Solutions to Problem Set #11 Physics 151

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1 Solutions to Problem Set #11 Physics 151 Problem 1 The Hamiltonian is p k H = + m Relevant Poisson brackets are H p [ H, ] = =, p m H [ p, H] = = k We can now write the formal solution for (t as 3 t t t ( = 0 + th [, ] 0 + [[ H, ], H] 0 + [[[ H, ], H], H] 0 +!! 3! p0 t k0 t kp0 t k 0 t k p0 = 0 + t + +! m 3! m 3! m 4! m t! m p = 0 1 ω t + ω t! + ωt ω t + ω t!! 4! mω 3! 5! p0 = 0 cosωt+ sin ωt, mω where ω = k/ m Problem Let s define the transformation from Cartesian to polar coordinates by = lsin cos φ, y = lsin sin φ, z = lcos The kinetic energy of the system is m ml T = ( " + y" + z" = (" + sin φ" from which we can derive the canonical momenta p = ml ", = ml sin φ" The Hamiltonian, epressed in terms of the polar coordinates and their conjugate momenta, is 1 p φ H = p mglcos + + ml sin

2 Now, let s consider the angular momentum L= r p We must epress the linear momentum p in terms of the polar coordinates p = m" = ml(cos cosφ" sinsin φφ" py = my" = ml(cos sinφ" + sin cos φφ" pz = mz" = mlsin " We can now eplicitly calculate L and epress it in terms of, φ, p and p φ L = ypz zpy = ml sin sinφ" ml cos (cossinφ" + sin cos φφ" = ml sinφ" ml cos sin cosφφ" = p sinφ cot cos φ, Ly = zp pz = ml cos (cos cosφ" sinsin φφ" + ml sin cosφ" = ml cosφ" ml cos sin sinφφ" = p cosφ p cotsin φ, and φ Lz = py yp = ml sin cos φ(cos sinφ" + sin cos φφ" ml sin sin φ(cos cosφ" sinsin φφ" = ml sin φ" = p φ It is now trivial (albeit tedious to calculate the Poisson brackets L L y L L y L L y L L y [ L, Ly] p p cosφ sinφ = cosφ + sinφ sin sin ( p cosφ + cot sin φcot sinφ cot cos φ( p sinφ + cot cos φ = p cot φ sin = = L, z and Ly L L z y L L z y L L z y Lz [ Ly, Lz] p p p p Ly = φ = p sinφ p cot cosφ = L, φ φ φ

3 Lz L Lz L Lz L Lz L [ Lz, L] p p L = φ = p cosφ cotsinφ = L y These three Poisson brackets satisfy [ Li, Lj] = εijklk The other combinations are obvious due to the Poisson brackets identities So, why is it that p and p φ can be used as canonical momenta, although they are perpendicular components of the angular momentum? As discussed in the lecture, given the Poisson bracket relations between the three components of the angular momentum, at most one of them can be a canonical momentum However, we are using p and p φ as two canonical momenta in this problem, and they are in fact the components of the angular momentum around two perpendicular aes The Poisson bracket relations [ Li, Lj] = εijklk are satisfied for the Cartesian coordinates, y, and z, which are defined independently from each other More specifically, the directions of the aes do not depend on the coordinates themselves This is not true for the polar coordinates The ais around which the coordinate is defined rotates as we vary φ Although, for eample, at φ = 0, the momenta p and p φ equal to L y and L z, their dependences on φ differ, making the Poisson brackets different Problem 3 (a Let s model the system by an array of masses m, spaced by, connected by springs of a constant k The vibration displaces the mass at position by η( in a direction 1 perpendicular to the string The kinetic energy of this mass is mη" Since there is a constant tension T, the spring is stretched even at the equilibrium by T k When the string vibrates, the spring between and + is further stretched due to the slope between the two masses by η( + η( 1 + Assuming that this slope is small, one can approimate the increase of the length of the spring as η( + η(, which makes the total stretch of the spring from its natural length T η( + η( + k This potential energy of this spring is given by

4 k T η( + η( + k which can, for small slopes, be approimated to T T η( + η( + k Since the first term does not depend on η, we can eliminate it from our problem by redefining the zero-point of the potential energy Now we can epress the total Lagrangian as an infinite sum: m T η( + η( L= {" η( } m η( + η( = {" η( } T Bringing to 0 gives dη L= µη T d " d The above construction of the Lagrangian works even if the density of the string is not uniform We just have to use different mass for each position, ending up with the linear mass density µ( as a function of position Once we know the Lagrangian density L, we can write down the equation of motion as d L d L L d d dη + d ( µη T η = " dt " η d d η dt d d d η = µη"" T = 0 d This is the usual 1-dimensional non-dispersive wave equation Even if the density is position dependent, the wave equation at each point is unchanged (b The easiest way of handling this problem is to consider it as a special case of (a in which the tension T is not constant but a linear function of the height from the lowest end of the string Let s define the ais vertically and the lowest end of the string is = 0 The mass of the string below is µ, which makes the tension T = µ g This gives us the Lagrangian density dη L = µη " µg d Alternatively, we can start from an infinite pendulums, each hanging from the mass of the pendulum just above it We may still label each pendulum using the nominal height of the mass The horizontal displacement of the mass is given by η( The

5 1 kinetic energy is obviously mη" For small displacement, the mass is higher than the nominal position due to the slope by η( + η( This pulls up not only the mass m of the pendulum, but also the masses of all the pendulums below it The total mass below this pendulum is given by m = µ The potential energy is therefore µ g ( ( η + η We can now write the Lagrangian m µ g η( + η( L= " η η( + η( = µη " µ g Taking the limit 0 results in the same Lagrangian density as above The equation of motion is d L d L L d d dη + d ( µη µ g η = " dt " η d d η dt d d dη d η = µ "" η g g = 0 d d (3 I m very curious to see if anybody managed to get the solution for this nasty differential equation