Cyclotron resonances
Consider a particle of charge q and mass m in a constant, uniform magnetic field B = B 0 ẑ and in
an electric field of constant amplitude E0 rotating with frequency ω in the (x, y) plane.
Show that:
a) depending on the sense of rotation of the electric field, i.e. on the sign of ω with respect to B, a
resonance in the motion of the particle appears;
b) if no friction terms are present in the equations of motion, the solution of the latter is not periodic
and determine such solution.
c) Find the periodic solution of the equations of motion if a friction term is inserted, and calculate
the absorbed power as a function of ω.
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Solution
a) The rotating field can be written as
E = E0 (x̂ cos ωt ± ŷ sin ωt)
where the positive (negative) sign indicates counterclockwise (clockwise) rotation. The clockwise case is shown in the figure.
From the equation of motions
m
dv
= q(E + v × B)
dt
we immediately find that dvz /dt = 0, thus we may assume that the motion in in the (x, y) plane.
The equations of motion along the x anf y axes are
½
v̇x = +qvy B0 /m + (qE0 /m) cos ωt
(1)
v̇y = −qvx B0 /m ± (qE0 /m) sin ωt
We may proceed by differentiating both equations with respect to time and then substituting for
v̇x,y ; in this way one obtains two uncoupled equations of the forced harmonic oscillator type.
We take a more “elegant” approach by introducing the complex variable ζ = vx + ivy . The
velocity is thus represented by a complex vector in the (Reζ, Imζ) plane.
Summing up Eqs.(1) after multiplicating the second one by i we obtain
ζ̇ = −iωc ζ + (qE0 /m)e±iωt ,
(2)
where ωc = qB0 /m is the cyclotron (or Larmor) frequency.
The homogeneous solution of Eqs.(2) is
ζ(t) = ζ(0)e−iωc t
which represents the particle motion if E0 = 0: the velocity rotates clockwise with frequency ω in
the plane.
The inhomogeneous solution can be searched in the form
ζ = ζ̃e±iωt ,
so that by direct substitution we find
ζ̃ =
−iqE0 /m
.
ωc ± ω
We thus find a resonance at ω = ωc only if the field rotates clockwise, because the electric field thus
tends to accelerate the particle along the direction of its “natural” motion.
b) At resonance (ω = ωc ) we search for a non-periodic solution of the type
ζN S = ζ̃N S (t)e−iωt
from which, substituing in Eqs.(1), we obtain
(ζ̃˙ N S − iω ζ̃N S )e−iωt = −iωc ζ̃N S e−iωt + (qE0 /m)e−iωt .
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Recalling that ωc = ω, the latter equation reduces to ζ̃˙ N S = qE0 /m which has a solution
ζ̃N S (t) = ζ(0) + (qE0 /m)t = ζ(0) + (E0 /B0 )ωc t .
The trajectory is a spiral with the radial velocity increasing linearly with time.
c) Introducing a viscous force Fv = −mνv and considering only the resonant case, we obtain the
following equation for ζ
ζ̇ = −iωc ζ − νζ + (qE0 /m)e−iωt .
The solution is thus of the same form of point a) if we replace ωc → ωc − iν:
ζ=
−iqE0 /m −iωt
e
+ ζ(0)e−iωc t−νt
ωc − ω − iν
The second term decays exponentially, so that there is “no memory of homoegeneous solutions” after
a transient phase. Thanks to the factor iν in the denominator the “particular”, oscillating solution
is well defined also at resonance.
The absorbed power is given by P = hqE · vi where “hi” indicates the temporal average over a
period:
Z
1 T
f (t)dt ,
hf (t)i =
T 0
where T = 2π/ω. The componnents of the velocity are
qE0 /m
[ν cos ωt − (ωc − ω) sin ωt] ,
(ωc − ω)2 + ν 2
qE0 /m
[−ν sin ωt − (ωc − ω) cos ωt] .
vy = Imζ =
(ωc − ω)2 + ν 2
­
®
Thus, being Ex = E0 cos ωt, Ey = −E0 sin ωt, and hcos2 ωti = sin2 ωt = 1/2, hcos ωt sin ωti = 0,
we obtain
vx = Reζ =
P =
q 2 E02 ν/m
.
(ωc − ω)2 + ν 2
The “frequency spectrum” P = P (ω) of the absorbed power is a Lorentzian centered around the resonance; its width is
proportional to the “viscous” frequency ν, and its maximum is
proportional to ν −1 . Such a tupical “absorption profile” is shown
in the figure for ν = 0.1ωc .
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