Synchrotron Radiation -...

Synchrotron Radiation

D.Maino

Physics Dept., University of Milano

Radio Astronomy II

D.Maino — Synchrotron Radiation 1/42

Synchrotron from relativistic particles

Consider an ensemble of rel. e− (or charged particles ingeneral): we need their energy spectrum N(E )

Measured from Cosmic-Rays (CRs) showers in the atmosphere

The energy spectrum is well fitted by

N(E ) = A× E−g

where A is (almost - nearby the Earth) constant and g ' 2.4.



Galactic synchrotron from 1 <∼ E GeV <∼ O(10GeV)

Low E : influence from solar winds (deviation from power-law)

E ' 1015eV: change in N(E) called knee

E ' 1018eV: change called ankle

E > 1020eV: Ultra High Energy Cosmic Rays (UHECR), origin totallyunknown



To get the intensity of N(E ) distributed CRs we need first theemissivity

4πεν =

∫ E2

E1

P(ν)N(E ) dE

and the intensity along the line-of-sight (assuming no bckgr)

Iν =

∫ s0

0εν ds =

1

4π

∫ s0

0

∫ ∞0

P(ν)N(E ) dEds

Assuming P and N not depending on position and usingWallis approx

Iν =s04π·√

3e3

m0c2· B⊥A · 1.78 ·

∫ ∞0

(ν

νc

)0.3

· e−ν/νc E−g dE



Rewrite ν/νc in terms of E since νc = 3/(4π)e/(m3c5)B⊥E2

νc = η B⊥ E2 ⇒

√νcν≡ x =

√ηB⊥ν

E

Hence the integral becomes∫ ∞0

x−(g+0.6) e−1/x2dx

g=2.4===⇒

∫ ∞0

x−3 e−1/x2dx =

1

2

and the total intensity:

Iν = 2.4× 10−10( s0cm

)( A

erg1.4cm−3

)(B⊥G

)1.7 ( ν

Hz

)−0.7



Close to the Earth A = 8.2× 10−17erg1.4cm−3

If this is constant along a 10 kpc line-of-sight with B = 10µGwe expect Iν ≈ 10−18erg s−1 cm−2Hz−1 sr−1 at observingν = 1GHz consistent with observations along the galacticplane

In general assuming

N(E ) ∝ E−g ⇒ Iν ∝ B1+α⊥ ν−α

with α = (g − 1)/2

Iν depends on the B component perperdicular to the l.o.s.and on ν with a power-law index linked to the power-lawindex of CRs


Synchrotron Maps

Given the ν−0.7 term, synch. dominates sky @ ν <∼ 1GHz

Antenna dimensions are inverse of frequency → large antennaapertures → ground-based measurements → atmosphericemission

For a synch. full-sky map: several identical (?) instrumentsoperating in different Earth location for long time period(duty cycles ' 0.3 year)


Synchrotron Maps


Losses and particle lifetime

Synchrotron radiation leads to energy loss of radiatingparticles → cut-off in their energy spectrum → steepening ofthe spectrum above a certain ν

Use these properties to infer particle lifetimes i.e. time elaspedfrom when their sources ceased to “work”



Emitted power from a single particle i.e. its energy loss

P =2

3

e2

c3v2(

E

m0c2

)4

Using the Larmor radius rL and frequency ωL we can rewrite v

v =v2

rL= ωLv =

eB

m0c

1

γv =

eB

m0c

m0c2

E· v

Since v ≈ c

P =2

3

e4

m40c

7B2 E 2

Hence

P = −dE

dt= −1.48

(B

µG

)2 ( E

eV

)2

eV s−1



Want an expression for particle half-time life

dE

E 2= −a B2 dt

integration======⇒ 1

E− 1

E0= a B2(t − t0)

where E0 and t0 are initial energy/time before any energy loss

Half-life time t1/2 is when E (t1/2) = E0/2 and taking t0 = 0

E (t) =E0

1 + a B2 E0 t

Thus

E0

2=

E0

1 + aB2E0 t1/2⇒ t1/2 = 8.34×109

(B

µG

)−2( E

GeV

)−1yr



Assuming B = 10µG for particles with E = 1GeV we havet1/2 ≈ 108 yr

Typical radio source evolution: once energy supply switchedoff

spectrum with an energy cut-off Ec

Ec moves towards lower and lower energiesfast at high-energy and slow at lower ones



Synchrotron emission exhibit similar cut-off at frequency νc

νc =3

4π

e

m30c

5B E 2

c

Such νc gives hints on the age of the source or duration of theremnant phase

τsource = 1.06× 109(

B

µG

)−3/2 ( ν

GHz

)−1/2yr



NGC1569: dwarf starbust galaxy

Black: measured points. Red and blue: fits for thermalemission (free-free α = −0.15) and synch

From low-frequency part (pure synch) estimateB ≈ 20µG→ τsource = 5Myr


Magnetic Field Diagnostic

From measurements of radio emission, assuming synch.dominates we can infer B strength, better B⊥

