Download The EoS, together with the thermodynamic equation, allows to

Equation of State and Thermodynamic Equilibrium
Goal of the Chapter: derive the equation of state (or the mutual
dependencies among local thermodynamic quantities such as
P, T , ρ, and Ni ), not only for the classic ideal gas, but also for
photons and fermions.
Equation of State and Thermodynamic Equilibrium
Goal of the Chapter: derive the equation of state (or the mutual
dependencies among local thermodynamic quantities such as
P, T , ρ, and Ni ), not only for the classic ideal gas, but also for
photons and fermions.
The EoS, together with the thermodynamic equation, allows to
study how the stellar material properties react to the heat,
changing density, etc.
Equation of State and Thermodynamic Equilibrium
Goal of the Chapter: derive the equation of state (or the mutual
dependencies among local thermodynamic quantities such as
P, T , ρ, and Ni ), not only for the classic ideal gas, but also for
photons and fermions.
The EoS, together with the thermodynamic equation, allows to
study how the stellar material properties react to the heat,
changing density, etc.
The thermodynamic equation with chemical reactions:
dQ = TdS = dE + PdV −
X
µi dNi
i
where µi is the chemical potential of the ith species
∂E
µi =
.
∂Ni S,V
Chemical reactions determine the species number density Ni ,
which may have the units of number per gram of material; i.e.,
Ni = ni /ρ.
Outline
The Local Thermodynamic Equilibrium
The Distribution Function
Black Body Radiation
Fermi-Dirac EoS
The Complete Degenerate Gas
Application to White Dwarfs
Temperature Effects
Ideal Gas
The Saha Equation
“Almost Perfect” EoS
Adiabatic Exponents and Other Derivatives
The Local Thermodynamic Equilibrium
LTE means that in a sufficiently small volume at any position in
a star, the complete thermodynamic (mechanical, thermal, and
chemical) equilibrium is very nearly true.
The Local Thermodynamic Equilibrium
LTE means that in a sufficiently small volume at any position in
a star, the complete thermodynamic (mechanical, thermal, and
chemical) equilibrium is very nearly true.
The equilibrium of chemical reactions means
X
µi dNi = 0,
i
appropriate for a star, since the mean free path of particles is
generally short, compared to the variation scales of the
thermodynamic quantities; e.g., λph = (κρ)−1 is on order of 1
cm ( except for stellar photospheres).
The Local Thermodynamic Equilibrium
LTE means that in a sufficiently small volume at any position in
a star, the complete thermodynamic (mechanical, thermal, and
chemical) equilibrium is very nearly true.
The equilibrium of chemical reactions means
X
µi dNi = 0,
i
appropriate for a star, since the mean free path of particles is
generally short, compared to the variation scales of the
thermodynamic quantities; e.g., λph = (κρ)−1 is on order of 1
cm ( except for stellar photospheres).
The time scale to establish the LTE is also typically short
enough, except for nuclear reactions, compared to the
evolutionary time scale of a star with the exception of various
explosions.
As an example, let’s consider the ionization-recombination
reaction:
H + + e − ↔ H 0 + χH
where χH = 13.6 eV is the ionization potential from the ground
state of H. For simplicity, we have assumed that H has only one
bound level and that involve the above constituents and that the
gas is pure hydrogen.
As an example, let’s consider the ionization-recombination
reaction:
H + + e − ↔ H 0 + χH
where χH = 13.6 eV is the ionization potential from the ground
state of H. For simplicity, we have assumed that H has only one
bound level and that involve the above constituents and that the
gas is pure hydrogen.
Consider a classic black body cavity filled with radiation in
thermodynamic equilibrium with the wall. The equilibrium of a
chemical reaction with photons can be written as
X
µi dNi + µγ dNγ = 0
i
Since photon number is not strictly conserved, in general
dNγ 6= 0. While dNi = 0 for the cavity wall (as a whole), we
have µγ = 0. Therefore, we can neglect the photons here.
The ionization-recombination reaction can then be written as
1H + + 1e− − 1H 0 = 0
And in general, a reaction can be written in such a symbolic
form:
X
νi Ci = 0,
i
where Ci is the names of the participating particle (except for
photons). Clearly, dNi ∝ νi . Thus
X
µi νi = 0
i
This is the equation of chemical equilibrium.
The ionization-recombination reaction can then be written as
1H + + 1e− − 1H 0 = 0
And in general, a reaction can be written in such a symbolic
form:
X
νi Ci = 0,
i
where Ci is the names of the participating particle (except for
photons). Clearly, dNi ∝ νi . Thus
X
µi νi = 0
i
This is the equation of chemical equilibrium.
