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?

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    $\begingroup$ 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 $\endgroup$ – 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.

  • $\begingroup$ Hello Dr. Shifrin. I am reading your excellent book "Multivariable Mathematics", and I have a question about one of your statements regarding the equality of mixed partials. I was wondering if you could take a look? :) math.stackexchange.com/questions/3262481/… $\endgroup$ – Ovi Jun 14 at 17:51

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