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 (
) which is given by
![{\displaystyle {\boldsymbol {F}}={\frac {\partial \mathbf {x} }{\partial \mathbf {X} }}={\boldsymbol {\nabla }}_{\circ }\mathbf {x} ~;~~\det {\boldsymbol {F}}>0~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d8105a1884493bbb655ddda28726dd00a9dd29f4)
A thermoelastic material is one in which the internal energy (
) is a function only of
and the specific entropy (
), that is
![{\displaystyle e={\bar {e}}({\boldsymbol {F}},\eta )~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3b66a695432e2e545d424022338d31579573b8a8)
For a thermoelastic material, we can show that the entropy inequality can be written as
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At this stage, we make the following constitutive assumptions:
1) Like the internal energy, we assume that
and
are also functions only of
and
, i.e.,
![{\displaystyle {\boldsymbol {\sigma }}={\boldsymbol {\sigma }}({\boldsymbol {F}},\eta )~;~~T=T({\boldsymbol {F}},\eta )~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fe5a6670d7976b4d22fde8bec73f3a6c445fa380)
2) The heat flux
satisfies the thermal conductivity inequality and if
is independent of
and
, we have
![{\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} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/41e46f4b4dae41e3bed4161c728cc648c0e5f751)
i.e., the thermal conductivity
is positive semidefinite.
Therefore, the entropy inequality may be written as
![{\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~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f824298d4f4d7c0852bac8e7103c5a398fbd1d02)
Since
and
are arbitrary, the entropy inequality will be satisfied if and only if
![{\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}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/eadca12785a91f0233ffa3eb44003b35e995a6ee)
Therefore,
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Given the above relations, the energy equation may expressed in terms of the specific entropy as
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If a thermoelastic body is subjected to a rigid body rotation
, then its internal energy should not change. After a rotation, the new deformation gradient (
) is given by
![{\displaystyle {\hat {\boldsymbol {F}}}={\boldsymbol {Q}}\cdot {\boldsymbol {F}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8e427e35e42bd4c2251bdd026051dc87dda9f492)
Since the internal energy does not change, we must have
![{\displaystyle e={\bar {e}}({\hat {\boldsymbol {F}}},\eta )={\bar {e}}({\boldsymbol {F}},\eta )~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ab46ace5f4d8c0741faaa2f7682df3e8c0116473)
Now, from the polar decomposition theorem,
where
is the orthogonal rotation tensor (i.e.,
) and
is the symmetric right stretch tensor. Therefore,
![{\displaystyle {\bar {e}}({\boldsymbol {Q}}\cdot {\boldsymbol {R}}\cdot {\boldsymbol {U}},\eta )={\bar {e}}({\boldsymbol {F}},\eta )~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a4e6be24498231b71d08b8b799a929c2fb9357b2)
We can choose any rotation
. In particular, if we choose
, we have
![{\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 )~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5b1955900d3fb320d3d20b05de65fc2a77cdaf08)
Therefore,
![{\displaystyle {\bar {e}}({\boldsymbol {U}},\eta )={\bar {e}}({\boldsymbol {F}},\eta )~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1d56136b995f214eccbf8e38f2f6a1bca1254fe2)
This means that the internal energy depends only on the stretch
and not on the orientation of the body.
