ThermalThermal BremsstrahlungBremsstrahlung

''Radiation due to the acceleration of a charge in the Coulomb feld of another charge is called bremsstrahlung or free-free emission”

A full understanding of the process requires a quantum treatment, since photons of energies [may be] are comparable to that of the emitting particle can be produced

However, a classical treatment is justifed in some (most of ) regimes, and the formulas so obtained have the correct functional dependence for most of the physical parameters. Therefore, we frst give a classical treatment and then state the quantum results as corrections (Gaunt factors) to the classical formulas''

George B. Rybicky & Alan P. Lightman in ''Radiative processes in astrophysics'' p. 155

Processi di radiazione e MHD – THERMAL BRESSTRAHLUNG – Daniele Dallacasa (Thermal) Bremsstrahlung

on textbooks

G. Ghisellini: “Radiative Processes in High Energy Astrophysics” § 2

M. Longair: “High Energy Astrophysics” § 6.3 – 6.6

G. B. Rybicky & A.P. Lightman: “Radiative Processes in Astrophyiscs” § 5

C. Fanti & R. Fanti: “Lezioni di radioastronomia” § 5

T. Padmanabhan: “Theoretical Astrophysics” § 6.9 Radiative processes: foreword

Continuum from radiative processes from free-free transitions h √1−v2 /c2 Classical physics holds when λ = smaller thand=the typical scale of the interaction dB m v → Quantum mechanics(Δ x Δp≥h/2 π)can be avoided h For non relativistic velocities ≪ d m v h=Planck 'sconstant m=massof the(lightest)particle v=velocity of the(lightest)particle d = characteristic scale of the interaction or λ of the electromagnetic wave if it is involved)

λdB = de Broglie wavelength Luis (7th duke of) de Broglie 1892 -1987 (Nobel 1929)

Implication → Particles act like points when h ν≪Eparticle Bremsstrahlung Foreword

Free electrons are accelerated (decelerated) in the Coulomb feld of atomic (ionized) nuclei (free-free) and radiate energy

If Electrons and nuclei are in thermal equilibrium at the temperature T, their emission Is termed “thermal bremsstrahlung”

Electrons move at velocity v wrt the ions and during ''collisions'' are defected. The electrostatic interaction decelerates the electron which emits a photon bringing away part of its kinetic energy

N.B. e and p have also homologous interactions (e with e; p with p) but they do not emit radiation since the electric dipole is zero.

Astrophysical examples: 1. HII regions (104 oK) 2. Galactic bulges & hot-coronae (107 oK) 3. Intergalactic gas in clusters/groups (107-8 oK)

Bremsstrahlung is the main cooling process for high T (> 107 K) plasma b Bremsstrahlung single event

Radiated power (single particle) is given by the Larmor's formula: v v' 2 _ dE 2e 2 _ e ) e

t b ( P = − = a dt 3c3 x acceleration is from Coulomb's law Ze+ Ze2 Ze2 F = ma ≃ − → a = x2 mx2 where x= actual distance between electron (e-) and ion (Ze+)

more appropriately x ∼ x(t) then

2e2 2Z2 e6 1 1 P(t) = a2(t) = 3c3 3c3 m2 [x(t)]4 Bremsstrahlung single event

Radiated power (single particle) is given by the Larmor's formula: v v' 2 2 6 2e 2 Z e 1 1 _ _ e ) e t b ( P(t) = a(t) = x 3c3 3c3 m2 [x(t)]4

Remarks: Ze+ −1 −2 1. a ∼ m → P ∼ m −2 −4 2. a ∼ x → P ∼ x

Radiated power is relevant 1. for electrons (positrons) only i.e. light particles 2. when the two charges are at the closest distance: (x ∼ b)

2b The interaction lasts very shortly: Δ t ∼ where v = velocity v Bremsstrahlung Pulses and spectrum

The emitted energy in a single “collision” (consider x ~ b) 2 2 e2 Ze2 2b 4 Z2 e6 1 P Δ t = = x≃b P(t) 3 c3 (m x2 ) v 3 c3 m2 b3 v Radiation comes in pulses Δ t long Asymmetric profle ( decrease of the e– velocity) t Fourier analysis of the pulse →Spectral distribution of the radiated energy P(ν) 1 v Flat up to a frequency ν ≃ ≈ max 2Δ t 4b Related to the kinetic energy of the electrons

(exponential decay beyond νmax ) ν νmax P(t) t

P(ν) ν max ν Bremsstrahlung radiated energy P(ν) The radiated energy per unit frequency (assuming fat spectrum) from a single event can be written as

2 6 P Δ t P Δ t 2 16 Z e 1 1 ν (Δ ) νmax ≈ = 2P t ≃ 3 2 2 2 ≈ 2 2 Δ ν νmax 3 c me b v b v

