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.

However, we know that bound orbits for attractive inverse square forces are generally ellipses. Sommerfeld (1916) attempted to generalize Bohr's work to the case of elliptical orbits, by introducing the quantization condition

(1) where p is the generalized momentum conjugate to the generalized coordinate q, defined as

(2) the integral is over one period of motion, n is an integer, and h is Planck's constant. For the hydrogen atom the Lagrangian is

(3) where, and the total energy (relative to the center of mass) is

(4) with the reduced mass of the system.

(a) Start with the angular momentum,

. (5) Since the angular momentum is conserved, show that this implies that, where and m is a positive integer.

(b) Substitute your result from part (b) into the energy, and show that

(6)

(c) Sketch the effective potential for this problem, and draw the energy

and turning points for a bound orbit. Show that the turning points r min and r max can be determined by solving a quadratic equation, and find expressions for the turning points as a function of the energy E.

(d) Now we need to quantize the radial motion. Solve Eq. (6) to obtain pr as a function of r. Substitute the result into

(7) with n an integer, and perform the integral. You might find the following integral to be useful:

(8) Substitute your expressions for r min and r max as a function of the energy E into this expression, and solve the resulting equation for E to show that the allowed energy levels are

(9) Note that now we can have circular orbits (n = 0) or elliptical orbits (n 6= 0). Notice that we can have many deferent orbits with the same energy; for instance, the circular orbit with m = 3 and n = 0 has the same energy as the elliptical orbits with m = 2 and n = 1 and m = 1 and n = 2. This is a case of degeneracy.

Solution:

(a) First, calculate the momenta conjugate to the generalized

coordinates. We have

,

Since the angular momentum is conserved, it is time-independent, and we have

Therefore, .

(b) Substituting and into Eq. (4), we obtain

(c) The effective potential is

A plot of this function is shown below. The turning points are determined by

setting :

Multiplying by and rearranging, we obtain

which can be solved to obtain the two turning points, r min and r max. Note that

(d) First, solve Eq. (6) to obtain pr as a function of E:

Now we need to integrate this over one period; in one period the electron would go from r min to r max and back to r min, so the integral over one period is twice the integral from r min to r max:

Solving for E, we obtain Eq. (9). Note that m=1,2,3,..., while n=0,1,2,...

11.1 Canonical transformation. Construct the Hamiltonian for the

one-dimensional harmonic oscillator of mass m and spring constant. Determine the value of the constant C such that the following equations define a canonical transformation from the old variables to the new variables

where What is the generating function, for this transformation?

Find Hamilton’s equations of motion for the new variables and integrate them. Hence find the solution to the original problem.

Solution:

must satisfy the Possion bracket. Therefore

So

With a type-2 generating function, ,we have

Using

Now,

Hence,

The new Hamiltonian is given by

Now, for the 1d harmonic oscillation

Since

Hamilton’s equations of motion become:

are determined from initial condition Therefore, become

11.2 Canonical transformation Find the values of and for which the transformation

is canonical. For these values of and , construct a generating function F3(p,Q) for the canonical transformation.

Solution: Require (Q,P) to satisfy the fundamental Poisson bracket

,

Then,

,

,

Now,

Therefore, so