of $\psi$ passing through $x$. Then, we obtain a smooth vector field $e_1$ which, together with $N$, forms a global orthonormal frame. Now $${\langle{\nabla_NN},{N}\rangle}=\dfrac 12 N\cdot{\langle{N},{N}\rangle}=0.$$ On the other hand, as the Hessian is a symmetric tensor: $${\langle{\nabla_NN},{e_1}\rangle}=\nabla^2\psi(N,e_1)=\nabla^2\psi(e_1,N)={\langle{\nabla_{e_1}N},{N}\rangle}=\dfrac 12e_1\cdot{\langle{N},{N}\rangle}=0.$$ Hence $\nabla_NN=0$ as asserted. As each integral curve of $N=\nabla\psi$ is a geodesic meeting $\Sigma_1$ orthogonally, we see that $\psi$ is actually the distance function to $\Sigma_1$. We introduce coordinates on $\Omega$ as follows. For a fixed point $p\in\Omega$ consider the unique integral curve $\gamma$ of $N$ passing through $p$ and let $x\in\Sigma_1$ be the intersection of $\gamma$ with $\Sigma_1$ (note that $x$ is the foot of the unique geodesic which minimizes the distance from $p$ to $\Sigma_1$). Let $r$ be the distance of $p$ to $\Sigma_1$. We then have a map $
\Omega\to [0,a]\times\Sigma_1
$ which sends $p$ to $(r,x)$. Its inverse is the map $F: [0,a]\times\Sigma_1\to\Omega$ defined by $$F(r,x)=\exp_x(rN).$$ Note that $F$ is a diffeomeorphism; we call the pair $(r,x)$ the [*normal coordinates*]{} based on $\Sigma_1$. We introduce the arc-length $t$ on $\Sigma_1$ (with origin in any assigned point of $\Sigma_1$) and recall that $L$ is length of $\Sigma_1$ (which is also the length of $\Sigma_2)$). Let us compute 