For thermoelastic materials, show that the specific heats are related by the relation
C
p
−
C
v
=
1
ρ
0
(
S
−
T
∂
S
^
∂
T
)
:
∂
E
~
∂
T
.
{\displaystyle C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}\left({\boldsymbol {S}}-T~{\frac {\partial {\hat {\boldsymbol {S}}}}{\partial T}}\right):{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}~.}
Proof:
Recall that
C
v
:=
∂
e
^
(
E
,
T
)
∂
T
=
T
∂
η
^
∂
T
{\displaystyle C_{v}:={\frac {\partial {\hat {e}}({\boldsymbol {E}},T)}{\partial T}}=T~{\frac {\partial {\hat {\eta }}}{\partial T}}}
and
C
p
:=
∂
e
~
(
S
,
T
)
∂
T
=
T
∂
η
~
∂
T
+
1
ρ
0
S
:
∂
E
~
∂
T
.
{\displaystyle C_{p}:={\frac {\partial {\tilde {e}}({\boldsymbol {S}},T)}{\partial T}}=T~{\frac {\partial {\tilde {\eta }}}{\partial T}}+{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}~.}
Therefore,
C
p
−
C
v
=
T
∂
η
~
∂
T
+
1
ρ
0
S
:
∂
E
~
∂
T
−
T
∂
η
^
∂
T
.
{\displaystyle C_{p}-C_{v}=T~{\frac {\partial {\tilde {\eta }}}{\partial T}}+{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}-T~{\frac {\partial {\hat {\eta }}}{\partial T}}~.}
Also recall that
η
=
η
^
(
E
,
T
)
=
η
~
(
S
,
T
)
.
{\displaystyle \eta ={\hat {\eta }}({\boldsymbol {E}},T)={\tilde {\eta }}({\boldsymbol {S}},T)~.}
Therefore, keeping
S
{\displaystyle {\boldsymbol {S}}}
∂
η
~
∂
T
=
∂
η
^
∂
E
:
∂
E
∂
T
+
∂
η
^
∂
T
.
{\displaystyle {\frac {\partial {\tilde {\eta }}}{\partial T}}={\frac {\partial {\hat {\eta }}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\frac {\partial {\hat {\eta }}}{\partial T}}~.}
Noting that
E
=
E
~
(
S
,
T
)
{\displaystyle {\boldsymbol {E}}={\tilde {\boldsymbol {E}}}({\boldsymbol {S}},T)}
C
p
−
C
v
=
T
∂
η
^
∂
E
:
∂
E
~
∂
T
+
1
ρ
0
S
:
∂
E
~
∂
T
.
{\displaystyle C_{p}-C_{v}=T~{\frac {\partial {\hat {\eta }}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}+{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}~.}
Recalling that
∂
η
^
∂
E
=
−
1
ρ
0
∂
S
^
∂
T
{\displaystyle {\frac {\partial {\hat {\eta }}}{\partial {\boldsymbol {E}}}}=-{\cfrac {1}{\rho _{0}}}~{\frac {\partial {\hat {\boldsymbol {S}}}}{\partial T}}}
we get
C
p
−
C
v
=
1
ρ
0
(
S
−
T
∂
S
^
∂
T
)
:
∂
E
~
∂
T
.
{\displaystyle {C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}\left({\boldsymbol {S}}-T~{\frac {\partial {\hat {\boldsymbol {S}}}}{\partial T}}\right):{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}~.}}
For thermoelastic materials, show that the specific heats can also be related by the equations
C
p
−
C
v
=
1
ρ
0
S
:
∂
E
∂
T
+
∂
E
∂
T
:
(
∂
2
ψ
∂
E
∂
E
:
∂
E
∂
T
)
=
1
ρ
0
S
:
∂
E
∂
T
+
T
ρ
0
∂
E
∂
T
:
(
∂
S
∂
E
:
∂
E
∂
T
)
.
{\displaystyle C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\frac {\partial {\boldsymbol {E}}}{\partial T}}:\left({\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right)={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\cfrac {T}{\rho _{0}}}~{\frac {\partial {\boldsymbol {E}}}{\partial T}}:\left({\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right)~.}
We can also write the above as
C
p
−
C
v
=
1
ρ
0
S
:
α
E
+
T
ρ
0
α
E
:
C
:
α
E
{\displaystyle C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\boldsymbol {\alpha }}_{E}+{\cfrac {T}{\rho _{0}}}~{\boldsymbol {\alpha }}_{E}:{\boldsymbol {\mathsf {C}}}:{\boldsymbol {\alpha }}_{E}}
where
α
E
:=
∂
E
∂
T
{\displaystyle {\boldsymbol {\alpha }}_{E}:={\frac {\partial {\boldsymbol {E}}}{\partial T}}}
thermal expansion tensor and
C
:=
∂
S
∂
E
{\displaystyle {\boldsymbol {\mathsf {C}}}:={\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {E}}}}}
stiffness tensor.