Ni is closely linked to µi and depends on T and ρ or equivalent
thermodynamic quantities, as well as on a catalog of the
possible reactions.
Outline
The Local Thermodynamic Equilibrium
The Distribution Function
Black Body Radiation
Fermi-Dirac EoS
The Complete Degenerate Gas
Application to White Dwarfs
Temperature Effects
Ideal Gas
The Saha Equation
“Almost Perfect” EoS
Adiabatic Exponents and Other Derivatives
The Distribution Function
For a particular species of elementary nature in the LTE, the
occupation number at a certain quantum state in the
coordinate-momentum phase space is
F (j, E(p)) =
1
,
exp[−µ + Ej + E(p)]/kT ± 1
where Ej is the internal energy state j referred to some reference
energy level, E(p) is the kinetic energy as a function of the
momentum p, E(p) = (p2 c 2 + m2 c 4 )1/2 − mc 2 , µ is the chemical
potential of the species and is to be determined, and “+” or “−” is for
Fermi-Dirac or Bose-Einstein particles.
The Distribution Function
For a particular species of elementary nature in the LTE, the
occupation number at a certain quantum state in the
coordinate-momentum phase space is
F (j, E(p)) =
1
,
exp[−µ + Ej + E(p)]/kT ± 1
where Ej is the internal energy state j referred to some reference
energy level, E(p) is the kinetic energy as a function of the
momentum p, E(p) = (p2 c 2 + m2 c 4 )1/2 − mc 2 , µ is the chemical
potential of the species and is to be determined, and “+” or “−” is for
Fermi-Dirac or Bose-Einstein particles.
The number of states in the phase space, accounting for the
degeneracy of the state gj , is
gj d 3 pd 3 r
.
h3
So the phase-space density can be defined as
n(p) =
gj
1 X
.
exp[−µ + Ej + E(p)]/kT ± 1
h3
j
We can then calculate the physical space density n, kinetic
pressure P, and internal energy E as
Z
n = n(p)4πp2 dp,
p
which gives the connection between n, T , and µ.
Z
1
P=
n(p)vp4πp2 dp,
3 p
where v =
∂E(p)
.
∂p
Z
E=
p
n(p)E(p)4πp2 dp
Outline
The Local Thermodynamic Equilibrium
The Distribution Function
Black Body Radiation
Fermi-Dirac EoS
The Complete Degenerate Gas
Application to White Dwarfs
Temperature Effects
Ideal Gas
The Saha Equation
“Almost Perfect” EoS
Adiabatic Exponents and Other Derivatives
Black Body Radiation
Photons are mass-less bosons of unit spin, in particular. µ = 0,
as discussed earlier. Since they travel at c (hence no rest
frame), they have only two states (two spin orientations or
polarizations). Hence,
Z
p2 dp
8π ∞
∝ T 3.
nγ = 3
h 0 exp(pc/kT ) − 1
aT 4
and
3
4
Erad = aT = 3Prad , where a is the radiation constant.
The nice thing about the LTE is that all these quantities depend
only on T .
So radiation follows a γ-law EoS P = (γ − 1)E with γ = 4/3.
Similarly, one can easily get Prad =
The specific energy density per unit
frequency (ν) is
uν =
8πhν 3
1
erg cm−3 Hz−1 .
3
exp(hν/kT ) − 1
c
and the frequency-dependent Planck
function
c
uν .
Bν (T ) =
4π
The integrated Planck function is
Z
σ
B(T ) = Bν (T )dν = T 4 erg cm−2 s−1 sr−1 .
π
where σ = ca/4 is the Stefan-Boltzmann
constant.
B(x) = x 3 /[exp(x) − 1].
The maximum is at
x = hν/kT = 2.82.
Outline
The Local Thermodynamic Equilibrium
The Distribution Function
Black Body Radiation
Fermi-Dirac EoS
The Complete Degenerate Gas
Application to White Dwarfs
Temperature Effects
Ideal Gas
The Saha Equation
“Almost Perfect” EoS
Adiabatic Exponents and Other Derivatives
Fermi-Dirac EoS
Elections, protons, and neutrons are fermions. They all have
spin one half. Hence g = 2.
The occupation number is
1
,
exp ([E(p) − µ]/kT ) + 1
p 2
)1/2 − 1
where E(p) = mc 2 (1 +
mc
−1/2
2
p
p
and v (p) =
1+
.
m
mc
When T → 0, the integral tends to be
either zero or unity, depending on whether
the kinetic energy E(p) < µ or E(p) > µ.
