University:
Santa Fe College
Course:
PHY 2004 | Applied Physics 1
Academic year:

2024
Views:

308

Pages:

5
Author:

Jabin F.

Conserved Quantity: In physics, any quantity is said to be conserved if its total time derivative is zero. If df/dt = 0, then f is conserved or a constant of motion. Note: If a question is asked related to a conserved quantity, then use the fact that its total time derivative is zero and also use EOM if required. P-92: A system is governed by the Hamiltonian H = (1/2)(px - ay)² + (1/2)(py - bx)², where a and b are constants and px, py are momenta conjugate to x and y respectively. For what values of a and b will the quantities (px - 3y) and (py + 4x) be conserved? Solution: Hamilton's EOMs: ẋ = px - ay, ẏ = py - bx, ṗx = b(py - bx), ṗy = a(px - ay) For (px - 3y) to be conserved, d/dt(px - 3y) = ṗx - 3ẏ = b(py - bx) - 3(py - bx) = (b - 3)(py - bx) = 0 For (px - 3y) to be satisfied, b = 3 or (py - bx) = 0 For (py + 4x) to be conserved, d/dt(py + 4x) = ṗy + 4ẋ = a(px - ay) + 4(px - ay) = (a + 4)(px - ay) = 0 For (py + 4x) to be satisfied, a = -4 Continuation of the previous image: For this to be satisfied: (a + 4) = 0 or (px - ay) = 0 a = -4 D-18-19/8/19: Q. Lagrangian of a system is L = (1/2)m(ẋ² + ẏ²) + k(x + 2y). If (px + by) is conserved, find the value of b. Solution: If (px + by) is conserved, then its total time derivative is zero. d/dt(px + by) = ẋ + ẏ = 0 k + k = 0 k = -k k = 0 Conclusion: If L = (1/2)m(ẋ² + ẏ²) + k(x + 2y), then (px - 3y) and (py + 4x) are conserved. P-24: Q. A particle in 2-D is in a potential V(x, y) = x + 2y. Which of the following (apart from the total energy of the particle) is also a constant of motion? (a) px - 3y (b) px + py (c) px + 3y (d) py + 9x Solution: Using the above conclusion, (px - 3y) is conserved. P-18: Hamiltonian of a particle is H = (p²/2m) + kx⁴. If energy of the particle is doubled, then area of phase space trajectory will become →? Solution: H = (p²/2m) + kx⁴ → n = 4 A ∝ E^(1/n) = E^(1/4) A_after / A_before = (E_after / E_before)^(1/4) = (2E / E)^(1/4) = 2^(1/4) A_after = 2^(1/4) * A_before