3 Stellar Atmospheres

1
3 - Stellar Atmospheres
In addition to these ppt presentation, there are
more detailed chapters in English (on request,
.doc file), and lecture notes in Italian (pdf
file) collected by the students Lucia Marchetti
and Benedetta Vulcani in the AY 2006/07.
2
Summary stellar atmospheres theory

• The atmosphere of a star contains less than
  1x10-9 of its total mass, but it is that what we
  can see, measure, and analyze.
• Spectroscopic analyses provide elemental
  abundances and show us results of
  cosmo-chemistry, starting from the earliest
  moments of the formation of the Universe to
  present day.
• Photometric analyses are used to put a star from
  the observed color-magnitude diagram (e.g. B-V,
  V) into the theoretical HR Diagram (L,Teff) and,
  hence, to guide the theories of stellar structure
  and evolution.
• The study of stellar atmospheres is a very
  difficult task. The atmosphere is that region,
  where the transition between the thermodynamic
  equilibrium of the stellar interior into the
  almost empty circumstellar space occurs. It is a
  region of extreme non-equilibrium states.

3
Stellar Atmospheres
We call stellar atmosphere the ensemble of the
outer layers to which the energy, produced in the
deep interior of the star, is carried, either by
radiation, convection or conduction. Interacting
with the matter present in the outer layers, this
energy finally produces the observed
electromagnetic radiation, particle flux and
magnetic field. By analogy with the terminology
adopted for the Sun, a typical atmosphere can be
divided very schematically in several regions, as
in the figure.
The abscissa is the outward radial distance. The
ordinate is the temperature. Both scales vary
with the stellar spectroscopic type.
4
Schematics of temperature and density in the
solar atmosphere
Schematics of the solar temperature profile
(thick line) with height z. The zero level is the
conventional surface. Notice the sudden increase
of T in the transition region between
chromosphere and corona. The dotted line gives
the matter density profile ?(z). See also in
later slides.
5
Energy transport mechanisms

• A more physical approach to this subdivision
  would make use of the main energy transport
  mechanism in each region
• in the photosphere, the energy is transported by
  radiation (this assumption is equivalent to an
  outward decreasing temperature), and the geometry
  is well approximated by plane-parallel
  stratification
• in the chromosphere, there is also dissipation of
  waves (acoustic, hydrodynamic)
• above a very sharp transition layer, in the
  corona, the magnetic field energy is very
  important (magneto-hydrodynamics). Large scale
  motions with velocities larger than the escape
  velocity give origin to a loss of particles known
  as stellar wind, and the plane parallel
  approximation is certainly untenable.

6
Limitations of the present treatment
In the following, we will concentrate our
attention essentially on the photosphere and the
chromosphere, and in the visible region of the
spectrum. This limitation can be justified a
posteriori by the overall energy coming out of
the various layers, as schematically indicated in
Table 1, valid for the Sun. Moreover, the
present treatment will be appropriate to the
formation of the continuum of the star. Later
on, we will simply touch the much more difficult
problem of the formation of spectral lines and
bands.
7
Table 1- Solar Energy Output
In the Sun, the convective flux (granulation) is
of the order of 1 and the conductive flux is of
the order of 10-5 of the radiative flux. The
solar wind velocities range from 400 to 800 km/s.
The mass loss is about 1036 particles per
second ? 1012 g/s ? 10-14 M?/y.
On other spectral type stars, the situation can
be very different (for instance, see the lectures
by Y. Nazé).
8
Basics on Transport of radiation - 1
Consider a surface area of size , having normal
n, and an elementary solid angle d? in direction
(?, ?), where
The light passing through d? in the unit time, of
wavelength between (?, ?d?), carries an energy
E? that can be written as
(dimensions erg/s)
where the factor cos? comes from the surface
projection effect.
9
Basics on Transport of radiation - 2

• The quantity I?(? , ?) is called intensity of
  the radiation field it is the energy that flows
  each second through the unit area d?, in the
  wavelength interval d? centered at ?, into the
  unit solid angle d? in direction n(? , ?) to
  the normal n.
• The units of I in our mixed cgs system are
  erg?cm-2?s-1A-1?sr-1.
• In addition to the spherical coordinate system, a
  Cartesian representation is sometimes useful. The
  axis z is along the normal n to d?.
• In general I? I?(x, y, z, ?, ?, t) I?(x, y,
  z l,m, n t) where (l, m, n) are the direction
  cosines.
• For simplicity, in the following we shall assume
• azimuthal symmetry of the radiation field,
  namely independence from ? ,
• stationarity, namely time constancy.
• The radiation field is said isotropic if the
  intensity is independent of direction in that
  point, and homogeneous if it is the same in all
  points.
• Furthermore, no account will be taken of a
  possible polarization of the beam.
• Caveat in planetary atmospheres none of these
  simplifying assumptions is really fulfilled!

10
An idealized experiment - 1
Consider now, in a fairly idealized experiment, a
collimated beam of radiation which passes through
a volume of gas, contained in a column with
rectangular cross-section ?, length s, and
perfectly transparent walls (to avoid reflection
effects), perpendicularly to one of the faces.
The shape of the column (here shown as a
rectangular box) is irrelevant, it could be a
cylinder.