We call this critical kinetic energy, EF = µ,
as the “Fermi energy”, though we are yet to
determine µ.
F (E) =
The shaded area and the
dashed line show how F (E)
is changed by raising the
temperature from kT = 0 to
EF /20.
The Complete Degenerate Gas
When T = 0, fermions fill the energy range, 0 ≤ E ≤ EF .
Correspondingly, we have pF , as definedby pF
EF = mc 2 (1 + xF2 )1/2 − 1 , where xF =
.
mc
The number density can be calculated as
Z
8π
h −3 3
8π pF 2
p dp =
xF ,
n= 3
3 mc
h 0
h
(= 2.4 × 10−10 cm for electrons) is the Compton
mc
wavelength.
For the complete degenerate electron gas, we then have
ρ
= BxF3 ,
µe
−3
8πma
h
where B =
= 9.7 × 105 g cm−3 and µe is
3
me c
electron mean molecular weight.
ρ
Once
is known, we can derive xF , and hence pF and EF .
µe
where
The pressure and internal energy density can be calculated as
P=
xF
8π m4 c 5
3 h3
Z
−3
Z
0
x4
dx,
(1 + x 2 )1/2
and
E = 8π
h
mc
mc
2
xF
x 2 [(1 + x 2 )1/2 − 1]dx,
0
which have limiting forms of P ∝ E ∝ xF5 ∝ ρ5/3 for xF 1 and
P ∝ E ∝ xF4 ∝ ρ4/3 for xF 1.
It is easy to show E/P = 3/2 (hence γ = 5/3, assuming a
γ-law EoS, P = (γ − 1)E) for xF 1 and E/P = 3 (γ = 4/3)
for xF 1.
xF ∼ 1 or pF ∼ me c corresponds to the demarcation between
non-relativistic and relativistic mechanics.
The corresponding density is
∼ 106 g cm−3 , which, incidentally, is a
typical central density of white dwarfs.
For typical densities in a neutron star
(comparable to the nuclear densities of
ρ ≈ 2.7 × 1014 g cm−3 ), it is easy to show
xF ≈ 0.35. Thus neutron stars are
non-relativistic.
The completely degenerate gas in a
non-relativistic limit acts like a
monatomic ideal gas whereas, in the
extreme relativistic limit, it behaves like a
photon gas.
The equation of state for
completely degenerate
electrons. The transition
between the non-relativistic to
the extremely relativistic
cases is smooth, but takes
place at densities around
ρtr ≈ 106 µe g cm3 .
Application to White Dwarfs
The above P(x(ρ)) relation together with the hydrostatic
equation forms a second-order differential equation (e.g., ρ vs.
Mr ). Its solution would give the structure of a WD (see Chapter
7); The WD radius can be approximately expressed as
R ≈ 0.013R
2
µe
5/3 M
M
−1/3 "
1−
M
MCh
4/3 #1/2
,
where the Chandrasekhar limiting mass is
MCh = 1.4M
2
µe
2
.
When M → MCh , R → 0! The interpretation is that an electron
degenerate object (of fixed µe ) cannot have the mass
exceeding the Chandrasekhar limit. The EoS is too soft!
Increasing density and pressure cannot halt the collapse
because the relativistic limit has already been reached.
A similar limit (∼ 3M ) can be found for a neutron star mass.
Temperature Effects
Suppose the temperature is low — on some scale yet to be
determined — but not zero. Fermions near the surface of the
sea, within a depth of ∼ kT , may find themselves elevated into
energies greater an EF .
The net effect is that the occupation
distribution smooths out to higher
energies.
Apparently, the criterion for the transition
between degeneracy and near- or
degeneracy is EF ∼ kT .
The criterion gives a ρ/µe − T relation that roughly separates
the two regimes.
For a non-relativistic electron gas
(EF = me c 2 x 2 /2),
ρ
= 6.0 × 10−9 T 3/2 g cm−3 .
µe
For a relativistic electron gas
(EF = me c 2 x),
ρ
= 4.6 × 10−24 T 3 g cm−3 .
µe
Figure: The location of the
center of the present-day sun
in these coordinates is
indicated by the sign.
