components of normal field are Jacobi fields? Given a minimal surface $\Sigma$ in $R^3$ with associated normal field $N$, I am told that each of the components of $N$ is a Jacobi field, meaning that $Lu=0$ where L is the stability (Jacobi) operator and $u$ is the function $u(x) = \langle N(x), e_i \rangle$, with $e_1 = (1,0,0)$, etc.
Supposedly this follows from looking at a one parameter family of isometries and the second variation formula, as given in e.g. Minicozzi and Colding's course on minimal surfaces, which I am studying from.
I am just exploring this material for the first time, I guess I'm missing something obvious but I just can't see the conclusion.  As in the reference, the motivation is in obtaining a proof that any minimal graph is stable.  
Intuition and rigor both welcome!  
 A: The other answer has given a good intuitive explanation. Let me just give a computational one. The Jacobi operator is (1.147 in the book)
$$L u = \Delta u + |A|^2 u + \text{Ric}_M(N, N)u = \Delta u + |A|^2 u .$$
as $M = \mathbb R^3$ has $\text{Ric}_M = 0$. To see this $u = \langle N, v\rangle $ satisfies $Lu_ = 0$ (where $v$ is a fixed vector in $\mathbb R^3$), note that if we fix a normal coordinate at $x$ in the minimal surface, then (note that the $e_i$'s orthonormal basis of the minimal surface, not that of $\mathbb R^3$)
$$\begin{split}
\Delta u &= \nabla_{e_i} \nabla_{e_i} u \\
&= \nabla_{e_i} \nabla_{e_i} \langle N, v\rangle  \\
&=  \nabla_{e_i} \langle \nabla_{e_i}N, v\rangle \\
&= - \nabla_{e_i}\langle A_{ij} e_j, v\rangle \\
&= -\langle \nabla_{e_i}A_{ij} e_j + A_{ij} \nabla^{\mathbb R^3} _{e_i } e_j, v\rangle \\
&=  -\langle \nabla_{e_j}A_{ii} e_j + A_{ij}A_{ij} N, v\rangle 
\end{split}$$
Note that in the last equaliy we used the Codazzi equation $\nabla_{e_k} A_{ij} = \nabla_{e_i} A_{kj}$ and the fact that $\nabla^{\mathbb R^3} _{e_i } e_j = A_{ij} N$. Then the first term is zero as the surface is minimal, thus 
$$\Delta u = - A_{ij}^2 \langle N, v\rangle = -|A|^2 u.$$
This is the same as $Lu= 0$. 
A: Consider the variation of $\Sigma$ generated by translation in the direction $e_i$; i.e. $X(x,t) = X_0(x) + te_i$ for $X_0$ a parametrization of $\Sigma$. This is a family of isometries so the area element is constant in time, and it has normal speed $u = \langle e_i, N\rangle$; so the second variation formula tells us $Lu = 0$.
