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

2024
Views:

79

Pages:

5
Author:

Jabin F.

# Poisson's Theorem: If A and B are two quantities which are constants of motion or conserved, then if their Poisson bracket gives another quantity, then that quantity is also constant of motion or conserved. Proof: If A is conserved, then dA/dt = 0. Similarly, dB/dt = 0. Now, d/dt {A, B} = (∂{A, B}/∂qi) * dq_i/dt + (∂{A, B}/∂pi) * dp_i/dt + ∂{A, B}/∂t = (∂{A, B}/∂qi) * (∂H/∂pi) + (∂{A, B}/∂pi) * (-∂H/∂qi) + ∂{A, B}/∂t = {0, B} + {A, 0} + {0, 0} = 0 (Jacobi identity) Therefore, d/dt {A, B} = 0 → {A, B} is constant/conserved. Question 1: A particle moves in a potential V = 1/2(x^2 + y^2 + z^2). Which component of angular momentum is a constant of motion? Solution: L = (m^2)(x^2 + y^2 + z^2) - 1/2(x^2 + y^2 + z^2) For Lx, (x^2 + y^2 + z^2) appears in x. For Ly, (x^2 + y^2 + z^2) appears in y. For Lz, (x^2 + y^2 + z^2) appears in z. Therefore, Lz is a constant of motion. Question 2: The Lagrangian of a system is L = 1/2m(x^2 + y^2 + z^2) - V. Which of the following is conserved? (a) Lx (b) Ly (c) Lz (d) energy Solution: Lz is present but V is not cyclic in z. So Lz is not conserved. V is not present explicitly. So energy is conserved. We know that on rotation about the z-axis, x and y don't change. So Lz is invariant under rotation about the z-axis. Therefore, Lz is conserved.