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Bremsstrahlung

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

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Bremsstrahlung

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  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 e- v Electron moves past ion, assumed to be stationary. b= “impact parameter” b R Ze ion - 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 where is 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 ni = ion density (# ions/vol.) ne = electron density (# electrons / vol) The # of electrons incident on one ion is d/t # e-s /Vol 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 ergs /s /cm3 /Hz Free-free emission coefficient

  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

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

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

  17. Examples: Important in hot plasmas where the gas is mostly ionized, so that bound-free emission can be neglected. T (oK) Obs. of

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

  19. Important Characteristics of (1) (e.g. X-rays) Because of term, is very small unless ne is very large. in X-rays, thermal bremsstrahlung emission can be treated as optically thin (except in stellar interiors)

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

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

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

  23. Spherical source of X-rays, radius R distance L=10 kpc flux F= 10 -8 erg cm-2 s-1 R&L Problem 5.2 (a) What is T? Assume optically thin, thermal bremsstrahlung. Turn-over in the spectrum at log hν (keV) ~ 2

  24. (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. Vol. 1/r2

  25. - Since T=109 K, the gas is completely ionized - Assume it is pure hydrogen, so ni = ne, then ρ=mass density, g/cm3 Z=1 since pure hydrogen (1)

  26. - Hydrostatic equilibrium  another constraint upon ρ, R Virial Theorem: For T=109 K (2) - Eqn (1) & (2)  Substituting L=10 kpc, F=10-8 erg cm-2 s-1

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