Effect of scattering on the emitted spectrum
S.M.Lea
Here we are going to consider what happens when a system is effected by scattering as
well as emission and absorption. For simplicity we consider a simple model with only one
emission mechanism. Many astrophysical systems may be understood using this simple
model. Much of the early work was done to understand the spectra of compact x-ray sources
and supernovae.
1
Coherent scattering
Consider a slab of ionized hydrogen. The emission and absorption is due to
brensstrahlung (free-free), and we also have Thomson scattering. Then:
´
¡
¢ n2 ³
®º = 4 £ 10 8 cm ¡1 3 1=2 1 ¡ e¡h º=kT g
º T
where g is the Gaunt factor, and
®s = n¾ T
1.1
large cloud, ¿ s large
1.1.1 low frequency:
At a low enough frequency, the absorption optical depth will be much larger than the
scattering optical depth, since ® º increases
as º decreases. In this regime we can ignore the
p
effects of scattering, since ¿ ef f = ¿ º (¿ º + ¿ s ) ' ¿ º : Thus the emitted radiation is a
black body spectrum:
Iº = B º (T ) =
which is the Rayleigh-Jeans law.
2hº 3 =c2
2hº 3 =c2
º2
'
= 2 2 kT
¡1
hº=kT
c
ehº=kT
1.1.2 high frequency
At a high enough frequency, the absorption optical depth becomes small enough that
¡1=2
¿ º ¿ ¿ s : Then the }skin depth} of the source is approximately [®º (® º + ® s )]
and the
1
emitted spectrum is:
Iº = jº [®º (® º + ®s )]
¡1=2
=
jº
®º
r
®º
' Bº (T )
®º + ® s
r
®º
®s
a }diluted} black body.
If we still have hº ¿ kT in this regime, then Bº » º 2; ® º » 1=º 2 and Iº » º :
Eventually, at higher º; B º » e¡h º=kT and Iº » e¡ hº=kT :
The transition from the low frequency (Iº » º 2 ) to high frequency (Iº » º) regimes
occurs where
¿º ' ¿ s
or at a frequency º B where
³
´ n2 µ hº ¶
4 £ 10 8 cm5 s¡3 K 1=2
g ¼ n¾ T
º 3 T 1=2 kT
and so
v³
v³
´
´
u
u
u 4 £ 10 8 cm5 s ¡3 K1=2 n
u 4 £ 10 8 cm5 s¡ 3 K1=2 n
t
h
t
6:6 £ 10 ¡27 erg ¢ s
ºB =
g=
g
3=2
3=2
¡
25
2
T ¾T
k
T
(6:6 £ 10
cm ) 1:4 £ 10 ¡16 erg/K
r
n
1: 7 £ 10 11 cm 3=2 K 3=4 1=2 1=2
=
2:9 £ 10 22 cm 3 s¡2 K 3=2 3=2 g =
n g
T
T 3=4
The exact solution that is also valid at the }knee} is
Iº = Bº (1 ¡ ")1=2 H
where " = ® s = (®s + ® º ) and H is a weak function of "; varying between 1 and 3.
1.2
Thin slab, ¿ º < 1 at high frequencies, but ¿ s > 1:
The slab becomes translucent (not transparent) at high º : Photons escape without absorption
for ¿ ef f < 1;or
p
¿ º (¿ º 2+ ¿ s) < 1
Since ¿ º ¿ ¿ s ;this becomes:
¿ º¿ s < 1
The transition occurs at ¿ º » ¿ ¡1
at
frequency
ºt :
s
³
´ n2 µ hº ¶
1
4 £ 108 cm5 s¡3 K 1=2
g` ¼
n¾ T `
º 3 T 1=2 kT
Thus:
r
³
´ n3 ¾ g` 2 h
T
ºt ¼
4 £ 10 8 cm5 s ¡3 K1=2
kT 3=2
s
³
´
6:6 £ 10¡ 27 erg ¢ s n3 g`2
=
4 £ 10 8 cm5 s ¡3 K1=2 (6:6 £ 10 ¡25 cm2 )
1:4 £ 10¡16 erg/K T 3=2
sµ
r
¶
7
3 g` 2
cm
n
n3 g
3=2
¡13
7=2
3=4
=
1: 2 £ 10 ¡26 2 K
=
1:
1
£
10
Hz
¢
cm
K
`
s
T 3=2
T 3=2
1
Note that º B is de®ned by ¿ º » ¿ s while º t is de®ned by ¿ º » ¿ ¡
s ; and since
¿ º » º ¡2 ; then
º ¡2
1
t
» 2
¡2
¿s
ºB
or
º t=º B » ¿ s
Above º t ;the source is optically thin and the spectrum is a bremsstrahlung spectrum.
