Download Graduate Course The Saha Equation + Debye Length

Graduate Course
The Saha Equation + Debye Length
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The chemical potential
Chemistry (this is the route to the Saha equation!
The Saha equation
Debye Length
Plasma Parameter
Plasma Frequency
Chemical Reactions
Let us consider a chemical reaction, for example hydrogen and oxygen
reacting to form water vapour:
2H 2 + O2 ⇔ 2H 2O
−2H 2 − O2 + 2H 2O = 0
In a isolated box, atoms can move between molecules during a reaction, but the number of atoms is conserved. Let the coefficient of
the Bi molecule be bi, where we take the products to have positive b.
€
( for example, the first molecule in the reaction is H2, thus B1 is H2, b1=-2). Thus a general chemical reaction can be written
m
∑b B = 0
i
i=1
i
Conservation of atoms
Let there be Ni molecules of type Bi. During the chemical reaction
the number of molecules of a particular type can change, but the atoms
have to go somewhere!
dN i = λbi
€
where λ is some constant
that is the same for all molecules
Conditions for equilibrium
For a single component system the chemical equilibrium means that
the particle number is constant. This implies a constant chemical potential The condition for chemical equilibrium for a multi-component system
can be written as
∑ µ dN
i
i
=0
i
i.e.
€
€
∑b µ
i
i
i
=0
Reactions at constant temperature
and volume
 ∂F 
µi = 

∂N i V ,T ,N j≠i
Recall
F is the free energy
So our condition for equilibrium now reads
€
 ∂F 
=0
∑ bi ∂N 
i V ,T ,N j≠i
i
F = −k BT ln Z N
where, as we know, €
Chemical equilibrium states can be calculated if we know
the partition functions of the reactants and products.
€
Reactions in ideal gases
For systems of weakly interacting indistinguishable particles (like
ideal gases) we know that
ZN =
(Z )
sp
N!
∏ (Z sp )
N
=
Ni
i
∏N !
i
Zsp is the partition function
of a single particle
i
F = −k BT ( N i ln Z sp − ln N i!)
€
= −k BT ( N i ln Z spi − N i ln N i + N i )
∂
µi = −k BT
N i ln Z spi − N i ln N i + N i )
(
∂N i
µi = −k BT (ln Z spi − ln N i )
€
Law of mass action
Thus our condition for chemical equilibrium now becomes
∑ b (ln Z
i
spi
− ln N i ) = 0
i
Or to put it in a really compact form:
€
∏ (Z
i
bi
bi
)
=
(N
)
∏
spi
i
i
This is known as the law of mass action. If we know the partition
functions, we€know the concentrations of reactants and products.
Chemistry
Without going into detail, we know that the partition functions
must be some function of volume and temperature. Therefore
the law of mass action states:
N1b1 N 2b2 N 3b3 ...N ibi = K(T ,V )
where K is known as the ‘equilibrium constant’ of the reaction.
€
Plasma Physics
•  Rather than do any chemistry, let us do some physics that uses
the same ideas - plasma physics.
•  The ionization of atoms looks similar to a chemical reaction.
•  Let us consider the simplest case: the ionization of atomic
hydrogen.
H ⇔ p + + e−
p + + e− − H = 0
•  We will assume that the density is low enough, and the
temperature high enough, that the electrostatic energy can be
ignored
€ compared with the thermal
€ energy. Thus we assume a
very weakly interacting particles (ideal gas).
Saha Equation (1)
bi
bi
(Z
)
=
(N
)
∏ spi ∏ i
i
i
be− = b p+ = 1 , bH = −1
€
€
N e− N p +
NH
=
Z sp − Z sp +
e
p
ZH
But in a hydrogen plasma, the number of electrons is the number
of protons, so
N e2− Z spe− Z sp p+
€
=
NH
ZH
So now we need to work out the partition functions….
Partition functions (1)
•  To work out how ionized the gas is, we need to work
out the partition functions of electrons, protons, and
hydrogen atoms
•  In order to do this, we need to decide what we mean by
zero energy
•  We will take zero energy to be free electrons/protons at
rest
•  Note that a static recombined atom in the ground state
has an energy of -13.6eV (the Rydberg) on this energy
scale
Partition functions (2)
The electrons and protons look like free gas atoms…the partition function is
just the total volume divided by the volume of the particle
Z sp −
 2 πme k BT 3/2
= 2 ×V 

