9. Stellar Atmospheres Goals Goals: 1. Develop the basic equations of radiative transfer describing the flow of light through stellar atmospheres. 2. Examine.

9. Stellar Atmospheres9. Stellar AtmospheresGoalsGoals:1. Develop the basic equations of radiative

transfer describing the flow of light through stellar atmospheres.

2. Examine how stellar continua and spectral lines are affected by various parameters, and how stellar abundances are derived.

3. Derive some useful approximations for describing the radiative flux from stars.

4. Derive the fundamental equations describing the equilibrium conditions for stellar atmospheres, as used in stellar atmosphere models.

The Radiation FieldIn order to describe radiation from a star (or nebula) it is necessary to begin with some definitions of observable parameters, the first being specific intensity. Begin with radiation passing through an infinitesimally small area of a star (or nebula’s) surface, dA, into an infinitesimally small solid angle, dΩ, directed at an angle θ to the surface normal.

The dimensions of the rectangle subtended by solid angle dΩ are rdθ and r sin θ dφ, so dΩ = sin θ dθ dφ for r = 1. The average intensity of the light entering the solid angle dΩ originating from the surface area dA amounts to the energy Eλ dλ per unit time dt projected in that direction, i.e. dA cos θ.

The limit as dA, dλ, dΩ, and dt → 0 is referred to as the specific intensity Iλ. Defined in such fashion the intensity represents the amount of energy per unit time present along the ray path, which for dΩ → 0 does not spread out as distance increases (i.e. in comparison with flux). Also,

Iλ dλ = Iν dν, so

ddAdtd

dEI

cosi.e.

I

cI

2

The specific intensity may vary with direction, so one defines a mean intensity <Iλ> (sometimes referred to as Jλ) as:

If the radiation is isotropic, i.e. the same intensity in all directions, then <Iλ> = Iλ.

Black body radiation is isotropic, in which case:

<Iλ> = Bλ.

Now consider the energy carried by the radiation field:

where we define dL in the following illustration.

2

0 0

sin4

1

4

1i.e. ddIdII

dLdA

dE Energy

Consider the energy associated with the radiation flow through a perfectly “reflective” measuring cylinder (depicted) placed symmetrically about an axis normal to the radiating surface.The transit time for the radiation is:

So the energy carried by the flow is given by:

c

dLdt

cos

c

ddI

cdLdA

ddAdLdI

dLdA

ddAdtdI

coscoscosEnergy

The energy density uλ of the radiation flow is found by integrating the energy of the flow over all solid angles, i.e.:

And, for black body radiation, which is isotropic, we expect:

or

The total energy density is obtained through integration over all wavelengths or frequencies, i.e.:

dIc

dddIc

ddIc

du4

sin11 2

0

d

e

hcdB

cdI

cdu

kThc

1

844 5

de

hdB

cdI

cdu

kTh

1

844 3

0

duu

For black body radiation we have:

Radiative flux is a measure of the net energy flow across dA. Thus,

measures the flow of radiation through the surface dA in the direction of the z-axis.

4315

444

0

BB

Kcmerg10566.7where

444

a

aTc

TT

cdB

cu

ddId cosF

Typically the radiation field is isotropic, i.e. Iλ does not depend upon direction. Then:

For a flow through only one hemisphere, i.e. 0 ≤ θ ≤ ½π:

Sometimes an “astrophysical flux” is defined. It is the true flux Fλ divided by π, i.e.:

Be careful, since the usage varies from one textbook to another!

0cossinsince

,0cossin2cossini.e.

0

0

2

0 0

d

dIddIF

IIdI 2cossin2 21

0

2

F

FF

FandF

The Difference Between Intensity and Flux?The specific intensity of a source is independent of distance from the source, whereas the radiative flux varies with distance according to the inverse square law, i.e. 1/r2. For a distant source it is only possible to measure intensity Iλ if the source is resolved, otherwise radiative flux Fλ is measured. In the example shown one measures specific intensity in (a) when the source subtends an angle larger than the resolution of the telescope/detector system, otherwise radiative flux (b).

The Radiation Field, 2A photon of energy E carries a momentum p = E/c, which means that it can exert radiation pressure. For photons incident on a reflecting surface (image at right) the momentum exchange with the surface is simply the change in momentum upon reflection:

ddAdtdc

I

dc

E

dc

E

c

E

dppddp

2cos2

cos2

coscos

finalinitial

The radiation pressure Prad is equivalent to the force exerted by the photons, i.e. the rate of change of momentum per unit area.