2 Incoherent scattering
The frequency of the scattered photon may differ from that of the incoming photon if
3
1. the incoming photon has a high frequency (hº & mc2 ) (Compton scattering) or
p
2. the electron is relativistic, moving with speed v = ¯c and large ° = 1= 1 ¡ v 2 =c2 :
In calculating the frequency of the scattered photon in the second case, we ®rst move into
the rest frame of the incoming electron. In this frame, the incoming photon is doppler shifted
to a frequency » °º 0 , but the frequency shift is probably small (Thomson scattering). Then
we shift back to the lab frame, Doppler shifting the frequency by a factor ° again. (For an
exact derivation of this result, see Rybicki and Lightman p 195-199) The net power radiated
by the scattered photons is approximately
P » u ph c¾ T ° 2
where u ph = hv 0 nph is the energy density in the photon ®eld. More exactly (R&L eqn
7.16a)
4
P = u ph c¾ T ° 2 ¯ 2
(1)
3
This process is called Inverse Compton Scattering.
In this section wezll consider how the frequency shifts in each scattering can effect the
emergent spectrum from a source.
2.1
Single scattering
2.1.1 Approximate treatment
(For a more exact analysis, see R&L §7.3)
We consider a cloud with a particle distribution n (°) particles per unit volume per unit
energy and an incident photon spectrum f (") photons per unit volume per unit energy. We
may write an approximate emission coef®cient due to scattered photons as:
Z
¡
¢
1
j (" 0) =
± " 0 ¡ °2 " n (°) f (") P (°; ") d"d°
4¼
where P (°; ") is the power radiated in a single scattering,
4
P (°; ") = c¾ T ° 2 ¯ 2 "
(2)
3
Now we want to do the integral over the electron energy spectrum, so we rewrite the
±¡function as:
Ã
Ã
r !
r !
¡ 0
¢
1
"0
1
"0
2
± " ¡° " =
± °¡
= p ± °¡
0
2°"
"
"
2 ""
and then after integration we get
0
j (" ) =
=
Ãr !
1
"0
4
"0
p n
f (") c¾ T ¯ 2 "d"
0
"
3
"
2 ""
Z r 0 Ãr 0 !
c¾ T
"
"
n
f (") ¯ 2 d"
6¼
"
"
1
4¼
Z
4
Now let n (°) be a power law
n (°) = K° ¡p
so
Ãr !¡p
"0
"0
j (" ) =
K
f (") ¯ 2 d"
"
"
Z Ã r 0 ! ¡p+1
c¾ T
"
= K
f (") ¯ 2 d"
6¼
"
Z
c¾ T 0¡(p¡1)=2
= K
"
" (p¡ 1)=2 f (") d"
6¼
where in the last step I took ¯ ¼ 1: The limits of the integral are the maximum and minimum
values of the incident spectrum. Thus the emission coef®cient leads to a power law spectrum
with index ¡® = ¡ (p ¡ 1) =2:
A more exact analysis gives almost the same result. There is a multiplicative factor of:
9
A (p)
16
where A (p) is a slowly varying function of p.
p A (p)
0 1.96
1 1.33
2 1.4
and thus the difference between my result and the exact result is a factor of between 0.9
and 1.6.
R
The integral " (p¡ 1)=2 f (") d" can be done and gives an analytical closed form solution
when f (") is a black body spectrum, as discussed in R&L (p208) : The result is important
when the scattered photons are the microwave BB radiation (cosmic background radiation).
In some sources the photons are also produced within the same source, e.g. by
synchrotron radiation. In this case we have f (") » " ¡®¡1 (more about this later) leading to
Z
j (" 0) _ "0 ¡(p¡ 1)=2 " ®"¡ ®¡1 d" = "0¡(p¡1) =2 ln ("max ="min)
0
c¾ T
6¼
Z r
which diverges unless "max is ®nite. This is always the case due to rapid radiation losses
at the high energy end of the electron spectrum.
