For thermoelastic materials, show that the following relations hold:
∂
ψ
∂
E
=
1
ρ
0
S
^
(
E
,
T
)
;
∂
ψ
∂
T
=
−
η
^
(
E
,
T
)
;
∂
g
∂
S
=
1
ρ
0
E
~
(
S
,
T
)
;
∂
g
∂
T
=
η
~
(
S
,
T
)
{\displaystyle {\frac {\partial \psi }{\partial {\boldsymbol {E}}}}={\cfrac {1}{\rho _{0}}}~{\hat {\boldsymbol {S}}}({\boldsymbol {E}},T)~;~~{\frac {\partial \psi }{\partial T}}=-{\hat {\eta }}({\boldsymbol {E}},T)~;~~{\frac {\partial g}{\partial {\boldsymbol {S}}}}={\cfrac {1}{\rho _{0}}}~{\tilde {\boldsymbol {E}}}({\boldsymbol {S}},T)~;~~{\frac {\partial g}{\partial T}}={\tilde {\eta }}({\boldsymbol {S}},T)}
where
ψ
(
E
,
T
)
{\displaystyle \psi ({\boldsymbol {E}},T)}
is the Helmholtz free energy and
g
(
S
,
T
)
{\displaystyle g({\boldsymbol {S}},T)}
is the Gibbs free energy.
Also show that
∂
S
^
∂
T
=
−
ρ
0
∂
η
^
∂
E
and
∂
E
~
∂
T
=
ρ
0
∂
η
~
∂
S
.
{\displaystyle {\frac {\partial {\hat {\boldsymbol {S}}}}{\partial T}}=-\rho _{0}~{\frac {\partial {\hat {\eta }}}{\partial {\boldsymbol {E}}}}\qquad {\text{and}}\qquad {\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}=\rho _{0}~{\frac {\partial {\tilde {\eta }}}{\partial {\boldsymbol {S}}}}~.}
Proof:
Recall that
ψ
(
E
,
T
)
=
e
−
T
η
=
e
¯
(
E
,
η
)
−
T
η
.
{\displaystyle \psi ({\boldsymbol {E}},T)=e-T~\eta ={\bar {e}}({\boldsymbol {E}},\eta )-T~\eta ~.}
and
g
(
S
,
T
)
=
−
e
+
T
η
+
1
ρ
0
S
:
E
.
{\displaystyle g({\boldsymbol {S}},T)=-e+T~\eta +{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\boldsymbol {E}}~.}
(Note that we can choose any functional dependence that we like, because
the quantities
e
{\displaystyle e}
,
η
{\displaystyle \eta }
,
E
{\displaystyle {\boldsymbol {E}}}
are the actual quantities and not any
particular functional relations).
The derivatives are
∂
ψ
∂
E
=
∂
e
¯
∂
E
=
1
ρ
0
S
;
∂
ψ
∂
T
=
−
η
.
{\displaystyle {\frac {\partial \psi }{\partial {\boldsymbol {E}}}}={\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}~;\qquad {\frac {\partial \psi }{\partial T}}=-\eta ~.}
and
∂
g
∂
S
=
1
ρ
0
∂
S
∂
S
:
E
=
1
ρ
0
E
;
∂
g
∂
T
=
η
.
{\displaystyle {\frac {\partial g}{\partial {\boldsymbol {S}}}}={\cfrac {1}{\rho _{0}}}~{\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {S}}}}:{\boldsymbol {E}}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {E}}~;\qquad {\frac {\partial g}{\partial T}}=\eta ~.}
Hence,
∂
ψ
∂
E
=
1
ρ
0
S
^
(
E
,
T
)
;
∂
ψ
∂
T
=
−
η
^
(
E
,
T
)
;
∂
g
∂
S
=
1
ρ
0
E
~
(
S
,
T
)
;
∂
g
∂
T
=
η
~
(
S
,
T
)
{\displaystyle {{\frac {\partial \psi }{\partial {\boldsymbol {E}}}}={\cfrac {1}{\rho _{0}}}~{\hat {\boldsymbol {S}}}({\boldsymbol {E}},T)~;~~{\frac {\partial \psi }{\partial T}}=-{\hat {\eta }}({\boldsymbol {E}},T)~;~~{\frac {\partial g}{\partial {\boldsymbol {S}}}}={\cfrac {1}{\rho _{0}}}~{\tilde {\boldsymbol {E}}}({\boldsymbol {S}},T)~;~~{\frac {\partial g}{\partial T}}={\tilde {\eta }}({\boldsymbol {S}},T)}}
From the above, we have
∂
2
ψ
∂
T
∂
E
=
∂
2
ψ
∂
E
∂
T
⟹
−
∂
η
^
∂
E
=
1
ρ
0
∂
S
^
∂
T
.
