Physics 351 Wednesday, April 22, 2015

Transcription

1 Physics 351 Wednesday, April 22, 2015 HW13 due Friday. The last one! You read Taylor s Chapter 16 this week (waves, stress, strain, fluids), most of which is Phys 230 review. Next weekend, you ll read Feynman s two lectures (Feynman Lectures II-40 and II-41) that cover fluids. I haven t yet written up questions for that last reading assignment. Final exam is Wednesday, May 6, from 9am-11am, in DRL A4. You can bring a sheet of your own handwritten notes. Remember, final exam will just cover Lagrangians, accelerating/rotating frames, rotation, and Hamiltonians, and will (like midterm) be mainly based on homework problems. Today: Let s go over one more rotation problem from HW13 (q8). Connecting the classical action with the Schrödinger equation. Next time: Fluids, e.g. Bernoulli s equation.

2 (a) Show that I = I 0 and find the constant I 0. (b) Calculate L at t = (c) Sketch ê 3, ω, and L at t = 0. (d) Draw/label body cone and space cone on your sketch.

3 (e) Calculate precession frequencies Ω body and Ω space. Indicate directions of precession vectors Ω body and Ω space on drawing. (f) You argued in HW11 that Ω space = Ω body + ω. Verify (by writing out components) that this relationship holds for the Ω space and Ω body that you calculate for t = 0.

4 (g) Find the maximum angle between ẑ and ê 3 during subsequent motion of the plate. Show that in the limit α 1, this maximum angle equals α. (h) When is this maximum deviation first reached? video: watch?v=oh-dlrifo10

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6 In last week s reading, Feynman argued that the classical action S cl = tf t i L(ẋ(t), x(t), t) dt is proportional to the trajectory s quantum-mechanical phase: phase = S cl / Most of you noticed that Feynman s Problems 2-4 and 2-5 suggest a way to prove, using calculus of variations, that ( ) L (p) x=xf = = + S cl and E = H = S cl ẋ x f t f x=x f Here s another route to that result: Remember that H = pẋ L L = pẋ H So we can rewrite the classical action as S cl = tf t i (pẋ H) dt = tf t i pẋ dt tf t i H dt = xf x i tf p dx H dt t i

7 L dt = (pẋ H) dt = t t S = t i t i Therefore, ( ) S t = H and fixed x x x i p dx t t i ( ) S x fixed t H dt = p S/ t + H = 0 is the Hamilton-Jacobi equation. If we plug in H = p2 2m + U(x) we can write this differential equation for the classical action: S t + 1 ( ) S 2 + U(x) = 0 2m x or in three dimensions, What is this telling us? S t + 1 2m ( S)2 + U(r) = 0

8 S t = E S x = p x S y = p y S z = p z S = p For constant energy, an action of the form S(r, t) = p r Et satisfies these equations. Notice that momentum p is to surface of constant S. Near the classical path, moving to the trajectory does not change the action as we expect from the principle of stationary action. In Physics 250, you may have described matter waves using the de Broglie relations p = k and E = ω. This suggests S(r, t)/ = k r ωt which describes the phase of a plane wave.

9 Meanwhile, the Hamilton-Jacobi equation S t + 1 2m ( S)2 + U(r) = 0 is starting to smell vaguely similar to Schrödinger s equation: ) ψ(r, t) i = ( 2 t 2m 2 + U(r) ψ(r, t) Let s try plugging (into Schrödinger) a wavefunction ψ(x, t) = ψ 0 (x, t) e iσ(x,t)/ where ψ 0 (x, t) and Σ(x, t) are real (i.e. not complex) functions. So ψ 0 2 tells us about probability, and Σ/ tells us about phase. ψ t = ψ ( ) 0 i Σ t eiσ/ + ψ 0 e iσ/ t 2 ψ x 2 = 2 ψ 0 x 2 eiσ/ + 2i Σ ψ 0 x x eiσ/ + i 2 Σ x 2 ψ 0e iσ/ 1 2 ( ) 2 Σ ψ 0 e iσ/ x

10 Plugging in and canceling common factor ψ 0 e iσ/ gives real part Σ t + 1 2m ( ) Σ 2 + U = 2 x 2m 1 2 ψ 0 ψ 0 x 2 which equals the Hamilton-Jacobi equation, in 0 limit. So evidently in some classical limit, the phase Σ of Schrödinger s ψ(x, t) satisfies the same diffeq. as does the classical action S. The imaginary part gives (skip the math here) ψ 0 t + 1 Σ ψ 0 2m x x + 1 2m ψ 2 Σ 0 x 2 = 0 which can be turned into (multiply by 2ψ 0, use Σ/ x p if Σ S) t (ψ2 0) + x (ψ2 0 1 m Σ x ) = 0 t (ψ2 0) + (ψ 2 0 v) = 0 which is (Taylor ) just the continuity equation expressing conservation of probability (ψ 2 0) as the particle travels.

11 The Schrödinger equation gives us Σ t + 1 2m ( Σ x ) 2 + U = ψ 0 2m ψ 0 x 2 while the classical Hamilton-Jacobi equation gave us S t + 1 ( ) 2 S + U = 0 2m x What does it mean for ψ 0 2m ψ 0 to be small? Consider a x 2 gaussian distribution ψ 0 (x) e x2 /2σ 2. Then ( 1 2 ) ψ 0 ψ 0 x 2 = 1 σ 2 1 L 2 x=0 where L is the length over which the probability for finding the particle varies appreciably, e.g. slit size, or distance over which U(x) varies considerably. Then classical limit means p L L λ de Broglie

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13 Morin A bead is free to slide along a frictionless hoop of radius R. The hoop is forced to rotate with constant angular speed ω around a vertical diameter. Find H in terms of θ and p θ, then write down Hamilton s equations. Is H the energy? Is H conserved?

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15 Physics 351 Wednesday, April 22, 2015 HW13 due Friday. The last one! You read Taylor s Chapter 16 this week (waves, stress, strain, fluids), most of which is Phys 230 review. Next weekend, you ll read Feynman s two lectures (Feynman Lectures II-40 and II-41) that cover fluids. I haven t yet written up questions for that last reading assignment. Final exam is Wednesday, May 6, from 9am-11am, in DRL A4. You can bring a sheet of your own handwritten notes. Remember, final exam will just cover Lagrangians, accelerating/rotating frames, rotation, and Hamiltonians, and will (like midterm) be mainly based on homework problems. Today: Let s go over one more rotation problem from HW13 (q8). Connecting the classical action with the Schrödinger equation. Next time: Fluids, e.g. Bernoulli s equation.