Set 4:
Thermal Equilibrium Applications
Saha Equation
• What is the equilibrium ionization state of a gas at a given
temperature?
• Hydrogen example: e + p ↔ H + γ
• Define ntot = np + nH and an ionization fraction xe ≡ np /ntot
x2e
np ne
=
nH ntot
1 − xe
• Number densities defined by distribution function in thermal
equilibrium. e and p are non-relativistic at the eV energy scales of
recombination
• Maxwell-Boltzmann distribution
−(mc2 −µ)/kT −q 2 /2mkT
f =e
e
Saha Equation
• Number density:
Z
3
−(mc2 −µ)/kT Z ∞
dq
ge
2
−q 2 /2mkT
q dqe
n=g
f=
3
3
2
(2πh̄)
2π h̄
0
√ Z ∞
−(mc2 −µ)/kT
e
π
3/2
2
−x2
=g
(2mkT )
x dxe
=
3
2
4
2π h̄
0
3/2
√
mkT
−(mc2 −µ)/kT
= ge
, (x = p/ 2mkT )
2
2πh̄
• Hydrogen recombination (ntot = np + nH )
2 3/2
−(mp c2 −µp )/kT
mp kT /2πh̄
np = gp e
−(me c2 −µe )/kT
ne = ge e
nH = gH e
2 3/2
me kT /2πh̄
−(mH c2 −µH )/kT
2 3/2
mH kT /2πh̄
Saha Equation
• Hydrogen binding energy B = 13.6eV: mH = mp + me − B/c2
x2e
gp ge −B/kT µp +µe −µH
np ne
=
≈
e
e
nH ntot
1 − xe
gH ntot
me kT
2πh̄2
3/2
• Spin degeneracy: spin 1/2 gp = 2 , ge = 2; gH = 4 product
• Equilibrium µp + µe = µH
x2e
1 −B/kT
≈
e
1 − xe
ntot
me kT
2πh̄2
3/2
• Quadratic equation involving T and the total density - explicit
solution for xe (T )
• Exponential dominant factor: ionization drops quickly as kT drops
below B - exactly where the sharp transition occurs depends on the
density ntot
Saha Equation
• Photon perspective: compare photon number density at T to ntot
3
2ζ(3) kT
nγ = 2 3
c
π h̄
3/2
3
c
me kT
nγ
−B/kT π h̄
=
e
1 − xe
ntot
2ζ(3) kT
2πh̄2
1/2
2 3/2
π
nγ
me c
−B/kT
=
e
ntot
25/2 ζ(3) kT
x2e
2 3
• Photon-baryon ratio controls when recombination occurs:
typically a very large number since baryon number is conserved
(µ 6= 0) -a low baryon density medium is easy to keep ionized
with the high energy photons in tail of the black body
• Cosmologically, recombination occurs at an energy scale of
kT ∼ 0.3eV
Saha Equation
• Electron perspective: the relevant length scale is the (“thermal”) de
Broglie wavelength for a typical particle
q 2 ∼ m2e v 2 ∼ (me kT )
2 1/2
h
h
2πh̄
= =
=
1/2
q
(2πme kT )
me kT
me v 2 ∼ kT,
λT e
which is the factor in the Saha equation
x2e
1
−B/kT
=
e
1 − xe
ntot λ3T e
NT e = ne λ3T e = # electrons in a de Broglie volume and is 1 for
non-degenerate matter
• Equivalently, occupation number fe 1 at average momenta
Saha Equation
• Saha equation
xe
1 −B/kT
=
e
1 − xe
NT e
• Electron chemical potential
NT e = 2e
−(me c2 −µe )/kT
xe
1 −[B−(me c2 −µe )]/kT
= e
1 − xe
2
• Transition occurs when Beff = B − me c2 + µe = kT - chemical
potential or number density determines correction to B ∼ kT rule
• However equilibrium may not be maintained - 2 body interaction
may not be rapid enough in low density environment - e.g.
freezeout cosmologically
Cosmic Recombination
• Rates insufficient to maintain equilibrium - due to Lyα opacity
cosmic recombination relies on forbidden 2 photon decay and
redshift
104
redshift z
103
102
ionization fraction
1
10–1
Saha
10–2
2-level
10–3
10–4
10–3
scale factor a
10–2
Kirchhoff’s Law
.
• Infer a relationship
between absorption and emission
based on thermodynamic equilibrium
• Consider a source at temperature
T emitting with a source
function Sν
• The general radiative transfer
equation says the specific
intensity evolves as
.
dIν
= −Iν + Sν
dτ
T
Iν(τ=0)=Bν
T,δτ
Iν(τ=δτ)=Bν
Kirchhoff’s Law
• Recall that the source function is the ratio of emission and
absorption coefficients
jν
Sν =
αν
• Consider the source to be in a black body enclosure of the same
temperature. Then Iν (τ = 0) = Bν (T )
• Radiative transfer must preserve Iν (τ ) = Bν (T ) so Sν = Bν or or
the emission coefficient jν = αν Bν
• Since αν and jν are properties of the source and not the initial
radiation field, this relationship is general for a black body source
Einstein Relations
.
