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A Gradient divergence

  1. Jan 17, 2017 #1
    Nabla operator is defined by

    [tex]\nabla = \sum^3_{i=1} \frac{1}{h_i}\frac{\partial}{\partial q_i}\vec{e}_{q_i}[/tex]
    where ##q_i## are generalized coordinates (spherical polar, cylindrical...) and ##h_i## are Lame coefficients. Why then
    [tex]div(\vec{A})=\sum^3_{i=1} \frac{1}{h_i}\frac{\partial}{\partial q_i}\vec{e}_{q_i} \cdot \sum_j A_j\vec{e}_{q_j}=\sum_i\frac{1}{h_i}\frac{\partial}{\partial q_i}A_i[/tex]
    where I am making the mistake?
    here is different definition.
    https://www.jfoadi.me.uk/documents/lecture_mathphys2_05.pdf
     
  2. jcsd
  3. Jan 17, 2017 #2

    pasmith

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    Homework Helper

    The unit vectors are generally functions of position; only in cartesian coordinates are they constants. This is why we generally write [itex]
    \nabla = \sum_i \mathbf{e}_i h_i^{-1} \partial_i[/itex] to make it clear that the derivative does not operate on [itex]\mathbf{e}_i[/itex] or [itex]h_i[/itex]. However for divergence we get [tex]
    \sum_i \mathbf{e}_i h_i^{-1} \partial_i \cdot \left( \sum_j A_j \mathbf{e}_j \right) =
    \sum_i \sum_j \mathbf{e}_i h_i^{-1} \cdot \left( A_j \partial_i\mathbf{e}_j + \mathbf{e}_j \partial_i A_j \right)[/tex]
     
  4. Jan 18, 2017 #3
    Thanks a lot. It makes sense of course. What is the easiest way to calculate ##A_j \partial_i \vec{e}_j##? How to write that in order to get real expression of divergence with the Lame coefficients and derivatives.
     
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