The Stellar Opacity

The mean absorption coefficient, κ, is not a constant; it is de-pendent on frequency, and is therefore frequently written as κν .Inside a star, several different sources of opacity are important,and each has its own frequency dependence. To compute the ap-propriate “average” opacity, κ recall that Fick’s law states thatthe flux at any point is related to the energy density U by

Fν = −D∇U = −1

3vl∇n = −1

3

c

κνρ∇U

and that, when integrated over all energies,

F = −4acT 3

3κρ

dT

dr(3.1.4)

But how is κ related to κν? To answer this question, we start byexplicitly writing in the frequency dependence of U , i.e.,

Uν =4π

cBν(T ) =

4π

c

2hν3

c2

{1

ehν/kT − 1

}and compute the monochromatic flux distribution

Fν = −D∇Uν = − c

3ρ

1

κν∇{4π

cBν(T )

}= −4π

3ρ

1

κν

dBν

dT

dT

dr

The total flux, integrated over all frequencies is then

F = −4π

3ρ

dT

dr

∫ ∞

0

1

κν

dBν

dTdν (6.1.1)

Comparing (6.1.1) and (3.1.4), we see that

1

κ=

π

acT 3

∫ ∞

0

1

κν

dBν

dTdν (6.1.2)

or, since ∫ ∞

0

dBν

dTdν =

d

dT

∫ ∞

0

Bνdν =dB

dT=

ac

πT 3

1

κ=

∫ ∞

0

1

κν

dBν

dTdν∫ ∞

0

dBν

dTdν

(6.1.3)

The average opacity that you should use is therefore that whichis harmonically weighted over the temperature derivative of thePlanck curve. This is called the Rosseland mean opacity.Note that the function weights the high frequencies more thanthe low frequencies. To see this, we can take the temperaturederivative of the Planck function

dBν

dT=

2h2ν4

c2kT 2

ehν/kT{ehν/kT − 1

}2=

(2k3T 2

c2h2

)(x4ex

{ex − 1}2

)(6.1.4)

where x = hν/kT , and find its peak by taking the derivative.Then (

2k3T 2

c2h2

){x4ex + 4x3ex

(ex − 1)2 − 2x4e2x

(ex − 1)3

}= 0

x3ex (ex − 1)−3

(xex − x+ 4ex − 4− 2xex) = 0

x3ex (ex − 1)−3

(4ex − 4− x− xex) = 0

Mathematically, this works out to x ≈ 3.83. In other words, thehigh energy photons count much more than low energy photons.

Although in real models, the Rosseland mean opacity is com-puted numerically, it is instructive to evaluate the form of κwhen κν ∝ ν−n. From the definition of the Rosseland mean

1

κ∝

2h2

c2kT 2

∫ ∞

0

νn+4 ehν/kT(ehν/kT − 1

)2 dν

2h2

c2kT 2

∫ ∞

0

ν4ehν/kT(

ehν/kT − 1)2 dν

which, in terms of x is

1

κ∝

∫ ∞

0

(kT

h

)n+4

xn+4 ex

(ex − 1)2

(kT

h

)dx∫ ∞

0

(kT

h

)4

x4 ex

(ex − 1)2

(kT

h

)dx

or

1

κ∝(kT

h

)n

∫ ∞

0

xn+4 ex

(ex − 1)2 dx∫ ∞

0

x4 ex

(ex − 1)2 dx

∝ Tn (6.1.5)

Thus, if the opacity law is κ ∝ ν−n, the temperature dependenceof the Rosseland mean opacity will be κ ∝ T−n. This followssince the mean opacity must drop as the number of high energyphotons increases.

ELECTRON SCATTERING

When free electrons have thermal motions that are much lessthan their rest mass energy, i.e., mec

2 ≫ kT , or T ≪ 5.9 ×109 K, the cross-section for scattering of photons off electrons isfrequency independent. The opacity per unit mass of material inthis case is

κe =neσe

ρ

where σe is the Thomson cross-section of the electron

σe =8π

3

(e2

mec2

)2

= 0.665× 10−24 cm2

and ne is the density of electrons

ne = ρNA

∑i

(xi

Ai

)fiZi (5.1.4)

If the gas is fully ionized, i.e., if the ionization fractions fi ≈ 1 (asit will be in stellar interiors), and if we approximate the atomicweight (Ai) of a metal as twice its atomic number (Zi), then

ne = ρNA

∑i

Zixi

Ai≈ ρNA

{X

1+

Y

2+

Z

2

}where X, Y , and Z are the mass fractions of hydrogen, helium,and metals, respectively. Since X + Y + Z = 1,

ne = ρNA

{1

2+

1

2X

}(6.1.6)

and

κe = NAσe(1 +X)/2 = 0.2 (1 +X) cm2 g−1 (6.1.7)

Note again that as long as the ionization is complete (and matteris not degenerate), the result is independent of temperature anddensity.