2.2
Multiple scatterings
2.2.1 Energy change per scattering
The small energy change that occurs in Compton scattering can become important when
multiple scatterings occur. we have:
4"
"
'¡ 2
"
mc
5
in the electron rest frame. If we have a thermal distribution of electrons, some of them will
have less energy than " and some will have more. The less energetic electrons gain energy
from the photon, but the more energetic electrons lose energy to the photon. Averaging over
the entire distribution of electrons we expect to ®nd:
4"
kT
"
'® 2 ¡
(3)
"
mc
mc2
in the lab frame. If the system is isolated, it should eventually come to equilibrium at some
temperature. If the only interaction between photons and matter is scattering (no absorption
and emission) then the equilibrium photon distribution is NOT a black body spectrum, but
rather a |relativistic Maxwellian}
N (") _ "2 e¡"=kT
For this distribution
Letting y = "=kT ; we get
R
R 3 ¡"=kT
"N (") d"
" e
d"
< " >= R
= R 2 ¡"=kT
N (") d"
" e
d"
R 3 ¡y
y e dy
R
3
y 2 e¡ y dy
(kT )
4
< " >=
(kT )
Now integrate by parts:
Z 1
Z
¯1
y 3 e¡y dy = ¡y 3 e¡ y ¯ 0 +
0
1
3y 2 e ¡y dy =
0
Z
1
3y 2 e¡ y dy
0
and thus
< " >= 3kT
R 2
R 4 ¡"=kT
" N (") d"
" e
d"
2
2
< " >= R
= R 2 ¡"=kT
= 4 £ 3 (kT )
N (") d"
" e
d"
Now once equilibrium is established, there is no further net energy transfer between
electrons and photons, so averaging equation 3, we have:
Also:
< 4" >' ®
kT
< "2 >
<
"
>
¡
=0
mc2
mc2
and thus
2
®kT (3kT ) ¡ 3 £ 4 (kT ) = 0 ) ® = 4
So we may rewrite equation 3 as
"
4" '
(4kT ¡ ")
mc2
Now we may reinterpret equation 1 for thermal electrons with < ¯ 2 >= 3kT =mc2 :
4
3kT
< P >= uph c¾ T
3
mc2
kT
= 4
< " > nphc¾ T
mc2
= (average energy shift per scattering) (rate of scattering)
(4)
As an aside, we note here that this scattering process also acts as a heating or cooling
6
The heating rate ¡ may be written as:
"
¡ = F (") ne ¾ T
(" ¡ 4kT )
mc2
where F (") is the photon -ux to which the electrons are exposed. (If ¡ is negative the
result is a net cooling of the plasma.)
process for the electrons.
2.2.2 Emergent spectrum
The net change in photon energy due to multiple scatterings is determined using the
Compton "y" parameter:
¡
¢
4kT
y =
max ¿ ; ¿ 2
2
mc
= (fractional energy change per scattering) (mean # of scatterings)
(5)
where " ¿ 4kT has been assumed in the interpretation in words, and ¿ is the scattering
optical depth.
If absorption is important, we need the number of scatterings within one mean free path
of the surfaceu that is, the |¿ } that we need in y is
¿y
= ® s (mfp between emission and absorption)
¡ 1=2
= ® s [® abs (® abs + ® s)]
¿s
=
(¿ ef f & 1)
¿ ef f
(6)
Emergent spectrum for small ¿
Let A be the ampli®cation factor per scattering. (For
example, A = °2 for relativistic electrons). Consider an input beam of photons at energy
hv = "i : After k scatterings, the incidents photon is scattered to energy
" k » "iAk
(7)
k
For ¿ s < 1; the probability of k scatterings is ¿ ; and so
I ("k ) ' I0 ("i) ¿ k
and from equation 7,
k=
To proceed, note that
so
ln ("k ="i)
ln A
³
´ ln b
³
´
ln a
ln aln b= ln c =
ln a =
ln b = ln bln a= ln c
ln c
ln c
¿ ln(" k="i)= ln A = ("k ="i )ln ¿ = ln A
Since ¿ < 1 in this example, it is useful to de®ne ® ´ ¡ ln ¿ = ln A; so that ® > 0: Then the
emitted spectrum has the form:
µ ¶¡ ®
"k
I ("k ) ' I0 ("i )
"i
i.e. the spectrum is a power law of index ®: Note that this derivation makes sense only if
7
A > 1 (e.g. scattering off relativistic electrons.) The result is important since so many
astronomical objects have power law spectra over a fairly large range of frequencies.
One of the other common models- synchrotron radiation- can run into trouble because
of the |Compton catastrophe}. (Photons are Compton scattered up to higher energy,
drastically reducing the lifetime of the relativistic electrons, and thus increasing the energy
requirements for the source.)