2
• Generalize Kirchhoff’s law
• Consider a 2 level atom with
energies separated by hν: in equilibrium
the forward transition balances
the backwards transition leaving the level
distribution with a Boltzmann distribution
hν
emission
absorption
• Ignoring stimulated emission for
the moment, spontaneous emission
balances absorption
• Analog to jν for a single atom: A21 the emission probability per
unit time [s−1 ]
1
Einstein Relations
• Analog to αν is B12 where B12 Jν is the absorption probability per
unit time in an isotropic radiation field
• Transition rate per unit volume depends on number densities in
states
1→2:
n1 B12 Jν ;
2→1:
n2 A21
• Detailed balance requires
n1 B12 Jν = n2 A21
n1
A21
= Jν
→
B12
n2
Einstein Relations
• Atoms follow the non relativistic Maxwell-Boltzmann distribution
(with µ1 = µ2 ), radiation a Planck distribution
n1 ∝ g1 exp[−E1 /kT ],
n2 ∝ g2 exp[−(E1 + hν)/kT ]
ν3
2h
Jν = 2 hν/kT
c e
−1
• So ignoring stimulated emission would imply
A21
g1 hν/kT 2h
ν3
= e
B12
g2
c2 ehν/kT − 1
• But the rates should not depend on temperature and so something
is missing.
Einstein Relations
• Clue: photons become Maxwell Boltzmann in the Wien tail where
there is on average < 1 photon at the line frequency
2h 3 −hν/kT
Jν ≈ 2 ν e
c
• Then
A21
g1 2h 3
=
ν
2
B12
g2 c
• Missing term involves a condition where there is a large number of
photons at the transition frequency: stimulated emission
• Suppose there is an additional emission term whose transition rate
per unit volume
2→1:
n2 Jν B21
Einstein Relations
• Then the balance equation becomes
n1 B12 Jν = n2 A21 + n2 Jν B21
ν3
A21
2h
=
Jν = 2 hν/kT
c e
−1
(n1 /n2 )B12 − B21
A21
=
B21 [(g1 B12 /B21 g2 )ehν/kT − 1]
• Matching terms
g1 B12 = g2 B21 ,
2h 3 A21
ν =
2
c
B21
Einstein Relations
• Import: given spontaneous emission rate, measured or calculated,
A21 → stimulated emission rate B21 → absorption rate, fully
defining the radiative transfer for this process independent of the
radiation state Iν
• Usage: oscillator strength defined against a classical model for
absorption, via semiclassical (quantized atomic levels, classical
radiation) calculation of absorption and stimulated emission, or
line width measurement deterimining the spontaneous emission
rate
Einstein Relations
• Relation to jν : multiply by energy hν, divide into 4π and put a
R
normalized line profile dνφ(ν) = 1
dΩ
dt
dEem = jν dV dΩdνdt = hνφ(ν)dνn2 A21 dV
4π
hν
jν =
n2 A21 φ(ν)
4π
• Relation to absorption αabs : similarly
dEabs
αabs
dΩ
= hν φ(ν)dν n1 B12 Jν dV
dνdt
4π
= −[dJν = −αabs Jν ds]dtdνdAdΩ
hν
=
n1 B12 φ(ν)
4π
Einstein Relations
• Add stimulated term in the emission
dΩ
dEem = hνφ(ν)dνn2 B21 Jν dV dt ,
4π
• Absorption and emission coefficient
αem
hν
= − n2 B21 φ(ν)
4π
hν
αν =
φ(ν)[n1 B12 − n2 B21 ]
4π
2
hν
n1 g2
c
=
φ(ν)
− n2
A21
3
4π
g1
2hν
hν
jν =
n2 A21 φ(ν)
4π
Einstein Relations
• Source function
2hν 3
1
Sν = jν /αν =
(n1 g2 /n2 g1 − 1) c2
• In thermal equilibrium n1 g2 /n2 g1 = ehν/kT and Sν = Bν ,
Kirchoff’s law
Maser/Laser
• Net absorption coefficient becomes negative if
n1 g2 /g1 − n2 < 0
n1 /g1 < n2 /g2
• Requires a population inversion: higher energy state is more
populated than lower energy state
Rosseland Approx (Tight Coupling)
.
• Radiative transfer near
equilibrium where the source function
Sν = Bν . Recall plane parallel case
dIν
µ
= −αν (Iν − Bν )
dz
• If interaction
is strong then the difference between
Iν and Bν is small - solve iteratively
µ dBν
µ dIν
Iν − Bν = −
≈−
αν dz
αν dz
µ dBν dT
Iν = Bν −
αν dT dz
θ
dz
Iν=
+
ds=dz/cosθ
+ ...
Rosseland Approx (Tight Coupling)
• Specific flux follows the temperature gradient
Z
Z 1
Fν = Iν (z, µ)µdΩ = 2π
Iν (z, µ)µdµ
−1
2π dBν dT
=−
αν dT dz
Z
1
4π dBν dT
µ dµ = −
3αν dT dz
−1
2
and is inhibited by high absorption
• Net flux
Z
F (z) =
0
∞
4π dT
Fν dν = −
3 dz
Z
0
∞
1 dBν
dν
αν dT
and is dominated by the frequencies that have the lowest
absorption – generally true: energy transport is dominated by the
lowest opacity channel – e.g. lines (dark) vs continuum (bright)
Rosseland Approx (Tight Coupling)
• Flux for a constant αν involves
Z ∞
Z
σ 3
d
dB
Bν dν = 4 T
dν =
dT
dT 0
π
• Define the Rosseland mean absorption coefficient
R −1 dBν
αν dT dν
−1
αR = R dBν
dν
dT
• Net flux becomes
4σT
4π dT
16σT 3 dT
−
=−
F (z) =
παR
3 dz
3αR dz
3