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Inverse Curl

  1. Jan 31, 2010 #1
    Hello everyone!

    Having a field [tex] \bf B = \nabla \times \bf A [/tex] , how is it possible to get [tex] \bf A [/tex] ?
    For constant fields, the answer is easy, but is there a general approach to find A ?
    Some algorithm to do it numerically would help me immensly, too.
    If anyone knows some book or reference that could help me, I'd be appreciated.
    Thanks in advance.
  2. jcsd
  3. Feb 1, 2010 #2
    In usual classroom cases, there is ussually only one component for the field, making it an ordinary differential equation. Antoher choice is to use [tex]\int_{S}BdS=\int_{\partial S}Adl[/tex], and exploit possible symmetries of the fields. In the most general case, you might use the fact that [tex]\nabla \times B=\mu_{0}j[/tex], and assuming the gauge condition [tex]\nabla \cdot A=0[/tex], you have to solve the Poisson equation: [tex]\nabla^{2}A=-\mu_{0}j[/tex]
  4. Feb 1, 2010 #3
    Thanks for your answer

    In my case, J is not known (actually it's one of the things I will want to find). B is not uniform , but for a first approximation I would want to find B(x,y,z) = (0 , 0 , B0 + k*z) , B0 >> k It's a constant magnetic field with a small gradient on the top of it.

    k is only some small constant.

    or something like B = (k*x, 0, B0)
  5. Feb 1, 2010 #4


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    Be careful. The form of B that you are looking for doesn't seem to satisfy div B = 0 anywhere, so it isn't a legit magnetic field.

    The general way to approach this is with Helmholtz's Theorem. IT is a purely mathematical result that is often invoked in EM. For a bounded region it states that for an arbitrary vector field,

    \bf P(r)= \nabla \left[ \int_V \frac{-\nabla \prime \cdot P(r\prime)\, dV\prime}{4 \pi |r-r\prime|} + \int_S \frac{P(r\prime)\cdot dS\prime}{4 \pi |r-r\prime|}\right]
    + \nabla \times \left[ \int_V \frac{\nabla \prime \times P(r\prime)\, dV\prime}{4 \pi |r-r\prime|} + \int_S \frac{P(r\prime) \times dS\prime}{4 \pi |r-r\prime|}\right]

    where the surface integrals are over closed surfaces. So you need to specify the divergence and the curl and the boundary conditions in order to uniquely determine a vector field. For magnetic fields, div B = 0 always, which simplifies things a bit. If your boundary conditions are also zero, then you basically just need to perform an integration and curl operation.

  6. Feb 2, 2010 #5
    Thanks a lot Jason!
    I just wrote some B field that somehow resembled what i wanted, didn't even check its divergence.
    I'll try to figure it out now
  7. Feb 2, 2010 #6
    You should be aware that there is a constant of integration, phi. Phi is a vector.

    [tex]A \leftarrow A' = A + \phi[/tex]

    I think you should find that the best approach is to write out the definition of the cross product of A, in terms of the parital derivatives with respect to your coordinates, and then integrate.
    Last edited: Feb 2, 2010
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