Integration over one hemisphere gives the radiation pressure exerted by the flow from the source, i.e. a “photon gas” that does not reflect from the surface:

For isotropic radiation the formula becomes;

2

0 0

2

2rad

sincos2

cos2

dddIc

dIcdAdt

ddpdP

ddIc

dP 2

rad cos1

2

0 0

2rad cossin

1dddI

cdP

But:

So:

The total radiation pressure is found by integration over all wavelengths:

which for isotropic black body radiation becomes:

By way of comparison, the pressure exerted by an ideal monotonic gas is 2/3 of its energy density.

3

2cossin

2

0

2

d

dIc

dIc

dP3

4

3

22rad

0

rad dPP rad

33

1

3

4

3

4

3

4 BB444

0

rad

uaT

c

TT

cdB

cP

LTEThe definition of temperature for a star depends upon how T is derived:

For gas in a box the various temperatures should match, since thermodynamic equilibrium (TE) applies. Stars cannot be in perfect TE since there is an outward flow of energy producing a temperature gradient in their atmospheres. If the distance over which T changes significantly is small relative to the distances traveled by atoms and ions between collisions, then local thermodynamic equilibrium (LTE) is a good approximation.

Tex = temperature derived using the Boltzmann equation to establish a match to the observed energy level populations of atoms.

Tion = temperature derived using the Saha equation to establish a match to the observed ionization states of atoms.

Tkin = temperature as inferred from the Maxwell-Boltzmann equation to describe the velocity distribution of particles.

Tcolor = or TBB is the temperature established by matching the unreddened flux distribution to a Planck function.

In the Sun, T varies from 5650 K to 5890 K over a distance of 27.7 km (1 K/0.1 km) according to the Harvard-Smithsonian Reference Solar Atmosphere. The resulting temperature scale height is:

over which the temperature changes by one scale factor and where T = 5770 K has been used as a typical region of the solar atmosphere.

Clearly it is safe to assume that most regions of the solar atmosphere are in LTE. Exceptions are restricted to regions where the temperature changes rapidly.

km666

cm1066.6

cm1077.256505890

5770

6

6

KK

K

drdT

THT

Stellar OpacityThe mean free path of particles is calculated as follows:

Typical densities in the solar atmosphere where T = Teff are of order ρ ≈ 2.5 × 10–7 g/cm3. For pure hydrogen gas, the corresponding number density is n = ρ/mH = 1.5 × 1017 /cm3. Two atoms will collide if their centres pass within two Bohr radii 2a0. In time t a single atom sweeps out a volume given by:

π(2a0)2vt = σvt,

where σ = π(2a0)2 is the collisional cross-section.

There are nV atoms in the volume = nσvt atoms. The average distance traveled between collisions is therefore:

The mean free path for a hydrogen atom is therefore l = 1/nσ. For hydrogen, a0 = 0.53 × 10–8 cm so σ = π(2a0)2 = 3.52 × 10–16 cm2.

The mean free path is much smaller than the distance over which T changes by 1 K. Gas atoms in the solar atmosphere, and typical stellar atmospheres, are therefore confined to a reasonably isolated region within which LTE can be assumed to be valid.

The same is not true for photons, since they are able to escape freely into space.

nvtn

vtl

1

atomsofnumber

traveleddistance

.K1whichoverkm,1.0cm1089.11 2 T

nl

Photon AbsorptionAbsorption refers to scattering and pure absorption of photons by particles, anything that removes photons from a beam of light. The amount of absorption dIλ is related to the initial intensity Iλ of the beam, the distance traveled ds, the gas density ρ and the opacity of the gas as defined by its absorption coefficient κλ:

The negative sign indicates that the intensity of the beam decreases in the presence of absorption. Note the form of the relationship:

Integration of both sides of the equation gives:

or

dsIdI

dsI

dI

seIIsI

0

orln

s

ds

eII 0

0

Because of the exponential drop-off, the intensity decreases by a factor of 1/e = 1/2.718 = 0.368 when the exponent is unity, i.e. over a scale length of l = 1/ρκλ = 1/nσλ.

In the case of the Sun, for the parameters used earlier and κ5000Å = 0.264 cm2/gm the implied scale length for photons is:

which is comparable to the temperature scale height. In other words, photons travel through regions of different T.

