A set of constitutive equations is required to close to system of balance laws. These are relations between appropriate kinematic quantities and stress measures that can be assigned a physical meaning.
In thermoelasticity we assume that the fundamental kinematic quantity is the deformation gradient (
F
{\displaystyle {\boldsymbol {F}}}
) which is given by
F
=
∂
x
∂
X
=
∇
∘
x
;
det
F
>
0
.
{\displaystyle {\boldsymbol {F}}={\frac {\partial \mathbf {x} }{\partial \mathbf {X} }}={\boldsymbol {\nabla }}_{\circ }\mathbf {x} ~;~~\det {\boldsymbol {F}}>0~.}
A thermoelastic material is one in which the internal energy (
e
{\displaystyle e}
) is a function only of
F
{\displaystyle {\boldsymbol {F}}}
and the specific entropy (
η
{\displaystyle \eta }
), that is
e
=
e
¯
(
F
,
η
)
.
{\displaystyle e={\bar {e}}({\boldsymbol {F}},\eta )~.}
For a thermoelastic material, we can show that the entropy inequality can be written as
ρ
(
∂
e
¯
∂
η
−
T
)
η
˙
+
(
ρ
∂
e
¯
∂
F
−
σ
⋅
F
−
T
)
:
F
˙
+
q
⋅
∇
T
T
≤
0
.
{\displaystyle {\rho ~\left({\frac {\partial {\bar {e}}}{\partial \eta }}-T\right)~{\dot {\eta }}+\left(\rho ~{\frac {\partial {\bar {e}}}{\partial {\boldsymbol {F}}}}-{\boldsymbol {\sigma }}\cdot {\boldsymbol {F}}^{-T}\right):{\dot {\boldsymbol {F}}}+{\cfrac {\mathbf {q} \cdot {\boldsymbol {\nabla }}T}{T}}\leq 0~.}}
At this stage, we make the following constitutive assumptions:
1) Like the internal energy, we assume that
σ
{\displaystyle {\boldsymbol {\sigma }}}
and
T
{\displaystyle T}
are also functions only of
F
{\displaystyle {\boldsymbol {F}}}
and
η
{\displaystyle \eta }
, i.e.,
σ
=
σ
(
F
,
η
)
;
T
=
T
(
F
,
η
)
.
{\displaystyle {\boldsymbol {\sigma }}={\boldsymbol {\sigma }}({\boldsymbol {F}},\eta )~;~~T=T({\boldsymbol {F}},\eta )~.}
2) The heat flux
q
{\displaystyle \mathbf {q} }
satisfies the thermal conductivity inequality and if
q
{\displaystyle \mathbf {q} }
is independent of
η
˙
{\displaystyle {\dot {\eta }}}
and
F
˙
{\displaystyle {\dot {\boldsymbol {F}}}}
, we have
q
⋅
∇
T
≤
0
⟹
−
(
κ
⋅
∇
T
)
⋅
∇
T
≤
0
⟹
κ
≥
0
{\displaystyle \mathbf {q} \cdot {\boldsymbol {\nabla }}T\leq 0\qquad \implies \qquad -({\boldsymbol {\kappa }}\cdot {\boldsymbol {\nabla }}T)\cdot {\boldsymbol {\nabla }}T\leq 0\qquad \implies \qquad {\boldsymbol {\kappa }}\geq \mathbf {0} }
i.e., the thermal conductivity
κ
{\displaystyle {\boldsymbol {\kappa }}}
is positive semidefinite.
Therefore, the entropy inequality may be written as
ρ
(
∂
e
¯
∂
η
−
T
)
η
˙
+
(
ρ
∂
e
¯
∂
F
−
σ
⋅
F
−
T
)
:
F
˙
≤
0
.
