# Bremsstrahlung by yurtgc548

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```									  Bremsstrahlung

Rybicki & Lightman Chapter 5
Bremsstrahlung                “Free-free Emission”

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
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

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
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

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