It is convenient to introduce the term optical depth τλ such that:

for integration from the outermost layer of a star inwards.

km152cm1052.1105.2264.0

111 77

nl

s

dsdsd0

thatso

thinopticallydefines1

andthick,opticallydefines1

Application to Atmospheric ExtinctionConsider the case of observations of stars made from ground-based telescopes, where the light traverses the Earth’s atmosphere and suffers extinction.For starlight traversing Earth’satmosphere the distancetraveled is ds = dt/cos z, wheredt is the thickness of theatmosphere, i.e. ds = sec θ dt in the diagram.So:

and:

or:

zdtzdtzdstts

secsecsec0

000

zIIeIIz

secconstantloglogor0

o

0

sec

zmm secconstanto

Because of the curvature of the Earth, the value of sec z is not quite equivalent to the air mass X, which measures the total amount of atmospheric extinction between the star and the observer. The best formula for air mass X is:

X = sec z (1 – 0.0012 tan2 z)

The “constant” term kλ varieswith wavelength λ and can varyfrom night to night.

i.e. mλ = mλ(0) + kλ X

An example of an atmospheric extinction plot for a standard star used for photometric standardization, this time plotting Fν versus sec z rather than mλ versus X.

The strong 1/λ4 dependence of extinction in Earth’s atmosphere means that blue stars fade more rapidly than red stars with increasing air mass X. It also gives rise to colour terms in the extinction coefficients.

Opacity Sources in Stars1. Bound-bound transitions, involving photon absorption and reemission in random directions resulting in a net loss of light in the direction of the original photon.2. Bound-free transitions, involving photoionizations from the ground state. For hydrogen the cross-section for bound-free absorption is:

3. Free-free absorption, in which free electrons passing near hydrogen atoms absorb energy from photons. The process can occur for all ranges of wavelengths, so κλ(ff) is a contributor to the continuous opacity along with κλ(bf). The process is also referred to as bremsstrahlung, or braking radiation.4. Electron scattering, or Thomson scattering, in which photons are scattered by free electrons, a wavelength-independent mechanism ~2 × 109 larger than σbf. The formula is:

23

A50005

15bf cm

11031.1 o

n

225

2

2

2

T cm10655.63

8

cm

e

e

4. Electron scattering, part 2. Photons can also be scattered by electrons that are loosely bound to atoms. Compton scattering describes photon scattering where λ < size of the atoms. Since most atoms and molecules have dimensions of ~1 Å, Compton scattering applies mainly to X-rays and gamma rays. Rayleigh scattering describes photon scattering where λ > size of the atoms. The latter process is highly wavelength dependent, typically varying as 1/λ4, as in atmospheric scattering (below).

An example of various absorption sources in the atmospheres of stars: hydrogen and ionized helium bound-free absorption (early-type stars), and the H– ion (late-type stars). The former is highly λ–dependent, the latter almost λ–independent.

The continuum of the B7 V star Regulus (α Leo) showing the signature of hydrogen bound-free absorption in its spectral energy distribution.

Black body curves: what the continuous energy distributions of stars would look like in the absence of continuous opacity sources in their atmospheres.

Be careful how stellar spectral energy distributions are plotted. They appear different when different parameters are used.

Atomic bound-bound absorption by various metal lines in the continuous spectra of late-type stars, relative to H– absorption.

Typical formulae are, for Rayleigh scattering:

And for Thomson scattering:

Atomic hydrogen absorption is strongest where the population of the n = 2 level of hydrogen maximizes relative to all hydrogen particles, i.e. near 9800 K. H– opacity is the dominant opacity source in cool stars. The ionization potential for the H– ion is only 0.754 eV, so any photon with

ionizes it.Molecules are also opacity sources (in cool stars) because they can be dissociated and also give rise to bound-bound and bound-free absorptions of photons. Molecular absorptions produce large numbers of closely-spaced lines, much like bands.

3442

40

4

cm1

3

8

cm

Ne

eN2510655.6

oo

A400,16eV754.0

AeV400,12

hc

The total opacity in a star is the sum of the various individual opacity sources, i.e.:

Where the first three terms are wavelength dependent.

