1 Radiative transfer etc
Last time we derived the transfer equation
dIº
= S º ¡ Iv
d¿ º
R
where Iº is the intensity, Sº = jº =®º is the source function and ¿ º = ® º d` is the optical
depth.
The formal solution is
Z ¿
0
¡¿
Iº (¿ ) = Iº (0) e +
Sº (¿ 0 ) e ¡(¿ ¡¿ ) d¿ 0
(1)
0
1.1
Special cases: No emission
If Sº ´ 0; then the intensity is given by the ®rst term in equation 1. The intensity decreases
exponentially. The Earthzs atmosphere provides an example. The optical depth varies with
angle: ¿ = ¿ 0 sec µ where ¿ 0 is the otpical depth along a vertical path. Scattering is very
important in this situation, so wezll defer further discussion for the moment.
1.2
Optically thin source
When ¿ ¿ 1; the solution reduces to:
Z ¿
Iº (¿ ) = Iº (0) +
Sº (¿ 0 ) d¿ 0 ¼ Iº (0) + Sº ¿ = Iº (0) + j º L
0
where L is the total path length through the source.
Example: HII region in the radio. The emission is due to thermal bremsstrahlung (more
later)
n2
j º = 6 £ 10 ¡39 1=2 e¡ hº=kT erg/(cm3 ¢ s ¢ ster ¢ Hz)
(2)
T
and in the radio (hº ¿ kT , T ¼ 10 4 K)
®º ' 0:01
n2
T 3=2 º 2
cm¡ 1
(3)
Thus for typical values of n » 7000 cm¡3 ; T » 10 4 K, and º » 2 GHz, we have
7000 2
cm¡1 = 1: 2 £ 10 ¡19 cm¡1
106 £ 4 £ 10 18
and for a path length of 1/2 pc,
® º » 0:01
¿ = 1:2 £ 10¡ 19 £ 1:5 £ 10 18 = 0: 18
Thus we can compute the emitted intensity by a straightforward multiplication of emission
coef®cient times path length.
1
1.3
Optically thick source
If the source function is a constant, and ¿ À 1; then
Z ¿
¡
¢
0
¡¿
Iº (¿ ) = Iº (0) e +
Sº (¿ 0) e¡( ¿¡ ¿ ) d¿ 0 = Iº (0) e¡¿ + Sº 1 ¡ e¡¿
0
'
Sº
and so the intensity equals the source function in this case.
1.4
Thermal emission
In thermal equilibrium, we reach a situation in which the intensity does not change within
the source, and thus
Iº = Sº
Thus very large optical depth corresponds to thermal equilibrium. We know what the
intensity is in this case: it is just the black body (Planck) function:
2hº 3 =c2
eh º=kT ¡ 1
and thus when we have thermal radiation, the source function equals the planck function.
This is a fundamental relation, since j º and ®º involve only the microphysics of the emission
and absorption processes. It follows that
Iº = Bº (T ) = Sº =
j º = ® º Bº (T )
(4)
Thus once we ®nd j; we can ®nd ® immediately (or vice versa). That is how I obtained
equation (3) from equation (2).
Now the transfer equation for a medium in thermal equilibrium has become
dIº
= Bº (T ) ¡ Iv (¿ )
d¿ º
Now R&L equation 1.13 for a uniformly bright sphere becomes:
µ ¶2
R
Fº = ¼Bº
r
and integrating over frequency
µ ¶2
R
F = ¾T 4
r
where ¾ = 5:67 £ 10¡ 5 erg=(cm2 ¢ deg 4 ¢s)
Sometimes we use the relation Iº ´ Bº (T b ) to de®ne a temperature called the brightness
temperature Tb : Radio astronomers like to do this. The brightness tempertaure equals the
thermodynamic temperature only in the optically thick case. The general solution 1 may be
written as:
¡
¢
T b = T bo e¡¿ + T 1 ¡ e¡¿
or, if there is no background source,
¡
¢
Tb = T 1 ¡ e¡ ¿
2
from which it is clear that the brightness temperature can never exceed the thermodynamic
temperature, and T b approaches T as ¿ ! 1:
(Never is a strong word. Perhaps we shouldbe more careful. Exceptions may occur if
the source is not at least locally in thermodynamic equilibrium. More later)
low frequency limit:
hv=kT ¿ 1: We obtain the Rayleigh-Jeans law:
Bº (T ) '
2º 2
kT
c2
high frequency limit:
hv=kT À 1: Here we obtain the Wien law
2hº 3 ¡hº=kT
e
c2
which shows that the spectrum decreases exponentially at high frequencies.
Wien displacement law
The peak of the spectrum is found by setting the derivative to zero:
Bº (T ) '
c2 dBº
2h dº
=
=
3º 2
hº 3 ehº=kT
¡
¢2
ehº=kT ¡ 1
kT ehº=kT ¡ 1
µ ³
¶
´ hº
º2
hº=kT
hº=kT
¡1 ¡
e
=0
¡
¢2 3 e
kT
eh º=kT ¡ 1
¡
We can solve this equation numerically:
x
3
is small for x > 1; we expect that x ' 3: So use this as a
1 ¡ e¡x =
where x = hº=kT : Since e¡ x
®rst guess, and then evaluate
The results are:
try
x
1
3
2
2: 8506
3
2: 8266
4
2: 8224
5
2: 8216
6
2: 82147
7
2: 8214
so the result is hº
¡
¢
xnext = 3 1 ¡ e¡ x
xnext
¡
¢
¡3
3
1
¡
e
=
¡
¢ 2: 8506
3 ¡1 ¡ e¡2: 8506 ¢ = 2: 8266
3 ¡1 ¡ e¡2: 8266 ¢ = 2: 8224
3¡ 1 ¡ e¡2: 8224¢ = 2: 8216
3¡ 1 ¡ e¡2: 8216 ¢ = 2: 82147
3 ¡1 ¡ e¡2: 82147 ¢ = 2: 82144
3 1 ¡ e¡2: 82144 = 2: 82144
= 2:82144kT : The algorithm converges quickly.
