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Vector calculus identities proof using suffix notation

  1. Apr 25, 2014 #1
    I must become good at this ASAP.

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

    prove [itex] \vec{\nabla}\cdot (\vec{a}\times\vec{b} ) = \vec{b} \cdot(\vec\nabla\times\vec{a}) - \vec{a}\cdot(\vec\nabla\times\vec{b})[/itex]

    2. Relevant equations
    [itex] \vec a \times \vec b = \epsilon_{ijk}\vec a_j \vec b_k [/itex]
    [itex] \vec\nabla\cdot = \Large\frac{\partial}{\partial x_i} [/itex]
    summation over [itex] i [/itex]
    3. The attempt at a solution

    I don't know where to start. I'm sure it must involve some product rule.
    but I'm not 100% sure whether or not [itex] \vec\nabla\cdot(\vec a \times \vec b) = (\vec b \times\vec\nabla\cdot\vec a) + (\vec a \times \vec\nabla\cdot\vec b) [/itex] (or something resembling it)

    ....
    ....Right now I have no decent book with me and searching on the internet has done more harm than good.

    If that identity correct I might (probably) be able to do the rest.
     
  2. jcsd
  3. Apr 25, 2014 #2

    Matterwave

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    Look at the types of quantities you are using in your attempt at a solution. What type of quantity is ##\nabla\cdot \vec{a}##? Does it make sense to take a cross product of a vector with this quantity?

    As for how to prove the necessary assertion. You can always go back and use components! The brute force method is always doable (and it's not so hard for this problem).

    For a more elegant method of proof, none immediately come to mind. Perhaps with some more thought, one can find a more elegant proof.
     
  4. Apr 25, 2014 #3
    Nope. it doesn't make sense.
    what was I thinking...
     
  5. Apr 25, 2014 #4

    Ray Vickson

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    Letting ##\partial_i \equiv \partial / \partial x_i##, we have
    [tex] \vec{\nabla}\cdot (\vec{a}\times\vec{b} ) = \partial_i (\epsilon_{ijk} a_j b_k)
    = \epsilon_{ijk} \partial_i (a_j b_k)[/tex]
    (using the Einstein summation convention). Now use the differentiation product rule, and simplify.
     
  6. Apr 26, 2014 #5

    lurflurf

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    $$\vec{\nabla}\cdot (\vec{a}\times\vec{b} ) = \vec{b} \cdot(\vec\nabla\times\vec{a}) - \vec{a}\cdot(\vec\nabla\times\vec{b})=(\vec{b} \times\vec\nabla)\cdot\vec{a} - (\vec{a} \times\vec\nabla)\cdot\vec{b}=[\vec{b} \vec\nabla\vec{a}]-[\vec{a} \vec\nabla\vec{b}]$$
    are several notations for the same thing

    Though I personally like to include the parentheses some people do not. The justification for omitting the parentheses is that only one interpretation is possible. It is silly to make the nonsensical interpretation.

    The problem you have is with parity of a permutation. Each switch in order changes the sign.
    (123)=-(132)=(312)=-(213)=(231)=-(321)
    we see that
    $$\vec{b} \cdot(\vec\nabla\times\vec{a})$$
    corresponds to (312) and thus has a positive sign
    and
    $$\vec{a}\cdot(\vec\nabla\times\vec{b})$$
    corresponds to (213) and thus has a negative sign
     
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