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    Hello there,

    really stuck on this problem, would be very grteful if anyone can give me some help on it, thanks!

    Using suffix notation, verify the following identities

    a. Nabla (del) dot ( Nabla cross A ) = 0 (div curl = 0)

    b. Nabla dot ( A cross B ) = B dot (nabla cross A) - A dot (nabla cross B)

    Where dot = dot product, cross = cross product , Nabla (del) is the upside down triangle symbol is the vector differential operator
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    A

    (\nabla \times A)_{i} = \epsilon_{ijk}\nabla_{j}A_{k}
    \nabla_{k}(\nabla \times A)_{k} = \nabla_{k}\epsilon_{ijk}\nabla_{  j}A_{k}

    Zero by the symmetry of the nablas but antisymmetry of the epsilon.

    \nabla_{i}(A \times B)_{i} = \nabla_{i}(\epsilon_{ijk}A_{j}B_  {k}) = \epsilon_{ijk}\nabla_{i}(A_{j}B_  {k}) = \epsilon_{ijk}\nabla_{i}(A_{j})B  _{k} + \epsilon_{ijk}A_{j}\nabla_{i}B_{  k} =  \epsilon_{ijk}\nabla_{i}(A_{j})B  _{k} - \epsilon_{jik}A_{j}\nabla_{i}B_{  k} = answer

    I used the product rule in that.
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    Thank you mate, im still struggling to get to grip with vector calculus and suffix notation, much appreciated!
    • Thread Starter
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    Actually if anyone could also help me with this 1 as well, i cant seem to get the answer

    show the following:

    nabla cross ( a cross r) = 2a

    where a=constant vector
    r= xi + yj + zk = xiei (i being in suffix notation)
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    [\nabla \times (a \times r)]_{i} = \epsilon_{ijk}\nabla_{j}(a \times r)_{k} = \epsilon_{ijk}\nabla_{j}(\epsilo  n_{klm}a_{l}r_{m})
    = \epsilon_{ijk}\epsilon_{klm}\nab  la_{j}(a_{l}r_{m}) since epsilon is a constant
    = \epsilon_{ijk}\epsilon_{klm}(r_{  m}\nabla_{j}a_{l} + a_{l}\nabla_{j}r_{m}) product rule
     = \epsilon_{ijk}\epsilon_{klm}a_{l  }\nabla_{j}r_{m} a is constant
     = (\delta_{il}\delta_{jm}-\delta_{im}\delta_{jk})a_{l}\nab  la_{j}r_{m}

    Can you go from here? Use \nabla_{j}r_{m}=\delta_{jm} (for that form of r) and \delta_{ij}\delta_{ij}=3.
 
 
 

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