SynchrotronSynchrotron RadiationRadiation
''Particles accelerated by a magnetic feld will radiate. For non – relativistic velocities the complete nature of the radiation is rather simple and is called cyclotron radiation. The frequency of emission is simply the frequency of gyration in the magnetic feld.
However, for extreme relativistic particles the frequency spectrum is much more complex and can extend to many times the gyration frequency. This radiation is known as synchrotron radiation''
George B. Rybicky & Alan P. Lightman in ''Radiative processes in astrophysics'' p. 167
Daniele Dallacasa Non-thermal Synchrotron Radiation on textbooks
G. Ghisellini: “Radiative Processes in High Energy Astrophysics” § 4
M. Longair: “High Energy Astrophysics” § 8
G. B. Rybicky & A.P. Lightman: “Radiative Processes in Astrophyiscs” § 6 (+4.7 & 4.8)
C. Fanti & R. Fanti: “Lezioni di radioastronomia” § 4
T. Padmanabhan: “Theoretical Astrophysics” § 6.10 & 6.11 Radiation from a charged particle the Lorentz force
A charge q is defected by a uniform magnetic feld B according to: d⃗p q = ⃗v×H⃗ → v = v cosθ v = v sinθ dt c ∥ ⊥
θ is the ''pitch angle'' between the magnetic feld and the velocity of the charge θ Radiation from a charged particle the Lorentz force (2)
d(mv ∥ ) = 0 → v is constant → The pitch angle remains constat dt ∥
d(mv ⊥ ) q = v × H dt c ⊥
In the direction perpendicular to the B feld we have a circular motion and in the non-relativistic case we have the well known cyclotron relations:
mc r = v L qH ⊥
2 πrL 2πmc TL = = v ⊥ qH x 2π qH ωL = = [c.g.s.units] TL mc
Side view Top view Cyclotron radiation
''A charge moving in a magnetic feld is accelerated and then radiates [CYCLOTRON]'' - dipole in the plane of the circular orbit (dipole angular distribution) - linear polarization or circular or elliptical polarization
Max amplitute 0.5 * Max amplitute
45o No radiation!
0.5 * Max amplitute
Max amplitute Cyclotron radiation frequency of emission d⃗p q Lorentz Force: = m⃗a = ⃗v×H⃗ dt c q in modulus ma = v H sin θ = qβHsin θ c
4 dW 2 q 2 therefore in in the Larmor's formula ( − ) = β2 H2 sin θ dt 3 m2 c3
ωL qH at the frequency ν = = L 2π 2πmc
H If we consider an electron, its gyration frequency is: ν ≃ 2.5 [MHz] L Gauss
N.B. In general, the magnetic felds are weak, except in very particular stars, and collapsed bodies. Example: 34keV feature in Her-X1 (see Longair), → radiation emitted at low energies Towards synchrotron radiation: relativistic Larmor parameters
The relativistic case is dealt with by considering the appropriate expression of the particle mass:
mo m = = mo⋅γ √1−β2
γ mo c r = v = γ r L ,rel qH ⊥ L
2πrL ,rel 2πmo c γ TL ,rel = = = γ TL v ⊥ qH
2π qH ωL ωL ,rel = = = TL,rel γmo c γ if we consider v ∼ c ,we have:
2 γmo c ε r ≃ = L ,rel qH qH
N.B. ''classical'' physics is ok, since for typical B felds strengths, h/p≪rL, rel Cyclotron radiation towards relativistic particles (1)
[(ULTRA) - Relativistic] Charge ( γ≫1) moving in a uniform magnetic feld Recall Larmor's Formula in the invariand form (scalars are not affected) dt d τ = γ
= [⃗p ,(i/c)W] FYI, i Computations never asked W2 = p2 c2+m2 c4
2 dW 2q dpi dpi P = = dt 3m2 c3 [d τ d τ ] 2 2 2 2 dpi dpi d⃗v 1 dW d⃗v dp = − = −β2 [d τ d τ ] (d τ ) c2 (d τ ) (d τ ) (d τ ) Already seen earlier
2 2 2 dW 2q 2 d⃗v 2 dp Two options: Prel = = γ −β dt 3m2 c3 [(d τ ) (d τ ) ] Cyclotron radiation towards relativistic particles (2)
Classical, non relativistic, Larmor case → linear acceleration (no ac ) d⃗p dp ≃ ∣d τ ∣ d τ dW 2q2 1 dp 2 P = = dt 3m2 c3 γ2 (d τ )
d⃗p dp 1 dW → Centripetal acceleration ≫ β = ∣d τ ∣ ( d τ ) c ( d τ ) dW 2q2 d⃗p 2 2q2 d⃗p 2 P= = = γ2 dt 3m2 c3 (d τ ) 3m2 c3 (dt )
Relevant in relativistic cyclotron and synchrotron emission! (relativistic) Cyclotron radiation frequency of emission
The energy is radiated in various harmonics of the gyration frequency:
v ∥ cosθ ν = k ν 1 − k=1,2,3,4,....n Doppler shift along the LoS k rel ( c )
dW dW The energy in the various harmonics follows: ≈ β2 [ dt ] l+1 [ dt ] l (relativistic) Cyclotron radiation frequency of emitted radiation dW dW The energy in the various harmonics follows: ≈ β2 [ dt ]n+1 [ dt ]n
As a result of a relativistic effect on electron motion and its emitted radiation...
