601exam1

# 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) ˆ L = ˆ 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 ˆ 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: ∂ ˆ 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 )[ ∂ ˆ L/∂ ˙ p i ] = ( d/dt ) q i (from part a above) = ˙ q i But also ∂ ˆ 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 ) ∂ ˆ L/∂ ˙ p i = ˙ q i = ∂ ˆ L/∂p i PROBLEM 1(c) From part b above we see that ˆ 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 ˆ 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, ˆ 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, θ, 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 θ . Note then that ˙ θ = ∂H/∂L = ( L + ar 3 ) /mr 2 , so...
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601exam1 - PHYSICS 601 EXAM 1 SOLUTIONS Disclaimer These...

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