The intensity I?o entering on the left side will
be changed by absorption and emission along the
path s, and will leave on the right with an
intensity I? which depends on the optical depth
?? of the box (see later for the definition of
optical depth) and on its emission properties.
We want to determine the intensity I? of the
radiation exiting the vessel on the right.
11
Absorption and Emission
12
Elementary energy and intensity variations
(Notice we might have used equally well
frequency instead of wavelength)
13
Optical depth and Source Function
Let us introduce now the a-dimensional variable
elementary optical depth
The previous equation for the intensity variation
becomes
where the function S? ?? / ?? is called the
source function.
The total optical depth of the column is obtained
by
which clearly shows that the same geometrical
thickness can correspond to very different
optical depths at the different ?s.
14
The Source Function at equilibrium
Let us assume that the gas is in perfect
thermodynamic equilibrium, and that the passage
of the radiation field does not alter this
condition. The intensity of the beam cannot
change either, so that
In other words, under these ideal conditions, the
intensity of the radiation (and so also the
source function) is expressed by Plancks
function B?(T)
The second equality is another way of expressing
Kirchoffs law the ratio of the emissivity to
the absorption coefficient is independent from
the chemical composition of the gas. Notice that
the Planck function is isotropic we shall
maintain isotropy of S (and ?) even when the
identification of S with B is not justified.
15
Local thermodynamic equilibrium -1
Indeed, the assumption S B is very convenient,
and provides very useful indications, but it is
not always justified. In the general case, the
source function must be derived by the detailed
knowledge of the physical conditions of the
atmosphere. In particular, in a stellar
atmosphere, the strict condition of thermodynamic
equilibrium never applies. Following Milne, we
shall assume its local validity (a condition
known as LTE), with a temperature T having a well
determined value at any depth z, but changing
along the column. In other words, the source
function along the path is equal to Plancks
function, but with changing T S?B? (T) The
consequence is not entirely intuitive even if
the absorption would take place in only one
absorption line, nevertheless the emissions would
be distributed over all wavelengths according to
B? . Furthermore, the emission will be isotropic.
16
Local thermodynamic equilibrium -2
The previous expression can be written as
If moreover B? is assumed constant along the
path (as in laboratory conditions), the intensity
at the exit face of the column will be
The first term on the right-hand side is the
percentage of energy that entered the volume at x
0, and left from the front face at x s (whose
optical depth is ?? (s)). The second term is the
contribution of the emissivity of the gas.
17
Case 1 no input radiation
No input radiation means I?(0) 0, the column of
hot gas shines with intensity given by
This case can be subdivided in two limiting
situations 1.1 when the optical depth ?? (s) is
very small (optically thin gas), then
The intensity will be large only at the
wavelengths where ?? is large, namely at the
resonance lines typical of the gas at that
temperature, lines which we see in emission. 1.2
when the optical depth ?? (s) is very large
(optically thick gas), then
The intensity becomes totally independent from
the length of the column and also from the
chemical composition of the gas (namely from ??
). We observe a black body of a given temperature
T.
18
Case 1.1 no input radiation and very thin gas
Case 1.1 is typical of many astrophysical
situations, such as emission and planetary
nebulae, or the solar corona observed outside the
solar limb (e.g. during an eclipse, or with a
coronagraph occulting the disk). These gases are
all very hot, as it can be judged by the high
ionization, but we see emission lines because
their optical depth is small, not because they
are hot. Notice also that the small optical
thickness condition certainly prevails in the
continuum, and in the wings of the line.
However, at the very peak of the line the depth
can become high. The brightness will then
approach that of the black body having the
temperature of the gas.
19
Case 2 Appreciable input radiation
Appreciable input radiation I?(0) gtgt 0 (this
would be the case of a stellar atmosphere).
Again, two limiting cases can be considered 2.1
optically thin case
If the sign of the second term is negative (I?(0)
gt B?), we observe the spectral distribution at
the entrance of the column minus a fraction which
is larger where ?? is larger, namely absorption
lines superimposed on the entrance spectrum. The
interpretation is fairly straightforward assuming
that I and B are both black body functions the
temperature of the entrance radiation is higher
than the temperature of the gas in the column.
If the sign is positive (I?(0) lt B?), emission
lines, superimposed over the entrance continuum,
are observed where ?? is larger (see next
slide). 2.2 optically thick case
20
Case 2.1 - Emission lines
In case 2.1, if the sign is positive (I?(0) lt
B?), emission lines, superimposed over the
entrance continuum, are observed. This is the
case for instance of the solar spectrum seen at ?
lt 1600 A all lines are in emission, not in
absorption. Evidently, the UV absorption
coefficient becomes so large (large optical
thickness) that we only see the upper layers of
the atmosphere (the chromosphere), that must have
a source function (namely a temperature)
increasing toward the exterior, therefore higher
than that of the visible photosphere (say 12.000
K instead of 6.000 K). Notice that this
conclusion of outwards increasing temperature
doesnt come from the ionization, but simply by
the lines being in emission, instead than in
absorption as in the visible region. The observed
ionization simply confirms the conclusion. (We
defer further discussion of objects with emission
lines to the end of this chapter).
21
Absorption line spectra
Concluding this first discussion, an absorption
line spectrum is formed in
A deep optically thick gas surmounted by a
thinner layer, with source function S decreasing
outwards, as in the solar photosphere in the
present simplified interpretation, source
function means temperature. Therefore, the bottom
of the lines comes from regions higher on the
photosphere (and cooler) than the wings and the
adjacent continuum.
Absorptions can also form in an optically thin
gas penetrated by a background radiation whose
intensity is larger than the source function of
the column. This can be the case of a thin shell
around a star, or of the interstellar medium
between us and a hot star.
22
Normal stars
The previous approximate discussion has shown
that even in laboratory conditions a variety of
cases are possible. Normal stars usually
display (at least in the visible region), an
absorption line spectrum the situation in their
atmospheres must therefore correspond to the case
I?,0 gt S? (the intensity coming from the
interior is higher than the source function of
