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Stellar Atmospheres

I Parte del Corso di Astrofisica Stelle e

Pianeti 2006-07 Approfondimenti sono contenuti

in vari capitoli (in inglese, file word) che non

sono ancora in forma definitiva ma che possono

essere già utilizzati da chi fosse interessato

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.

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.

Temperature in the solar atmosphere

Schematics of the solar temperature profile

(thick line) with height. 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.

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 (notice that 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, magneto-hydrodynamic) - above a very sharp transition layer, in the

corona, the magnetic field energy is very

important. 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 approximations is certainly

untenable. - 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.

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.

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.

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? into the unit solid angle

d? in direction ? to the normal n. - The units of I are erg?cm-2?s-1A-1?sr-1 (in our

mixed cgs system) - In general I? I?(x,y,z,?,?,t)

I?(x,y,zl,m,nt) 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.
- Furthermore, no account will be taken of a

possible polarization of the beam. - 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. - Caveat in planetary atmospheres none of these

simplifying assumptions is fulfilled!

Basics on Transport of radiation - 3

Consider now, in a idealized experiment which

we could perform in the laboratory, a beam of

radiation of intensity I?,0 which enters from the

left in a vessel of gas at a given temperature T.

The cross-section of the vessel is ?, its length

is s. The shape of the column (here shown as a

rectangular box) is irrelevant, it could be a

cylinder. We also suppose that the walls of the

vessel are transparent, in order not to have

reflection effects.

We want to determine the intensity of the

radiation exiting the vessel on the right.

Transport of radiation - 4

Absorption after a trajectory ds along the path

of the beam, some energy of the beam will be lost

due to absorption (here, absorption has a loose

significance, to be specified later)

where ?? is a linear absorption coefficient

(cm-1) appropriate to that particular gas.

Emission On the other hand, each elementary

volume dVd? ?ds of gas will emit, contributing

some energy to the beam according to an

appropriate emissivity coefficient ??

which we assume here independent from the

incoming radiation field. The units of ?? are

erg?cm-3?sr-1. Notice that the emission is

(assumed) isotropic.

Transport of radiation - 5

Therefore, the total energy variation can be

expressed as

The intensity variation along the beam is then

Notice that the sign of dI?/ds is determined by

that of the difference

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. Notice that these definitions

are appropriate to the laboratory experiment,

where the radiation comes into the volume at x

0 and exits at x s. Later on, we shall reverse

the direction, the radiation from the stellar

surface will come from the deep interior and will

exit at z 0.

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

Under these 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.

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 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.

Local thermodynamic equilibrium -2

The previous expression can be written as

which can be integrated over the interval

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.

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.

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 prevail 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.

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

). 2.2 optically thick case

independent from the entrance spectrum.

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 doesnt come from the ionization, but

simply by the lines being in emission, instead

than in absorption as in the visible region.

Real stars

In the visible region, the stars usually show an

absorption line spectrum, they 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, for real stars, the discussion is much

more complex, even assuming LTE and radiative

transport only, because nor the density nor the

source function are constant inside the

atmosphere. Although the concept of a reversing

layer maintains a considerable intuitive

validity, a deeper physical and mathematical

analysis is therefore warranted, as done in the

following.

Absorption line spectra

Concluding this fairly approximate discussion, an

absorption line spectrum is formed in

- A deep optically thick gas surmounted by a

thinner layer, with source function decreasing

outwards, as in the solar photosphere in the

present simplified interpretation, source

function means temperature.

- 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.

The radiative transfer equation - 1

The previous approximate discussion has shown

that a variety of cases are possible, even in

laboratory conditions. 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. As already done 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).

The radiative transfer equation - 2

As in the previous considerations, along the path

ds inside the atmosphere the following equation

will be satisfied

but

where

because of the assumed isotropy of the source

function. Let us assume now a given plane as

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. It is

convenient at this point to introduce a change of

perspective, because the observer sees the

radiation from the outside, as explained in the

following figure

The radiative transfer equation - 3

Be z a linear coordinate (say, in km) increasing

from the surface inward, and ?? the perpendicular

optical depth, also increasing 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.

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

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 can 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.

A first approximation for S(?)

Let us make the simple assumption that the

unknown source function S is a linear function of

the optical depth

(this assumption can 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)

which is also written as

As already said, in the LTE hypothesis S

coincides with the Plancks function B at a

z-dependent temperature

Limb Darkening

The previous equation tells us that at the center

of the disk we see the source function S? at a

depth ?? 1, at the border of the disk 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

Because in the photosphere the temperature

increases inwards, the temperature measured at

the center of the disk must be higher than at the

limb.