FREE-FREE ABSORPTION

The energy emitted per unit volume by a single electron scatteredby ions of charge Z (as a function of velocity, v) is

dE

dωdV dt=

16πe6

3√3c3m2

evnenIZ

2gff(v, ω)

where e is the charge of the electron, and gff is the gaunt factor.To get the emissivity of thermal bremsstrahlung as a function offrequency, we integrate this over a Maxwellian distribution, notingthat to get a photon with frequency ν, the incident electron musthave mev

2/2 ≥ hν. Thus, vmin = (2hν/me)1/2 and

ϵff(ν) = 4πjff =

∫ ∞

vmin

dE

dωdV dt4π( me

2πkT

)3/2v2e−mev

2/2kT dv

=25πe6

3mec3

(2π

3mekT

)1/2

gff ne nI Z2 e−hν/kT

Next, to go from emission to absorption, we can use the fact thatin thermodynamic equilibrium, Kirchoff’s law must hold, hencethe source function is Sν = jν/aν = Bν(T ), where aν , is theabsorption coefficient per electron (with units of inverse distance).Thus

aff(ν) = ρκν =4e6

3mehc

(2π

3mekT

) 12

gff ne nI Z2 ν−3

{1− e−hν/kT

}(6.1.8)

or, upon substituting ρ = neµe/NA and ρ = nIµI/NA

κν =4e6

3mehc

(2π

3mek

) 12 N2

AZ2

µeµIgff ρ T−1/2ν−3

{1− e−hν/kT

}(6.1.9)

The first term in the equation gives the total absorption coefficient;the exponential term at the end represents a negative contributionfrom stimulated emission.

If we combine the constants, letting

C =4e6

3mehcN2

A

(2π

3mek

)1/2

= 1.34× 1056 (cgs units)

then

κν = C

(Z2

µeµI

)gff ρ T− 1

2 ν−3{1− e−hν/kT

}(6.1.10)

To calculate the Rosseland mean free-free opacity, the inverse ofthis expression, weighted by the derivative of the Planck Function,must be integrated over all frequencies. This is tedious, but notethat if we neglect the effects of stimulated emission, the frequencydependence of the opacity is ν−3, which, through (6.1.5), impliesa temperature dependence of T−3. This, plus the T−1/2 depen-dence which is already in the equation, gives a total temperaturedependence of T−7/2. Numerically,

κff ∼ 1023Z2

µeµIρ T−7/2 cm2 g−1 (6.1.11)

Opacities of the form κ ∝ ρ T−7/2 are called Kramers opacities.

BOUND-FREE ABSORPTION

For a hydrogenic atom, the cross section for absorption from state(n, l) to the continuum is

σbf(Z, n, l) =29π7mee

10Z4

3√3 c h6n5

gbf

(2πν)3 =

64π4mee10

3√3 c h6

Z4

n5

gbfν3

(6.1.12)where the energy of the incident photon must be great enough toionize the atom, i.e.,

hν > χn =2π2mee

4

h2

Z2

n2

From this, it is clear that the opacity law for a single species ofatom will have a number of absorption “edges” below which theabsorption cross section will drop quickly. The total bound-freeopacity, of course, will be the sum over all elements, Z, all ioniza-tion states, i, and and all excitation states, n. In the hydrogenicion approximation, the opacity is

κν =∑Z

∑n

σbf(Z, n, l)

(Ni,n

Ni

)Z

(Ni∑Ni

)Z

ρNA

AZ(6.1.13)

This summation, along with the process of taking the Rosselandmean, will act to smear out many of the edges.

