Clausius-Duhem inequality for thermoelasticity
For thermoelastic materials, the internal energy is a function only of the deformation gradient and the temperature, i.e.,
e
=
e
(
F
,
T
)
{\displaystyle e=e({\boldsymbol {F}},T)}
. Show that, for thermoelastic materials, the Clausius-Duhem inequality
ρ
(
e
˙
−
T
η
˙
)
−
σ
:
∇
v
≤
−
q
⋅
∇
T
T
{\displaystyle \rho ~({\dot {e}}-T~{\dot {\eta }})-{\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} \leq -{\cfrac {\mathbf {q} \cdot {\boldsymbol {\nabla }}T}{T}}}
can be expressed as
ρ
(
∂
e
∂
η
−
T
)
η
˙
+
(
ρ
∂
e
∂
F
−
σ
⋅
F
−
T
)
:
F
˙
≤
−
q
⋅
∇
T
T
.
{\displaystyle \rho ~\left({\frac {\partial e}{\partial \eta }}-T\right)~{\dot {\eta }}+\left(\rho ~{\frac {\partial e}{\partial {\boldsymbol {F}}}}-{\boldsymbol {\sigma }}\cdot {\boldsymbol {F}}^{-T}\right):{\dot {\boldsymbol {F}}}\leq -{\cfrac {\mathbf {q} \cdot {\boldsymbol {\nabla }}T}{T}}~.}
Proof:
Since
e
=
e
(
F
,
T
)
{\displaystyle e=e({\boldsymbol {F}},T)}
, we have
e
˙
=
∂
e
∂
F
:
F
˙
+
∂
e
∂
η
η
˙
.
{\displaystyle {\dot {e}}={\frac {\partial e}{\partial {\boldsymbol {F}}}}:{\dot {\boldsymbol {F}}}+{\frac {\partial e}{\partial \eta }}~{\dot {\eta }}~.}
Therefore,
ρ
(
∂
e
∂
F
:
F
˙
+
∂
e
∂
η
η
˙
−
T
η
˙
)
−
σ
:
∇
v
≤
−
q
⋅
∇
T
T
or
ρ
(
∂
e
∂
η
−
T
)
η
˙
+
ρ
∂
e
∂
F
:
F
˙
−
σ
:
∇
v
≤
−
q
⋅
∇
T
T
.
{\displaystyle \rho ~\left({\frac {\partial e}{\partial {\boldsymbol {F}}}}:{\dot {\boldsymbol {F}}}+{\frac {\partial e}{\partial \eta }}~{\dot {\eta }}-T~{\dot {\eta }}\right)-{\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} \leq -{\cfrac {\mathbf {q} \cdot {\boldsymbol {\nabla }}T}{T}}\qquad {\text{or}}\qquad \rho \left({\frac {\partial e}{\partial \eta }}-T\right)~{\dot {\eta }}+\rho ~{\frac {\partial e}{\partial {\boldsymbol {F}}}}:{\dot {\boldsymbol {F}}}-{\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} \leq -{\cfrac {\mathbf {q} \cdot {\boldsymbol {\nabla }}T}{T}}~.}
Now,
∇
v
=
l
=
F
˙
⋅
F
−
1
{\displaystyle {\boldsymbol {\nabla }}\mathbf {v} ={\boldsymbol {l}}={\dot {\boldsymbol {F}}}\cdot {\boldsymbol {F}}^{-1}}
. Therefore, 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
=
σ
:
(
F
˙
⋅
F
−
1
)
=
(
σ
⋅
F
−
T
)
:
F
˙
.
{\displaystyle {\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} ={\boldsymbol {\sigma }}:({\dot {\boldsymbol {F}}}\cdot {\boldsymbol {F}}^{-1})=({\boldsymbol {\sigma }}\cdot {\boldsymbol {F}}^{-T}):{\dot {\boldsymbol {F}}}~.}
Hence,
ρ
(
∂
e
∂
η
−
T
)
η
˙
+
ρ
∂
e
∂
F
:
F
˙
−
(
σ
⋅
F
−
T
)
:
F
˙
≤
−
q
⋅
∇
T
T
{\displaystyle \rho \left({\frac {\partial e}{\partial \eta }}-T\right)~{\dot {\eta }}+\rho ~{\frac {\partial e}{\partial {\boldsymbol {F}}}}:{\dot {\boldsymbol {F}}}-({\boldsymbol {\sigma }}\cdot {\boldsymbol {F}}^{-T}):{\dot {\boldsymbol {F}}}\leq -{\cfrac {\mathbf {q} \cdot {\boldsymbol {\nabla }}T}{T}}}
or,
ρ
(
∂
e
∂
η
−
T
)
η
˙
+
(
ρ
∂
e
∂
F
−
σ
⋅
F
−
T
)
:
F
˙
≤
−
q
⋅
∇
T
T
.
{\displaystyle \rho ~\left({\frac {\partial e}{\partial \eta }}-T\right)~{\dot {\eta }}+\left(\rho ~{\frac {\partial e}{\partial {\boldsymbol {F}}}}-{\boldsymbol {\sigma }}\cdot {\boldsymbol {F}}^{-T}\right):{\dot {\boldsymbol {F}}}\leq -{\cfrac {\mathbf {q} \cdot {\boldsymbol {\nabla }}T}{T}}~.}