# Bertrand’s theorem

In classical mechanics, Bertrand’s theorem states that among central-forcepotentials with bound orbits, there are only two types of central-force (radial) scalar potentials with the property that all bound orbits are also closed orbits.[1][2]

The first such potential is an inverse-square central force such as the gravitational or electrostatic potential:

${displaystyle V(r)=-{frac {k}{r}}}$

with force

${displaystyle f(r)=-{frac {dV}{dr}}=-{frac {k}{r^{2}}}}$

.

The second is the radial harmonic oscillator potential:

${displaystyle V(r)={frac {1}{2}}kr^{2}}$

with force

${displaystyle f(r)=-{frac {dV}{dr}}=-kr}$

.

The theorem is named after its discoverer, Joseph Bertrand.

## . . . Bertrand’s theorem . . .

All attractive central forces can produce circular orbits, which are naturally closed orbits. The only requirement is that the central force exactly equals the centripetal force, which determines the required angular velocity for a given circular radius. Non-central forces (i.e., those that depend on the angular variables as well as the radius) are ignored here, since they do not produce circular orbits in general.

The equation of motion for the radius r of a particle of mass m moving in a central potentialV(r) is given by motion equations

${displaystyle m{frac {d^{2}r}{dt^{2}}}-mromega ^{2}=m{frac {d^{2}r}{dt^{2}}}-{frac {L^{2}}{mr^{3}}}=-{frac {dV}{dr}},}$

where

${displaystyle omega equiv {frac {dtheta }{dt}}}$

, and the angular momentumL = mr2ω is conserved. For illustration, the first term on the left is zero for circular orbits, and the applied inwards force

${displaystyle {frac {dV}{dr}}}$

equals the centripetal force requirementmrω2, as expected.

The definition of angular momentum allows a change of independent variable from t to θ:

${displaystyle {frac {d}{dt}}={frac {L}{mr^{2}}}{frac {d}{dtheta }},}$

giving the new equation of motion that is independent of time:

${displaystyle {frac {L}{r^{2}}}{frac {d}{dtheta }}left({frac {L}{mr^{2}}}{frac {dr}{dtheta }}right)-{frac {L^{2}}{mr^{3}}}=-{frac {dV}{dr}}.}$

This equation becomes quasilinear on making the change of variables

${displaystyle uequiv {frac {1}{r}}}$

and multiplying both sides by

${displaystyle {frac {mr^{2}}{L^{2}}}}$

${displaystyle {frac {d^{2}u}{dtheta ^{2}}}+u=-{frac {m}{L^{2}}}{frac {d}{du}}Vleft({frac {1}{u}}right).}$

As noted above, all central forces can produce circular orbits given an appropriate initial velocity. However, if some radial velocity is introduced, these orbits need not be stable (i.e., remain in orbit indefinitely) nor closed (repeatedly returning to exactly the same path). Here we show that stable, exactly closed orbits can be produced only with an inverse-square force or radial harmonic oscillator potential (a necessary condition). In the following sections, we show that those force laws do produce stable, exactly closed orbits (a sufficient condition).

Define J(u) as

${displaystyle {frac {d^{2}u}{dtheta ^{2}}}+u=J(u)equiv -{frac {m}{L^{2}}}{frac {d}{du}}Vleft({frac {1}{u}}right)=-{frac {m}{L^{2}u^{2}}}fleft({frac {1}{u}}right),}$

where f represents the radial force. The criterion for perfectly circular motion at a radius r0 is that the first term on the left be zero:

${displaystyle u_{0}=J(u_{0})=-{frac {m}{L^{2}u_{0}^{2}}}fleft({frac {1}{u_{0}}}right),}$

(1)

where

${displaystyle u_{0}equiv 1/r_{0}}$

.

The next step is to consider the equation for u under small perturbations

${displaystyle eta equiv u-u_{0}}$

from perfectly circular orbits. On the right, the J function can be expanded in a standard Taylor series:

${displaystyle J(u)approx J(u_{0})+eta J'(u_{0})+{frac {1}{2}}eta ^{2}J”(u_{0})+{frac {1}{6}}eta ^{3}J”'(u_{0})+cdots }$

Substituting this expansion into the equation for u and subtracting the constant terms yields

${displaystyle {frac {d^{2}eta }{dtheta ^{2}}}+eta =eta J'(u_{0})+{frac {1}{2}}eta ^{2}J”(u_{0})+{frac {1}{6}}eta ^{3}J”'(u_{0})+cdots ,}$

which can be written as

${displaystyle {frac {d^{2}eta }{dtheta ^{2}}}+beta ^{2}eta ={frac {1}{2}}eta ^{2}J”(u_{0})+{frac {1}{6}}eta ^{3}J”'(u_{0})+cdots ,}$

