Lorentz Force. Acceleration of electrons due to the magnetic field gives rise to synchrotron radiation Lorentz force.

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1 Set 10: Synchrotron

2 Lorentz Force Acceleration of electrons due to the magnetic field gives rise to synchrotron radiation Lorentz force 0 E x E y E z dp µ dτ = e c F µ νu ν, F µ E x 0 B z B y ν = E y B z 0 B x E z B y B x 0 c 0 E x E y E z c d dτ γm v x = e E x 0 B z B y v x γ v y c E y B z 0 B x v y E z B y B x 0 v z v z

3 Lorentz Force d dτ γm c v x v y v z = e c γ E v E x c + B z v y B y v z E y c B z v x + B x v z E z c + B y v x B x v y ( d c dτ γm v ( d c dt γm v ) = ec γ ( E v Ec + v B ) ( ) = e E v c Ec + v B ) [dτ = dt/γ]

4 Energy-Momentum Equations Energy and momentum equations d dt γmc2 = ee v d dt γmv = e(e + v c B) Consider a B field with no E field d dt γmc2 = 0, d dt γmv = ev c B where the former ignores radiation energy losses Then the velocity is constant dγ dt = 0 γ = const v = const

5 Cyclical Motion. Define components v B, v B. Momentum equation becomes dv dt = 0 dv dt = e γmc v B v v v α B so that v =const. v 2 = v 2 + v2 =const. and thus v =const. Acceleration is orthogonal to v with a = e γmc v B

6 Power Equation of motion solved by v (t) = v ( sin(ωb t + δ) cos(ω B t + δ) ), ω B = eb γmc Radiated power P = 2e2 3c 3 γ4 a 2 = 2 3 e 4 v 2 B2 m 2 c 5 γ2 = 1 4π v 2 B2 c γ 2 σ T, [ σ T = 8π 3 ] e 4 m 2 c 4 Synchrotron emission can be viewed as Compton scattering off of virtual B photons. Defining the pitch angle α between B and v, v = v sin α sin 2 α = sin 2 α dω 4π = 2 3 P synch = 1 v 2 B 2 γ 2 σ T 6π c

7 Virtual Photons The energy density associated with the magnetic field is u B = B 2 /8π P synch = 4 3 σ T c γ2 u B = 4 3 σ T cβ 2 γ 2 u B Power in Compton scattering v 2 P comp = 4 3 σ T cβ 2 γ 2 u γ so that synchrotron power= Compton power with incident energy density replaced by the magnetic field energy density P synch P comp = u B u γ Synchrotron radiation produces photons for inverse Compton scattering - removes energy from electrons - self regulation process

8 Frequency Spectrum. The electron gyrates at a frequency E t cyclotron ω B = eb γmc P ω ω Β ω Non relativistic, or cyclotron limit, given by the dipole approximation. The electric field of the radiation follows a sinusoid: the frequency structure is a near delta function at ω B P ω P ω ω Β ω c ω Β E 2ω Β 3ω Β F(ω/ω c ) t ω ω pulse sharpening synchrotron (blended)

9 Frequency Spectrum As the velocities become relativistic beaming sharpens the time profile while remaining periodic Appears as a series of delta functions at integer multiples of ω B The beaming sets the width of the pulses and hence a cut off in the frequency spectrum of ω c 1/ t In the ultrarelativistic limit, the frequency content extends to ω c ω B and forms a continuum due to the range of γ s

10 Frequency Spectrum. The profile width is given by beaming by determining the length of time the emission is observable. Take the ultra-relativistic limit where the radiation makes a cone of angle α 1/γ R θ π/2 1/γ ds 1/γ observer The observer first sees the radiation when the velocity (tangent to the spiral) is 1/γ from the line of sight and continues until it is 1/γ on the other side These two points and the center form an equilateral triangle

11 Frequency Cutoff The angle along the spiral traversed during the emission + 2 angles θ + 2 ( π 2 1 ) γ = π θ = 2 γ The arclength traversed given a gyration radius R combined with velocity gives the duration of emission s = R θ = v t em t em = R θ v = 2R γv

12 Frequency Cutoff Now eliminate the gyration radius R by expressing the angular speed as θ t em = v R and extracting the angular speed from the centripetal acceleration γm v t em = γmv θ t em = e c v B = ev c B sin α [ v = v θ]

13 Frequency Cutoff Reexpress using θ/ t em = v/r γmv 2 1 R = ev c B sin α R = t em = γmvc eb sin α = 2 γω B sin α v ω B sin α The arrival time is further shortened in that light must travel across the path difference between beginning and end of the emission t trav = s c = 2 γc v ω B sin α

14 Frequency Distribution The emission only beats the particle by a small amount leading to a smaller observed pulse duration 2 t = t em t trav = γω B sin α 1 t γ 3 ω B sin α = mc γ 2 eb sin α Define a critical frequency ω c = 3 2 t = 3 eb γ2 2 mc sin α ( 1 v ) c [ 1 v c 1 ] 2γ 2

