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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.

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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

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Depends on what you mean by 'central force'.

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).

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That is what I thought. Thanks! –  n3rd Oct 3 '12 at 5:29

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