Astro 300B: Jan. 24, 2011
Optical Depth
Eddington Luminosity
Thermal radiation and Thermal Equilibrium
Radiation pressure: why cos2?
Each photon of energy E=hn has
momentum hn/c
Want the component of the momentum normal
To direction defined by dA, this will be
hn
cos
c
Pressure = net momentum normal to dA/time/area
or the net energy x cos /c /time/area

The net energy = flux in direction n, i.e.
F
 In cosd
n
OPTICAL DEPTH
It’s useful to rewrite the transfer equation in terms of the optical depth:
d


dl
n
n
l
n(l)
l)d
l
n(
or
lo
L
2



n


I
(
L
)

I
(
L
)
e

j
(
l
)
e
d
l
n 2 n1
n


n
L
1
emergent
incident
n
>1: optically thick
opaque, typical photon will be absorbed
n
<1: optically thin
transparent, typical photon will traverse
the medium without being absorbed
n
Sn
Source Function
Define
Sn 
jn
n
so that the Equation of Radiative Transfer is
Which has solution

n

dI
n

In
S
n
d
n





(



)
n
n
n


I
(
)

I
(
0
)
e

e
(
)
d
nn n
n
n
n
 S
0
In S
n
If
In S
n
If
so
In
S
n
then
dIn
0
dn
Specific intensity decreases
along path
then
dIn
0
dn
Specific intensity increases
along path
If
 >>1,
In S
n
Mean Free Path
The mean free path,
n
n
is the average distance a photon travels before
being absorbed
Or in other words, the distance through the absorbing material
corresponding to optical depth = 1
n 1
Recall
so
also
d


dlwhere  n
n
n


1
n
n
n
or

N

n
n
#absorbers/Vol
Cross-section for absorption
= absorption coefficient
n 
1
n
hence
1
n 
N
n
1
n 
N
n
Makes sense:
If N increases,
If σν increases,
n
n
decreases
decreases
Radiation Force: The Eddington Limit
see R&L Problem 1.4 and p. 15
•
•
Photons carry momentum
When radiation is absorbed by a
medium, it therefore exerts a force
upon it
Consider a source of radiation, with luminosity L
(ergs/sec)
And a piece of material a distance r from the source
Each photon absorbed imparts momentum = E / c
Specific flux = Fν
ergs s-1 cm-2 Hz-1
Momentum flux = Fν / c
momentum s-1 cm-2 Hz-1
Momentum imparted by absorbed photons =
Where
n
n Fn
c
= absorption coefficient, cm-1
n Fn
Momentum /area /time /Hz /pathlength
through absorber
c
Now, area x pathlength = volume
so
n Fn
c
Is momentum/time /Hz /volume
But momentum /time = Force
So, integrating over frequency, the Force/volume imparted by the
absorbed photons is
F
n


d
n
n

c
Likewise, in terms of the mass absorption coefficient, κ
1
f 

n
nF
nd
c
An important application of this concept is
the Eddington Luminosity, or Eddington Limit
This is the maximum luminosity an object can have before it ejects
hydrogen by radiation pressure
Eddington Luminosity
c.f. Accretion onto a black hole
When does fgrav = fradiation?
r
M,L
Force per unit mass = force per unit mass
due to gravity
due to absorption of radiation
fradiation =
 F
c
 L

c 4r2
F = radiation flux, integrated over frequency
L = luminosity of radiation, ergs/sec
r = distance between blob and the source
Κ = mass absorption coefficent
fgravity =
So…
GM
2
r
M = mass of the source
GML 
 2
2
r 4
r c
M 

L 4Gc
Define Eddington Luminosity = the L at which f(gravity) = f(radiation)
Led d 
4

GcM
Led d 
4

GcM
Maximum luminosity of
a source of mass M
Note: independent of r
A “minimal” value for κ is the Thomson cross-section,
For Thomson scattering of photons off of free electrons,
assuming the gas is completely ionized and pure hydrogen
Other sources of absorption opacity, if present, will contribute to
larger κ, and therefore smaller L
Thomson cross-section
σT = 6.65 x 10-25 cm2
Independent of frequency (except at very high frequencies)
T 
T
mH
Where mH= mass of hydrogen atom
GMcm
H
L