It is often useful to use an opacity averaged over all wavelengths under consideration, one that depends upon density, temperature, and chemical composition. Such an average opacity is known as the Rosseland mean opacity, or simply the Rosseland mean. Although there is no simple formula for the various contributors, approximations have been developed, namely:

esffbfbb

/gmcm11034.4 25.3

bf25bf T

XZt

g

/gmcm111068.3 25.3ff

22ff T

XZg

/gmcm12.0 2es X

where X and Z are the fractional abundances of hydrogen and the heavy elements, respectively, by mass. Typical values for the Sun are X = 0.75 and Z = 0.02. The terms gbf and gff are quantum mechanical correction factors calculated by J. A. Gaunt, hence their names: Gaunt factors. Generally gbf ≈ gff ≈ 1 for the visible and ultraviolet regions of interest for stellar atmospheres. The factor “t” is an additional correction factor called a guillotine factor, and describes the cutoff for κ once an atom or ion has been ionized. Generally 1 < t < 100.

Also:

The Rosseland mean opacity is usually represented graphically.

KramerA.H.afterLaw,sKramer'..

,asvaryandboththatNote

5.30bf

5.3bf

Tei

T

ff

ff

esffbfbb

From Rogers and Iglesius (1992) for X = 0.70 andZ = 0.02. Value of ρ, in units of gm/cm3, are indicated above each curve. The opacities are calculated for a specific mixture of elements known as the Anders-Grevesse abundances.

Note:1. κ ↑ as ρ ↑.2. κ ↑ as T ↑ initially from the ionization of H and He.3. κ ↓ with further T ↑ results from the 1/T3.5 dependence of Kramer’s opacity.4. κ flattens at large T as electron scattering dominates.

Radiative TransferConsider the flow of photons out of a star as a random walk problem. If a photon has a mean free path ― average distance traveled before absorption and reemission or scattering from an atom ― of l, then a photon undergoing a sequence of N random walks undergoes a net displacement d where:

Nlllld

321

The net displacement as an absolute value is given by:

since the term in brackets ≈ 0 for large N. Random angular displacements generate an average value of θ ≈ π/2, i.e. cos θ = 0.

i.e. d = 10 l requires N = 100d = 100 l requires N = 10,000

d = 1000 l requires N = 1,000,000

But d is also related to optical depth since l = 1/ρκλ = 1/nσλ and

So optical depth τλ = 1 implies a photon has suffered only one scatter before escaping the star (actually τλ = ⅔ for light we see).

2

1131222

212

221212111

coscoscos

Nl

lNl

llllllll

lllllllllldd

NN

NNNNN

N

NldNld or22

Nl

Nld

d

d

ds

so1

Textbook Example:What is the mean free path length and average time between collisions for atoms in the Orion Nebula where n ≈ 100 /cm3?

Solution (see textbook):For hydrogen, σ = π(2a0)2 = π(2 × 0.53 × 10–8)2 cm2 ≈ π × 10–16 cm2

So the mean free path is:

l = 1/nσλ = 1/(100 × π × 10–16) ≈ 3 × 1013 cm ≈ 2 A.U.

the root-mean-squared velocity is:

vRMS = (3kT/m)½ = (3 × 1.38 × 10–16 × 10,000/1.66 × 10–24)½ ≈ 1.6 × 106 cm/s

and the average time between collisions is:

t = l/v = (3 × 1013 cm)/(1.6 × 106 cm/s) ≈ 2 × 107 s ≈ 8 months

When viewing the Sun the light originates from τλ = ⅔ for all parts of the visible disk. Near disk centre the light originates from deeper, hotter regions than for the solar limb, where the light originates from shallow, cooler regions. The result is an apparent limb darkening of the Sun.

Equation of TransferThe emission of light by material along a specific line of sight is proportional to the emission coefficient jλ of the material, the density ρ, and the distance traversed ds, i.e.:

dIλ = jλρds

for photons created by emission processes. The light beam is also affected by the opacity of the gas, which scatters and absorbs photons from the line of sight. For absorption we have:

dIλ = –κλρIλds

so for the combined processes we must have the equation of radiative transfer: dIλ = –κλρIλds + jλρds

or

and the former equation is the equation of transfer.

function,sourcethecallediswhere

1

Sj

SIj

Ids

dI

The source function Sλ = jλ/κλ describes the proportionality between the emitting and absorbing properties of the medium. Clearly, Sλ has identical units to Iλ (cm s–3 steradian–1). The form of the transfer equation leaves very simple expectations:

If dIλ/ds = 0, the light intensity is constant and Iλ = Sλ.

If dIλ/ds < 0, Iλ > Sλ, and with increasing s, Iλ → Sλ.

If dIλ/ds > 0, Iλ < Sλ, and with increasing s, Iλ → Sλ.