(Radiation) felds on the approaching side are amplifed, while they are attenuated on the receding side.
The effective feld can therefore be represented as the superposition of a number of harmonics. Synchrotron radiation relativistic aberration
sin(α')(√1−β2) tan(α) = cos α'+β 1 α o → (α) β 2 ' = 90 tan ≈√ 1− ≈ 2 γ ⇒ Amplifcation along the velocity direction ⇒(relativistic beaming) 2 Half of the radiation is within a cone wide γ Synchrotron radiation
It is generated by ultra-relativistic electrons, for which the curvature of the trajectory plays the major role in determining the emitted power (single electron with its γ in a H feld)
4 dW 2q 2 − = β2 γ2 H2 sin θ H energy dt 3m2 c3 density UH
2 ifβ≃1 and ε=mo c γ then it can be rewritten as
4 2 dW 2q 2 2 2 2 H 2 −15 2 2 2 −1 − = ε H sin θ = 2c σ γ sin θ ≃ 1.62⋅10 γ H sin θ ergs 4 7 T π dt 3mo c 8
and σT is known as electron Thomson cross section defned as 2 2 8 π e 8 π 2 −25 −2 σT = = ro = 6.6524586...⋅10 cm 3 m c2 3 ( e ) Synchrotron radiation pulse duration (1)
Pulse duration in the electron reference frame is
me c γ 2 2 Δ t = Δ θ = = ωrel γ eH ωL In the observer frame, the duration of the pulse is shortened by propagation (Doppler) effects: the electron is closer to the observer in 2 travelling nearly at same speed of radiation emitted in 1
−8 v 1 1 1 5⋅10 Δ τ = (1− )Δ t = (1−β)Δ t ≈ 2 Δ t = 2 = 3 ≃ 2 s c γ γ ωL γ ωrel γ H[G]
2 = Synchrotron radiation pulse duration (2)
−8 v 1 1 1 5⋅10 Δ τ ( )Δ ( β)Δ Δ = 1− t = 1− t ≈ 2 t = 2 = 3 ≃ 2 s c 2 γ 2 γ ωL 2 γ ωrel γ H[G]
Foreword/consequences:
2 1. The duration of the pulse is ≈1/ γ shorter thanΔ t=TL 2. The factor (1 − v/c) is the same as appears in the L-W potentials and accounts for the beaming (more generally Doppler effect) in case of relative motion wrt the observer (pp. 200-202 Longair) Synchrotron radiation pulse angular distribution
The radiation is emitted in a narrow cone aligned with v
A given LoS receive pulsed fashes Istantaneous illumination Synchrotron radiation single electron spectral distribution (1)
algebra is very complex; a detailed discussion is in Rybicki-Lightman p.175
Fourier analysis of the pulse provides the spectrum of the radiated energy.
The spectrum is the same as the relativistic cyclotron but with an infnite number of harmonics.
