Coordinates that map vertical planes to the planes of constant value of a smooth function

Question

Suppose $ f: \mathbb{R}^3 \to \mathbb{R}$ is a smooth function and $V = \nabla f$ is a smooth vector field. Let $\theta : \mathbb{R} \times \mathbb{R}^3 \to \mathbb{R}^3$ be the integral curves of $V$. It would be very good for my problem if I could construct a chart $\psi = (t,\theta,\varphi)$ with the property that $(f\circ \psi)^{-1}(a) = \{p\} \times [0,2\pi]\times [0,\pi]$ is some vertical two dimensional plane in the domain of $\psi$ for each $a$ in the image of $f$. I would very much like to determine the Laplacian in these coordinates. I have taken an introductory course to smooth manifolds during my master studies, so I have a basic understanding of smooth manifolds. I feel this should be possible for the particular $f$ that I am using, but I am not sure how to do it. (Please let me know if the question is not well-posed, I am a beginner, so that would be helpful.) If you know how it is possible to solve my problem, it would be great if you can provide some details, such that I can read up on it and see if it works.

Answer 1

You can do it away from the zero of $V$ (i.e. the critical point of $f$).

Let $q \in \mathbb R^3$ be a point where $V(q)\neq 0$. By the Straightening theorem for vector fields, there is an open set $U$ with coordinates $(t, x, y)$ and a local coordinates $\varphi : U \to \mathbb R^3$ so that $V = \frac{\partial }{\partial t}$ in $\varphi(U)$. By shrinking $U$, we assume $U = I \times V$ for some interval $I$. Under this coordinates, the integral curve $\theta$ is given by

$$ \theta (t_0, (t, x, y )) = (t_0+ t, x, y)$$

and after a scaling in the $(x, y)$ direction, this coordinates has the properties you need.