# Probs2 - Physics 316, Problem Set 2, due Tuesday October 11...

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Unformatted text preview: Physics 316, Problem Set 2, due Tuesday October 11 in class 1. Phase space area: For one dimensional periodic motion in an external potential, with energy E , let S ( E ) be the area enclosed by the phase curve during one period. a)Show that the period T (the time required for the motion to repeat) is given by dS/dE . ( cf. Problem p. 20) Suppose T varies smoothly with E . If the initial position x is displaced infinitessimally, the displaced orbit x d ( t ) is altered from the unperturbed orbit x ( t ) Moreover, ( x d ( t )- x ( t )) ≡ Δ( t ) grows with time. If Δ(0) is very small, it must remain much smaller than the original orbit for many orbital periods even though it may grow by a large factor. The problem concerns how Δ grows during this range of time. b) Express the shift in energy E d- E as a function of Δ(0) and the potential energy U ( x ). c) Determine how Δ grows by considering times t that are large integer multiples of the original period T as a function of dT/dE . Consider times such that Δ remains small compared to the original orbit. Solution: a) The area enclosed by the phase curve during one period can be expressed as: S ( E ) = 2 Z x max x min ˙ xdx = 2 Z x max x min p 2( E- U ) dx. And the period can be expressed as: T = 2 Z x max x min dx ˙ x = 2 Z x max x min dx p 2( E- U ) . So dS dE = 2 Z x max x min 1 p 2( E- U ) dx = T. b) Assume the initial position is infinitesimally displaced to y = x +Δ with Δ << x and the period is correspondingly shifted to T y = T + dT dE dE = T + dT dE dU dx | x Δ . Since both are periodic motions, we have for generic point x and its corresponding y that x ( nT ) = x (0) and y ( nT y ) = x (0) + Δ. Notice that since it’s a infinitesimal change, we have Δ = Δ + O (Δ 2 )....
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## This note was uploaded on 03/02/2010 for the course PHYSICS 316 taught by Professor Pucker during the Spring '10 term at King's College London.

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Probs2 - Physics 316, Problem Set 2, due Tuesday October 11...

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