2
 h

Z sp +
 2 πm p k BT 3/2
= 2 ×V 

2
 h

e
spin
p
The volume occupied by a
particle is given by the deBroglie wavelength
€
So do the hydrogen atoms, but a static H atom has an energy of -Ry
compared with a static electron/protron:
€
+Ry   2 πm H k BT 3/2
Z sp = 2 × 2 × exp

V 
2
H

h
 k BT  
Saha Equation (2)
N e2− Z spe− Z sp p+
=
NH
ZH
3/2
3/2


2
π
m
k
T

 2 πme k BT
p B
2V
2V




2
2
N e2−
 h

 h

€=
 Ry   2 πm H k BT 3/2
NH
4 exp

V 
2

h
 k BT  
assume mp=mH, then
€
 −Ry  2 πme k BT 3/2
N e2−
= V exp


2

NH
 k BT  h
Saha Equation
Solving Saha for Hydrogen
Define the degree of ionization, ξ which is defined in terms of the total number of atoms (ionized +unionized), N0, such that €
N e = N p = ξN 0
N H = (1− ξ )N 0
so the Saha equation can be written as
3/2




ξ € V
−Ry 2 πme k BT
=
exp
 = f (N 0 ,V,T )

2

(1− ξ ) N 0
 k BT  h
2
ξ=
€
f2+4f − f
2
Typical Numbers for Hydrogen
As an example, take a density of atoms/ions of 1020 m-3
(typical Tokamak plasma)
T (eV)
f(N0,V,T)
ξ
0.6
2x10E-3
4.4E-02
0.8
0.9
0.6
1.0
37.4
0.975
Plasmas are ‘easy’ to make
•  Notice that the hydrogen is 97% ionized at a
temperature of 1.0 eV
•  But the ionization energy was 13.6 eV •  The plasma is created at a ‘low’ temperature. Why?
•  The answer is in the statistical weight…ionization
occurs by those few high energy electrons on the tail of
the distribution function •  The continuum is ‘heavy’ compared with the
recombined atoms – explains why it is slightly easier
to ionize as the density decreases
Multi-electron atoms (1)
Let AZ represent the ion A with Z out of a total of Z’ electrons removed. A0 ⇔ e− + A1
....................
AZ ⇔ e− + AZ +1
....................
AZ ' −1 ⇔ e− + AZ '
Z sp(e− ) Z sp(Z +1)
€
Z sp(Z )
N e− N (Z +1)
€
=
N (Z )
Can no longer assume
N e− = N (Z +1)
Multi-electron atoms (2)
Z'
N e− = N1 + 2N 2 + ...+ ZN Z + ...+ Z ' (N Z ' ) = ∑ iN i
i=0
€
 IZ 
Z sp(Z +1) g(Z +1)
=
exp

Z sp(Z )
gZ
 k BT 
Procedure: (1) For given ion density guess Ne. (2) Use this guess to
€ all the ion density ratios computationally. (3) Check if
determine
this set of ratios of ion densities is consistent with the initially
guessed electron density. (4) If not adjust guess. (5) Iterate until
convergence achieved.
Example
Aluminum at
a density of
1020 cm-3
Debye Shielding (1)
+
-
+
-
-
+
-
+
-
+
+
-
+ Excess of electrons close
to test charge, deficit of ions
close to test charge.
+
-
-
+
-
+
+
+
-
-
+
+
-
Debye Shielding (2)
Assume potential at a distance r from the test charge is φ (r)
 eφ (r) 
 eφ (r) 
ne (r) = n0 exp
 ≈ n0 1+

 kT 

kT 
€
Therefore there is an excess of electrons of order
eφ (r)
n (r) ≈ n0
kT
+
e
€
By similar reasoning there is a deficit of ions at the same position
€
eφ (r)
n (r) ≈ −n0
kT
−
i
Debye Shielding (3)
Therefore the excess charge density at the point r is given by
eφ (r)
ρ (r) = e(n (r) − n (r)) ≈ −2n0
kT
−
i
+
e
For self consistency the potential itself is related to the charge density
by Poisson’s equation
€
2



1
∂
∂φ
ρ
(r)
2n
e
∇ 2φ (r) = 2 r 2  = −
=  0 φ (r)
r ∂r  ∂r 
ε 0  ε 0 kT 
€
Debye Shielding: solution
 2r 
A
φ (r) = exp−

r
 λD 
where the Debye length is given by
€
ε 0 kT
λD =
ne2
 2r 
Q
€ φ (r) = 4 πε r exp− λ 