the external layers, like having a reversing
layer on top of a hotter surface). In the LTE
assumption, this also means that the temperature
of the external layers is smaller than the
temperature of the interior layers (outwards
decreasing temperature). However, although the
concept of a reversing layer maintains a
considerable intuitive validity, for real stars
the discussion is much more complex, even
assuming LTE and radiative transport, because nor
the density nor the source function are constant
inside the atmosphere.
23
The radiative transfer equation - 1
Therefore, to describe and understand the stellar
(and planetary) atmospheres, we must put the
above considerations on firmer physical and
mathematical grounds. We shall assume that the
energy coming from the interior of the star is
transported through the atmosphere by radiation
only, an assumption which is not always
justified, because other mechanisms, such as
convection and conduction are possible, but not
treated at present. Another simplification is
the assumption of a plane parallel atmosphere. In
the case of the Sun, this assumption is well
justified in the visible domain indeed, the
geometrical thickness of the solar atmosphere to
visible radiation (photosphere) is much smaller
than the solar radius (see the first slides). As
before, the radiation field is assumed stationary
and unpolarized, with azimuthal symmetry
(dependence on ? only, independence from ? . We
have already commented that planetary atmospheres
are more complex).
24
The perpendicular optical depth
It is convenient at this point to introduce a
change of perspective, because the observer sees
the radiation from the outside. In the previous
discussion appropriate to laboratory experiments,
the radiation comes into the volume at x 0 and
exits at x s. In treating the stellar
atmosphere, it is more convenient to reverse the
direction of z and ? , because the radiation
exiting from the stellar surface comes from the
deep interior (we might even consider from z ? ?)
and exits at an ideal surface z 0 toward the
observer. Therefore, the line of sight of the
observer enters the stellar atmosphere at z 0,
and proceeds as much as possible toward the inner
layers.
25
The radiative transfer equation - 2
Therefore, let us assume a given plane as the
surface of the stellar atmosphere the
geometrical position of this surface is at moment
immaterial. The unit vector n indicates the
outward normal to the plane atmosphere, and ?
the angle of a given radiation pencil with n.
As in the previous considerations, along the
path ds inside the atmosphere the following
equation will be satisfied
because of the assumed isotropy of the source
function.
26
The radiative transfer equation - 3
But now the linear coordinate z (say, in km)
increases from the surface inward, and ?? the
perpendicular optical depth, also increases
inwards along the perpendicular to the surface.
The coordinate s instead increases outwards, so
that along the beam of radiation, ??s increases
outwards at an angle ? (see figure).
where the quantity cos? is usually designated
with ?.. Notice the change in sign and the
presence of cos? with respect to previous
discussion.
27
The radiative transfer equation - 4
The radiative transfer equation for the plane
parallel case with azimuthal symmetry is then
In order to derive the intensity of the radiation
exiting the surface in a given direction ?,
namely I?(0,?), multiply both sides by
and obtain
28
The radiative transfer equation - 5
Integrating in ?? from 0 to ?, and taking into
account that for ?? going to ? the exponential
term vanishes, we finally get
Notice that in this integral equation, ? cos?
is not a variable, but a parameter the equation
provides a family of solutions, one for each
direction with respect to the normal n to the
surface of the star. The interpretation is
straightforward the intensity leaving the
surface at an angle ? results from the summation
of all the contributions of the volume elements
along the path of the light. If we measure
I?(0,cos?) , as is possible for the Sun, then by
inverting the previous equation we could derive
S. However, mathematically the inversion is
always a difficult task, not necessarily
single-valued and very sensitive to measurement
errors. Here we treat only the direct problem, by
assuming a particular functional form for S, and
deriving I.
29
A first approximation for S?(??) - 1
Let us make the simple assumption that the
unknown source function S is a linear function of
the optical depth
(this assumption could be seen as the result of
the usual technique of expanding a function in
Taylors series and considering the first order
only, but later on we will justify it on the
basis of Eddingtons approximation). We then get
the following result (using cos? for clarity)
Thus, measuring the intensity over the disk of
the star at a series of wavelengths, we derive
the two coefficients a? , b? , and therefore S?.
Notice that if I is measured in physical units,
then also S will be determined in the same units.
30
A first approximation for S?(??) - 2
In the LTE hypothesis, S coincides with the
Plancks function B at a z-dependent temperature
Therefore, measuring I we derive also the
temperature at the different optical depths.
Sometimes it is convenient (or necessary) to
loose the calibration in physical units,
normalizing to intensity at the center of the
disk
Due to the normalization, it must obviously be
w(1-w)1 .
31
Limb Darkening
The previous equations tell us that at the center
of the disk (? 0) we see the source function
S? at a depth ?? 1. At the border of the disk
(? 90) we see S? at the surface but
remembering that in the plane parallel
approximation z/s cos? , we see that ?? cos?
when ??s 1 in conclusion, at any point on the
stellar disk we always see down to a depth
corresponding to ??s 1, as in the Figure
??s 1
If the temperature in the photosphere increases
inwards, the center must be brighter than the
limb, and the temperature measured at the center
of the disk must be higher than that at the limb.
32
Solar limb darkening - 1
Let us observe the Sun, which is effectively at
infinite distance but resolved as a disk. The
radiation coming to the observer from the center
of the disk leaves the star perpendicular to the
surface, so that
The radiation coming from the borders of the
solar disk leaves the surface at ? 90, so that
The observations prove indeed that we see less
light from the border, a fact named limb
darkening. See the following figures. We then
conclude that
Therefore, in the solar photosphere, the
temperature must decreases outwards. The solar
limb darkening gave indeed the first convincing
proof ot the validity of the previous
assumptions, in particular of radiative transport
of the energy through the photosphere.
33
Images of Solar LD
The Sun is darker and redder at the limb. The
temperature is 6050 K at the center, and 4550 K
at the limb (the effective temperature Teff being
intermediate, ? 5800 K). In the visible at
5010A, I(0,0) 4x108 erg?cm-2?s-1?A-1.
34
Two reasons for the photospheric limb darkening
Therefore, we have identified two reasons for the
limb darkening 1 - optical depth 2 - temperature
gradient in the photosphere