{\displaystyle \rho ~\left({\frac {\partial {\bar {e}}}{\partial \eta }}-T\right)~{\dot {\eta }}+\left(\rho ~{\frac {\partial {\bar {e}}}{\partial {\boldsymbol {F}}}}-{\boldsymbol {\sigma }}\cdot {\boldsymbol {F}}^{-T}\right):{\dot {\boldsymbol {F}}}\leq 0~.}
Since
η
˙
{\displaystyle {\dot {\eta }}}
and
F
˙
{\displaystyle {\dot {\boldsymbol {F}}}}
are arbitrary, the entropy inequality will be satisfied if and only if
∂
e
¯
∂
η
−
T
=
0
⟹
T
=
∂
e
¯
∂
η
and
ρ
∂
e
¯
∂
F
−
σ
⋅
F
−
T
=
0
⟹
σ
=
ρ
∂
e
¯
∂
F
⋅
F
T
.
{\displaystyle {\frac {\partial {\bar {e}}}{\partial \eta }}-T=0\implies T={\frac {\partial {\bar {e}}}{\partial \eta }}\qquad {\text{and}}\qquad \rho ~{\frac {\partial {\bar {e}}}{\partial {\boldsymbol {F}}}}-{\boldsymbol {\sigma }}\cdot {\boldsymbol {F}}^{-T}=\mathbf {0} \implies {\boldsymbol {\sigma }}=\rho ~{\frac {\partial {\bar {e}}}{\partial {\boldsymbol {F}}}}\cdot {\boldsymbol {F}}^{T}~.}
Therefore,
T
=
∂
e
¯
∂
η
and
σ
=
ρ
∂
e
¯
∂
F
⋅
F
T
.
{\displaystyle {T={\frac {\partial {\bar {e}}}{\partial \eta }}}\qquad {\text{and}}\qquad {{\boldsymbol {\sigma }}=\rho ~{\frac {\partial {\bar {e}}}{\partial {\boldsymbol {F}}}}\cdot {\boldsymbol {F}}^{T}~.}}
Given the above relations, the energy equation may expressed in terms of the specific entropy as
ρ
T
η
˙
=
−
∇
⋅
q
+
ρ
s
.
{\displaystyle {\rho ~T~{\dot {\eta }}=-{\boldsymbol {\nabla }}\cdot \mathbf {q} +\rho ~s~.}}
Effect of a rigid body rotation of the internal energy
edit
If a thermoelastic body is subjected to a rigid body rotation
Q
{\displaystyle {\boldsymbol {Q}}}
, then its internal energy should not change. After a rotation, the new deformation gradient (
F
^
{\displaystyle {\hat {\boldsymbol {F}}}}
) is given by
F
^
=
Q
⋅
F
.
{\displaystyle {\hat {\boldsymbol {F}}}={\boldsymbol {Q}}\cdot {\boldsymbol {F}}~.}
Since the internal energy does not change, we must have
e
=
e
¯
(
F
^
,
η
)
=
e
¯
(
F
,
η
)
.
{\displaystyle e={\bar {e}}({\hat {\boldsymbol {F}}},\eta )={\bar {e}}({\boldsymbol {F}},\eta )~.}
Now, from the polar decomposition theorem,
F
=
R
⋅
U
{\displaystyle {\boldsymbol {F}}={\boldsymbol {R}}\cdot {\boldsymbol {U}}}
where
R
{\displaystyle {\boldsymbol {R}}}
is the orthogonal rotation tensor (i.e.,
R
⋅
R
T
=
R
T
⋅
R
=
1
{\displaystyle {\boldsymbol {R}}\cdot {\boldsymbol {R}}^{T}={\boldsymbol {R}}^{T}\cdot {\boldsymbol {R}}={\boldsymbol {\mathit {1}}}}
) and
U
{\displaystyle {\boldsymbol {U}}}
is the symmetric right stretch tensor. Therefore,
e
¯
(
Q
⋅
R
⋅
U
,
η
)
=
e
¯
(
F
,
η
)
.
{\displaystyle {\bar {e}}({\boldsymbol {Q}}\cdot {\boldsymbol {R}}\cdot {\boldsymbol {U}},\eta )={\bar {e}}({\boldsymbol {F}},\eta )~.}
We can choose any rotation
Q
{\displaystyle {\boldsymbol {Q}}}
. In particular, if we choose
Q
=
R
T
{\displaystyle {\boldsymbol {Q}}={\boldsymbol {R}}^{T}}
, we have
e
¯
(
R
T
⋅
R
⋅
U
,
η
)
=
e
¯
(
1
⋅
U
,
η
)
=
e
~
(
U
,
η
)
.