Outline
The Local Thermodynamic Equilibrium
The Distribution Function
Black Body Radiation
Fermi-Dirac EoS
The Complete Degenerate Gas
Application to White Dwarfs
Temperature Effects
Ideal Gas
The Saha Equation
“Almost Perfect” EoS
Adiabatic Exponents and Other Derivatives
Ideal Gas
The Boltzmann distribution for an ideal gas is characterized by
µ/kT −1, i.e., the occupation number is very low. For
simplicity, we assume that the gas particles are non-relativistic
with E = p2 /2m, v = p/m. The number density of the gas is
then
Z p→∞
4π
2
e(µ−p /2m)/kT p2 dp.
(1)
n = 3g
h
0
Ideal Gas
The Boltzmann distribution for an ideal gas is characterized by
µ/kT −1, i.e., the occupation number is very low. For
simplicity, we assume that the gas particles are non-relativistic
with E = p2 /2m, v = p/m. The number density of the gas is
then
Z p→∞
4π
2
e(µ−p /2m)/kT p2 dp.
(1)
n = 3g
h
0
Taking logarithmic differentials of the above equation on both
sides and noticing
Z ∞
2
e−p /(2mkT ) p2 dp = (2πmkT )3/2 /(4π)
(2)
0
yield the distribution function for a Maxwell-Boltzmann ideal
gas:
4π
dn(p)
2
=
e−p /2mkT p2 dp.
3/2
n
(2πmkT )
From this equation, it is easy to get the pressure P and to prove
P = nkT , which is true even if the particles are relativistic.
Furthermore, the combination of Eqs. 1 and 2 yields
eµ/kT =
nh3
g(2πmkT )3/2
Furthermore, the combination of Eqs. 1 and 2 yields
eµ/kT =
nh3
g(2πmkT )3/2
Now consider particles (e.g., ions) with multiple energy levels,
which are populated or depopulated by photon absorption or
emission, for example. We need to have the internal energy
level Ej restored in Eq. 1. But because the photon chemical
potential is zero, µ1 = µ2 for any two levels.
Furthermore, the combination of Eqs. 1 and 2 yields
eµ/kT =
nh3
g(2πmkT )3/2
Now consider particles (e.g., ions) with multiple energy levels,
which are populated or depopulated by photon absorption or
emission, for example. We need to have the internal energy
level Ej restored in Eq. 1. But because the photon chemical
potential is zero, µ1 = µ2 for any two levels.
Then we have
n1
g1 −(E1 −E2 )/kT
=
e
,
n2
g2
which is the Boltzmann population distribution.
Outline
The Local Thermodynamic Equilibrium
The Distribution Function
Black Body Radiation
Fermi-Dirac EoS
The Complete Degenerate Gas
Application to White Dwarfs
Temperature Effects
Ideal Gas
The Saha Equation
“Almost Perfect” EoS
Adiabatic Exponents and Other Derivatives
The Saha Equation
The LTE and hence the chemical equilibrium also allow us to
determine the number densities of individual species.
Again consider the above ionization-recombination reaction:
H + + e − ↔ H 0 + χH .
We establish the reference energy levels for all species by
taking the zero energy as the just-ionized to be the H + + e−
state. Thus E0 is zero for both electrons and protons, whereas
= −χH for neutral hydrogen at the ground state.
The Saha Equation
The LTE and hence the chemical equilibrium also allow us to
determine the number densities of individual species.
Again consider the above ionization-recombination reaction:
H + + e − ↔ H 0 + χH .
We establish the reference energy levels for all species by
taking the zero energy as the just-ionized to be the H + + e−
state. Thus E0 is zero for both electrons and protons, whereas
= −χH for neutral hydrogen at the ground state.
To avoid the double counting, we may choose g − = 2 and
g + = 1 (or vise verse) as well as g 0 = 2.
The Saha Equation
The LTE and hence the chemical equilibrium also allow us to
determine the number densities of individual species.
Again consider the above ionization-recombination reaction:
H + + e − ↔ H 0 + χH .
We establish the reference energy levels for all species by
taking the zero energy as the just-ionized to be the H + + e−
state. Thus E0 is zero for both electrons and protons, whereas
= −χH for neutral hydrogen at the ground state.
To avoid the double counting, we may choose g − = 2 and
g + = 1 (or vise verse) as well as g 0 = 2. We have
ne =
2[2πme kT ]3/2 µ− /kT + [2πmp kT ]3/2 µ+ /kT
e
;n =
e
;
h3
h3
n0 =
2[2π(me + mp )kT ]3/2 µ0 /kT χH /kT
e
e
.
h3
Notice that for the LTE, µ− + µ+ − µ0 = 0 and
[me mp /(me + mp )] ≈ me , we have the Saha equation:
n + ne
=
n0
2πme kT
h2
3/2
e−χ
H /kT
.