Beaming effects move the peak to frequencies higher than νL
3 1 3 2 3 2 eH The characteristic frequency is: νs ≃ = γ νL = γ = 4 π τ 4π 4 π me c
3 eH 2 18 2 −9 2 3 5 ε ≃ 6.24⋅10 ε H ≃ 4.2⋅10 γ H[μG] GHz 4 π me c
4 ⇒example: an electron with γ ∼ 10 in a H ∼ 1 μG feld has νs ∼ 0.4GHz
Synchrotron radiation is generally emitted in the radio window H felds are generally 0.1 - 10 μG in space. Higher felds in ''condensed'' objects (e.g. stars) Synchrotron radiation single electron spectral distribution (2)
The full expression of the emitted energy as a function of frequency is
3 dW(ν) dW∥ dW ⊥ √3e H sinθ ν = + ≈ F dt dt dt 8 π2 cm ν e ( s ) ν ν ∞ where F ν (y)dy = ∫ K5/ 3 ν ν νs ( s ) s
K = modifed Bessel function 1/3 ν≪νs → F(ν) ∼ ν
−ν/ νs ν≫νs → F(ν) ∼ e ν peak at ∼ 0.3 νs
Nearly monochromatic emission, at 2 FYI, νm 0.3 νs depending on γ and H Computations never asked Synchrotron radiation typical radio spectrum of a radio source
Which is the nature of the emitted energy?
The cases of 3C219 and 3C273
HST + radio radio Synchrotron radiation the full sky at 408 MHz
Synchrotron emission from relativistic electrons (positrons) within the Milky Way. In a feld of 1 μG, their energies correspond to γ γ 104.
Sky @ 408 MHz (from skatelescope.org) Fast charges within a spiral galaxy: do they escape?
Compute the Larmor radius of a relativistic proton in our galaxy B ∼ 10μ G
mv c rL = γrL = γ rel qH alternative formulas γmc2 v rL = γrL = rel qH c
γmc2
6 [GeV] v rL = 3.3×10 cm rel q H ( c ) [e] [Gauss] Synchrotron radiation a SNR
Synchrotron emission from relativistic electrons (positrons) in the crab nebula Synchrotron radiation power – law energy distribution
Let's consider an ensemble of relativistic electrons with energies distributed according to a power – law: −δ N(ε)dε = No ε dε
The specifc emissivity of the whole population is dW (ν,ε) J (ν)d(ν) = s N(ε)dε (****) s dt ν ≈ F N ε−δ dε ν o ( s )
There are various (approximate) ways to derive the total emissivity:
1. all the energy is radiated at the characteristic frequency
2. the energy is radiated at a constant rate over a small frequency range The synchrotron spectrum
Let's choose approach 1: all the energy is radiated at the characteristic frequency: ε 2 eH ν ≃ ν ≈ γ2 ν = ν where ν = then s L 2 L L π (mc ) 2 me c 1/2 2 ν 2 2 1/2 −1/2 ε = γ me c = me c = me c ν νL then, getting the differential ν ( L ) 2 me c 1 ε ν−1/2 ν 2 ν−1/2 ν−1/2 ν d = 1/2 d = me c L d 2 νL 2 4 dW 2 q 2 2 Larmor's formula for synchrotron radiation (−) = β2 γ2 H2 sin θ ≈ ε2 H2 sin θ dt 3 m2 c3
Then, for a relativistic (= power-law ) population of electrons, (****) becomes......
dWs (ν ,ε) (δ+1)/2 −(δ−1)/2 J (ν) = N(ε) dε ≈ constant N H ν s dt o
(δ+1)/2 −α Js(ν) ∼ NoH ν The synchrotron spectrum
−9 2 νs ≃ 4.2⋅10 γ H[μ G] GHz → Given a magnetic feld H,
2 2 the emitted frequency is detemined by γ (i.e. the energy of the charge, being ε = γ me c ) ) ) ε ν ( ( S N
g g o o L L γ 2
– ε ν α – δ γ 2
γ 2
Log (ε) Log (ν) The synchrotron spectrum
The total spectrum is interpreted as the superposition of many contributions from the various electrons emitting each at its own characteristic frequency νs in the same feld H δ − 1 α = 2 spectral index of the synchrotron radiation
ν – α Synchrotron self-absorption (1): (SSA)
“(Low energy) Photons cannot escape regions with high electron densities” NOT in thermal equilibrium, then Kirchoff's law does not apply.