• This figure shows two possible polar diagrams
• On the left a? b? 1
• On the right a? 0.5, b? 2
• The larger the ratio b? / a? , the more the
  radiation is in forward direction.

35
Solar limb darkening - 2
The solar limb darkening observed at wavelengths
from 3000 A to 2.6 ?m, for several values of cos?
from 0.2 (? 79) to 0.9 (? 25). Notice
the decrease with ? The limb darkening becomes
smaller in the near IR because the larger optical
depth allows to observe the regions of minimum
temperature. Notice also the irregularities in
the curve (e.g. at 3600A and 13000A) due to sharp
variations in the absorption coefficient caused
by H and H- (negative H ion).
(Adapted from Pierce and Waddell, 1961)
36
Table of solar limb darkening
37
Solar limb darkening - 3
The previous curves and table show that the first
approximation requires a (small) correction For
instance, at ? 5010, the observations provide
a 0.26 , b 0.87 plus a smaller term, which
can be expressed as 2c -0.13cos2?. A better
mathematical representation of the limb darkening
would be
38
Solar Photospheric Source Function S?(?)
The observations prove that the limb darkening
requires at least a second order term in ? .
However, it can be demonstrated (see Exercises)
that if
Therefore, the main result remains valid by
measuring the limb darkening law (namely the A?i
) we can derive the source function at each
optical depth.
The figure shows the result of this calculation
for the continuum at 5010 A, which is a region in
the solar spectrum devoid of strong absorption
lines. The calculation was made taking into
account the second-order term (in ? 2) in Pierce
and Waddell data. For ? gt 1.5 the reconstructed
curve becomes very uncertain, because we receive
very little radiation from that optical depth.
39
Solar Photospheric Temperature T(??)
In the LTE assumption, source function is
equivalent to temperature, which can thus be
derived by the same procedure. Notice that at
this stage we have only T(?? ), not T(z), in this
particular figure in the continuum at ? 5010
A. Our assumptions though require that each layer
z in the parallel stratified atmosphere has a
unique temperature.
40
Solar Photospheric ??(z)
Therefore, we calculate a family of curves T(??)
for each measured wavelength, as in this example.
A given temperature T (in this case 6300K) must
belong to the same height z . This method gives
us the empirical observational mean to determine
the different optical depths corresponding to the
same geometrical position. In our example, the
optical depth at 5010 is slightly smaller than at
3737 A, and decidedly smaller than at 8660 (see
the abscissae).
The next step is then to determine the thickness
s of the photosphere .
41
Determination of photospheric thickness s
Recall that the optical depth is the integral of
the linear absorption coefficient ?? over the
geometrical depth z . As a first approximation,
suppose that ?? is constant with z,
therefore ? ? ??? s where s is the
thickness of the photospheric columns of
gas. Later on (when considering the conditions
for the hydrostatic equilibrium) we shall derive
that an indicative value of the linear absorption
coefficient in the visible around 10-18 cm-1
therefore, to produce ? ? 1 we need a thickness s
of approximately 100 km.
42
Determination of the absorption coefficients ??
From the family of curves T(??) of the previous
figure, we can determine the different optical
depths at the different layers z (each layer has
the same T) for each measured ?.
From these values we can derive the function ??
at each T , as was done for the first time by
Chalonge and Kourganoff in 1946 (see figure).
These curves show that the optical depth
increases with T (outwards decrease of
photospheric temperature). Moreover, the shape
of the curves confirms the validity of Wildts
assumption of the importance of H- as main source
of visible opacity.
43
The radius of the Sun
As a consequence, the optical depth enters in the
definition of the radius of the star. For the
Sun, a good approximation for the decrease of
density ? with the height z above the surface is
At about z 3H, the density of matter becomes
extremely small. If we remember that at the Suns
distance, 1 arcsec corresponds approximately to
700 km, we see that only very good seeing
conditions (and adaptive optics techniques) will
permit to detect a difference in solar radius
according to the wavelength. It is also clear
that the radius of the Sun can have very
different values at wavelengths where the
absorption coefficient is very different from
that at visible wavelength, for instance in the
radio domain the Sun has a much larger radius
(approximately 1.8 times).
44
The radiation flux - 1
In addition to the intensity, we wish to
determine the flux, namely the total amount of
radiation leaving the unit area per unit time per
unit bandwidth in all directions, a quantity
usually indicated with ?F? (notice the factor ?
usually entering in the definition, however not
all authors have it indeed this flux is often
referred to as astrophysical flux)
Recalling the azimuthal symmetry
From the mathematical point of view, notice that
the flux is the first moment of the intensity
with respect to ? .
45
The radiation flux - 2
If the intensity were a strictly isotropic
function, independent of ? , the integral would
vanish no net flux would be observed, in any
direction (this is the case for instance inside a
cavity in thermodynamic equilibrium). Therefore,
the flux measures the anisotropy of the radiation
field. To see this more clearly, let us split
the integral in two parts, one for the radiation
going outwards (? lt ?/2), and one for the
radiation going inwards (? gt ?/2)
In particular, at the surface of the star no
radiation will enter from above
(this would not be true for a planet, nor for a
close binary star!).
46
The radiation flux - 3
It must be underlined that in general the flux is
a vector with a direction. In the
