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

AI Thread Summary
The discussion focuses on expanding the vector cross product $\vec{v}\times({\nabla}{\times}\vec{A})$. The user attempted to use Lagrange's formula but encountered issues with the results, specifically regarding the treatment of the differential operator $\nabla$. It was clarified that one cannot simply commute the differential operator with the fields, and that proper manipulation is necessary for applying the bac-cab rule. The conversation emphasizes the need to correctly handle the components of $\vec{v}$ and $\nabla$ during the expansion process. Understanding these nuances is crucial for accurately expanding the vector cross product in this context.
Muthumanimaran
Messages
79
Reaction score
2

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.
 

Attachments

  • Screenshot 2016-09-28 20.51.10.png
    Screenshot 2016-09-28 20.51.10.png
    92.8 KB · Views: 516
Physics news on Phys.org
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:
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top