Concept of temperature must be widened, it does not imply thermal equilibrium
The ''electron (kinetic) temperature'' Te can be defned based on the electron energy
FYI, SSA applies when TB ∼ Te Computations never asked TB can NEVER EXCEED Te
The absorption coeffcient is effectively determined by making use of the Einstein's coeffcients (not given here, see further lectures)
−(δ+4)/2 (δ+2)/2 μs ≈ No ν H ⊥ Synchrotron self-absorption (2)
−(δ+4)/2 (δ+2)/2 ⇒ Synchrotron absorption μs ≈ No ν H ⊥ δ−1 δ+1 − 2 2 ⇒ Synchrotron emissivity Js (ν) ≈ ν H ⊥
To be inserted in the radiative transfer equation of an emitting cloud (partially) opaque J (ν) s −τs (ν) Bs= (1−e ) 4 πμs (ν) Which becomes δ−1 δ+1 − 2 2 Bs(ν) ≈ ν H ⊥ τ≪1 optically thin regime
5/2 −1/2 Bs(ν) ≈ ν H ⊥ τ≫1 optically thick regime
N.B. The frst case is that derived from the earlier description which provides a power-law distribution Synchrotron self-absorption (3): understanding the 2.5 dependence
Use a trick: the power-law distribution can be seen as the superposition of many Maxwellian 2 distributions at different T: γme c = 3kT (for a relativistic plasma)
Let's suppose that each electron emits at its own characteristic frequency νs FYI, Computations −9 2 never asked νs ≃ 4.2⋅10 γ H[μG] GHz 1/2 2 2 ν −1/2 1/2 From this we get kT ∼ γme c ∼ me c H → T ∼ ν ν ( L )
At loww energies, in a fully opaque source (black body!) the brightness temperatureis defned as ν2 I(ν) = B(ν , T) ≝ 2kTB TB must be = to the kinetic temperature of electrons c2 1/2 2 5/2 2 ν −1/2 ν ν I(ν) ∼ me c H ∼ ν c2 H1/2 ( s ) Synchrotron self-absorption (4)
The brightness temperature of a region does NEVER exceed the electron temperature: 2 λ S(ν) −α 2 TB ≃ but S(ν)≈ ν and 3kTe = γme c 2k θ1⋅θ2
1/2 −1/2 νmax H T =3.0⋅1010⋅ ⋅ [K] B,max [GHz] [G]
Observations may fnd TB in excess of this maximum value → Further considerations are necessary (later on...) Synchrotron self-absorption (5):
Let's derive the expression for Js to get the peak values:
2/5 −4/5 1/5 νp Sp θ H ⊥ 1/5 ≃ const (1+z) ( GHz ) ( Jy ) (mas ) ( mG )
With these units, the constant is ≈ 2 ) ν ( S g o S L p
ν 5/2 ν – α
ν P Theory .vs. Observations Log ν Synchrotron self-absorption (6): A peaked radio spectrum Synchrotron self-absorption (7): Other examples of compact (young) radio sources
2/5 −4/ 5 1/ 5 1/5 p ≈ S p H ⊥ 1z
Can help in measuring the magnetic feld in the relativistic plasma!!!!
Dallacasa & Orienti 2016 Spectra of radio sources: (none of these is “compact”!)
Guess the mechanism! Energetics of a radio source (1)
Total energy of a synchrotron emitting body must take into account both particles and the H feld
Utot = εel+εpr+UH = (1+k)εel+UH = Up+UH
k is a measure of our ''ignorance'' of the plasma composition Let's consider relativistic electrons: their energy is given by ε ε max max No V 2−δ 2−δ ε = V εN(ε)dε = V N ε ε−δ dε = ε −ε δ≠2 el ∫ε o∫ε ( max min ) min min 2−δ