{\displaystyle {\bar {e}}({\boldsymbol {R}}^{T}\cdot {\boldsymbol {R}}\cdot {\boldsymbol {U}},\eta )={\bar {e}}({\boldsymbol {\mathit {1}}}\cdot {\boldsymbol {U}},\eta )={\tilde {e}}({\boldsymbol {U}},\eta )~.}
Therefore,
e
¯
(
U
,
η
)
=
e
¯
(
F
,
η
)
.
{\displaystyle {\bar {e}}({\boldsymbol {U}},\eta )={\bar {e}}({\boldsymbol {F}},\eta )~.}
This means that the internal energy depends only on the stretch
U
{\displaystyle {\boldsymbol {U}}}
and not on the orientation of the body .
Other strain and stress measures
edit
The internal energy depends on
F
{\displaystyle {\boldsymbol {F}}}
only through the stretch
U
{\displaystyle {\boldsymbol {U}}}
. A strain measure that reflects this fact and also vanishes in the reference configuration is the Green strain
E
=
1
2
(
F
T
⋅
F
−
1
)
=
1
2
(
U
2
−
1
)
.
{\displaystyle {{\boldsymbol {E}}={\frac {1}{2}}({\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}}-{\boldsymbol {\mathit {1}}})={\frac {1}{2}}({\boldsymbol {U}}^{2}-{\boldsymbol {\mathit {1}}})~.}}
Recall that the Cauchy stress is given by
σ
=
ρ
∂
e
¯
∂
F
⋅
F
T
.
{\displaystyle {\boldsymbol {\sigma }}=\rho ~{\frac {\partial {\bar {e}}}{\partial {\boldsymbol {F}}}}\cdot {\boldsymbol {F}}^{T}~.}
We can show that the Cauchy stress can be expressed in terms of the Green strain as
σ
=
ρ
F
⋅
∂
e
¯
∂
E
⋅
F
T
.
{\displaystyle {{\boldsymbol {\sigma }}=\rho ~{\boldsymbol {F}}\cdot {\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}\cdot {\boldsymbol {F}}^{T}~.}}
Also, recall that the first Piola-Kirchhoff stress tensor is defined as
P
=
J
(
σ
⋅
F
−
T
)
where
J
=
det
F
{\displaystyle {\boldsymbol {P}}=J~({\boldsymbol {\sigma }}\cdot {\boldsymbol {F}}^{-T})~{\text{where}}~J=\det {\boldsymbol {F}}}
Alternatively, we may use the nominal stress tensor
N
=
J
(
F
−
1
⋅
σ
)
{\displaystyle {\boldsymbol {N}}=J~({\boldsymbol {F}}^{-1}\cdot {\boldsymbol {\sigma }})}
From the conservation of mass, we have
ρ
0
=
ρ
det
F
{\displaystyle \rho _{0}=\rho ~\det {\boldsymbol {F}}}
. Hence,
P
=
ρ
0
ρ
σ
⋅
F
−
T
and
N
=
ρ
0
ρ
F
−
1
⋅
σ
{\displaystyle {{\boldsymbol {P}}={\cfrac {\rho _{0}}{\rho }}~{\boldsymbol {\sigma }}\cdot {\boldsymbol {F}}^{-T}~~{\text{and}}~~{\boldsymbol {N}}={\cfrac {\rho _{0}}{\rho }}~{\boldsymbol {F}}^{-1}\cdot {\boldsymbol {\sigma }}}}
The first P-K stress and the nominal stress are unsymmetric. Also recall that we can define a symmetric stress measure with respect to the reference configuration called the second Piola-Kirchhoff stress tensor (
S
{\displaystyle {\boldsymbol {S}}}
):
S
:=
F
−
1
⋅
P
=
N
⋅
F
−
T
=
ρ
0
ρ
F
−
1
⋅
σ
⋅
F
−
T
.