Proof:
Recall that
S
=
ρ
0
∂
ψ
∂
E
=
ρ
0
f
(
E
(
S
,
T
)
,
T
)
.
{\displaystyle {\boldsymbol {S}}=\rho _{0}~{\frac {\partial \psi }{\partial {\boldsymbol {E}}}}=\rho _{0}~{\boldsymbol {f}}({\boldsymbol {E}}({\boldsymbol {S}},T),T)~.}
Recall the chain rule which states that if
g
(
u
,
t
)
=
f
(
x
(
u
,
t
)
,
y
(
u
,
t
)
)
{\displaystyle g(u,t)=f(x(u,t),y(u,t))}
then, if we keep
u
{\displaystyle u}
g
{\displaystyle g}
t
{\displaystyle t}
∂
g
∂
t
=
∂
f
∂
x
∂
x
∂
t
+
∂
f
∂
y
∂
y
∂
t
.
{\displaystyle {\frac {\partial g}{\partial t}}={\frac {\partial f}{\partial x}}~{\frac {\partial x}{\partial t}}+{\frac {\partial f}{\partial y}}~{\frac {\partial y}{\partial t}}~.}
In our case,
u
=
S
,
t
=
T
,
g
(
S
,
T
)
=
S
,
x
(
S
,
T
)
=
E
(
S
,
T
)
,
y
(
S
,
T
)
=
T
,
and
f
=
ρ
0
f
.
{\displaystyle u={\boldsymbol {S}},~~t=T,~~g({\boldsymbol {S}},T)={\boldsymbol {S}},~~x({\boldsymbol {S}},T)={\boldsymbol {E}}({\boldsymbol {S}},T),~~y({\boldsymbol {S}},T)=T,~~{\text{and}}~~f=\rho _{0}~{\boldsymbol {f}}.}
Hence, we have
S
=
g
(
S
,
T
)
=
f
(
E
(
S
,
T
)
,
T
)
=
ρ
0
f
(
E
(
S
,
T
)
,
T
)
.
{\displaystyle {\boldsymbol {S}}=g({\boldsymbol {S}},T)=f({\boldsymbol {E}}({\boldsymbol {S}},T),T)=\rho _{0}~{\boldsymbol {f}}({\boldsymbol {E}}({\boldsymbol {S}},T),T)~.}
Taking the derivative with respect to
T
{\displaystyle T}
S
{\displaystyle {\boldsymbol {S}}}
∂
g
∂
T
=
∂
S
∂
T
=
ρ
0
[
∂
f
∂
E
:
∂
E
∂
T
+
∂
f
∂
T
∂
T
∂
T
]
{\displaystyle {\frac {\partial g}{\partial T}}={\frac {\partial {\boldsymbol {S}}}{\partial T}}=\rho _{0}~\left[{\frac {\partial {\boldsymbol {f}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\frac {\partial {\boldsymbol {f}}}{\partial T}}~{\frac {\partial T}{\partial T}}\right]}
or,
0
=
∂
f
∂
E
:
∂
E
∂
T
+
∂
f
∂
T
.
{\displaystyle \mathbf {0} ={\frac {\partial {\boldsymbol {f}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\frac {\partial {\boldsymbol {f}}}{\partial T}}~.}
Now,
f
=
∂
ψ
∂
E
⟹
∂
f
∂
E
=
∂
2
ψ
∂
E
∂
E
and
∂
f
∂
T
=
∂
2
ψ
∂
T
∂
E
.
{\displaystyle {\boldsymbol {f}}={\frac {\partial \psi }{\partial {\boldsymbol {E}}}}\qquad \implies \qquad {\frac {\partial {\boldsymbol {f}}}{\partial {\boldsymbol {E}}}}={\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}\quad {\text{and}}\quad {\frac {\partial {\boldsymbol {f}}}{\partial T}}={\frac {\partial ^{2}\psi }{\partial T\partial {\boldsymbol {E}}}}~.}
Therefore,
0
=
∂
2
ψ
∂
E
∂
E
:
∂
E
∂
T
+
∂
2
ψ
∂
T
∂
E
=
∂
∂
E
(
∂
ψ
∂
E
)
:
∂
E
∂
T
+
∂
∂
T
(
∂
ψ
∂
E
)
.
