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Vector Integration in ND

  1. Feb 11, 2016 #1
    In N D I want to do an integral of the flux through an N-1 D surface. The usual vector calculus integration theorems say I can integrate around the perimeter of the surface. OK, but that perimeter is now N-2 D. In 4D it could be a cube or a 2-sphere. I can't use Gauss' theorem because the surface isn't closed. Now what?
  2. jcsd
  3. Feb 12, 2016 #2


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    The surface doesn't have to be closed. The boundary of the surface is closed.
  4. Feb 12, 2016 #3
    (If you're not familiar with differential forms, this is a good opportunity to learn about them, since they are exactly the appropriate objects to use in the answer to your question. This may seem a bit technical, but once you are familiar with differential forms, it should make perfect sense.)

    Let's say we're talking about a volume-preserving flow, generated by a vector field V on Euclidean space n. Let ω denote the volume form.

    That V preserves ω means that the Lie derivative LVω = 0.

    In other words,

    d(ιVω) + ιV(dω) = 0, ​

    where ιV denotes the contraction of a differential form by V.

    Since ω is already an n-dimensional form on n, we have

    dω = 0 ​

    automatically, which means that

    d(ιVω) = 0​

    Now the flux form τ, i.e., the (n-1)-form that must be integrated over an (n-1)-dimensional manifold M to get the flux of V through M, is given by just contracting ω by V:

    τ = ιVω​

    and so from the previous equation, the exterior derivative of the flux form vanishes everywhere:

    dτ = 0.​

    In Euclidean space (and other spaces with sufficiently simple topology), when the exterior derivative d of any differential form σ vanishes everywhere, this implies by the de Rham theorem that σ itself is the exterior derivative of some form of one lower dimension. Hence we have that there exists an (n-2)-dimensional form μ such that

    τ = dμ.​

    Now, the general version of Stokes's theorem that works in any dimension states that for a compact manifold M and a differential form ρ of dimension = dim(M) - 1, we have

    M dρ = ∫∂M ρ.​

    (In fact, all the "usual vector calculus integration theorems" you refer to can be shown to be special cases of this general Stokes's theorem for differential forms.)

    Applying this Stokes's theorem to the flux form τ with

    τ = dμ,​

    we have

    M τ = ∫M dμ​

    and so we get

    flux through M = ∫M τ = ∫∂M μ.​

    This is exactly what you asked about: How to calculate the flux through an (n-1)-dimensional manifold by integrating something else over its boundary.
  5. Feb 12, 2016 #4
    Golly! Thanks for going to all this effort for lil' ole me.
  6. Feb 12, 2016 #5
    thanks for sharing even I was in need of this equation..
  7. Feb 13, 2016 #6


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    One thing to underscore in zinq's post is that if you start with an arbitrary vector field then its divergence may not be zero. In this case it is not possible to find an N-2 form such that

    τ = dμ

    This is true even in three dimensions.

    zinq's point is that if you can solve this equation, then the flow must be volume preserving. The physical term for a volume preserving flow is that it is "incompressible".

    A formal way to look at this is to use the identity d(dμ)=0 which is actually true for any differential form, then one has dτ = d(dμ) = 0.
    Last edited: Feb 13, 2016
  8. Feb 13, 2016 #7
    Are you thinking of Stokes' Theorem? For Gauss' Theorem

    " the outward flux of a vector field through a closed surface is equal to the volume integral of the divergence over the region inside the surface." -- Wikipedia
  9. Feb 13, 2016 #8


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    I think closed manifold here means a manifold without boundary. An N-1 manifold with boundary - as you assumed - is not closed but its boundary is.

    For instance, a disk in 3 space is a 2 - manifold with boundary a circle. So it is not a closed manifold. But its boundary, the circle, is closed.

    In general, the boundary of a boundary is empty

    Gauss's Theorem would apply if the N-1 manifold did not have a boundary and itself was the boundary of an N-dimensional volume. So for instance, it would apply if instead of a disk, the surface was a sphere.
    Last edited: Feb 13, 2016
  10. Feb 15, 2016 #9


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    Can you think of a case when it is not? I think in any chain complex you have ##\partial^2 =0##, for a general boundary operator. So a counter would come from somewhere else.
  11. Feb 17, 2016 #10


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    Good question. Don't know of any.
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