{\displaystyle {{\boldsymbol {S}}:={\boldsymbol {F}}^{-1}\cdot {\boldsymbol {P}}={\boldsymbol {N}}\cdot {\boldsymbol {F}}^{-T}={\cfrac {\rho _{0}}{\rho }}~{\boldsymbol {F}}^{-1}\cdot {\boldsymbol {\sigma }}\cdot {\boldsymbol {F}}^{-T}~.}}
In terms of the derivatives of the internal energy, we have
S
=
ρ
0
ρ
F
−
1
⋅
(
ρ
F
⋅
∂
e
¯
∂
E
⋅
F
T
)
⋅
F
−
T
=
ρ
0
∂
e
¯
∂
E
{\displaystyle {\boldsymbol {S}}={\cfrac {\rho _{0}}{\rho }}~{\boldsymbol {F}}^{-1}\cdot \left(\rho ~{\boldsymbol {F}}\cdot {\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}\cdot {\boldsymbol {F}}^{T}\right)\cdot {\boldsymbol {F}}^{-T}=\rho _{0}~{\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}}
Therefore,
P
=
ρ
0
F
⋅
∂
e
¯
∂
E
.
{\displaystyle {\boldsymbol {P}}=\rho _{0}~{\boldsymbol {F}}\cdot {\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}~.}
and
N
=
ρ
0
∂
e
¯
∂
E
⋅
F
T
.
{\displaystyle {\boldsymbol {N}}=\rho _{0}~{\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}\cdot {\boldsymbol {F}}^{T}~.}
That is,
S
=
ρ
0
∂
e
¯
∂
E
;
P
=
ρ
0
F
⋅
∂
e
¯
∂
E
;
N
=
ρ
0
∂
e
¯
∂
E
⋅
F
T
{\displaystyle {{\boldsymbol {S}}=\rho _{0}~{\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}}~;~~{{\boldsymbol {P}}=\rho _{0}~{\boldsymbol {F}}\cdot {\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}}~;~~{\boldsymbol {N}}=\rho _{0}~{\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}\cdot {\boldsymbol {F}}^{T}}
The stress power per unit volume is given by
σ
:
∇
v
{\displaystyle {\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} }
. In terms of the stress measures in the reference configuration, we have
σ
:
∇
v
=
(
ρ
F
⋅
∂
e
¯
∂
E
⋅
F
T
)
:
(
F
˙
⋅
F
−
1
)
.
{\displaystyle {\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} =\left(\rho ~{\boldsymbol {F}}\cdot {\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}\cdot {\boldsymbol {F}}^{T}\right):({\dot {\boldsymbol {F}}}\cdot {\boldsymbol {F}}^{-1})~.}
Using the identity
A
:
(
B
⋅
C
)
=
(
A
⋅
C
T
)
:
B
{\displaystyle {\boldsymbol {A}}:({\boldsymbol {B}}\cdot {\boldsymbol {C}})=({\boldsymbol {A}}\cdot {\boldsymbol {C}}^{T}):{\boldsymbol {B}}}
, we have
σ
:
∇
v
=
[
(
ρ
F
⋅
∂
e
¯
∂
E
⋅
F
T
)
⋅
F
−
T
]
:
F
˙
=
ρ
(
F
⋅
∂
e
¯
∂
E
)
:
F
˙
=
ρ
ρ
0
P
:
F
˙
=
ρ
ρ
0
N
T
:
F
˙
.
{\displaystyle {\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} =\left[\left(\rho ~{\boldsymbol {F}}\cdot {\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}\cdot {\boldsymbol {F}}^{T}\right)\cdot {\boldsymbol {F}}^{-T}\right]:{\dot {\boldsymbol {F}}}=\rho ~\left({\boldsymbol {F}}\cdot {\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}\right):{\dot {\boldsymbol {F}}}={\cfrac {\rho }{\rho _{0}}}~{\boldsymbol {P}}:{\dot {\boldsymbol {F}}}={\cfrac {\rho }{\rho _{0}}}~{\boldsymbol {N}}^{T}:{\dot {\boldsymbol {F}}}~.}
We can alternatively express the stress power in terms of
S
{\displaystyle {\boldsymbol {S}}}
and
E
˙
{\displaystyle {\dot {\boldsymbol {E}}}}
. Taking the material time derivative of
E
{\displaystyle {\boldsymbol {E}}}
we have
E
˙
=
1
2
(
F
T
˙
⋅
F
+
F
T
⋅
F
˙
)
.
