# Proof check for critical point definition with mean curvature

I'm currently trying to prove:

"Definition: We say that a surface $S \in R^3$ is minimal if it is a critical point for the area functional"

Starting with this:

"If we consider a family of smooth (differentiable at every point) regular surfaces patches $\sigma^{\tau} : U \rightarrow R^3$, such that $\tau \in (-\delta, \delta) : \delta > 0$ and $\sigma^0 = \sigma$. This family of surface patches is known as the variation of $S$; such that the surface variation is defined as the function $\phi: U \rightarrow R^3$ given by:

$$\displaystyle{\phi = \dot{\sigma}^{\tau}|_{\tau=0}} \hspace{2cm} where \quad \cdot = \dfrac{d}{d\tau}$$

Let $\psi$ be a simple closed curve that is contained within $U$, then $\psi$ corresponds to a simple closed curve $\gamma^{\tau} = \sigma^{\tau} \circ \psi$ in the surface patch $\sigma^{\tau}$, and we consequently can define the area function $A(\tau)$ to be the area of $\sigma^{\tau}$ within $\gamma^{\tau}$:

$$A(\tau) = \iint_{\psi} d A_{\sigma^{\tau}}$$

Note: if we consider a family of surface patches with a fixed boundary curve, $\gamma$, then $\gamma^{\tau} = \gamma$ for all $\tau$, and hence $\phi^{\tau}(u,v)=0$ when $(u,v)$ is a point on $\psi$. \"

Then if we Let $\phi^{\tau}= \dot{\sigma}^{\tau}$ so that $\phi^0=\phi$, and let $N^{\tau}$ be the standard unit normal of $\sigma^{\tau}$. There are smooth functions $\alpha^{\tau}, \beta^{\tau} \text{and} \gamma^{\tau}$ of $(u,v,\tau)$ such that:

$$\phi^{\tau} = \alpha^{\tau}N^{\tau} + \beta^{\tau}\sigma_u^{\tau} + \gamma^{\tau} \sigma_v^{\tau}$$

If we take $\beta=\gamma=0$ how do you prove:

$$\dot{A}(0) = -2 \iint_{\psi} H(EG-F^2)^{\frac{1}{2}} \alpha\,du\,dv$$

-
You'd better explain some of your notations. From your context, I guess that $\sigma^{\tau}$ is a family of surfaces, but then what does $\dot{\sigma^{\tau}}$ mean? OK, I assume that it means the tangent vector $\frac{\partial}{\partial \tau}(\sigma^{\tau})$. I think I got your notation and question now. What you're asking is: why do we need only to care the normal part of a variation? – Xipan Xiao Mar 19 '14 at 23:49
What you're asking is: what is simplified if we only consider the normal part of a variation? – Xipan Xiao Mar 19 '14 at 23:55
Okay I've actually added the start of the proof (where I've got up to) and the objective, but I'm lost now... – Sarah Jayne Mar 20 '14 at 9:14