Docstoc

Bremsstrahlung

Document Sample
Bremsstrahlung Powered By Docstoc
					  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
  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
Emission from Single-Speed Electrons
               e-
                    v       Electron moves past ion, assumed
                            to be stationary.
   b
           R                b= “impact parameter”

   Ze
   ion


- Suppose the deviation of the e- path is negligible
     small-angle scattering
                          
   The dipole moment     e
                        d  is a function of time during
                           R
     the encounter.
                                       energy   dW 8 4 ˆ      2
- Recall that for dipole radiation                     d ( )
                                     frequency d   3c 3

                                                         
       where    ˆ )
                d(       is the Fourier Transform of   d
After some straight-forward algebra, (R&L pp. 156 – 157), one can
derive
            dW(b)
             d         in terms of impact parameter, b.
Now, suppose you have a bunch of electrons, all with the same
speed, v, which interact with a bunch of ions.
Let ni = ion density      (# ions/vol.)
    ne = electron density (# electrons / vol)



The # of electrons incident on one ion is

          nv
            electro
          e # area/



                   d/t
 # e-s /Vol




                   
         b around one
       Area ion, in terms of b
        
        2db
 So total emission/time/Vol/freq is

               dW                   dW (b)
                       ne ni 2v         bdb
             d dV dt                d

     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     16e 6 1
                             n e n i Z 2 g ff (v, )
          ddVdt 3 3c 3 m 2 v

                   n electrons/
                    e       vol.
                   n ions/vol
                   i

                   Z#ofcharges
                              inion
                   gff (v Gaunt
                         , )   factor

        Gaunt factors are quantum mechanical corrections
          function of e- energy, frequency

        Gaunt factors are tabulated (more later)
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:


           v  vv
         dW 2  m
          (, )
          v                                              2

        ddVdt 2 
     dW 
                  
                exp d
                  kT 
       
      
    dVdtd 2  m v2

           exp v
           v  
               2 
                kT
                    d

   The result, with     
                       
                       
                      d2d
dW
                                2
                                    /
                                    1


         
                 56
  e
 22                                   2 
  Z
    Tn g
     ne                               /
                                      1
                                      2
                                        e
                                        i
                                          h
                                          /kT
                                            ff
  
  3
 33
 mckm
dVdtd

where     
         fvelo
            Ga
             fa
         g ver
          f

In cgs units, we can write the emission coefficient



    dW 
     dVdtd
        .  e /
          n g
       381 / ff
        ff
        82
         10h
        6 Z2kT
           nT
            i e


                                              ergs /s /cm3 /Hz
   Free-free emission coefficient
     Integrate over frequency:


                     2 e 2kT  2
                             5   6           1/ 2
               dW
                        3      Z n e n i gB
               dVdt 3hmc  3km 

     where      gB  frequency average of the
                velocity averaged Gaunt factor

     In cgs:                 dW
                  ff
                                         27 2
                                  1.4 10 Z ne niT gB
                                                   1/ 2

                             dVdt
                                     Ergs sec-1 cm-3
The Gaunt factors                      
                                     f(,
                                    gfT )

 - Analytical approximations exist to evaluate them
 - Tables exist you can look up

 - For most situations,

                  ~ for
                       h
                       1
                 g 1 10        
                                4      
                 ff
                       kT
      so just take
                          g .
                           ff 12
Handy table, from Tucker: Radiation Processes in Astrophysics
Important Characteristics of Thermal Bremsstrahlung Emissivity  
                                                                  ff


                                           ff
  (1) Usually optically thin. Then




   (2)                                                   
           ff is ~ constant with hν at low frequencies h kT
                                                          

    (3)    ff   falls of exponentially at        kT
                                                  h~
   Examples:
    Important in hot plasmas where the gas is mostly ionized, so
    that bound-free emission can be neglected.

                        T (oK)                 Obs. of    ff
Solar flare            107 (~ 1keV)            radio flat
                                               X-ray  exponential
H II region            105                     radio flat

Orion                  104                     radio-flat

Sco X-1                108                     optical-flat
                                               X-ray  flat/exp.
Coma Cluster ICM       108                     X-ray  flat/exp.
 Bremsstrahlung (free-free) absorption

                                                    photon           Brems emission


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:

                                      T
                                     j  )
                                       (
                                       B
      ff   is isotropic, so      
                                  j
                                   ff
                                  4
                                             ff
                                                  and thus
                                             1/ 2
                                 4e 6  2 
                       ff                  n e n i Z T   e   1  h / kT g ff
                                                        2 1/ 2 3
                                     
                                3mhc 3km
in cgs:             ff  3.7 108 ne ni Z 2T1/ 2 3  eh / kT g ff
                                                        1
Important Characteristics of    ff

      
                                               
(1)    
      h kT(e.g. X-rays)

               ff
                   n
                   T
                   .
                   3 n g
                    10
                   7Z            8 2 
                                 ei
                                    
                                    1
                                    /3
                                    2
                                      ff

      Because of    
                 T/2  term, 
                  
                  1  3         ff

       is very small unless ne is very large.

             in X-rays, thermal bremsstrahlung emission can be
             treated as optically thin

              (except in stellar interiors)
(2)   
      h kT              e.g. Radio: Rayleigh Jeans holds




      
      ff
         n g
         0Z
         .nT
         018              e
                          i
                            
                           2 
                            / 2
                            3
                            2
                               ff          
       Absorption can be important, even for low ne
       in the radio regime.
From Bradt’s book: BB spectrum is optically thick limit of
                   Thermal Bremss.
HII Regions, showing free-free absorption in their radio spectra:
                       Spherical source of X-rays, radius R
 R&L Problem 5.2
                              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

                                 E ax 9o
                                
                               T m K
                                    10
                                 k
(b) Assume the cloud is in hydrostatic equilibrium around a
    central mass, M.

         Find M, and the density of the cloud, ρ



               F
                 14
                 2
                     f f
                    R           3


                 
                 4 3
                  L
                      

                                       Vol. emission coeff.
                  1/r2       Vol.

               1 4R 3
           F
              4L 3
                 2     1.4  1027 T 1/ 2 ne ni Z 2 gB 
                       



- Since T=109 K, the gas is completely ionized

- Assume it is pure hydrogen, so ni = ne, then

        n e
                  2
        in n     H                               ρ=mass density, g/cm3
               
                       2

                       
                .  47 2
                   6
                3 10
               H
               m 
   Z=1 since pure hydrogen

    gB 1.2

               
               2 T
               F R
                .
                010L               20 
                                    2 3 (1)
                                    1 2
                                     /
                                     2
                                       
 - Hydrostatic equilibrium  another constraint upon ρ, R

      Virial Theorem:         K.E.  grav.energy 
                         2                    
                             particle   particle 

                                  GMm
                               3 
                               kT   H

                                 R
   For T=109 K              M  
                        R5 
                          10 
                             cm (2)
                                
                                  8

                                
                             sun
                             M
                                                  
                                                  32
                                                   /
                              
   - Eqn (1) & (2)     4 LF 
                         10  
                            
                              M   
                                  261
                                    /2

                                
                              sun
                             M
Substituting L=10 kpc, F=10-8 erg cm-2 s-1                      
                                                                32
                                                                 /
                                                   
                                              g  
                                             10  
                                              4 cm
                                                   M
                                                    7       3
                                                            -

                                                     
                                                   sun
                                                  M

				
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
views:8
posted:6/13/2012
language:English
pages:29