{\displaystyle {\dot {\boldsymbol {E}}}={\frac {1}{2}}({\dot {{\boldsymbol {F}}^{T}}}\cdot {\boldsymbol {F}}+{\boldsymbol {F}}^{T}\cdot {\dot {\boldsymbol {F}}})~.}
Therefore,
S
:
E
˙
=
1
2
[
S
:
(
F
T
˙
⋅
F
)
+
S
:
(
F
T
⋅
F
˙
)
]
.
{\displaystyle {\boldsymbol {S}}:{\dot {\boldsymbol {E}}}={\frac {1}{2}}[{\boldsymbol {S}}:({\dot {{\boldsymbol {F}}^{T}}}\cdot {\boldsymbol {F}})+{\boldsymbol {S}}:({\boldsymbol {F}}^{T}\cdot {\dot {\boldsymbol {F}}})]~.}
Using the identities
A
:
(
B
⋅
C
)
=
(
A
⋅
C
T
)
:
B
=
(
B
T
⋅
A
)
:
C
{\displaystyle {\boldsymbol {A}}:({\boldsymbol {B}}\cdot {\boldsymbol {C}})=({\boldsymbol {A}}\cdot {\boldsymbol {C}}^{T}):{\boldsymbol {B}}=({\boldsymbol {B}}^{T}\cdot {\boldsymbol {A}}):{\boldsymbol {C}}}
and
A
:
B
=
A
T
:
B
T
{\displaystyle {\boldsymbol {A}}:{\boldsymbol {B}}={\boldsymbol {A}}^{T}:{\boldsymbol {B}}^{T}}
and using the
symmetry of
S
{\displaystyle {\boldsymbol {S}}}
, we have
S
:
E
˙
=
1
2
[
(
S
⋅
F
T
)
:
F
˙
T
+
(
F
⋅
S
)
:
F
˙
]
=
1
2
[
(
F
⋅
S
T
)
:
F
˙
+
(
F
⋅
S
)
:
F
˙
]
=
(
F
⋅
S
)
:
F
˙
.
{\displaystyle {\boldsymbol {S}}:{\dot {\boldsymbol {E}}}={\frac {1}{2}}[({\boldsymbol {S}}\cdot {\boldsymbol {F}}^{T}):{\dot {\boldsymbol {F}}}^{T}+({\boldsymbol {F}}\cdot {\boldsymbol {S}}):{\dot {\boldsymbol {F}}}]={\frac {1}{2}}[({\boldsymbol {F}}\cdot {\boldsymbol {S}}^{T}):{\dot {\boldsymbol {F}}}+({\boldsymbol {F}}\cdot {\boldsymbol {S}}):{\dot {\boldsymbol {F}}}]=({\boldsymbol {F}}\cdot {\boldsymbol {S}}):{\dot {\boldsymbol {F}}}~.}
Now,
S
=
F
−
1
⋅
P
{\displaystyle {\boldsymbol {S}}={\boldsymbol {F}}^{-1}\cdot {\boldsymbol {P}}}
. Therefore,
S
:
E
˙
=
N
T
:
F
˙
{\displaystyle {\boldsymbol {S}}:{\dot {\boldsymbol {E}}}={\boldsymbol {N}}^{T}:{\dot {\boldsymbol {F}}}}
.
Hence, the stress power can be expressed as
ρ
0
ρ
σ
:
∇
v
=
P
:
F
˙
=
N
T
:
F
˙
=
S
:
E
˙
.
