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Lecture 8
Optical depth
Review: opacity
Opacity (k) is a cross-section per unit mass (units m2/kg)
of material for absorbing photons of a specific
wavelength.
dI   k  I  ds
l
1
k 
is the mean free path
How does the intensity of radiation depend on opacity and distance
travelled through a homogeneous medium?
I   I  , 0 e k   s
After the photon has traveled one mean free path its intensity will have
decreased by a factor e-1=0.37.
Optical depth
The optical depth is a unitless quantity that relates to
how much light is absorbed along its path s.
d  k  ds  ds / l
The difference in optical depth between the initial and
final position of a light ray is then given by:
     , f    ,0
s
   k  ds
0
Note the optical depth is greater at the initial position
than at the final. i.e. the optical depth decreases as
the photon moves toward the observer.
Optical depth
For a stellar atmosphere, we can set    0
at the surface.
  0
Thus, for a photon which traveled a distance s through
the stellar atmosphere, the initial optical depth is
s
    k  ds
0
and the intensity of the light ray can be expressed as:
I   I  ,0ek  s  I  ,0e 
Optical depth
If  and k are approximately constant over the photon’s path,
s
    k  ds
0
 k  s
 s/l
where
l
1
k 
is the mean free path
The optical depth can be thought of as the number of mean free
paths a photon takes between its initial position and the surface.
Random walk
As a photon moves through a dense gas, it interacts with
atoms being scattered, absorbed, and re-emitted.
If each scattering process is
random, how far does the
photon move (on average)?
s  Nl
s
Random walk
Recall that the optical depth is approximately the
number of mean free paths traveled:
  s / l
The number of interactions a photon has on the way to the surface
is
2
s
N      2
l
Thus a photon can escape to the surface
 1
when
and the number of steps required increases as the square of
the distance travelled
A more careful calculation (coming soon) will show that the average
optical depth from which photons originate is    2 / 3
Optical depth: example
Using the numbers we used
before for the solar
photosphere, approximately
how far into the Sun
corresponds to an optical
depth of =1?
The photons we see
represent the
conditions very close
to the Sun’s surface,
0.00022RSun.
Sources of opacity
The opacity is determined by the physical processes
that can remove photons from a beam of radiation. In
general this includes true absorption processes, as well
as scattering.
1.
2.
3.
4.
Bound-bound transitions
Bound-free transitions
Free-free absorption
Electron scattering
The Balmer jump
The energy of an electron in the n=2 orbit of a hydrogen
atom is E   13.6  3.40eV
22
So the atom can be photoionized by any photon with
wavelength
hc

2
 364.7nm
This gives rise to the
Balmer jump in stellar
spectra
Note the jump is not
present in hotter stars,
where most of the
hydrogen is already ionized
The Lyman limit
Similarly, the energy of an electron in the n=1 orbit of a
hydrogen atom is E  13.6eV
So the atom can be photoionized by any photon with
wavelength   91.2 nm
This is a very efficient process, and it takes very little neutral hydrogen to
absorb all photons with this energy.
For stars more than a few pc away, there is sufficient neutral hydrogen
between them and us to almost all the light at these UV wavelengths. We
therefore know little about the far-UV flux of hot stars.
Typical sources of opacity
In most stellar atmospheres, the primary source of
continuum opacity is the photoionization of H- ions
In cool stars, molecules can form. These complex arrangements of
atoms can make many bound-bound and bound-free transitions
which greatly increase the opacity
Rosseland mean opacity
The total opacity is the sum of all these sources. It
depends on the wavelength of the light being absorbed,
as well as the composition, density and temperature of
the gas.
• A weighted average over all
wavelengths gives the
Rosseland mean opacity k
Rosseland mean opacity
•
•
•
Opacity increases with density at
fixed T
At fixed density and low
temperatures, opacity rises
steeply as T increases
After the peak, the opacity
decreases like
k  T 3.5
•
•
The bump at T~40,000 K is due to
the second ionization of He
At the highest temperatures, the
opacity is due to electron
scattering
log10 
Break
Absorption lines and Kirchoff’s laws
Photons we see originate (on average) from an optical depth of 2/3.
Recall the optical depth is related to the opacity along the photon path by
s
    k  ds
0
If the opacity is much higher at some wavelength 0 (for example due to
bound-bound absorption) the photons we see will originate from closer
to the surface
0
Low opacity High opacity

Low opacity
Absorption lines and Kirchoff’s laws
This doesn’t make any difference if the temperature of the gas is uniform:
the luminosity radiated from all points in the gas is the same, so it
doesn’t matter where the photon originates from
However – if the temperature is lower near the surface, the luminosity will
be much lower:
4
L T
Thus the luminosity will be lowest at wavelengths where the opacity is the
highest: this gives rise to absorption lines.
0
Low opacity High opacity

Low opacity
Emission lines
What would be the result if the star’s temperature increased outward?
In this case – light from regions of high opacity would arise from higher in
the atmosphere, where the temperature (and therefore luminosity) is
higher. We would therefore see an emission line
0
Low opacity High opacity

Low opacity
The Solar Atmosphere
• The solar atmosphere extends thousands
of km above the photosphere (from which
the optical radiation is emitted)
T~106 K
The temperature increases above the
photosphere (in the chromosphere and
corona)
We can therefore see emission lines from
these regions
T~25000 K
T~5770 K
Core
T~107 K
HeI emission
from the
chromosphere
(~20000K)
Line profiles
• The region near the central
wavelength is the core of the line
• The sides which join up with the
continuum are the wings.
• The area of the line is related to
the amount of absorption
wings
core
 Fc  F 
W   at higher(and
d
• The central regions of the line must be formed
cooler) regions of the atmosphere. The wings probe
c
 Fdeeper
 into
the atmosphere.
Line profile
Absorption lines are due to bound-bound atomic
transitions of fixed energy. Why are the lines not
infinitely narrow?
•
•
•
Natural broadening
Pressure or collisional broadening
Doppler broadening
Voigt profile
The absorption line profile is due to the Doppler and
damping (pressure and natural) broadening. The
resulting profile is known as the Voigt profile.
Pressure broadening: example
For supergiant stars, the atmospheres are ~1 million times less dense
than in the sun. So the pressure broadening is weaker by this
factor, which is why more luminous stars have much narrower lines
(the cores are still dominated by Doppler broadening).
Reminder: Spectroscopic parallax
In principle, you can identify
both the spectral class and the
luminosity class from the
spectrum.
• This can be used to determine
the distance to the star. This
method is known as spectroscopic
parallax
Solar abundances
From an analysis of spectral lines, the following are the most abundant
elements in the solar photosphere
Element
Hydrogen
Atomic
Number
1
Log Relative
Abundance
1
Column Density
kg m-2
11
Helium
2
-1.01
43
Oxygen
8
-3.07
0.15
Carbon
6
-3.4
0.053
Neon
10
-3.91
0.027
Nitrogen
7
-4
0.015
Iron
26
-4.33
0.029
Magnesium
12
-4.42
0.01
Silicon
14
-4.45
0.011
Sulfur
16
-4.79
0.0057