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

2024
Views:

358

Pages:

5
Author:

Jabin F.

# Lagrangian EOM: It was obtained accidentally from Euler's equation. d/dt (∂L/∂qi) - ∂L/∂qi = 0 where yi = qi which is the condition for extremization of functional ∫(f(qi, qi', x))dx. By doing the following replacement: t → x, yi → qi, yi' → qi', and f → L = T - U, we get d/dt (∂L/∂qi) - ∂L/∂qi = 0 → It gives the same result as Newtonian equations. Action: S = ∫L dt → Lagrangian equation is the condition of extremization of action. → A particle always moves in a path in which action is extremum. Dimension of action is the same as angular momentum or Planck's constant (h). → Any question related to Lagrangian can be solved using Euler-Lagrangian equations. Q. If Q = αpq, P = q - αp in a CT, then find the value of α. Solution: {Q, Q} = 0 ⇒ {αpq, αpq} = 0 ⇒ α²{p, q} = 0 ⇒ α² = 0 ⇒ α = 0 Q. If transformation Q = α²pq + βp² + γq², P = q - αp is canonical, then the value of ω + β is? Solution: {Q, P} = 1 ⇒ {α²pq + βp² + γq², q - αp} = 1 ⇒ α²{p, q} + β{p², q} + γ{q², q} - α{pq, p} = 1 ⇒ α² + 0 + 0 - α² = 1 ⇒ 0 = 1 For this to be satisfied, α² - 1 = 0 ⇒ α = ±1 If α = 1, then β = 0 If α = -1, then β = 1 Therefore, ω + β = 1 or 0.