{\displaystyle {{\frac {\rho _{0}}{\rho }}~{\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} ={\boldsymbol {P}}:{\dot {\boldsymbol {F}}}={\boldsymbol {N^{T}}}:{\dot {\boldsymbol {F}}}={\boldsymbol {S}}:{\dot {\boldsymbol {E}}}~.}}
If we split the velocity gradient into symmetric and skew parts using
∇
v
=
l
=
d
+
w
{\displaystyle {\boldsymbol {\nabla }}\mathbf {v} ={\boldsymbol {l}}=\mathbf {d} +{\boldsymbol {w}}}
where
d
{\displaystyle \mathbf {d} }
is the rate of deformation tensor and
w
{\displaystyle {\boldsymbol {w}}}
is the spin tensor,
we have
σ
:
∇
v
=
σ
:
d
+
σ
:
w
=
tr
(
σ
T
⋅
d
)
+
tr
(
σ
T
⋅
w
)
=
tr
(
σ
⋅
d
)
+
tr
(
σ
⋅
w
)
.
{\displaystyle {\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} ={\boldsymbol {\sigma }}:\mathbf {d} +{\boldsymbol {\sigma }}:{\boldsymbol {w}}={\text{tr}}~({\boldsymbol {\sigma }}^{T}\cdot \mathbf {d} )+{\text{tr}}~({\boldsymbol {\sigma }}^{T}\cdot {\boldsymbol {w}})={\text{tr}}~({\boldsymbol {\sigma }}\cdot \mathbf {d} )+{\text{tr}}~({\boldsymbol {\sigma }}\cdot {\boldsymbol {w}})~.}
Since
σ
{\displaystyle {\boldsymbol {\sigma }}}
is symmetric and
w
{\displaystyle {\boldsymbol {w}}}
is skew, we have
tr
(
σ
⋅
w
)
=
0
{\displaystyle {\text{tr}}~({\boldsymbol {\sigma }}\cdot {\boldsymbol {w}})=0}
.
Therefore,
σ
:
∇
v
=
tr
(
σ
⋅
d
)
{\displaystyle {\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} ={\text{tr}}~({\boldsymbol {\sigma }}\cdot \mathbf {d} )}
. Hence, we may also
express the stress power as
ρ
0
ρ
tr
(
σ
⋅
d
)
=
tr
(
P
T
⋅
F
˙
)
=
tr
(
N
⋅
F
˙
)
=
tr
(
S
⋅
E
˙
)
.
{\displaystyle {{\frac {\rho _{0}}{\rho }}~{\text{tr}}~({\boldsymbol {\sigma }}\cdot \mathbf {d} )={\text{tr}}~({\boldsymbol {P}}^{T}\cdot {\dot {\boldsymbol {F}}})={\text{tr}}~({\boldsymbol {N}}\cdot {\dot {\boldsymbol {F}}})={\text{tr}}~({\boldsymbol {S}}\cdot {\dot {\boldsymbol {E}}})~.}}
Helmholtz and Gibbs free energy
edit
Recall that
S
=
ρ
0
∂
e
¯
∂
E
.
{\displaystyle {\boldsymbol {S}}=\rho _{0}~{\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}~.}
Therefore,
∂
e
¯
∂
E
=
1
ρ
0
S
.
{\displaystyle {\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}~.}
Also recall that
∂
e
¯
∂
η
=
T
.
{\displaystyle {\frac {\partial {\bar {e}}}{\partial \eta }}=T~.}
Now, the internal energy
e
=
e
¯
(
E
,
η
)
{\displaystyle e={\bar {e}}({\boldsymbol {E}},\eta )}
is a function only of the Green strain and the specific entropy. Let us assume, that the above relations can be uniquely inverted locally at a material point so that we have
E
=
E
~
(
S
,
T
)
and
η
=
η
~
(
S
,
T
)
.