Radio Luminosity:

Pν ∝ V AB1+αν−α

where V is the volume of the emitting source

The total energy of the source (equipartition)

Etot = Epart + Emag = V

∫ ∞0

E N(E )dE +B2

8πV

Account for protons (do not radiate but contribute to Epart)

Ep = βEe ⇒ Epart = (1 + β)Ee = ηEe



Recall the relation between ν and E

ν =3

4π

e

m30c

5B E 2 = C B E 2

we can change integration range between νmin ≈ 10MHz andνmax = 100GHz (usual range of radio frequencies)

We use radio luminosity to derive A and the integration gives

Epart ∝ηV

g − 2

(Pν

V B1+αν−α

)(CB)g/2−1

[ν1−g/2min − ν1−g/2max

]= Gη PνB

−3/2

with G = G (α, νmin, νmax)



The total energy becomes

Etot = Gη PνB−3/2 + V

B2

8π



Compute B at minimum of energy E

∂E

∂B= −Gη Pν

3

2B−5/2 +

V

4πB = 0⇒ Bmin =

(6Gη Pνπ

V

)2/7

Subst. Bmin into Epart we get

Emag =3

4Epart

The minimum E required for the observed radio emissioncorresponds very closely to equipartition

Assuming α = 0.75 we get

Emin = 6× 1041( ν

100MHz

)2/7( Pν

WHz−1

)4/7( R

kpc

)9/7

η4/7 erg



Ratio of energy carried by protons wrt from electrons isη ' 102 (with shock diffusive acceleration η ' 20− 40)

Assume equipartition (not far from reality) ⇒ use synchrotronintensity to estimate B

Estimate brightness temperature Tb from (RJ i.e.ν < 50 GHz)

Sν =2ν2k

c2TbΩs

Flux Sν is related to the total radio luminosity at ν

Sν =Pν

4πD2

where D is source distance



Let’s call L the size of the source. The solid angle

ΩS =

(L

D

)2

The minimum-energy B field for an L disk galaxy and radiothickness H is

Bmin = 1.7η2/7(Tb

K

)2/7 ( ν

GHz

)4/7( H

kpc

)−2/7µG

With η ∼ 40, Tb = 1K, ν = 1GHz and H = 1kpc we get

Bmin ≈ 5µG


Faraday Rotation - Rotation Measure Diagnostic

What does happen to linear polar. radiation entering into a Bfield region?

Process called Faraday Rotation: informative on the B field(at least on B‖)

Remember:linear polar. wave is the superposition of 2opposite circular polar. waves (LHC and RHC)


Faraday Rotation

Consider all forces acting on free-electrons ⇒ motion alonghelical path: centrifugal force −mω2r, Lorentz: ±(e/c)ωB‖ rand e.m. eE(r, t)

± in Lorentz is for RHC and LHC; B‖ is along the e.m. wavepropagation


Faraday Rotation

Forces on e− induce electrical polarization (displacementbetween induced field D and external field E)

This gives a relation for dielect. const ε

ε− 1

4πE = nee r

From EOM−mω2 r = ±e

cωB‖ r + eE

we solve for r to get

r = − e

m

(1

ω2 ± ωωc

)E

with ωc = eB‖/(mc) cyclotron freq.


Faraday Rotation

Subst. r we get

ε = 1− ωp

ω(ω ± ωc)ω2p =

4πnee2

m

and the refractive index n is then

n ≡√ε =

√1−

ω2p

ω(ω ± ωc)

Difference in n⇒ different phase velocities for Er and El ⇒phase shift between Er and El ⇒ change in linear polarization


Faraday Rotation

Consider a path dr and rotation of El by dθl and of Er by dθrE rotation by ψ each El and Er should rotate 2ψ

d(∆ψ) =kl − kr

2dr

Phase velocity vp = ω/k = c/n⇒ k = ω/c · n and hence

d(∆ψ) =∆k

2dr =

ω

2c∆n dr

Difference in nl and nr

n2l − n2r = εl − εr = (nl + nr )(nl − nr ) = 2n∆n

since nl ≈ nr and finally

∆n =ω2pωc

nω

1

ω2 − ω2c


Faraday Rotation

Remember that

n =

√1−

ω2p

ω(ω ± ωc)

ωωp ,ωc=====⇒ n ≈ 1− 1

2

(ωp

ω

)2We can compute ∆n and then obtain

d(∆ψ) =ω2pωc

2cω2dr

The total rotation is obtained by integration along the totalpath lenght and restoring the expression for ωp

∆ψ =2πe2

m2c2ω2

∫ r0

0ne B‖dr


Rotation Measure

The last eq. is usually written as

∆ψ = RM λ2

RM is called Rotation Measure, it is expressed in radm−2 andλ is in meters

In astrophysics application we usually have

RM = 0.81

∫ r0

0

( necm−3

)( B‖µG

)(dr

pc

)radm−2


Rotation Measure

The λ2 dependence means RM decreases quickly as frequencyincreases

Measuring RM by means of measuring ∆ψ over a wide rangeof λ allows to get information on B‖ if ne is known

