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Vector Calculus Theorems - Duality Question

  1. Aug 14, 2013 #1
    I'm trying to go over some vector analysis using forms & kind of noticed what look like random vector identities are more appropriately thought of, to me at least, as differential analogues of the classical integral theorems in the way Maxwell's equations can be cast in differential & integral form. However I'm missing a theorem:


    Integral Gradient Theorem: [itex] \smallint_{\vec{a}}^{\vec{b}} \nabla f \cdot d \vec{r} = f(\vec{b}) - f(\vec{a})[/itex]

    Integral Curl Theorem: [itex]\smallint_S (\nabla \times \vec{F}) \cdot \hat{n} dS = \smallint_{\partial S} \vec{F} \cdot d \vec{r}[/itex]

    Integral Divergence Theorem: [itex]\smallint_V \nabla \cdot \vec{F} dV = \smallint_{\partial V} \vec{F} \cdot \hat{n} dS[/itex]

    ---

    Derivative Gradient Theorem:

    Derivative Curl Theorem: [itex]\nabla \times (\nabla \phi) = 0[/itex]

    Derivative Divergence Theorem: [itex]\nabla \cdot (\nabla \times \vec{F}) = 0[/itex]

    ---

    Any ideas as to what I should put in there?

    Also, could one bluff an appropriate differential & integral theorem for the scalar or vector laplacians?
     
  2. jcsd
  3. Aug 14, 2013 #2
    Well I'm not really sure, but by looking at the Derivative Curl and Divergence theorems, the pattern seems to be:

    Derivative [operator] Theorem: [operator]( some expression ) = 0,

    where (some expression) is replaced by something that guarantees the right-hand integral is 0 (for any general F of this form) in the above 3 equations.
    For the curl and the divergence theorems, the form of [itex]\vec{F}[/itex] is respectively a gradient and a curl.

    The most obvious continuation (for the gradient theorem) would be the constant vector functions, since [itex]f(\vec{b})-f(\vec{a}) = 0[/itex] then. So my guess would be that the Derivative gradient theorem is:
    [itex]\nabla (const) = \vec{0}[/itex] which is fairly trivial.

    By the way, I don't think the derivative [operator] theorems are equivalent to their integral counterparts, because the derivative theorems are simply vector identities that can be checked from the definition to be 0, while the integral theorems must work even if the integrals are not 0.
    EDIT: This page has a lot of identities. Maybe you'll find a more satisfactory choice here http://en.wikipedia.org/wiki/Vector_calculus_identities
     
    Last edited: Aug 14, 2013
  4. Aug 14, 2013 #3
    Maybe you are thinking of Maxwell's equations in their integral and differential versions (which are equivalent).
    But they are proven equivalent because of the integral theorems (Green's, Stokes's and Divergence Theorems), not the 3 other identities. Or at least, for the major part of the proof.

    Oh and for the laplacian integral theorems, I think Green's identities are relevant.
     
    Last edited: Aug 14, 2013
  5. Aug 15, 2013 #4
    One moves between the differential & integral form of Maxwell's equations via Stokes & Gauss, so on face value if you can't think of the theorems in my post as dual to each other you can't think of the differential & integral form of Maxwell's equations as dual either right?

    Apparently either a constant function or the zero function work, depending on whether you're thinking in terms of cohomologies or reduced cohomologies :tongue: I was just hoping there was something a bit more substantial you could put in it's place I guess...

    If we could find a way to fit the Laplacian into all of this it would be really nice, but I don't see what to do with it right now...
     
  6. Aug 15, 2013 #5

    lurflurf

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    This a a problem in the traditional vector calculus. We have many variations of each formula.
    Your first three are examples of the fundamental theorem of calculus (Stokes theorem)
    $$\int_{\partial \Omega} \omega=\int_\Omega \mathrm{d}\omega$$
    The other two are examples of exact forms being closed
    $$\mathrm{d}^2\omega=0$$
    you can dress it as an integral if you like
    $$\int_{\partial \Omega} \mathrm{d}\omega=\int_{\partial^2 \Omega} \omega=\int_{ \Omega} \mathrm{d}^2\omega=0$$
    I do not think of these as dual because the second set have one more derivative.
    That is the reason there are two instead of three.
    The Laplacian arises form
    $$\Delta=d\delta+\delta d$$
    where δ is the adjoint of d
     
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