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Helmholtz decomposition and relative magnitudes of irrotational/solenoidal components

  1. Jun 8, 2012 #1

    Consider [itex]P = \boldsymbol{\nabla} f +\boldsymbol{\nabla}\times \bold{A} [/itex]

    where f and A are scalar and vector potentials, respectively, and [itex]P[/itex] is strictly positive and well behaved, and only nonzero in a domain [itex]\mathcal{D}[/itex].

    I want to find how the magnitude of

    [itex]\int \boldsymbol{\nabla} f dV[/itex] and see how it compares to the size of

    [itex]\int P dV[/itex] where we are integrating over all of space [itex]\mathbb{R}^n[/itex] for n the number of dimensions we are working in.

    I can use Green's functions to find [itex] f[/itex] in terms of [itex] P[/itex]. I test my results with examples, but when I integrate over all of space to find the magnitude of this term, it appears as if the value I find is dependent on the shape of the integration as I let it go to infinity. This seems non-physical since I have a fixed total input [itex]P[/itex].

    Any suggestions would be greatly appreciated.

    PS This problem comes from trying to figure out the momentum partitioning between irrotational and rotational fluid flows, if that adds any context.


  2. jcsd
  3. Jun 10, 2012 #2
    Re: Helmholtz decomposition and relative magnitudes of irrotational/solenoidal compon

    I don't think there's an easy trick here, so perhaps you could elaborate on this apparent dependence on the shape of the integration domain to see if that can be rooted out directly instead?
  4. Jun 11, 2012 #3
    Re: Helmholtz decomposition and relative magnitudes of irrotational/solenoidal compon

    The problem I was having was happening when I was dealing with integration of surfaces at infinity. This confuses me for a variety of reasons and I'm not sure that I even need to adress this issue. Instead, let's discuss what I am really interested:

    Assume [itex]\vec{P}[/itex] is nonzero, is in a finite domain [itex] \mathcal{D}[/itex], is only acting in the [itex]\hat{x}[/itex] direction, and goes to 0 (in a suitable mathematical way) as we approach [itex] \partial \mathcal{D}[/itex].

    I want to find
    [tex]I_1 =\int_{\mathcal{D}} \vec{\nabla} f \ dV, [/tex]

    where V is an integration over [itex]\mathcal{D}[/itex]. By the divergence theorem, I can rewrite the above integral as

    [tex] \int_{\partial \mathcal{D}} f \ d\vec{A}, [/tex]

    [itex]\nabla f=0[/itex] on the boundary which implies that f is a constant here. This implies f must be constant for all of region outside of [itex] \mathcal{D}[/itex], else it would induce a nonzero gradient and would violate P being non zero outside of [itex]\mathcal{D}[/itex].

    My intuition tells me that we should be even more restrictive with f, and say it is zero on the boundary. This comes from the fact that as we let the domain get large, if f is constant the integral [itex]I_1[\itex] keeps getting larger, which is nonsensical because the total integral of P is fixed. So this would imply that f must be identically zero on the boundary and for the rest of the domain.

    Again, this is all very hand wavy, so any hints as to actually showing this is rigorously the case would be most desirable.

    Last edited: Jun 11, 2012
  5. Jun 11, 2012 #4
    Re: Helmholtz decomposition and relative magnitudes of irrotational/solenoidal compon

    The points of logic that give me trouble: does [itex]\nabla f = 0[/itex] on the boundary necessarily imply that [itex]f[/itex] is constant outside the domain? ...it probably does, I guess, given the assumptions you've made, but that leads me to question the assumption that [itex]P[/itex] is zero outside the domain. I'm afraid I don't see why this is a useful property (even knowing how it plays into realizing that [itex]f[/itex] is constant outside).

    Anyway, if you see the wiki page on Helmholtz decomposition, it suggests that the form of the scalar potential is

    [tex]f(r) = -\int_{\mathcal D} \frac{\nabla' \cdot P(r') \; dV'}{4\pi |r - r'|} + \int_{\partial \mathcal D} \frac{P(r') \cdot dS'}{4\pi |r - r'|}[/tex]

    And that as long as [itex]P[/itex] is sufficiently well-behaved and falls off toward infinity, the surface integral vanishes.
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