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.