{\displaystyle {\boldsymbol {E}}={\tilde {\boldsymbol {E}}}({\boldsymbol {S}},T)\qquad {\text{and}}\qquad \eta ={\tilde {\eta }}({\boldsymbol {S}},T)~.}
Then the specific internal energy, the specific entropy, and the stress can also be expressed as functions of
S
{\displaystyle {\boldsymbol {S}}}
and
T
{\displaystyle T}
, or
E
{\displaystyle {\boldsymbol {E}}}
and
T
{\displaystyle T}
, i.e.,
e
=
e
¯
(
E
,
η
)
=
e
~
(
S
,
T
)
=
e
^
(
E
,
T
)
;
η
=
η
~
(
S
,
T
)
=
η
^
(
E
,
T
)
;
and
S
=
S
^
(
E
,
T
)
{\displaystyle e={\bar {e}}({\boldsymbol {E}},\eta )={\tilde {e}}({\boldsymbol {S}},T)={\hat {e}}({\boldsymbol {E}},T)~;\qquad \eta ={\tilde {\eta }}({\boldsymbol {S}},T)={\hat {\eta }}({\boldsymbol {E}},T)~;\qquad {\text{and}}\qquad {\boldsymbol {S}}={\hat {\boldsymbol {S}}}({\boldsymbol {E}},T)}
We can show that
d
d
t
(
e
−
T
η
)
=
−
T
˙
η
+
1
ρ
0
S
:
E
˙
or
d
ψ
d
t
=
−
T
˙
η
+
1
ρ
0
S
:
E
˙
.
{\displaystyle {\cfrac {d}{dt}}(e-T~\eta )=-{\dot {T}}~\eta +{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\dot {\boldsymbol {E}}}\qquad {\text{or}}\qquad {\cfrac {d\psi }{dt}}=-{\dot {T}}~\eta +{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\dot {\boldsymbol {E}}}~.}
and
d
d
t
(
e
−
T
η
−
1
ρ
0
S
:
E
)
=
−
T
˙
η
−
1
ρ
0
S
˙
:
E
or
d
g
d
t
=
T
˙
η
+
1
ρ
0
S
˙
:
E
.
{\displaystyle {\cfrac {d}{dt}}(e-T~\eta -{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\boldsymbol {E}})=-{\dot {T}}~\eta -{\cfrac {1}{\rho _{0}}}~{\dot {\boldsymbol {S}}}:{\boldsymbol {E}}\qquad {\text{or}}\qquad {\cfrac {dg}{dt}}={\dot {T}}~\eta +{\cfrac {1}{\rho _{0}}}~{\dot {\boldsymbol {S}}}:{\boldsymbol {E}}~.}
We define the Helmholtz free energy as
ψ
=
ψ
^
(
E
,
T
)
:=
e
−
T
η
.
{\displaystyle {\psi ={\hat {\psi }}({\boldsymbol {E}},T):=e-T~\eta ~.}}
We define the Gibbs free energy as
g
=
g
~
(
S
,
T
)
:=
−
e
+
T
η
+
1
ρ
0
S
:
E
.
{\displaystyle {g={\tilde {g}}({\boldsymbol {S}},T):=-e+T~\eta +{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\boldsymbol {E}}~.}}
The functions
ψ
^
(
E
,
T
)
{\displaystyle {\hat {\psi }}({\boldsymbol {E}},T)}
and
g
~
(
S
,
T
)
{\displaystyle {\tilde {g}}({\boldsymbol {S}},T)}
are unique. Using
these definitions it can be shown that
∂
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 {\hat {\psi }}}{\partial {\boldsymbol {E}}}}={\cfrac {1}{\rho _{0}}}~{\hat {\boldsymbol {S}}}({\boldsymbol {E}},T)~;~~{\frac {\partial {\hat {\psi }}}{\partial T}}=-{\hat {\eta }}({\boldsymbol {E}},T)~;~~{\frac {\partial {\tilde {g}}}{\partial {\boldsymbol {S}}}}={\cfrac {1}{\rho _{0}}}~{\tilde {\boldsymbol {E}}}({\boldsymbol {S}},T)~;~~{\frac {\partial {\tilde {g}}}{\partial T}}={\tilde {\eta }}({\boldsymbol {S}},T)}
and
∂
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}}}}~.}
The specific heat at constant strain (or constant volume) is defined as
C
v
:=
∂
e
^
(
E
,
T
)
∂
T
.
{\displaystyle {C_{v}:={\frac {\partial {\hat {e}}({\boldsymbol {E}},T)}{\partial T}}~.}}
The specific heat at constant stress (or constant pressure) is
defined as
C
p
:=
∂
e
~
(
S
,
T
)
∂
T
.