Exercise: Determine the range of distances (order of mag)in which the released energy is signifcant −10 −28 e = 4.8⋅10 statC me = 9⋅10 g v = 100km/s −13 Remember.... Bohr's radius ao ∼5⋅10 cm −3 Typical ISM densities nISM ∼ 1 cm Bremsstrahlung (un)real plasma

Unphysical case:a cloud with ion number density nZ and free electron number density ne

In a unit time, each electron experiences a number of collisions with an impact parameter between b and b + db is

2 πnZ v bdb

The total number of collisions is 2πne nZ v bdb

Total emissivity from a cloud emitting via bremsstrahlung (single velocity!)

b 2 6 dW max 16 Z e 1 J (v , ν) = = 2 πn n v bdb = br e Z ∫ 3 2 2 2 ν bmin d dV dt 3 c me b v 6 6 bmax 32πe db 32 πe 1 bmax = n n Z2 = n n Z2 3 2 e Z ∫ 3 2 e Z ln bmin b v b 3c me v 3c me ( min ) Bremsstrahlung impact parameter

Evaluation of bmax : at a given ν, we must consider only those interactions where electrons have impact parameters corresponding to vmax > v and then v b ≤ is a function of frequency max 4 ν

Evaluation of bmin : a) classical physics requires that Δ v ≤ v Ze2 2b 2 Ze2 Δ v ≃ a Δ t = 2 ≤ v → bmin−c≥ 2 me b v me v b) quantum physics and the Heisemberg principle ΔpΔ x ≥ h/2 π h h h Δ p=me Δ v ≈ me v ≥ ≈ → bmin−q≥ 2π Δ x 2 πbmin 2πme v

we must consider which bmin has to be used Bremsstrahlung evaluatingbmin ,bmax v b ≃ frequency dependent!....velocity dependent! max 4 ν 6 −1 3 10 10 ms −5 e.g. if ~v=10 km/ s and ν =10 Hz then bmax ≃ 10 −1 ≈ 2.510 m 410 s Thermal distribution → range of velocities FYI, 3 1 2 Order of magnitude kT = m〈v 〉 Computatcions 2 2 never asked −23 2 −2 o −1 3kT 3⋅1.38⋅10 kgm s K T 4 〈v〉≃ = −31 = 0.68⋅10 √T m/s m 9.11⋅10 kg √ e √ 4 For a minimum temperature of 10 K we get a typical velocity of 680km/ s. 5 10 v 6.8⋅10 m/s −5 At 10 Hz we get:bmax≃ ≃ 10 −1 ≃ 1.7⋅10 m 4 ν 4⋅10 s

−34 2 −1 h 6.62610 kmm s −10 bmin(c) ≥ ≈ −31 5 −1 ≈0.17⋅10 m 2πme v 2π9.11⋅10 kg6.8⋅10 ms

2Ze2 bmin(c) ≥ 2 .... me v Thermal bremsstrahlung electron velocities / total emissivity

A set of particles at thermal equilibrium follow Maxwell-Boltzmann distribution: 3/2 2 me −m v /2kT f (v)dv = 4 π e e v2 dv (2πk T )

Then ne(v) = ne f(v)dv is the number density of electrons whose velocity is between v and v+dv and ne(v) replaces ne in Jbr (ν) When the plasma is at equilibrium, the process is termed thermal bremsstrahlung

Total emissivity: must integrate over velocity (energy) distribution (0≤v<∞ ?)

dW ∞ (ν ) (ν ) ( ) Jbr ,T = = ∫v Jbr ,v f v dv = dt dV d ν min 2 Plasma density (ne )

−38 −1/2 −h ν/kT 2 −1 −3 −1 = 6.8⋅10 T e ne nZ Z gff (ν, T) ergs cm Hz

T ~ particle energy high energy cut-off (T) Bremsstrahlung classical/quantum limit which [classical/quantum] limit prevails: >1 when v≥0.01c bmin−q h/(2 πme v) h v 137 v ≈ 2 2 = 2 c ≈ bmin−c 2 Ze /(me v ) 4 π Ze c Z c

bmin−q 137 3k 1/2 but v = 3k T /m then ≈ T √ e b Zc m min−c √ e Examples: HII region (T~104) Hot IGM cluster gas (T~108) Bremsstrahlung Gaunt factor

3 bmax g (ν, T) = √ ln ff b π ( min ) The Gaunt Factor ranges from about 1 to about 10 in the radio domain

