Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $\vec{B}$ is a differentiable vector field defined everywhere such that $\nabla\cdot \vec{B}=0$. Define $\vec{A}$ by the integral $$A_1=\int_0^1 \lambda(zB_2(\lambda x,\lambda y,\lambda z)- yB_3(\lambda x,\lambda y,\lambda z)) d\lambda$$ Together with its two cyclic permutations for $A_2,A_3$

I'm trying to work out two things here:

$1.$ What is $\frac{d}{d\lambda}B_i(\lambda x,\lambda y,\lambda z)$

$2.$ How we can use $1.$ to determine $\frac{\partial A_2}{\partial x}-\frac{\partial A_1}{\partial y}=B_3$

From this we can deduce the existance of the magnetic potential by extending $2.$, this is what I have so far:

Is $\frac{d}{d\lambda}B_i(\lambda x,\lambda y,\lambda z)=(x,y,x) \cdot \nabla B_i$? And can we bring the partial derivative on $A_i$ inside the integral? I have proceeded along these lines but have not found a way to substitute.

Any help would be greatly appreciated!

share|cite|improve this question

Apply the chain rule in 1 to get started.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.