Download Cyclotron radiation

Cyclotron radiation
An electron moves in the xy plane under the action of a constant and uniform magnetic field B = B0 ẑ.
Assume the motion to be non-relativistic and the initial velocity to have modulus v (≪ c).
a) Characterize the radiation emitted by the electron, specifying its frequency and its polarization
along either the axis parallel to B or in the perpendicular plane.
b) Calculate the ratio between the total irradiated power Prad and the total energy E of the electron.
c) Discuss how the preceding answers a)-b) change for a positron and for a proton (mass mp ≃
1836me ), making a comparison for the same energy of the particles.
d) Because of radiation the electron loses energy and performs a spiral motion. Assuming the amount
of energy lost per period to be small with respect to the total energy, find the temporal variation of
the orbit radius r = r(t).
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Solution
a) The electron performs a circular orbit with the cyclotron (Larmor) frequency ωc = eB0 /me . The
electric dipole moment p = −er rotates in the xy plane with frequency ωc which is also the frequency
of the emitted radiation.
The rotating dipole can be written as p = p0 (x̂ cos ωc t+ ŷ sin ωc t). For dipole radiation E ∼ −(p×
n̂) × n̂ with n̂ the direction in which we observe the radiation. If n̂ = ẑ, then E ∼ x̂ cos ωc t + ŷ sin ωc t
(circular polarization); if n̂ = x̂, E ∼ −ŷ sin ωc t (linear polarization).
b) Since r̈ = v × ω c (being ω c = ωc ẑ), we obtain
Prad =
2k0 |er̈|2
2k0 e2 v 2 ωc2
=
.
3 c3
3
c3
(1)
The electron energy is E = me v 2 /2 and then
4k0 e2 ωc2
4k0 rc ωc2
Prad
=
,
=
E
3 me c3
3 c
(2)
where rc is the classical electron radius.
c) The positron has charge +e and mass me , thus with respect to the electron only the direction of
the orbit is changed; with respect to the magnetic field, the circular polarization is reversed.
For the proton the frequency is Ωc = (me /Mp )ωc . The polarization is the same as for the positron,
while (Prad /E)p = (me /Mp )3 (Prad /E)e .
e) Since v = rωc , the equation for energy loss dE/dt = −Prad can be rewritten as
d mωc2 r2
2k0
= − 3 e2 ωc4 r2 .
(3)
dt
2
3c
Since d(r2 )/dt = 2rdr/dt we obtain
dr
2k0 e2 ωc2 r
2rc ωc2
r
=−
=
−
r
≡
−
,
dt
3mc3
3c
τ
(4)
r(t) = r(0)e−t/τ .
(5)
whose solution is
The particle spiralizes with a characteristic time τ = 3c/(2rc ωc2 ).
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