Values around 10-15 in the radio domain, slightly higher than 1 at higher energies bremsstrahlung – interlude: photon frequency .vs. temperature Consider an electron moving at a speed of v =1000 km s-1. In case it would radiate all its kinetic energy in a single interaction 1  E = h  = m v 2 = 0.5 ×9.1×10−31 kg ×10 6 m s−1 2 = 4.55 ×10−19 J 2 −19 2 −2  E 4.55 ×10 kg m s 14 the frequency of the emitted radiation is  = = = 6.87 ×10 Hz h 6.626 ×10−34 kg m2 s−1 in case this electron belongs to a population of particles at thermal equilibrium for which v = 1000 km s−1 is the typical speed, determine which is the temperature of the plasma

−19 2 −2  E 4.55 ×10 kg m s 4 T = = = 3.30 ×10 K k 1.38 ×10−23 kg m2 s−2 K −1

k = 1.3806503 × 10 -23 m2 kg s -2 K -1 h = 6.626068 × 10 -34 m2 kg s -1 -31 me = 9.10938188 × 10 kg bremsstrahlung – interlude (2): the cut-off frequency The cut-off frequency in the thermal bremsstrahlung spectrum depends on the temperature only, and is conventionally set when the exponential equals 1/e.

−23 2 −2 −1 h  k 1.38 ×10 kg m s K 10 = 1   = T = 2 × T = 2.08×10 ×T Hz  kT h 6.626 ×10−34 kg m s−1

It is interesting to determine the cut-off frequency for a warm plasma (HII region, T ~ 10 4 K) and for a hot plasma (IGM in clusters of galaxies, T ~ 10 8 K). ⇨ In observations, the measure of the cut-off frequency is a way to determine the plasma temperature Thermal bremsstrahlung total emissivity(2)

Total emissivity does not depend on frequency (except the cut – off) and the spectrum is fat

−38 −1/2 −h ν/kT 2 −1 −3 −1 Jbr (ν , T)=6.8⋅10 T e ne nZ Z gff (ν , T) ergs cm Hz

Total emitted energy per unit volume:must integrate over velocity (energy) distribution(0≤v<∞ ?) at a given frequency ν, the velocity of the particle v must be as such 1 2h ν mv2 ≥ h ν → v ≥ 2 min m √ e and integrating over the whole spectrum:

dW −27 1/2 2 −1 −3 J (T) = ≃ 2.4⋅10 T n n Z ḡ (T) ergs cm br dt dV e Z ff ḡff (T) average gaunt factor Thermal bremsstrahlung cooling time

Let's consider the total thermal energy of a plasma and the radiative losses: the cooling time is tot Eth 3/2(ne+np)kT tbr = = we can assume ne = np Jbr (T) Jbr (T) 11 3 3ne kT 1.8⋅10 1/2 6⋅10 1/2 ≃ −27 1/2 2 = T sec → = T yr 2.4⋅10 T ne ḡff ne ḡff ne ḡff

N.B. The“average”Gaunt Factor is about afactor of afew

Astrophysicalexamples:

2 3 −3 4 HIIregions: ne ∼ 10 −10 cm ; T ∼ 10 K

−3 −3 7 8 Galaxy clusters ne ∼ 10 cm ; T ∼ 10 −10 K (thermal) bremsstrahlung is the main cooling process at temperatures above T∼107 K N.B.-2- all galaxy clusters have bremsstrahlung emission Bremsstrahlung practical applications(1)

A galactic HII region: M42 (Orion nebula) Distance: 1.6 kly ~ 0.5 kpc angular size ~ 66 ' → 1.1o 4 o T=10 K , S1.4GHz=410 Jy

Let's get some astrophysical information k T ν ∼ T = 2.08⋅1010 Hz cutoff h ( oK ) J (ν, T)⋅V S = br 4πD2 3 6⋅10 1/2 tbr = T yr ne ḡff −27 1/2 2 −1 −3 courtesy of NRAO/AUI jbr (T) = 2.4⋅10 T ne ḡff erg s cm −1 Lbr = jbr⋅V erg s Bremsstrahlung practical applications(2)

A cluster of galaxies: Coma (z=0.0232), size ~ 1 Mpc

Typical values of IC hot gas have radiative cooling time exceeding 10 Gyr

All galaxy clusters are X–ray emitters Bremsstrahlung practical applications(3) Cluster of galaxies: a laboratory for plasma physics Bremsstrahlung self-absorption (1)

Thermal free-free absorption: energy gained by a free moving electron. For a Maxwellian velocity (energy) distribution, i.e at thermal equilibrium, Kirchhoff's law holds j(ν, T) h ν3 1 (ν) (ν ) S = = B , T = 2 2 h ν/kT μ(ν, T) c e −1

j(ν , T) = μ(ν , T)B(ν, T) [Kirchhoff 'slaw] For an isotropically emitting plasma cloud via thermal bremsstrahlung dW = 4 π J (ν , T) dt dV d ν br

