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UMD PHYS 601 - Homework #12

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Physics'601'Homework'12333Due'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 =12I˙ θ 2+ V0cos(θ)where!I!is!the!moment!of!inertia!and!V0!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.!!Consider!the!following!power!counting!scheme!in!terms!of!λ:!€ L =12I˙ θ 2+V0cos(λ12θ)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=θ0+λθ1+λ2θ2+ ...!!!.!Find!explicitly!the!form!for!€ θ1!and!€ θ2.!!2. The!approximate!solution!found!in!problem!with!naïve!perturbation!theory!does!not!impose!periodicity.!!!!However!on!general!grounds!we!expect!that!€ θ(t) = cncos((2n + 1)ωt)n∑!.!!Match!this!form!onto!the!perturbative!solution!using!the!power!counting!scheme!€ ω=ω0+λω1+λ2ω2+ ...,!!!!to!compute!the!frequency!as!a!function!of!amplitude!up!to!the!order!computed.!!Show!that!this!gives!€ ω(θ0)!as!a!!series!in!€ θ0!with!accurate!terms!up!to!2nd!order.!!!!!3. Consider!a!particle!of!mass!m!and!charge!q!particle!moving!in!a!spatial!constant!(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!potential!in!the!form! €  A =ˆ y B(t)x!so!that!the!Hamiltonian!for!this!system!is!given!by!€ H(x, y,z, px, py, pz) =px2+ py− qB(t)x( )2+ pz22m.!!Thus!€ py!and!€ pz!are!conserved.!!a. Find!€ vx,vy!and!€ vz!in!terms!of!€ px, py, pz, x,q,m!and!€ B(t).!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) =px2+ py− qB(t)x( )2+ pz22m.!!!Show!that!this!can!be!written!in!terms!of!action!angle!variables:!€ H(ω,J) =pz22m+JqB(t)2πm!where€ J = qB(t)πx −pyqB(t)⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2+πpx2qB(t)!!and!€ ω=12πtan−1qB(t) x −pyqB(t)⎛ ⎝ ⎜ ⎞ ⎠ ⎟ px⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ⎟ .!!c. Show!that!!where!€ v2− vz2B!(where!€ v!is!the!speed!of!the!particle)!is!an!adiabatic!invariant.!d. Note!that!in!the!adiabatic!limit!any!given!time!the!particle!is!making!circular!orbits!(with!the!radius!varying!in!time!as!the!magnetic!field!strength!changes)!show!that!the!magnetic!flux!threading!through!the!circular!orbit!is!an!adiabatic!invariant.!!e.


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UMD PHYS 601 - Homework #12

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