electromagnetic theory of the radiation, the
vector ?F is identical to the Poynting vector,
and we can apply to it the usual vector operators
'div' and 'grad' . However, because of the
spherical symmetry of the star we have a flux of
radiation going radially outward, therefore the
flux has only a radial component ?Fz. In the
plane-parallel approximation we then have
It can be shown (see later) that the radiative
equilibrium condition is equivalent to
47
The radiation flux and the spectrum - 1
Let us ask now the following question When we
take the spectrum of a very distant star, what do
we measure, I? or F?? To answer, recall that
each elementary surface of the star ?? (e.g. each
cm2) radiates into the solid angle??? , making an
angle ? to the radial direction, an amount of
energy equal to I? ? cos? ? ?? . For a given
angle ? , we then receive radiation from an
elementary area equal to d? 2?R?Rsin? ? d? as
shown in the figure.
The dashed region indicates the area with ?
between (? ,? d?). The observer at infinity
receives the radiation coming from all directions
? between 0 and ?/2.
48
The radiation flux and the spectrum - 2
On the other hand, an elementary area ?? (e.g.1
cm2 )at the Earth subtends a solid angle ??
??/d2 where d is the distance to the star. The
total amount of received energy is thus
or else
49
The radiation flux and the spectrum - 3
Notice that the star behaves as a luminous disk
of area ?R2 , and ltI?gt is the mean intensity
coming from a unit area on the disk.
In conclusion, F?(0) is the mean intensity
radiated by the stellar disk. So the spectra of
the stars give the energy distribution of the
flux.
50
Average Intensity
In addition to the flux through a surface, we
can define, in a given point inside the
atmosphere, the average intensity
where the integral is extended to the effective
solid angle of the source. From the mathematical
point of view, the average intensity is the
zero-th order moment of the intensity with
respect to ? . For a truly isotropic radiation
field it would be
This condition is fairly well satisfied in the
deep interiors of the star, where the temperature
gradient is very small, but only very
approximately so in the photosphere. Notice
that J can be defined even outside the stellar
atmosphere. For instance, the average intensity
of the solar radiation at the Earth is
51
Energy density and radiation pressure
The average intensity is connected to the energy
density u? by
The energy density in its turn is connected to
the radiation pressure, because any photon of
frequency ? has an associated momentum p h?
/c, and the arrival rate of photons on the walls
of the column is
Therefore, the net impulse transferred by
radiation to the volume element is
From the mathematical point of view, the
radiation pressure is the 2-nd order moment of
the intensity with respect to ? .
52
Moments of the intensity
Collecting the previous definitions, we see that
the average intensity J, the astrophysical flux
?F and the radiation pressure P can be considered
the first three moments of the intensity I
However, given the different physical meaning of
P and the presence of the velocity of light, we
shall introduce as second moment a fourth
radiation quantity, namely the function K
53
Some mathematical considerations -1
Let us write the basic equation of the radiation
transfer in a slightly different, more general
form, with I I(r,?)
which is a partial differential equation.
Instead of working directly with such equation,
it is useful to consider its first three moments
with respect to cos ? . So, multiply the basic
equation by 1, cos ? , cos2? respectively, and
integrate over all ? s. The first two moments
(by 1 and cos ? ) provide two ordinary
differential equations
for the three unknown functions F, J and K.
54
Some mathematical considerations -2
This insufficiency (2 equations for 3 unknown
functions) cannot be removed by considering the
moment of order 3 of the intensity, because a
fourth unknown radiation function would be
introduced. This insufficiency, which is almost
always encountered when a partial differential
equation is substituted by a set of ordinary
differential equations via the moments method,
can be overcome by some physical reasoning tying
together two of the unknown functions. For
instance, if the radiation field would be very
isotropic (as in the stellar interior), we could
insert into the basic equation the approximate
form for the intensity I I0 I1 cos? and
show that K F/3. In the stellar atmosphere
the isotropy is not so good, and the conclusions
less firm, as shown later on.
55
Exit Flux and Temperature in LTE
At this point, we wish to determine how the exit
flux through the surface is connected to the
effective temperature of the star. Let us
introduce again the hypothesis that S? is a
linear function of the optical depth ?? . The
intensity is then a linear function of ?, and we
obtain
a most important result known as
Eddington-Barbier relation the flux that exits
the surface at each wavelength, equals the source
function at an optical depth ?? 2/3 at that
wavelength. In particular, in the LTE hypothesis
S? B? (T)
56
The grey atmosphere
If in addition, the absorption coefficient ?
could be assumed independent of ? (namely, if the
stellar atmosphere could be considered a grey
atmosphere), the resulting outward flux would be
that of a black body with the temperature
occurring at ? 2/3, T T(? 2/3). Moreover,
each linear coordinate z would have the same
optical depth. Since, by definition, the
integral of the outward flux over all wavelengths
is proportional to the 4-th power of the
effective temperature (Stefan-Boltzmann law)
Although we have reached this conclusion using
drastic approximations, however the observations
prove that the spectral energy of the Sun is
reasonably similar to that of a black body at
5800 K, which is therefore the temperature at ?
2/3.
57
The main processes originating the continuum

