Physics 308: Advanced Classical MechanicsBryn Mawr College, Fall 2011

Problem Set 6

Distributed: Thursday, October 27, 2011.Due: Thursday, November 3, 2011 at the start of class.

Reading

For Tuesday, please read Sections 8.1 through 8.4 (inclusive) in Taylor. And for Thursday,please finish Chapter 8 (8.5 through 8.8).

As always, I encourage you to work collaboratively on these problems (with the exception ofthe problem labeled “INDIVIDUAL PROBLEM,” which must be completed by you alone).

Problems

1. Forces of constraint: A great advantage of Lagrangian mechanics is that you don’thave to “worry” about forces of constraint (normal forces, tension, etc.) when derivingequations of motion. But in cases where you actually want to know what the forces ofconstraint are, you need to get clever. In class, we found two ways to calculate the forcesof constraint within the Lagrangian framework. The first (not covered in Taylor’s book)entailed introducing a very steep potential function V (~r), and the second made use of theso-called Lagrange multiplier λ(t) and the constraint equation f(x, y).

(a) Using the steep potential approach, show that for a mass m sliding down a frictionlessplane inclined at an angle θ, the normal force from the plane is given by the familiarmg cos θ. Note: the inclined plane is rigidly attached to the floor.

(b) Solve the same problem using Lagrange Multiplier approach.

(c) In Example 7.8 (p. 279) Taylor calculates the tension in the string for a simple Atwoodmachine. He uses the constraint equation: f(x, y) = x + y = constant. Anotherperfectly legitimate expression of constraint would be f(x, y) = 7(x + y) = constant(if x+y is a constant, then surely 7(x+y) is also a constant). If you use this alternateexpression for f(x, y), do you get the correct tension? Comment on this.

2. Coordinate Choices Can Obfuscate Physics: In class, we calculated the Hamiltonianfor a free particle in 2D using Cartesian coordinates. We found that neither x nor yappeared in the Hamiltonian and so both were “ignorable” coordinates, which impliedthat their conjugate momenta (px and py) were conserved. We’ll now repeat this exercisebut instead use 2D polar coordinates (r, θ).

(a) Find the Lagrangian for the system.

(b) Find q(q, p) from p = ∂L /∂q and calculate H(r, θ, pr, pθ)

(c) Using Hamilton’s Equations, find the time-evolution of the coordinates and momenta.

(d) What quantity does pθ correspond to? Explain using Newtonian reasoning why pθ = 0.

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(e) Why isn’t pr = 0?

(f) Since H doesn’t explicitly depend on t, we know that H is conserved. And sincethe relationship between (r, θ) and (x, y) is independent of time, we also know thatH = T + U is the total energy of the system. Show explicitly that H = 0. Does thishelp explain why pr 6= 0?

(g) Comment on the benefit of a wise choice of coordinates.

3. Ignorable Cordinates: Do Taylor 13.21. This problem is also good preparation for ournext topic: central force motion.

4. Mixing Coordinates: Do Taylor 13.25. The Hamiltonian formalism allows you to forma coordinate transformation that involves both the position and the momentum. Youcannot do this in Lagrangian mechanics.

5. (INDIVIDUAL) Phase Space Orbits: Do Taylor 13.28. Note: this force is not thesame as the spring force (Hooke’s Law).

6. Liouville’s Theorem and a Beam of Particles: Do Taylor 13.35 to get a sense ofwhat’s going on inside the particle beam pipe at colliders like the Tevatron (well, whenthe Tevatron was still running that is...) and the LHC. For what it’s worth, Liouville’stheorem is also applicable in optical systems, which is especially important if you’re de-signing a telescope or microscope. The so-called etendue (a.k.a. the AΩ product), whichcharacterizes how “spread out” light is in area and angle, is directly analagous to a phase-space volume. Liouville tells you that AΩ cannot increase in free space or via reflectionor refraction in an optical system (though you certainly can decrease it if you design theoptics poorly and truncate some light rays). The etendue is a fundamental quantity thatyou cannot circumvent no matter how good of an optical engineer you are.

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