In other words, over any distance ds, the intensity of light approaches the local source function. If the conditions for LTE are satisfied, dIλ/ds = 0, so Iλ = Sλ. And Iλ = Bλ for black body radiation, so Sλ = Bλ.

But Iλ ≠ Bλ unless τλ >> 1, i.e. the photons are able to interact many times with matter in the star’s atmosphere.

SI

ds

dI

1

Textbook Example:Imagine a beam of light of intensity Iλ,o at s = 0 entering a volume of gas of constant density ρ, constant opacity κλ, and constant source function Sλ. What is Iλ(s)?

Solution (see textbook):The result is:

ss eSeIsI

1o,

Equation of Transfer, 2Recall that dτλ = –κλ ρ ds, if s is measured outwards radially in a star but optical depth is measured inwards, so the equation of transfer:

Can be rewritten as:

Consider a plane parallel stellaratmosphere and define dτλ in terms of a reference direction, z. Define:

SI

ds

dI

1

SI

d

dI

0

,

z

v dzz

But:

for any direction s. Thus:

and the transfer equation becomes:

A simple approximation that can be made at this point is to remove the wavelength dependence of the opacity κλ . An atmosphere that is approximated by a constant opacity κ as a function of λ, i.e. the same opacity throughout the spectrum, is referred to as a gray atmosphere, and is a good approximation for some stars. Thus, τλ,v becomes τv and it is possible to generate values for:

v,cos

cosandsec

cos,

,, v

vv d

d

SI

d

dI

v

,

cos

00

and dSSdII

The equation of transfer then becomes:

The resulting radiation field originating from such a plane-parallel atmosphere can be established by integration over all solid angles, i.e.:

which reduces the transfer equation to:

SId

dI

v

cos

radcosand,4,4but

cos

F

dIdIId

dSdIdId

d

v

SId

d

v

4radF

The same equation of transfer can also be multiplied by cos θ and integrated, resulting in the first moment:

[The same equation in a spherical co-ordinate system is:

]

cd

dP

cPd

ddI

d

dd

dSdIdId

d

v

vv

v

radrad

rad2

2

:giveswhich

cosand,0cosbut

coscoscos

F

radrad F

cdr

dP

The interpretation of the first moment equation is that the net radiative flux is driven by the natural gradient in radiation pressure within a star.

When LTE applies:

<I> = S so that Frad = constant = Fsurface = σTeff4

The situation requires flux conservation throughout the stellar atmosphere:

cd

dP

v

radrad F

constant1

or

1so

constant

radrad

radrad

radrad

v

v

v

cP

dc

dP

cd

dP

F

F

F

The Eddington ApproximationThe great astrophysicist Sir Arthur Eddington took the same equations a step further by adopting a simple approximation for the flux from a star that separated it into an outwards directed flux and an inwards directed flux at each point in the atmosphere. The intensity of the light originatingat each depth z in the atmosphere therefore had two components:

Iin = intensity of the radiation inthe inward direction

Iout = intensity of the radiation inthe outward direction

Note: Iin = 0 at the top of theatmosphere.

So, for any point in the atmosphere:

At the surface of the star the situation is simplified by the fact that Iin = 0.

Ic

IIc

Ic

Ic

dIc

dIc

dIc

P

IIII

dIdIdI

IIdIdI

dIdIdI

3

4

3

23

123

12

sincos2

sincos2

cos1

4224

22

sincos2sincos2cos

sinsin

sinsin4

I

inoutinout

2/

2in

2/

0

2out

2rad

inoutinout

2/

in

2/

0

outrad

inout21

2/

in21

2/

0

out21

2/

in21

2/

0

out21

F

In such circumstances:

<I> = ½(Iout + Iin)

Frad = π(Iout – Iin)

Prad = 2π/3c(Iout + Iin) = 4π/3c <I>

The condition of flux constancy inthe atmosphere implies thatIout > Iin at all levels of theatmosphere.

constant1

3

4

thatfollowsit

constant1

From

rad

radrad

v

v

cI

c

cP

F

F

At the top of the atmosphere τv = 0 and Iin = 0, so:

<I> = ½Iout = ½(Frad/π) = Frad/2π

and the constant in the radiation equation is evaluated from:

So:

A simple substitution for the flux at the top of the atmosphere, i.e. Fsurface = σTeff

4 gives:

For LTE:

cccI

c 3

20

23

41

3

4constant radrad

toprad

FFF

32radradrad

3

2

3

4 vv ccc

Ic

FFF

324eff

4

3 v

TI

44

orT

IT

BS

Substitution then gives:

or:

An obvious result of the Eddington approximation is that, for the gray atmosphere approximation, the temperature in a stellar atmosphere is T = Teff when τv = ⅔, so can be thought of as the point of origin for the light from a star (rather than, say, τv = 0 or τv = 1).