• A particular way of defining an average
  absorption coefficient is Rosselands mean. To
  understand how it is calculated, we first recall
  the main processes that give rise to the opacity
  in the continuum and in the lines (details are
  available on word files, for interested
  students)
• Photo-ionizazion by photon absorption from a
  bound to a free state (b-f) the inverse process
  is recombination
• Scattering by free electrons (Thomson), atoms and
  molecules (Rayleigh)
• Absorption of a photon by an electron transition
  between two free levels (free-free). It can take
  place only with the presence of a ion
  (conservation of energy and momentum). The
  inverse process is also called thermal
  bremsstrahlung.
• The negative H ion H-
• Resonant scattering in absorption lines of atoms
  and molecules
• Raman scattering (inelastic) by molecules
  (usually not considered in stellar atmospheres,
  but there are exceptions)
• Mie scattering by large particles (usually not
  considered in stellar atmospheres)

58
The continuous absorption coefficient - 1
Summing up all the contributions of the different
processes applied for each chemical species
properly weighted for their relative abundances,
taking into account the temperature and
electronic pressure, one finally obtains the
overall opacity in the continuum of the gas
having a given chemical composition at each
wavelength. The process is legitimate, because
opacities sum up, and the total coefficient is
simply the sum of the partial ones. However, the
calculation certainly it is not simple,
especially if molecules have to be taken into
account. The following figure shows examples
for two different temperatures, one slightly
cooler than solar and one much hotter. The
horizontal line is Rosseland mean opacity, namely
an average value of the opacity useful in
calculating stellar atmosphere models. Because
the calculations make use of Saha formula, the
values of ?? depend on the electronic pressure in
the gas the solar curve was computed with logPe
0.5, the hot star with logPe 3.5. It is to
be expected therefore that the importance of the
several discontinuities (e.g. at the Balmer
limit) will be different for different luminosity
classes. These expectations are born out by the
observations.
59
Two graphs of the continuous absorption
coefficient
Left, a star slightly cooler than the Sun. Notice
the importance of the negative H ion (H-) in the
visible and near IR. Right, a B0 type, whose
opacity in the visible is about 20 times larger
than for a solar type star (from Ünsold)
60
Rosseland Mean
Rosseland Mean (log-log plot) calculated as
function of the gas pressure, for several values
of the temperature parameter ? 5040/T (labels
on each curve). Ordinate in (particle)-1.
61
The continuous absorption coefficient - 2
The previous slide has shown that the solar
continuous opacity has a minimum in the near IR
at 16 micrometers (the matter is more
transparent), while it is maximum at radio
frequencies and in the near UV going towards the
Lyman limit. Notice how different is the
situation for hotter stars. We can also say
that photons come out of the photosphere from
very different regions, according to their
wavelength ?. If, at that ? , the matter is
transparent, we see deep in the atmosphere, if
opaque, we see only the outer layers. Roughly
speaking, photons of a given ? are the result of
absorption and emission processes taking place in
regions extending from ?? ? 100 to ?? ? 0.001,
the value ?? ? 1 being a very useful indicative
value. The treatment of the spectral lines and
bands would require a much more complicated
discussion.
62
Temperature and Optical Depth - 1
Let us consider again the fundamental equation of
the radiative transfer
which we want to solve in order to determine the
source function S at each optical depth. The
assumptions are - energy is generated from
below, and is transported by radiation only,
without sources or sinks in the atmosphere, so
that the temperature decreases outwards, - ETL
is satisfied, S is identified with B(T), and each
layer z maintains a well defined T. In general,
the optical depth of z depends on the wavelength,
but for the gray atmosphere each z has a unique
optical depth ? . This approach is of course a
very drastic approximation of the real atmosphere.
63
The radiative equilibrium
At this point, integrate over all ?s in order
to have bolometric quantities. The previous
requirements amount to say that the bolometric
energy flux must remain constant with the depth,
namely
We underline that the condition of thermal (or
radiative) equilibrium is not equivalent to
thermodynamic equilibrium only the overall flux
remains constant with z , not the temperature,
proceeding outwards the radiation color becomes
redder and redder.
64
Temperature and Optical Depth - 2
At his point, multiply the fundamental equation
by ? , integrate over all ?s and exchange order
of integration and derivation. Recalling the K
function
The first integral on the right side is ?F , the
second one is zero because the source function is
isotropic, therefore
65
Temperature and Optical Depth - 3
On the other hand
Therefore, we have the first result that in the
gray, plano-parallel atmosphere in radiative
equilibrium, the source function equals the mean
intensity. For the K function
a result which is known as K-integral. Consider
now that while a small anisotropy is crucially
important for I, it will be almost negligible for
K, because of the presence of the cos2? factor.
Therefore
which is known as Eddington approximation.
66
Temperature and Optical Depth - 4

