Solve \nabla x A = B for A

1. Feb 1, 2009

Nikratio

Hello,

Is there a general, analytic way to obtain a vector potential for a prescribed magnetic field? I.e. to solve

$$\vec \nabla \times \vec A = \vec B$$

for $$\vec A$$?

I'm looking for something like what e.g. the Green's function method is for Poisson's equation, so the answer might be a complicated integral but it is a definite, closed solution.

Best,
Nikratio

2. Feb 1, 2009

confinement

Yup, we can write:
$$\vec A(x) = \int d^3x \frac{\vec \nabla \times \vec B(x')}{|x-x'|}$$

where the curl in the integrand is with respect to the primed coordinates, and I have neglected the displacment current proportional to the time derivative of the electric field.

Last edited: Feb 1, 2009
3. Feb 1, 2009

jostpuur

Some time ago I was wondering the some problem. I had one difficulty with it, so I posted this integral formula for vector potential in hope of getting some help. However, I solved it myself quickly, posted a clarifying reply myself, and seemingly nobody considered it useful to respond anymore.

It seems that confinement's formula was lacking a $4\pi$ constant somewhere, but now I started to wonder my own minus sign there too. Is my formula in contradiction with the formula in Wikipedia because of the minus sign?

4. Feb 1, 2009

confinement

Jostpuur is right that there should be a 1/4Pi in front of the integral in the formula I gave, although I am reasonably confident that there is no missing minus sign.

5. Feb 1, 2009

Griffiths suggests this: compare $$\vec\nabla \cdot \vec A = 0, \vec\nabla \times \vec A = \vec B$$ with Maxwell's equations for B (in magnetostatics): $$\vec\nabla \cdot \vec B = 0, \vec\nabla \times \vec B = \mu_0 \vec J$$. Then, in the same way the Biot-Savart law gives
$$\vec B(\vec r) = \frac{\mu_0}{4\pi} \int \frac{\vec J(\vec r') \times (\vec r - \vec r')}{\lvert \vec r - \vec r' \rvert^3} \,d^3\vec r',$$
$$\vec A(\vec r) = \frac{1}{4\pi} \int \frac{\vec B(\vec r') \times (\vec r - \vec r')}{\lvert \vec r - \vec r' \rvert^3} \,d^3\vec r'.$$