The gray atmosphere approximation can be further tested using the transfer equation:

Multiplication of both sides by exp(–τλ) gives:

324eff

4

4

3 v

TT

324eff4

34 vTT

SI

d

dI

eSeIe

d

dI

or:

which becomes:

The equation can be integrated from an initial position of a ray of light at an optical depth τλo to the top of the atmosphere, τλ = 0, to give:

which yields:

Namely,the intensity at the top of the atmosphere = intensity at any depth reduced by attenuation less any further contribution along the line of sight less attenuation.

eSIe

d

d

deSIed

0

0

oo

deSIe

0

o

o0

deSeII

If we next return to a discussion of the intensity emerging at the surface of the atmosphere from any direction, the optical depth and intensity equation become:

The observed intensity at the top of theatmosphere is the result of allcontributions along the line of sight, i.e. to τv,o = ∞. And e–∞ = 0, so:

The dependence of the source function S on optical depth is unknown, but a reasonable first approximation is:

0

secseco

,

o,

o, sec0

andsec

v

vv

v

v

deSeII

0

secsec0 vdeSI v

vbaS

Next evaluate the integral using the above approximation, in which case the integral becomes:

But:

And:

which gives a source function described by:

and an emergent intensity described by the equation:

0

sec0

sec0

sec

o,o,o,

secsecsecv

v

v

v

v

vvvvv debdeadeS

10sec 0

0

sec0

sec

o,

eede v

v

vv

cos1000sec

1

sec

1

sec

1sec

00

0sec

o,

xxx

vv exedxexdev

v

cosbaS

cos0 baI

The Solar Limb DarkeningThe assumption of a gray atmosphere with the Eddington approximation can now be used to generate a formula describing the limb darkening of the Sun, i.e.:

So:

The solar limb darkening is bestexpressed relative to the intensityat disk centre, i.e.:

vv baTT

IS

32

4eff

4

4

3

4

3and

24

3

3

2

4eff

4eff

4eff

Tb

TTa

o0 I

I

I

I

Substitution gives:

cos6.04.0.,.

cos5

3

5

2cos

coscos

o

53

524

eff45

53

524

eff45

4eff4

34eff2

4eff4

34eff2

o

I

Iei

T

T

TT

TT

ba

ba

I

I

Modeling Stellar AtmospheresThe parameters for the gas in all stellar atmospheres must obey certain relationships with one another in order to preserve equilibrium in the outer layers of the star,namely:

A temperature distribution T(τ) to account for limb darkening.

Flux constancy, since there is no net loss or gain of energy in a star’s atmosphere, i.e.:

Hydrostatic equilibrium (right):

which gives:

0or

constant

0

rad

o

0

rad

dd

d

d

F

FF

Substitution gives:

For hydrostatic equilibrium:

So:

2c

4eff

0

rad σTd

d

dP

gdz

dPg

d

dP from

o

4effrad

4eff

o

gas

ifc

T

d

dP

c

Tg

d

dP

Structure of Spectral LinesTerminology: Fλ = radiant flux at wavelength λ

Fc = continuum flux expectedλo = wavelength of line centre

The equivalent width of a line corresponds to the width of a box in Å of continuum light absorbed that is identical in area to the integrated area of the spectral line.Full-width at half maximum: (Δλ)½ = width of a spectral line measured between line depths corresponding to one half the line depth at line centre λo.

linetheofdepthc

c FFF widthequivalent

c

c

dWFFF

Line Broadening Mechanisms1. Natural BroadeningAccording to the Heisenberg uncertainty principle, ΔE = ħ/Δt.Electrons spend almost infinite time in ground states of atoms and ions, so ΔE ≈ 0 for n = 1, but very little time in excited levels.

o

2

2

2photon

photon

1

2or

11

2

2

Since

21

ΔtcΔ

ΔtΔtcΔ

Δt

hhcΔΔE

hcE

fi

Textbook Example:The average time spent by an electron in the 1st and 2nd excited levels of hydrogen is Δt = 10–8 s. What is the corresponding expected half-width for the Balmer Hα line, which corresponds to a transition between levels n = 2 and n = 3?

