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Radiation Processes High Energy Astrophysics [email protected] http://www.mssl.ucl.ac.uk/ Absorption Processes So far, considered the production of X-rays. Now, will consider X-ray absorption. Emission processes Recombination Absorption process Photoionization Inverse Compton electron scattering e-/p+ annihilation e-/p+ pair production synchrotron emission synchrotron self absorption Photoionization e- Atom absorbs photon EI h 3 Atom, ion or molecule h Cross-section () characterized by edges corresponding to ionization edges. Photoelectric Absorption Cross-section The photoelectric absorption cross-section for photons with E > EI and h << mec2 is given by K = 4√2 T a4 Z5 (moc2/)7/2 where EI is the electron binding energy, a is the fine structure constant and T is the Thomson cross-section Note dependence on Z5 and on -7/2 Example of photoelectric absorption eg. soft X-rays from a star absorbed by ISM star interstellar cloud observer I I How much passes through? Take a path of length dl (metres) nZ is the number density (m 3 ) of element Z. Cross-section offered by element Z at energy 2 Z ( E )( m ) E is given by: dl (m) dV The fraction of volume dV which is blocked by the presence of element Z is : nZ Z ( E )dl Thus fraction of flux F lost in volume dV is: dF FnZ Z ( E)dl or : dF nZ Z ( E )dl F Integrating over length from source... dF F nZ Z ( E )dl Z ( E ) nZ dl F F0 exp( Z ( E ) nZ dl ) Including all elements in the line of sight: n H F F0 exp Z ( E ) nZ dl n Z H Optical depth This becomes: F0 exp eff ( E ).N H This is ‘t’, the optical depth, which has no dimensions nZ eff ( E ) Z ( E ) nH Z This is the effective cross-section, weighted over the abundance of elements with respect to hydrogen Column density The column density given by : N H nH dl is the number of H – atoms per m2 column Column density is measured from the 21cm atomic hydrogen line - but not foolproof. There is a factor of 2 uncertainty, wide beams, molecular hydrogen contamination... Clumping of the ISM Take an example at low energies, e.g. at ... 24 2 h 0.1keV , eff 10 Average ISM density H 10 m 6 3 m At a distance, d=100 pc 310 m 18 Smooth versus clumpy star observer smooth 6 10 / m 3 clumpy Hot medium 0.110 / m 6 3 Cold dense clouds 4 10 / m 6 3 Numerical example • Through the smooth medium - N H H d 3 10 / m 24 F F0 exp 3 10 10 24 24 2 F0 0.05F0 20 • Through the clumpy medium - N H 3 10 0.110 0.3 10 / m 18 6 F F0 exp 0.3 10 10 24 24 24 2 0.75F 0 Electron scattering • Thomson scattering - the scattering of a photon by an electron where the photon energy is much less than the rest mass of the electron. • Compton scattering - photons have a much higher energy in this case and lose some of their energy in the scattering process. Thomson Scattering low-E photon scattered by electron - h electron h Thomson cross-section is given by - 8 2 re , where re 2.82 1015 m 3 29 2 e 6.65 10 m Thomson scattering cont. If 3 N = number of particles per m then fraction of area 1m blocked by a square metre of path = 6.65 10 1m If R is the extent of the absorbing region along the line of sight, 29 N /m t 6.65 10 29 ( = optical depth) and NR F F0 exp t Compton scattering In Compton scattering, wavelength increases and frequency decreases i.e. photon energy decreases. electron h 0 frequency change q h h 1 cos q 2 0 mo c 1 1 Compton scattering (cont.) On average, 0 h 2 0 mo c 0 h 2 mo c Electron-positron pair production e g-ray y q x photon e+ Two photons, one of which must be a g-ray, collide and create an electron-positron (e-/e+) pair. This is therefore a form of g-ray absorption Minimum g-ray energy required Must first demonstrate that relativistic invariant. E pc is a Rest energy of particle, m gm0 g 2 2 E mo c 1 v 1 2 c 2 2 Thus, from E mc m c 2 2 1 v /c m0vc 2 0 2 and