The Kompaneets equation
(Reference: Kompaneets, JETP 4, 730, 1957)
In the event that the ampli®cation factor A is not large, the previous analyis fails. We
must develop a more exact formalism. We derive the Boltzmann equation, which describes
the change in the distribution function n (!) of the photons in phase space due to scattering
off electrons.
Let n (!) be the number of photons/cm3 =momentum space volume at frequency !:
Let fe (~
p ) be the number of electrons/cm3 /momentum space volume d3 ~p at momentum
~p:
n (!) is changed because photons are scattered to a new frequency ! 0:
Z Z
d¾
# scattered out/time =
f e (~p) n (!) c (1 + n (!0 ))
d-d3 ~p
dwhere n (!) c is the photon -ux and the factor n (!0 ) is due to induced scattering. (Induced
scattering is similar to stimulated emission, and occurs because photons are bosons.) The
values of ~p ; ! and ! 0 are related by the physics of the scattering event. Similarly:
Z Z
d¾
# scattered in/time =
fe (~p 1 ) n (! 0) c (1 + n (!))
d-d3 ~p 1
dThe rate of change of n (!) is due to these two processes:
·Z
¸
Z
Z
@n (!)
d¾
0
3
0
3
=c
df e (~p1 ) n (! ) c (1 + n (!)) d ~p 1 ¡ fe (~p ) n (!) c (1 + n (! )) d ~p
dt
d(8)
To simplify this equation, we make 2 approximations:
1. Electrons are non-relativistic
fe (~
p) =
ne
3=2
(2¼mkT )
and
¡
¢
exp ¡p2 =2mkT
2. Energy transfer per scattering is small.
h (! 0 ¡ !)
4´
¿1
kT
Then we expand n (! 0) in a Taylor series:
n (! 0)
2
@n
1
2 @ n
+ (! 0 ¡ !)
+¢¢¢
@!
2
@! 2
@n
1
@2n
= n (!) + 4
+ 42 2 + ¢ ¢ ¢
@x
2
@x
= n (!) + (! 0 ¡ !)
where x = h!=kT :
8
(9)
Now we need to relate ~p 1 to ~p and 4: By conservation of energy:
p21
p2
+ h!0 =
+ h!
2m
2m
so
p21
p2
p2
=
+ h! ¡ h! 0 =
¡ kT 4
2m
2m
2m
Thus
¡
¢
¡
¢
fe (~
p 1 ) _ exp ¡p 21 =2mkT = exp ¡p 2 =2mkT e4
and expanding e4 to second order, we get:
¡
¢
fe (~p 1 ) = fe (~p ) 1 + 4 + 4 2 =2
Then:
fe (~p 1 ) n (! 0 ) (1 + n (!)) =
=
(10)
¡
¢¡
¢
fe (~p ) 1 + 4 + 4 2 =2 n + 4n0 + 4 2 n0 0=2 (1 + n)
µ
¶
1
fe (~p ) n + 4 (n0 + n) + 42 (n00 + 2n0 + n) (1 + n)
2
and
µ
¶
1 2 00
f e (~p ) n (!) (1 + n (! )) = f e (~p ) n 1 + n + 4n + 4 n
2
In our primary equation 8, we need
Z
µZ
Z
¶
@n (!)
d¾
= c
df e (~p1 ) n (! 0) c (1 + n (!)) d3 ~p 1 ¡ fe (~p ) n (!) c (1 + n (! 0 )) d3 ~p
dt
d·
¡
¢
¸
Z
Z
d¾
fe (~p ) n + 4 (n0 + ¡n) + 12 42 (n00 + 2n0 + n)
(1 + n)
¢
= c
d- d3 ~p
¡f e (~p) n 1 + n + 4n0 + 12 4 2 n00
dµµ
¶¶
Z
Z
¡
¢ 1
d¾
= c
d- d3 ~p fe (~p )
4 n0 + n + n2 + 4 2 (n0 0 + (1 + n) (n + 2n0 )) (11)
d2
0
0
To proceed further we need 4 in terms of ~p : So we use conservation of momentum. In
the non-relativistic case:
h! ^
h! 0 ^ 0
k + ~p =
k + ~p 1
c
c
where ^k; ^k0 are unit vectors in the direction of photon propagation (unit wave vectors).