Notice that for the LTE, µ− + µ+ − µ0 = 0 and
[me mp /(me + mp )] ≈ me , we have the Saha equation:
n + ne
=
n0
2πme kT
h2
3/2
e−χ
H /kT
.
The electrical neutrality gives ne = n+ . Using y to parametrize
n+
(where
the fraction of all hydrogen that is ionized, i.e., y =
n
0
+
n = n + n ), the equation then becomes
y2
1
=
(1 − y)
n
2πme kT
h2
3/2
e−χ
H /kT
.
Notice that for the LTE, µ− + µ+ − µ0 = 0 and
[me mp /(me + mp )] ≈ me , we have the Saha equation:
n + ne
=
n0
2πme kT
h2
3/2
e−χ
H /kT
.
The electrical neutrality gives ne = n+ . Using y to parametrize
n+
(where
the fraction of all hydrogen that is ionized, i.e., y =
n
0
+
n = n + n ), the equation then becomes
y2
1
=
(1 − y)
n
2πme kT
h2
3/2
e−χ
H /kT
.
The dominant factor here is the exponential, while the
dependence on the density is weak.
The characteristic temperature for ionization-recombination is
∼ 1.6 × 104 K , or roughly y ≈ 1/2 when χ/kT ∼ 10.
The thermal pressure can be expressed as
P = (ne + n+ + n0 )kT = (1 + y)NρkT ,
where N is the total ion plus neutral atom number density per
unit mass and is a constant, independent of density.
The thermal pressure can be expressed as
P = (ne + n+ + n0 )kT = (1 + y)NρkT ,
where N is the total ion plus neutral atom number density per
unit mass and is a constant, independent of density.
The specific internal energy to be used later is
E = (1 + y )N
3kT
+ yNχH erg g−1 ,
2
including the ionization energy. P and E clearly depend only on
the temperature and density.
The thermal pressure can be expressed as
P = (ne + n+ + n0 )kT = (1 + y)NρkT ,
where N is the total ion plus neutral atom number density per
unit mass and is a constant, independent of density.
The specific internal energy to be used later is
E = (1 + y )N
3kT
+ yNχH erg g−1 ,
2
including the ionization energy. P and E clearly depend only on
the temperature and density.
For real calculation, of course, all species, energy levels, and
reactions must be considered. For example, one also has the
ionization-recombination for Helium at high temperatures. The
presence of such zones of ionization has profound
consequences for the structure of a star.
The thermal pressure can be expressed as
P = (ne + n+ + n0 )kT = (1 + y)NρkT ,
where N is the total ion plus neutral atom number density per
unit mass and is a constant, independent of density.
The specific internal energy to be used later is
E = (1 + y )N
3kT
+ yNχH erg g−1 ,
2
including the ionization energy. P and E clearly depend only on
the temperature and density.
For real calculation, of course, all species, energy levels, and
reactions must be considered. For example, one also has the
ionization-recombination for Helium at high temperatures. The
presence of such zones of ionization has profound
consequences for the structure of a star.
Later we will further use similar Saha equations for some fast
nuclear reactions under certain states where the
thermodynamic equilibrium can be assumed.
Outline
The Local Thermodynamic Equilibrium
The Distribution Function
Black Body Radiation
Fermi-Dirac EoS
The Complete Degenerate Gas
Application to White Dwarfs
Temperature Effects
Ideal Gas
The Saha Equation
“Almost Perfect” EoS
Adiabatic Exponents and Other Derivatives
“Almost Perfect” EoS
Sometimes, interactions need to be accounted for that modify
the above “perfect” results. In addition, a stellar EoS might
consist of many components with radiation,
Maxwell-Boltzmann, and degenerate gases competing in
importance.
“Almost Perfect” EoS
Sometimes, interactions need to be accounted for that modify
the above “perfect” results. In addition, a stellar EoS might
consist of many components with radiation,
Maxwell-Boltzmann, and degenerate gases competing in
importance.
For example, the demarcation between the radiation and ideal
gas pressures (aT 4 /3 = ρkT /µmA ) for completely ionized
hydrogen gas is ρ = 1.5 × 10−23 T 3 g cm−3 .
“Almost Perfect” EoS
Sometimes, interactions need to be accounted for that modify
the above “perfect” results. In addition, a stellar EoS might
consist of many components with radiation,
Maxwell-Boltzmann, and degenerate gases competing in
importance.
For example, the demarcation between the radiation and ideal
gas pressures (aT 4 /3 = ρkT /µmA ) for completely ionized
hydrogen gas is ρ = 1.5 × 10−23 T 3 g cm−3 .
