Set 4: Thermal Equilibrium Applications

23
Set 4: Thermal Equilibrium Applications

Transcript of Set 4: Thermal Equilibrium Applications

Page 1: Set 4: Thermal Equilibrium Applications

Set 4:Thermal Equilibrium Applications

Page 2: 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

npnenHntot

=x2e

1− xe• Number densities defined by distribution function in thermal

equilibrium. e and p are non-relativistic at the eV energy scales ofrecombination

• Maxwell-Boltzmann distribution

f = e−(mc2−µ)/kT e−q2/2mkT

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Saha Equation• Number density:

n = g

∫d3q

(2πh)3f =

ge−(mc2−µ)/kT

2π2h3

∫ ∞0

q2dqe−q2/2mkT

= ge−(mc2−µ)/kT

2π2h3 (2mkT )3/2

[∫ ∞0

x2dxe−x2

=

√π

4

]= ge−(mc2−µ)/kT

(mkT

2πh2

)3/2

, (x = p/√

2mkT )

• Hydrogen recombination (ntot = np + nH)

np = gpe−(mpc2−µp)/kT

(mpkT/2πh

2)3/2

ne = gee−(mec2−µe)/kT

(mekT/2πh

2)3/2

nH = gHe−(mHc

2−µH)/kT(mHkT/2πh

2)3/2

Page 4: Set 4: Thermal Equilibrium Applications

Saha Equation• Hydrogen binding energy B = 13.6eV: mH = mp +me −B/c2

npnenHntot

=x2e

1− xe≈ gpgegHntot

e−B/kT eµp+µe−µH(mekT

2πh2

)3/2

• Spin degeneracy: spin 1/2 gp = 2 , ge = 2; gH = 4 product

• Equilibrium µp + µe = µH

x2e

1− xe≈ 1

ntot

e−B/kT(mekT

2πh2

)3/2

• Quadratic equation involving T and the total density - explicitsolution for xe(T )

• Exponential dominant factor: ionization drops quickly as kT dropsbelow B - exactly where the sharp transition occurs depends on thedensity ntot

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Saha Equation• Photon perspective: compare photon number density at T to ntot

nγ =2ζ(3)

π2h3

(kT

c

)3

x2e

1− xe=

(nγntot

)e−B/kT

π2h3

2ζ(3)

( c

kT

)3(mekT

2πh2

)3/2

=

(nγntot

)e−B/kT

π1/2

25/2ζ(3)

(mec

2

kT

)3/2

• 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 ionizedwith the high energy photons in tail of the black body

• Cosmologically, recombination occurs at an energy scale ofkT ∼ 0.3eV

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Saha Equation• Electron perspective: the relevant length scale is the (“thermal”) de

Broglie wavelength for a typical particle

mev2 ∼ kT, q2 ∼ m2

ev2 ∼ (mekT )

λTe =h

q=

h

(2πmekT )1/2=

(2πh2

mekT

)1/2

which is the factor in the Saha equation

x2e

1− xe=

1

ntotλ3Te

e−B/kT

NTe = neλ3Te = # electrons in a de Broglie volume and is 1 for

non-degenerate matter

• Equivalently, occupation number fe 1 at average momenta

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Saha Equation• Saha equation

xe1− xe

=1

NTe

e−B/kT

• Electron chemical potential

NTe = 2e−(mec2−µe)/kT

xe1− xe

=1

2e−[B−(mec2−µe)]/kT

• Transition occurs when Beff = B −mec2 + µe = kT - chemical

potential or number density determines correction to B ∼ kT rule

• However equilibrium may not be maintained - 2 body interactionmay not be rapid enough in low density environment - e.g.freezeout cosmologically

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Cosmic Recombination• Rates insufficient to maintain equilibrium - due to Lyα opacity

cosmic recombination relies on forbidden 2 photon decay andredshift

Saha

2-levelioni

zatio

n fr

actio

n

scale factor a

redshift z

10–4

10–3

10–2

10–1

1

10–3

103104 102

10–2

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Kirchhoff’s Law.

Iν(τ=0)=Bν

Iν(τ=δτ)=Bν

T

T,δτ

• Infer a relationshipbetween absorption and emissionbased on thermodynamic equilibrium

• Consider a source at temperatureT emitting with a sourcefunction Sν• The general radiative transfer

equation says the specificintensity evolves as

.

dIνdτ

= −Iν + Sν

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Kirchhoff’s Law• Recall that the source function is the ratio of emission and

absorption coefficients

Sν =jναν

• Consider the source to be in a black body enclosure of the sametemperature. Then Iν(τ = 0) = Bν(T )

• Radiative transfer must preserve Iν(τ) = Bν(T ) so Sν = Bν or orthe emission coefficient jν = ανBν

• Since αν and jν are properties of the source and not the initialradiation field, this relationship is general for a black body source

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Einstein Relations.