{\displaystyle {C_{p}:={\frac {\partial {\tilde {e}}({\boldsymbol {S}},T)}{\partial T}}~.}}
We can show that
C
v
=
T
∂
η
^
∂
T
=
−
T
∂
2
ψ
^
∂
T
2
{\displaystyle C_{v}=T~{\frac {\partial {\hat {\eta }}}{\partial T}}=-T~{\frac {\partial ^{2}{\hat {\psi }}}{\partial T^{2}}}}
and
C
p
=
T
∂
η
~
∂
T
+
1
ρ
0
S
:
∂
E
~
∂
T
=
T
∂
2
g
~
∂
T
2
+
S
:
∂
2
g
~
∂
S
∂
T
.
{\displaystyle C_{p}=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}}~.}
Also the equation for the balance of energy can be expressed in terms
of the specific heats as
ρ
C
v
T
˙
=
∇
⋅
(
κ
⋅
∇
T
)
+
ρ
s
+
ρ
ρ
0
T
β
S
:
E
˙
ρ
(
C
p
−
1
ρ
0
S
:
α
E
)
T
˙
=
∇
⋅
(
κ
⋅
∇
T
)
+
ρ
s
−
ρ
ρ
0
T
α
E
:
S
˙
{\displaystyle {\begin{aligned}\rho ~C_{v}~{\dot {T}}&={\boldsymbol {\nabla }}\cdot ({\boldsymbol {\kappa }}\cdot {\boldsymbol {\nabla T)}}+\rho ~s+{\cfrac {\rho }{\rho _{0}}}~T~{\boldsymbol {\beta }}_{S}:{\dot {\boldsymbol {E}}}\\\rho ~\left(C_{p}-{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\boldsymbol {\alpha }}_{E}\right)~{\dot {T}}&={\boldsymbol {\nabla }}\cdot ({\boldsymbol {\kappa }}\cdot {\boldsymbol {\nabla T)}}+\rho ~s-{\cfrac {\rho }{\rho _{0}}}~T~{\boldsymbol {\alpha }}_{E}:{\dot {\boldsymbol {S}}}\end{aligned}}}
where
β
S
:=
∂
S
^
∂
T
and
α
E
:=
∂
E
~
∂
T
.
{\displaystyle {\boldsymbol {\beta }}_{S}:={\frac {\partial {\hat {\boldsymbol {S}}}}{\partial T}}\qquad {\text{and}}\qquad {\boldsymbol {\alpha }}_{E}:={\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}~.}
The quantity
β
S
{\displaystyle {\boldsymbol {\beta }}_{S}}
is called the coefficient of thermal stress and the quantity
α
E
{\displaystyle {\boldsymbol {\alpha }}_{E}}
is called the coefficient of thermal expansion .
The difference between
C
p
{\displaystyle C_{p}}
and
C
v
{\displaystyle C_{v}}
can be expressed as
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}}~.}
However, it is more common to express the above relation in terms of the elastic modulus tensor 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 the fourth-order tensor of elastic moduli is defined as
C
:=
∂
S
^
∂
E
~
=
ρ
0
∂
2
ψ
^
∂
E
~
∂
E
~
.
{\displaystyle {\boldsymbol {\mathsf {C}}}:={\frac {\partial {\hat {\boldsymbol {S}}}}{\partial {\tilde {\boldsymbol {E}}}}}=\rho _{0}~{\frac {\partial ^{2}{\hat {\psi }}}{\partial {\tilde {\boldsymbol {E}}}\partial {\tilde {\boldsymbol {E}}}}}~.}
For isotropic materials with a constant coefficient of thermal expansion that follow the St. Venant-Kirchhoff material model, we can show that
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]~.}
T. W. Wright. The Physics and Mathematics of Adiabatic Shear Bands . Cambridge University Press, Cambridge, UK, 2002.
R. C. Batra. Elements of Continuum Mechanics . AIAA, Reston, VA., 2006.
G. A. Maugin. The Thermomechanics of Nonlinear Irreversible Behaviors: An Introduction . World Scientific, Singapore, 1999.