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I am really confused here: Why one of these functions is conservative, while the other not?

$F_{1} = \frac{-y \hat i + x \hat j}{x^2+y^2}$

$F_{2} = \frac{x \hat i + y \hat j}{x^2+y^2}$

Suppose both these vector functions are applied on a region $D=R^{2}-(0,0)$.

$\nabla \times F_{1} =\nabla \times F_{2} = 0 $

Both apply at the same region.

Even so, every exercise i see about this region says something like:: "F1 isn't a gradient, the region isn't simple connected/ F2 is a gradient". So:

Why (1) in one case the fact that D is "a ill region" import, and in the other not?

(2)As i pointed, both vectors have the same properties, but one is gradient, and the other not. So what is preciselly being the differential here that says when one is a gradient and when one isn't?

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  • $\begingroup$ hint: the first de Rham cohomology group of $\mathbb{R}^2 - (0,0)$ is 1-dimensional. $\endgroup$
    – David Lui
    Jun 25, 2021 at 2:12

2 Answers 2

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You are checking whether a vector field is a gradient by seeing if it is curl free. Over a simply-connected domain, being curl free and being a gradient are equivalent. The purpose of this exercise is that these two concepts, equivalent over nice domains, are no longer equivalent on the punctured plane. Just because something is curl free, does not mean it is a gradient. To prove that $F_2$ is a gradient, instead of checking the curl, you should try and explicitly find what it is the gradient of. To see that $F_1$ isn't a gradient, just integrate it around a circle of radius 1 and you'll get a nonzero value (if $F_1$ was a vector field, then of course the line integral around any closed curve would be zero by the Fundamental Theorem of Calculus).

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  • $\begingroup$ "To see that F1 isn't a gradient, just integrate it around a circle of radius 1 and you'll get a nonzero value" While that works in this case, a space with more than one hole can have zero intergral around one of them and a nonzero value around another. $\endgroup$ Jun 25, 2021 at 3:34
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In a simply connected region a vector field $F$ is a gradient of some function iff $\nabla \times F = 0$.

But if a region is not simply connected, this is no longer true. The reason is the same as if we would consider $S^1$. You can easily prove, that for any smooth function $f$ on $S^1$ $\frac{\partial f}{\partial \varphi}$ is again a smooth function on $S^1$, but the converse is not true. It is impossible to find a continuous function on $S^1$ which has a non-vanishing derivative everywhere. Equivalently, there’s no solution to a differential equation $y’ = \lambda$, $\lambda \in \mathbb R \setminus \{0\}$, among smooth functions on $S^1$. Constant functions are bad functions on $S^1$ from this point of view. It turns out, that they are the only bad functions: all other functions are good modulo constant ones, because for any smooth function $f$ on $S^1$ $$ \int_{S^1} (f-C_f) \, d\varphi = 0, $$ when $C_f = \frac1{2\pi}\int_{S^1} f \, d\varphi$.

The case of $S^1$ is in a certain sense a limit of the punctured real plain in your question. But the formal statements and their proofs require the de Rham cohomology technic.

Edit: In order to make the answer more complete, I’d like to add, that $F_1$ is an $\mathbb R^2$-analog of the constant function. The main difference between $S^1$ and the punctured plain is their dimensions. However, one can easily notice, that the restriction of $F_1$ to any circle centered at the origin is exactly the field corresponding to a certain constant function. It can be shown also, that $\lambda F_1$, $\lambda \in \mathbb R \setminus \{0\}$, are the only bad fields in this case. In fact, this is what the complex analysis is all about.

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