Lecture 7 : Opacity

Lecture 7 : Opacity

7.1  Bound-Free Opacity

Bound-free opacity takes place when a photon interacts with an electron which is bound to an atom. Unbound electrons have energy E > E¥ while bound electrons have E £ E¥. The energy of a level n is given by En

En = - me Z2 e4
2 \hbar2 n2
(1)

From an astrophysical point of view, H is the most important atom since it makes up the major fraction of matter both in stars and in the insterstellar gas. In this case the energy levels are arranged so that

En = - Z2 me e4
2 n2 k2
= - Z2 \cal R
n2
(2)

where \cal R is the Rydberg constant. In the case n = 1 we get the binding energy,

-E1 = 1
2
Z2me e4
\hbar2
= 13.6 Z2 eV
(3)

and hence \cal R = 13.6 eV.

Important upper wavelengths produced by capture of an electron into the three lowest levels of the H atom are shown in Table 7.1, and are called the Lyman, Balmer and Paschen series. They are in the ultra-violet (UV), optical and infra-red (IR) regions repsectively.

The most important of these for historical reasons is in the optical region, the Balmer series. The first few Balmer lines are shown in Table 7.2.

These lines are very clear in hot stars of spectral type from O to F and strongest at about spectral type A (see figure 7.2). We'll look at the reasons for this later on in the lecture.

We consider the two possible processes of electron capture and electron photoionisation. In both cases the energy is conserved:

1
2
me v2 = hn+ En
(4)

where the electron has velocity v. The major difference between the bound-free and free-free processes is that we now need to take into account the quantisation of the energy levels of the atom.

It turns out that most of the electron's kinetic energy is radiated away as it approaches the ion, as in the free-free case, and so the form of the opacity for the bound-free case is quite similar to free-free emission. The opacity has the form

sCN = 32
3
p4 Z4 e10
me c3 h4 v2 nn3
(5)

In a stellar context, in order to compute the opacity of the plasma, we need to know the relative amounts of the various ions, which may be parameterised conveniently by X, Y and Z, and what ``state'' of ionisation they are in i.e. in which energy levels the electrons are sitting.

Table 7.1: Important lower level transitions and their wavelengths for Hydrogen
Lower Series Symbol Wavelength Energy
level name l eV
n = 1 Lyman La 1216 Å 0
n = 2 Balmer Ha 6563 Å 10.19
n = 3 Paschen Pa 18750 Å 12.07

Table 7.2: Some Balmer lines in the optical region
Name transition wavelength l
Ha n = 3 ® 2 6563 Å
Hb n = 4 ® 2 4861 Å
Hg n = 5 ® 2 4340 Å
Hd n = 6 ® 2 4101 Å
H¥ n = ¥® 2 3647 Å

7.2  The Boltzmann Distribution and Saha Equation

In order to determine the amount of Bound-Free opacity we need to know the state of the gas - i.e. which ions are present, in what quantities and the state (energy level) of the electrons. This is determined by the Boltzmann equation

NA
NB
= gA
gB
 exp æ
ç
è
- EB-EA
kT
ö
÷
ø
(6)

where NA and NB are number densities of atoms in state A and state B, EA and EB are the energies of the states, and gA and gB are the statistical weights of the levels (i.e. the number of states permitted at that energy level).

The Boltzmann equation allows us to compute the ratio of atoms of a certain species, such as Hydrogen, in various possible energy states as a function of temperature.

For Hydrogen gn = 2 n2, which is the number of spin and angular momentum states for each energy level En. For a given energy level n, the quantum number l can take values from 0 to n-1, while for a given l the quantum number m takes values from -l to l. This makes n2 states, and the electrons themselves can be either spin-up or spin-down, so that the total number of states is 2n2.





Problem 7.1 Consider the solar surface to be a plasma at Teff = 5800 K. Show that the typical electron energy at this temperature is kT » 0.5 eV.



The previous problem shows that there is a very small energy per electron, much less than the energy required to get from the ground state to the first excitation level (see Table 7.1). This indicates that the H is not strongly ionised. The amount of inonisation can be computed from the Boltzmann equation.





