Transcript Continuous Opacity Sources

Continuous Opacity Sources
Continuous Opacity Sources
Principal Sources:
–
–
–
–
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Bound-Bound Transitions
Bound-Free
Free-Free (Bremstralung)
Electron Scattering (Thompson & Compton)
Molecular Transitions
We consider only H or H-Like cases
– We will do this classically (and correct to the QM result)
– Stars are mostly H (in one form or another).
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Dominant Opacity Sources
Type of Star
Cool Stars (GM) of Normal
Composition
Warmer A-F
stars
Interiors and
hottest stars
Species
Type
H-
b-f
Dominant
H-
f-f
Secondary
H
b-f
Dominant
H
f-f
Secondary
H
f-f
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Caveats and Details
At High Temperatures: (1-e-hν/kT) → 0 so all of the
bb, bf, and ff sources go to 0!
Electron scattering takes over (is always there and
may be important).
– Free-Free is not the same as electron scattering:
Conservation of momentum says a photon cannot be
absorbed by a free particle!
In principal we start with a QM description of the
photon - electron interaction which yields the cross
section for absorption/scattering of the photon of
energy hν, call the cross section ai(ν).
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The Opacity Is
The opacity (g/cm2) for the process is
Κi(ν) = niai(ν)/ρ
ni is the number density (#/cm3) of the operant
particles
The subscript i denotes a process/opacity
The total opacity is
Κtotal = ∑Κi(ν)
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Compton Scattering
This process is important only for high energy
photons as the maximum change is 0.024Å.
Reference: Eisberg -- Fundamentals of
Modern Physics p. 81 ff.
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Electron Scattering Conditions
Collision of an electron and a photon
– Energy and Momentum Must be conserved
In stellar atmospheres during photon-electron
collisions the wavelength of the photon is
increased (assume the Eelectron < Photon (hν))
–
–
–
–
At 4000Å: hν = 4.966(10-12) ergs
½ mv2 < 5(10-12) ergs ==> v < 108 cm/s
At 5000K vRMS = 6.7(105) cm/s
At 100000K vRMS = 3(106) cm/s
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Thompson Scattering
Classical Electron Scattering
Reference: Marion - Classical Electromagnetic
Radiation p. 272 ff
Low Energy Process: v << c
Energy Absorbed from the EM field is
dE 2 e2 2

a
3
dt 3 c
(#)
a = acceleration of the electron: a = eE/me and E is
the magnitude of the electric field.
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Thompson Scattering
Field Energy Density =
<E2/4π> (Time Average)
Energy Flux Per Electron
= c <E2/4π>
Now Take the Time
Average of (#):
But that has to be the
energy flux per electron
times the cross section
which is σT c <E2/4π>.
2
dE
2e
2

a 
3
dt
3c
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2
2e e

E
3
2
3 c me
2
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The Thompson Cross Section
E
T c
4
2
2
2
2e e
2

E
3
2
3 c me
8
T 
3
2
 e 
25
2
 6.65(10 ) cm

2 
 mec 
2
At High Temperatures this breaks down: T ≥ 109 K
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Electron Scattering Opacity
Κe = σTNe/ρ
There is no frequency dependence!
Scattering off other ions is unimportant:
Cross Section Goes as (1/m)2 so for ions
(1/AmH)2 while for electrons it goes as (1/me)2
Ratio: Ions/Electrons = (me/AmH)2 =
(1/A(1840))2 < 10-6
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Rayleigh Scattering: σR
Scattering of a low energy photon by a bound
electron.
Classically: Rayleigh scattering occurs when a
photon of energy less than the atomic energy
spacing is absorbed.
– The electron then oscillates about the unperturbed
energy level (harmonically).
– The electron reradiates the same photon but
remains in the same energy state.
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The Rayleigh Cross-Section
The cross section is: σ = σT/(1-(ν0/ν)2)2
– Where hν is the photon energy
– hν0 is the restoring force for the oscillator
When ν << ν0: σR = σT (λ0/λ)4
– Now since ν << ν0 we have E << kT
– Which implies T ~ 1000 K for this process.
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Free-Free Opacities
Absorption Events
Bremstralung
– Electron moving in the field of an ion of charge Ze emits or
absorbs a photon:
Acceleration in field produces a photon of hν
De-acceleration in field consumes a photon of hν
Consider the Emission Process
– Initial Electron Velocity: v′
– Final Electron Velocity: v
Conservation of Energy Yields
½ me v2 + hν = ½ me v′2
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Energy Considerations
Energy Absorbed: dE/dt = 2/3 (e2/c3) a2 where a =
eE/me (a is the acceleration)
 dE
2 e2  2
E
dt 
a dt
3

 dt
3 c 
Most energy is absorbed during the time t  b / v′
when the electron is close to the ion
– b is called the impact parameter
– b is the distance of closest approach
Acceleration is ~ Ze2/meb2
Eabs  2/3 (e2/c3) (Ze2/meb2)2 (b / v′)
 2/3 ((Z2e6)/(me2 c3 b3 v′))
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Frequency Dependence
Expand dE/dT in a Fourier Series
dE ( )
1