4

edd

T
GMcm
H
L

4

edd

T
If M = M(Sun), then Ledd = 1.25 x 1038 erg/sec
Compared to L(sun) = 3.9 x 1033 erg/sec
Another example of a cross-section for absorption:
Photoionization of Hydrogen from the ground state
H atom + hν  p + e-
Only photons more energetic than threshold χ can ionize hydrogen,
where
χ = 13.6 eV
912 Å
1 Rydberg
Lyman limit
ν1 = 3.3x1015 sec-1
The cross-section for absorption is a function of frequency,
n


where ν =3.3x10


6
.
63

10
  cm
n
n


3

18
1
2
1
15
sec-1
More energetic photons are less likely to ionize hydrogen than photons at
energies near the Lyman Limit
Note: αν : photon-particle cross-sections; σν: particle-particle cross-sections
Similarly, one can consider the ionization of
He I  He II
He II  He III
Thresholds:
Hydrogen hν = 13.6 eV 912 Å
Helium I
24.6 eV 504 Å
Helium II
54.4 eV 228 Å
For HeI α(504 Å) = 7.4 x 10-18 cm2
declines like ν2
For HeII α(228 Å) = 1.7 x 10-18 cm2
declines like ν3
Thermal Radiation, and
Thermodynamic Equilibrium
Thermal radiation is radiation emitted by matter in thermodynamic
equilibrium.
When radiation is in thermal equilibrium, Iν is a universal function of
frequency ν and temperature T – the Planck function, Bν.
Blackbody Radiation:
In B
n
In a very optically thick media, recall the SOURCE FUNCTION
jn
S

In
n

n
So thermal radiation has

S

B
an
j

B
d
n
n
n
n
n
And the equation of radiative transfer becomes
 
d
I
d
I
n
n


I

B
or


I

B
(
T
)
n
n
n
n
n
n
dl
d
n
THERMODYNAMIC EQUILIBRIUM
When astronomers speak of thermodynamic equilibrium, they mean a lot
more than dT/dt = 0, i.e. temperature is a constant.
DETAILED BALANCE: rate of every reaction = rate of inverse reaction
on a microprocess level
If DETAILED BALANCE holds, then one can describe
(1) The radiation field by the Planck function
(2) The ionization of atoms by the SAHA equation
(3) The excitation of electroms in atoms by the Boltzman distribution
(4) The velocity distribution of particles by the Maxwell-Boltzman distribution
ALL WITH THE SAME TEMPERATURE, T
When (1)-(4) are described with a single temperature, T, then the system is
said to be in THERMODYNAMIC EQUILIBRIUM.
In thermodynamic equilibrium, the radiation and matter have the same
temperature,
i.e. there is a very high level of coupling between matter and radiation
 Very high optical depth
By contrast, a system can be in statistical equilibrium,
or in a steady state, but not be in thermodynamic equilibrium.
So it could be that measurable quantities are constant with time, but there
are 4 different temperatures:
T(ionization)
T(excitation)
T(radiation)
T(kinetic)
given by the Saha equation
given by the Boltzman equation
given by the Planck Function
given by the Maxwell-Boltzmann distribution
Where
T(ionization) ≠ T(excitation) ≠ T(radiation) ≠ T(kinetic)
LOCAL THERMODYNAMIC EQUILIBRIUM (LTE)
If locally, T(ion) = T(exc) = T(rad) = T(kinetic)
Then the system is in LOCAL THERMODYNAMIC EQUILIBRIUM,
or LTE
This can be a good approximation if the mean free path for
particle-photon interactions << scale upon which T changes
Example: H II Region (e.g. Orion Nebula, Eagle Nebula, etc)
Ionized region of interstellar gas around a very hot star
Radiation field is essentially a black-body at the temperature of the central
Star, T~50,000 – 100,000 K
However, the gas cools to Te ~ 10,000 K
(Te = kinetic temperature of electrons)
O star
H II
HI
Q.: Is this room in thermodynamic equilibrium?