1. Bremsstrahlung Rybicki & Lightman Chapter 5

2. 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

3. 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)

4. A qualitative picture

5. Emission from Single-Speed Electrons v e- R Ze ion b 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 is a function of time during the encounter. - Recall that for dipole radiation whereis the Fourier Transform of

6. After some straight-forward algebra, (R&L pp. 156 – 157), one can derive in terms of impact parameter, b.

7. Now, suppose you have a bunch of electrons, all with the same speed, v, which interact with a bunch of ions.

8. 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

9. So total emission/time/Vol/freq is Again, evaluating the integral is discussed in detail in R&L p. 157-158. We quote the result 

10. Energy per volume per frequency per time due to bremsstrahlung for electrons, all with same velocity v. Gaunt factors are quantum mechanical corrections  function of e- energy, frequency Gaunt factors are tabulated (more later)

11. 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

12. where In cgs units, we can write the emission coefficient Free-free emission coefficient ergs /s /cm 3 /Hz

13. Integrate over frequency: where In cgs: Ergs sec -1 cm -3

14. The Gaunt factors - Analytical approximations exist to evaluate them - Tables exist you can look up - For most situations, so just take

16. Handy table, from Tucker: Radiation Processes in Astrophysics

17. Important Characteristics of Thermal Bremsstrahlung Emissivity (1) Usually optically thin. Then (2) is ~ constant with hν at low frequencies (3) falls of exponentially at

19. Examples : Important in hot plasmas where the gas is mostly ionized, so that bound-free emission can be neglected. Solar flare10 7 (~ 1keV)radio  flat X-ray  exponential H II region10 5 radio  flat Orion10 4 radio-flat Sco X-110 8 optical-flat X-ray  flat/exp. Coma Cluster ICM10 8 X-ray  flat/exp. T ( o K)Obs. of

20. Bremsstrahlung (free-free) absorption Recall the emission coefficient, jν, is related to the absorption coefficient αν for a thermal gas: is isotropic, soand thus in cgs: Brems emission Inverse Bremss. free-free abs. e- ion e- photon collateral

21. 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)

22. (2)e.g. Radio: Rayleigh Jeans holds Absorption can be important, even for low n e in the radio regime.

23. From Bradt’s book: BB spectrum is optically thick limit of Thermal Bremss.

24. HII Regions, showing free-free absorption in their radio spectra:

26. 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 hν (keV) ~ 2

27. (b) Assume the cloud is in hydrostatic equilibrium around a central mass, M. Find M, and the density of the cloud, ρ Vol. 1/r 2 Vol. emission coeff.

28. - 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 (1)

29. - Hydrostatic equilibrium  another constraint upon ρ, R Virial Theorem: For T=10 9 K(2) - Eqn (1) & (2)  Substituting L=10 kpc, F=10 -8 erg cm -2 s -1