Then:
´
h! ^
h!0 ^ 0
h³ ^
~p 1 =
k + ~p ¡
k = ~p +
! k ¡ !0 k^ 0
c
c
c
and
´
´
h2 ³
h³ ^
p21 = p 2 + 2 ! 2 + ! 02 ¡ 2!! 0~k ¢ ^k0 + 2
! k ¡ ! 0k^ 0 ¢ ~p
(12)
c
c
We use relation 12 to eliminate p 1 from equation 10:
³
´
³
´
h2
2
02
0~ ^ 0
h
^k ¡ ! 0 ^k0 ¢ ~p
2
2
!
+
!
¡
2!!
k
¢
k
+
2
!
2
p ¡ p1
c
c
4=
=¡
2mkT
2mkT
and then using equation 9
h! 0 = h! + 4kT
9
so to ®rst order in 4 :
2
4
= ¡
= ¡
2 hc 2 !2 + 24kT hc !2 ¡ 2 ch2 ! (h! + 4kT ) ~k ¢ ^k0 +
2
2 hc 2 !2
³
2
c
³
´
h!k^ ¡ (h! + 4kT ) ^k0 ¢ ~p
2mkT
´
h!
h
0
~
^
1 ¡ k ¢ k + 24kT c 2 ¡ 2 c2 !4kT ~k ¢ ^k0 +
2
c
³
³
´
´
h! k^ ¡ ^k0 ¡4kT ^k0 ¢ ~p
2mkT
µ
¶
´
³
´
h! h! ³
h!
~p
h! ³^ ^ 0´ ~p
0
0
0
~
^
~
^
^
= ¡
1¡k ¢k ¡ 4
1¡k¢k ¡k ¢
¡
k¡k ¢
2
2
mc kT
mc
mc
kT
mc
Collecting terms:
µ
¶
´
´
³
´
h! ³
h! h! ³
~k ¢ ^k0 ¡ k^ 0 ¢ ~p
~k ¢ ^k0 ¡ h! ^k ¡ ^k0 ¢ ~p
4 1+
1
¡
=
¡
1
¡
mc2
mc
mc2 kT
kT
mc
For photons with h! ¿ mc2 = 511 keV, the bracket on the LHS is approximately = 1 ,
and we may ignore the ®rst term on the RHS, to get
³
´
h! ³^ 0 ^ ´ ~p
^0 ¡ k
^ ¢ ~p
4=
k ¡k ¢
=x k
kT
mc
mc
Now we return to equation 11. The differntial scattering cross section is:
¢
d¾
3¾ T ¡
=
1 + cos2 µ
d16¼
where µ is the angle between ^
k and ^
k0 : Evaluating the 4 2 term ®rst, we have:
Z
Z
Z
Z
d¾
d¾
p 2 ³ 0 ^ ´2
3
2
I1 =
d- d ~p f e (~p) 4 =
d- d3 p~f e (~p) x2 2 2 ^
k ¡k cos2 »
ddm c
^ We do the integral over p~ ®rst, taking » as the
where » is the angle between p~ and ^k0 ¡ k:
polar angle in a spherical coordinate system, so that
d3 ~p = p 2 dp sin » d»dÁ0 = p 2 dpd¹»dÁ0
where ¹» = cos» : Then
I1
=
3¾ T
n ex 2
16¼ (2¼mkT )3=2
=
3¾ T
16¼
=
3¾ T
16¼
Z
¢ ³ 0 ´2
1 + cos 2 µ ^
k ¡^
k sin µdµdÁ
Z
¡
¢
p2
exp ¡p2 =2mkT p2 dp¹2» d¹» dÁ0
2
2
m c
¯+1
Z
Z
¡
¢ ³ 0 ´2
¡ 2
¢
n ex 2
p4
1 3 ¯¯
2
^
^
1
+
cos
µ
k
¡
k
sin
µdµdÁ
exp
¡p
=2mkT
dp
¹
2¼
3=2
m2 c2
3 » ¯¡ 1
(2¼mkT )
Z
Z 1
¡
¢ ³ 0 ´2
¡
¢ 4¼
n ex 2
1
2
^
^
1
+
cos
µ
k
¡
k
sin
µdµdÁ
p 4 exp ¡p2 =2mkT dp
3=2
2
2
m c 0
3
(2¼mkT )
¡
10
Change variables to u = p 2 =2mkT Then du = pdp=mkT ; and
Z 1
Z 1
¡
¢
p 4 exp ¡p2 =2mkT dp =
(2mkT )3=2 u 3=2 exp (¡u) mkT du
0
0
Z 1
1
5=2
=
(2mkT )
u 3=2 exp (¡u) du
2
0
1
=
(2mkT )5=2 ¡ (5=2)
2
Now
µ ¶
µ ¶
µ ¶
5
3
3
31
1
3p
¡
= ¡
=
¡
=
¼
2
2
2
22
2
4
So:
Z
¡
¢ ³ 0 ´2
p
3¾ T
ne x 2
^ ¡k
^ sin µdµdÁ 1 3 ¼ (2mkT )5=2 4¼
I1 =
1 + cos2 µ k
3=2
2
2
16¼ (2¼mkT )
m c 8
3
Z
³
´
2
¡
¢ 0
3 ne ¾ T
=
kT x2
1 + cos 2 µ ^
k ¡^
k sin µdµdÁ
16¼ mc2
where
³
´2
^ 0¡ k
^ = 2 ¡ 2k
^ ¢k
^ 0 = 2 (1 ¡ cos µ)
k
Making the usual substitution ¹ = cos µ;the integral is:
Z
Z +1
¡
¢
¡
¢
2
2
1 + ¹ (1 ¡ ¹) d¹dÁ = 4¼
1 ¡ ¹ + ¹2 ¡ ¹3 d¹
¡1
= 4¼
µ
¹¡
µ ¶
8
= 4¼
3
and then
¶ ¯+1
¹2
¹3
¹4 ¯¯
+
¡
2
3
4 ¯¡1
µ ¶
8
3 n e¾ T
kT 2
I1 = 4¼
