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Does this theorem have a name?

  1. Aug 28, 2011 #1
    For a surface S bounded by a curve C,

    [tex]\int_{C} \mathbf{t} \times \mathbf{n}ds = -2\int_{S} H \mathbf{n}da[/tex]

    t is the unit tangent to C, and n is the unit normal to the surface S. H is the mean curvature of S. It can be derived from Stokes' theorem, but it seems like the kind of result that might have been known earlier.
     
  2. jcsd
  3. Sep 2, 2011 #2
    For any constant vector c,

    [tex] \mathbf{c} \cdot \int_C \mathbf{t} \times \mathbf{n} ds = \int_C (\mathbf{n} \times \mathbf{c}) \cdot \mathbf{t} ds =\int_S \nabla \times (\mathbf{n} \times \mathbf{c}) \cdot \mathbf{n}da [/tex]

    Using the vector identity

    [tex]\nabla \times (\mathbf{a} \times \mathbf{b}) = \mathbf{b} \cdot \nabla \mathbf{a} - \mathbf{a} \cdot \nabla \mathbf{b} - (\nabla \cdot \mathbf{a} ) \mathbf{b} + (\nabla \cdot \mathbf{b} )\mathbf{a} [/tex]

    and the fact that c is constant, the above integral becomes

    [tex]\int_S [\mathbf{c}\cdot \nabla \mathbf{n} - (\nabla \cdot \mathbf{n})\mathbf{c} ] \cdot \mathbf{n} da
    =
    \mathbf{c} \cdot \int_S (\nabla \mathbf{n} ) \cdot \mathbf{n} - (\nabla \cdot \mathbf{n }) \mathbf{n} da [/tex]

    The first term in the integrand is zero, since it is just the gradient of the length of the unit normal. Also, since the vector c was arbitrary, its dot product with the integral can be dropped:


    [tex]\int_C \mathbf{t} \times \mathbf{n} ds = -\int_S (\nabla \cdot \mathbf{n} ) \mathbf{n} da [/tex]

    Depending on how you define the normal (outward or inward), div n = -2H up to a sign*. So that gives the result

    [tex]\int_C \mathbf{t} \times \mathbf{n} ds = \int_S 2H \mathbf{n} da [/tex]

    I had an extra minus sign in the first post.


    * The mean curvature H of a surface is equal to half the trace of the second fundamental form B: 2H = gjkbjk. The components of the second fundamental form can be defined as

    [tex]b_{jk} = -\mathbf{x}_j \cdot \mathbf{n}_{,k} = -\mathbf{x}_j \cdot \mathbf{n}_{,k} = - n^i_{{ };k}\mathbf{x}_j \cdot \mathbf{x}_i = -n^i_{{ };k}g_{ij}[/tex]

    Multiplying both sides by the inverse of g_ij and then summing the diagonal elements gives

    [tex]g^{jk}b_{jk} = - n^k_{{ };k} = -\nabla \cdot \mathbf{n}[/tex]
     
    Last edited: Sep 2, 2011
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