分析力学第四次作业解答

7.3 Laplace-Runge-Lenz vector. Show form the Poisson bracket condition for conserved quantities that the Laplace-Runge-Lenz vector

Is a constant of motion for a mass moving under the Kepler potential. Also, show this by taking the time derivative of this quantity Solution:

(a) In terms of Possion brackets, the time rate of change of a dynamical variable, in this case the Laplace-Runge-Lenz vector, is given by

Since, ,to show that is conserved, we need

First, the Hamiltonian H is given by

The canonical coordinates and moment are

Write in polar coordinates

The Poisson bracket is given by

Writing as a function of the canonical coordinates

And recalling

We have

Hence

So,

(b) We can also show that is conserved by direct differentiation

Where we have used

Since:

Thus

7.4 Orbits on a cone. A particle of mass m is constrained to move on the surface of a cone of half-angle (see Example 7-4, p. 244 of Marion and Thornton). Work in cylindrical coordinates ().

(a) Show that the e_ective potential for this problem is

(1) where l is the angular momentum.

(b) Using the effective potential, discuss the types of possible orbits for the particle, assuming that it has a total energy E and angular momentum l. Make sure to include a carefully labeled figure. What is the equation which determines the turning points? What is the condition for a circular orbit?

(c) Investigate the stability of circular orbits on the surface of the cone. Solution:

(a) In cylindrical coordinates, the kinetic energy is

The constraint for motion on the cone is that . Substituting this into the kinetic energy, we obtain

The potential energy is

The lagrangian is

We see immediately tha is cyclic (a consequence of angular momentum conservation), so that

constant

The total energy, which is conserved, is

Eliminating using Eq. , we obtain

The last two terms comprise the “effective potentail”.

(b) A figure showing the effective potential and the turning points is shown below. The turning points for energy E1 are and , and a circular orbit of radius occurs at energy E0.

The equation which determines the turning points is

A circular orbit of radius occurs at the minimum of the effective potential:

Solving for , we have

(c) Circular orbits are always stable. To see this, calculate the second derivative of the effective potential at the minimum:

so that the curvature of the potential is always positive and the orbits are stable.

7.5Extensions of the Bohr model of the hydrogen atom. The Bohr model was developed by assuming that the electron orbit is circular.