# “How many” vector fields are conservative?

A vector field $$F:\mathbb R^n\to \mathbb R^n$$ is conservative if for some "potential function" $$f:\mathbb R^n\to \mathbb R$$, we have $$F=\nabla f$$.

I am intuitively wondering "how many" vector fields are conservative. Obviously this can be interpreted in multiple ways, which is why I have multiple questions:

• if we define a "uniform" measure on the space of such vector fields $$\mathcal F$$ for say $$n=2$$, is the set of conservative vector fields then measured larger than $$0$$?

• can we put certain unrestrictive assumptions, or "natural" assumptions, on $$F$$ to ensure that they are conservative?

• how often do we encounter nonconservative vector fields in practice, e.g. in physics?

• Given that we essentially compare smooth functions $\to \Bbb R$ with smooth functions $\to \Bbb R^n$, the conservative fields certainly form a zero-set – Hagen von Eitzen Jun 12 at 12:53

In $$\Bbb R^3$$, $$F$$ must satisfy $$\text{curl}\,F = 0$$, and so this is three (closed) conditions on the space of vector fields. In higher dimensions, you convert the vector field $$F$$ to a $$1$$-form $$\omega$$ and it must satisfy $$d\omega = 0$$, which is again $$\binom n2$$ closed conditions. So you have closed conditions, which certainly define closed submanifolds (in the infinite dimensional function space). I'm not sure how you would put a measure on this space, though.
Thermodynamics is certainly full of path-dependent line integrals. Heat and work arise as non-exact $$1$$-forms, for example.