# Non-conservative field with zero curl

I'm looking for examples of this, the only one I've found so far is: $f(x,y) = \left( \frac{-y}{x^2+y^2}, \frac{x}{x^2+y^2} \right)$

Is there another example of this? Also, what should I follow if I wanted to "build" a field that is non-conservative and has zero curl. Is both partial derivatives being the same and the line integral of the force along some closed curve being different than zero enough? Thank you.

• Thank you for the edit @GFR – Thums Apr 29 '14 at 17:42

The existence of non-conservative vector fields with zero curl is a topological matter, i.e. it only depends on the topology of the space on which the vector fields is defined. This is studied by de Rham cohomology using the language of differential forms.

In your case, the vector field is defined on $R^2$ but, because it is singular at the origin, the domain is actually $R^2$ with the origin removed. Such a space is topologically non-trivial. However, it can be shown that the vector space of non-conservative, irrotational vector fields on punctured $R^2$ is one dimensional. So the vector field you have written is in the appropriate sense unique, where in the appropriate sense means up to addition of any conservative vector field.

Note that the vector field you have written is basically $\mathrm{d}\theta$ expressed in Cartesian coordinates. If you denote by $\theta_i$ the angle $\arctan((y-y_i)/(x-x_i))$ you will see that $\sum_i \mathrm{d}\theta_i$, where the sum is over an arbitrary collection of points $p_i\in R^2$, has zero curl (think of $\mathrm{d}$ as the gradient operator) but is not exact.

In a more general language, irrotational vector field translates to closed differential form, and conservative vector field translates to exact differential form. Vector fields translates to differential 1-forms (to do this properly you need a metric to be defined on your space). Conservative => irrotational translates to exact=> closed. The space of closed 1-forms modulo the exact 1-form is the first de Rham cohomology group.

The first cohomology group is always trivial for a simply-connected space. In the case of $R^2$ with k holes it is $k$-dimensional.

This is essentially'' the only example on $\mathbb{R}^2 - \{(0,0)\}$: every other example is just this one plus a conservative vector field. In other words, your example generates the de Rham cohomology of $\mathbb{R}^2-\{(0,0)\}$. If you want examples on other domains, you can use the same idea to get that there are basically two "essentially different" examples on $\mathbb{R}^2 - \{p_1,p_2\}$. Things get more interesting when your domain is a more interesting manifold, like a ball or torus.

I would recommend the book "from Calculus to Cohomology" if you are interested in this kind of thing.

Is the vector field being defined on a simply connected area a necessary condition for it to be conservative? In other words, is it possible to have a conservative vector field (in two dimensions) which would be undefined in (0,0)?

The field I have in mind is $(\frac{-x}{(x^2+y^2)^{3/2}},\frac{-y}{(x^2+y^2)^{3/2}})$. It is undefined if $x=y=0$ but seems to be conservative as it can be written as the gradient of $\frac{1}{\sqrt{x^2+y^2}}$.