Problem 7.2 Use the Boltzmann Equation to show that the expected number of atoms in the first excited state n = 2 relative to the ground state n = 1 in the Solar atmosphere (T = 5800 K) is

NA
NB
= g2
g1
 exp æ
ç
è
- E2-E1
kT
ö
÷
ø
» 10-8
(7)

This shows that at solar surface temperatures a very small fraction of Hydrogen is in the excited state.



For atoms of different species, one makes use of the Saha equation:

Ni ne
N
= 2 gn
g
æ
ç
è
2pme kT
h2
ö
÷
ø
[(3)/2]

 
 exp æ
ç
è
- c
kT
ö
÷
ø
(8)

where N is the number of atoms, Ni is the number of ions, ne is the number density of electrons and c is the energy of the level.

Note that the Saha equation is the equivalent of the Boltzmann equation applied to ionisation states, rather than energy states within the same ionisation state.

Expressing the electron number density in cm-3, temperature T in K, and c in eV, Saha's equation takes the form

NI+1
NI
= 4.829 ×1015 UI+1
UI
ne-1 T[3/2]  exp æ
ç
è
- c
kT
ö
÷
ø
(9)

where U is called the partition function and is given by

Ui = gi0 + gi1 e-Ei1/kT + + gi2 e-Ei2/kT + ...
(10)

By starting from the ground state (I = 0) one can thus determine the occupancy of the ionised states (I = 1, 2, etc). For example, for the ionisation of Hydrogen we get

N1
N0
= 2.4 ×1015 T[3/2]  exp( -1.58 ×105 T-1 ) ne-1
(11)

and for the first ionisation of Helium we get

N1
N0
= 9.6 ×1015 T[3/2]  exp( -2.85 ×105 T-1 ) ne-1.
(12)

where ne is in cm-3.





Problem 7.3 In the so-called line forming regions of the Solar atmosphere, where the absorption typically takes place to form the lines in the spectrum, we have

T = 6390  \mathrm K
(13)

and

ne = 6.4 ×1013  cm-3
(14)

Show that for Hydrogen

N1
N0
» 4 ×10-4
(15)



A similar analysis for Calcium shows that for first ionised electron for which c = 6.113 eV, we have

N1
N0
» 103
(16)

and for the second ionised state (c = 11.871 eV)

N2
N1
» 10-2
(17)

Practically all the Ca is in the singly ionised state. It is for this reason that the Calcium lines (and many other low ionisation energy ``metals'' such as Na, Mg, Cl, Cr, Mg, Fe etc) show prominent lines in the Solar spectrum. These lines are called the Fraunhofer lines. Some prominent lines are shown in Figure 7.1 for a G5 dwarf star, which is quite similar to the Sun.


Figure 7.1: Some of the prominent Fraunhofer lines in a G5 dwarf star

The strength of the main lines in stellar spectra varies strongly with spectral type, or in other words, temperature. The spectral types at which various lines peak in their strength, and the range of spectral types in which they are strong, is shown in Table 7.1.

The problem above suggests that in hotter stars than the Sun, Hydrogen lines should become much more prominent. Figure 7.2 shows the spectrum of an A0 dwarf star, for which the effective surface temperature is about 10,000 K, showing that the H lines are in fact much more prominent. The next two problems examine this in mode detail by computation of the strength of the Hydrogen line as a function of temperature.


Figure 7.2: Illustration of the very prominent Hydrogen lines in a hot, A0 star.





Problem 7.4 Use Gnuplot or write a program to plot the fraction of Hydrogen atoms in the first excited state, as a function of temperature T in the range 5000 < T < 100,000 K. Show that at a temperature of about 85,000 K the excitation ratio is about 1:1. What fraction of the Hydrogen is not actually ionised at this high temperature?







Problem 7.5 Write a program or use Gnuplot to plot the fraction of un-ionised Hydrogen as a function of temperature in the range 5000 < T <30,000 K for an assumed electron density of ne = 6.4 ×1013 cm-3. Combine the curve obtained with the curve from problem 7.4 to obtain a plot of the relative strength of the Balmer lines series as a function of temperature. The curve should peak at about 10,000 K.