We can get rid of No V if we consider the source Lumonosity ν ε max max dε L = 4 πD2 S(ν)d ν = V N(ε) dε ∫ν ∫ε min min dt 2 ε 2 max 2 2 2 −3 No VH sin θ 3−δ 3−δ L ≈ N V ε−δ H ε θdε = 2.4⋅10 ε −ε o ∫ε sin ( max min ) min 3−δ 1/2 νmin/max ε where min/max = 18 Now let's get No V out of this relation (6.24⋅10 H ) Energetics of a radio source (2)
2 2 −3 No V H sin θ 3−δ 3−δ (3−δ) L L = 2.4⋅10 ε −ε ⇒ N V = 3−δ ( max min ) o −3 2 2 3−δ 3−δ 2.4⋅10 H sin θ ( εmax −εmin ) 2−δ 2−δ 3−δ εmax −εmin L 1 −10 −1/2 −1/2 If ε = ≈ ε ≃ 4.0⋅10 H ν el 2−δ 3−δ 3−δ 2 2 −3 εmax −εmin H sin θ 2.4⋅10 (2−δ)/2 (2−δ)/2 −(2−δ)/2 −(2−δ) 3−δ [νmax −νmin ] H sin θ L ≈ 2−δ (3−δ)/2 (3−δ)/2 −(3−δ)/2 −(3−δ) 2 2 [νmax −νmin ] H sin θ H sin θ
Once chosen the minimum & maximum energy, considering an isotropic pitch angle distribution
2 2 → H2 sin θ = H2 3 −3/2 The above becomes εel = Cel H L
It is the energy associated with the relativistic particles (electrons) radiating synchrotron emission Energetics of a radio source (3)
Energy is stored in the magnetic feld as well H2 U = dV = C H2 V H ∫ 8π H
−3/2 2 The total energy budget is Utot = (1 + k)Cel H L + CH H V
By deriving wrt H U →a minimum in the total energy content Utot of a synchrotron emitting region! 4 (1 + k)εel = UH 3 UH
Uel H ''Equipartition'' Magnetic Field
2/7 2/7 3 Cel L H = H(ε) = (1+k) eq min 4 C V [ H ] ( )
Such equipartition feld provides the value of the feld intensity for which the total energy content in the radio source is minimum (related to the relativistic plasma), which amounts to
7 7 −3/2 4/7 3/7 4/7 3/7 U = (1+k)ε = (1+k)C H L = 2[(1+k)C ] C L V tot ,min 4 el 4 el eq el H
It can be calculated from observations, although a number of assumptions are hidden
1. The particle density is homogeneous ( flling factor ϕ = 1 ) 2. The magnetic feld is uniform all over the relativistic plasma ''Equipartition'' Magnetic Field Total energy Utot ,min
The total energy can can be written in a specifc expression for radio emission at 1.4 GHz as
4/7 3/7 41 4/7 L1.4GHz V Utot ,min = 2⋅10 (1+k) [erg] ([Watt] ) ([kpc]3 )
This can be used to defne the energy density and the internal pressure as
3 4/7 Utot,min L1.4GHz [kpc] u = = 6.8⋅10−24 [erg cm−3] min V ( [Watt] V ) H2 11 P = P + P = + (Γ−1)u = u min H rel 8π rel 21 min
E.g. ⇒ Such pressure can be compared to that of the surrounding ambient medium Energetics of a radio source application to 3C219
➢ Physical quantities can be calculated either globally for the whole source, or locally, at various sites within the same radio source:
➢ Determination of the “equipartition feld” along the radio source structure (plasma evolution?)
➢ Pressure balance: Pint and PISM / PICM provide the ram pressure and then the espansion/growth velocity (supersonic growth?)
2/7 2/7 3 Cel L H = H(ε) = (1+k) eq min 4 C V [ H ] ( ) 3 4/7 Utot ,min −24 L1.4GHz [kpc] −3 u = = 6.8⋅10 [erg cm ] min V ([Watt] V ) H2 11 P = P + P = + (Γ−1)u = u min H rel 8π rel 21 min Is equipartition Magnetic Field representative?
