Review of Thermodynamics
The equations of stellar structure involve derivatives of thermodynamic variables such as pressure, temperature, and density.
To express these derivatives in a useful form, we will need to review the basic thermodynamic relations. First, let’s define the
variables:
ρ:
P:
T:
µ:
the
the
the
the
gas density
gas pressure
gas temperature
mean molecular weight
q:
u:
s:
V:
the
the
the
the
specific
specific
specific
specific
heat content
internal energy
entropy
volume (1/ρ)
Note that q, u, s, and V are all per unit mass. From these
variables come the specific heats
µ ¶
µ
¶
dq
∂s
cV =
=T
(1.1)
dT V
∂T V
µ ¶
µ
¶
dq
∂s
cP =
=T
(1.2)
dT P
∂T P
the ratio of the specific heats
γ=
cP
cV
the adiabatic temperature gradient
¶
µ
∂ ln T
∇ad =
∂ ln P s
an isothermal compressibility coefficient
µ
¶
µ
¶
P ∂V
∂ ln ρ
α=−
=
V ∂P T,µ
∂ ln P T,µ
(1.3)
(1.4)
(1.5)
a volume coefficient of expansion
T
δ=
V
µ
∂V
∂T
¶
=−
P,µ
µ
∂ ln ρ
∂ ln T
¶
µ
∂ ln ρ
∂ ln µ
¶
(1.6)
P,µ
and a chemical potential coefficient
µ
ϕ=−
V
µ
∂V
∂µ
¶
=
P,T
(1.7)
P,T
(For the following, we will assume the chemical composition is
fixed.)
We will also need the first law of thermodynamics:
dq = T ds = du + P dV
(1.8)
Note that although there are four variables in this equation (s,
T , P , and V ), only two are independent.
To derive the relationships between the various thermodynamic
variables, first take s and V as independent, and re-write (1.8)
as
du = T ds − P dV
(1.9)
However, when written in terms of s and V , du is formally
du =
µ
∂u
∂s
¶
ds +
V
which means that
µ ¶
∂u
=T
∂s V
µ
∂u
∂V
µ
and
¶
∂u
∂V
dV
s
¶
= −P
(1.10)
s
Now, mathematically
∂2u
∂2u
=
∂V ∂s
∂s∂V
so
µ
∂u
∂V
¶ µ
s
∂u
∂s
¶
=
V
µ
∂u
∂s
¶ µ
V
∂u
∂V
¶
s
or
µ
∂T
∂V
¶
=−
s
µ
∂P
∂s
¶
(1.11)
V
Similarly, if we choose s and P as the independent variables, and
add d(P V ) to each side of (1.9), then the first law of thermodynamics becomes
dH = d(u + P V ) = T ds − P dV + P dV + V dP = T ds + V dP
The total first derivative of H is then
¶
µ
¶
µ
∂H
∂H
ds +
dP
dH =
∂s P
∂P s
which implies that
µ
¶
∂H
=T
∂s P
and
µ
∂T
∂P
µ
and
¶
=
s
µ
∂V
∂s
∂H
∂P
¶
¶
=V
s
(1.12)
P
If we subtract d(T s) from each side of the first law of thermodynamics, then T and V are the free parameters, via
dF = d(u − T s) = T ds − P dV − T ds − sdT = −P dV − sdT
We then get the relations
µ
¶
∂F
= −s
∂T V
and
µ
∂F
∂V
which leads via the second derivatives to
¶
µ
¶
µ
∂s
∂P
=
∂V T
∂T V
¶
= −P
T
(1.13)
Finally, if T and P are chosen to be independent, and d(P V −T s)
are added to (1.8), then we can derive the relation
µ
¶
µ
¶
∂s
∂V
=−
(1.14)
∂P T
∂T P
Thus, we have Maxwell’s relations
µ
∂T
∂V
µ
µ
¶
∂T
∂P
=−
s
¶
=
s
µ
µ
