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

In summary, the conversation is about understanding how to expand the vector cross product in the context of deriving the potential of a charged particle in electric and magnetic fields. The individual is reading from a book and has attempted to use Lagrange's formula, but has encountered an issue with commuting the differential operator with the fields. They are seeking help and have been instructed to write out the expansion in detail, using partial derivatives to represent the components of the vectors involved.
  • #1
Muthumanimaran
81
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.
 

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  • #2
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
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:

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

1. What is the vector cross product?

The vector cross product, also known as the vector or cross product, is a mathematical operation that takes two vectors as inputs and produces a third vector that is perpendicular to both input vectors. It is denoted by the symbol "×" and is commonly used in physics and engineering to describe the direction and magnitude of physical quantities such as force and torque.

2. What is the significance of expanding $\vec{v}\times({\nabla}{\times}\vec{A})$?

Expanding $\vec{v}\times({\nabla}{\times}\vec{A})$ allows us to simplify and better understand the relationship between the vector v, the gradient operator ∇, and the vector potential A. It also helps us to solve problems involving vector fields and their derivatives.

3. How do you expand $\vec{v}\times({\nabla}{\times}\vec{A})$?

To expand $\vec{v}\times({\nabla}{\times}\vec{A})$, we can use the identity $\vec{a}\times(\vec{b}\times\vec{c}) = \vec{b}(\vec{a}\cdot\vec{c}) - \vec{c}(\vec{a}\cdot\vec{b})$ and the properties of the gradient and cross product operators. This will result in a new vector expression that is equivalent to the original one.

4. What is the physical interpretation of $\vec{v}\times({\nabla}{\times}\vec{A})$?

The physical interpretation of $\vec{v}\times({\nabla}{\times}\vec{A})$ is the curl of the vector potential A, which describes the rotation or circulation of a vector field. This quantity is important in electromagnetism and fluid dynamics, where it represents the vorticity of a flow.

5. What are some applications of $\vec{v}\times({\nabla}{\times}\vec{A})$?

$\vec{v}\times({\nabla}{\times}\vec{A})$ has various applications in physics and engineering, such as computing the magnetic field generated by a current, describing the motion of charged particles in a magnetic field, and solving problems involving rotational motion and fluid flow. It is also used in mathematical models of electromagnetism, fluid dynamics, and quantum mechanics.

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