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.

The 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 leave the surface at ? 90, so that

The observations prove 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 indeed 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.

Solar Limb Darkening - 2

The Sun is darker (cooler, redder) at the limb.

Notice how well the observations are described by

the first approximation. The temperature is 6050

K at the center, and 4550 K at the limb (the

effective temperature being intermediate, ? 5800

K). In the visible at 5010A, I(0,0) 4x108

erg?cm-2?s-1A-1.

Solar limb darkening - 3

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). The

previous first approximation requires a

correction. For instance, at ? 5010, the

observations provide a 0.26, b 0.87 plus a

smaller term, which can be expressed as 0.13

cos2? , which we will consider further at a

later stage. Notice the irregularities in the

curve (e.g. at 3600 A and 13000) due to sharp

variations in the absorption coefficient due to H

and H-. The limb darkening becomes smaller in the

near IR because the larger optical depth allows

to observe the regions of minimum

temperature. From these observations we can

determine ?? (see also later).

(Adapted from Pierce and Waddell, 1961)

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

where H ? 200 km. 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 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).

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 ? .

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!).

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. Inside the stellar

atmosphere this angle ? is 4?. for an 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. From the

mathematical point of view, the average intensity

is the zero-th order moment of the intensity with

respect to ? . Notice that J can be defined even

outside the atmosphere. For instance, the average

intensity of the solar radiation at the Earth is

Average intensity, 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 h? has an associated momentum p h?/c,

and the arrival frequency 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 ? .

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)

The gray atmosphere

If in addition, the absorption coefficient ?

could be assumed independent of ? (namely, if the

stellar atmosphere could be considered a gray

atmosphere), the resulting outward flux would be

that of a black body with the temperature at ?

2/3, and each linear coordinate z would have the

same optical depth. A particular way of defining

an average absorption coefficient is Rosselands

mean (see next threes). 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)

the most important result is obtained

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.

The Rosseland mean

- We 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 - Mie scattering by large particles (usually not

considered in stellar atmospheres) - Raman scattering (inelastic) by molecules (again,

usually not considered in stellar atmospheres)

The total absorption coefficient

Summing up all the contributions of the different

processes applied for each chemical species, and

with proper weights that take into account the

relative abundances, one finally obtains the

overall opacity of the gas having a given

chemical composition, e.g. the solar composition,

a given temperature and a given electron

pressure. 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 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 log Pe 0.5, the hot

star with log Pe 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.

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 Unsold)

The continuous absorption coefficient

The previousshows 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 geometrical

radial coordinates z1 and z2 of the atmosphere

therefore are such that

in the most transparent wavelength

in the most opaque wavelength

The treatment of the spectral lines would require

a much more careful discussion.

Temperature and Optical Depth - 1

Let us consider again the fundamental equation of

the radiative transfer in LTE

which we want to solve in order to determine the

source function at each optical depth. To do so,

let us examine again the consequences of the

initial assumptions that energy is transported by

radiation only, and that ETL is satisfied, namely

that each layer maintains a well defined T, and

that the temperature must decrease outwards. We

shall simplify furthermore the problem assuming

that S B (so that source function is equivalent

to temperature), and that the atmosphere is gray.

(This approach is of course a very drastic

approximation of the real atmosphere).

The radiative equilibrium

These requirements amount to say that the

(bolometric) energy flux must remain constant

with the depth, or else

Notice two points - 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. - Each depth z has the same

temperature T, but its optical depth depends on

the wavelength. Only for the grey atmosphere each

z has a unique optical depth ?.

Temperature and Optical Depth - 2

At this point, multiply the fundamental equation

by ? , and integrate over all ?s and over all

?s. This procedure gives (as shown by Eddington,

see file word for the function K and the

demonstration)

Therefore, in the gray atmosphere, the source

function equals the mean intensity.

But it is also

so that finally

a result which is is known as Milne-Eddington

equation.

The Milne-Eddington equation

To determine the value of the constant, consider

again the equation

where T0 is the value of the temperature at the

boundary ?? 0. The total emergent flux is twice

that of a black body at temperature T0 (because

the inward flux must be 0)

so that finally

With more general assumptions, one finds

with q slowly varying between q(0) 0.58 and

q(?) 0.71.

Solar Photospheric Source Function S(?)