pc mvc, 2 2 1 v 2 /c 2 m02 c 2 c 2 v 2 2 2 1 v / c m c c v 2 4 m c 0 2 2 c v 2 And this is a c relativistic invariant 2 0 2 2 2 p pg p p Total initial momentum, pc 2 thus px c p y c 2 2 pg c p p c cosq p p c sin q 2 2 pg c p c cos q 2 2 2 2 2 pg p p c cosq p p c sin q 2 2 2 2 p 2 2 pg c p c 2 pg p p c cosq 2 2 2 2 p 2 pg c Eg , But since pc 2 Eg E 2Eg E p cosq 2 2 p and - [ E pc ]initial Eg E p 2 2 2 Eg E 2Eg E p cosq 2 2 p 2 Eg E p 1 cosq Calculating the minimum energy Assuming e+ and e- have no momentum… [ E pc ] final 2moc 2 2 and since 2 Eg E p 1 cosq Which gives us this expression for the energy of the g-ray photon Eg 2 2 , 2m c 2 2 o 2 E p 1 cos q And this is... found by simply making the denominator as large as possible, ie when cos(q)=-1, ie when q=180 degrees. g-ray And the minimum g-ray energy is given by: e-/e+ photon mc 2 2 Eg min o Ep Photon-nucleus pair production • In the laboratory, it is more usual to consider photon-nucleus production. So why do we ignore it in space? • Photons and nuclei have a similar crosssection, and the g-ray does not differentiate much between another photon or a nucleus. • Then we must compare the photon density with the particle density in space. Photon versus particle density e.g. for 3 K m-wave background photons - E h 3 10 eV 4 U ph 5 10 14 3 3 Jm 310 eVm 5 Corresponding to about 109 photons / m 3 6 No of nuclei in space is about 10 / m3 Synchrotron Self-Absorption e- e- Relativistic electrons moving in a magnetic field Synchrotron Emission Electrons, mainly responsible for emission at frequency , have energy, E, given by: 1 2 2mo c 2 1 E~ mo c . c eB and for a power law electron spectrum logF log Blackbody turnover Assume Synchrotron power-law cut off, max, is given by: max 2 E eB 3 4 2mo c And assume each electron emits & absorbs only at this peak frequency. Then, we will replace this with the mean energy per particle for a thermal source, ~ kT. On the Rayleigh-Jeans side... impossible logF R-J blackbody synchrotron log Rayleigh-Jeans approximation to blackbody... 2kT 2 I d 2 d c Source distance For d=source distance and R=source size, R d 2 R 2 d Total flux at Earth... So total energy flux at Earth is given by: 2E 2 F I 2 c 8m Be 3 o 5 1 2 SSA spectrum SSA log F Optically-thick regime a log Optically-thin a lies at the point where the observed synchrotron flux equals the blackbody limit. … and SSA frequency Substituting for then: 8m F Be 3 o 5 1/ 2 2 R 2 d and R 3 10 F B d 17 1/ 2 1/ 4 5 / 4 SSA in Compact X-ray sources X-ray frequency, =1018 Hz -29 If F ~ 10 J m-2 s-1 Hz - typical X-ray source value d = 10 kpc and B = 108 Tesla (the field for a neutron star) This gives a maximum for R of ~1 km for SSA of X-rays to occur (ie for a to be observable in the X-ray band). but a neutron star diameter is 10 to 20km Radiation processes (summary) • Thermal - Bremsstrahlung electron energies ~ photon energies to produce X-rays, b = v/c ~ 0.1 • Non-thermal - Synchrotron and Inverse Compton Synchrotron Emission For an electron spiralling in a magnetic field B with energy E, the peak radiated frequency, m is m = g2 B e/2 mo = E2 B e/2 mo3 c4 But Hence E = g mo c2 - for a relativistic electron g2 = 2 mo m/B e Electron energies required • Synchrotron emission depends on the magnetic field strength. Assuming equipartition of energy - starlight, cosmic rays + magnetic fields have all the same energy density in Galaxy 2 B -10 U PH , => B=6x10 Tesla and from 2m 0 To produce X-rays of m ~ 1018 Hz, we need g S2 ~ 5 1016 Inverse Compton Scattering For a relativistic electron colliding with a low energy photon, gIC2 ≈ hfinal/hinitial For X-ray production consider: - starlight: <h> ~ 2eV (l~6000A) -4 - 3K background: <h> ~3x10 eV then 3 g 2 IC 8keV h = 410 for stars 7 = 3 10 for the 3K background We need cosmic rays!!! RADIATION PROCESSES END OF TOPIC