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Inverse/Anti - Divergence? Maxwells Eqns.

  1. Jun 15, 2010 #1
    So I have a simple/easy to answer question for any physics buffs out there. I think i'm doing something fundamentally flawed.

    Can you take the inverse of a divergence? analagous to antiderivative-integral?

    e.g., I want to find J from the continuity equation with a known [tex] \rho(\vec{r},t) [/tex]

    like

    [tex] \frac{\partial \rho}{\partial t} + \nabla \cdot \mathbf{J} = 0[/tex]

    so i'm trying to do

    [tex] invdiv(\frac{\partial \rho}{\partial t}) = invdiv(- \nabla \cdot \mathbf{J}) [/tex]

    where "invdiv" would be some inverse divergence operation.

    I think trying to get J this way may be fundamentally flawed. As extra information [tex] \rho(\mathbf{r},t) = -e \delta(x) \delta(y) \delta(z - \frac{\Delta Z}{9}sin(\omega t))[/tex] where e is electron charge. How do you handle the Dirac functions in there?

    THanks for the help.
     
  2. jcsd
  3. Jun 15, 2010 #2

    K^2

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    You need to know divergence and curl of a field to completely define it. For the special case where curl vanishes, you can use linearity and Gauss Theorem to derive result familiar from electrodynamics:

    [tex]\mathbf{a}(\mathbf{x}) = \int d^3y \frac{1}{4\pi} \frac{\nabla\cdot\mathbf{a}(\mathbf{y})}{(\mathbf{x}-\mathbf{y})^2}[/tex]
     
  4. Jun 15, 2010 #3

    king vitamin

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    You have a full-blown partial differential equation on your hands. My advice is to take the divergence of the continuity equation, plug in your charge distribution, and then use whatever methods you can to solve Poisson's equation (which is the form you will obtain).

    The dirac delta's derivative is [tex]- \frac{\delta(x)}{x}[/tex] if I remember correctly.
     
  5. Jun 23, 2012 #4
    I have written a recent paper on this, which shows how to construct a (non-unique) tensor field whose divergence is equal to a tensor density of one rank less. See Section 4 of
    International Journal of Engineering Science 49 (2011) 1486–1493. Joe Goddard
     
  6. Jun 23, 2012 #5
    It should be clear that there's no general way to do this; the divergence doesn't give you enough information. If the charge in a small volume is decreasing, then current must be leaving it, but you have no idea which direction it's leaving in.

    You need more information to actually solve this problem, and that information is encoded in [itex]\nabla \times J[/itex], exactly as K^2 said.

    Why is this so? Because neither the divergence nor the curl are individually invertible, but there's a solution to this. Geometric algebra defines a derivative operator [itex]\nabla \wedge J[/itex] which is like the curl, but it formally produces a bivector field when it acts on a vector (whatever you would get from the curl, the [itex]\nabla \wedge[/itex] operator produces a field whose planes are perpendicular to the curl.

    When considered together with the divergence, this "wedge" derivative helps form a fully invertible derivative operator. Define [itex]\nabla J \equiv \nabla \cdot J + \nabla \wedge J[/itex]. It's perfectly fine to add scalars and bivectors; this is no different from adding real and imaginary numbers to get complex numbers. Together, this makes [itex]\nabla[/itex] fully invertible, for it admits a Green's function.

    When you have a differential equation of the form [itex]\nabla F = J[/itex] (which is what Maxwell's equations are when there's no matter), you get the simple result,

    [itex]F(r) = \int_{V'} J(r') G(r-r')|dV'|[/itex]

    And the free space Green's function is [itex]G(r) = r/4\pi|r|^3[/itex], which ought to be familiar from, for example, the point charge field.
     
  7. Jun 23, 2012 #6
    Nobody said it was unique. As pointed out in my article, it is unique up to a solenoidal field. More mathematically put, we are finding a particular solution to a linear PDE whose general solutions involves an additive homogeneous solution.

    By the same token, there is no unique inverse to the vector equation a.x = b where a and x are vectors. In Fourier space, the equation div f = b takes on a similar form. ( I have a second paper in progress which is invited for publication in the Revs. Appl. Mech. which provides a number of different versions in Fourier space.)

    I make no claims as to "simplicity" (which like "beauty" may lie in the eye of the beholder).

    Joe Goddard
     
  8. Jun 23, 2012 #7
    Yeah, you can add any analytic (i.e. [itex]\nabla F = 0[/itex]) field to match a boundary condition--and there are plenty of those, for example the fields of point charges that lie outside the volume of interest, and so on.

    Geometric algebra also allows for inversion of vectors when [itex]a \cdot b[/itex] and [itex]a \wedge b[/itex] are known, and it gives the inverse of a vector under the "geometric product" [itex]ab \equiv a \cdot b + a \wedge b[/itex]. In this case, [itex]b^{-1} = b/b^2[/itex] such that [itex]a b b^{-1} = a b (b/b^2) = ab^2/b^2 = a[/itex].
     
  9. Jun 23, 2012 #8
    Or, possibly, differential forms and representations based on the Helmholtz-Hodge decomposition...
     
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