# Derivative of area gives mean curvature?

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!

• Uhm...what was the contest? What was he talking about in particular? – geodude Mar 17 '14 at 19:05
• en.wikipedia.org/wiki/Minimal_surface – Xipan Xiao Mar 17 '14 at 19:09
• I'm doing a course on geometry of surfaces, and we're learning about Gaussian, Mean, Principal curvatures...and he said that ^^ but I don't see how that really related. I'm having trouble actually finding the mathematical 'equation' so to speak that summarises what he said? And what significance this has? – Sarah Jayne Mar 17 '14 at 19:10
• @XipanXiao I am aware of what minimal surfaces are (well that H=0 implies a minimal surface) but what does this formula have to do with it? And what is said formula? – Sarah Jayne Mar 17 '14 at 19:11

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 $S$

• Okay I think I follow that, however I'm not really sure I understand what you mean by a 'normal variation with variational vector'. Does the last statement that you wrote have relevance to this definition: "A surface M in R^3 is minimal if and only if it is a critical point of the area functional for all compactly supported variations" in that it is a critical point when the derivative = 0? Or am I misunderstanding this? – Sarah Jayne Mar 18 '14 at 13:28
• It does not matter the variation is normal or not. But choosing a normal variation simplify the calculation. One can prove that a tangent variation does not contribute to area change, just like re-parameterizing a curve does not change its length. Yes a minimal surface is a critical point when the derivative vanishes( for any variations), that is, H vanishes. – Xipan Xiao Mar 18 '14 at 13:37
• So what exactly happens in this formula "$\left.\tfrac{d}{dt}\right|_{t=0} dS^t = 2 \phi H \, d S$" to define it as a minimal surface? Sorry if this is simple, I think there is something obvious that I'm not seeing. (It probably doesn't help that I am not actually familiar with what a 'variation' is) – Sarah Jayne Mar 18 '14 at 13:52
• Integral both sides, you'll get $\frac{d}{dt}|_{t=0}S^t=\int_S 2\phi H dS$. If a surface is minimal, the derivative vanishes for any kinds of normal vector field V(the variation), this has to be the case when H is identically zero on S( otherwise you can always choose a $\phi$ to make the integral non-zero. – Xipan Xiao Mar 18 '14 at 15:02
• @Takashi Hi Takashi! This is all quite standard, I can't come up with the best reference now, but maybe you should take a look at my thesis where I give a more detailed exposition and bibliography. See section 1.3 "Variations of Riemannian hypersurfaces" there. – Yuri Vyatkin Jan 18 at 6:26

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: $$(\dot{g_t})_{ij}=\frac{d}{dt}\langle f_i,f_j\rangle=\langle \frac{d}{dt}f_i,f_j\rangle+\langle f_i,\frac{d}{dt}f_j\rangle$$ $$= \langle \frac{\partial}{\partial_{u^i}}f_t,f_j\rangle+\langle f_i,\frac{\partial}{\partial_{u^j}}f_t\rangle$$ $$= \langle \frac{\partial}{\partial_{u^i}}\phi N,f_j\rangle+\langle f_i,\frac{\partial}{\partial_{u^j}}\phi N\rangle$$ $$= \phi \langle \frac{\partial}{\partial_{u^i}} N,f_j\rangle+\phi \langle f_i,\frac{\partial}{\partial_{u^j}} N\rangle$$ $$= \phi \langle N_i,f_j\rangle+\phi \langle f_i,N_j\rangle=2\phi\mathrm{I\!I}_{ij}$$

• Thank you for this calculation. It it is very instructive and should be accessible for @SarahJayne – Yuri Vyatkin Mar 18 '14 at 19:31
• @YuriVyatkin I dont understand the concept of the family $S^t$. Why can't we just use $S$? – Tesla Jan 3 at 18:29