∂P
∂s
∂V
∂s
¶
¶
(1.11)
V
(1.12)
P
¶
µ
¶
∂s
∂P
=
∂V T
∂T V
µ
¶
µ
¶
∂s
∂V
=−
∂P T
∂T P
(1.13)
(1.14)
To derive a relation between the specific heats, start by letting
T and P be independent, and write the specific heat content as
dq = T ds = T
and dP as
dP =
µ
·µ
∂P
∂T
∂s
∂T
¶
¶
dT +
P
dT +
V
µ
µ
∂P
∂V
∂s
∂P
¶
¶
dV
dP
T
¸
T
This gives
dq = T
µ
∂s
∂T
¶
dT + T
P
µ
∂s
∂P
¶ ·µ
T
∂P
∂T
¶
dT +
V
µ
∂P
∂V
¶
dV
T
¸
We can now evaluate (dq/dT ) while holding V constant, i.e.,
with dV = 0
µ ¶
µ
¶
µ
¶ µ
¶
dq
∂s
∂s
∂P
=T
+T
dT V
∂T P
∂P T ∂T V
The term on the left side of the equation is cV , the first term
on the right is cP , and (ds/dP )T = −(dV /dT )P by a Maxwell
relation (1.14). Thus,
cP − cV = T
µ
∂V
∂T
¶ µ
P
∂P
∂T
¶
V
The first partial differential can immediately be written in terms
of the volume coefficient of expansion (1.6)
µ
∂V
∂T
¶
=
P
Vδ
T
(1.15)
The second partial differential can also be re-written, if one first
notes that the total derivative for dV is
µ
¶
µ
¶
∂V
∂V
dV =
dT +
dP
∂T P
∂P T
Thus, when V is held constant, dV = 0, and
µ
∂P
∂T
¶
=−
V
µ
∂V
∂T
¶ Áµ
P
∂V
∂P
¶
T
The numerator on the right side is again (δ/ρT ), while the denominator is related to the compressibility coefficient by
µ
Thus
and
∂V
∂P
µ
¶
∂P
∂T
=−
T
¶
=
V
αV
P
Pδ
Tα
P V δ2
P δ2
cP − cV =
=
Tα
ρT α
Note that this reduces to R = k/mA for an ideal gas.
(1.16)
(1.17)
Finally, to express the change in the heat content of a system,
dq, in terms of the intensive parameters only, choose V and T as
the independent variables, and write the change in entropy as
¶
¶
µ
µ
∂s
∂s
ds =
dT +
dV
∂T V
∂V T
Using the definition of heat capacity (1.1) and the Maxwell relation (1.13), this becomes
µ
¶
∂P
cV
dT +
dV
ds =
T
∂T V
If we now substitude (1.16) for (∂P/∂T )V , and convert dV to dρ
using dV = −1/ρ2 dρ, we get an expression for dq
dq = T ds = cV dT −
P δ dρ
ρα ρ
This can then be further simplified by noting that
µ
¶
µ
¶
dρ
∂ ln ρ
∂ ln ρ
dP
dT
=
d ln P +
d ln T = α
−δ
ρ
∂ ln P T
∂ ln T P
P
T
Thus
P δ2
δ
dq = cV dT +
dT − dP
ρT α
ρ
or
δ
dq = cP dT − dP
ρ
(1.18)
This equation also leads directly to an expression for the adiabatic temperature gradient. If dq = 0, then
cP dT =
δ
dP
ρ
which implies that
µ
and
∇ad =
∂T
∂P
µ
¶
=
δ
ρcP
¶
=
s
∂ ln T
∂ ln P
s
Pδ
T ρcP
(1.19)
Note that for an ideal gas, the definition of an adiabat implies
that
µ ¶γ
P
P ∝ ργ ∝
=⇒ T ∝ P (γ−1)/γ
T
Hence for a monotonic gas with γ = 5/3,
∇ad
Á
= (2/3) (5/3) = 0.4
(1.20)
Note also that (1.19) can then be substituted back into (1.18)
to yield an equation for dq in terms of P , T , cP , and adiabatic
temperature gradient
·
¸
dT
dP
T cP ∇ad
dP = cP T
− ∇ad
dq = cP dT −
P
T
P
(1.21)