Figure 7.3: Illustration of the change in the strength of the Balmer series as a function of spectral type, or temperature, along the main sequence.

Table 7.3: Strength of various prominent species in stellar spectra, showing the spectral types at which they are most prominent and the range of spectral types over which they can be seen. Note that spectral types (OBAFGKM) are divided into 10 subclasses indicated by an integer from 0 to 9, going from earlier to later spectral type (e.g. A3, B8, G2). Neutral species are indicated by ``I'' and singly ionised species by ``II''.
Species Peak Range
H I A0 O - K
He I B5 B0 - A0
He II O O - B
Fe II F0 A - M
Ca II K5 A0 - M5
Sr II K5 F0 - M5
TiO M M0 - M9

From the Saha Equation, a list of the relative abundances of different elements in the Sun and the energy levels of these elements, we can calculate the degrees of ionisation of all atomic species. This gives us the density of the species as a function of the electron density ne and the temperature T. Assuming that the Sun is electrically neutral, means that the source of the electrons is just the ionised atoms themselves, allowing us to determine ne. In the atmosphere of the Sun, where the lines are actually formed, one can construct a model of the temperature, density and pressure as a function of height (or optical depth), and from the line strength and electron density derive the amount of each element or abundance present in the Sun. This calculation is in principal straight forward but in practice it is a lot of work. Table 7.4 shows the relative abundances of the elements in the Solar atmosphere. This relative pattern of abundances is very typical of stars near the Sun (i.e. the so-called Population I stars).

Table 7.4: Relative abundances of the first 40 elements in the Sun, normalised to Si for which the abundance is set to 1×106.
Element Abundance Element Abundance
H 3.2×1010 Sc 35
He 2.1×1010 Ti 2.8 ×103
Li 49.5 V 262
Be 0.81 Cr 1.3 ×104
B 350 Mn 9.3 ×103
C 1.2 ×107 Fe 8.3 ×105
N 3.7 ×106 Co 2.2 ×103
O 2.2 ×107 Ni 4.8 ×104
F 2.5 ×103 Cu 540
Ne 3.4 ×106 Zn 1.2 ×103
Na 6.0 ×104 Ga 48
Mg 1.1 ×106 Ge 115
Al 8.5 ×104 As 6.6
Si 1.0 ×106 Se 67.2
P 9.6 ×103 Br 13.5
S 5.0 ×105 Kr 46.8
Cl 5.7 ×103 Rb 5.9
Ar 1.2 ×105 Sr 26.9
K 4.2 ×103 Y 4.8
Ca 7.2 ×104 Zr 28


Figure 7.4: Relative abundances of the first 40 elements in the Sun, normalised to Si at 1 ×106. The up-down pattern of adjacent elements is readily apparent.

7.3  Gas State and the Bound-Free Opacity

Following on from section 7.1, we have the cross-section for photon emission in the bound-free case in terms of the electron velocity v for a state of a Hydrogen-like atom n.

sCN = 32
3
p4 Z4 e10
me c3 h4 v2 nn3
(18)

The reverse of this process is photo-ionisation, or the ejection of an electron by the incoming photon. In thermodynamic equilibrium, the capture and photo-ionisation processes must balance. Furthermore, the electron velocity distribution is Maxwellian. These taken together can be used to show that the cross-section of level n is

s(n,n) = 64p4
3 Ö3
mZ4 e10
h6 c n3 n5
gbf(n,n)
(19)

where gbf is the bound-free Gaunt factor, and similarly to the free-free case, is a slowly varying function of order unity taking into account corrections for quantum effects.

Photons involved in this process must have an energy greater than the binding energy of the level n considered. Hence, for hn< En, the cross-section is identically zero. In terms of the Hydrogen-like atom, we may express this as

hn< IH Z2
n2
(20)

or that the energy level must satisfy

n >   æ
Ö

IH Z
h n
 
.
(21)


Figure 7.5: Bound-free cross-section sbf near an absorption edge. Below the frequency n0 = IHZ2/hn2 the opacity is zero, while above it the opacity falls off like n-3.