kT x 2 = 2ne ¾ T
x
(13)
2
3 16¼ mc
mc2
³Next we
´ turn to the term in ¢: In principle, we could proceed the same way, but the term
0 ^
^
in k ¡k ¢ p
~ is much harder to work with than its square was. So instead we use the
method from Kompaneetsz original paper (also your text, p 215).
First note that since we are considering scattering only, the total number of photons is
conserved. The total number of quanta is found by integrating the phase space density over
momentum space.
µ
¶2
Z
Z
h!
h
N = n (!) 4¼
d! _ nx 2 dx
c
c
So the conservation of photon number is expressed as
Z
d
nx 2 dx = 0
dt
11
or, equivalently,
@n
@ ¡ 2 ¢
x2
+
x j =0
(14)
@t
@x
where j is the -ux of photons in momentum space. (This is like the conservation law for
charge or mass. Interpret x 2 dx as the volume element dV in spherical coordinates. The
density can change only by -ux of particles into or out of the volume.) Now we need j: Note
that in equilibrium j must vanish identically. But we know what n looks like in equilibrium.
Kompaneets just assumed neq to be a Planck funtion, but thatzs not quite right. When the
photon number is conserved, the equilibrium distribution is a Bose-Einstein distribution
1
neq = ®+x
(15)
e
¡1
where ® depends on the occupation number N as ® _ 1=N : So when the occupation number
is small ® À 1 and
neq _ e¡ x
the Wien law.
kT
Next we write the as-yet-undetermined integral as ne ¾ T mc
2 I; so the Boltzmann
equation becomes:
¡
¢¢
1 @n
kT ¡ 2 00
= ne ¾ T
x n + 2x 2 n0 (1 + n) + n0 I + n (1 + n) x2 + I
(16)
2
c dt
mc
which must look like equation 14:
@n
j
= ¡j 0 ¡ 2
(17)
@t
x
and thus j has the form:
j = g (x) (n0 + h (n; x))
so that j 0 has a term in n00 nut no higher derivatives. We can determine the function h by
looking at the equilibrium form (equation 15). For this function:
µ
¶
¡e®+x
1
0
neq =
+ 1 n2eq = ¡neq (1 + neq )
2 = ¡
neq
(e®+x ¡ 1)
If j is to be zero in equilibrium, then h (n; x) = n (1 + n) : Now we compare equations 16
and 17 in detail.
j 0 = g 0 (n0 + n (1 + n))+g (n00 + n0 (1 + n) + nn0) = g 0 (n0 + n (1 + n))+g (n00 + n0 + 2nn0)
and so
¡j 0 ¡
2j
x
=
=
2g 0
¡g 0 (n0 + n (1 + n)) ¡ g (n00 + n0 + 2nn0 ) ¡
(n + n (1 + n))
x
·
µ
µ
¶
¶¸
³
2
g´
¡ gn00 + n0 g0 + g 1 + 2n +
+ n (1 + n) g 0 + 2
x
x
Comapring the n00 terms in the two equations:
g = ¡ne¾ T c
12
kT 2
x
mc2
and comparing the n0 term:
µ
¶
¢
2
kT ¡
g 0 + g 1 + 2n ¡
= ¡ne ¾ T c 2 2x2 (1 + n) + I
x
mc
and the n (1 + n) term:
ne ¾ T c
¢
kT ¡ 2
x +I
2
mc
(18)
³
g´
= ¡ g0 + 2
x
kT
= ne ¾ T c 2 (2x + 2x)
mc
From this last relation, we ®nd
I = (4 ¡ x) x
You should convince yourself that this also satis®es the n0 term condition (18).
