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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
up vote 25 down vote accepted

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
Note that the English Wikipedia article explicitly defines a central force as being spherically symmetric ("a force field is central if and only if it is spherically symmetric"), while the German Wikipedia article explicitly states that it will be considering radial force fields without spherical symmetry ("In diesem Artikel werden jedoch auch nichtkonservative Zentralkräfte behandelt, die insbesondere keine Radialsymmetrie aufweisen müssen.") Or, at least, they now contain these sentences (they may not have three years ago...) – Michael Seifert Jul 8 '15 at 14:51
Haha, good call – n3rd Nov 26 '15 at 6:32

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.

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