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Divergence of (A x B)

  1. Apr 16, 2006 #1
    Hi, I'm having trouble proving that:
    [tex] \nabla \cdot (\textbf{A} \times \textbf{B}) = \textbf{B}\cdot
    (\nabla \times \textbf{A}) - \textbf{A}\cdot (\nabla \times
    \textbf{B}) [/tex]

    This is how I proceeded:
    [tex]\textbf{A} \times \textbf{B} = \overrightarrow{i}(A_y B_z - A_z
    B_y) + \overrightarrow{j} (A_z B_x - A_x B_z) + \overrightarrow{k}
    (A_x B_y - A_y B_x)[/tex]

    [tex]\nabla \cdot (\textbf{A} \times \textbf{B}) = \frac{\partial A_y B_z - A_z
    B_y}{\partial x} + \frac{\partial A_z B_x - A_x B_z}{\partial y} +
    \frac {\partial A_x B_y - A_y B_x}{\partial z}[/tex]

    [tex]\nabla \times \textbf{A} = \overrightarrow{i}(\frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z})
    + \overrightarrow{j}(\frac {\partial A_x}{\partial z} - \frac
    {\partial A_z}{\partial x}) + \overrightarrow{k}(\frac {\partial
    A_y}{\partial x} - \frac {\partial A_x}{\partial y}) [/tex]

    [tex] \textbf{B} \cdot (\nabla \times \textbf{A}) = (\frac{\partial A_z B_x}{\partial y} - \frac{\partial A_y B_x}{\partial z})
    + (\frac {\partial A_x B_y}{\partial z} - \frac {\partial A_z
    B_y}{\partial x}) + (\frac {\partial A_y B_z}{\partial x} - \frac
    {\partial A_x B_z}{\partial y}) = [/tex]

    [tex] = \frac{\partial A_y B_z - A_z
    B_y}{\partial x} + \frac{\partial A_z B_x - A_x B_z}{\partial y} +
    \frac {\partial A_x B_y - A_y B_x}{\partial z} = \nabla \cdot
    (\textbf{A} \times \textbf{B}) [/tex]

    [tex] \textbf{A} \cdot (\nabla \times \textbf{B}) = (\frac{\partial B_z A_x}{\partial y} - \frac{\partial B_y A_x}{\partial z})
    + (\frac {\partial B_x A_y}{\partial z} - \frac {\partial B_z
    A_y}{\partial x}) + (\frac {\partial B_y A_z}{\partial x} - \frac
    {\partial B_x A_z}{\partial y}) = -\nabla \cdot (\textbf{A} \times
    \textbf{B}) [/tex]

    Which finally yields:
    [tex] \textbf{B}\cdot (\nabla \times \textbf{A}) - \textbf{A}\cdot (\nabla \times \textbf{B}) = 2\nabla \cdot (\textbf{A} \times \textbf{B}) [/tex]

    Where did I make a mistake?
     
    Last edited: Apr 16, 2006
  2. jcsd
  3. Apr 16, 2006 #2

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    The obvious problem is that you've written B.(\/xA) and somehow got the components of the B's past the differential symbols, which is worrying

    [tex]B_x\partial_yA_z[/tex]

    is not the same as

    [tex]\partial_yA_zB_x[/tex]

    though it is the same as

    [tex] (\partial_yA_z)B_x[/tex]

    that might be one problem.
     
  4. Apr 16, 2006 #3
    Thanks a lot, I didn't realize that.
     
  5. Apr 16, 2006 #4

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    It's just the product rule. if you differentiate f(x)g(x) the answer is not f'(x)g'(x), which I'm sure you do know.
     
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