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Del inverses

  1. Sep 2, 2008 #1
    If you're given a vector q and have that del x p=q (i.e curl(p)=q), how would you find p?
    Also for divergences.
  2. jcsd
  3. Sep 2, 2008 #2


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    Right down the equations for the individual components:
    [tex]\nabla\times p= q[/tex]
    means that
    [tex]\frac{\partial p_z}{\partial y}- \frac{\partial p_y}{\partial z}= q_x[/tex]
    [tex]\frac{\partial p_x}{\partial z}- \frac{\partial p_z}{\partial x}= q_y[/tex]
    [tex]\frac{\partial p_y}{\partial x}- \frac{\partial p_x}{\partial y}= q_z[/tex]
    Solve those equations.
  4. Sep 2, 2008 #3


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    You might get some mileage out of Stokes' theorem. (And some real analysis to figure out how to extract information about p) Keep mind mind that p is not uniquely determined, not even up to a constant. You can add any irrotational field to p and get a new solution. I'm pretty sure there's some ugly integral you can write down that gives you a particular solution -- hopefully someone will remember it and post it here.
  5. Sep 2, 2008 #4


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    Hurkly, that's funny!

    paddo, you may like to compare Ampere's circuital law and the Biot Savart law:

    You may also like to compare Gauss's law with Coulomb's law for a continuous charge distribution:
  6. Sep 3, 2008 #5
    You can use homotopy operators to do this type of thing.
  7. Sep 3, 2008 #6


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    Do you happen to know the divergence of p, and the projection of p onto the outward pointing normal vector of the boundary of the volume you're solving for p in?

    In general, if you know

    [tex]\nabla \times \mathbf{u} = \mathbf{C}(\mathbf{r})[/tex]

    [tex]\nabla \cdot \mathbf{u} = s(\mathbf{r})[/tex]

    and the value of [itex]\mathbf{\hat{n}}\cdot \mathbf{u}[/itex] on the boundary of the volume you're solving in, [itex]\partial V[/itex], then there is a unique solution for [itex]\mathbf{u}[/itex]. Writing

    [tex]\mathbf{u} = -\nabla \phi + \nabla \times \mahtbf{A}[/tex],


    [tex]\phi(\mathbf{r}) = \frac{1}{4\pi}\int d^3\mathbf{r'} \frac{s(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} + \mbox{const.}[/tex]

    [tex]\mathbf{A}(\mathbf{r}) = \frac{1}{4\pi}\int d^3\mathbf{r'} \frac{\mathbf{C}(\mathbf{r}')}{|\mathbf{r}-\mathbf{r}'|} + \nabla f(\mathbf{r})[/tex]
  8. Sep 4, 2008 #7
    I should probably elaborate on my previous post. You want a linear homotopy operator [tex]h:\Lambda^k \rightarrow \Lambda^{k-1}[/tex] to be such that:

    [tex] \theta = \mathrm{d}h (\theta) + h (\mathrm{d}\theta) [/tex]

    for a k-form [tex]\theta[/tex]. Clearly if [tex]\theta[/tex] is closed you have [tex]\theta = \mathrm{d}\eta[/tex] with [tex]\eta = h(\theta)[/tex]. If we work on star-shaped domains, then the following holds:

    [tex] h(\theta) = \int_0^1 (\iota_X \theta )[\lambda x] \frac{\mathrm{d}\lambda}{\lambda} [/tex]

    where [tex]X= x^i \partial /\partial x^i[/tex] is the scaling vector field. You might like to try it out with some examples.
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