At ρ ∼ 1 g cm−3 , such gas can also get pressure-ionized
because the Wigner-Seitz radius a ∼ [3/(4πnI )]1/3 is
comparable to the first Bohr orbit.
“Almost Perfect” EoS
Sometimes, interactions need to be accounted for that modify
the above “perfect” results. In addition, a stellar EoS might
consist of many components with radiation,
Maxwell-Boltzmann, and degenerate gases competing in
importance.
For example, the demarcation between the radiation and ideal
gas pressures (aT 4 /3 = ρkT /µmA ) for completely ionized
hydrogen gas is ρ = 1.5 × 10−23 T 3 g cm−3 .
At ρ ∼ 1 g cm−3 , such gas can also get pressure-ionized
because the Wigner-Seitz radius a ∼ [3/(4πnI )]1/3 is
comparable to the first Bohr orbit.
A measure of the interaction energy between ions is the
Z 2 e2
characterizes the
Coulomb potential. The ratio ΓC ≡
akT
importance of the Coulomb effects relative to the thermal
agitation. When Γc is much greater than 1, the gas settles down
into a crystal.
“Almost Perfect” EoS
This composite figure shows
how the ρ − T plane is broken
up into regions dominated by
pressure ionization,
degeneracy, radiation, ideal
gas, crystallization, and
ionization-recombination. The
gas is assumed to be pure
hydrogen.
“Almost Perfect” EoS
This composite figure shows
how the ρ − T plane is broken
up into regions dominated by
pressure ionization,
degeneracy, radiation, ideal
gas, crystallization, and
ionization-recombination. The
gas is assumed to be pure
hydrogen.
In general, the EoS here is approximated as P ∝ T χT ρχρ ,
where the exponents typically need to be evaluated numerically.
Outline
The Local Thermodynamic Equilibrium
The Distribution Function
Black Body Radiation
Fermi-Dirac EoS
The Complete Degenerate Gas
Application to White Dwarfs
Temperature Effects
Ideal Gas
The Saha Equation
“Almost Perfect” EoS
Adiabatic Exponents and Other Derivatives
Adiabatic Exponents and Other Derivatives
With the EoS, we can in principle construct a simplified stellar
model. But to construct realistic ones and to evolve them, we
need several thermodynamic derivatives, for example, to study
the stability of a model and to determine weather or not the
convection needs to be considered.
Adiabatic Exponents and Other Derivatives
With the EoS, we can in principle construct a simplified stellar
model. But to construct realistic ones and to evolve them, we
need several thermodynamic derivatives, for example, to study
the stability of a model and to determine weather or not the
convection needs to be considered.
Such derivatives include the specific heats (cρ and cp ),
Adiabatic Exponents and Other Derivatives
With the EoS, we can in principle construct a simplified stellar
model. But to construct realistic ones and to evolve them, we
need several thermodynamic derivatives, for example, to study
the stability of a model and to determine weather or not the
convection needs to be considered.
Such derivatives include the specific heats (cρ and cp ),
power-law
exponents
of the
EoS:
∂lnP
∂lnP
χT =
and χρ =
,
∂lnT ρ
∂lnρ T
Adiabatic Exponents and Other Derivatives
With the EoS, we can in principle construct a simplified stellar
model. But to construct realistic ones and to evolve them, we
need several thermodynamic derivatives, for example, to study
the stability of a model and to determine weather or not the
convection needs to be considered.
Such derivatives include the specific heats (cρ and cp ),
power-law
exponents
of the
EoS:
∂lnP
∂lnP
χT =
and χρ =
,
∂lnT ρ
∂lnρ T
and adiabatic
exponents: ∂lnP
Γ2
∂lnP
1
∂lnT
Γ1 =
,
=
=
, Γ3 − 1 =
.
∂lnρ S Γ2 − 1
∂lnT S
∇S
∂lnρ S
Specific heats
Recall the thermodynamic equation with chemical reactions:
X
dQ = TdS = dE + PdV −
µi dNi .
i
As an example, we consider the specific heat, cρ =
∂E
=
for the above pure hydrogen gas Recall
∂T ρ
E = (1 + y )N
∂Q
∂T
ρ
3kT
+ yNχH erg g−1 .
2
We have
∂E
∂T
∂E
∂y
∂y T ∂T ρ
y
"
#
1
∂y
3kT 1
2 3 χH
= (1 + y)N
+
+
.
