601exam1 - PHYSICS 601, EXAM 1, SOLUTIONS Disclaimer: These...

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Unformatted text preview: PHYSICS 601, EXAM 1, SOLUTIONS Disclaimer: These solutions were cobbled together in a somewhat helter- skelter fashion, and likely contain errors. Also, detailed steps have been left out in some places in favor of mere gestures at what to do. Thanks to various people (most notably Tom Cohen and Ben Dreyfus) for help. -Don Perlis PROBLEM 1(a) L = L ( p, p ) = p i q I + H ( q, p ) where implicitly the vector q is a function of p and p (otherwise the definition of L would not make sense, as a function of p and p ) and where the p i are the conjugate momenta of the q i ; and thus we have p i = L/ q i and the Hamilton equations: q i = H ( q, p ) /p i- p j = H ( q, p ) /q j We then calculate: L/ p i = [ p j q j + H ( q, p )] / p i = ( p j q j ) / p i + H ( q ( p, p ) , p ) / p i [summing over j] = ( p j / p i ) q j + p j ( q j / p i ) + ( H/q j )( q j / p i ) + ( H/p j )( p j / p i ) = q i + p j ( q j / p i )- p j ( q j / p i ) + q j = q i PROBLEM 1(b) ( d/dt )[ L/ p i ] = ( d/dt ) q i (from part a above) = q i But also L/p i = ( p j q j ) /p i + H ( q, p ) /p i = ( p j /p i ) q j + p j q j /p i + ( H/q j )( q j /p i ) = ( p i /p i ) q i + p j ( q j /p i )- p j ( q j /p i ) = q i So ( d/dt ) L/ p i = q i = L/p i PROBLEM 1(c) From part b above we see that L satisfies the E-L equations, which means that the time-integral of L over a given time interval (i.e., the action) achieves an extreme value (max or min) with the function p . PROBLEM 1(d) We are given L ( q, q ) = (1 / 2) m q 2- (1 / 4) aq 4 , in 1-dim. 1 2 PHYSICS 601, EXAM 1, SOLUTIONS Then p = dL/ q and H ( q, p ) = p q- L = p q- (1 / 2) m q 2 + (1 / 4) aq 4 So L ( p, p ) = pq + H = pq + p q- (1 / 2) m q 2 + (1 / 4) aq 4 But p = L/ q = m qso q = p/m and ( d/dt ) L/ q = L/q ( d/dt ) m q =- aq 3 m q =- aq 3 m ( p/m ) =- aq 3 q =- [ p/a ] 1 / 3 Finally, L =- p [ p/a ] 1 / 3 + p 2 /m- (1 / 2) m ( p/m ) 2 + (1 / 4) a [ p/a ] 4 / 3 = [- 1 /a 1 / 3 + 1 / (4 a 1 / 3 )]( p ) 4 / 3 + p 2 / 2 m =- 3 / (4 a 1 / 3 )( p ) 4 / 3 + p 2 / 2 m PROBLEM 2(a) We have planar motion given by H ( r, , p r , L ) = ( p r ) 2 / 2 m + ( L + ar 3 ) 2 / 2 mr 2 where L is not the Lagrangian but rather is a generalized momentum conjugate to . Note then that = H/L = ( L + ar 3 ) /mr 2 , so...
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This note was uploaded on 12/29/2011 for the course PHYSICS 601 taught by Professor Hassam during the Fall '11 term at Maryland.

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601exam1 - PHYSICS 601, EXAM 1, SOLUTIONS Disclaimer: These...

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