TITLE: When a vector field can be scaled to form a conservative vector field
QUESTION [19 upvotes]: Consider a vector field given by its components $g_i(x_1, \dots, x_n)$. It is well known that necessary and sufficient condition for a following system
$$
\frac{\partial f}{\partial x_i} = g_i(x_1, \dots, x_n)
$$
to have a solution is circulation of $\vec g$ being zero around any closed path:
$$
\oint \vec g\, d\vec r = 0.
$$
I wonder if a given non-conservative field $\vec g$ could be scaled some way so the new field will be conservative. In other words there exists $\mu(x_1, \dots x_n) \neq 0$ such the following system has a solution
$$
\frac{\partial f}{\partial x_i} = \mu(x_1,\dots,x_n) g_i(x_1, \dots, x_n).
$$
The scalar field $\mu$ acts like a kind of an integrating factor. For $n = 2$ such $\mu$ always exists, but for higher $n$ it does not seem to do.
For $n=2$ the $\mu$ may be obtained by solving the equation
$$
0 = \begin{vmatrix}
\partial_x & \partial _y\\
\mu g_x & \mu g_y
\end{vmatrix} = 
\mu \left(\frac{\partial g_y}{\partial x} - \frac{\partial g_x}{\partial y}\right)
+ g_y \frac{\partial \mu}{\partial x} -  g_x \frac{\partial \mu}{\partial y}.
$$
Then $\operatorname{rot} \mu \vec g = \vec 0$.
For $n > 2$ this approach (zerowing the rotor) leads to a system of PDE's which is almost never consistent.

REPLY [2 votes]: There's a very good answer to an equivalent question here.
Take the example of 3D, where I'll restate things avoiding the language of differential forms. The existence of $\mu$ is locally equivalent to $\nabla\times \vec{g}$ being orthogonal to $\vec{g}$ (at least if we relax the requirement that $\mu$ must be nonzero everywhere). To show that this condition is necessary, suppose we have found $f$ and $\mu$ such that $\nabla f=\mu\vec{g}$, and take the curl:
$$
0=\nabla\times\nabla f=\nabla\times(\mu \vec{g})=\mu \nabla\times\vec{g}+(\nabla\mu) \times\vec{g}\\
\implies \vec{g}\cdot\nabla\times\vec{g}=\mu^{-1}\vec{g}\cdot(\vec{g}\times\nabla\mu)=0
$$
A simple counterexample is $\vec{g}=(y,0,1)$, so $\nabla\times\vec{g}=(0,0,-1)$, and these are not perpendicular. There's no $\mu$ such that $\mu\vec{g}$ is conservative, even locally.