So, we have
Thanks to ETL
a result which is is known as Milne-Eddington
equation.
67
The Milne-Eddington equation
To determine the value of the constant, consider
that the total emergent flux must twice that of a
black body at temperature of the boundary ? 0
(T0 T(?0)), because the inward flux must be 0
so that finally
with w slowly varying between w(0) 0.58 and
w(?) 0.71.
68
Milne's problem Exact solution
The table shows that the differences between the
first approximation and a more exact solution
are not large.
69
Temperature of the photosphere
The limit of the previous discussion is reached
when we encounter the chromosphere. The figure
shows that going from high to small optical
depths (right to left), the temperature decreases
until it reaches a minimum around 4300 K, and
then increases again toward the chromospheric
values (above 104 K). Notice the differences in
the different theoretical models.
70
Temperature of the deeper photosphere
This figure shows theoretical results for the
deeper photosphere. The optical depth increases
to the right. Notice the differences among the
different authors.
71
Structure of the solar temperaturewith height in
the outer layers
The previous discussion has put on firmer
theoretical and mathematical grounds the
temperature values, at least near the
photosphere. Notice how thin is the photosphere,
say 500 km. We need now to discuss with more
insight the behavior of the density .
72
The hydrostatic equilibrium condition - 1
To derive other values of the physical conditions
in the solar photosphere, let impose a condition
of hydrostatic equilibrium for a spherical, non
rotating gaseous star the variation of pressure
P with depth r will equal the gravitational
attraction of the matter inside r, through the
differential equation
The gas pressure can be expressed in terms of
density and temperature by the perfect gas law
where mH is the Hydrogen atom mass, and ? is the
mean molecular weight. If the gas were composed
only by ionized H, ? would be equal to 0.5 for
He II, ? 4/3 heavy metals of charge Z produce
Z1 particles, and we can assume their atomic
weight equal to 2Z, so that ? 2. Finally
where X, Y, Z are the number density of H, He and
metals respectively.
73
The hydrostatic equilibrium condition - 2
For example, a fully ionized stellar mixture,
like the solar corona, or the deep interior,
provides ? ? 0.6. In the solar photosphere at
6000 K, the matter is certainly only slightly
ionized, because H and He are essentially all
neutral, and only low ionization potential metals
can provide electrons, so that ? ? 1 . Let us
introduce the surface gravity
where the scale height H is function of ? and
T we have reached the important conclusions
(valid also for planetary atmospheres) that the
hydrostatic equilibrium condition fixes the scale
height H once we know the mean molecular weight ?
and the temperature T. Using the previous values,
in the solar photosphere we have H ? 200 km .
The matter density can be obtained at this stage
by knowing the average opacity per gram of
matter the total density must be ? ? 1017.5
atoms/cm3, and the (very low) electron density Ne
? 10-4? ? 1013.5 electrons/cm3. These must be
roughly the values at optical depth ? 1.
74
Solar temperature and density as function of
geometric depth in the photosphere
These two figures give another (and very
schematic) representation of the behavior of
temperature and density with the linear
coordinate z (increasing outwards). Zero is the
conventional bottom of the visible photosphere.
Using an average optical depth, the geometric
thickness of the photosphere is approximately 400
km between 6000 and 4400 K (photosphere). A
thickness of 500 km give an optical depth ?vis ?
10.
75
Emission line spectra

• The previous discussion has shown that there
  are two limiting cases leading to the formation
  of an emission line spectrum, namely
• An optically thick volume of gas with the source
  function increasing toward the exterior, such is
  seen in the solar spectrum below 1600 A.
• A gas optically thin in the continuum, with no
  background light, such as the solar chromosphere
  and corona seen outside the disk (the same would
  apply to emission nebulae, see later slides).

