I've just started writing dynamical systems and I was trying to find an example to show that if we are in a $2$ or $3$ dimensional real space "positional do not implies conservative".
In order to do this I'd like to better understand the relation betweeen the differential forms and a force field because I was thinking that if we are in $\mathbb R^2$ and we consider a vector field $F(x,y):=\begin{cases}\dot x=f_1(x,y)\\\dot y=f_2(x,y) \end{cases}$ and two points $a,b\in\mathbb R^2$ we could connect them through, for example, a rectangle and if we find a vector field such that $\partial_xf_1\ne\partial_xf_2$ we should have found a counterexample.
This method immediately reminded me the closeness of differential forms and I know there is a relation between the fact of being conservative for a force and the exactness of the form.
I've seen differential forms only in analysis and I'd like to better understand how these objects have a role in such a context.
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$\begingroup$ conservative means rot $F\neq 0$, so if by positional you mean "depending on the coordinates only" you just write rot F components in cartesian coordinates and make sure all but one term = 0. $\endgroup$– sleepyCommented Apr 14, 2021 at 15:54
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$\begingroup$ You want a counterexample that show that positional is not the same as conservative? Simply think to a ring force field (by the way, it is not artificial or pedantic: think to induced electric fields). Clearly it is positional and clearly $\oint \mathbf{F} \cdot d \mathbf{r} \neq 0$. $\endgroup$– Fausto VezzaroCommented Aug 20, 2021 at 21:30
1 Answer
We can view the force field as a one-form $F$ on the tangent bundle of $\mathbb{R}^n$. Saying that $F$ is conservative implies that there exists a zero-form $U$ on $\mathbb{R}^n$ (i.e., a scalar) such that $F = -\mathrm{d}U$. If this is the case, then along any curve $\gamma$ going from $a$ and $b$ we have $$ \int_\gamma F = - \int_\gamma \mathrm{d}U = U(a) - U(b) $$ by the definition of integration of differential forms. In other words, if $F$ is an exact one-form, then the integral of $F$ between two points $a$ and $b$ is independent of the path $\gamma$ taken, and is equal to minus the difference of the values of $U$ (the potential energy) at the endpoints.
We also know, however, that all exact forms are closed. Thus, if $U$ exists, then the two-form $\mathrm{d}F = \mathrm{d}^2 U$ vanishes. Moreover, since $\mathbb{R}^n$ is topologically trivial, all closed forms are exact; which implies that for any one-form $F$ having $\mathrm{d}F = 0$, there exists a potential zero-form $U$ such that $\mathrm{d}U = - F$.
In 2-D and 3-D, these statements are usually discussed in terms of the curl of $F$ and the gradient of $U$. In 3-D, it turns out that the components of the one-form $\star\mathrm{d}F$ in terms of some Cartesian basis (where $\star$ is the Hodge dual) are exactly the coordinate components of $\vec{\nabla} \times \vec{F}$; and the components of the one-form $dU$ are (of course) equal to coordinate components of $\vec{\nabla}U$. So in the more common "vector" language, the statement is that in $\mathbb{R}^n$, $\vec{\nabla}\times \vec{F} = 0$ if and only if $\vec{F} = - \vec{\nabla}U$ for some $U$.
