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Math Help - Mean Curvature

  1. #1
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    Mean Curvature

    On page 256 http://zakuski.utsa.edu/~jagy/papers/Michigan_1991.pdf there is a formula for the mean curvature of a level set of a function. Im interested in the case n=3. How do you prove this formula? I can prove it for the case of a graph, so i thought i would use the implicit function theorem to make it a graph locally, but i dont see how I will get the original function in the final answer. Alternatively, I'm aware this equation is just the divergence of the normal vector, but how do you show that H can be defined this way? I seem to need to refer to a chart, but i can't seem to find one that gives a nice answer. Any help would be much appreciated
    Tom
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  2. #2
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    Re: Mean Curvature

    Yes since grad(F) is not zero we can assume that( without loss of generality) F_{n+1} \ne 0, according to the implicit function theorem x_{n+1}=f(x_1, x_2, \ldots , x_n) where f is a smooth function. Then the remaining is trivial.

    To see the relationship between mean curvature and the normal vector field, still let n be the unit normal vector of the hyper-surface M in R^{m+1}. \mathrm{I\!I}(X, Y)=\langle D_X n, Y \rangle be the second fundamental form of M, where D is the ordinary differential operator of R^{m+1}. The mean curvature H is the mean value of the principal curvatures, that is, eigenvalues of \mathrm{I\!I}.
    So H = \frac{1}{m} \mathrm{tr}(\mathrm{I\!I}).
    Let e_1, e_2, \ldots, e_m be an orthonormal basis of the tangent plane, we have
    H = \frac{1}{m} \mathrm{tr}(\mathrm{I\!I}) = \frac{1}{m} \sum_{i=1}^m \mathrm{I\!I}(e_i, e_i) = \frac{1}{m} \sum_{i=1}^m \langle D_{e_i} n, e_i\rangle .
    The last term is nothing but \frac{1}{m}\mathrm{div}(n). You may find a proof in some standard text book. But if you couldn't find one let me know, and I'll post one here.
    Thanks from tommyjbaby
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  3. #3
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    Re: Mean Curvature

    Hi. Thanks for your reply, and sorry for my late reply, I only just realised there was a comment! Im still a bit confused. Say for example I have M as a level set, so that n = gradf \ |gradf|. How does the mean curvature expression give me the "divergence in the flat metric" ie the standad \nabla = (d\dx, d\dy,... etc.). Wouldnt my basis have to be the standaard basis for R^3? but then, this isnt neccessarily a basis for the tangent plane of M? indeed, there would only be two vectors in that case. This is where im getting quite confused. any help would be much appreciated
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  4. #4
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    Re: Mean Curvature

    together with n the normal vector of the surface you got 3, which is a basis of R^3.
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  5. #5
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    Re: Mean Curvature

    So then why is it that i can then use the normal divergence from multivariable calculus? is it because the divergence doesent depend on the orthonormal basis chosen, so that i can then take the standard orthonormal basis on R^3? Because using the levi civita connection with the 3 basis vectors (n, e_1, e_2) where e_1, e_2 are the orthonormal basis for the tangent plane of M gives me a different expression than the "standard" divergence of R^3
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  6. #6
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    Re: Mean Curvature

    let e_3=n, div(v)=\sum <D_{e_i}v, e_i>
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