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    curl (a × x )=2 a,
    curl ((r^2).a) = 2(x × a),

    x and a are vectors
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    (Original post by CammieInfinity)
    curl (a × x )=2 a,
    curl ((r^2).a) = 2(x × a),

    x and a are vectors
    these proofs are trivial from the basic definitions
    however am I right in saying you trying to prove them using tensor notation (Einstein summation convention)?
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    (Original post by TeeEm)
    these proofs are trivial from the basic definitions
    however am I right in saying you trying to prove them using tensor notation (Einstein summation convention)?
    I forgot to mention that r=magnitude of x

    I have done the proofs normally but now we have to do it using index notation.
    We have used Einstein summation convention and the kronecker to help do previous questions.
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    I don't think you've told us all of the information you have.

    Is it possible that a is a constant vector??

    (Ps I can't help with the index notation, I'm not good enough at it - sorry )
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    (Original post by CammieInfinity)
    curl (a × x )=2 a,
    curl ((r^2).a) = 2(x × a),

    x and a are vectors
    I'll work through the first one, because I don't really see that there's anything more to the question than just mechanical application of rules.

    \left(\nabla \times (a \times x) \right)_i = \epsilon_{ijk} \dfrac{\partial (a \times x)_k}{\partial x_j} = \epsilon_{ijk} \epsilon_{krs} a_r \dfrac{\partial (x_s)}{\partial x_j}.

    \epsilon_{ijk} = \epsilon_{kij}, so \epsilon_{kij} \epsilon_{krs} = \delta_{ir} \delta_{js} - \delta_{is} \delta_{jr}.

    Substituting that in yields a_i \dfrac{\partial (x_j)}{\partial x_j} - a_j \dfrac{\partial (x_i)}{\partial x_j}. First term is 3 a_i; second term is a_i.
 
 
 
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