Synchrotron Radiation D.Maino Physics Dept., University of Milano Radio Astronomy II D.Maino — Synchrotron Radiation 1/42
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.
D.Maino — Synchrotron Radiation 2/42
Synchrotron from relativistic particles
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
D.Maino — Synchrotron Radiation 3/42
Synchrotron from relativistic particles
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
D.Maino — Synchrotron Radiation 4/42
Synchrotron from relativistic particles
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
D.Maino — Synchrotron Radiation 5/42
Synchrotron from relativistic particles
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
D.Maino — Synchrotron Radiation 6/42
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)
D.Maino — Synchrotron Radiation 7/42
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”
D.Maino — Synchrotron Radiation 9/42
Losses and particle lifetime
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
D.Maino — Synchrotron Radiation 10/42
Losses and particle lifetime
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
D.Maino — Synchrotron Radiation 11/42
Losses and particle lifetime
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
D.Maino — Synchrotron Radiation 12/42
Losses and particle lifetime
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
D.Maino — Synchrotron Radiation 13/42
Losses and particle lifetime
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
D.Maino — Synchrotron Radiation 14/42
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
D.Maino — Synchrotron Radiation 15/42
Magnetic Field Diagnostic
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)
D.Maino — Synchrotron Radiation 16/42
Magnetic Field Diagnostic
The total energy becomes
Etot = Gη PνB−3/2 + V
B2
8π
D.Maino — Synchrotron Radiation 17/42
Magnetic Field Diagnostic
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
D.Maino — Synchrotron Radiation 18/42
Magnetic Field Diagnostic
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
D.Maino — Synchrotron Radiation 19/42
Magnetic Field Diagnostic
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
D.Maino — Synchrotron Radiation 20/42
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)
D.Maino — Synchrotron Radiation 21/42
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
D.Maino — Synchrotron Radiation 22/42
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.
D.Maino — Synchrotron Radiation 23/42
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
D.Maino — Synchrotron Radiation 24/42
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
D.Maino — Synchrotron Radiation 25/42
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
D.Maino — Synchrotron Radiation 26/42
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
D.Maino — Synchrotron Radiation 27/42
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
D.Maino — Synchrotron Radiation 28/42
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)
D.Maino — Synchrotron Radiation 29/42
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
D.Maino — Synchrotron Radiation 30/42
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
D.Maino — Synchrotron Radiation 31/42
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
D.Maino — Synchrotron Radiation 32/42
Magnetic Field Diagnostic
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
D.Maino — Synchrotron Radiation 33/42
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
D.Maino — Synchrotron Radiation 35/42
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)
D.Maino — Synchrotron Radiation 36/42
Supernova Remnants - SNRs
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)
D.Maino — Synchrotron Radiation 37/42
Supernova Remnants - SNRs
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
D.Maino — Synchrotron Radiation 38/42
Supernova Remnants - SNRs
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
D.Maino — Synchrotron Radiation 39/42
Supernova Remnants - SNRs
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
D.Maino — Synchrotron Radiation 40/42
Supernova Remnants - SNRs
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
D.Maino — Synchrotron Radiation 41/42