Solution (see textbook):The calculated value for the natural line width is:

which is much too narrow relative to the actual observed line widths for the hydrogen Balmer lines. It can be concluded that natural broadening is not the source of line broadening for the hydrogen lines, although it is presumably important for the spectral lines of heavy elements.

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2. Doppler BroadeningAs a result of the thermal motions of atoms, they are moving relative to one another at fairly large speeds, given by the Maxwell-Boltzmann distribution. Recall the value for the most probable speed which produces Doppler shifts :

So we expect line widths to vary as:

Where the factor of 2 is introduced by the positive and negative velocity shifts. An detailed analysis taking into account contributions to the line across the stellar disk and the true distribution of motions gives:

m

kTv

2mp c

vΔ r

m

Tk

cΔ

22

m

kT

cΔ

2ln222

1

Textbook Example:What is the Doppler width for spectral lines from hydrogen (m = 1.6735 × 10–24 gm) in the Sun, where Teff = 5779 K?

Solution (see textbook):The calculated value for the Doppler line width is:

which is much larger than the natural broadening, although still smaller than the observed line widths for the hydrogen Balmer lines. It can be concluded that Doppler broadening is not the main source of line broadening for the hydrogen lines.

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cm10224

2216

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5

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3. Doppler Broadening with Turbulent MotionsIf there is an additional component of turbulence in the gas, the actual velocities of the gas atoms will be in excess of those predicted by the Maxwell-Boltzmann equation, i.e.:

So we expect line widths to vary as:

2turb

2true

2v

m

kTv

2ln22 2

turb21

v

m

kT

cΔ

Textbook Example:What is the Doppler width for spectral lines from hydrogen (m = 1.6735 × 10–24 gm) in the Sun, where Teff = 5779 K, when the turbulent velocities are ~2 km/s?

Solution (see textbook):The calculated value for the Doppler line width is:

Rotation and pulsation also generate large-scale mass motions that result in line broadening and add to the Doppler broadening of spectral lines.

2211

22524

162turb

s/cm10938.9

s/cm102gm101.6735

577910381.122

vm

kT

A363.0beforeas~2

1 Δ

4. Pressure (and Collisional) BroadeningAn additional source of line broadening is produced by the perturbing actions of passing atoms and ions. Effects can be produced by electric field effects (Stark broadening) or pressure effects (van der Waal’s broadening). The primary result is a damping profile in the spectral line shape that produces broadened line wings.

For pressure broadening the effect can be estimated using for Δt the average time between collisions:

The expected line width should therefore vary as:

Where n is the number density of the atoms and σ is the collisional cross-section.

kT

m

nv

lΔt

2

1o

m

kTn

cΔtcΔ

21 2

o

2

21

Textbook Example:What is the van der Waal’s broadening for Balmer Hα lines in the Sun where n = 1.5 × 1017 cm–3 and for hydrogen σ = π(2ao)2 = π(2 × 0.5292 × 10–8 cm)2 = 3.5189 × 10–16 cm2?

Solution (see textbook):The calculated value line width for van der Waal’s broadening is:

which is approximately the same size as the natural broadening discussed earlier.

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5. Rotational BroadeningAlthough the rotational speed of the Sun is only 2 km/s, in some stars that value can exceed 400 km/s! The effect on the width of a spectral line resulting from the smearing of gas motions on opposite hemispheres of a star reaching ±400 km/s can be estimated using the Doppler effect, i.e.:

For a line at 5000 Å the resulting broadening is:

a very significant amount. The effect of rotational broadening of spectral lines dominates over all other factors at high rotational speeds.

c

vΔ R2

A3.13km/s102.9979

km/s4002A5000

25

c

vΔ R

Rapid rotation affects mainly early-type stars. In late-type dwarfs there is a “break” at spectral type F5 that appears to mark the onset of chromospheres in cooler stars.

SummaryThe effects of the various line broadening mechanisms can be summarized as follows:

Natural broadening: ~0.0001 Å

Van der Waal’s broadening: ~0.0002 Å

Doppler broadening: ~0.4 Å

Turbulent broadening: ~0.4 Å

Rotational broadening: ≤13 Å

Stark broadening: ~5–50 Å

The quadratic sum of all “natural” broadening mechanisms — the damping profile for natural and pressure broadening, as well as Doppler broadening — is referred to as the Voigt profile. In general, Voigt profiles for spectral lines have Doppler cores and damping wings.

