# Locally every force admits a potential?

I have a little doubt about a force being or not conservative. Well, as I understood, some forces cannot be expressed as exterior derivative of some scalar potential because the work done by the force is path dependent, in other words, it's not possible to determine the work based only on some value at the end points.

But my question is: locally every force admits some potential? In other words, if $F$ is a force field defined on some subset $U\subset\mathbb{R}^n$, and if $a\in U$, is there always some $\epsilon$-neighborhood of $a$, $B_\epsilon(a)$ such that the restriction of $F$ to that $B_\epsilon(a)$ admits potential? Because for me it seems that for any force, it's possible to find some small region in which the "path-dependance" won't interfere that much. Is this correct?

And in this case, I'm assuming that space is flat. If we're working with some curved manifold $M$, this idea yet applies?

Sorry if that's silly, it's just a thought that I wanted to confirm. Thanks in advance for your aid.

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OP might find it interesting that in 1D (with one physical d.o.f., and, say, with forces with no explicit time dependence), it is always possibly to locally find a Hamiltonian formulation, cf. this Phys.SE answer. –  Qmechanic Feb 26 '13 at 16:05

No, most force fields refuse to be conservative. The path-independence is a nontrivial constraint in an arbitrarily small region of space, an arbitrary neighborhood.

If the force field is conservative, it must be $$\nabla\times \vec F = 0$$ because $F = -\nabla\Phi$. It's clear that the curl of $\vec F$ may be nonzero even if you look at a small neighborhood. For example, assume that $\vec F = (0,x,0)$. The curl of this field is $(0,0,1)$. It's just nonzero, even at the very point $(x,y,z)=(0,0,0)$ or any other point, for that matter. If you draw an arbitrarily small area $dS$ in the $xy$-plane, e.g. a small square in the $z=0$ plane, the integral of $\vec F$ over the boundary of the area will be nonzero so, equivalently, the work will be path-dependent. The work will be proportional to $dS$, namely $1\cdot dS$ in my case, but that's the normal scaling.

To make the force field "overlook" that it's not conservative, you have to look at "so small regions" that the derivatives of $\vec F$ are completely invisible. For example, one may be strict and notice that if you only know $\vec F$ at one point only, there is no provable obstruction that would prevent $\vec F$ from being conservative. However, a fairer description of the situation is that if you only know $\vec F$ at one point, you cannot say whether the field is conservative – instead of saying that it is. If the force field were constant, it would be conservative. But to meaningfully say whether the force field is constant, you have to look at nearby different points, too. And as soon as you may measure some derivatives of $\vec F$, the curl of $\vec F$ may be measured as well and it may be nonzero.

You could discuss forces that are "conservative within some error margin". For example, if $\vec F$ were almost constant in your small neighborhood, there would be a sense in which $|\Delta F|\ll |F|$, and as long as you would be satisfied with the approximation $\vec F+\Delta \vec F\approx \vec F$, you could say that the force field is conservative in that region within the same precision. But this whole conclusion only emerges because we threw the baby out with the bath water. Whether a field is conservative may typically be found by looking at small regions of space. A field is "locally" conservative if $\nabla\times \vec F=0$ at the nearby points. One doesn't need to know the behavior of the force field everywhere.

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