# Are all central forces conservative?

It might be just a simple definition problem but I learned in class that a central force does not necessarily need to be conservative and the German Wikipedia says so too. However, the English Wikipedia states different on their articles for example:

A central force is a conservative field, that is, it can always be expressed as the negative gradient of a potential

They use the argument that each central force can be expressed as a gradient of a (radial symmetric) potential. And since forces that are gradient fields are per definition conservative forces, central forces must be conservative. As far as I understand, a central force can have a (radial symmetric) potential but this is not necessarily always the case.

Update Sep 2017: The english Wikipedia has updated its text and now explicitly states

Not all central force fields are conservative nor spherically symmetric. However, it can be shown that a central force is conservative if and only if it is spherically symmetric.[2]

• In principle, every central force/field with complete azimuthal symmetry, as long as it is also time-independent, can be integrated to find the potential associated with it. That potential may not be analytic, but it exists, and its existence is equivalent to stating that the force is conservative. – Benji Remez Oct 2 '12 at 4:02
• Re: update: In Sep 2017, English Wiki used two different definitions of central force. The definition in Wolfram ScienceWorld includes the constraint that the the force be spherically symmetric (i.e. depend only on r), whereas the definition in Taylor's Classical Mechanics does not. I've since amended English Wiki to adhere to Taylor's definition. – LastStar007 Oct 17 '17 at 6:29

If your central force is of the form ${\vec F} = f(r){\hat r}$ (the force points radially inward/outward and its magnitude depends only on the distance from the center), then it is easy to show that $\phi = - \int dr f(r)$ is a potential field for the force and generates the force. This is usually what I see people mean when they say "central force."
If, however, you just mean that the force points radially inward/outward, but can depend on the other coordinates, then you have ${\vec F} = f(r,\theta,\phi){\hat r}$, and you're going to run into problems finding the potential, because you need $f = - \frac{\partial V}{\partial r}$, but you will also need to have $\frac{\partial V}{\partial \theta} = \frac{\partial V}{\partial \phi} = 0$ to kill the non-radial components, and this will lead to contradictions.
It's logical that a field of this form is gong to be nonconservative, because if the force is greater at $\theta = 0$ than it is at $\theta = \pi/2$, then you can do net work around a closed curve by moving outward from $r_{1}$ to $r_{2}$ at $\theta = 0$ (positive work), then staying at $r_{2}$ constant, going from $\theta =0$ to $\theta = \pi/2$ (zero work--radial force), going back to $r_{1}$ (less work than the first step), and returning to $\theta = 0$ (zero work).
Take curl in spherical polar coordinates of a central force ,you will see that since there is no component of the force in the $\theta$ and $\phi$ direction and $f(r)$ doesn't depend on $\theta$ and $\phi$ ,so curl of central force is zero. Hence central forces can be represented as gradient of some scalar,i.e. central forces are conservative.