I found in Goldstein's Classical Mechanics that the condition for closed orbits is given by $\frac{d^2 V_{eff}}{dr^2}>0$.(bertrand's theorem). Can somebody explain to me, how this inequality is related to the boundness of the orbits? I do not see it. Furthermore, I was wondering, whether this property is the only property that an effective potential has to fulfill in order to produce bounded orbits`?

  • 3
    $\begingroup$ Your definition of bertrand's theorem seems incorrect. Read the Wikipedia article. $\endgroup$
    – Trimok
    Commented Jun 21, 2013 at 20:20

2 Answers 2


I) Assume that

$$ \mathbb{R}_{\geq 0}~\ni~ r \quad\stackrel{V_{\rm eff}}{\mapsto} \quad V_{\rm eff}(r)~\in~ \mathbb{R} $$

is a continuous function. Then energy conservation reads

$$E~=~ \frac{m}{2} \dot{r}^2+ V_{\rm eff}(r).$$

A necessary and sufficient condition for the existence of a radially bounded orbit (that is not necessarily periodic/closed or stable) is that there exists an energy level $E$ such that at least one of the connected components (which are necessarily intervals or points) of the preimage

$$V_{\rm eff}^{-1}(]-\infty,E]) ~\subseteq~ \mathbb{R}_{\geq 0}$$

is a radially bounded set.

II) It seems that the only place where Goldstein (in his book Classical mechanics) considers the concavity condition

$$V_{\rm eff}^{\prime\prime}(r)~>~0$$

is in the beginning of Section 3.6, where he discusses stable circular orbits. He uses concavity to conclude the relatively mild inequality $p>-3$ for a power-law force $f(r) \propto r^p$.

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    $\begingroup$ Correction to the answer(v2): The word concave should be concave upwards according to Wikipedia. $\endgroup$
    – Qmechanic
    Commented Jan 17, 2014 at 19:41

At a more pedestrian level than Qmechanic...

As used by Goldstein, the positive concavity condition:

$$ \frac{d^2 V_{\rm eff}}{dr^2} > 0 $$

is required for stability of a circular orbit. (The second derivative is evaluated at the orbit radius.)

First, such an orbit can exist only at an extremum of $V_{\rm eff}$, where the radial force (the negative of the first derivative of $V_{\rm eff}$) is 0:

$$ -\frac{dV_{\rm eff}}{dr} = 0$$

(Otherwise it wouldn't be an equilibrium orbit.)

To examine the stability, imagine that the orbit deviates slightly, by $\Delta r$, from the equilibrium position. Then the radial force $f_r$ is no longer exactly zero: by a Taylor expansion of the potential around the equilibrium (remembering that the linear term is 0 by assumption) one finds a version of Hooke's law:

$$f_r = - \frac{d^2 V_{\rm eff}}{dr^2} \Delta r$$

This force is restoring (stabilizing) if the concavity is positive, but destabilizing if the concavity is negative.

This situation is analogous to a vertical pendulum. There are two equilibria: 1) the pendulum at rest at the bottom and 2) at rest at the top. The bottom equilibrium point is stable, but the top equilibrium is unstable.

As noted by others, this condition is not Bertrand's theorem, which is concerned with closed orbits in general (not just circular orbits).


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