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Unformatted text preview: Lecture 13: Action, constraints (2 Oct 09) A. Action integral exs 1. Evaluate S = Z t 2 t 1 L [ q, q ] dt subject to q ( t 1 ) = q 1 ; q ( t 2 ) = q 2 for 3 potentials. 2. Free particle in 1D x = 0 x ( t ) = x 1 + x 1 ( t t 1 ) and set the initial velocity by the match at t 2 : x 1 = ( x 2 x 1 ) / ( t 2 t 1 ) S = Z dtm x 2 / 2 = [ m x 2 1 / 2] Z dt = m ( x 2 x 1 ) 2 2( t 2 t 1 ) 3. Particle in 1D with constant force. x ( t 1 ) x 1 ; x ( t 2 ) x 2 = fx m x = f x ( t ) = x 1 + v 1 ( t t 1 ) + f ( t t 1 ) 2 / 2 x 2 x 1 = v 1 ( t 2 t 1 ) + ( f/ 2)( t 2 t 1 ) 2 v 1 = x 2 x 1 t 2 t 1 ( f/ 2)( t 2 t 1 ) Calculate the action (let t 2 t 1 t ). Algebra leads to: S = Z t 2 t 1 [ 1 2 m x 2 + fx ] dt = 1 2 t ( x 2 x 1 ) 2 + 1 2 ft ( x 2 + x 1 ) t 3 24 f 2 4. Particle in 1D oscillator potential, angular frequency ; measure times from t 1 = 0 and define = t 2 : x ( t ) = 1 sin [ x 1 sin (  t ) + x 2 sin t ]...
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This note was uploaded on 12/14/2009 for the course PHYS 711 taught by Professor Bruch during the Fall '09 term at Wisconsin.
 Fall '09
 BRUCH

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