The bound-free opacity must be computed numerically, but we canget a rough idea of how bound-free opacity behaves by making theassumption that in a star’s interior, all ions will be very highlyionized and thus will have at most one electron. The Saha equationstates that the number of atoms in ionization state i is related tothe number in ionization state i+ 1 by

Ni

Ni+1=

Ne

2

ui(T )

ui+1(T )

(h2

2πmekT

)3/2

eχi/kT (5.2.5)

where u(T ) are the partition functions and χi is the ionizationenergy. If the gas is highly ionized, Ni+i ≈ 1, ui+1 = 1, andui = 2, and

Ni ≈ Ne

(h2

2πmekT

)3/2

eχi/kT

We can then estimate the number of these atoms in excitationstate n from the Boltzmann equation. For hydrogenic ions, thestatistical weight of a level is n2, so if we let χi,n = χi + χn, thenthe total number of absorbing ions is

Ni,n ∼ Ne n2

(h2

2πmekT

)3/2

eχi,n/kT

The total absorption coefficient is therefore

aν ≈ σbf Ni,nρXZNA

AZ

≈ 64π4mee10

3√3 c h6

Z4

n5

gbfν3

(ρ2N2

AXZ

µeAZ

)n2

(h2

2πmekT

) 32

eχi,n/kT

≈ 16πe6Z2gbf3mec h nν3

(2π2mee

4

h2kT

Z2

n2

)ρ2N2

AXZ

µeAZ

(1

6πmekT

) 12

eχi,n/kT

≈ 16πe6Z2gbf

3mec h n√6πmek

(χi,n

kTeχi,n/kT

)(ρ2N2AXZ

µeAZ

)T− 1

2 ν−3

and the opacity is

κbf = aν/ρ ≈∑z,n

C(χi,n

kTeχi,n/kT

)(Z2XZ

µeAZ

)ρ T− 1

2 ν−3

(6.1.14)

where we have dumped the many constants into the variable C(and have ignored the stimulated emission term).

Now examine the factor χi,n/kT . If for a particular n and Z,χi,n ≪ kT , then that term will contribute little to the opacity,since the species will have too few bound electrons. Similarly, ifχi,n ≫ kT , there will not be many photons of the requisite energyto be absorbed, and the contribution to the opacity will again besmall. Thus, only those terms with χi,n ∼ kT will contribute.This leaves κ ∝ ρ T−1/2ν−3, which is the same as free-free opac-ity. Bound-free absorption will therefore also follow the Kramersopacity law.

A rough numerical estimate for the Rosseland mean for bound-freeopacity is

κbf ∼ 1025 Z(1 +X) ρ T−7/2 cm2 g−1 (6.1.15)

Note that this is a factor of ∼ 100 greater than the Rosselandmean for free-free opacity.

To first order, the cross-section for ionization of most species goesas ν−3 for frequencies above the ionization energy.

BOUND-BOUND ABSORPTION

At extremely high temperatures, T > 106 K, the radiation field willconsist mostly of ionizing photons, hence bound-bound absorptionwill be minimal (< 10%). However, at cooler temperatures theonset of bound-bound transitions in the UV and far UV can doublethe stellar opacity.

Obviously, the computation of bound-bound opacity must be donenumerically, since it involves the summation of millions of individ-ual absorption lines. It should be noted, however, that in theinterior of a star, the pressure (and thermal) broadening termswill be substantial, so the net result will be continuum absorp-tion, rather than a series of absorption lines. Also, the expressionfor the absorption coefficient of hydrogenic bound-free transitions(6.1.12) is the same as that for bound-bound transitions. Thus, wecan expect the same ∼ ν−3 dependence and therefore a Kramer-like opacity law. The strength of the total absorption, however,will be critically dependent on temperature.

H− OPACITY

A free electron passing by a neutral hydrogen atom can producea dipole moment in the atom. At that time, an extra electron canattach itself to the system; the result is an H− ion. This ion isfragile – it has an ionization energy of only 0.754 eV, and a largecross section for absorption, either through bound-free absorption

H− + hν ⇀↽ H + e−(v)

or free-free absorption

H− + e−(v) + hν ⇀↽ H− + e−(v)

In the envelopes of cool stars, H− can be the dominant source ofopacity.

The contribution of H− to the star’s opacity will depend on thenumber of H− ions, which can be calculated from the Saha equa-tion.

nH−

nH0

=ne

2

uH−

uH0

(h2

2πmekT

)3/2

eχ/kT (5.2.5)

where uH− = 1, uH0 = 2, and χ = 0.754 eV. If x is the fractionof ionized hydrogen (x ∼ 0 for cool stars), and aν is the frequencydependency of H− absorption, then the opacity due to H− is

κH− =aνnH−

ρ= nH (1− x)

(nH−

nH0

)(aνρ

)

= aνXNA (1− x)ne

4

(h2

2πmekT

) 32

eχ/kT

= aνXNA(1− x)

4

Pe

µekT

(h2

2πmekT

) 32

eχ/kT(6.1.16)

Note that the total opacity is directly proportional to the electrondensity. In extremely metal poor stars, the free electrons mustcome from ionized hydrogen, hence the opacity will rise with tem-perature, until the number of neutrals starts to decline. However,for normal metal rich stars, common, low ionization metals (suchas Ca, Na, K, and Al) contribute to the supply of free electrons,thereby lessening this dependence. Consequently, H− opacity ismuch stronger in metal rich stars than in metal poor stars, and isstronger in dwarfs than in giants (due to the pressure term).

