Spectroscopy – Continuous Opacities

Spectroscopy – Continuous Opacities

I. Introduction: Atomic Absorption CoefficentsII. Corrections for Stimulated EmissionIII. HydrogenIV. Negative Hydrogen IonV. Negative Helium IonVI. MetalsVII. Electron ScatteringVIII. OthersIX. Summary

In order to calculate the transer of radiation through a model stellar atmosphere, we need to know the continuous absorption coefficient, .

This shapes the continous spectrum → more absorption, less light.

It also influences the strength of stellar lines → more continous absorption means a thinner photosphere with few atoms to make spectral lines.

Also, before we compute a theoretical spectrum, you need to compute an atmospheric model, and this also depends on .

= –I j/ = –I + S

dI

d

Recall the radiative transfer equation:

To solve this you need to know the opacity. You can have a nice solution, but it will not reproduce observations. The grey atmosphere has a simple opacity, but no bearing with reality.

The problem is that there are a lot of opacity sources which are temperature dependent

B1I V

F0 V

G2 V

I. The atomic absorption coefficent

The total continuous absorption coefficient is the sum of absorption resulting from many physical processes. These are in two categories:

• bound-free transition: ionization

• free-free transition: acceleration of a charge when passing another charge

bound-bound transitions result in a spectral line and are not included in but in cool stars line density is so great it affects the continuum

I. The atomic absorption coefficent

The atomic absorption coefficient, , has units of area per absorber.

The wavelength versus frequency question does not arise for:

=

is not a distribution like I and I, but the power subtracted from I in interval d is d. This is a distribution and has units of erg/(s cm2 rad2 Hz)

d =(c/2) d

II. Corrections for Stimulated Emission

Recall that the stimulated emission (negative absorption) reduces the absorption:

= NℓBℓuh – NuBuℓh

= Nℓ Bℓuh(1– NuBuℓ/NℓBℓu)

= NℓBℓuh[1– exp(–h/kT)]

BulBlu

Nu

Nℓ

T cm K T 1– exp(h/kT) % decrease

0.2 4000 0.999 0.08

0.4 8000 0.973 2.8

0.6 12000 0.909 10.0

0.8 16000 0.834 19.8

1.0 20000 0.763 31.1

Teff (10% reduction)

30000 2500 Ang

3000 2m

· · ·

0.00

10.20

12.08

12.75

13.60

Lym

an

n=1

n=2

n=3

n=4

n=∞B

alm

er

Pach

en

Bra

cket

t

= 13.6(1–1/n2) eV

½ mv2 = h–hRc/n2

R = 1.0968×105 cm–1

III. Atomic Absorption Coefficient for Hydrogen

H

2 3647 Balmer

3 8206 Paschen

4 14588 Brackett

5 22790 Pfund

At the ionization limit v=0, =Rc/n2

½ mv2 = h–hRc/n2

R = 1.0968×105 cm–1

n Ang Name

1 912 Lyman

Absorption edges:

III. Neutral Hydrogen: Bound-Free

Original derivation is from Kramers (1923) and modified by Gaunt (1930):

n = 6.16e6

hR

n53gn

∕

n = 6.16e6

h3c3 R gn∕3

n5=

0 gn∕

n5

a0 = 1.044×10–26 for in angstroms

e = electron charge = 4.803×10–10 esu

gn∕ = Gaunt factor needed to make Kramer´s result in agreement

with quantum mechanical results

Per neutral H atom

0,

10×

–17 c

m2 /

H a

tom

Wavelength (Angstroms)

n =

1 n

=1

n =

2

n =

1

n =

2

n =

3

n =

1

n =

2

n =

3

n =

4

n =

1

n =

2

n =

3

n =

4

n =

5

n =

1

n =

2

n =

3

n =

4

n =

5

n =

6

n =

1

n =

2

n =

3

n =

4

n =

5

n =

6

n =

7

n =

1

n =

2

n =

3

n =

4

n =

5

n =

6

n =

7

~3/n5

~3

The sum of absorbers in each level times n is what is needed.

Recall:

Nn

N=

gn

u0(T) ( kT ) –exp

gn=2n2

=I – hRc/n2 = 13.6(1-1/n2) eVu0(T) = 2


The absorption coefficient in square centimeters per neutral hydrogen atom for all continua starting at n0

(Hbf) = Σn0

∞ nNn

N= 0 Σ

n0

∞ 3

n3 gn∕ ( kT ) –exp

= 0 Σn0

∞ 3

n3 gn∕ 10–


= 5040/T, in electron volts

Unsöld showed that the small contributions due to terms higher than n0+2 can be replaced by an integral:

Σn0+3

∞1n3 ( kT ) –exp

= ½ ∫

n0+3

∞

( kT ) –exp

d(1/n2)

=I – hRc/n2 => d = –Id(1/n2)

Σn0+3

∞1n3 ( kT ) –exp

= ½ ∫3

I

( kT ) –exp d

I

3 = I [1–1

(n0 +3)2 ]


=kTI

[ ( kT–exp ) ( kT

–exp ) ]–

We can neglect the n dependence on gn∕ and the final

answer is:

This is the bound free absorption coefficient for neutral hydrogen

(Hbf) = 03 [Σn0

n0+2 gn∕

n3 10– +log e2I

(10–3 – 10 –I) ]


n =

1

n =

2

n =

3

n =

4

n =

5

n =

6

n =

7


bf(>3647)bf(<3647)

=bf(n=3) + ...

bf(n=2) + bf(n=3) + ...

