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CHEMICAL EQUILIBRIUM IN THEATMOSPHERE OF COOL WHITE DWARFS
Simon Blouin,PhD candidate at Université de Montréal
Current Challenges in the Physics of White Dwarf StarsSanta Fe – June 13, 2017
Part IMotivation
1. Cool white dwarfs are interesting
The coolest WDs are the oldest and thus the most constrainingfor cosmochronology.
[Kowalski 2006]
2
2. Cool white dwarfs have a dense photosphere
Cool He-rich WD atmospheres are non-ideal gases.
102 103 104 105 106
T (K)
10 8
10 6
10 4
10 2
100
102
104(g
/cm
3 )
Jupiter
Sun
WD (Teff =5000K)
WD (Teff =5000K)
He
H
rs = rHevdW
3
2. Cool white dwarfs have a dense photosphere
Towards observer
ρ ≈ 1 g/cm3
ρ ≈ 0.001 g/cm3
"Warm" He-rich WD
He− ff
Hee
Shallow photosphere
"Cool" He-rich WD
Deep photosphere
4
3. Everything relies on chemical equilibrium
Structure: p(τ), T(τ)
EOS: ρ(τ), ni(τ), eint(τ)
Opacities: κν(τ), χν(τ)
Radiative transfer + convection:Iν(τ), Sν(τ), Fν(τ), Fconv
Synthetic spectrum
5
Outline
1. Motivation2. Problem statement
◦ Chemical picture3. Classical approaches
◦ Occupation probability formalism4. Current state of the art
◦ Ionization potentials◦ Classical fluid theory◦ Ab initio techniques
5. Current challenges in the chemical equilibrium of coolwhite dwarfs
6
Part IIProblem statement
What needs to be fixed
# Saha equation predicts a neutral gas at any temperature ifthe density is very large,
n j+1
n jne �
(2πme kT
h2
)3/2 2Z j+1
Z je−χ j/kT
# To solve theses issues→ take interactions into account# How?
8
Physical vs chemical pictures
Physical picture
# Basic particles: nuclei andelectrons
# Formally exact# Hard to include in
atmosphere models(population of thousandsof levels required tocompute opacities)
Chemical picture
# Basic particles: includebound species
# Not exact, physicscontained in pair potentials
# The notion of boundspecies can be problematic
# More physical insight# Easier to relate to opacities
9
Physical vs chemical pictures
Physical picture
# Basic particles: nuclei andelectrons
# Formally exact
# Hard to include inatmosphere models(population of thousandsof levels required tocompute opacities)
Chemical picture
# Basic particles: includebound species
# Not exact, physicscontained in pair potentials
# The notion of boundspecies can be problematic
# More physical insight# Easier to relate to opacities
9
Physical vs chemical pictures
[Holst+2008]
9
Physical vs chemical pictures
Physical picture
# Basic particles: nuclei andelectrons
# Formally exact
# Hard to include inatmosphere models(population of thousandsof levels required tocompute opacities)
Chemical picture
# Basic particles: includebound species
# Not exact, physicscontained in pair potentials
# The notion of boundspecies can be problematic
# More physical insight
# Easier to relate to opacities
9
Physical vs chemical pictures
Physical picture
# Basic particles: nuclei andelectrons
# Formally exact# Hard to include in
atmosphere models(population of thousandsof levels required tocompute opacities)
Chemical picture
# Basic particles: includebound species
# Not exact, physicscontained in pair potentials
# The notion of boundspecies can be problematic
# More physical insight# Easier to relate to opacities
9
Free energy minimization
Free energy minimization recipe
1. Given a model for F(T,V, {N}), minimize F givenstoichiometric constraints, T and V .
dF � 0⇒ ∂F∂N j− ∂F∂N j+1
− ∂F∂Ne
� 0
2. The solution yields populations {N} and allthermodynamic properties (p, E, cv).
⇒ The problem boils down to finding a good model forF(T,V, {N}).
10
How does F look like?
F � Fide +
∑j
∑k
Fidj,k +
∑j
∑k
Fintj,k + Fexc
# Internal structure: destruction of highest energy levels dueto interactions
# Excess free energy:◦ Electrostatic interaction Fee + Fie + Fii
◦ Neutral interaction (van der Waals)
11
Part IIIOccupation probability formalism
Occupation probability
Key insightUnder given physical conditions, an electron as a finiteprobability of being bound and a corresponding probabilityof being ionized.
N jk
N j� w jk
g jk
Z je−ε jk/kT , Z j �
∑k
w jk g jk e−ε jk/kT
1. w jk allow Z j to converge2. w jk are continuous→ thermodynamic properties are
continuous3. w jk are combined to account for various interactions
(e.g., w � wneutral × wcharged)13
w jk for neutral interactions
# Atoms are considered as hard spheres with a radius rk foran excitation level k
# Using the second virial coefficient in the van der Waalsequation of state,
wk � exp
(−4π
3
∑k′
n′k(rk + rk′)3)
InterpretationWhen a state occupies a volume of the order of the meanvolume allowed per particle, it is gradually destroyed.
14
Limitations of the occupation probability formalism
1. The excluded volume effect is only a caricature of the realinteraction potential between two neutral particles.
2. No theoretical prescription for rk (rn � f n2a0Z ?)
0.5 1.0 1.5 2.0 2.5 3.0 3.5Separation (Å)
10 4
10 3
10 2
10 1
100
101
102
103
V (e
V)He-He pair potential
Experiment (Young+1981)Van der Waals radius (Bondi 1964)Hydrogenic approximation
[Bergeron+1991]
15
Limitations of the occupation probability formalism
1. The excluded volume effect is only a caricature of the realinteraction potential between two neutral particles.
2. No theoretical prescription for rk (rn � f n2a0Z ?)
[Bergeron+1991] 15
Part IVCurrent state of the art
Ionization potentials
In the context of ionization equilibrium, we are interested in thevariation of F when ionization occurs.
