3.7 Non-adiabatic motion in symmetric geometry 99In this cylindrical coordinate system the Lagrangian, Eq.(3.12), has the formL=
m
2(
r ̇^2 +r^2 θ ̇2
+ ̇z^2)
+qrθA ̇ θ−qφ(r,z,t). (3.155)Sinceθis an ignorable coordinate, the canonical angular momentum is a constant of the
motion, i.e.
Pθ=∂L
∂θ ̇=mr^2 θ ̇+qrAθ=const. (3.156)or, in terms offlux,
Pθ=mr^2 θ ̇+q
2 π
ψ(r,z,t) =const. (3.157)Thus, the Hamiltonian is
H =
m
2(
r ̇^2 +r^2 θ ̇2
+ ̇z^2)
+φ(r,z,t)=
m
2(
r ̇^2 + ̇z^2)
+
(Pθ−qψ(r,z,t)/ 2 π)^2
2 mr^2+φ(r,z,t)=
m
2(
r ̇^2 + ̇z^2)
+χ(r,z,t)(3.158)
where
χ(r,z,t) =1
2 m[
Pθ−qψ(r,z,t)/ 2 π
r] 2
(3.159)
is aneffectivepotential. For purposes of plotting, the effective potential can be writtenin a
dimensionless form as
χ(r,z,t)
χ 0=
2 πPθ
qψ 0−
ψ(r,z,t)
ψ 0
r/L
2(3.160)
whereLis some reference scale length,ψ 0 is some arbitrary reference value for theflux,
andχ 0 =qψ^20 / 8 π^2 L^2 m.For simplicity we have setφ(r,z,t) = 0,since this term gives
the motion of a particle in a readily understood, two-dimensional electrostatic potential.
Suppose that for timest <t 1 the coil currents are constant in which case the associated
magnetic field andflux are also constant. Since the Lagrangian does not explicitly depend
on time, the energyHis a constant of the motion. Hence there are two constants of the
motion,HandPθ.Consider now a particle located initially on the midplanez= 0with
r < rmax.The particle motion depends on the sign ofqψ/Pθand so we consider each
polarity separately.