Physics 351 Friday, April 24, 2015

Transcription

1 Physics 351 Friday, April 24, 2015 HW13 median report time = 5 hours. You ve solved 145 homework problems this term (not counting XC). Whew! This weekend, you ll read Feynman s two lectures (Feynman Lectures II-40 and II-41) that cover fluids. Online questions are in the usual place. Final exam is Wednesday, May 6, from 9am-11am, in DRL A4. You can bring a sheet of your own handwritten notes. Final exam includes Ch7 (Lagrangians), Ch9 (non-inertial frames), Ch10 (rigid-body rotation), Ch13 (Hamiltonians), and will (like midterm) be mainly based on HW problems. You can submit XC work as late as Tuesday, May 12. Today / next week: Connecting the classical action with the Schrödinger equation. Fluids, e.g. Bernoulli s equation (for fun). Perhaps a couple of Mathematica-based chaos illustrations

2 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

3 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

4 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.

5 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

6 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 m 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.

7 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 2 2m 2 2mL 2 p L L λ de Broglie

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9 How does pressure P vary with depth z beneath water surface? What is the difference in pressure forces exerted on the bottom of the tube by the water vs. on the top of the tube by the atmosphere, as a function of immersed depth z? What is the condition for static equilibrium?

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11 A sphere floats in water with 60% of its volume submerged. The same sphere floats in oil with 70% of its volume submerged. What is the density of the oil?

12 A sphere floats in water with 60% of its volume submerged. The same sphere floats in oil with 70% of its volume submerged. What is the density of the oil? By the way, what is the density of water in SI units?

13 The gauge pressure (i.e. pressure w.r.t. P atm ) in the tires of a car is 200 kpa. (That s plausible: about 2 atm.) The area of each tire in contact with the road is 120 cm 2. What is the mass of the car? (We re assuming that the tire is initially perfectly round, then deforms until the pressure force balances the weight. I m not sure how realistic this is.)

14 The gauge pressure (i.e. pressure w.r.t. P atm ) in the tires of a car is 200 kpa. (That s plausible: about 2 atm.) The area of each tire in contact with the road is 120 cm 2. What is the mass of the car? (We re assuming that the tire is initially perfectly round, then deforms until the pressure force balances the weight. I m not sure how realistic this is.) By the way, what is 1 atm in SI units? What s the density of mercury? How tall a column of mercury has its weight-per-area balanced by atmospheric pressure?

15 The gauge pressure (i.e. pressure w.r.t. P atm ) in the tires of a car is 200 kpa. (That s plausible: about 2 atm.) The area of each tire in contact with the road is 120 cm 2. What is the mass of the car? (We re assuming that the tire is initially perfectly round, then deforms until the pressure force balances the weight. I m not sure how realistic this is.) By the way, what is 1 atm in SI units? What s the density of mercury? How tall a column of mercury has its weight-per-area balanced by atmospheric pressure? 1 atm = Pa. ρ Hg 13.6ρ H2 O kg/m /( ) = m = 760 mm.

16 A container of liquid has a vertical acceleration of a upward. How does the absolute pressure vary as a function of depth in the container?

17 Water flows through an old plumbing pipe that is partially blocked by mineral deposits along the wall of the pipe. Through which part of the pipe is the fluid speed largest? (A) Fastest in the narrow part. (B) Fastest in the wide part. (C) The speed is the same in both parts.

18 A water-borne insect drifts along with the flow of water through a pipe that is partially blocked by deposits. As the insect drifts from the narrow region to the wider region, it experiences (A) an increase in pressure (P wide > P narrow ) (B) no change in pressure (P wide = P narrow ) (C) a decrease in pressure (P wide < P narrow ) Hint: is the speed the same or different? Is there a net force (per unit area) that causes this change in speed?

19 (P 1 A 1 )(v 1 t) (P 2 A 2 )(v 2 t) = 1 2 m(v2 2 v 2 1) + mg(y 2 y 1 ) (P 1 P 2 )(Volume) = (P 1 P 2 ) m ρ = 1 2 m(v2 2 v 2 1) + mg(y 2 y 1 )

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21 This experiment combines the equation of continuity with Bernoulli s equation. When I open the valve, compressed air will flow from left to right through the horizontal tube. The tube is wide on the left and right, but narrow in the middle. Where is the speed largest? Where is the pressure lowest? How will height of the green liquid respond to changes in pressure in the horizontal tube?

22 Once I turn on the air flow, the horizontal speed of the flowing air will be (A) fastest above tube B (B) slowest above tube B (C) the same above all tubes

23 Once I turn on the air flow, the height of the green liquid will be (A) lowest in tube B (B) highest in tube B (C) the same in all tubes

24 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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26 Physics 351 Friday, April 24, 2015 HW13 median report time = 5 hours. You ve solved 145 homework problems this term (not counting XC). Whew! This weekend, you ll read Feynman s two lectures (Feynman Lectures II-40 and II-41) that cover fluids. Online questions are in the usual place. Final exam is Wednesday, May 6, from 9am-11am, in DRL A4. You can bring a sheet of your own handwritten notes. Final exam includes Ch7 (Lagrangians), Ch9 (non-inertial frames), Ch10 (rigid-body rotation), Ch13 (Hamiltonians), and will (like midterm) be mainly based on HW problems. You can submit XC work as late as Tuesday, May 12. Today / next week: Connecting the classical action with the Schrödinger equation. Fluids, e.g. Bernoulli s equation (for fun). Perhaps a couple of Mathematica-based chaos illustrations