Vector cross product

  • #1

Homework Statement


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 $$\vec{v}\times({\nabla}{\times}\vec{A})=?$$

Homework Equations


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

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.
 

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Answers and Replies

  • #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.
 
  • #3
Ray Vickson
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Homework Statement


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 $$\vec{v}\times({\nabla}{\times}\vec{A})=?$$

Homework Equations


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

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.

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:

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