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Mean curvature

by hedipaldi
Tags: curvature
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hedipaldi
#1
Feb16-12, 10:26 AM
P: 206
Hi,
I know that the mean curvature at an extremum point where the function vanishes must be nonpositive.can this say someting about the sign of the mean curvature at the farthest point on a close surface from the origin?
Thank's
Hedi
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lavinia
#2
Feb17-12, 07:43 AM
Sci Advisor
P: 1,716
Quote Quote by hedipaldi View Post
Hi,
I know that the mean curvature at an extremum point where the function vanishes must be nonpositive.can this say someting about the sign of the mean curvature at the farthest point on a close surface from the origin?
Thank's
Hedi
You need to explain what you are talking about more clearly. What function?
hedipaldi
#3
Feb17-12, 09:25 AM
P: 206
I mean a real smooth function of two variables whose graph is a closed surface in R3

lavinia
#4
Feb17-12, 10:40 AM
Sci Advisor
P: 1,716
Mean curvature

Quote Quote by hedipaldi View Post
I mean a real smooth function of two variables whose graph is a closed surface in R3
so you mean the mean curvature of the graph?

Isn't the mean curvature of the standard sphere strictly positive - in fact for any surface of positive Gauss curvature?

There is a therem that says that any closed surface in 3 space must have a point of positive Gauss curvature. At this point the mean curvature is positive.
hedipaldi
#5
Feb17-12, 01:43 PM
P: 206
But Gauss curvature is positive also if the two principal curvatures are negative.at the farthest point from the origin, the surface is enclosed within a ball whose boundary shares a common tangent plane with the surface at this point.It seems that this implies that the mean curvature of the surface at this point must have a definite sign.I don't know what sign and how to explain it.
lavinia
#6
Feb17-12, 01:47 PM
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P: 1,716
Quote Quote by hedipaldi View Post
But Gauss curvature is positive also if the two principal curvatures are negative.at the farthest point from the origin, the surface is enclosed within a ball whose boundary shares a common tangent plane with the surface at this point.It seems that this implies that the mean curvature of the surface at this point must have a definite sign.I don't know what sign and how to explain it.
right. My mistake. But for a sphere or an ellipsoid or any convex surface of positive curvature, the principal curvatures should both be positive. Yes?.

I am having trouble visualizing the case of both negative principal curvatures. May it can happen at a single point but in a region?can you give an example?
hedipaldi
#7
Feb17-12, 01:59 PM
P: 206
The surface must be convex at the farthest point so if this forces positive mean corvature'we are done,but i am not sure of it.
hedipaldi
#8
Feb17-12, 03:51 PM
P: 206
At a saddle point Gauss curvature is negative so one of the principal directions is negative.this may help for imagining a negative principal direction.
lavinia
#9
Feb17-12, 06:38 PM
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Quote Quote by hedipaldi View Post
At a saddle point Gauss curvature is negative so one of the principal directions is negative.this may help for imagining a negative principal direction.
Ok. Now I see what your question is.

A classical argument says - surround the surface with a very large sphere centered at the origin and let it's radius shrink until it first touches the surface.At this point the surface and the sphere are tangent and the entire surface lies on the inside of the sphere. Therefore the surface must be convex at this point.
hedipaldi
#10
Feb18-12, 10:42 AM
P: 206
Right.Doed it implies something about the sign of the mean curvature at the farthest point?
lavinia
#11
Feb18-12, 01:29 PM
Sci Advisor
P: 1,716
Quote Quote by hedipaldi View Post
Right.Doed it implies something about the sign of the mean curvature at the farthest point?
I think so. In order for the surface to be tangent it must curve away from the surrounding sphere in all directions so the principal curvatures must both be positive.


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