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Tensor algebra, divergence of cross product

  1. Jul 30, 2014 #1
    Hi there. I wanted to demonstrate this identity which I found in a book of continuum mechanics:

    ##curl \left ( \vec u \times \vec v \right )=div \left ( \vec u \otimes \vec v - \vec v \otimes \vec u \right ) ##

    I've tried by writting both sides on components, but I don't get the same, I'm probably making some mistake.

    ##\displaystyle curl \left ( \vec u \times \vec v \right )= \epsilon_{ijk} \frac{\partial}{\partial x_j}(\epsilon_{kji}u_j v_i)= \epsilon_{ijk} \left [ \epsilon_{kji} \frac{\partial u_j}{\partial x_j}v_i +u_j \epsilon_{kji} \frac{\partial v_i}{\partial x_j} \right ] ##

    I've used that ##(\vec u \times \vec v )_k=\hat e_k \cdot \epsilon_{ijk} u_j v_k \hat e_i=\delta_{ki} \epsilon_{ijk} u_j v_k=\epsilon_{kji}u_j v_i## I'm not sure if this is right.

    Then, using the epsilon delta identity I've got ##rot \left ( \vec u \times \vec v \right )=6 \left [ \frac{\partial u_j}{\partial x_j}v_i +u_j \frac{\partial v_i}{\partial x_j} \right ]=6 \left [ (div \vec u) \vec v + (grad \vec v) \vec u \right ]##

    Expanding the right hand side of the identity in components:

    ##div \left ( \vec u \otimes \vec v - \vec v \otimes \vec u \right ) = \frac{\partial}{\partial x_j}(u_i v_j -v_i u_j)=v_j \frac{\partial u_i}{\partial x_j}+ u_i \frac{\partial v_j}{\partial x_j} -u_j \frac{\partial v_i}{\partial x_j}- v_i \frac{\partial u_j}{\partial x_j}=(grad \vec u) \vec v+\vec u div \vec v - (grad \vec v) \vec u- \vec v div \vec u##

    PD: The title should be curl of cross product instead of divergence. Sorry.
     
  2. jcsd
  3. Jul 30, 2014 #2

    vela

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    You shouldn't have indices that appear more than twice. ##i## appears 3 times; ##j##, 4 times.
     
  4. Jul 30, 2014 #3
    ups, I see. Thank you vela.
     
  5. Aug 1, 2014 #4

    Fredrik

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    It is, but I wouldn't do those first two steps. I think of ##(\vec u\times\vec v)_k=\varepsilon_{kij}u_jv_k## as the definition of ##\vec u\times\vec v##, so it would have been my starting point. It makes sense to take this as the definition, since it implies that ##\vec u\times\vec v=(\vec u\times\vec v)_k e_k =\varepsilon_{kij}u_jv_k e_k##.
     
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