A particle of charge moves in a uniform magnetic field

PHYSICS · Multi-part · Fermi University Physics · Mechanics & Dynamical Systems · Hamiltonian mechanics and canonical structure · Hamiltonian mechanics and canonical structure

Action and Symmetry, Ch Canonical transformations, generating functions and symplectic structure

Problem

A particle of charge QeQ_{\rm e} moves in a uniform magnetic field B=Bz^\mathbf B=B\hat{\mathbf z}. The canonical momentum of a charged particle (Chapter 13) in the gauge A2=(0,Bx,0)\mathbf A_2=(0,Bx,0) is p=mv+QeA2\mathbf p=m\mathbf v+Q_{\rm e}\mathbf A_2, giving px=mvxp_x=mv_x, py=mvy+QeBxp_y=mv_y+Q_{\rm e}Bx. A second gauge, A1=(−12By,12Bx,0)\mathbf A_1=(-\tfrac12By,\tfrac12Bx,0), describes the same field and gives canonical momentum P=mv+QeA1\mathbf P=m\mathbf v+Q_{\rm e}\mathbf A_1. The two gauges are related by A1=A2+∇Λ\mathbf A_1=\mathbf A_2+\nabla\Lambda with Λ(x,y)=−12Bxy\Lambda(x,y)=-\tfrac12Bxy, so the new variables are X=x,Y=y,Px=px+Qe∂Λ∂x=px−12QeBy,Py=py+Qe∂Λ∂y=py−12QeBx.\begin{aligned} X &= x, \qquad Y = y,\\ P_x &= p_x+Q_{\rm e}\frac{\partial\Lambda}{\partial x}=p_x-\tfrac12Q_{\rm e}By,\\ P_y &= p_y+Q_{\rm e}\frac{\partial\Lambda}{\partial y}=p_y-\tfrac12Q_{\rm e}Bx. \end{aligned} ((a)) Using the fundamental Poisson brackets of the old variables (x,y,px,py)(x,y,p_x,p_y) (Chapter 14), compute {X,Px}\{X,P_x\}, {Y,Py}\{Y,P_y\} and {Px,Py}\{P_x,P_y\}, treating QeQ_{\rm e} and BB as constants. The last is the bracket in which the field terms could spoil the result. ((b)) Say why the remaining fundamental brackets need no calculation, state whether the gauge change is a canonical transformation, and say which earlier result (Chapter 2's point transformations) already guarantees that X=xX=x, Y=yY=y alone pass the coordinate part of the test.

Browse the problem library