Notice that I (the term in ¢) gives us the secular change in frequency, which is positive
for 4 > x (4kT > h!) and negative for 4 < x (4kT < h!); in agreement with equation 4.
The term in ¢2 may be thought of as the random walk change in energy. The ®nal result is
the Kompaneets equation:
¢¢
@n
kT 1 @ ¡ 4 ¡ 0
= ne ¾ T c 2 2
x n + n + n2
(19)
@t
mc x @x
It is valid under the conditions that
1. The electrons are non-relativistic (kT =mc2 ¿ 1) and
2. we are in the Thomson limit h!=mc2 ¿ 1:
The Kompaneets equation can always be solved numerically, but a few important cases
can be treated analytically.
Evolution of total energy
If the occupation number is small, n ¿ 1; then the
Kompaneets equation (19) can be simpli®ed:
¢
@n
kT 1 @ ¡ 4 0
= ne ¾ T c 2 2
x (n + n)
@t
mc x @x
The total energy radiated is:
Z
4
(kT )
E=
nxx 2 dx4¼ 3
c
where x2 dx represents the volume element in momentum space. Then the rate of change of
the energy is:
Z
4
@E
@n 3
(kT )
=
x dx4¼ 3
@t
@t
c
At low frequencies, x ¿ 1; we also have n0 » n=x À n: This is true throughout most of the
13
spectrum, and so
Z
4
kT 1 @ ¡ 4 0 ¢ 3
(kT )
x n x dx4¼ 3
mc2 x2 @x
c
5 Z
(kT )
@ ¡ 4 0¢
= ne ¾ T 4¼
x n xdx
4
mc
@x
Z 1
¶
5 µ
¯
(kT )
5 0 ¯1
4 0
= ne ¾ T 4¼
x
n
¡
x
n
dx
0
mc4
0
µ
¶
Z 1
5
¯
(kT )
4 ¯1
3
= ne ¾ T 4¼
0¡ x n 0 +
4x ndx
mc4
0
kT
= ne ¾ T c 2 4E
mc
kT 4E
=
mc2 tc
where tc = 1=ne¾ T c is the mean time between scatterings. Thus the energy increases
exponentially:
µ
¶
kT t
E (t) = E0 exp 4 2
mc tc
The result is true provided that the input frequencies satisfy h!=kT ¿ 1: The quantity
inside the exponential is:
(mean fractional energy change per scattering)£(mean number of scatterings)
@E
@t
=
n e¾ T c
Frequency shifts in bremsstrahlung plus scattering.
We have previously de®ned two
signi®cant frequencies: º B and º t ; such that:
º < ºB
optically thick
Iº _ B º _ º 2 in R-J limit
º = ºB
¿B = ¿ s
ºB < º < ºt
Iº _ º
º = ºt
¿ B = 1=¿ s
º > ºt
optically thin, translucent
Iº _ exp (¡hº kT )
Now we introduce a third frequency º coh which de®nes the region in which incoherence
is important. The de®nition is
y (º coh ) ´ 1
(20)
For º > º coh there are enough scatterings per mfp for a substantial energy shift to occur.
