Homework 12 - Physics 601 Homework 12   Due Friday...

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Unformatted text preview: Physics 601 Homework 12   Due Friday Dec. 10 1. Last week, I had you consider the case of a physical pendulum in the regime the high energy regime where the kinetic energy is much large then the potential. This time I want to consider the opposite regime ­ ­ ­extremely low energies. Again the ˙ Lagrangian is given by L = 1 Iθ 2 + V0 cos(θ ) where I is the moment of inertia and V0 2 is the maximum value of the potential energy (i.e. m g L) where L is length from the pivot point to the center of mass. The problem under consideration is this: suppose that at t=0 the system is at rest with θ = θi. I wish to develop a systematic € power counting scheme for this problem. 1 V0 cos( λ2θ ) 1 ˙2 Consider the following power counting scheme in terms of λ: L = 2 Iθ + λ . a. Assuming this system is at low energy and is θ small and one can expand out the cosine. Show that this is sensible in that leading order term with the € harmonic term in leading order, a quartic term as a perturbation and a sextic term as a smaller perturbation and so on. b. One can write θ = θ 0 + λθ1 + λ2θ 2 + ... . Find explicitly the form for θ1 and θ 2 . 2. The approximate solution found in problem with naïve perturbation theory does not impose periodicity. However on general grounds we expect that € € € θ ( t ) = ∑ c n cos((2 n + 1)ωt ) . Match this form onto the perturbative solution using n € the power counting scheme ω = ω 0 + λω1 + λ2ω 2 + ... , to compute the frequency as a function of amplitude up to the order computed. Show that this gives ω (θ 0 ) as a series in θ 0 with accurate terms up to 2nd order. € € 3. € onsider a particle of mass m and charge q particle moving in a spatial constant C (but time varying magnetic field). For the sake of simplicity assume that the magnetic field, which we will take to be in the z ­direction. Write the vector ˆ potential in the form A = yB( t ) x so that the Hamiltonian for this system is given by ( ) 2 2 px + py − qB( t ) x + pz2 . Thus py and pz are conserved. 2m a. Find €x , v y and v z in terms of px , py , pz , x, q, m and B( t ) . v H ( x, y, z, px , py , pz ) = € € € € € € € b. Since py and pz are conserved one can treat them as constants and consider a Hamiltonian in terms of the x degrees of freedom only: € H ( x, px ) = € ( ) 2 2 px + py − qB( t ) x + pz2 2m . Show that this can be written in terms of pz2 JqB( t ) + where 2 m 2π m ȹ ȹ p ȹ ȹ ȹ qB( t )ȹ x − y ȹ ȹ 2 2 ȹ qB( t ) Ⱥ ȹ py ȹ 1 π px ȹ ȹ tan −1ȹ J = qB( t )π ȹ x − + and ω = ȹ ȹ . 2π px ȹ €qB( t ) Ⱥ qB( t ) ȹ ȹ ȹ ȹ ȹ Ⱥ v 2 − v z2 c. Show that where (where v is the speed of the particle) is an B€ adiabatic invariant. d. Note that in the adiabatic limit any given time the particle is making circular orbits (with the radius v€ arying in time as the magnetic field strength € how that the magnetic flux threading through the circular orbit is changes) s an adiabatic invariant. e. Suppose a particle makes a circular orbit with radius ro in a magnetic field Bo. What is the radius if the magnetic field is slow increased to 2 Bo. action angle variables: H (ω , J ) = € € ...
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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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