Derivative of area gives mean curvature?

Question

My lecturer made a comment today about how 'the derivative of the area gives you the mean curvature' but I'm not really sure what he meant?

I guess what I mean is that I don't understand this definition:

"A surface $M \in \mathbb R^3$ is minimal if and only if it is a critical point of the area functional for all compactly supported variations"

I understand that a critical point is a stationary point (ie $d/dt=0$) and that if $H=0$ a surface is minimal. That is all I currently understand in this definition.

If anyone could explain this simply (as possible) I would really appreciate it!

Answer 1

First let us recall that the area $A$ of a parametrized regular surface $S$ given by the parametrization $f \colon U \to \mathbb{R}^3$ is the integral $$ A = \int_U d S $$ where $d S$ is the surface's volume form, known also as the surface element. Of course, $U \subseteq \mathbb{R}^2$. A standard fact is that the area does not depend on the parametrization, so we can think about the area on an unparametrized surface, if we want.

To talk about the derivative of $A$ you need to have a smooth family of (parametrized regular) surfaces $f^t \colon U \to \mathbb{R}^3$ such that $f^0 = f$, and $t \in (-\varepsilon,\varepsilon)$. Such a family $S^t$ is called a variation of $S$. In this case we have a function $A \colon (-\varepsilon,\varepsilon) \to \mathbb{R}$.

Thus the question is what the following derivative would be: $$ \tfrac{d}{d t} A^t = \tfrac{d}{d t} \int_U d S^t = \int_U \tfrac{d}{d t} d S^t $$

The volume forms $d S^t$ are quantities that are well defined at each point of $S$, equivalently at each $u \in U$, and therefore swapping the operators in the second equality of the last display makes sense.

Now, each point $p = f(u)$ on $S$ gives rise to a curve $p^t$ of point on $S^t$. The velocity $\dot{p} = \left.\tfrac{d}{dt}\right|_{t=0} p^t$ of the curves $p^t$ at $t = 0$ is called the variational vector field $V$ along $S$.

The variation $S^t$ is called normal if $V = \phi N$ where $N$ is the unit normal vector field along $S$.

The precise statement is that for a normal variation with variational vector $V = \phi N$ the derivative of the volume form at $t=0$ will involve the mean curvature as follows: $$ \left.\tfrac{d}{dt}\right|_{t=0} dS^t = 2 \phi H \, d S $$ where $H$ is the mean curvature of $S$

Answer 2

Thank Yuri who stated the formula. Here comes a simple proof. Using Yuri's notation, $$dS^t=\sqrt{\det(g_t)}du^1\wedge du^2$$ Where ${(g_t)}_{ij}=\langle f_i, f_j\rangle=\langle \frac{\partial}{\partial u^i}, \frac{\partial}{\partial u^j}\rangle$ is the first fundamental form. From wiki, we have $\frac{d}{dt}(\det(g_t))=\det(g_t)tr(g_t^{-1}\dot{g_t})$, hence $$\frac{d}{dt}dS^t=\frac{d}{dt}\sqrt{\det(g_t)}du^1\wedge du^2$$ $$=\frac{1}{2\sqrt{\det(g_t)}}\det(g_t)tr(g_t^{-1}\dot{g_t})du^1\wedge du^2$$ $$=\frac{1}{2}tr(g_t^{-1}\dot{g_t})dS^t$$

Also from wiki, $H=\frac{1}{2}tr(g_0^{-1}\mathrm{I\!I})$, so if we can show that $\dot{g_0}=2\phi\mathrm{I\!I}$ we'll get the desired result: $$\frac{d}{dt}dS^t|_{t=0}=\frac{1}{2}tr(g_0^{-1}\dot{g_0})dS=\phi tr(g_0^{-1}\mathrm{I\!I})dS=2\phi H dS$$ To show that $\dot{g_0}=2\phi\mathrm{I\!I}$, it's a direct computation: $$(\dot{g_t})_{ij}=\frac{d}{dt}\langle f_i,f_j\rangle=\langle \frac{d}{dt}f_i,f_j\rangle+\langle f_i,\frac{d}{dt}f_j\rangle$$ $$= \langle \frac{\partial}{\partial_{u^i}}f_t,f_j\rangle+\langle f_i,\frac{\partial}{\partial_{u^j}}f_t\rangle$$ $$= \langle \frac{\partial}{\partial_{u^i}}\phi N,f_j\rangle+\langle f_i,\frac{\partial}{\partial_{u^j}}\phi N\rangle$$ $$= \phi \langle \frac{\partial}{\partial_{u^i}} N,f_j\rangle+\phi \langle f_i,\frac{\partial}{\partial_{u^j}} N\rangle$$ $$= \phi \langle N_i,f_j\rangle+\phi \langle f_i,N_j\rangle=2\phi\mathrm{I\!I}_{ij}$$