Heq measured from Luminosity, Volume
Peaked Spectrum (Self-absorption) measured from νp , Sp ,θ (angular size)
Orienti & Dallacasa Synchrotron radiation Time evolution
Particles radiate at expenses of their kinetic energy. Hence, the energy distribution of a synchrotron emitting region within a given volume V fully flled by magnetized relativistic plasma will change with time
Particle Flow Energy Losses Leakage Injection ∂N(ε, t) ∂ dε N(ε, t) + N(ε,t) + = Q(ε, t) ∂ t ∂ε (dt ) Tconf In particular −δ N(ε,0) = No ε Q(ε, t) = A ε−δ
First considerations in a simple case: Tconf = ∞ & Q(ε,t) = 0 (pure) Radiative losses: Tconf = ∞ & Q(ε, t) = 0
dε 2 2e ε2 2 θ − = Csync H sin where Csync = 4 7 dt 3me c ⇒ It is possible to derive the energy of each particle as a function of time:
dε 2 1 1 2 2 θ 2 θ − 2 = Csync H sin dt → − = Csync H sin t ε ε(t) εo 2 ε 2 θ ε 1 1 2 2 1+ o Csync H sin t o = +Csync H sin θ t = → ε(t) = 2 2 ε(t) εo εo 1 + εo Csync H sin θ t
Defne the cooling time t* as the ratio between the total particle energy and its loss rate:
∗ εo εo 1 t = = = ε 2 2 2 2 2 d /dt Csync εo H sin θ Csync εo H sin θ
∗ With such t we can rewrite the particle energy with time
εo ε(t) = 1+(t/ t ∗ ) particles from the energy distribution: Radiative losses: evolution of the spectrum The emission spectrummodifed is accordingly due to thethe of depletion high energy High energyparticles have t*energyshorter lowthan ones
Log S() it is possibledefne to defnes ν *, to be considered a signature ofageing Log E remains unchanged The energy lowspectrum ε
* which inturn ν Time evolution: A case study (more general)
∂N(ε,t) ∂ dε N(ε, t) + N(ε,t) + = Q(ε,t) T ∂ t ∂ε ( dt ) conf
1. The confnement time is extremely large Tconf →∞
2. There is continuous injection of fresh electrons: Q(ε,t) = A ε−δ
A balance between dying and refurbished particles is achieved at a particular energy, corresponding to a particular frequency, and, in turn, to a characteristic time; one solution of the equation is such that the synchrotron emissivity becomes:
−(δ−1)/2 −α Js (ν) ≈ ν = ν when → ν≪ν∗
−δ /2 −(α+1/2) Js (ν) ≈ ν = ν when → ν>ν∗ 1. distribution Initial power-law Spectral“ageing”: How does work?it
Log N(ε)
ε
–
δ Log ε 2. distribution Initialhigh depletes energies power-law asitages Spectral“ageing”: How does work?it
Log N(ε)
ε
–
δ cut-off Log ε 3. Agedandgeneration initialpower-lawnew plasma(1) distribution of Spectral“ageing”: How does work?it
Log N(ε)
ε
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δ (cut-off) Log ε 4. Agedand of initialpower-law(asecond)generation distribution new plasma(2). Spectral“ageing”: How does work?it
Log N(ε)
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δ 2 cut-offs Log ε 5. generation Agedand(third) plasma initialpower-lawnew distributions of Spectral“ageing”: How does work?it
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δ 2 cut-offs Log ε 6. Agedand of initialpower-lawagedgeneration distributions new plasma Spectral“ageing”: How does work?it
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1 due tomany cut-offs Change inslope Log ε Spectral “ageing” The integrated spectrum of a radio source Spectral “ageing”
The search for a break in the high frequency spectrum is a tool to estimate the radiative age of the dominant population of relativistic electrons
)
ν
(
S Different regions of the g ν ν 1 2
o same radio source may L have different radiative age based on: Spectral break
1. injection time 2. location within the r-source (i.e. hystory of the plasma) 3. information on individual radio source evolution Observations can measure the local spectral index Measurements
Log ν Spectral “ageing”
The “age” of electrons is different in various region of a radio source
Info on the radio source formation and evolution
Image of the spectral index α (color scale) overimposed to iso-brightness contour levels
S(ν) ≈ ν−α S(ν ) log 2 S(ν ) α = 1 (ν ) log 2 (ν1) Spectral “ageing”
The “age” of electrons is different in various region of a radio source
Info on the radio source Young formation and evolution
Middle age Image of the spectral index Oldish α (color scale) computed Old between two frequencies, overimposed to iso-brightness contour levels
S(ν) ≈ ν−α S(ν ) log 2 S(ν ) α = 1 (ν ) Orru' et al. 2010 log 2 (ν ) 1 Spectral “ageing”
The “age” of electrons is different in various region of a radio source
Info on the radio source formation and evolution Fresh electrons in an active nuclei & old electrons in dead AGN Other energy losses Ionization losses: relativistic particles can interact with (neutral) matter with density no and let electrons free as a consequence of electrostatic force
dε − ≈ no ln(ε) (dt )i FYI, ε[erg] Computatcions which has a characteristic time scale t* = 9.4⋅1017 sec never asked no
Relativistic bremsstrahlung losses:
dε 2 − ≈ no (ε) (dt )rel−br The characteristic time of this phenomenon does not depend on the energy of the particles. Both these processes are more relevant for low energy electrons since they scale with ln(ε) and ε while synchrotron losses scale with ε2.
