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Vector cross product

  1. Sep 28, 2016 #1
    1. The problem statement, all variables and given/known data
    This is not a homework problem, I am currently reading the Derivation of potential of a charged particle in Electric and Magnetic field from the book Mechanics by Symon (I attached the image of the page), I need to know how to expand the vector cross product
    such as $$\vec{v}\times({\nabla}{\times}\vec{A})=?$$

    2. Relevant equations
    ax(bxc)=(a.c)b-(a.b)c (Lagrange's formula)

    3. The attempt at a solution
    I tried to expand the vector triple product using Lagrange's formula, but what i am getting is,
    $$(\vec{v}.\vec{A})\nabla-(\nabla.\vec{A})\vec{v}$$
    As far as I know $$(\nabla.\vec{A})\vec{v}\neq(\vec{v}.\nabla\vec{A})$$ Is there any mistake I did during my calculation or should I use some other formula? please help.
     

    Attached Files:

  2. jcsd
  3. Sep 28, 2016 #2

    Orodruin

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    You cannot just commute the differential operator with the fields in the way you have done. In order to apply the bac-cab rule to an expression with a ##\nabla## you must manipulate the end expression in such a way that the nabla acts on the same vectors before and after applying it.
     
  4. Sep 28, 2016 #3

    Ray Vickson

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    Write it out in detail. Using ##D_u f## to stand for ##\partial f/\partial u## for ##u = x,y ,z##, the ##x##-component of ##\vec{v} \times (\nabla \times \vec{A})## is
    $$ \left(\vec{v} \times (\nabla \times \vec{A}) \right)_x = v_y (\nabla \times \vec{A})_z - v_z (\nabla \times \vec{A})_y \\
    = v_y (D_x A_y - D_y A_x) - v_z (D_z A_x -D_x A_z)
    $$.
    Expand it out and see what you get, then do the same type of thing for the ##y##- and ##z##-components. In particular, note that the components of ##\vec{v}## are not differentiated, so cannot lie to the right of the ##D##-sign, unless you close it out by putting it inside parentheses, like this: ##(D_x A_y)##.
     
    Last edited: Sep 28, 2016
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