Vector Cross Product Homework: Expand $\vec{v}\times({\nabla}{\times}\vec{A})$

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SUMMARY

The discussion focuses on expanding the vector cross product $\vec{v}\times({\nabla}{\times}\vec{A})$ using Lagrange's formula. The user initially applied the formula but encountered discrepancies in their results, specifically with the expression $$(\vec{v}.\vec{A})\nabla-(\nabla.\vec{A})\vec{v}$$. It was clarified that the differential operator $\nabla$ cannot be commuted with the fields without proper manipulation, and the correct approach involves carefully expanding the components of the vector cross product while maintaining the integrity of the differential operators.

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  • Understanding of vector calculus, specifically vector cross products.
  • Familiarity with Lagrange's formula for vector triple products.
  • Knowledge of the nabla operator ($\nabla$) and its applications in vector fields.
  • Basic proficiency in manipulating partial derivatives and vector components.
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Muthumanimaran
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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.
 

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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.
 
Muthumanimaran said:

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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