76
The solar spectrum below 1600 A - 1
Photographic spectra taken from rockets showed
for the first time that below 1600 A the spectrum
of the solar disk turns to emission (being
photographic, the emission lines are black in the
figure).
The strongest lines are Ly-? 1216 A C II 1336
A Si IV 1406 A C IV 1550 A O III 1660 A namely
lines of high excitation and ionization.
77
The solar spectrum below 1600 A - 2
At 6000 K the photosphere does not radiate much
UV, the entire emission below 1500 A is only 1/20
of that in 1A at 5000 A! Therefore, when we look
at the Sun in those wavelengths we do not see the
continuum of the photosphere, but only emission
lines originating in the chromosphere and corona
(the lines seen there are resonance lines, the
most important in the spectrum). To understand
this fact, recall that the previous discussion
has shown that the opacity increases as we go
into the UV, so our line of sight terminates
higher in the photosphere, until at 1800 A we
reach the temperature minimum of about 4000 K
(which is the color temperature of the radiation
at 1800 A). The emission lines seen at ? ? 1800
A must then come from higher, hotter regions the
exponential increase of excitation with
temperature overweighs the falloff in density,
resulting in emission lines (absorption lines
could be present but have not been detected below
1500 A).
78
The solar spectrum below 1600 A - 3
The intensities of the UV lines are determined by
excitation conditions, abundances, and atomic
term structure peculiarities. HI Ly-? 1216 A is
as strong as all the other UV lines put
together He II Ly-? 304 A is as strong as all UV
lines below 500 A Because of the low density,
collisional ionization is not balanced by their
counterparts, namely triple collisions. As a
result, the most common ions have ionization
potentials five or ten times higher the thermal
energy. In other words, we cannot even use a LTE
approximation. Going form the chromosphere into
the corona, also the plane-parallel approximation
will start to fail.
79
The chromosphere
The thickness of the chromosphere is
approximately 3000 km from T 4300 K to T
30000 K, as derived from direct spatial
resolution during a solar eclipse. During an
eclipse, we see for few minutes a bright reddish
solar limb (reddish because dominated by the Ha).
All lines are in emission (flash spectrum). The
main limitation of spatial resolution from ground
observations can be derived by the following
considerations at the distance of the Sun, 1
arcsec equals approximately 700 km, which is more
or less the geometrical thickness of the
photosphere. Therefore, very good atmospheric
seeing conditions, and very good optical quality
of instruments, are required. After Lyot
invented the coronagraph (see dedicated lecture),
there is no need to wait for an eclipse, but sky
conditions must anyhow be very good.
80
The steep rise of temperature in the chromosphere
and corona
In the transition region, due to the steep
temperature gradient we cannot ignore the
conductive flux
with K 1.1x10-6Te5/2
81
The steep increase of ionization
The figures show the steady decrease of matter
density with the height h (left panel, right to
left), and the sudden increase in ionization
(temperature) at h? 500 km. Notice that the
total density has a very rapid and regular
(almost exponential) decline, while the electron
density reaches a long plateau corresponding to
the region where the hydrogen becomes ionized,
thus releasing a great number of free electrons
to the gas.
82
The spectrum of the solar corona - 1
Some visible lines 4471 He I 4686 He II 4713 He
I H-beta 5303 Fe XIV 5876 He I 6374 Fe
X H-alpha 6678 He I
This spectrum was obtained by Davidson and
Stratton in 1927, during a solar eclipse. We
clearly see the image of the corona in the Fe XIV
line at 5303 another strong coronal line is 6374
FeX. The other lines are emitted in chromosphere
and in the transition region between the
chromosphere and the corona.
83
The spectrum of the solar corona - 2
84
The solar corona in X rays
Not only the Sun, other stars have very hot
coronae, as seen from their intense X-ray
emission.
Image taken from the Yohkok Japanese satellite.
85
Identification of the forbidden lines of the
solar corona - 1
How B. Edlén (1942) could identify the coronal
lines. The strongest coronal line is 5303A of Fe
XIV. To remove 13 electrons from the Fe atom, the
energy of approximately 1 KeV is required (or a
temperature of about 1 MK). These conditions can
be obtained in the laboratory, but not the very
low densities required to see forbidden lines.
The density of the corona is instead so weak that
the population of metastable levels can remain
undisturbed for fractions of seconds. B. Edlén
had succeeded to obtain in the laboratory far UV
spectra of highly ionized Fe atoms, and he and
Bowen (1939) were able to show that bright
emission lines of the nova RR Pictoris agreed
with those of FeVII. In the same year, Grotrian
pointed out that high temperatures should be
present both in the novae envelopes and in the
solar corona, and that there was an almost exact
correspondence between the red coronal line at
6374A and that one could derive from Edlén's
measurements of FeX.
86
Identification of the forbidden lines of the
solar corona - 2
Let us see the principle of this procedure.
Although the forbidden lines are not observable
in the laboratory, their wavelengths can be
calculated by those of the permitted lines.
Suppose we have a 3-level atom, 1, 2
(metastable), 3 normal. The ionized atom can then
recombine capturing a free electron in n 3, so
we observe the two permitted lines 3?1 and 3?2,
with frequencies ?31 , ?32 respectively. Therefore
, the frequency of the forbidden line 2?1 is
obtained by ?21 (?31 - ?32). For the
specific case of FeX according to Edlén, there
are 4 permitted lines, two at 96.8 and 96.6 A
deriving from the cascade 2P1/2, 2P3/2 ?
1P1/2, and two at 96.1 and 95.4 from the cascade
2P1/2, 2P3/2 ? 1P3/2. Therefore, the
forbidden line corresponds to the transition
1P1/2 ? 1P3/2, with an excellent agreement of the
wavelength (hint it is easier f you use
wavenumbers).
87
Identification of the forbidden lines of the
solar corona - 3
Then, Grotrian identified the red line at 7892A
with a magnetic dipole transition of the ground
term of FeXI. So, Edlén went on identifying
other strong lines in the corona with forbidden
transitions of CaXII and CaIII. However, to
identify the strongest line at 5303, he had no
direct proof, but he got the answer by
extrapolating the Z4 law we have already pointed
out. The Einstein coefficient for 5303 FeXIV
is A 60 s-1.
88
Spectra of Planetary Nebulae
Short- ward of the Balmer limit, starts a faint
blue continue which has its peak around 2400 A.
This continuum is due to two-photon emission, an
H I mechanism discovered by M. Goppert-Mayer in
1932 between n 1 and n 2 there might be a
'phantom' level giving rise to two UV continuum
photon instead of one Ly-?. The sum of energies
of the two must equal that of Ly-?.
89
The far IR (ISO) spectrum of NGC 7027
90
Exercize on the electrons of the solar corona
Visual observations taken during solar eclipses
show a faint corona starting at the solar limb
(conventionally, at 1.003 solar radii R0 from the
center of the disk) with an intensity of 10-5
that of the disk (one half of that of the full
Moon), and dropping to 10-8 after 1 solar
radius. Assuming that the main mechanism of
opacity is Thomson scattering, determine the
volume and column density of free electrons.
Solution in a drastically simplified
discussion, the light we see is scattered to ? ?
90, the Thomson cross-section is 3.3x10-25 cm2 ,
the total column is about 1R0 ? 7x1010 cm, so
that
We can also derive the scale height H by the
condition of hydrostatic equilibrium
where T ? 1.2x106 K, ? is the mean molecular
weight (? 0.6), mH the mass of the proton, and g
the surface gravity. Further notions will be
given on the chapter on the Sun.
91
Exercises
1 - Calculate the term
for I?,0 B? (5900 K), and S? B? (4500 K) and
S? B? (10000 K) respectively, in the interval
3000 A lt ? lt 10000 A.
92
Stellar Atmospheres Some Literature

• S. Chandrasekhar, Radiative Transfer, Dover
• E. Bohm-Vitense, Introduction to stellar
  astrophysics, vol. 2, Cambridge Un. Press
• L. Gratton, Introduzione all'Astrofisica, 2voll.
  (in Italian), Zanichelli
• D.F.Gray, The Observation and Analysis of stellar
  photospheres, Cambridge University Press
• D. Mihalas, Stellar Atmospheres, W.H. Freeman,
  San Francisco
• R. Rutten, Lecture Notes Radiative Transfer in
  Stellar Atmospheres http//www.fys.ruu.nl/rutten/
  node20.html
• M. Schwarzschild, Structure and Evolution of the
  Stars, Dover
• A. Unsöld, Physik der Sternatmosphären, Springer
  Verlag (in German), and The New Cosmos