0
D 
Plasma Parameter
A ‘good’ plasma is one that has a large number of particles within
a Debye Sphere:
4
N D = n 0 πλ3D >> 1
3
Also note that the Debye length is the typical distance that a thermal
electron travels during a plasma period…
€
ε 0 kT
kT ε 0 m vth
λD =
=
=
2
2
ne
m ne ω p
€
Plasma Oscillations
•  Ions are heavy compared with electrons - assume they are
stationary
•  Displace the electrons from the ions and ‘let go’ - the resulting
oscillation occurs at what is known as the plasma frequency
•  Note we assume no collisions take place •  We also assume that the electrons are cold - their motion is due
to the electrostatic restoring force – we ignore thermal motion
Plasma Frequency
l
Apply Gauss’ law:
2
nel
x
2
El =
ε0
nex
E=
ε0
2
+
+
+ - + - + - + - + - + -
-
- +
+
+ + -
+ - + - + - + - -
-
-
+
+
+ - +
-
-
+
+
+ + -
- + - + - + - + overall neutral
+ - + - + - + - -
-
-
+
+
+ -
+ -
-
-
+
+
+ + -
+ - + - + - + - -
-
-
+ -
+ -
+ -
Ex
E
2
+ -
x
Ex = 0
(electrons flow to
left)
€
d x
ne x
Equation of motion
∴ F = m 2 = −eE = −
dt
ε0
2
ne
2
ω
Oscillation at frequency
p =
ε0 m
€
Typical plasma frequencies
Useful aide memoire : fp ~ 9000 n 1/2 (n in cm-3).
Ionosphere , n ~ 104 cm-3 fp ~ 1 MHz
Tokamak n ~ 1012 cm-3 fp ~ 10 GHz
Laser plasma n ~ 1021 cm-3 fp ~ 1 THz
Note about plasma frequency
1/2
ε 0 k BT  k BT  1 vth
λD =
=
≈

2
 m  ωp ωp
ne
€
Note that the Debye length represents the distance travelled by
a typical thermal electron during the oscillation period of one
plasma wave.
Summary
•  All equilibrium chemistry is in the partition functions
of the reactants and products
•  The density of reactants and products is determined by
the law of mass action
∏ (Z ) = ∏ (N )
•  Ionization is a little like a chemical reaction - applying
the law of mass action leads to the Saha equation.
bi
bi
spi
i
€
i
i
 −Ry  2 πme k BT 3/2
N e2−
= V exp


2


NH
h
 k BT 
•  The ground state of the atom dominates the partition
function
€
Summary
•  Charges in a plasma are shielded over a distance of
order the Debye length
ε 0 kT
λD =
ne2
•  Electrostatic waves occur in plasmas, in a cold plasma
they are independent of wavelength, and oscillate with
a frequency given by
€
ne2
ω =
ε0 m
2
p
€
Graduate Course
Opacity and X-ray Spectra
•  Einstein A and B Coefficients
•  Introduction to Opacity
•  Various forms of equilibrium
Blackbody Radiation
•  In undergrad classes we studied blackbody radiation – i.e. a
thermal distribution of photons.
•  But how do photons get into thermal equilibrium? To put it
another way, why does a sodium or hydrogen lamp emit a line
spectrum, but a tungsten filament a ‘blackbody’ continuum? Or
when do you see line radiation from a plasma, and when a
continuum?
The Blackbody/Line Spectrum Problem
•  To answer this riddle we will use a circular
argument (!)
•  We will first assume Blackbody radiation can
exist
•  We will use this fact to determine the Einstein
A and B coefficients (as in Atomic Physics)
•  We will then explain the difference between the
line radiation source and the blackbody source
Reminder about Blackbody Radiation
V
dω
3
ρ(ω )dω = 2 × 2 3 ω
2π c
[exp(ω / k BT ) − 1]
ω 3 dω
ρ(ω )dω = C
[exp(ω / kBT ) −1]
where
V
C= 2 3
π c
Einstein A and B Coefficients (1)
•  Put an atom at temperature T in a blackbody
radiation bath also of temperature T
•  Assume the atom only has two levels, populated
by the Boltzmann distribution
•  Equate the transition rates between the levels
(i.e. stimulated absorption = stimulated
emission + spontaneous emission)
Einstein A and B Coefficients (2)
N2 atoms, energy ε2
Degeneracy g2
ε 2 − ε1 = ω 0
N1 atoms, energy ε1
Degeneracy g1
 −ω 
N2 g2
0
= exp 