2
T
3 2 kT (1 + y) ∂T ρ
=
ρ
∂E
∂T
+
Specific heats
Now recall the Saha equation
y2
1
=
(1 − y)
n
2πme kT
h2
3/2
e−χ
H
/kT
.
of it andthen take
the partial derivative of T , we get
Take a log 1
∂y
3 χH 1
2
+
=
+
→
y
1 − y ∂T 2 kT T
1
∂y
3 χH 1
y (1 − y )
= D(y )
+
, where D(y ) =
.
(1 + y ) ∂T
2 kT " T
(2 −#y )(1 + y )
2
3k
2 3 χH
+
1 + D(y)
erg g−1 K−1 ,
Therefore, cρ = (1 + y)N
2
3 2 kT
where the second term arises from the consideration of the
ionization-recombination reaction.
Specific heats
Now recall the Saha equation
y2
1
=
(1 − y)
n
2πme kT
h2
3/2
e−χ
H
/kT
.
of it andthen take
the partial derivative of T , we get
Take a log 1
∂y
3 χH 1
2
+
=
+
→
y
1 − y ∂T 2 kT T
1
∂y
3 χH 1
y (1 − y )
= D(y )
+
, where D(y ) =
.
(1 + y ) ∂T
2 kT " T
(2 −#y )(1 + y )
2
3k
2 3 χH
+
1 + D(y)
erg g−1 K−1 ,
Therefore, cρ = (1 + y)N
2
3 2 kT
where the second term arises from the consideration of the
ionization-recombination reaction.
Note that D(1) = D(0) = 0 and, for general 0 ≤ y ≤ 1,
D(y ) ≥ 0. Thus the term is greater than zero, meaning that an
increase of the temperature, keeping the density fixed, requires
more energy when the reaction occurs than when it does not.
Variable Conversions
In realistic (complicated) situation, the calculations of the
derivatives need to be done numerically. Thus it is desirable to
know the analytic conversions of the derivatives from a minimal
set of calculations.
Variable Conversions
In realistic (complicated) situation, the calculations of the
derivatives need to be done numerically. Thus it is desirable to
know the analytic conversions of the derivatives from a minimal
set of calculations.
Among the five variables in the first law of thermodynamics
(E, S, P, T , and V ), only two are independent.
Variable Conversions
In realistic (complicated) situation, the calculations of the
derivatives need to be done numerically. Thus it is desirable to
know the analytic conversions of the derivatives from a minimal
set of calculations.
Among the five variables in the first law of thermodynamics
(E, S, P, T , and V ), only two are independent.
The law can be expressed in various forms by replacing E with
H = E + PV (enthalpy),
Variable Conversions
In realistic (complicated) situation, the calculations of the
derivatives need to be done numerically. Thus it is desirable to
know the analytic conversions of the derivatives from a minimal
set of calculations.
Among the five variables in the first law of thermodynamics
(E, S, P, T , and V ), only two are independent.
The law can be expressed in various forms by replacing E with
H = E + PV (enthalpy),
F = E − TS (free energy),
Variable Conversions
In realistic (complicated) situation, the calculations of the
derivatives need to be done numerically. Thus it is desirable to
know the analytic conversions of the derivatives from a minimal
set of calculations.
Among the five variables in the first law of thermodynamics
(E, S, P, T , and V ), only two are independent.
The law can be expressed in various forms by replacing E with
H = E + PV (enthalpy),
F = E − TS (free energy),
or Φ = E + PV − TS (the Gibbs free energy),
Variable Conversions
In realistic (complicated) situation, the calculations of the
derivatives need to be done numerically. Thus it is desirable to
know the analytic conversions of the derivatives from a minimal
set of calculations.
Among the five variables in the first law of thermodynamics
(E, S, P, T , and V ), only two are independent.
The law can be expressed in various forms by replacing E with
H = E + PV (enthalpy),
F = E − TS (free energy),
or Φ = E + PV − TS (the Gibbs free energy), and
correspondingly replacing the independent variables S and V
with S and P, T and V , or T and P, respectively.
Variable Conversions
In realistic (complicated) situation, the calculations of the
derivatives need to be done numerically. Thus it is desirable to
know the analytic conversions of the derivatives from a minimal
set of calculations.
Among the five variables in the first law of thermodynamics
(E, S, P, T , and V ), only two are independent.
The law can be expressed in various forms by replacing E with
H = E + PV (enthalpy),
F = E − TS (free energy),
or Φ = E + PV − TS (the Gibbs free energy), and
correspondingly replacing the independent variables S and V
with S and P, T and V , or T and P, respectively.