{\displaystyle \mathbf {0} ={\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\frac {\partial ^{2}\psi }{\partial T\partial {\boldsymbol {E}}}}={\frac {\partial }{\partial {\boldsymbol {E}}}}\left({\frac {\partial \psi }{\partial {\boldsymbol {E}}}}\right):{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\frac {\partial }{\partial T}}\left({\frac {\partial \psi }{\partial {\boldsymbol {E}}}}\right)~.}
Again recall that,
∂
ψ
∂
E
=
1
ρ
0
S
.
{\displaystyle {\frac {\partial \psi }{\partial {\boldsymbol {E}}}}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}~.}
Plugging into the above, we get
0
=
∂
2
ψ
∂
E
∂
E
:
∂
E
∂
T
+
1
ρ
0
∂
S
∂
T
=
1
ρ
0
∂
S
∂
E
:
∂
E
∂
T
+
1
ρ
0
∂
S
∂
T
.
{\displaystyle \mathbf {0} ={\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\cfrac {1}{\rho _{0}}}~{\frac {\partial {\boldsymbol {S}}}{\partial T}}={\cfrac {1}{\rho _{0}}}~{\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\cfrac {1}{\rho _{0}}}~{\frac {\partial {\boldsymbol {S}}}{\partial T}}~.}
Therefore, we get the following relation for
∂
S
/
∂
T
{\displaystyle \partial {\boldsymbol {S}}/\partial T}
∂
S
∂
T
=
−
ρ
0
∂
2
ψ
∂
E
∂
E
:
∂
E
∂
T
=
−
∂
S
∂
E
:
∂
E
∂
T
.
{\displaystyle {\frac {\partial {\boldsymbol {S}}}{\partial T}}=-\rho _{0}~{\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}=-{\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}~.}
Recall that
C
p
−
C
v
=
1
ρ
0
(
S
−
T
∂
S
∂
T
)
:
∂
E
∂
T
.
{\displaystyle C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}\left({\boldsymbol {S}}-T~{\frac {\partial {\boldsymbol {S}}}{\partial T}}\right):{\frac {\partial {\boldsymbol {E}}}{\partial T}}~.}
Plugging in the expressions for
∂
S
/
∂
T
{\displaystyle \partial {\boldsymbol {S}}/\partial T}
C
p
−
C
v
=
1
ρ
0
(
S
+
T
ρ
0
∂
2
ψ
∂
E
∂
E
:
∂
E
∂
T
)
:
∂
E
∂
T
=
1
ρ
0
(
S
+
T
∂
S
∂
E
:
∂
E
∂
T
)
:
∂
E
∂
T
.
{\displaystyle C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}\left({\boldsymbol {S}}+T~\rho _{0}~{\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right):{\frac {\partial {\boldsymbol {E}}}{\partial T}}={\cfrac {1}{\rho _{0}}}\left({\boldsymbol {S}}+T~{\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right):{\frac {\partial {\boldsymbol {E}}}{\partial T}}~.}
Therefore,
C
p
−
C
v
=
1
ρ
0
S
:
∂
E
∂
T
+
T
(
∂
2
ψ
∂
E
∂
E
:
∂
E
∂
T
)
:
∂
E
∂
T
=
1
ρ
0
S
:
∂
E
∂
T
+
T
ρ
0
(
∂
S
∂
E
:
∂
E
∂
T
)
:
∂
E
∂
T
.
{\displaystyle C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+T~\left({\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right):{\frac {\partial {\boldsymbol {E}}}{\partial T}}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\cfrac {T}{\rho _{0}}}~\left({\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right):{\frac {\partial {\boldsymbol {E}}}{\partial T}}~.}
Using the identity
(
A
:
B
)
:
C
=
C
:
(
A
:
B
)
{\displaystyle ({\boldsymbol {\mathsf {A}}}:{\boldsymbol {B}}):{\boldsymbol {C}}={\boldsymbol {C}}:({\boldsymbol {\mathsf {A}}}:{\boldsymbol {B}})}
C
p
−
C
v
=
1
ρ
0
S
:
∂
E
∂
T
+
T
∂
E
∂
T
:
(
∂
2
ψ
∂
E
∂
E
:
∂
E
∂
T
)
=
1
ρ
0
S
:
∂
E
∂
T
+
T
ρ
0
∂
E
∂
T
:
(
∂
S
∂
E
:
∂
E
∂
T
)
.
{\displaystyle {C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+T~{\frac {\partial {\boldsymbol {E}}}{\partial T}}:\left({\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right)={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\cfrac {T}{\rho _{0}}}~{\frac {\partial {\boldsymbol {E}}}{\partial T}}:\left({\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right)~.}}
Proof:
Recall that,
C
p
−
C
v
=
1
ρ
0
S
:
α
E
+
T
ρ
0
α
E
:
C
:
α
E
.