{\displaystyle {\frac {\partial ^{2}\psi }{\partial T\partial {\boldsymbol {E}}}}={\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial T}}\qquad \implies \qquad -{\frac {\partial {\hat {\eta }}}{\partial {\boldsymbol {E}}}}={\cfrac {1}{\rho _{0}}}{\frac {\partial {\hat {\boldsymbol {S}}}}{\partial T}}~.}
and
∂
2
g
∂
T
∂
S
=
∂
2
g
∂
S
∂
T
⟹
∂
η
~
∂
S
=
1
ρ
0
∂
E
~
∂
T
.
{\displaystyle {\frac {\partial ^{2}g}{\partial T\partial {\boldsymbol {S}}}}={\frac {\partial ^{2}g}{\partial {\boldsymbol {S}}\partial T}}\qquad \implies \qquad {\frac {\partial {\tilde {\eta }}}{\partial {\boldsymbol {S}}}}={\cfrac {1}{\rho _{0}}}{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}~.}
Hence,
∂
S
^
∂
T
=
−
ρ
0
∂
η
^
∂
E
and
∂
E
~
∂
T
=
ρ
0
∂
η
~
∂
S
.
{\displaystyle {{\frac {\partial {\hat {\boldsymbol {S}}}}{\partial T}}=-\rho _{0}~{\frac {\partial {\hat {\eta }}}{\partial {\boldsymbol {E}}}}\qquad {\text{and}}\qquad {\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}=\rho _{0}~{\frac {\partial {\tilde {\eta }}}{\partial {\boldsymbol {S}}}}~.}}
For thermoelastic materials, show that the following relations hold:
∂
e
^
(
E
,
T
)
∂
T
=
T
∂
η
^
∂
T
=
−
T
∂
2
ψ
^
∂
T
2
{\displaystyle {\frac {\partial {\hat {e}}({\boldsymbol {E}},T)}{\partial T}}=T~{\frac {\partial {\hat {\eta }}}{\partial T}}=-T~{\frac {\partial ^{2}{\hat {\psi }}}{\partial T^{2}}}}
and
∂
e
~
(
S
,
T
)
∂
T
=
T
∂
η
~
∂
T
+
1
ρ
0
S
:
∂
E
~
∂
T
=
T
∂
2
g
~
∂
T
2
+
S
:
∂
2
g
~
∂
S
∂
T
.
{\displaystyle {\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}}=T~{\frac {\partial ^{2}{\tilde {g}}}{\partial T^{2}}}+{\boldsymbol {S}}:{\frac {\partial ^{2}{\tilde {g}}}{\partial {\boldsymbol {S}}\partial T}}~.}
Proof:
Recall,
ψ
^
(
E
,
T
)
=
ψ
=
e
−
T
η
=
e
^
(
E
,
T
)
−
T
η
^
(
E
,
T
)
{\displaystyle {\hat {\psi }}({\boldsymbol {E}},T)=\psi =e-T~\eta ={\hat {e}}({\boldsymbol {E}},T)-T~{\hat {\eta }}({\boldsymbol {E}},T)}
and
g
~
(
S
,
T
)
=
g
=
−
e
+
T
η
+
1
ρ
0
S
:
E
=
−
e
~
(
S
,
T
)
+
T
η
~
(
S
,
T
)
+
1
ρ
0
S
:
E
~
(
S
,
T
)
.