From equation 20,
4kT 2
¿ =1
(21)
mc2 y
where (equation 6)
®s
®
p s
¿y = p
=
® º (®º + ® s)
® º 1 + ®s =®º
®s
®s
'
¿ 1 for
¿1
®
®º
rº
®s
®s
'
À 1 for
À1
®º
®º
14
Notice that if kT < mc2 ; we only need to consider the case ¿ y > 1; which means
®s =®º > 1:
If x coh À 1; then since most photons are produced at x . 1 ¿ xcoh ; the impact of
incoherence on the spectrum is minimal and we need not consider it. Thus the interesting
case has x coh < 1:
Recall
® s = 6:6 £ 10 ¡25 n
and
´
n2 ³
¡hº=kT
1
¡
e
in cgs units
v 3 T 1=2
¢
n2 ¡
= 4 £ 10¡ 23 3 1=2 1 ¡ e¡ x
x T
For x < 1 we may approximate the exponential:
®º
= 4 £ 108
® º = 4 £ 10 ¡23
n2
x 3 T 1=2
(x) = 4 £ 10 ¡23
n2
x2 T 1=2
Thus
®s
6:6 £ 10 ¡25 n 2 1=2
0:0 165 2 1=2
=
x T
=
x T
¡23
2
®º
4 £ 10
n
n
For a ®xed n and T ; we may write this as:
µ
¶2
®s
x
=
®º
xB
since ® s =® B = 1 at x = x B : Then equation 21 becomes:
4kT ®s
= 1
mc2 ® º
µ
¶2
4kT x coh
= 1
mc2 x B
or
mc2
x coh = x B
À xB
4kT
Now as frequency increases above xcoh ; y increases. For y À 1 there will be enough
scatterings for the spectrum to }saturate} , i.e. we get a Wien spectrum, n (x) _ e¡x : Then
Iº =
2hº 3
2hº 3 ¡® ¡h º=kT
n
(x)
=
e e
c2
c2
and the spectrum looks like:
15
The height of the peak is determined by the factor e¡® (= 25 in this plot) which we have
yet to ®nd.
Total -ux:
The photons are all generated by Bremssrahlung, but the energy of each
is increased by scattering. For large y; almost all the photons end up with an energy near
3kT ; so
Z
dE
4¼jº
» 3kT
dº
dtdV
hº
³
p ´
where jº is the emission coef®cient for bremsstrahlung, jº _ n2 = T exp (¡hº=kT ) :
Integrating over º; we ®nd
Z
p
"f f =
4¼j º dº = "o n2 k T =h
and so
Z
dE
g (º ) exp (¡hº=kT )
» 3" f f
dº = A" f f
dtdV
º
where the ampli®cation factor A is
µ
¶
Z 1
e¡x
3 2 2:25
A=3
g (x)
dx = ln
x
4
x coh
xcoh
as discovered by Kompaneets in 1957.
For a thin cloud with x t ¿ 1;
while for x t À 1
where R ¤ is de®ned by
I»
dE R
dtdV 4¼
I»
dE R¤
dtdV 4¼
¿ eff (R¤ ; x = 3) = 1
Since all photons end up around x = 3; that is the appropriate frequency at which to
compute the optical depth.
Unsaturated Comptonization
If y & 1 but xcoh ¼ 1; we do not get a Wien spectrum.
16
Wezll look for a steady state solution in the case that photons are produced by a source
within the cloud and ultimately escape after scattering. Thus we set @=@t ´ 0 in equation
19, but add source and sink terms.
¢¢
kT 1 @ ¡ 4 ¡ 0
0 = ne ¾ T c 2 2
x n + n + n2 + Q (x) ¡ S (x)
mc x @x
In addition, if n ¿ 1; we may neglect the terms in n2: The sink term S (x) has the form
n
n
n
=
=
tescape
tc ompt £ (number of scatterings)
tcompt £ max (¿ ; ¿ 2)
where tcompt = 1=ne ¾ T c is the mean time between scatterings. Then from the de®nition of
y (equation 5) we get:
n e¾ T c
4kT n
S (x) = n
=
n e¾ T c
max (¿ ; ¿ 2 )
mc2 y
and the equation becomes:
¢
1 @ ¡ 4 0
4n
x (n + n) + q (x) ¡
=0
x 2 @x
y
where q = Qtcompt :
If the source is soft, q (x) ! 0 for x > x s .: Then for xs ¿ x ¿ 1 we ®nd
¢ 4n
1 @ ¡ 4 0
1 @ ¡ 4 0 ¢ 4n
x (n + n) ¡
' 2
x n ¡
=0
2
x @x
y
x @x
y
and the equation simpli®es to
4n
x2 n00 + 4xn0 ¡
=0
y
This equation has a power law solution, n / xp where
4
p (p ¡ 1) + 4p ¡ = 0
y
with
p
¡3 § 9 + 16=y
p=
2
There is a particularly nice result for y = 1
¡3 § 5
p=
= ¡4; 1
2
The positive power is not physically possible, because it implies in®nite energy in the
spectrum. Thus the solution we need is
n / x ¡4
or
F º / x 3 n / x ¡1
This is a power law that ®ts the observed spectrum of a number of sources.
17