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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). 1 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 ). 2