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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.
 Fall '11
 Hassam
 mechanics, Energy, Work

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