{\displaystyle {\tilde {g}}({\boldsymbol {S}},T)=g=-e+T~\eta +{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\boldsymbol {E}}=-{\tilde {e}}({\boldsymbol {S}},T)+T~{\tilde {\eta }}({\boldsymbol {S}},T)+{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\tilde {\boldsymbol {E}}}({\boldsymbol {S}},T)~.}
Therefore,
∂
e
^
(
E
,
T
)
∂
T
=
∂
ψ
^
∂
T
+
η
^
(
E
,
T
)
+
T
∂
η
^
∂
T
{\displaystyle {\frac {\partial {\hat {e}}({\boldsymbol {E}},T)}{\partial T}}={\frac {\partial {\hat {\psi }}}{\partial T}}+{\hat {\eta }}({\boldsymbol {E}},T)+T~{\frac {\partial {\hat {\eta }}}{\partial T}}}
and
∂
e
~
(
S
,
T
)
∂
T
=
−
∂
g
~
∂
T
+
η
~
(
S
,
T
)
+
T
∂
η
~
∂
T
+
1
ρ
0
S
:
∂
E
~
∂
T
.
{\displaystyle {\frac {\partial {\tilde {e}}({\boldsymbol {S}},T)}{\partial T}}=-{\frac {\partial {\tilde {g}}}{\partial T}}+{\tilde {\eta }}({\boldsymbol {S}},T)+T~{\frac {\partial {\tilde {\eta }}}{\partial T}}+{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}~.}
Also, recall that
η
^
(
E
,
T
)
=
−
∂
ψ
^
∂
T
⟹
∂
η
^
∂
T
=
−
∂
2
ψ
^
∂
T
2
,
{\displaystyle {\hat {\eta }}({\boldsymbol {E}},T)=-{\frac {\partial {\hat {\psi }}}{\partial T}}\qquad \implies \qquad {\frac {\partial {\hat {\eta }}}{\partial T}}=-{\frac {\partial ^{2}{\hat {\psi }}}{\partial T^{2}}}~,}
η
~
(
S
,
T
)
=
∂
g
~
∂
T
⟹
∂
η
~
∂
T
=
∂
2
g
~
∂
T
2
,
{\displaystyle {\tilde {\eta }}({\boldsymbol {S}},T)={\frac {\partial {\tilde {g}}}{\partial T}}\qquad \implies \qquad {\frac {\partial {\tilde {\eta }}}{\partial T}}={\frac {\partial ^{2}{\tilde {g}}}{\partial T^{2}}}~,}
and
E
~
(
S
,
T
)
=
ρ
0
∂
g
~
∂
S
⟹
∂
E
~
∂
T
=
ρ
0
∂
2
g
~
∂
S
∂
T
.
{\displaystyle {\tilde {\boldsymbol {E}}}({\boldsymbol {S}},T)=\rho _{0}~{\frac {\partial {\tilde {g}}}{\partial {\boldsymbol {S}}}}\qquad \implies \qquad {\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}=\rho _{0}~{\frac {\partial ^{2}{\tilde {g}}}{\partial {\boldsymbol {S}}\partial T}}~.}
Hence,
∂
e
^
(
E
,
T
)
∂
T
=
T
∂
η
^
∂
T
=
−
T
∂
2
ψ
^
∂
T
2
{\displaystyle {{\frac {\partial {\hat {e}}({\boldsymbol {E}},T)}{\partial T}}=T~{\frac {\partial {\hat {\eta }}}{\partial T}}=-T~{\frac {\partial ^{2}{\hat {\psi }}}{\partial T^{2}}}}}
and
∂
e
~
(
S
,
T
)
∂
T
=
T
∂
η
~
∂
T
+
1
ρ
0
S
:
∂
E
~
∂
T
=
T
∂
2
g
~
∂
T
2
+
S
:
∂
2
g
~
∂
S
∂
T
.
{\displaystyle {{\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}}=T~{\frac {\partial ^{2}{\tilde {g}}}{\partial T^{2}}}+{\boldsymbol {S}}:{\frac {\partial ^{2}{\tilde {g}}}{\partial {\boldsymbol {S}}\partial T}}~.}}