For example,
dE = TdS − PdV → dΦ = −SdT + VdP
Variable Conversions
Using the above expressions and such mathematical relations
as
∂2Φ
∂2Φ
=
,
∂P∂T
∂T ∂P
one can readily do various variable transformations of the
derivative.
Variable Conversions
Using the above expressions and such mathematical relations
as
∂2Φ
∂2Φ
=
,
∂P∂T
∂T ∂P
one can readily do various variable transformations of the
derivative.
For example, one can easily drive the Maxwell’s relations
(giving it a try!):
∂P
∂T
∂V
=−
,
=
,
∂S V
∂P S
∂S P
S
∂S
∂P
∂S
∂V
=
,
=−
.
∂V T
∂T V
∂P T
∂T P
∂T
∂V
These relations are very useful to express the derivatives in
certain desirable forms and to know their relationships.
Variable Conversions
For variable changes, one may find useful such Jacobian
transformation as
∂P
∂(V , P)
∂T V
∂V
∂(V , T )
=
= ∂(T , P)
∂P
∂T P
−
∂(V , T )
∂V T
This relation can also be derived from
∂V
∂V
dV =
dT +
dP
∂T P
∂P T
by holding V constant, dV = 0.
Let’s see a couple of examples for the use of the conversions.
Example 1: Various γs:
∂lnP
I P = K ργ , where γ = Γ1 =
.
∂lnρ S
I γ ≡ cP /cV . This γ = Γ1 is good for ideal gas, but does not
hold, when the radiation pressure is important, for
example.
I
γ in the γ-law EoS: P = (γ − 1)ρE.
Let’s try to prove that the last γ equals to Γ3 = 1 +
This means that we need to prove
∂P
= (Γ3 − 1)ρ
∂E ρ
∂lnT
∂lnρ
.
S
(3)
∂E
∂P
∂S
∂P
From dE = TdS − PdV , we have
=T
and
ρ
ρ
∂E
.
T =
∂S ρ
∂T
1 ∂P
Furthermore,
= 2
.
∂ρ S
∂S ρ
ρ
We can then express the left side of Eq. 3 as
∂P
1 ∂P
ρ2 ∂T
=
=
= (Γ3 − 1)ρ.
∂E ρ T ∂S ρ
T ∂ρ S
But, γ = Γ3 in the γ-law EoS is generally not equal to
γ ≡ cP /cV = 1 +
χT
(Γ3 − 1).
χρ
Try to prove this latter relation! Clearly, when χT = 1 and
χρ = 1 (for ideal gas), then this latter γ is also equal to Γ3 .
Example 2: Relation between cP and cρ or cVρ . Start by letting
T and P be independent and write the specific heat content as
∂S
∂S
dq = TdS = T
dT +
dP .
∂T P
∂P T
and
dP =
∂P
∂T
dT +
V
∂P
∂V
dV .
T
This gives
∂S
∂S
∂P
∂P
dq = T
dT + T
dT +
dV .
∂T P
∂P T
∂T V
∂V T
While holding V constant,
∂q
∂S
∂S
∂P
=T
+T
∂T V
∂T P
∂P T ∂T V
∂S
∂P
cp − cV = −T
.
∂P T ∂T V
Now let’s express the right-hand side with the exponents of the
E.o.S.
Using one of the Maxwell’s relations
∂V
∂S
=−
∂P T
∂T P
and then the relation we have just derived:
∂P
∂T V
∂V
= ∂P
∂T P
−
∂V T
we have
∂P −1 ∂P 2
cp − cV = −T
∂V T
∂T V
−1 VP ∂lnP
∂lnP 2
P χ2T
=−
=
.
T
∂lnV T
∂lnT V
ρT χρ
Review
Key Concepts: local thermodynamic equilibrium, chemical potential
µ, chemical equilibrium, Fermi energy, Chandrasekhar limiting mass
1. What is an equation of state?
2. Why should the chemical potential of photon gas be equal to
zero?
3. What is the degenerate pressure?
4. What is the ρ − T relation corresponding to the condition under
which the degenerate and thermal energies are comparable?
5. What is the Saha equation? Under which circumstance can the
equation be used?
6. In reality, what may be various interactions that may make an
EoS imperfect?
In addition, you should be able to derive such quantities as density,
pressure, and thermal energy for photon, ideal (Maxwell-Boltzmann),
and completely degenerate gases, starting from the basic occupation
distribution (with different µ limits).
Get familiar with derivations of various adiabatic exponents and other
derivatives.