Inverse Compton scattering losses: ....let's wait for a while..... Adiabatic expansion:
Γ−1 3(4/3−1) Poisson's law: TV = const → Tr = Tr=const 1 ε ∼ T ∼ εr = ε r = const r o o
ro ro ε(t) = εo = εo r ro+vexp t
dε vexp = − ε FYI, dt r Computatcions never asked
∗ ro εo t = = const ε(t) = ad v ∗ exp 1+t/ tad
r 2 magnetic feld: H(t) = H o o(r )
The shape of the spectrum does not change, but it is shifted at lower frequencies Effects of adiabatic expansion:
1st epoch (red squares)
2nd epoch (color bullets)
The shape of the spectrum does not change, but it is shifted at lower frequencies The source expansion makes self-absorption less effective and the optically thick nd ν Spectrum gets stronger. The peak in the 2 epoch spectrum has lower P e weaker SP Non uniform static magnetic feld (in particle frame)
Let's consider the RF in which the curvature radius has uniform motion(v ∥ =const): the particle ''feels'' an electric feld which is indeed the varying magnetic feld, which modifes v ⊥ 1 ∂H⃗ ∇⃗ ×E' = − c ∂ t For each projected orbit the variation of energy can be written as ⃗ 1 2 ⃗ ⃗ ⃗ ⃗ ⃗ q ∂H ⃗ Δ mv ⊥ = ∮ q E '⋅d l = q∫ ∇×E '⋅d S = − ∫ d S (2 ) S c S ∂ t If the magnetic feld changes smoothly within a single orbit ∂H⃗ ΔH ≈ ∂ t TL
1 2 q Δ H 2 q Δ H 2 Δε = Δ m v ⊥ = (π r ) = v ⊥ (π r ) → ⊥ L π L ( 2 ) c TL c 2 rL ε v ⊥ ΔH ε ΔH But r = → q ≈ β⊥ ε ⊥ L qH c 2 qH H
Δ ε⊥ Δ H ε⊥ FYI, ≈ → = const Computations ε⊥ H H never asked Non uniform magnetic feld (observer frame):
In the observer's frame – the motion takes place within a static non uniform feld; – the Lorentz force does not make work and the total kinetic energy of the particle does not change with time.
FYI, Computatcions never asked
If v ⊥ increases (decreases) going into stronger (fainter) H regions, thenv ∥ must decrease (increase).
If we consider the energy (scalar), if ε ⊥ increases (decreases) thenε ∥ must decrease (increase) of the same amount Non uniform magnetic feld (observer frame)[2]
Let's consider velocity and feld along and orthogonally to the mean H direction axis. The motion is within a static feld, the Lorentz force does not make work and the total kinetic energy of the particle does not change: ⃗ ⃗v×H = (v ∥ +v ⊥ )×(H ∥ +H ⊥ ) =
= v ∥ × H ∥ + v ⊥ × H ∥ + v ∥ × H ⊥ + v ⊥ × H ⊥
0 centripetal acceleration
tangent to crf, (in/de)crease v as H ⊥ along helix, against v v || FYI, H ⊥ || responsible for reversal Computatcions of versus of v never asked || Magnetic mirrors
Δ ε ⊥ Δ H ε ⊥ ≈ → = const FYI, ε ⊥ H H Computatcions 2 sin θ never asked It is equivalent to: = const H if Ho and θo (pitch angle) are known at a given point on the particle trajectory, then: H sinθ = sinθ o H √ o If H increases, then also the pitch angle increases, up to reach the maximum value of 90o and cannot penetrate further regions where H is stronger. It is then induced to move in the reverse direction
Charged particles may be trapped into regions where the magnetic feld is strong enough (example, terrestrial magnetic feld) Example: Earth Magnetic Field and charged particles
FYI,
never asked If a particle is initially very energetic, the magnetic bottle will not be able to confne the particle and it will escape. In exactly the same way, the aurora borealis/australis (Northern and Southern Lights) occurs when charged particles escape from the Van Allen radiation belt. They interact with the atmosphere through optical excitations of the gaseous atoms. Auroras are produced when the magnetosphere is suffciently disturbed by the solar wind that the trajectories of charged particles in both solar wind and magnetospheric plasma, mainly in the form of electrons and protons, precipitate them into the upper atmosphere (thermosphere/exosphere) due to Earth's magnetic feld, where their energy is lost. Auroras form from the interaction of high energy particles (mostly electrons) with neutral atoms in the high atmosphere. These particles may axcite (via collisions) bound electronsin the neutal atom. After a characteristic time the excited electrons decay to their initial state, producing radiation. Such process is similar to that in the neon bulbs. The various colors are determined by the composition of the atmosphere and by the colliding particles energy. Atomic oxygen is generally responsible for the red (630 nm) and green (558 nm) light, while molecular nitrogen (both neutral and ionized) causes blue (428 nm)light. Magnetic trap (bottle)
