilon_{ijk}(A\times B)_{j}(C\times D)_{k} \\
    &= \varepsilon_{ijk} \varepsilon_{jmn} A_{m}B_{n} \varepsilon_{kpq} C_{p}D_{q} \\
    &= -\varepsilon_{jik} \varepsilon_{jmn} A_{m}B_{n} \varepsilon_{kpq} C_{p}D_{q} \\
    &= -(\delta_{im}\delta_{kn} - \delta_{in}\delta_{km})A_{m}B_{n} \varepsilon_{kpq} C_{p}D_{q}\\
    &= (-A_{i}B_{k} + A_{k}B_{i}) \varepsilon_{kpq} C_{p}D_{q}\\
    &= -A_{i} \varepsilon_{kpq} B_{k}C_{p}D_{q} + B_{i} \varepsilon_{kpq} A_{k}C_{p}D_{q} \\
    &= (\mathbf{A} \cdot \mathbf{C} \times \mathbf{D})\mathbf{B} - (\mathbf{B} \cdot \mathbf{C} \times \mathbf{D})\mathbf{A}\end{align*}[/itex]

    This has to be the right approach...too many things went right while I was working through it, but then I came to then end and compared it to the identity:

    [itex](\mathbf{A} \cdot \mathbf{B} \times \mathbf{D})\mathbf{C} - (\mathbf{A} \cdot \mathbf{B} \times \mathbf{C})\mathbf{D}[/itex]

    So close, yet so far. Anybody see any mistakes? My brain is fried.
     
  6. Sep 7, 2014 #5

    vela

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    You didn't make a mistake, but to get the identity you want, try contracting ##\varepsilon_{ijk}## with ##\varepsilon_{kpq}## instead.
     
  7. Sep 7, 2014 #6

    Thanks, that does it.
     
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