Inverse of the curl

  1. Aug 14, 2010 #1
    I want to express A as a function of B in the following equation:

    curl{A}=B

    So I need the inverse of the curl operator, but I don't know if it exist.

    Thanks.
     
  2. jcsd
  3. Aug 14, 2010 #2
    It's called vector potential
     
  4. Aug 14, 2010 #3

    Redbelly98

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    There is no unique solution for A. You can always add a vector field of zero curl to one solution and get another solution.
     
  5. Aug 15, 2010 #4

    HallsofIvy

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    If A= f(x,y,z)i+ g(x,y,z)j+ h(x,y,z)k then curl A = (h_y- g_z)i+ (f_z- h_x)j+ (g_x- f_y)k.

    If you are given that curl A= B= p(x,y,z)i+ q(x,y,z)j+ r(x, y, z)k then you must solve the system of equation h_y- g_z= p, f_z- h_x= q, g_x- f_y= r.

    Since those are partial differential equations, the "constants of integration" will be functions of x, y, z. That is why, as RedBelly98 says, "You can always add a vector field of zero curl".
     
  6. Aug 15, 2010 #5

    arildno

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    Note that this non-uniqueness is not something that is a strange facet by the vector potential only.

    You know of it from before, as the so-called "constant of integration".
    When doing partials, functions of the other variables will be "constants" with respect to that variable you have differentiated with respect to.


    A far more important question, though, is, not uniqueness vs. non-uniqueness, but existence vs. non-existence of the vector potential(s)!

    Do you know, given B, how to be certain that at least one "A" exists?
     
  7. Aug 16, 2010 #6
    Consider the vector field defined by:

    [tex] \mathbf{A}(\mathbf{x}) = \int_0^1 \mathbf{B}(\lambda \mathbf{x}) \wedge (\lambda\mathbf{x})\, \mathrm{d}\lambda[/tex].

    You might like to show that if [tex]\nabla\cdot\mathbf{B}=0[/tex], then [tex]\nabla \wedge \mathbf{A} = \mathbf{B}[/tex]. Obviously this [tex]\mathbf{A}[/tex] is not unique.
     
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