The determination of the frequency dependent cross-section of theH− ion involves the quantum mechanical computation of overlap-ping wave functions. It is not a power law: it increases linearly atshort wavelengths until ∼ 8500 A, then declines until the ioniza-tion threshold at ∼ 16, 500 A. After that, that opacity increasesagain due to free-free opacity. As a result, the Rosseland mean ofH− opacity is not a Kramers law opacity. Very roughly

κH− ∼ 2.5× 10−31(Z/0.02) ρ1/2 T 9 cm2 g−1 (6.1.17)

The combined bound-free and free-free opacity from the H− ionat T = 6300 K. The y-axis plots the cross-section (×10−26 perneutral H atom per unit electron pressure, Pe = nekT ). FromMihalas, Stellar Atmospheres (2nd edition), pg. 103.

This figure illustrates the importance of the various opacities, asa function of temperature and density, for a normal (PopulationI) mix of abundances. The lines are labeled in units of electronscattering opacity, κ0 = κe = 0.2(1 + X). From Hayashi, Hoshi,& Sugimoto 1962, Suppl. Prog. Theo. Phys. Japan, Vol. 22.

The Los Alamos National Labs radiative opacities for X = 0.7,Y = 0.28, and Z = 0.02. The dashed line shows the half-ionizationcurve for pure hydrogen.

The Rosseland mean opacity (in cm2 g−1) as a function of density(in g cm−3) and temperature for X = 0.739, Y = 0.24, and Z =0.24. The range of values plotted is applicable for the outer regionsof stars; the dots represent values in the solar model.

Tabulated Opacities

When computing real stellar models, simple power-law represen-tations of the Rosseland mean opacity are inadequate. Instead,one uses tables created either by The Opacity Project

http://cdsweb.u-strasbg.fr/topbase/TheOP.html

or OPAL

http://opalopacity.llnl.gov

One chooses the table appropriate for the metallicity being consid-ered, and then interpolates in density and temperature (or relatedvariables). Note that most of these tables assume a fixed metaldistribution (i.e., the amount of given element, such as oxygen,scales with that of iron). However, a few do allow for independentchanges in CNO, or α-process elements.

Opacity versus temperture for various measures of R = ρ/T 36 . The

plots are for solar composition of 17 elements, including H, He, C,N, O, Ne, Na, Mg, Al, Si, S, Ar, Ca, Cr, Mn, Fe, and Ni. Fromthe Opacity project.

The Eddington Luminosity

If energy transport at the surface of a star is through radiation,then equation (3.1.4) holds, and the star’s luminosity is related toits temperature gradient by

F =LT

4πR2= − 4ac

3κρT 3 dT

dr(6.2.1)

or

LT = −16πac

3κρR2T 3 dT

dr(6.2.2)

Now assume that the dominate source of pressure is from radiation.(At the surface of a bright star, this will certainly be true.) Forradiation pressure,

Prad =1

3aT 4 (6.2.3)

which impliesdP

dT=

4

3aT 3 (6.2.4)

and

LT =

(−16πacR2T 3

3κρ

)dT

dr

(3

4aT 3

)dP

dT= −4πc

κρR2 dP

dr(6.2.5)

Now if the star is in hydrostatic equilibrium,

dP

dr= −GMT

R2ρ (2.2.2)

which implies

LT =4πcGMT

κ(6.2.6)

This is the Eddington luminosity. If the luminosity of the star isgreater than this, the surface cannot be stable, and will be ejectedfrom the star via radiation pressure.

Note that equation (6.2.6), coupled with the mass-luminosity re-lation on the main sequence, implies that there is an upper limitto the mass of main sequence stars. If, for main sequence stars,LT ∝ M3.5 and electron scattering is the dominant source of opac-ity at the surface, then(

MM⊙

)2.5<∼

4πGc

0.2(1 +X)

M⊙

L⊙(6.2.7)

This equation implies that the maximum mass of a main-sequencestar is Mmax ∼ 60M⊙. Stars more massive than this must havestrong stellar winds and will lose mass rapidly.