Radiation Transport in a Gas (Continued)
Semi-Thick Plasmas Diffusion Approximation
An LTE plasma is such that its collisional processes are fast and equilibrate, but coupling
´L
s · VIν = k ' (Bν − Iν ). The quantity k ' dε is
to radiation is incomplete. Then, as we saw, Ω
ν
ν
0
called the optical depth, and if it is large, then Iν r Bν and we have black-body radiation as
well, e.g., full equilibrium. In many cases, plasmas (or other bodies) are not quite so thick
s · VBν which is still semi-local.
(optically); the next approximation is simply Iν r Bν − k1� Ω
ν
This is the diffusion approximation.
s
directions
To calculate energy flux at a point, we need to look at all Ω
ˆ
s dΩ
s
s =
Iν Ω
4π
and then V · sν measure the net local cooling rate by radiation. In the diffusion approxima­
tions,
ˆ
ˆ
1
s
1
s −
s
s
sndΩ
sν =
1 %
B%
Ω
(Ω · VBν )
νn
eΩ
edΩ
|{z}
|{z}
kν' e
1
41π
4π
cosθ sinθcosϕ sinθsinϕ sinθdθdϕ
|VBv |cosθ
ˆ
π
Only the component along VBν survives →
sν =
ν
− VB
2π
kν�
cos
2 θsinθdθ
0
e
Or, since uν =
4π
Bν ,
c
2
3
ν
sν = − 3c Vu
. This should be OK near the center of resonant lines.
k�
ν
´∞
This is for one frequency. For all frequencies, s =
sν dν. Using
0
8πhν 3 /c3
uν = hν/kT
e
−1
,
hν
8πhν 3 ex kT
2
Vuν =
VT
3
x
c (e − 1)2
16
Then, integrating,
s =
(lν )σT 3 VT
3
ˆ∞
,
lν =
0
For kν' = const., (lv ) =
1
,
kν�
x =
hν
kT
15 1 x4 e−x
dx
4π 4 kν' (1 − e−x )2
otherwise it is sensitive to T .
Very often plasmas are only thick near the centers of strong absorption lines, and are fairly
thin in between. Let us examine the nature of kν .
1
Bound-Bound electronic transitions. Line radiation. Broadening.
The cross-section for absorption of a photon of hν = Em −En by an n-level atom is expressed
as an integrated value (over frequency) times a shape factor φ(ν):
αnm = Qνn m =
ˆ
∞
e2
4f m c
e 0{z e }
φ(ν)dν ≡ 1
fnm φ(ν)
0
2.65×10−6 m2 /sec
where fnm is the absorption oscillator strength (non dimensional, between 0 and 1). The
oscillator strength is mostly empirical, but it is well known for many transitions. For the H
atom (and roughly for all alkalis),
fnm r
1.96gbb
n5 m 3
1
n2
−
1 3
m2
(gbb ∼ 1, especially at high m, n)
From Qνn m we easily calculate absorption coefficients:
kνn m = Nn Qνn m
The line shape φ(ν) integrates to unity by definition:
´∞
φ(ν)dν = 1. For low T , low P , only
0
natural line shape. “Lorentzian”, meaning,
φ(ν) =
1
γ/4π
π (ν − νc )2 + (γ/4π)2
φ(νc ) =
4
γ
where νc = Em h−En is the line-center frequency and there is a natural line-width at halfmaximum of ΔνN = γ/2π, due to Heisenberg uncertainly in energy of the levels because of
their finite lifetime:
γ =
Amj +
Anj
j<m
j<n
These γ’s and ΔνN are very small (Δλ = λc Δνcν ∼ 10−4 Å)
Collisions can be seen as modulations on the atomic oscillators, and they therefore create
“sidebands”, or broadening. Natural collisions produce Lorentz broadening (or pressure
broadening) if it is with unlike atoms, and Holtzmark (or resonance) broadening if it is with
the same species. Same line shapes as natural, but replace γ by,
Γ = γ + 2ν opt
ν opt =
np gop Qop → (close to (but larger) than cross-section for momentum exchange)
p
1 atm, ΔλLorentz r 0.05Å, so much more important than natural. But tiny at nν10−18 −
1020 m−3 .