(2)

where

${displaystyle beta ^{2}equiv 1-J'(u_{0})}$

is a constant. β2 must be non-negative; otherwise, the radius of the orbit would vary exponentially away from its initial radius. (The solution β = 0 corresponds to a perfectly circular orbit.) If the right side may be neglected (i.e., for small perturbations), the solutions are

${displaystyle eta (theta )=h_{1}cos(beta theta ),}$

where the amplitude h1 is a constant of integration. For the orbits to be closed, β must be a rational number. What’s more, it must be the same rational number for all radii, since β cannot change continuously; the rational numbers are totally disconnected from one another. Using the definition of J along with equation (1),

${displaystyle J'(u_{0})={frac {2}{u_{0}}}left[{frac {m}{L^{2}u_{0}^{2}}}fleft({frac {1}{u_{0}}}right)right]-left[{frac {m}{L^{2}u_{0}^{2}}}fleft({frac {1}{u_{0}}}right)right]{frac {1}{fleft({frac {1}{u_{0}}}right)}}{frac {d}{du_{0}}}fleft({frac {1}{u_{0}}}right)=-2+{frac {u_{0}}{fleft({frac {1}{u_{0}}}right)}}{frac {d}{du_{0}}}fleft({frac {1}{u_{0}}}right)=1-beta ^{2}.}$

Since this must hold for any value of u0,

${displaystyle {frac {df}{dr}}=(beta ^{2}-3){frac {f}{r}},}$

which implies that the force must follow a power law

${displaystyle f(r)=-{frac {k}{r^{3-beta ^{2}}}}.}$

Hence, J must have the general form

${displaystyle J(u)={frac {mk}{L^{2}}}u^{1-beta ^{2}}.}$

(3)

For more general deviations from circularity (i.e., when we cannot neglect the higher-order terms in the Taylor expansion of J), η may be expanded in a Fourier series, e.g.,

${displaystyle eta (theta )=h_{0}+h_{1}cos beta theta +h_{2}cos 2beta theta +h_{3}cos 3beta theta +cdots }$

We substitute this into equation (2) and equate the coefficients belonging to the same frequency, keeping only the lowest-order terms. As we show below, h0 and h2 are smaller than h1, being of order

${displaystyle h_{1}^{2}}$

. h3, and all further coefficients, are at least of order

${displaystyle h_{1}^{3}}$

. This makes sense, since

${displaystyle h_{0},h_{2},h_{3},ldots }$

must all vanish faster than h1 as a circular orbit is approached.

${displaystyle h_{0}=h_{1}^{2}{frac {J”(u_{0})}{4beta ^{2}}},}$

${displaystyle h_{2}=-h_{1}^{2}{frac {J”(u_{0})}{12beta ^{2}}},}$

${displaystyle h_{3}=-{frac {1}{8beta ^{2}}}left[h_{1}h_{2}{frac {J”(u_{0})}{2}}+h_{1}^{3}{frac {J”'(u_{0})}{24}}right].}$

From the cos(βθ) term, we get

${displaystyle 0=(2h_{1}h_{0}+h_{1}h_{2}){frac {J”(u_{0})}{2}}+h_{1}^{3}{frac {J”'(u_{0})}{8}}={frac {h_{1}^{3}}{24beta ^{2}}}left(3beta ^{2}J”'(u_{0})+5J”(u_{0})^{2}right),}$

where in the last step we substituted in the values of h0 and h2.

Using equations (3) and (1), we can calculate the second and third derivatives of J evaluated at u0:

${displaystyle J”(u_{0})=-{frac {beta ^{2}(1-beta ^{2})}{u_{0}}},}$

${displaystyle J”'(u_{0})={frac {beta ^{2}(1-beta ^{2})(1+beta ^{2})}{u_{0}^{2}}}.}$

Substituting these values into the last equation yields the main result of Bertrand’s theorem:

${displaystyle beta ^{2}(1-beta ^{2})(4-beta ^{2})=0.}$

Hence, the only potentials that can produce stable closed non-circular orbits are the inverse-square force law (β = 1) and the radial harmonic-oscillator potential (β = 2). The solution β = 0 corresponds to perfectly circular orbits, as noted above.

## . . . Bertrand’s theorem . . .

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