≈bf(n=3)bf(n=2)

8= 27

exp[ [

–(3 – 2)/kT

= 0.0037 at 5000 K and 0.033 at 10000 K

edge

(Ang)

T

3000

T

5000

T

10000

T

30000

Lyman 9×10–19 6×10–12 9×10–7 0.002

Balmer 3647 0.00021 0.00376 0.033 0.14

Paschen 8206 0.03 0.089 0.177 0.31

Brackett 14588 0.16 0.255 0.36 0.45

Pfund 22790 0.30 0.39 0.48 0.54


(red side)/(blue side)


III. Optical Depth and Height of Formation

Wavelength

Flu

x

912 3647 8602

Recall: = dx ~ 2/3 for Grey atmosphere

As increases, increases => dx decreases

You are looking higher in the atmosphere

continuumAcross a jump your are seeing very different heights in the atmosphere

<3647 A

=2/3

>3647 A

=2/3

z=0

Temperature profile of photosphere

100008000

6000

4000

z=0

Tem

pera

ture

z

zdx1 dx2

(<3647) > (>3647) =>

dx2 > dx1

B4 V

But wait, I just said that the Balmer jump should be larger for cooler stars. Why is this not the case?

For cooler stars other sources of opacity start to dominate, namely H–

Maximum of black body = T = 0.5099 cm K

But peak implies T=13400 K

But peak implies T=13400 K

The stronger opacity of on the blue side of the Balmer jump distorts the Planck curve. One cannot use the peak of the intensity, but must fit the full spectral energy distribution

Wavelength (Ang)

Am

plit

ude

(mm

ag)

Photometric Amplitude of rapidly oscillating Ap stars:

Different wavelengths probe different heights in atmosphere

III. Neutral Hydrogen: Free-Free

The free-free absorption of hydrogen is much smaller.

When the free electron has a collision with a proton its unbound orbit is altered.

The electron can absorbs a photon and its energy increases.

The strength of this absorption depends on the velocity of the electron

III. Neutral Hydrogen: Free-Free

proton

e–

Orbit is altered

The absorption of the photon is during the interaction

III. Neutral Hydrogen: Free-Free Absorption

According to Kramers the atomic coefficient is:

dff = 0.385he2 Rm3

13 v dv

This is the cross section in square cm per H atom for the fraction of the electrons in the velocity interval v to v + dv.

To get complete f-f absorption must integrate over v.

Using the Maxwell-Boltzmann distribution for v

ff = 0.385he2 Rm3

13 ( m kT)

23

exp( mv2

2kT )–v( 2

)2

1

∫0

∞

dv

( 2mkT)ff = 0.385

he2 Rm3 3

1 2

1

Quantum mechanical derivation by Gaunt is modified by f-f Gaunt factor gf


The absorption coefficient in square cm per neutral H atom is proportional to the number density of electrons, Ne and protons Ni:

(Hff) = fgfNiNe

N0 Density of neutral H

Recall the Saha Equation:

(23

25

=Ni

NPe

h3

2m ) ( kT) 2u1(T) u0(T) ( kT ) –exp

I

Pe = NekT


(Hff) = fgf

23

(2mkT) h3 ( kT ) –exp

I

I(Hff) = 03gf

log e10–I

Using: I=hcR R=22me4/h3c

=log e/kT = 5040/T for eV


III. Total Absorption Coefficient for Hydrogen

totalbound-freefree-free

IV. The Negative Hydrogen Ion

The hydrogen atom is capable of holding a second electron in a bound state.

The ionization of the extra electron requires 0.754 eV

All photons with < 16444 Ang have sufficient energy to ionize H – back to neutral H

Very important opacity for Teff < 6000 K

Where does this extra electron come from?

Metals!