2

1

absorptionemission

• Generalize Kirchhoff’s law

• Consider a 2 level atom withenergies separated by hν: in equilibriumthe forward transition balancesthe backwards transition leaving the leveldistribution with a Boltzmann distribution

• Ignoring stimulated emission forthe moment, spontaneous emissionbalances absorption

• Analog to jν for a single atom: A21 the emission probability perunit time [s−1]

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Einstein Relations• Analog to αν is B12 where B12Jν is the absorption probability per

unit time in an isotropic radiation field

• Transition rate per unit volume depends on number densities instates

1→ 2 : n1B12Jν ; 2→ 1 : n2A21

• Detailed balance requires

n1B12Jν = n2A21 →A21

B12

=n1

n2

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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 ]

Jν =2h

c2

ν3

ehν/kT − 1

• So ignoring stimulated emission would imply

A21

B12

=g1

g2

ehν/kT2h

c2

ν3

ehν/kT − 1

• But the rates should not depend on temperature and so somethingis missing.

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Einstein Relations• Clue: photons become Maxwell Boltzmann in the Wien tail where

there is on average < 1 photon at the line frequency

Jν ≈2h

c2ν3e−hν/kT

• Then

A21

B12

=g1

g2

2h

c2ν3

• Missing term involves a condition where there is a large number ofphotons at the transition frequency: stimulated emission

• Suppose there is an additional emission term whose transition rateper unit volume

2→ 1 : n2JνB21

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Einstein Relations• Then the balance equation becomes

n1B12Jν = n2A21 + n2JνB21

Jν =2h

c2

ν3

ehν/kT − 1=

A21

(n1/n2)B12 −B21

=A21

B21[(g1B12/B21g2)ehν/kT − 1]

• Matching terms

g1B12 = g2B21,2h

c2ν3 =

A21

B21

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Einstein Relations• Import: given spontaneous emission rate, measured or calculated,A21→ stimulated emission rate B21→ absorption rate, fullydefining the radiative transfer for this process independent of theradiation state Iν

• Usage: oscillator strength defined against a classical model forabsorption, via semiclassical (quantized atomic levels, classicalradiation) calculation of absorption and stimulated emission, orline width measurement deterimining the spontaneous emissionrate

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Einstein Relations• Relation to jν : multiply by energy hν, divide into 4π and put a

normalized line profile∫dνφ(ν) = 1

dEem = jνdV dΩdνdt = hνφ(ν)dνn2A21dVdΩ

4πdt

jν =hν

4πn2A21φ(ν)

• Relation to absorption αabs: similarly

dEabs = hν φ(ν)dν n1B12Jν dVdΩ

4πdνdt

= −[dJν = −αabsJνds]dtdνdAdΩ

αabs =hν

4πn1B12φ(ν)

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Einstein Relations• Add stimulated term in the emission

dEem = hνφ(ν)dνn2B21JνdV dtdΩ

4π, αem = −hν

4πn2B21φ(ν)

• Absorption and emission coefficient

αν =hν

4πφ(ν)[n1B12 − n2B21]

=hν

4πφ(ν)

(n1g2

g1

− n2

)c2

2hν3A21

jν =hν

4πn2A21φ(ν)

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Einstein Relations• Source function

Sν = jν/αν =1

(n1g2/n2g1 − 1)

2hν3

c2

• In thermal equilibrium n1g2/n2g1 = ehν/kT and Sν = Bν ,Kirchoff’s law

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Maser/Laser• Net absorption coefficient becomes negative if

n1g2/g1 − n2 < 0

n1/g1 < n2/g2

• Requires a population inversion: higher energy state is morepopulated than lower energy state

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Rosseland Approx (Tight Coupling).

θ

ds=dz/cosθdz

Iν= + + ...

• Radiative transfer nearequilibrium where the source functionSν = Bν . Recall plane parallel case

µdIνdz

= −αν(Iν −Bν)

• If interactionis strong then the difference betweenIν and Bν is small - solve iteratively

Iν −Bν = − µ

αν

dIνdz≈ − µ

αν

dBν

dz

Iν = Bν −µ

αν

dBν

dT

dT

dz

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Rosseland Approx (Tight Coupling)• Specific flux follows the temperature gradient

Fν =

∫Iν(z, µ)µdΩ = 2π

∫ 1

−1

Iν(z, µ)µdµ

= −2π

αν

dBν

dT

dT

dz

∫ 1

−1

µ2dµ = − 4π

3αν

dBν

dT

dT

dz

and is inhibited by high absorption

• Net flux

F (z) =

∫ ∞0

Fνdν = −4π

3

dT

dz

∫ ∞0

1

αν

dBν

dTdν

and is dominated by the frequencies that have the lowestabsorption – generally true: energy transport is dominated by thelowest opacity channel – e.g. lines (dark) vs continuum (bright)

Page 23: Set 4: Thermal Equilibrium Applications

Rosseland Approx (Tight Coupling)• Flux for a constant αν involves∫

dB

dTdν =

d

dT

∫ ∞0

Bνdν = 4σ

πT 3

• Define the Rosseland mean absorption coefficient

α−1R =

∫α−1ν

dBνdTdν∫

dBνdTdν

• Net flux becomes

F (z) =4σT 3

παR

(−4π

3

dT

dz

)= −16σT 3

3αR

dT

dz