UMD ASTR 601 - Cyclotron and Synchrotron Radiation

Unformatted text preview:

Cyclotron and Synchrotron RadiationInitial questions: What is one way that we can get a power law spectrum inastronomy? What information can we obtain? How can we measure the magnetic fieldsof individual objects? How can we determine that a particular spectrum is produced bysynchrotron radiation rather than other effects?When magnetic fields are present, charges can interact with them and radiate or absorbradiation. For slowly moving particles this happens at a single frequency, the cyclotronfrequency. For relativistically moving particles, the emission or absorption occurs over alarge range of frequencies, and is called in this case synchrotron radiation. Both names referto laboratory accelerators.Let’s go back to basics, and assume that we have a charge in a region that has amagnetic field but no electric field. Ask class: from the electromagnetic fields alone,does the particle energy change? No, because the acceleration due to a magnetic fieldis perpendicular to the motion. Therefore, a particle of Lorentz factor γ will evolve inmomentum and energy byddt(γmv) =qcv × Bddt(γmc2) = qv · E = 0 .(1)Ask class: what does this mean about the total speed? Constant, of course. Ask class:how does the speed along the magnetic field change? It doesn’t, because the acceleration isall perpendicular to magnetic field. Since the total speed is constant and the speed alongthe field is constant, this implies circular motion around the field, combined with (possibly)a uniform drift along the field. That is, the charge moves in a helix along the field. If wewrite just the perpendicular component of the equation of motion, we getdv⊥dt=qγmcv⊥× B . (2)Since by definition v⊥is perpendicular to B, this means that the rate of change of thedirection (i.e., the frequency of rotation or gyration) is the cyclotron frequencyωB= qB/(γmc) . (3)Now, this is interesting. It says that for a fixed Lorentz factor (or total speed), the frequencyof rotation is independent of the angle the charge makes to the magnetic field (the pitchangle). It also says that the frequency is lower when the momentum or mass or higher;you can think of this as meaning that it is more difficult for the magnetic field to bend thetrajectory.This rotation is an acceleration, with magnitude a⊥= ωBv⊥, so there is associatedradiation. From our previous formulae, the power radiated isP =2q23c3γ4q2B2γ2m2c2v2⊥=23r20cβ2⊥γ2B2.(4)As always, we can extend this to a distribution of directions. Suppose, in particular, that wehave a single speed β, but an isotropic distribution of velocities of charges. Then, averagingover directions and rewriting, we findPsynch=43σTcβ2γ2UB, (5)where UBis the magnetic energy density UB= B2/8π.Ask class: does this expression look familiar? It has exactly the same form as theCompton power, which was Pcompt=43σTcβ2γ2Uph, where Uphwas the photon energydensity. Therefore, the ratio of powers is just the ratio of energy densities:Psynch/Pcompt= UB/Uph. (6)Stuff like this doesn’t happen by accident. There is a strong relationship between“scattering” off of field lines and scattering off of photons.Now let’s think about the spectrum of radiation produced by motion around a magneticfield. First, consider a very slowly moving particle. Ask class: qualitatively, how doesthe electric field vary? The charge is moving in a circle, so the electric field variation issinusoidal. Ask class: what does this mean about the energy spectrum that we see? If wehave perfectly sinusoidal motion, then a Fourier transform gives us just a single frequency.Therefore, the energy spectrum would be a single line at the cyclotron frequency ωB. Thisalso means, incidentally, that a slowly moving particle can only absorb or emit photonsat this frequency; one consequence is that the scattering cross section goes way up forfrequencies close to the cyclotron frequency.Now suppose that the particle speed increases. Ask class: what happens to the electricfield variation? Recall that when relativity is involved, acceleration radiation is beamed inthe direction of motion of the charge. That means that instead of being perfectly sinusoidal,the electric field variation has sharper peaks. Ask class: what effect does that have on theFourier transform? Since the variation isn’t sinusoidal, we don’t just have a single frequencyany more. However, we do know that the motion is still periodic with frequency ωB, so wecan only have that frequency and multiples of it. The result is that the Fourier transforminvolves ωB, 2ωB, and so on, and hence the energy spectrum involves ¯hωB, 2¯hωB, and soon. There are therefore several lines, harmonically spaced.Now suppose that the particle speed is highly relativistic. Ask class: what’s the effecton the observed spectrum? In this case, the radiation is so strongly peaked forwards thatmany, many harmonics contribute. There are so many that the discreteness of the lines isdifficult to distinguish, so that we approach a continuum spectrum. In a real situation, ofcourse, particles of many different speeds are involved, which helps blend the spectrum if γis not constant.Let’s go into the highly relativistic case (γ ≫ 1) in more detail. We know that theradiation is beamed forward in a cone of approximate opening half-angle 1/γ. We thereforemight guess that since most of the radiation is emitted over an angle that is of order 2/γ ofa radian, the peak frequency in the synchrotron spectrum would be of order γωB. However,this is not the case, because of the effect of light travel times.Assume for simplicity that the pitch angle is α = 90◦, meaning that the circular motionof the particle is in the plane of our line of sight. In the cone of main emission, the particletravels a distance equal to 2/γ times the radius of curvature a of the path, a = v/ωB.Call point 1 the point in the orbit at which we are first in the beam, and point 2 the lastpoint at which we are in the beam (thus, the particle reaches point 2 after point 1). Asseen by someone on the side, the times at which these points are reached are related byt2− t1≈ 2/(γωB). But what is seen by the observer in the beam? If γ ≫ 1 then the lineardistance between these points is d ≈ v(t2− t1), where v is the speed of the particle. Here’sthe picture, then: at time t1in the frame of the center of gyration, a light pulse is sent outfrom the particle. At time t2another light pulse is sent out. But, remember,


View Full Document

UMD ASTR 601 - Cyclotron and Synchrotron Radiation

Download Cyclotron and Synchrotron Radiation
Our administrator received your request to download this document. We will send you the file to your email shortly.
Loading Unlocking...
Login

Join to view Cyclotron and Synchrotron Radiation and access 3M+ class-specific study document.

or
We will never post anything without your permission.
Don't have an account?
Sign Up

Join to view Cyclotron and Synchrotron Radiation 2 2 and access 3M+ class-specific study document.

or

By creating an account you agree to our Privacy Policy and Terms Of Use

Already a member?