Collisions with electrons or other charged particles can also produce broadening (Stark broad­
ening) in plasmas. See Griem (1964); can be very strong at ne > 1022 m−3 . The easiest
2
broadening to understand is the due to the thermal random motion of the atoms, since there
is Doppler shifting of the apparent line center depending on line-of-sight velocity. This is
the Doppler broadening, and depends only on T :
ν−νc 2
1
φ(ν) = √ e−( δ )
πδ
ΔνD = νc
;
ΔνD
√
2 ln 2
δ =
8kT ln 2
= 7.16 × 10−7 νc
2
ma c
T
Ma
For Na D-lines at 2000◦ K, ΔλD ∼ 0.04Å.
Lorentz and Doppler broadening are very often both important. In the center, Doppler
dominates, but the center is often black anyway; in the wings, Lorentz dominates. We can
combine the Lorentz and Doppler profiles into a Voigt profile:
1
φ(ν) = √ V (a, x)
πδ
ˆ∞
2
a
e−y dy
V (a, x) =
π
a2 + (x − y)2
ΔνD
√
2 ln 2
√
Γ/2π
a = ln 2
ΔνD
δ =
−∞
√
ν − νc
x = 2 ln 2
ΔνD
Bound-Free Radiation
A photon can be absorbed by a bound electron which is below ionization by less that hν. The
event causes ionization (photo-ionization). Similar physics applies if hν » eVi (essentially
the free electron Thompson scattering). The cross-section for a hydrogenic atom from a
principal quantum number m is
3
−22 m Emη
Qνphotoioniz = 7.91 × 10
gbf (Emη < hν)
z2
hν
e
|{z}
c1
where Emη is the energy below ionization (threshold for hν).
Free-Free (Bremsstrahlung)
An electron bouncing off a heavy particle emits radiation due to its acceleration. Conversely,
it can absorb a photon while in the vicinity of a heavy particle, which carries away for of the
momentum. The coefficient of free-free absorption (which gives the absorption rate when
Iν
s is
multiplied times the electron density, the heavy particle density and hν
dνdΩ,
βFree-Free = 180
z 2 gf f
ν 3 ce
(m5 ) (gf f r 1)
where ce is the electron speed. Integrating for a Maxwellian electron distribution, the effective
“cross-section” is
ne z 2 g¯f f
QνFree-Free = 230 3
ν c̄e
3
Escape factor for Resonant Radiation
Resonant radiation is radiation from m → ground transitions. Since ground is strongly
populated, resonant radiation is very likely to reabsorb, at least near the line center. So,
instead of ..., we probably see .... How much radiation does escape from a give radiating
volume? Take a slab geometry, assume uniform properties, Iν (0) = 0:
x
0�
Iν = Bν (1 − e−kν ε )
ε =
cos θ
'
Call s = kν =“optical depth”. Depends both on distance and on where in the spectrum (kν' ).
