TITLE: Intuition for mean curvature
QUESTION [23 upvotes]: I would like to get some intuitive feeling for the mean curvature. The mean curvature of a hypersurface in a Riemannian manifold by definition is the trace of the second fundamental form.
Is there any good picture to have in mind, when dealing with it?

REPLY [3 votes]: Let $M$ be an oriented hypersurface in an oriented euclidean space $E$.
The normal curvature at a point $x_0\in M$ measures how much it is displaced, in the positive normal direction, with respect to the points of $M$ that are near it, in the same way that the Laplacian of a function measures how much the value of a function at a point differs from the average of the points that are near. 
In fact, if $M$ is the graph of a nearly constant function $f:\mathbb R^n\to\mathbb R$ (for example, the vibrating membrane of a drum at some instant) then the Laplacian of $f$ is a good approximation of the normal curvature of $M$.
More generally, if $M$ is possibly not nearly flat, then for any $x_0\in M$ you can find the hyperplane $\Pi$ that is tangent to $M$ at $x_0$, and find a parametrization $f:\Pi\to M$ (only defined on a neighbourhood of $x_0$) such that $p\circ f=id_\Pi$, where $p:E\to\Pi$ is the orthogonal projection. This parametrization shows that $M$ is locally the graph of a function $g=q\circ p$ (where $q$ is the projection on the orthogonal complement of $\Pi$) such that $dg(x_0)=0$. Then, the normal curvature of $M$ at $x_0$ equals $\Delta g(x_0)$.