N1 g1
k
T
 B 
A21N2
ρ(ω 0 ) N2 B2 1
ρ(ω 0 ) N1 B1 2
N2 ( A2 1 + ρ (ω 0 ) B2 1) = N1 ρ (ω 0 ) B1 2
Einstein A and B Coefficients (3)












g2  
−ω 0 
−ω 0 
3
3




A
1
−
exp
+
CB
ω
exp
=
CB
ω
2
1
2
1
0
1
2
0


 k T 
 k T 
g1  
B
B

This equation must hold at all temperatures. Thus if T is very high we find B2 1 g1
=
B1 2 g2
Whereas if T is very low we see
g1
A2 1 = Cω 30 B1 2 = Cω 30 B2 1
g2
Line Radiation
The line emission has a finite width - perhaps due to Doppler
broadening, pressure broadening, or natural lifetime.
Gaussian Line Profile
1.2
1
0.8
0.6
0.4
0.2
0
Broadening of the line could
come about due to the thermal
motion of the atoms.
ω0
1
Frequency
∞
∫ φ (ω )dω = 1
0
Optically thin source
Gaussian Line Profile
1.2
1
∞
0.8
∫ φ (ω )dω = 1
0.6
0.4
0
0.2
0
1
ω0
Frequency
If the source is not very thick, the photons have little chance
of being reabsorbed on the way out, we see line radiation.
An optically thick source (1)
Gaussian Line Profile
1.2
1
0.8
0.6
0.4
0.2
0
ω0
1
Frequency
Gaussian Line Profile
1.2
1
0.8
0.6
L
0.4
0.2
0
1
ω0
Frequency
Absorption length L ~ 1/κω
Where κω is the absorption coefficient
An optically thick source (2)
•  Lots of photons are emitted at line centre, but the absorption depth is also
short - we only see photons from roughly the last absorption depth
•  Not so many photons are emitted from the wings of the line, but their chance
of being reabsorbed is corresponding less - i.e. the absorption depth is much
longer
•  These two effects cancel out, and the number of photons we see does not
depend on the line shape!
•  Observed energy density ~ ρ ω = N2 A2 1φ ω ω × L
( )
( )
An optically thick source (3)
N 2 A21φ (ω )ω jω
ρ(ω ) =
=
κω
κω
N 2 A21φ (ω )ω N 2 A21
ρ(ω ) =
=
N1B12φ (ω )ω N1B12
 ω 
ρ(ω ) = Cω exp −

 kB T 
3
Energy density looks similar to Blackbody Radiation in the
high temperature limit.
€
An optically thick source (4)
We forgot stimulated emission:-
κ ω = ( N1 B1 2 − N2 B2 1)φ (ω )ω
1
jω
ρ(ω ) = ρ BB (T ) = Cω
=
 ω

κω
exp 
−
1
kB T
3
ρ BB (T ) is the energy density of a blackbody at
temperature T
The optically thin to thick transition (1)
δρ(ω ) = [ jω − κ ω ρ (ω )]δx
δρ(ω ) = [ ρ BB (T ) − ρ (ω )]κ ωδx
Integrating over a length x:-
ρ(ω , x )
∫
0
δx
x
δρ(ω )
= − ∫κ ωδx
ρ(ω ) − ρ BB (T )
0
which results in
ρ(ω , x ) = ρ BB (T )[1− exp (−κ ω x )]
The optically thin to thick transition (2)
Tau = 0.1
Tau = 0.2
Tau = 0.5
Tau = 1.0
Tau = 2.0
Tau = 5.0
Tau = 20.0
1.2
1
Intensity
0.8
ρ(ω, x ) = ρ BB (T) if κ ω x >> 1
0.6
0.4
€
ρ(ω, x ) = jω = N 2 A21φ (ω )ω if κ ω x << 1
0.2
0
Frequency
€
τ=κx at the center of the line (Doppler Profile)
Hohlraums
Cavity at temperature T
Hohlraums are used to ensure that the surface of the
material (which loses the photons) stays at the same
temperature as the bulk of the material.
Non-LTE Physics
•  Radiation escapes from the plasma - therefore
the downward and upward rates are no longer in
balance
•  There is competition between collisional and
radiative processes
Complete Equilibrium
2
σ2,1
Β1,2
-
e
e-
Α2,1
hν
hν
Β2,1
σ1,2
1
N1ρ(ν1,2)B1,2+N1NeC1,2=N2ρ(ν1,2)B2,1+N2A2,1+N2NeC2,1
Detailed Balance (Collisions)
2
σ2,1
eeσ1,2
N1NeC1,2=N2NeC2,1
 ΔE 
C2,1 g2
= exp−

