Problem related to the vector potential of B

1. Jan 6, 2013

nuclear_dog

1. The problem statement, all variables and given/known data
A particle of mass m and charge q moves in a region with a magnetic field B(r) which is time independent. What is $\frac{dA}{dt}$ as seen by the particle.

2. Relevant equations

3. The attempt at a solution
Since the B field is not varying with time, I know that the change in A will be due to the change in position of particle with time. It should be (∂A/∂x)vx + (∂A/∂y)vy + (∂A/∂z)vz . So can it be (v.∇)A. But I cannot seem to be able to go furthur.

2. Jan 6, 2013

haruspex

A is a scalar, so (v.∇)A n= v.(∇A), right? And ∇A =?

3. Jan 6, 2013

nuclear_dog

I am sorry I forgot to add that in the question it was given that since ∇.B = 0 , therefore we can write B = ∇XA , where A is a vector. So I am not sure that we can use the transformation of ∇ in the expression.

4. Jan 6, 2013

haruspex

OK, I see. What makes you think (v.∇)A is an inadequate answer?

5. Jan 7, 2013

nuclear_dog

haruspex , I got hold of the original question paper today and have attached a screenshot of it. Please take a look at it to clearly understand the question , since I might not have explained it properly.

Since (dp/dt) = F = q(vXB) , hence if we can show that ∇X{d/dt( p + qA)} = 0 , it would mean that (d/dt(p + qA)) is a gradient of some function. But I haven't been able to prove that ∇X{d/dt( p + qA)} = 0 , by using dA/dt = (v.∇)A .

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6. Jan 7, 2013

haruspex

$\frac{d\vec{p}}{dt} =q\vec{v}\times\vec{B}$
$\frac{d}{dt}\left(\vec{p}+q \vec{A}\right) = q \vec{v}\times \vec{B}+q \left( \vec{v}.\vec{\nabla} \right) \vec{A} = q \vec{v}\times \left( \vec{\nabla}\times\vec{A}\right)+q \left( \vec{v}.\vec{\nabla} \right) \vec{A}$
Here I venture into areas I know even less about, based what I read at http://en.wikipedia.org/wiki/Curl_%28mathematics%29: [Broken]
$\vec{v}\times \left(\vec{\nabla}\times\vec{A}\right) = \vec{\nabla_A}\left(\vec{v}.\vec{A}\right) - \left(\vec{v}.\vec{\nabla}\right)\vec{A}$
So
$\frac{d}{dt}\left(\vec{p}+q \vec{A}\right) = q \vec{\nabla_A}\left(\vec{v}.\vec{A}\right)$
Does that help?

Last edited by a moderator: May 6, 2017
7. Jan 8, 2013

nuclear_dog

Thanks, that actually solves the problem. Although I will have to read up on the identities of vector calculus to feel satisfied on this question.