´
s
We are interested in the 1-D “radiant heat flux”, sν =
Iν ΩdΩ
Ω
π/2
ˆ
ˆπ/2
s
(ssν )x = qν =
(Iν cos θ)2π sin θdθ = Bν 2π (1 − e− (cos θ) ) cos θ sin θdθ
0
0
ˆ1
call u = cos θ
s
(1 − e− u )udu
qν = 2πBν
0
This can be done is therms of exponential integrals. But for an approximate solution, note
So, make,
u (u < s)
s (1 > u > s)
s
(1 − e− u )u r
Near,
ˆ
s
qν r 2πBv
ˆ
1
udu +
sdu
s
0 2
s
= 2πBν
+ s(1 − 2)
2
s
(s « 1)
= 2πBν s 1 −
2
ˆ
1
qν r 2πBν udu = πBν
(s » 1)
0
So,
qν
r
πBν
s(2 − s) s « 1
1
s»1
Now, assume line is Pressure broadened, at least in the wings:
kν =
kν (νc )
2
c
1 + 2 ν−ν
Δν
A
→s =
1 + δ2
r
A
δ =
−1
s
(Δν = full width at 1/2 height)
A = Lkν (νc ) ,
4
ν − vc
δ = 2
Δν
Re-plotting qν /πBν vs. frequency now,
⎧
⎨
A
qν
2−
1+δ 2
=
⎩
πBν
1
A
1+δ 2
√
|δ| > A − 1
√
|δ| < A − 1
√
when s = 1, δ = A − 1
(A > 1 assumed)
Then, for all frequencies in the line,
ˆ
∞
qν dν =
q =
Δν
2
ˆ
∞
−∞
0
√
A−1
ˆ
ˆ
∞
Δν
gν dδ =
πBν 2
1dδ +
s(2 − s)dδ
2
√
0
A−1
In general,
r
δ =
ˆ
∞
A
−1
s
dδ =
−A/s2 ds
A
s
2
ˆ
1
2 − s ds
√
√
ds = A
A
2 A − s s
√
2
−
1
1
0
s
A−1
√
√
For cases of interest A = zkν (νc ) » 1, while s here is in (0-1), so A − s r A
ˆ
∞
ˆ
0
(2s − s
2 )dδ =
−
(2s − s
2 )
−1
A
(2 − s
)dδ r √
2 A
ˆ
1
2
√
A−1
Also,
√
´A−1
dδ
=
A/s2
√ √
1
A
√
5 A
2 3/2
4 s − s
=
3
2
3
0
ds
(2 − s)
√
=
s
0
√
√
A − 1
r A
0
√
5
+1
A
q r πBv Δν
3
q r
�
8π
Bν Δν Lkν (0)
3
(Black Center)
For comparison, if the line were all of it “thin”, we would have,
ˆ
1 − us
1−e
e {z
qν = 2πBν
0
udu r 2πBν s
}
=s everywhere
ˆ
∞
Integrating over line,
q =
Δν
qν dν =
2
A
,
1+δ 2
´∞
sdδ = πA
Δν
2πBν
qν dδ =
2
−∞
0
And since s =
ˆ
∞
q = π 2 Lkνc ΔνBν (thin line).
−∞
5
ˆ
∞
sdδ
−∞
In terms of density and the total absorption cross-section QT OT =
´
αnm dv, we have
line
kν = Nn Qν = Nn QTOT
2
πΔν
1
1+
ν−νc
2 Δν/2
2
2
Nn QTOT
πΔν
2
LNn QTOT
so, Lkν (0) =
πΔν
kν (0) =
so, (q)Thin = 2πBν LNn QTOT
If this were a black body, we would have in Δν a radiation,
ˆ1
(1 − e−∞ )udu = πBν ;
(qν )BB = 2πBν
(qν )Thin
= 2s. At ν = νc , s = A = kνc L
(qν )BB
0
So
qνthin
qνBB
= 2Lkνc « 1
We can also define the “escape factor”, or fraction of light emitted which does escape. For
the “blackened-center” line,
q
8π
B
Δν
Lkν( 0)
ν
3
8
1
p
�
β =
=
«1
3π Lkνc
π 2 Lkν0 ΔνBν
Net radiant heat flux (to the right) in slab of thickness L, at a distance x from the left
boundary.
�
p
8π
q = q R − qL
qR =
Bν Δν xkν0
3
�
p
8π
Bν Δν (L − x)kν0
qL =
3
p
�
√
√
8π
q(x) =
Bν Δν kν0 ( x − L − x)
3
�
p
dq
4π
1
1
Radiant heat loss per unit volume:
=
Bν Δν kν0 √ + √
dx
3
x
L−x
For an almost “transparent” medium:
⎫
qR (x) = π 2 xkv0 ΔνBν
⎬
q(x) = π 2 (2x − l)kν0 ΔνBν
⎭
qL (x) = π 2 (1 − x)kν0 ΔνBν
Local Escape Factor:
dq
β(x) =
dx
dq
dx Thin
2
=
3π
1
1
�
p
p
+�
k ν0 x
kν0 (L − x)
6
;
dq
= 2π 2 kν0 ΔνBν
dx
!
(Must be limited to <1 near x = 0 and x = L)
MIT OpenCourseWare
http://ocw.mit.edu
16.55 Ionized Gases
Fall 2014
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