Therefore we can get

2 Jbr (ν, T) 1 −38 −1/2 2 c −h ν/kT μbr (ν , T) = = 6.8⋅10 T ne nZ Z ḡff (ν, T) (1−e ) 4 πB(ν, T) 4 π 2h ν3 −1/2 −3 −h ν/kT ⇒ μbr (ν, T) ≈ T ne nz ν (1−e ) ⇒ strong dependency on frequency Bremsstrahlung self-absorption (2)

– 3 –hv/kT – at high frequencies μ(η, T) is negligible(ν and e )

– at low frequencies (h ν≪kT)

−1/2 −3 −h ν/kT μbr (ν ,T) ≈ T ne nz ν (1−e ) h ν ≈T−1/2 n n ν−3 1−1+ = n n T−3/2 ν−2 e z ( kT ) e z 6 1/2 4e 2 π 2 −3/2 −2 μ (ν ,T) = n n Z T ν ḡ (T) br 3m hc 3k m e Z ff e ( e )

ḡff (T)≈10 therefore

−3/2 −2 2 −3/2 −2 2 μbr (ν, T) ≈ 0.018 T ν ne nZ Z gff (T) ≃ 0.2 T ν ne nZ Z Bremsstrahlung brightness/spectrum (1)

The total brightness from a free-free emitting cloud: let's consider emission + absorption

1 jbr (ν, T) −τ (ν ,T) −τ(ν, T) B(ν ,T) = (1−e )=BBB (ν , T)(1−e ) 4 π μbr (ν, T) according to radiative transfer Therefore: −1/2 2 T ne nZ Z ḡff (ν, T) −τ(ν, T) 2 −τ(ν, T) B(ν ,T) ≈ −3/2 −2 2 (1−e ) ≈T ν (1−e ) T ν ne nZ Z ḡff (ν, T) and the opacity [ τ(ν, T) = μ(ν, T)l ; where l = mean free path] determines the spectral shape and the regime:

1. optically thick τ(ν, T)≫1 2. optically thin τ(ν, T)≪1 Bremsstrahlung brightness/spectrum (fgures from Ghisellini)

7 15 Sphere with R = 10 15cm T=10 K, Sphere with R = 10 cm n = n =10 10 cm – 3 p e 10 14 – 3 ne =10 – 10 cm

2 −τ(ν, T) B(ν , T) ≈ T ν (1−e ) 1 τ≫1 → (1−e−τ) = 1− ≈ 1 → B(ν, T) ≈ T ν2 ( e−τ )

−τ 2 τ≫1 → (1−e ) = (1−1+τ−o(τ )) ≈ τ = μlo −2 2 0 but μ(ν) ∼ ν → B(ν, T) ≈ T ν τ ≈ T ν Bremsstrahlung : real X-ray spectra (3) Coma cluster: Conlon + 2016

The cut-off in these X-ray spectra is a measure of the plasma temperature

If distance is known, from the fux,

we can derive the density (ne) Relativistic Bremsstrahlung

Minor remark: the Gaunt Factor is slightly different wrt the case of ''thermal br.''

The emissivity for relativistic bremsstrahlung as a function of v for a given velocity v is: 32 π e6 1 183 j (v , ν) = n n Z2 ln rel ,br 3 2 3 v e Z 1/3 me c (Z )

N.B. T is not a relevant concept any more, v must be used

) E

The typical velocities of the particles are now relativistic and the (

N

E – g δ

energy (velocity) distribution is described by a power law: o

L

−δ

( )

ne E ≈ ne,o E

Integrating over velocities (energies)

Log E −δ+1 ∞ ∞ 2 −δ E −δ+1 jrel ,br (ν) ≈ ∫ ne(E)nZ Z dE ≈ ∫ E dE ≃ ≈ ν h ν h ν 1−δ The emitted spectrum is a power – law ! Bremsstrahlung the sun solar wind: -3 ne ~ 4 – 7 cm -1 vswind~ 300 – 900 km s T ~ 150 000 K (not consistent with v!)

fares: relativistic particles with power–law energy distribution Final remarks on bremsstrahlung emission: a photon is emitted when a moving electron deeply penetrates the Coulomb field of positive ion and then is decelerated the energy of the photon can never exceed the kinetic energy of the electron generally the ''classical'' description of the phenomenon is adequate, the quantum mechanics requires an appropriate Gaunt Factor only the individual interaction produces polarized radiation (fixed direction of E and B fields of the e-m wave wrt the acceleration) all the interactions produced in a hot plasma originate unpolarized emission (random orientation of the plane of interaction) we have either ''thermal'' or ''relativistic'' bremsstrahlung, depending on the population (velocity/energy distribution) of the electrons.