This is the general solution for \vec A, given a prescribed \vec B.

  • Thread starter Nikratio
  • Start date
In summary: But if we substitute the latter into the first equation we obtain\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'.This is the same as the equation in Wikipedia, except for the minus sign.
  • #1
Nikratio
13
0
Hello,

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

[tex]\vec \nabla \times \vec A = \vec B[/tex]

for [tex]\vec A[/tex]?

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.


Nikratio
 
Physics news on Phys.org
  • #2
Yup, we can write:
[tex]

\vec A(x) = \int d^3x \frac{\vec \nabla \times \vec B(x')}{|x-x'|}

[/tex]

where the curl in the integrand is with respect to the primed coordinates, and I have neglected the displacement current proportional to the time derivative of the electric field.
 
Last edited:
  • #3
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 [itex]4\pi[/itex] 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
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
Griffiths suggests this: compare [tex]\vec\nabla \cdot \vec A = 0, \vec\nabla \times \vec A = \vec B[/tex] with Maxwell's equations for B (in magnetostatics): [tex]\vec\nabla \cdot \vec B = 0, \vec\nabla \times \vec B = \mu_0 \vec J[/tex]. Then, in the same way the Biot-Savart law gives

[tex]\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',[/tex]

by analogy we should have

[tex]\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'.[/tex]
 
  • Like
Likes PabloAMC

FAQ: This is the general solution for \vec A, given a prescribed \vec B.

What does the notation "∇x" mean in this equation?

The notation "∇x" represents the gradient operator, which is used to calculate the directional derivative of a function in multivariable calculus.

How do I solve for A in this equation?

To solve for A in this equation, you can use matrix manipulation techniques such as row operations, inverse matrices, or Gaussian elimination. It is also helpful to have a good understanding of linear algebra.

Can you provide an example of solving this equation for A?

Sure, let's say we have the equation ∇x A = B, where A = [a b] and B = [c d]. Using Gaussian elimination, we can set up the augmented matrix [∇x | B] = [1 0 | c, 0 1 | d]. By performing row operations, we can solve for A and get the solution A = [c d].

Are there any special cases or restrictions when solving this equation for A?

Yes, there are a few special cases to consider when solving this equation. First, A and B must be matrices of the same size. Additionally, A must be an invertible matrix in order to have a unique solution. If A is not invertible, then there may be multiple solutions or no solution at all.

How is this equation used in scientific studies?

This equation is often used in scientific studies to solve for unknown variables in systems of equations. It is commonly used in physics, engineering, and other fields that involve mathematical modeling. It is also used in data analysis and machine learning to find optimal solutions to problems.

Back
Top