High Energy Astrophysics

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					  Radiation Processes

High Energy Astrophysics
      jlc@mssl.ucl.ac.uk
  http://www.mssl.ucl.ac.uk/
1. Interaction of radiation with matter:
   Photoelectric absorption and the ISM;
   Thomson and Compton scattering; Pair
   production; Synchrotron self-absorption;
   Inverse Compton scattering          [2]




                                              2
        Absorption Processes
       Photon emission processes have
      corresponding absorption processes
              We will consider
             X-ray absorption.
Emission processes             Absorption process
Recombination                     Photoionization
Inverse Compton                electron scattering
e-/e+ annihilation           e-/e+ pair production
synchrotron emission   synchrotron self absorption
                                                 3
            Photon Absorption Process Cross Sections
• Absorption coefficients are plotted
  against photon energy for the three
  processes:
        - Photoelectric absorption
        - Compton effect
        - Pair production
• Absorber is lead – plots shift
  up and down in energy with
  Z increasing or decreasing

• Photoelectric absorption is
  dominant at low energies
  and pair production at high      Z

  energies (E > 2moc2) while
  the Compton effect is
  dominant at intermediate                                4
                                              E (MeV) →
  energies
           Photoionization
                                   e-
Atom absorbs photon
                               (  EI )
                  Atom, ion or

                  molecule

      h   3.5
                         Cross-section ()
                         characterized by edges
                         corresponding to
                  h     ionization edges.
                                                  5
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
                                               6
Example of photoelectric absorption
eg. soft X-rays from a star absorbed by ISM
  star          interstellar cloud   observer




    I                          I


                                          
                                                7
    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
E is given by:       Z ( E )( m 2 )
                dl (m)




                         dV

                                                 8
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
                                                9
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:
                             nH  
F  F0 exp    Z ( E )  nZ dl  
           Z
                             nH  
                                                   10
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 ) 
                                  This is the effective
                                  cross-section,
              Z          nH     weighted over the
                                  abundance of
                 elements with respect to hydrogen
                                                      11
   Interstellar Medium Absorption Cross-section




The effective photoelectric
absorption cross-section, eff,
is plotted against wavelength
in Å for the interstellar medium
for an assumed set of interstellar
element abundances (Morison and
McCammon, 1983, Ap.J., 270, 119)



                                            12
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...
                                                  13
         Clumping of the ISM
 Take an example at low energies, e.g. at

h  0.1keV , eff  10 m            24         2



  Average ISM density     At a distance,
                            d=100 pc
  H  10 m   6    3
                          310 m   18

                                            14
        Smooth versus clumpy
 star                                 observer
                 smooth


                                  6         3
                            10 / m
             clumpy



Hot medium            Cold dense clouds
0.110 / m
         6   3
                       4 10 / m
                              6         3
                                                 15
        Numerical example
• Through the smooth medium -
     N H   H  d  3 10 / m
                          24       2


F  F0 exp( 3 10 10 ) 
                         24 F0
                  24
                                 0.05F0
                             20
• Through the clumpy medium -
 N H  3 10  0.110  0.3 10 / m
               18     6                24   2


           (
F  F0 exp  0.3 10 10
                     24     24
                                  )  0.75 F    0

                                                16
        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.

                                            17
        Thomson Scattering
low-E photon scattered by electron -
                   electron
                                 h
      h
Thomson cross-section is given by -
   8
  re , where re  2.82  10 15 m
       2

   3
    6.65  10
     e
                       29     2
                                       m
                                           18
Thomson scattering cont.
                                         3
If   N = number of particles per m
                    then fraction of area
             1m     blocked by a square
                    metre of path =
                                       29
                        6.65 10             N /m
      1m
If R is the extent of
the absorbing region
                        t  6.65 10         29
                                                   NR
                              ( = optical depth)

                              F  F0 exp ( t )
along the line of
sight,                  and
                                                    19
        Compton scattering
In Compton scattering, wavelength increases
and frequency decreases i.e. photon energy
decreases

                    electron
                                   h
       h 0                    q

 frequency      1
                 
                      h
                      1
                            (1  cosq )
 change          0 mo c 2


                                              20
Compton scattering (cont.)

                 0     h
  On average,          
                  0    mo c 2




                          h
                        
                    0     mo c 2




                                    21
    Electron-positron pair production
                                           e-
             g-ray
                                                y
               q
                                                    x
             photon                   e+

Two photons, one of which must be a g-ray with E > 2mec2,
collide and create an electron-positron (e-/e+) pair. This is
therefore a form of g-ray absorption
                                                         22
 Minimum g-ray energy required
Must first demonstrate that E  ( pc )
                                 2           2
                                                     is a
 relativistic invariant.
     Rest energy of particle,   E  mo c         2


                                     1
     m  gm0         g
                                 v     2
                                1  2 
                                 c 
                                      