15 Frequency Spectrum Scaling relations: the frequency spectrum a function of ω/ω c and the total power is known P = σ T 4π P ω = CF (ω/ω c ) v 2 B 2 γ 2 sin 2 α = c 0 P ω dω Solve for C up to an order unity coefficient F (x)dx = N P = Cω c C = σ T 4π 0 v 2 B 2 c F (x)dx = CN 3 eb γ2 2 mc sin α γ 2 sin 2 α N 3 2 γ2 eb = 4 e3 1 B sin α 9 mc 2 N sin α = σ T 6π mc [ σ T = 8π 3 v 2 B c mc Ne sin α ] e 4 m 2 c 4

16 Put it together Electron Distribution P ω = 4 e3 1 B sin α 9 mc 2 N F (ω/ω c) Power law electron distribution n e,γ γ p j ν dγn e,γ P ω n e,γ F (ω/ω c )dγ Transform to x = ω/ω c = ω/aγ 2 dx = 2ω Aγ dγ 3 γ p dγ = γ p+3 A ( ω 2ω dx = Ax j ν ω ( p+1)/2 α =polar angle wrt B F (ω/ω c )γ p dγ ) ( p+3)/2 A 2ω dx Just like inverse Compton scattering, the spectrum of radiation has a power law ω s with s = (p 1)/2

17 Full Calculation A detailed integration of the orbits similar to bremsstrahlung yields dw dtdω P ω = F (x) = x with asymptotic behavior x 3 2π e 3 B sin α mc 2 F (ω/ω c ) K 5/3 (y)dy lim F (x) = x 1 4π ( x ) 1/3 3Γ(1/3) 2 lim x 1 F (x) = ( π 2 and a maximum at x = 0.29 ) 1/2 e x x 1/2

18 Polarization Polarization in direction perpendicular to B vs complement with a component parallel Π = P,ω P,ω = G(x) P,ω + P,ω F (x) G(x) = xk 2/3 (x) The polarization is linear (when integrated over pitch angles) and integrated over frequencies for a single γ is 75% For a power law distribution of electron energies Π = p + 1 p + 7 3

19 Radiative Transfer Can think of the emission and absorption as the sum over a discrete set of states or lines with hν = E 2 E 1 j ν = hν 4π E 2 E 1 n 2 A 21 φ 21 (ν) = 1 4π hν E 1 A 21 φ 21 (ν) = 2πP ω P ν n 2 = 1 2πP ω n 2 4π E 2 E 2 where φ 21 (ν) the profile enforces energy conservation hν = E 2 E 1 α ν = hν 4π α ν = hν 4π [n 1 B 12 n 2 B 21 ]φ 21 (ν) E 1 E 2 c 2 [n 1 n 2 ] 2hν A 21φ 3 21 (ν) E 2 E 1

20 So that Synchrotron Self Absorption α ν = c2 4hν 3 E 2 [n 1 n 2 ]P ω Now switch to the continuous phase space distribution n g e E d 3 q (2π h) 3 f e α ν = c2 4hν g d 3 q 3 e (2π h) [f e(e hν) f 3 e (E)]P ω j ν = 1 2 g d 3 q e (2π h) f ep 3 ω = 1 4π g d 3 q e (2π h) f ep 3 ν

21 Synchrotron Self Absorption Check: for a thermal distribution f e = e (E µ)/kt then f e (E hν) f e (E) = e (E µ)/kt (e hν/kt 1) j ν α ν = In the limit that hν E then α ν c2 4ν 2 g e c2 1 2hν 3 e hν/kt 1 = B ν d 3 q (2π h) 3 f e E P ω For a power law distribution of relativistic electrons d 3 q q 2 n e = dγn e,γ = g e (2π h) f dq 3 e = g e 2π 2 h 3 f e γ 2 dγf e Thus f e γ 2 n e,γ

22 Synchrotron Self Absorption So that for a power law n e,γ γ p and f e / γ γ 3 p Combine with phase space integration q 2 dq γ 2 dγ and P ω F (x) and x = ω/ω c = ω/aγ 2 α ν ν 2 dγγ p 1 F (ω/ω c ) ν p/2 2 Compare to emission integral j ν n e,γ F (ω/ω c )dγ F (ω/ω c )γ p dγ ν p/2+1/2, Source function has extra ν 5/2 : ν 2 from absorption formula and ν 1/2 from extra γ S ν = j ν α ν ν 5/2

23 Synchrotron Self Absorption. At high synchrotron absorption optical depth (low frequencies) I ν S ν ν 5/2 At low optical depth I ν j ν ν (p 1)/2 The synchrotron spectrum has a characteristic turn over between the regimes I ν ν5/2 ν (1-p)/2 ν

24 Galactic Synchrotron HASLAM 408MHz synchrotron map

25 Galactic Synchrotron WMAP 23GHz emission and polarization map

Synchrotron Radiation: II. Spectrum

Synchrotron Radiation: II. Spectrum Massimo Ricotti ricotti@astro.umd.edu University of Maryland Synchrotron Radiation: II. Spectrum p.1/18 ds=v dt_em dt=ds cos(theta)/c=v/c cos(theta)dt_em Synchrotron