The solar photosphere is more complex that the

simplified model discussed so far, because 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, that by

measuring the limb darkening law we can derive

the source function at each optical depth. The

figure shows the result for the continuum at 5010

A taking into account the second-order term (in?2

) in Pierce and Waddell data. For ? gt 1.5 the

data become very uncertain.

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.

Solar Photospheric temperature ??(T)

Therefore, we calculate a family of curves for

each measured wavelength, as in this example. A

given temperature (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. The optical depth at

5010 is slightly smaller than at 3737 A, and

decidedly smaller than at 8660 (see the

abscissae).

Determination of the absorption coefficients ??

- Recalling that the optical depth is the integral

of the linear absorption coefficient ?? over the

geometrical depth z, we can therefore derive - the relative ratios of all linear absorption

coefficients ?? /??0

- the function ?? (T) , as was done for the first

time by Chalonge and Kourganoff in 1946 (see

figure), confirming the validity of Wildts

assumption of the importance of H- as main source

of visible opacity.

Temperature of the outer photosphere

The real limit of the previous discussion is

reached when we encounter the chromosphere. Going

from high to small optical depths (right to

left), the temperature decreases until it reaches

a minimum aroud 4300 K, and then increases again

toward the chromospheric values (above 10000 K).

Notice the differences in the different

theoretical models. To explore the upper

photosphere we can use the far infrared (100-200

micron), or the UV around 1600 A.

Temperature of the deeper photosphere

Finally, this figure shows theoretical results

for the deeper photosphere. Notice the

differences among the different authors.

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

The hydrostatic equilibrium condition - 2

(where X, Y, Z are the number density of H, He

and metals respectively), which for a typical

fully ionized stellar mixture, like the solar

corona, or the deep interior, provides ? ? 0.6.

Let us introduce the surface gravity

where the scale height H is function of ? and T.

In the solar photosphere at 6000 K, the matter

is certainly not fully ionized, in particular H

and He are essentially all neutral, and only

metals provide electrons, so that ? ? 1, and H

? 200 km (solar photosphere) The density

therefore must be ? ? 1017.5 atoms/cm3, and the

electron density Ne ? 10-4? ? 1013.5

electrons/cm3. These must be roughly the values

at optical depth ? 1.

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 . 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. The density is

obtained by the average opacity per gram of

matter.

Examples of emission line spectra

- We give now some examples of spectra differing

from those usually encountered on normal stars.

The previous discussion has shown that there are

two limiting cases leading to the formation of an

emission line spectrum, namely - A gas optically thin in the continuum, with no

background light, such as the solar chromosphere

and corona seen outside the disk, or emission

nebulae - an optically thick volume of gas with the source

function increasing toward the exterior, such as

the solar spectrum below 1600 A.

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 (upper figure). 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. The figure on the

left shows the steady decrease of matter density

with the height h, and the sudden increase in

ionization (temperature) at h? 500 km

The ionization of the solar upper atmosphere

The spectrum of the solar corona

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

The solar spectrum below 1600 A - 1

Strong lines Ly-? 1216 A C II 1336 A Si IV 1406

A C IV 1550 A O III 1660 A

From a photographic rocket spectrum (the emission

lines are black).

The solar spectrum below 1600 A - 2

At 6000 K the photosphere does not radiate much

UV, the entire emission below 1500 A is 1/20 of

that in 1 A at 5000 A. Therefore, we see only

chromosphere and corona when we look at the Sun

in those wavelengths (the lines seen there are

resonance lines, the most important in the

spectrum). 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 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).

The solar spectrum below 1600 A - 3

The intensities of the UV lines are determined

by excitation conditions, abundances, and atomic

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.

Solved exercize on 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, 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.

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-?.

The far IR (ISO) spectrum of NGC 7027

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.

Stellar AtmospheresLiterature

- S. Chandrasekhar, Radiative Transfer, Dover
- D. Mihalas, Stellar Atmospheres, W.H. Freeman,

San Francisco - A. Unsöld, Physik der Sternatmosphären, Springer

Verlag (in German), and The New Cosmos - E. Bohm-Vitense, Introduction to stellar

astrophysics, vol. 2, Cambridge Un. Press - L. Gratton, Introduzione all'Astrofisica, 2voll.

(in Italian), Zanichelli - R. Rutten, Lecture Notes Radiative Transfer in

Stellar Atmospheres http//www.fys.ruu.nl/rutten/

node20.html

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