C1,2 g1
 k BT 
Collisional equilibrium occurs when the electron density is high
and the radiative rates can be€neglected
Coronal Equilibrium
Assume weak radiation field, and density so low that
A2,1 >> NeC2,1
then we have ‘coronal equilibrium’ such that
N1NeC1,2 = N2A2,1
so N2/N1 = (NeC1,2)/A2,1
Note the upper state population is now proportional to the electron density,
and is clearly below the value given by Boltzmann’s law
Example
Aluminium - 400eV
0.001
Hy-2/Hy-1
1 0-5
1 0-7
1 0-9
1 0-11
1 014
1 016
1 018
Ne
1 020
1 022
Density dependence of ratios
of ionization stages
•  Saha (LTE) predicts weak dependence on density
•  Collisional ionization is balanced by three-body recombination
(if these are in balance we have Saha)
•  At lower densities collisional ionization is balanced by radiative
recombination, and ionization is density independent
Weak density dependence is used as a temperature diagnostic
Aluminium at 101 9cm-3
Aluminium - 400eV
1 05
100
1 04
Hy/13+
Hy/13+
1000
100
10
1
10
1 014
1 016
1 018
Ne
1 020
1 022
0.1
100
1000
Temperature (eV)
Use ratios of high lying optically thin lines in LTE with next ionization stage to find ionization ratios and hence temperatures. Thermal contact between levels in weak radiation field
If the plasma is thin, then radiation can escape. Under these conditions
N1NeC1,2 = N2NeC2,1+N2A2,1
For one level to be in ‘thermal contact’ with the next level we need
Ne >> A2,1/C2,1
In practice the numbers work out to be
Ne >> 1.6x1012 x (Tkelvin)1/2 (ΔEev)3 m-3
Temperature Diagnosis
•  High lying states are in good ‘thermal contact’ with the ground
state of the next ionization stage.
•  Therefore the ratio of lines from optically thin high lying states
from different ion states is a good measure of the ratio of the
ground state populations of successive ionization stages.
•  This ratio is strongly dependent upon temperature, but only
weakly dependent on density - thus we have a temperature
diagnostic.
Temperature Diagnosis continued…
Consider the intensity of radiation from level n in a hydrogenic ion
to the ground state.
I H (n,1) = AH (n,1)N H (n)
But the level is in good thermal contact with the next ionization stage…
 χ H (n,∞) 
NB
gB
=
exp−

€
N H (n) g H (n)
k
T


B
Similarly for a high lying level, n’, in a helium-like ion
€
 χ He (n ' ,∞) 
N H (1)
g H (1)
=
exp−

'
'
N He (n ) g He (n )
k
T


B
Temperature Diagnosis continued…
Hence the ratio of the intensity of the lines is given by
 χ H (n,∞) − χ He (n ' ,∞) 
I H (n,1)
AH (n,1)g H (n)g H (1)N B
=
exp

'
'
'
I He (n ,1) AHe (n ,1)g He (n )g B N H (1)
k BT


€
But the exponential is very close to unity (as the energy difference is so small – e.g. for Al n=4 of H, it is 144 eV, but for n=4 of He-like, is
123.3 eV, just over 20 eV different)
Thus the intensity of the lines is a good measure of the ionization stage ratios, and thus a good measure of temperature
Typical Spectrum
Density Diagnosis
•  An ion in a plasma experiences an electric field due to the
neighbouring electrons and ions
•  This field shifts the energy levels – the shift being some function
of the field and therefore distance to the perturbing charge
•  This shifts the transition energy of the radiation
•  As there are many ions, there are a range of fields being
experienced, leading to a broadening of the observed spectral
line
•  The simplest example is Stark broadening of emission from
hydrogenic ions, where linewidth ~ ne2/3
Stark Broadening
•  Shift of Degenerate levels in Hydrogen is
proportional to electric field.
•  Field (and range of fields) scales as 1/r2.
•  Electron density scale as r-3.
•  Therefore linewidth scales as ne2/3 Line-Width Calculation
Notice a full calculation
has a line width
that roughly scales with
density as we predict.
Summary
•  Blackbody radiation arises because in a thick source the
probability of being absorbed is proportional to the probability
of being emitted
•  In general plasmas are in collisional-radiative equilibrium
•  At low densities coronal equilibrium is appropriate. Ionization
ratios are density independent, but upper state levels are
proportional to density
•  Density insensitivity allows us to use high lying lines as
temperature diagnostics (as they are in good thermal contact
with the next ionization stage)
•  Line widths can be used as a density diagnostic