TITLE: In a contact manifold, is every tranverse 1-foliation given by some Reeb vector field?
QUESTION [9 upvotes]: Let $M$ be a contact manifold, and let $F$ be an oriented 1-dimensional foliation that is transverse to the contact structure. 
Is there a contact form $\alpha$ whose associated Reeb vector field generates the foliation $F$?

REPLY [3 votes]: If there is already a contact structure given say $\eta$ one might try to see if there exists a function $f$ so that Reeb vector field $R_f$ of $\eta_f = f\eta$ lies inside the given line bundle $\ell$. Assume that you can give $\ell$ as the kernel of two independent differential 1-forms $\eta_1,\eta_2$. Then we want $\eta_1(R_f)=\eta_2(R_f)=0$ which is if and only if $d\eta_f = \theta_1 \wedge \theta_2$. Let $E_i = ker(\theta_i)$ and $\Delta = ker(\eta)$ which is iff $d\eta_f|_{E_i}=0$ for both $i$. This condition could be written as a system of PDE.
Take $X_i \subset E_i \cap \Delta$ a vector field. Since $\ell$ is transverse to $\Delta$, $X_1$ and $X_2$ are distinct. And take $Z_i$ any other vector fields independent from $X_i$ inside $E_i$. Then the conditions above are true iff
$$(fd\eta + df \wedge \eta)(X_i,Z_i)=0$$
which is iff
$$X_i(ln(f)) = d\eta(X_i,\frac{Z_i}{\eta(Z_i)}) = \eta([X_i,\frac{Z_i}{\eta(Z_i)}])$$
or in more suggestive notation
$$\mathcal{L}_{X_i}(ln(f)) = \eta(\mathcal{L}_{X_i}W_i)$$
where $W_i=\frac{Z_i}{\eta(Z_i)}$. Hence $f$ must be a solution to this system of PDE.