For reference purposes, the cross-section for a harmonic oscillator is generally given as:

where the term in brackets is referred to as the Lorentz profile.

o22

o

2

when

2

mc

e

Curves of GrowthThe strength of a spectral line, as indicated by its equivalent width W, is determined by a variety of factors, namely:

i. the abundance of the element producing the line, the greater the abundance the stronger the line,

ii. the transition probability for the line, or the f-value, the higher the probability f the stronger the line,

iii. the population of the energy level where the line originates, the lower the population number for the energy level the weaker the line,

iv. the line broadening mechanism, since mechanisms that produce strong line wings do so at the expense of absorption in the line core,

v. the rotation velocity of the star, and

vi. the electron density Ne, since it governs the damping portion of the line.

Oscillator strength, f, represents the effective number of electrons per atom participating in a transition between energy levels. For abundance studies of stars we want A, the number of absorbing atoms per unit area of a star’s surface that have electrons in the proper energy level for producing a photon at the wavelength λ of the spectral line we are measuring. As more and more atoms contribute to the shape and area of an observed spectral line, the normalized equivalent width W/λ of the line changes in a specific fashion, referred to as the curve of growth.

The three portions of the curve of growth (COG) are:

i. linear portion, where W/λ increases with Aii. plateau, where W/λ is proportional to (ln A)½

iii. damping, where W/λ increases in proportion to A½

The variation of log W/λ with A relative to the shape of the spectral line.

The effect of different amounts of microturbulence ξ (in km/s).

Matching observations of solar titanium I lines to theoretical COG predictions to establish the amount of microturbulence ξ.

The iron I lines in HD 219134 indicate the presence of measurable microturbulence ξ in its atmosphere.

A star’s surface gravity g affects the density of atoms in its atmosphere, affecting the damping portion of the COG directly.

The effect of different excitation energies ξ (χ here) for spectral lines from the same element.

Recall the Boltzmann equation:

So:

For atomic spectral lines originating from the ionization state, location of log W/λ for the lines on the curve of growth depend upon the excitation potential of the level of origin ξm, the excitation temperature Tex, and the oscillator strengths f for the lines.

A plot of horizontal shift Δlog A versus excitation potential ξm for a series of lines then specifies Tex.

Tu

g

N

N mm

m loglog

fT

fTu

gN

fNfN

m

mmm

log5040

constant

logloglog

logloglog totaltotal

A plot of the values for different parameters can often be used to establish the effective temperature and surface gravity in individual stars.

Accurate estimates of stellar abundances are now done using spectral synthesis rather than curve of growth methods, but both depend upon a reliably established temperature distribution, like that for the Sun shown here.

Textbook Example:λ(Å) W(Å) f log W/λ log (fλ/5000 Å)

3302.38 0.088 0.0214 –4.57 –1.855889.97 0.730 0.6450 –3.91 –0.12

Given the data for solar sodium lines above (Na I), what is the abundance of sodium in the Sun?

Solution (see textbook):From the solar curve of growth, the corresponding values for the abundance of Na I atoms producing the lines is given below, along with derived values for the relative abundance of Na II to Na I, and the overall abundance of sodium.

λ(Å) log (fNaλ/5000Å) log (fλ/5000Å) log Na log Nion/Nn log N(Na)3302.38 13.20 –1.85 15.05 3.3856 18.3865889.97 14.83 –0.12 14.95 3.3856 18.386

The inferred abundance of sodium is 2.43 × 1016 cm–2, which implies a mass of 9.3 × 10–5 gm cm–2, (5.4 × 10–5 gm cm–2 from a more detailed analysis), relative to 1.1 gm cm–2 for hydrogen.

How one uses a model curve of growth with measurements of equivalent width to infer element abundance.

Textbook Example:mks

λ(Å) log W/λ log (fλ/5000 Å) log (Nfλ/5000 Å) log N3302.38 –4.57 –1.85 17.20 19.055889.97 –3.91 –0.12 18.83 18.95

λ(Å) log N Correction log (Natot I) Correction log N(Na)

3302.38 19.05 0.0002 19.05 3.386 22.445889.97 18.95 0.0064 18.97 3.386 22.36

The derived abundance corresponds to ~2.5 × 1022 atoms of sodium (Na) per square meter of atmosphere, or ~9.6 × 10–4 kg of sodium (Na) per square meter of atmosphere, compared with 11 kg/m2 for hydrogen (H).

The values (13.20 & rather than 17.20 & 18.83) on the previous slide refer to the abundance in cgs units (i.e. cm–2) rather than mks units (i.e. m–2).