The first order formalism and the transition to the

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1 Problem 1. Hamiltonian The first order formalism and the transition to the This problem uses the notion of Lagrange multipliers and Legendre transforms to understand the action in the Hamiltonian formalism. Previously we said that the action in the Hamiltonian formalism is ( S[q, p] = dt p dq ) H(p, q). (1) dt We showed that extremizing this action gives Hamilton s equations of motion and that these equations are equivalent to the Euler-Lagrange equations. We did not, however, derive S[q, p] directly from the action of the Lagrangian, S[q]. We will do this in this problem. The action principle says that the action is S[q] = dtl(q, q) (2) and the system will follow the rajectory q(t) which extremizes this action (with the endpoints fixed). Using a Lagrange multiplier called p(t), we may break the constraint between the velocity v(t) and q by defining Ŝ[q(t), v(t), p(t)] dt ˆL(q, v, p) ˆL L(q, v) p(v q). (3) and require that δŝ = 0 for independent variations of q, v, p. This theorist-gone-wild procedure is known as the first order formalism, and has been found to be useful in analyzing various rich theories (such as gravity) which have complicated constraints. (a) Consider the Lagrangian L = 1 2 m q2 U(q) (4) Show that the equations motion following from laws. δŝ[q, v, p] = 0, reproduce Newton s (b) Extremize Ŝ with respect to p, v first, leaving a reduced action S red[q] to be extremized later. Argue that this reduced action is S[q] in Eq. (2). (c) Extremize Ŝ with respect to v first, leaving a reduced action S red[q, p] to be extremized later, and argue that this reduced action is S[q, p] in Eq. (1). 1

2 Problem 2. Noether logic (Milton, de Raad, Schwinger) Virial theorem from The virial theorem says that for the periodic motion of a particle the time averaged kinetic energy is related to an average of the potential energy: 2T = r U(r) r For simplicity we will limit ourselves to the single particle Lagrangian. (5) L = 1 2 mṙ2 U(r), (6) but when many particles are involved, the theorem generalizes straightforwardly to 2T = a r a U(r) r a. (7) A direct proof of this theorem will be given in class. Here we will derive this useful result using Noether logic 1. Recall that we say that the trajectory is called onshell if it satisfies the equation of motion, and, when necessary, notate this by placing a bar underneath the coordinates r(t) (a) For a closed orbit of potential U(r) r β what is the statement of the virial theorem. What is the statement of the theorem for a harmonic oscillator U(r) r 2 and the gravitational potential U(r) r 1. (b) For a quantum mechanical particle in one dimension in an eigenstate ψ n (x) (which is analogous to the classical periodic trajectory) show that 2T = x U(x) (8) x by considering, [p, H]. (c) Now return to classical mechanics. Consider a specific variation of the trajectory consisting of an infinitessimal rescaling of the coordinate r r (1 + ɛ)r (9) What is the change of the onshell action S[r] for this specific variation over one complete period of a periodic classical trajectory r? (d) What is the change in the action δs[r, δr] for the specific variation in Eq. (9). Do not assume that r is onshell. (e) Using (c) and (d) prove the theorem in Eq. (5) 1 It is not exactly the Noether theorem, since there is no conserved charge and no symmetry. But the derivation is essentially the same as is used to derive Noether theorem. 2

3 Problem 3. The precession of Mercury due to Jupiter Recall that the trajectory of Mercury r(φ) is an ellipse with the sun at one focus as shown below. The perihelion (defined as the distance of closest approach) is rotated relative to the x-axis by an angle θ. The semi-major axis is denoted R M. The eccentricity of Mercury is small, ɛ = 0.2, although it is the most eccentric of the Sun s planets. Mercury r φ perihelion θ Sun R M Due to perturbations from the other planets, the angle of the perihelion θ changes (or precesses) as function of time. The precession rate is very small. The contribution of Jupiter to the precession rate is of order 150 arcsec/century, or (since the orbital period of Mercury is 88 days) approximately rad/turn. The goal of this problem is to estimate Jupiter s contribution to the precession rate 2. Specifically, we will model Jupiter as a ring of mass M J at the orbital radius of Jupiter R J, and compute how the ring perturbs Mercury s orbit and causes the perihelion of Mercury to precess. Jupiter s orbital radius is significantly larger than Mercury s, R J 10 R M. (a) Show that for R J R M the Lagrangian of Mercury interacting with the sun of mass M, and a ring of mass M J and radius R J is approximately and determine the coefficient α. L 1 2 mṙ mr2 φ2 + GmM r + αr 2, (10) (b) By introducing a dimensionless radius r = r/r M (r in units of Mercury s semi-major axis) and other suitable dimensionless variables to show that a dimensionless Lagrangian for the system is L 1 2 ( ) 2 dr + 1 dt 2 r2 ( dφ dt ) r + α r2, (11) 2 Famously, general relativity also perturbs the classical orbit and contributes 43 arcsecs/century to the total precession rate. This anomalous precession of Mercury was measured in the nineteenth century by le Verrier and finally explained by Einstein in The total precession rate is approximately 550 arcsec/century 3