For Teff > 6000 K, H– too frequently ionized to be an effective absorber

For Teff < 6000 K, H– very important

For Teff < 4000 K, no longer effective because there are no more free electrons


The bound free absorption coefficient can be expressed by the following polynomial

bf = a0 + a1 + a22 + a33 + a44 + a55 + a66

a0 = 1.99654

a1 = –1.18267 × 10–5

a2 = 2.64243 × 10–6

a3 = –4.40524 × 10–10

a4 = 3.23992× 10–14

a5 = –1.39568 × 10–10

a6 = 2.78701 × 10–23

is in Angstroms

IV. The Negative Hydrogen Ion: Bound-Free

Wavelength (angstroms)

bf, 1

0–18 c

m2 p

er H

– ion

IV. The Negative Hydrogen Ion: bound-free

The H – ionization is given by the Saha equation

log N(H)N(H –)

= –log Pe – 5040T I + 2.5 log T + 0.1248

in eV

(Hbf–) = 4.158 × 10–10 bf Pe

25

100.754

u0(T) = 1, u1(T) = 2

IV. The Negative Hydrogen Ion: free-free

(Hff–) = Peff = 10–26 × Pe 10 f0+f1log+f2log2

f0 = –2.2763–1.685 log+0.766 log2–0.0533464 log3

f1 = 15.2827–9.2846 log+1.99381 log2–0.142631 log3

f3 = –197.789+190.266 log–67.9775 log2+10.6913 log3–0.625151 log4

IV. The Negative Hydrogen Ion: Total

bound-free

free-free

V. The Negative Helium ion

The bound–free absorption is neglible, but free-free can be important in the atmospheres of cool stars and at longer wavelengths

VI. Metals

• In the visible a minor opacity source because they are not many around

• Contribute indirectly by providing electrons

• In the visible (metals) ~ 1% (Hbf–)

• A different story in the ultraviolet where the opacity is dominated by metals

VI. Metals

The absorption coefficient for metals dominate in the ultraviolet

VII. Electron (Thompson) Scattering

• Important in hot stars where H is ionized

• Only true „grey“ opacity source since it does not depend on wavelength

• Phase function for scattering ~ 1 + cos

• Stellar atmosphere people assume average phase ~ 0

The absorption coefficient is wavelength independent:

e =83 (

(e2

mc2

2

= 0.6648 x 10–24 cm2/electron

The absorption per hydrogen atom:

(e) =eNe

NH

ePe

PH

=

PH = Partial pressure of Hydrogen


PH is related to the gas and electron pressure as follows:

N = Nj + Ne = NH Aj + Ne

Nj particles of the jth element per cubic cm and Aj = Nj/NH

Solving for NH NH =

N–Ne

Aj

PH =Pg–Pe

Aj


(e) =

Electron scattering is important in O and Early B stars

Pg– Pe

ePeAj

If hydrogen dominates their composition Pe = 0.5Pg

(e) = e AjIndependent of pressure


Pe /

PT

ot

Teff

PTot = Pe + Pg

0.5

VIII. Other Sources of Opacity

• H2 (neutral) has no significant absorption in the visible

• H2+, H2

– do have significant absorption

• H2+(bf) important in the ultraviolet, in A-type stars it is

~ 10% of H– bound-free opacity

• Peaks in opacity around ≈ 1100 Å, is dominated by the Balmer continuum below 3600 Å in most stars

• H2– (free-free) important in the infrared (cool stars) and

fills the opacity minimum of H– at 16400 Å

H2 molecules


• Important only in O and B-type stars

• He II (bound-free) is hydrogenic → multiply hydrogen cross sections by Z4 or 16.

He I (bound-free), He II (bound-free)


• Important in cool stars

• Scattering by molecules and atoms

• Has a 1/4 dependence

Rayleigh Scattering


• Molecules and ions:

CN–, C2–, H2, He, N2, O2, TiO,....

Cool Stars

Basically Cool Stars are a mess and only for the bravest theoretical astrophysicist

IX. Summary of Continuous Opacities

Spectral Type Dominant opacity

O–B Electron scattering, He I,II (b-f),H(f-f)

B–A H I: b-f, f-fHe II: b-f, some electron scattering

A–F equal contributions from H I (b-b) H I (f-f), H– (b-f, f-f)

G–K H I (b-f), H– (b-f, f-f), RayleighScattering off H I

IX. Summary of Continuous Opacities

Spectral Type Dominant opacity

K–Early M H– (b-f, f-f), Rayleigh scattering

(UV) off H I and H2, molecular

opacities (line blanketing)

M: Molecules and neutral atoms,

H– (b-f, f-f), Rayleigh scattering

off other molecules

Changes in the continuous opacity is the cause of most, if not all pulsating stars

Mechanism: If in a region of the star the opacity changes, then the star can block energy (photons) which can be subsequently released in a later phase of the pulsation. Helium and and Hydrogen ionization zones of the star are normally where this works. Consider the Helium ionization zone in the interior of a star. During a contraction phase of the pulsations the density increases causing He II to recombine. Neutral helium has a higher opacity and blocks photons and thus stores energy. When the star expands the density decreases and neutral helium is ionized by the emerging radiation. The opacity then decreases.

He II/He III ionization zone

Contraction

During compression He II ionized to He III, He III has a higher opacity. This blocks radiation causing star to expand

Cepheid Pulsations are due to an opacity effect:

During expansion He zone cools, He III recombines, opacity decreases allowing photons to escape. Star then contracts under gravity.

Expansion

Most pulsating stars can be explained by opacity effects

The End….our only friend the End

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Spectroscopy – Continuous Opacities