# Mean Curvature

• May 18th 2013, 11:02 AM
tommyjbaby
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
• May 18th 2013, 05:10 PM
xxp9
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.
• June 2nd 2013, 02:06 PM
tommyjbaby
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
• June 2nd 2013, 04:14 PM
xxp9
Re: Mean Curvature
together with n the normal vector of the surface you got 3, which is a basis of R^3.
• June 3rd 2013, 01:24 AM
tommyjbaby
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
• June 3rd 2013, 06:05 AM
xxp9
Re: Mean Curvature
let e_3=n, div(v)=\sum <D_{e_i}v, e_i>