Inverse Square Orbits

Using the Laplace-Runge-Lenz Vector constant of the motion

Consider the motion of a particle of mass m that is subject to a force directed toward a fixed centre which depends only on the distance r from the centre. The motion takes place on a fixed orbital plane determined by the initial conditions. Bertrand (see References) showed that the path of motion, the orbit, is closed and stable only if the force is inversely proportional to the square of r, or directly proportional to r. We shall treat the first case, in which the orbits are conic sections, in this article. This case includes the important examples of gravitational and electrostatic forces, as well as the hydrogen atom.

Let the force F be derived from a potential energy V)r) = -k/r: F = -dV/dr = -kr/r3. Then the motion is determined by Newton's Law dp/dt = F, where p is the momentum mv. It is difficult to solve these three simultaneous second-order differential equations directly. We proceed by choosing as generalized coordinates in the orbital plane the radius r from the centre of force and the angle θ, as illustrated in the diagram. It is easy to express the kinetic energy T and the potential energy V in terms of these coordinates and their time derivatives, and so to form the Lagrangian function L = T - V. Then, Lagranges equations, d/dt(∂L/∂v) - ∂L/∂q = 0 for each coordinate, where v is the generalized velocity dq/dt, give the equations of motion, which are shown below.

The Lagrangian does not explicitly contain the angle θ, so the correspondiong generalized momentum ∂L/∂v is a constant. This quantity is the angular momentum L = mrxv. Its magnitude L should not be confused with the Lagrangian function. L is perpendicular to the plane of motion. The radial equation of motion is also easily obtained, and can be solved to determine the orbit. In this article, we will take a different route and find the orbit without solving any differential equations.

For a given L, there is an effective radial potential VL, shown in the figure. The minimum of this potential corresponds to a circular orbit, r = constant, whose total energy is the smallest of all orbits with this value of L, and whose eccentricity is zero. As the energy increases, the range of r increases as the eccentricity increases.

In the diagram at the left, we show that the vector A = p x L - mkr/r, called the Lagrange-Runge-Lenz vector, is a constant of the motion. This vector points in the direction of the major axis of an elliptical orbit. For a circular orbit, the vector vanishes and does not distinguish any special direction.

By taking the scalar product of A with r, and with itself, we can determine the equation of the orbit and its energy. This is done at the right. We see that the orbits are conic sections, and that elliptical orbits correspond to an eccentricity less than 1, parabolic orbits to e = 1, and hyperbolic orbits to e > 1. Negative total energy E corresponds to an eccentricity less than 1 and a positive semi-major axis a.

Let us choose rectangular coordinates with the x-axis in the direction of A = Ai and the z-axis in the direction of L. The y-axis then is in the direction of the minor axis of an elliptical orbit and x,y,z form a right-handed triplet. Then we can transpose and take the scalar product of mkr/r with itself, as shown in the figure. The result is that the hodograph of the motion is a circle in momentum space, as shown. Points 1 and 2 correspond to the ends of the major axis of the orbit. The momentum vector does not rotate at constant angular velocity, of course, moving much more slowly near point 2 than at point 1, where the speed of the particle is greatest.

When py = 0, px = ±po = ±√2m|E|. All the hodographs for orbits of this energy (E < 0) pass through these two points, differing in their angular momentum L. The vector A/po has the same dimensions as L, and its constancy is a result of symmetry under 4-dimensional rotations. In the Hydrogen atom, this has the consequence that orbits of the same principal quantum number but different angular momenta have the same energy.

The momentum in a circular orbit is of magnitude po, rotating uniformly around a circle, as can easily be seen by equating the attractive force to the mass times the centripetal acceleration. In this case, clealy A = 0 ane the eccentricity is zero.

For completeness, the diagram at the left shows the Hamiltonian function and the canonical, or Hamiltonian, equations, which are the same as those we derived from the Langrangian. These are the last four equations in the figure, two for each generalized coordinate. The Hamiltonian formalism is used in the quantum mechanical two-body problem. Note that the Lagrangian is expressed as a function of the generalized coordinates and their velocities, or total time derivatives, while the Hamiltonian is expressed as a function of the generalized coordinates and their canonical momenta. Note that the equivalent radial potential function appears naturally in this formalism. It is easy to show from the canonical equations that H is a constant of motion, dH/dt = 0, identified as the total energy.

The LRL vector has been virtually ignored, in spite of its useful and interesting properties, so it has been rediscovered several times since its recognition by Laplace in work by Bernoulli, who had it from even earlier work.


See the Wikipedia articles "Laplace-Runge-Lenz Vector" and "Bertrand's Theorem" for additional informatiion.

A good, concise introduction to Lagrangian and Hamiltonian mechanics will be found in L. D. Landau and E. M. Lifshitz, Mechanics (Reading, MA: Addison-Wesley Publ. Co., 1960).

The quantum mechanics of the hydrogen atom with the LRL vector and the O(4) dynamical symmetry is treated in L. I Schiff, Quantum Mechanics, 3rd ed. (New York: McGraw-Hill, 1968), pp. 234-239.

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Composed by J. B. Calvert
Created 1 January 2011
Last revised 22 January 2011