dt
2

dE (t ) 2 i t
 dt e dt
Greatest contribution is when 2πνt  1 during
time t  b / v′
– Therefore 2πν b / v′ = 1 or 2πν = v′ / b
The energy emitted per electron per ion in the
frequency range d is
– dqν = 2πb db Eabs
Why? 2πb db is an area!
– dqν = energy emitted per electron per ion per unit
frequency
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Bremstralung Energy
Total Energy Emitted: nine v′f(v′)dv′ dqν
– ni = Ion number density
– ne = electron number density (note that the electron
flux is ne v′f(v′)dv′)
The reverse process defines the Bremstralung
absorption coefficient aν giving the absorption
per ion per electron of velocity v from the
radiation field.
In TE: Photon Energy Density = Uνp = (4π/c)
Bν(T)
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The Absorption Coefficient
Net energy absorbed must be the product of
the photon flux (photon energy = hν) cUνpdν
and ninef(v)dvaν and (1-e-hν/kT) or
– cUνpdν ninef(v)dvaν (1-e-hν/kT)
But in TE that must be equal to the emission:
– cUνpdν ninef(v)dvaν (1-e-hν/kT) = nine v′f(v′)dv′ dqν
aν = π/3 (Z2e6)/(hc me2 ν3 v)
This is off by 4/3 from the exact classical
result.
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The Bremstralung Opacity

 ff ( )   ni ne f (V )a g ff ( ,V )dV
0
V is v in the previous equations
This reduces to
κff(ν) = 4/3 nine (2π/3mekT)1/2 ((Z2e6)/(hc meν3))
gff(ν)
gff(ν) is the mean Gaunt factor and the result
has been corrected to the exact classical result.
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Bound Free Opacities
Transition from a Bound State to Continuum or Visa Versa
This process differs from the free-free case due to the
discrete nature of one of the states
Nth Discrete State:
me Z e
IH Z
En 
 2
2 2
2 n
n
2 4
2
Then the electron capture/ionization process must
satisfy:
½ me v2 - En = hν
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Bound Free
V is the velocity of the ejected or absorbed electron.
Semiclassical treatment of electron capture
– Electron Initial Energy: ½ mev2 and is positive
– The energy decreases in the electric field as it accelerates
seeing ion of charge Ze
– Q: Why does it loose energy as it accelerates?
– A: It radiates it away
The energy loss per captured electron may be
estimated as:
– dqν = 2πbdbEabs = (8π2/3) ((Z2e6)/(me2 c3 v′2dν))
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Cross Section
The cross section for emission of photons into
frequency interval dν is defined by
– hν dσν = dqν = hν (dσν/dν) dν
The final state is discrete so define σcn as the cross
section for capture into state n (energy En)
characterized by n in the range (n, n+dn) so that dσν =
σcndn. Then
– hν (dσν/dν) = hν σcn dn/dν
– Solve for σcn and use En = -IHZ2/n2 to get dn/dν
Thus: σcn = ((2IHZ2/n2)/(h2νn3))(dqν/dν)
= (32/3) π4 ((Z2e10)/(mec3h4v2νn3))
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Photoionization
The reverse process is related by detailed balance
Let σνn = photoionization cross section
The number of photons absorbed of energy hν
with the emission of electrons of energy
½mev2 - En from the nth atomic state is
– (cUνp/hν) dν σνn Nn (1-e-hν/kT)
– Nn is the number of atoms in state n
The reverse process: the number of electrons
with initial velocities v captured per second
into state n is Ni σcn ne v f(v) dv
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State Population
The Boltzmann Equation for the system is:
N n gn

e
N1 g1
 I H Z 2 (1
1
)
n2
kT
where g1 = 2 and gn = 2n2 (H-like ions) and the Saha
equation is
3/ 2  I Z
Ni N e
gi  2 mekT 
kT
2
N
2
H

UI 
h
2


e
Detailed Balance says:
(cUνp/hν)dνσνnNn(1-e-hν/kT) = Niσcnnevf(v)dv
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The Bound Free Coefficient
We assume
– Maxwellian distribution of speeds
– Boltzmann and Saha Equations
σνn = (gi/gn) (mevc/hν)2 σcn
We assume the Z-1 electrons in the atom do not
participate, set gi = 1, and correct for QM then
 n
64 me Z e

g ( , n)
6
3 5 bf
3 3 h c n
4
4 10
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Limits and Conditions
Photon must have hν > En
– σ = 0 for hν < En (= IHZ2/n2)
Recombination
– Levels coupled to ν by bound free processes have
En for n > n* where n* = (IHZ2/hν)1/2
Total Bound Free Opacity:
 bf ( )  
n
Nn

Continuous Opacity
 n
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Other Opacities
Source
BF
FF
Other
HI
H2HHe I
He II
He-
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Rayleigh
Rayleigh
Low Temperature Atomic: T < 10000K
CI
Mg I
Si I
Al I
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Intermediate Atomic: 10000 < T < 20000
Mg II
Si II
Ca II
NI
OI
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Hot Atomic: T > 20000K
C II – IV
N II – V
Ne I - VI
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The Rosseland Mean Opacity

1 dB
d


0 
1
dT

 
dB

d

0 dT
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More Spectroscopic Notation
H Like Only
Electron and proton have
– Spin ½
– Angular momentum /2 and each has a
corresponding magnetic moment
The magnetic moment of the electron interacts
with both the orbital magnetic moment of the
atom (spin-orbit) and the spin of the proton
(spin-spin).
– Spin-orbit: fine structure (multiplets)
– Spin-spin: hyperfine structure
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Notation
Principal Quantum Number n
Orbital Quantum Number l: l ≤ n - 1
–
– l =
S P D F
0 1 2 3
Total angular momentum of a state specified by l = l
[(l+1)]½ 
Spin angular momentum (s=1/2): [s(s+1)]½  = 3 
/2
Orbital and spin angular momentum interacts to
produce a total angular momentum quantum number j
= l ± 1/2: S States have j = 1/2, P states have j = 1/2,
3/2; and D has j = 3/2, 5/2
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H Like Notation
n
2 s 1
j
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