Given the population of excited states in the star, the total opacity kbf due to bound-free absorption can be computed.

kbf =
å
n 
Nn
r
s(n,n)
(22)

where Nn is the number density of atoms in state n and r is the (mass) density. Substituting this expression into the form for the Rosseland mean opacity, one derives

kbf = k0 rT-3.5
(23)

where k0 is

k0 = 4.3 ×1025 Z (1+X)
_
g
 
bf

t
(24)

where t is a correction factor (called the guillotine factor) which takes into account low temperature effects of the number of electrons in bound states per atom not considered here.

7.4  Electron conduction

In a plasma of electrons and ions, the electrons have much greater typical velocities than the ions do, so that the electrons are able to transfer heat through the system. Opacity associated with electron conduction is in general much smaller than other sources of opacity, and only becomes significant at quite high densities or temperatures, or when the number of electron states available starts to become limited (i.e. the plasma starts to become degenerate).

Consider electrons of mean velocity [`v] transfering energy outward along a direction x, and that the electrons have a mean free path le. The electron flux is ne [`v] where ne is the electron density. The conducted flux Fe is then

Fe = -ne _
v
 
e
x
.
(25)

where e is the energy of the electrons, and e/x is the gradient of the energy over the distance traveled.

If we consider non-relativistic electrons, the energy is e = 3kT/2, so that

e
x
= 3k
2
T
x
.
(26)

We now define the thermal conductivity lc of a medium in terms of the temperature gradient along x

heat flow = lc T
x
(27)

and so the conducted flux is

Fcond = -lc DT.
(28)





Problem 7.6 Show that the thermal conductivity for electrons is given by

lc = 3k
2
ne _
v
 
le,
(29)

where le is a scale length for thermal conductivity.



The electrons are scattered by the ions. We may relate the kinetic energy me[`v]2/2 of the electrons to the size of the scattering region around each ion r0 by equating with the potential energy of the ion Ze2/r0,

1
2
me _
v
 
2 = Ze2
r0
(30)

so that the cross-section for scattering, s = pr02 is

s = 4p
9
Z2e4
k2T2
.
(31)

The electron density in an ionised gas of density r, and Hydrogen mass fraction X is

ne = r(1+X)
2 mH
(32)

and the ion density ni is given by

ni = r
A mH
(33)

where A is the atomic weight. Using the relationship between mean free path and cross-section and the above expressions for ne and ni, one can show that

lc = 27Ö3
16p
k7/2
me7/2 e4
A
Z2
( 1+X ) T5/2.
(34)

This can be expressed as an opacity for conduction kc, by defining

kc = 4acT3
3rlc
(35)

so that

kc = 5 ×103 Z2/A
1+X
[Ö(T/107)]
0.1r
(36)

where r is in g cm-3.

7.5  Total Opacity

In general, putting together all the sources of opacity is quite complicated, particularly in the outer stellar layers where the frequency dependence must be carefully considered. In stellar interiors, it is usually adequate to use frequency averaged opacity, such as the Rosseland mean. Also, in stellar interiors, the mostly ionised state of the gas means that the main sources of opacity are due to free-free, bound-free and electron scattering, for all of which relatively simple expressions can be computed. A summary of these is shown in Table 7.5.

Table 7.5: Simplified expressions for the major sources of stellar opacity. Temperature is in K and density is in g cm-3.
Opacity Source Expression

e- scattering

ke = 0.2(1+X)

free-free

kff = 3.9 ×1022 (1+X) (1-Z) [`g]ffrT-[7/2]

bound-free

kbf = 4.3 ×1025 Z (1+X) [([`g]bf)/t] rT-[7/2]

e- conduction

kc = 5 ×103[(Z2/A)/(1+X)][([Ö(T/107)])/(0.1r)]

References

[]
R. Bowers and T. Deeming. 1984. Astrophysics I, Stars, Jones and Bartlett Publishers.

[]
C.R. Kitchin. 1987. Stars, Nebulae and the Interstellar Medium. Bristol and Boston.

[]
E.Böhm-Vitense. 1992. Introduction to Stellar Astrophysics, Vol 3. Cambridge University Press.


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