{\displaystyle C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\boldsymbol {\alpha }}_{E}+{\cfrac {T}{\rho _{0}}}~{\boldsymbol {\alpha }}_{E}:{\boldsymbol {\mathsf {C}}}:{\boldsymbol {\alpha }}_{E}~.}
Plugging the expressions of
α
E
{\displaystyle {\boldsymbol {\alpha }}_{E}}
C
{\displaystyle {\boldsymbol {\mathsf {C}}}}
C
p
−
C
v
=
1
ρ
0
S
:
(
α
1
)
+
T
ρ
0
(
α
1
)
:
(
λ
1
⊗
1
+
2
μ
I
)
:
(
α
1
)
=
α
ρ
0
tr
S
+
α
2
T
ρ
0
1
:
(
λ
1
⊗
1
+
2
μ
I
)
:
1
=
α
ρ
0
tr
S
+
α
2
T
ρ
0
1
:
(
λ
tr
1
1
+
2
μ
1
)
=
α
ρ
0
tr
S
+
α
2
T
ρ
0
(
3
λ
tr
1
+
2
μ
tr
1
)
=
α
ρ
0
tr
S
+
3
α
2
T
ρ
0
(
3
λ
+
2
μ
)
=
α
tr
S
ρ
0
+
9
α
2
K
T
ρ
0
.
{\displaystyle {\begin{aligned}C_{p}-C_{v}&={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:(\alpha ~{\boldsymbol {\mathit {1}}})+{\cfrac {T}{\rho _{0}}}~(\alpha ~{\boldsymbol {\mathit {1}}}):(\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2\mu ~{\boldsymbol {\mathsf {I}}}):(\alpha ~{\boldsymbol {\mathit {1}}})\\&={\cfrac {\alpha }{\rho _{0}}}~{\text{tr}}{\boldsymbol {S}}+{\cfrac {\alpha ^{2}~T}{\rho _{0}}}~{\boldsymbol {\mathit {1}}}:(\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2\mu ~{\boldsymbol {\mathsf {I}}}):{\boldsymbol {\mathit {1}}}\\&={\cfrac {\alpha }{\rho _{0}}}~{\text{tr}}{\boldsymbol {S}}+{\cfrac {\alpha ^{2}~T}{\rho _{0}}}~{\boldsymbol {\mathit {1}}}:(\lambda ~{\text{tr}}{\boldsymbol {\mathit {1}}}~{\boldsymbol {\mathit {1}}}+2\mu ~{\boldsymbol {\mathit {1}}})\\&={\cfrac {\alpha }{\rho _{0}}}~{\text{tr}}{\boldsymbol {S}}+{\cfrac {\alpha ^{2}~T}{\rho _{0}}}~(3~\lambda ~{\text{tr}}{\boldsymbol {\mathit {1}}}+2\mu ~{\text{tr}}{\boldsymbol {\mathit {1}}})\\&={\cfrac {\alpha }{\rho _{0}}}~{\text{tr}}{\boldsymbol {S}}+{\cfrac {3~\alpha ^{2}~T}{\rho _{0}}}~(3~\lambda +2\mu )\\&={\cfrac {\alpha ~{\text{tr}}{\boldsymbol {S}}}{\rho _{0}}}+{\cfrac {9~\alpha ^{2}~K~T}{\rho _{0}}}~.\end{aligned}}}
Therefore,
C
p
−
C
v
=
1
ρ
0
[
α
tr
S
+
9
α
2
K
T
]
.
{\displaystyle {C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}\left[\alpha ~{\text{tr}}{\boldsymbol {S}}+9~\alpha ^{2}~K~T\right]~.}}
![{\displaystyle {\frac {\partial g}{\partial T}}={\frac {\partial {\boldsymbol {S}}}{\partial T}}=\rho _{0}~\left[{\frac {\partial {\boldsymbol {f}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\frac {\partial {\boldsymbol {f}}}{\partial T}}~{\frac {\partial T}{\partial T}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0321b722f0347cfe76a9c572b9ec092787b675e1)
![{\displaystyle C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}\left[\alpha ~{\text{tr}}{\boldsymbol {S}}+9~\alpha ^{2}~K~T\right]~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/93472149c43f1aeedd38d361ea6462c4a65dc22f)
![{\displaystyle {C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}\left[\alpha ~{\text{tr}}{\boldsymbol {S}}+9~\alpha ^{2}~K~T\right]~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e45735dd6a108445fbdce226d8a2dafabd0d4ac7)