FYI,
never asked (Linear) Polarization
It is the most distinctive property of the synchrotron radiation https://www.youtube.com/watch?v=HH58VmUbOKM https://www.youtube.com/watch?v=8YkfEft4p-w ↑ For a better understanding
When many photons have E and B felds oscillating in random (disconnected, but perperdicular each other) the radiation is unpolarized
In case radiation has been produced in such a way that, e.g., the E feld has a ''preferred'' direction, then the radiation becomes linearly polarized
Rybicky & Lightman; Fig. 6.7 (Linear) Polarization (2)
The radiation is emitted with the E feld oscillating perpendicularly to the local B feld direction
Fractional polarization never reaches 100% Reduction from 1/ γ opening angle, and propagation effects Rybicky & Lightman; Fig. 6.7
Two components can be separated P∥ ,P⊥ Algebra is very complex FYI,
never asked 3 P∥ √3e B F(ν/ νs)−G(ν/ νs) = 2 P⊥ 2m c F(ν/ ν )+G(ν/ ν ) ( ) e ( s s )
∞ (ν ν ) (ν ν ) ( ) where F / s = / s ∫ K5/3 y dy ν/ νs
G(ν/ νs) = (ν/ νs) K2/3(ν/ νs) K → modifed Bessel functions (Linear) Polarization (3)
The toal emitted power for monoenergetic electrons is
P(ν) = P∥(ν) + P⊥ (ν) ≈ F(ν)
Integrating over the energies a power-law electron energy distribution we get:
3 −(δ−1)/2 P∥(ν) √3 e H JF−JG 2 ν = No 2 P⊥ (ν) 2 m c J +J 3 νL ( ) ( ) e ( F G ) ( )
(δ+1)/2 2 p 19 p 19 where J = Γ + Γ − F δ+1 (4 12 ) (4 12 )
(p−3)/2 p 7 p 1 J = 2 Γ + Γ − G (4 12 ) (4 12 ) FYI, Computations never asked (Linear) Polarization (4)
For a relativistic plasma bubble with homogenous feld, the degree of polarization is defned by
P⊥ (ν) − P (ν) JG δ+1 ∥ = = P⊥ (ν) + P∥(ν) JF δ+7/3
The fractional polarization may reach 69.3%, 72.5% and 75.0% in case of FYI, δ= 2.0, 2.5 and 3.0, respectively. This is very large! Computatcions never asked
There are effects that may reduce the actual fractional polarizazion (see further, e.g. inhomogeneous feld, differential Faraday Rotation, insuffcient resolution, ...)
From Dallacasa, Radio Astronomy Lab, lecture notes (Linear) Polarization (5)
Polarization is a measure of the predominant direction of the E feld → a vector
FYI,
never asked
From Dallacasa, Radio Astronomy Lab, lecture notes U and Q images in the Stokes' parameters
For each pixel, the vector length P (side image) is given by ⇐ P = √ U2 + Q2
For each pixel the orientation of the vector (not shown) is given by 1 χ = atan(U,Q) 2 Synchrotron radiation polarization in M51
Example of polarized emission: M51 – vector length proportional to projected feld strength – vector direction (B feld shown here, but E is measured!) Synchrotron radiation polarization in 3C219
Example 2: polarized emission in the radio galaxy 3C219
➢ E-vector displayed, length proportional to polarization: it has 2 components defned by the Stokes'parameters U and Q
➢ B-vector (feld!) direction perpendicular to the vectors shown here.
➢ In case resolution is not adequate, the detected polarization emission may be severely reduced (beam depolarization).
➢ Often polarized emission is frequency dependent. Questions (1) What happens to equipartition as the relativistic plasma ages? (2) How long does a radio source live? (3) Are we sure to have the correct answer to (2)? (4) Which are the key points to fnd “old” radio sources?
Future work Test: compute the radiative age and the kinematic age E.g. Owsianik & Conway, 1998 Intrinsically small radio sources grow with time
Help with the summary: What are the relevant features of the Synchrotron radiation?