Now try the same thing to ®nd wavelength of the maximum. Write the spectrum as
B¸ =
x5
¡1
ex
3
where x = hc=¸kT : Use a similar numerical algorithm to show that
x max = 4:965
and thus
¸ max T = 0:29 cm ¢ K
This is the Wien displacement law.
Suppose we calculate the wavelength corresponding to the frequency maximum. Wezd
get
hc
¸ max T =
= 0:51 cm ¢ K
2:82144k
Why do we not obtain the same value for ¸ max in the two cases? Is one preferable to the
other?
2 The Einstein coef®cients
In applying our results to atomic absorption and emission, it is convenient to use the
Einstein coef®cients. They are de®ned as follows:
The Einstein B¡coef®cient B01 describes absorption:
hv
¾º =
B 01 Áv
4¼
the function Áº is the line pro®le function. It describes how the absorption is distributed
around the line center frequency. Broadening is due to|natural} broadening (radiation
damping), Doppler effect due to motion of the atoms, etc. We must also include stimulated
emission, described by B10: In terms of the Bs; the absorption coe®cient is:
hv
®º =
Á (n0 B01 ¡ n1 B10 )
4¼ v
where n0 and n1 are the number of atoms in the lower (upper) level respectively. The
Boltzmann relation relates the two populations:
n1
g1
= e¡ hº=kT
n0
g0
where g 1 and g0 are the statistical weights of the two levels. Thus
µ
¶
hv
g1 ¡hº=kT
®º =
n0 Áv B01 ¡ B10 e
4¼
g0
Emission is described by the Einstein A coef®cient:
hv
jº =
n1 A10 Áº
4¼
where A10 is the transition probability per unit time for spontaneous emission.
Now we know that
jº
n1 A10
A
2hº 3 =c2
³
´ = g0 hº=kT 10
= Bº (T ) =
= hº=kT
®º
e
B01 ¡ B10
e
¡1
n0 B 01 ¡ B10 g1 e¡h º=kT
g1
g0
4
Thus we obtain the relations
B01 =
and
A10 =
or
g1
B10
g0
2hv 3
B10
c2
¸2
A10
2hº
Interesting things can happen when the populations are not goverened by the Boltzmann
relation, i.e. when some non-thermal process operates. For example, if the population is
inverted (n1 > n0 ), the stimulated emission dominates and we can obtain maser emission.
B10 =
3 Scattering
As we have mentioned, scattering can increase the intensity (due to scattering from other
directions) and can decrease the intensity (due to scattering out). We can de®ne an emission
coef®cient for scattering as
jº;scatt = ns ¾ º Jº
where ns is the number density of scatterers and ¾ º is the scattering cross section. Notice
that the mean intensity appears here, since photons going in any direction can be scattered
into the beam. For the moment I shall assume that the scattering cross section has no
dependence on direction. This is close to the truth for Thomson (electron) scattering. Then
the transfer equation for scattering only becomes:
dIº
= ns ¾ º (J º ¡ Iº )
ds
Including both absorption and scattering, we have:
dIº
= ns ¾ º (Jº ¡ Iº ) + jº ¡ ® º Iº
ds
= ns ¾ º J º + jº ¡ (® º + ns ¾ º ) Iº
We can write
ns ¾ º = ® s
Then
dIº
= ®s Jº + ® º Bº ¡ (® º + ®s ) Iº
ds
Now letzs de®ne a source function
® sJ º + ®º Bº
Sº ´
®º + ®s
so that the equation takes the form
dIº
= (® º + ®s ) (Sº ¡ Iº )
ds
De®ning d¿ = (®º + ® s) ds; we retrieve the previous transfer equation in terms of ¿ and S:
The solution is more complicated, however, because Sº depends on Jº; and hence on Iº :
5
Note here that 1/(® º + ®s ) is the average distance a photon will travel before being
removed from the beam by either an absorption or a scattering event.
1
l=
® º + ®s
On the other hand, the photon will travel a longer distance
1
la =
®º
before being absorbed. The path will be a random walk. Each step of the walk will end by
a scattering, with probability
®s
ps =
® º + ®s
or with an absorption, with probability
®º
pa =
®º + ® s
Thus the source function is
Sº = (1 ¡ p a )J º + pa Bº
3.1
Theory of random walks
A particle has a mean step length `: The total displacement is
~ =
D
N
X
~i
D
i=1
and the mean distance travelled is
<
vÃ
! ÃN
!
u N
¯ ¯
p
u X
X
¯~ ¯
~ ¢D
~ >=< t
~i ¢
~i
D
D
¯D¯ >=< D
i= 1
=
v
u N
N
N
X
X
uX
~i ¢ D
~i +
~i ¢ D
~j >
<t
D
D
p
i= 1
p
i=1
i= 1 j=1; j6=i
N `2 = N ` = L
~ i and D
~ j are not correlated, and thus the average value of D
~i ¢ D
~ j is
since the directions of D
zero.
In our problem, the walk starts when the photon is created and ends when it is absorbed.
The step length is l = 1= (® º + ®s ) . The probability that the path ends, at any one event, is
pa : Thus the total number of steps expected is 1=p a : The mean distance travelled is:
r
r
1
1
® º + ®s
1
1
L=
=
= p
pa (® º + ®s )
®º
(® º + ®s )
® º (®º + ® s)
=
Thus the effective optical depth due to absorption and scattering is
p
1
¿ eff =
= ®º (® º + ®s )
L
6