                                                        23
Thus, from E  mc                              and pc  mvc,
                                       2





     (m c )      2 2
                            
                                   (m0vc)       2
                                                        
                                                          m c c v
                                                            2 2
                                                                  (   2   2
                                                                              )
    (1  v                 ) (1  v                     ) (               )
        0                                                   0
             2
                 /c    2                   2
                                               /c   2
                                                           1 v / c
                                                               2    2




     
       m c c v   2
                  0
                           2
                       m0 c
                               (
                               2 4
                                   2                2
                                                        )
         c v
          2     2

              2   And this is a
            c     relativistic invariant
                                                                              24
                                       
Total initial momentum,         p  pg  p p
       ( pc)      ( px c )  ( p y c )
             2              2           2
thus

 ( pg c  p p c cosq )  ( p p c sin q )
                            2                  2



 pg c  p c cos q
       2 2       2 2
                 p
                        2

        2 pg p p c cosq  p pc sin q
                   2         2 2   2



  pg c  p c  2 pg p pc cosq
       2 2        2 2
                  p
                                    2

                                                   25
But since         pg c  Eg ,
    ( 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 )

                                                26
Calculating the minimum energy
Assuming e+ and e- have no momentum…

    [ E  ( pc) ] final  2mo c
            2          2
                                  (       )
                                         2 2



and since        2 Eg E p (1  cosq )   ,

                                      (2m c )
  Which gives us
                                               2 2
  this expression
                           Eg           o
                                  2 E p (1  cosq )
  for the energy
  of the g-ray
  photon
                                                      27
              And this is...
found by simply making the denominator as
large as possible, ie when cos(q)= -1, or when
q=180 degrees.

      g-ray                     e-/e+ photon

  And the minimum
  g-ray energy is      Eg min   
                                  (m c )
                                      o
                                          2 2


  given by:                           Ep
                                                28
 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 cross-
  section, 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.

                                              29
  Photon versus particle density
e.g. for 3 K m-wave background photons -

        E  h  3 10 eV      4

               14     3                     3
U ph  5 10         Jm  310 eVm  5


   Corresponding to about 109 photons / m 3

                                    6
    No of nuclei in space is about 10 / m3
                                              30
Synchrotron Self-Absorption

                 e-

     e-



          Relativistic electrons moving
          in a magnetic field
                                          31
          Synchrotron Emission
Electrons, mainly responsible for emission at frequency ,
                 have energy, E, given by:
                               1
          2 mo c      2 1
                               2
       E~           mo c .
          eB               c
  and for a power law electron spectrum

            logF

                                      log           32
            Blackbody turnover
Assume Synchrotron power-law cut off, max, is
given by:

                                  2
                      E eB
              max 
                     2mo c
                         3 4


and assume each electron emits and absorbs only at
this peak frequency. Then, we will replace this with
the mean energy per particle for a thermal source or
E ~ kT

                                                       33
   On the Rayleigh-Jeans side...
                 impossible
    logF
                           blackbody
                R-J         synchrotron
                          log
Rayleigh-Jeans approximation to blackbody...

    I ( )d  2  d
              2kT 2
               c
                                           34
         Source distance
For d=source distance and R=source size,
              
                                 R
                   d

                        2
                 R
                2
                 d
                                           35
      Total flux at Earth...
So total energy flux at Earth is given by:

   F  I ( )  2  
                 2E 2
                 c
                              1
     8m         3    5
                             2
   
    
                   o
                             
                            
       Be                  
                                             36
            SSA spectrum
                  SSA
         log F

                                log 
                     a
   Optically-thick
   regime                 Optically-thin

 a lies at the point where the observed
synchrotron flux equals the blackbody limit.
                                               37
       … and SSA frequency
Substituting for  then:
                                  1/ 2
            8m      3    5
                                        R   2
      F  
            Be
                       o
                                  
                                          2
                                       d
and

  R  3  10 F B d
                  17       1/ 2       1/ 4       5 / 4


                                                          38
    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
                                                 39
 Radiation processes (summary)
• Thermal - Bremsstrahlung
  electron energies ~ photon energies
  to produce X-rays, b = v/c ~ 0.1

• Non-thermal - Synchrotron and Inverse
  Compton


                                          40
          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         E = g mo c2 - for a relativistic electron
Hence       g2 = 2  mo m/B e

                                                  41
       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
and from       U PH , => B=6x10 Tesla
         2m 0
To produce X-rays of   m ~ 1018 Hz, we need
                       g S ~ 5  10 16
                         2
                                               42
      Inverse Compton Scattering
For a relativistic electron colliding with a low
energy photon, gIC2 ≈ hfinal/hinitial
For X-ray production consider:
  - starlight: <h> ~ 2eV (l~6000A)
                                -4
  - 3K background: <h> ~3x10 eV

                                8keV
            then     g   2
                              
                                 h 
                         IC
        3
= 410 for stars
      7
= 310 for the 3K background
                 We need cosmic rays!!! 43
RADIATION PROCESSES

   END OF TOPIC



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posted:1/20/2013
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