Ionization Relaxation
⇒Must take configurational entropy change into account
17
Accounting for everything
There are 3 contributions to ∆F when ionization occurs,
1. Entropy change (exclusion volume effect)2. Ion interaction energy before/after3. Free electron interaction energy
18
1. Entropy change
# Given an interaction potential u(r), the Ornstein-Zernikeequation yields the radial distribution function g(r),
g(r) � 4πr2ndr
# ∆F associated with exclusion volume can be computedusing u(r) and g(r) [Kiselyov+1990].
0 1 2 3 4 5
r (A)
0.0
0.5
1.0
1.5
2.0
2.5
3.0
g(r
)
0 1 2 3 4 5
r (A)
−0.10−0.05
0.000.050.10
u(r
)(e
V)
19
1. Ion interaction energy
I0 Entropygain
0
1
2
3
4
5
6
7En
ergy
(eV)
6.11
-0.59
-3.88
+3.27 4.91
T = 6000 K = 0.5 g/cm3
Effective ionization potential of Ca in a dense He medium
20
2. Ion interaction energy
# When ionization occurs,1. The interaction potential changes2. The distribution of neighboring atoms changes
Computing the interaction energy
1. Extract atomic configurations from MD simulations2. Get the interaction energy for each configuration3. Compute the mean and the statistical error
� Eint
21
2. Ion interaction energy
I0 Entropygain
Ion excessenergy loss
0
1
2
3
4
5
6
7En
ergy
(eV)
6.11
-0.59
-3.66
+3.27 5.13
T = 6000 K = 0.5 g/cm3
Effective ionization potential of Ca in a dense He medium
22
3. Free electron interaction energy
# Add one electron to a neutral simulation box and computethe ground-state energy difference using DFT[Kowalski+2007]
# Good agreement with low-temperature measurements[Broomall+1976]
23
3. Free electron interaction energy
I0 Entropygain
Ion excessenergy loss
Electronexcess energy
0
1
2
3
4
5
6
7En
ergy
(eV)
6.11
-0.59
-3.66
+3.27 5.13
T = 6000 K = 0.5 g/cm3
Effective ionization potential of Ca in a dense He medium
24
Putting everything together
I0 Entropygain
Ion excessenergy loss
Electronexcess energy
Ieff0
1
2
3
4
5
6
7En
ergy
(eV)
6.11
-0.59
-3.66
+3.27 5.13
T = 6000 K = 0.5 g/cm3
Effective ionization potential of Ca in a dense He medium
25
Putting everything together
0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6(g/cm3)
2.5
2.0
1.5
1.0
0.5
0.0I(e
V)T = 6000K
Ionization potential depression of Ca in a dense He medium
[Blouin+ in prep]
26
Part VCurrent Challenges in the ChemicalEquilibrium of Cool White Dwarf
Atmospheres
How do we know if we are right?
Two ways to validate our theoretical predictions,
1. Fit cool helium-rich white dwarfs2. Comparison to experimental data
28
Comparisons to spectra are always indirect
0.5 1.0 1.5 2.0 2.5( m)
0.0
0.5
1.0
f(e
rgcm
2s
1Hz
1 ) 1e 25
Teff = 4450 K log g = 7.96
WD0552-041
3900 4000 4100 4200 4300 4400(Å)
0.0
0.2
0.4
0.6
0.8
1.0
1.2
Norm
alize
dflu
xlog Ca/He = 10.71
3900 4000 4100 4200 4300 4400(Å)
0.2
0.4
0.6
0.8
1.0
Norm
alize
d flu
x
log Ca/He = -10.7
Ca ICa II
ObservationsIdeal Ca ionization equilibrium
[Blouin+ in prep]
29
Comparisons to spectra are always indirect
0.5 1.0 1.5 2.0 2.5( m)
0.0
0.5
1.0
f(e
rgcm
2s
1Hz
1 ) 1e 25
Teff = 4450 K log g = 7.96
WD0552-041
3900 4000 4100 4200 4300 4400(Å)
0.0
0.2
0.4
0.6
0.8
1.0
1.2
Norm
alize
dflu
xlog Ca/He = 10.71
3900 4000 4100 4200 4300 4400(Å)
0.2
0.4
0.6
0.8
1.0
Norm
alize
d flu
x
log Ca/He = -10.7log Ca/He = -10.8
Ca ICa II
ObservationsIdeal Ca ionization equilibriumNon-ideal Ca ionization equilibrium
[Blouin+ in prep]
It is often hard to tell what is wrong (e.g., chemical equilibriumor line profiles?)
30
(Lack of) experimental data
# We are missing experimental data to validate ab initiopredictions.
# When experimental data are available, the comparison isoften inconclusive. For instance, concerning the He EOS ofKowalski+2007,◦ Excellent agreement with shock wave experiment data◦ Significant disagreement for conductivity data, indicating a
discrepancy for the density required to pressure ionize He
31
Summary
# Cool white dwarf atmospheres are dense mediums inwhich the chemical equilibrium is non-ideal.
# Although simple and easy to implement, the occupationprobability formalism has its limits.
# Ab initio simulation techniques can be used to addressthese shortcomings.
# Significant progress is being made, but some stars keepchallenging our models and experimental data is cruellylacking.
32
References
References – 1/4
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References – 3/4
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