Suppose $\vec{B}$ is a differentiable vector field defined everywhere such that $\nabla\cdot \vec{B}=0$. Define $\vec{A}$ by the integral $$A_1=\int_0^1 \lambda(xB_2(\lambda x,\lambda y,\lambda z)- yB_3(\lambda x,\lambda y,\lambda z)) d\lambda$$ 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!

  • $\begingroup$ when you try to write the expression for the curl of A components what are you left with? we need to see what you get in order to help you $\endgroup$ – lurscher Oct 19 '12 at 12:19

You are right in both your specific questions: your $\lambda$ derivative is right and the partial derivatives can go inside the integral.

You have, however, one crucial mistake in your original formula, which should read $$A_1=\int_0^1 \lambda(\quad z\quad B_2(\lambda x,\lambda y,\lambda z)- yB_3(\lambda x,\lambda y,\lambda z)) d\lambda$$ - i.e., replacing $x$ by $z$. This is needed to make "permutational sense": it now reads like "(1)=(3)(2)-(2)(3)", instead of "(1)=(1)(2)-(2)(3)", which is clearly wrong.

Plugging this and the cyclically-permuted expression for $A_2$ into the curl then gives (after applying $\nabla\cdot\mathbf{B}=0$, your formula for the $\lambda$ derivative, and an integration by parts) the desired $$\nabla\times\mathbf{A}=\mathbf{B}.$$

  • $\begingroup$ Ah, sorry about that error. So by making the purmutations we change the z,y coefficients to y,x etc? $\endgroup$ – Freeman Oct 19 '12 at 13:19
  • $\begingroup$ To make the permutations, change $x,y,z$ to $x_1,x_2,x_3$, and then change all 1s to 2s, 2s to 3s, and 3s to 1s. (In this example you should have something of the form "(2)=(1)(3)-(3)(1)".) If you do not also change the coordinates $y$ and $z$ above then you are fundamentally changing the geometrical situation. (Note, though, that the arguments $(\lambda x,\lambda y, \lambda z)$ to the magnetic field components are not altered!) $\endgroup$ – Emilio Pisanty Oct 19 '12 at 18:19

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