Bremsstrahlung Rybicki & Lightman Chapter 5 Bremsstrahlung Free-free Emission Braking Radiation Radiation due to acceleration of charged particle by the Coulomb field of another charge. Relevant for (i) Collisions between unlike particles: changing dipole emission e-e-, p-p interactions have no net dipole moment (ii) e- - ions dominate: acc(e-) > acc(ions) because m(e-) << m(ions) recall P~m -2 ion-ion brems is negligible 1
Method of Attack: (1) emission from single e- pick rest frame of ion calculate dipole radiation correct for quantum effects (Gaunt factor) (2) Emission from collection of e- thermal bremsstrahlung or non-thermal bremsstrahlung (3) Relativistic bremsstrahlung (Virtual Quanta) A qualitative picture 2
Emission from Single-Speed Electrons b Ze ion R e- v Electron moves past ion, assumed to be stationary. b= impact parameter - Suppose the deviation of the e- path is negligible small-angle scattering The dipole moment the encounter. - Recall that for dipole radiation is a function of time during energy frequency = dw dω = 8πω 4 3c 3 d ˆ (ω) 2 where is the Fourier Transform of After some straight-forward algebra, (R&L pp. 156 157), one can derive in terms of impact parameter, b. 3
Now, suppose you have a bunch of electrons, all with the same speed, v, which interact with a bunch of ions. Let n i = ion density (# ions/vol.) n e = electron density (# electrons / vol) The # of electrons incident on one ion is # e-s /Vol d/t around one ion, in terms of b 4
So total emission/time/vol/freq is dw dω dv dt = n e n i 2πv dw (b) dω bdb Again, evaluating the integral is discussed in detail in R&L p. 157-158. We quote the result Energy per volume per frequency per time due to bremsstrahlung for electrons, all with same velocity v. dw dωdvdt = 16πe6 1 3 3c 3 m 2 v n n Z 2 e i g ff (v,ω) Gaunt factors are quantum mechanical corrections function of e- energy, frequency Gaunt factors are tabulated (more later) 5
Naturally, in most situations, you never have electrons with just one velocity v. Maxwell-Boltzmann Distribution Thermal Bremsstrahlung Average the single speed expression for dw/dwdtdv over the Maxwell-Boltzmann distribution with temperature T: The result, with where In cgs units, we can write the emission coefficient Free-free emission coefficient ergs /s /cm 3 /Hz 6
Integrate over frequency: dw dvdt = 25 πe 6 3hmc 3 2πkT 3km 1/ 2 Z 2 n e n i g B where g B = frequency average of the velocity averaged Gaunt factor In cgs: ε ν ff = dw dvdt =1.4 10 27 Z 2 n e n i T 1/ 2 g B Ergs sec -1 cm -3 The Gaunt factors - Analytical approximations exist to evaluate them - Tables exist you can look up - For most situations, so just take 7
Handy table, from Tucker: Radiation Processes in Astrophysics 8
Important Characteristics of Thermal Bremsstrahlung Emissivity (1) Usually optically thin. Then (2) is ~ constant with hν at low frequencies (3) falls of exponentially at 9
Examples: Important in hot plasmas where the gas is mostly ionized, so that bound-free emission can be neglected. T ( o K) Obs. of Solar flare 10 7 (~ 1keV) radio flat X-ray exponential H II region 10 5 radio flat Orion 10 4 radio-flat Sco X-1 10 8 optical-flat X-ray flat/exp. Coma Cluster ICM 10 8 X-ray flat/exp. Bremsstrahlung (free-free) absorption photon Brems emission e- ion Recall the emission coefficient, jν, is related to the absorption coefficient αν for a thermal gas: e- e- photon collateral Inverse Bremss. free-free abs. is isotropic, so ff α ν = 4e6 2π 3mhc 3km 1/ 2 and thus n e n i Z 2 T 1/ 2 ν 3 hν / kt ( 1 e )g ff in cgs: α ν ff = 3.7 10 8 n e n i Z 2 T 1/ 2 ν 3 hν / kt ( 1 e )g ff 10
Important Characteristics of (1) (e.g. X-rays) Because of term, is very small unless n e is very large. in X-rays, thermal bremsstrahlung emission can be treated as optically thin (except in stellar interiors) (2) e.g. Radio: Rayleigh Jeans holds Absorption can be important, even for low n e in the radio regime. 11
4/11/11 From Bradt s book: BB spectrum is optically thick limit of Thermal Bremss. HII Regions, showing free-free absorption in their radio spectra: 12
R&L Problem 5.2 Spherical source of X-rays, radius R distance L=10 kpc flux F= 10-8 erg cm -2 s -1 (a) What is T? Assume optically thin, thermal bremsstrahlung. Turn-over in the spectrum at log h (kev) ~ 2 13
(b) Assume the cloud is in hydrostatic equilibrium around a central mass, M. Find M, and the density of the cloud, ρ Vol. emission coeff. 1/r 2 Vol. F = 1 4πR 3 4πL 2 3 ( 1.4 10 27 T 1/ 2 n e n i Z 2 g B ) - Since T=10 9 K, the gas is completely ionized - Assume it is pure hydrogen, so n i = n e, then ρ=mass density, g/cm3 Z=1 since pure hydrogen g B =1.2 (1) 14
- Hydrostatic equilibrium another constraint upon ρ, R Virial Theorem: K.E. 2 = grav.energy particle particle For T=10 9 K (2) - Eqn (1) & (2) Substituting L=10 kpc, F=10-8 erg cm -2 s -1 15