How many points? At least 3 since with only two rotation≈ ∆ψ + nπ with n integer


Rotation Measure

Typical value of external galaxies ' 100 radm−2

Radio galaxies can have RM <∼ 5000 radm−2

Object inside a cooling-flow cluster of galaxieslarge line-of sight (' 200 kpc)


Depolarization

Astrophysical objects with N(E )dE ∝ E−gde produce radioemission with Sν ∝ ν−α with α = (g − 1)/2 and a degree oflinear polarization

p =g + 1

g + 7/3=

α + 1

α + 5/3

p is general is a funcion p(r) and its maximum is neverreached → astro (FR) and instrumental effects


Depolarization - Bandwidth

Consider an instrument with bandwidth ∆λ = λ2 − λ1 and acentral λ0

The electric vector will be rotated across the bandwidth by

∆ψ = RM (λ22 − λ21) = (λ2 + λ1)(λ2 − λ1)RM = 2λ0 ∆λRM

Passing from λ to ν we get

∆ψ = −2λ0 RM∆ν

ν0

This “extra” RM has in impact on total polarization p

p(λ) = p(0)sin∆ψ

∆ψ


Depolarization - Beamwidth

Similarly we expect reduction in p due to the finite beamwidth

Different field orientations within beam FWHM give rise topolarization orientation that partially cancels out

Polarization from regions with B orientation perpendiculargive zero polarisation: 90 in the field → rotation of 180 of(Q,U) plane



Synchrotron difficult to use to infer B since

from I we require ne of the source which is unknownfrom P we have small depth due to de-polarisation

Pulsar estimation of RM together with dispersionmeasurement - DM to extract 〈B〉 weighted by ne

RM

DM= K

∫ rs0 neB‖ dr∫ rs0 ne dr

Hence

B‖ = 1.23RM

DMµG

DM are derived from dispersion of pulses from pulsars causedby time lag due to different refractive index of ionazed ISM


Magnetic Field - Summary

B are present everywhere in our Galaxy

Large-scale clock-wise B exists in Perseus arm

A reversal field is present in Sagittarius arm

B drops from 10µG at R ' 4 kpc to 4µG at R ' 10 kpc


Supernova Remnants - SNRs

SNRs are among the most brightest radio sources and amongthe first discrete ones

Two types of SNRs:

plerions i.e. “filled”: Crab Nebula (1054)shell-type: Cas A (1667)



From VLBI & spectroscopy: vexp ≈ 20000 km s−1

Very fast: higher density wrt surrounding medium

Collision shock expected when d ≈ protons mean-free-path

Protons with vexp has Ekin = 2GeV

MFP when p looses its Ekin via ionization of H2 molecules

σCR =2πe4Z 2

meχ0v20.285

ln2mev

2[1−

(vc

)2χ0

] + 3.04−(vc

)2with χ0 ioniz. energy (13.6 eV)



vexp ≈ 2× 109cms−1, Z = 1 → σCR = 10−17cm2

nH ' 1 cm−3 → path between two subsequent ionizationli ≈ 0.03pc

p looses ' 10−4 of Ekin → l ∼ 104 li = 300pc

No classical shock can develop to such large distances!!

With a B = 5µG the proton gyration radius is

rL =mvc

eB≈ 4× 1010cm = 10−8pc

Protons are tightly coupled with B → massless barrier

Different phases of expansion



Free-expansion: R ∝ t and massof swept-up gas is smaller thaninital ejected mass Ms

4

3πρextR

3s MS

ρext ∼ 2× 10−24grcm−3,Ms = 0.25M we get RS ≈ 1.3 pc

Time: t = RS/vexp = 60 yr

Monoch. emission decreases withtime moving at higher ν



Adiabatic expansion: when Msg ' MS dynamic described byadiabatic self-similar solution (Sedov,1958)

Gas temperature is very high and only energy loss is viaadiabatic expansion

In this phase R ∝ t2/5 n−1/5 and it ends when T ' 106K

For spherical-symmetric, 1D numerical solution exists(Chevalier 1974)

ESN = 5.3× 1043( n0cm−3

)1.12( v

km s−1

)1.4( R

pc

)3.12

erg

T = 1.5× 1011( nHcm−3

)−2/5( t

yr

)−6/5( E

4× 1050erg

)1/5

K

With typical value t ≈ 2× 104yr



Isothermal expansion: Main energy loss is radiative: C ,Nand O ions recombine with free e− → collected energy withline emission increases cooling rate

radiative shock: for nH ≈ 1 cm−3 → RS ≈ 15 pc andt ≈ 4× 104yr

effective cooling → p drops and then vexp ≈ 85 km s−1

snow plough phase when more and more interst. gas isaccumulated Msg MS

ISM mixing: shell breaks into individual clumps and SNRmixing up with ISM


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