4 where the dimensionless constant α is of order ( ) 3 M J RM (12) M R J To lighten the notation, stop underlining the variables in what follows. (c) Determine the Hamiltonian of the dimensionless system, and use the Hamiltonian to determine the equations of motion. (d) For α = 0 determine the trajectory of Mercury r(φ), in terms of the energy and the angular momentum. This is basically lecture stuff. (e) Determine the change in the orbital period and precession of the perihelion Mercury to first order in α. You may treat the eccentricity of Mercury as small so that the orbit is approximately circular. Neglect terms of order the (eccentricity) 2. Evaluate the precession rate numerically in radians per turn. You should find φ = 3π 2 ( ) 3 M J RM rad per turn, (13) M R J This should be compared to the experimental result of rad/turn. 4

5 Problem 4. A scattering cross section A particle of mass µ moves in the repulsive 1/r 2 potential U(r) = h r 2, h > 0. (a) Find equation for a generic trajectory r(φ) characterized with energy E and angular momentum l 0. Follow the convention that the direction φ = 0 points to the pericenter (point of closest approach). (b) Find the time dependence on this trajectory, taking the time t = 0 at the pericenter. (c) Find the differential scattering cross section dσ(θ) dω potential. for a particle with energy E in this 5

6 Problem 5. A Routhian tutorial and the effective potential Consider the Kepler Lagrangian again: L = 1 2 mṙ mr2 φ2 U(r) (14) There are two variables r and φ with associated momenta p r and p φ. The Hamiltonian is is formed by Legendre transforming with respect to r and φ H = p r ṙ + p φ φ L(r, ṙ, φ, φ). (15) It can be convenient to Legendre transform with respect to only some of the variables instead of all of them. We define the Routhian 3 : R(r, ṙ, φ, p φ ) p φ φ L(r, ṙ, φ, φ), (16) which serves as a Hamiltonian for φ, but a Lagrangian for r. This is especially helpful when some of the coordinates are cyclic (φ in this case). The p φ are then just constants (both in the equation of motion and in the action), and we have effectively a Lagrangian for the remaining (non-cyclic) coordinates. (a) From the Lagrange equations of motion, show that the Routhian equations of motion (for a generic Lagrangian not just Eq. (14)) are dφ dt = R p φ (17) dp φ dt = R φ (18) d R dt ṙ = R r (19) (b) Determine R(r, ṙ, φ, p φ ) for the Lagrangian in Eq. (14) and the Routhian equations of motion. You should find 4 R = 1 2 mṙ2 V eff (r, p φ ) (20) where V eff (r, p φ ) was defined in class and the equation of motions are m r = V eff r (21) p φ =const (22) Now might be a good time to review the appropriate comment (on page 5) in lecture to appreciate the how the Routhian can help. 3 Edward John Routh was a physicist of some repute. He was also an outstanding educator at Cambridge. 4 Note that the sign of R is conventional. The choice here is nice in that the Hamiltonian part of the equations (Eq. (17) and Eq. (18)) takes the form of Hamilton s equations. But then, R is minus the effective Lagrangian for the non-cyclic coordinates. We will get around this difficulty by presenting R. 6

7 (c) A particle of mass m is confined to move on the surface of a sphere. It moves freely on the surface but experiences the acceleration of gravity g: (i) Write down the Lagrangian for this system using the spherical angular variables θ, φ. (ii) Form the Routhian for this system by Legendre transforming with respect to the cyclic coordinate. (iii) Sketch the effective potential of θ for p φ small and large, after defining what large and small means. Determine the stationary point of θ at large p φ, and briefly describe the result physically. 7