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

2024
Views:

239

Pages:

5
Author:

Jabin F.

# Problem on evaluation of PB: Q. {xi, bj, bk} = {xi, (bjbk)} = {xi, bj(bk - bj)} = ∑i=1n bj(bk - bj) Q. {(a_i, b_i), (c_i, d_i)} = {a_i, b_i} + {c_i, d_i} = {a_i, (b_i + d_i)} + {c_i, (a_i + c_i)} = (a_i, b_i) + (a_i, d_i) + (c_i, a_i) + (c_i, c_i) = (a_i, b_i) + (a_i, d_i) + (c_i, a_i) + n(c_i, c_i) (Standard Result: {(a_i), (b_i)} = n(a_i, b_i)) Q. {xi, (xi + xk)} = {xi, xi} + {xi, xk} = {xi, xk} = ∑i=1n xk = nxk This is direct short step method. Now step by step method is: {xi, (xi + xk)} = {xi, ∑j=1n xk} = ∑i=1n ∑j=1n xk = ∑i=1n [∑j=1n xk] = ∑i=1n (nxk) = n²xk Note: Never write {∂L/∂pi, ∂L/∂qi} in PB directly. Use the result of PB with angular momentum. P-100 The Lagrangian for a simple pendulum is given by: L = (1/2)mℓ²θ̇² - mgℓ(1 - cos θ) Q-9 The Poisson bracket between θ and p is: Solution: First we have to express θ in terms of pθ: pθ = ∂L/∂θ̇ = mℓ²θ̇ ⇒ θ̇ = pθ / (mℓ²) θ = ∫θ̇ dt = ∫(pθ / (mℓ²)) dt = (pθ / (mℓ²))t {θ, pθ} = (∂θ/∂pθ)(∂pθ/∂θ) - (∂θ/∂θ)(∂pθ/∂pθ) = (1 / (mℓ²))(mℓ²) - 0 = 1 P-101 The co-ordinates and momenta xi, bi (i=1,2,3) of a particle satisfy canonical Poisson bracket relations {xi, bj} = δij. If Ci = xi + pi and Di = xi - pi are constants of motion and if C₃ = {C₁, C₂} = 0, then: (a) {C₂, C₃} = 0 and {C₃, C₄} = 0 (b) {C₂, C₃} = 0 and {C₃, C₄} = -1 (c) {C₂, C₃} = 1 and {C₃, C₄} = 0 (d) {C₂, C₃} = 1 and {C₃, C₄} = -1 Solution: {C₂, C₃} = {x₂ + p₂, x₃ + p₃} = {x₂, x₃} + {x₂, p₃} + {p₂, x₃} + {p₂, p₃} = 0 + 0 + 0 + {p₂, p₃} = {p₂, p₃} = (∂p₂/∂x₃)(-∂p₃/∂x₂) - (∂p₂/∂p₃)(∂p₃/∂x₂) = 0 - (-1) = 1 {C₃, C₄} = {x₃ + p₃, x₄ + p₄} = {x₃, x₄} + {x₃, p₄} + {p₃, x₄} + {p₃, p₄} = 0 + 0 + 0 + {p₃, p₄} = (∂p₃/∂x₄)(-∂p₄/∂x₃) - (∂p₃/∂p₄)(∂p₄/∂x₃) = -1 - 0 = -1