The internal energy depends on
only through the stretch
. A strain measure that reflects this fact and also vanishes in the reference configuration is the Green strain
![{\displaystyle {{\boldsymbol {E}}={\frac {1}{2}}({\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}}-{\boldsymbol {\mathit {1}}})={\frac {1}{2}}({\boldsymbol {U}}^{2}-{\boldsymbol {\mathit {1}}})~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/28311fa53d45e11e30ecc6db890b8c74a9043280)
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Recall that the Cauchy stress is given by
![{\displaystyle {\boldsymbol {\sigma }}=\rho ~{\frac {\partial {\bar {e}}}{\partial {\boldsymbol {F}}}}\cdot {\boldsymbol {F}}^{T}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8067238556bd70bd00fe04f056a2ca441c0a0ef0)
We can show that the Cauchy stress can be expressed in terms of the Green strain as
![{\displaystyle {{\boldsymbol {\sigma }}=\rho ~{\boldsymbol {F}}\cdot {\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}\cdot {\boldsymbol {F}}^{T}~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cb782d91f4257010c46cd5b08fdc51b1d66935f0)
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Also, recall that the first Piola-Kirchhoff stress tensor is defined as
![{\displaystyle {\boldsymbol {P}}=J~({\boldsymbol {\sigma }}\cdot {\boldsymbol {F}}^{-T})~{\text{where}}~J=\det {\boldsymbol {F}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4c1f3c5be29c0884fd5a05392c236489b7332bc9)
Alternatively, we may use the nominal stress tensor
![{\displaystyle {\boldsymbol {N}}=J~({\boldsymbol {F}}^{-1}\cdot {\boldsymbol {\sigma }})}](https://wikimedia.org/api/rest_v1/media/math/render/svg/98251911395b3ced7d503e32e97fb9e39e3267f4)
From the conservation of mass, we have
. Hence,
![{\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 }}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7f193f412643114ea681c6a0a1493159a8e6016a)
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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 (
):
![{\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}~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bf73e618875104ef30158793965f0525cf3fb96e)
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In terms of the derivatives of the internal energy, we have
![{\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}}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c46a40f0eef020d05f07b0697584dea822bda40a)
Therefore,
![{\displaystyle {\boldsymbol {P}}=\rho _{0}~{\boldsymbol {F}}\cdot {\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d9cda3f974df7f49ded42b79266e44605924f81e)
and
![{\displaystyle {\boldsymbol {N}}=\rho _{0}~{\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}\cdot {\boldsymbol {F}}^{T}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/115b2b03f27891541ce6023e53a8377148c40e91)
That is,
![{\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}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/42139588bc439f3eb198c1da0a87430963c7f425)
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The stress power per unit volume is given by
. In terms of the stress measures in the reference configuration, we have
![{\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})~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2a168ed7325c00d504ab9cbd0501f2b146b217d2)
Using the identity
, we have
![{\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}}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c1ee3e6b79185734244823169274459b25475d98)
We can alternatively express the stress power in terms of
and
. Taking the material time derivative of
we have
![{\displaystyle {\dot {\boldsymbol {E}}}={\frac {1}{2}}({\dot {{\boldsymbol {F}}^{T}}}\cdot {\boldsymbol {F}}+{\boldsymbol {F}}^{T}\cdot {\dot {\boldsymbol {F}}})~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a3d958ef410bced1ee40e5f9475f9ab66d2eaa50)
Therefore,
![{\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}}})]~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f1b9f7cfb6b7379f796ee7056c975524e72b6850)
Using the identities
and
and using the
symmetry of
, we have
![{\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}}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/792287f25bc6609e8446e2e7de3bbc2d43f49717)
Now,
. Therefore,
.
Hence, the stress power can be expressed as
![{\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}}}~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/be66eb576a5b3158822dd183c2860108f0e36956)
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If we split the velocity gradient into symmetric and skew parts using
![{\displaystyle {\boldsymbol {\nabla }}\mathbf {v} ={\boldsymbol {l}}=\mathbf {d} +{\boldsymbol {w}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e3df4fcb8ee2829187e9cc4f5b2b70f6d61d81bf)
where
is the rate of deformation tensor and
is the spin tensor,
we have
![{\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}})~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a9c8b2630d2a6cebba1446b8dbe94b688f430afb)
Since
is symmetric and
is skew, we have
.
Therefore,
. Hence, we may also
express the stress power as
![{\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}}})~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b71c630dfe1a1ecc5116083631838370a9d9aed7)
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Recall that
![{\displaystyle {\boldsymbol {S}}=\rho _{0}~{\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f6fb204a102f2287a2c913984209ff0e92c2ca76)
Therefore,
![{\displaystyle {\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cbb3aa8279ad0551f15fa3a0151436357edad833)
Also recall that
![{\displaystyle {\frac {\partial {\bar {e}}}{\partial \eta }}=T~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f93d49c354ebaaa3ee071fb200d9115264254e35)
Now, the internal energy
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
![{\displaystyle {\boldsymbol {E}}={\tilde {\boldsymbol {E}}}({\boldsymbol {S}},T)\qquad {\text{and}}\qquad \eta ={\tilde {\eta }}({\boldsymbol {S}},T)~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/76c895375d533927d5dac59ca78b31f65cd010a5)
Then the specific internal energy, the specific entropy, and the stress can also be expressed as functions of
and
, or
and
, i.e.,
![{\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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fc1e759ad187f748f9da6bcddb38c8f5d052ccc4)
We can show that
![{\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}}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/787d509bf992c625a3b21a796204d10b5326a595)
and
![{\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}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f18c7f50bf8bfa8c3f7a923ff37bf4daf67bf3f9)
We define the Helmholtz free energy as
![{\displaystyle {\psi ={\hat {\psi }}({\boldsymbol {E}},T):=e-T~\eta ~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e127b303bb3b8a0d00aed48b34a966b045d2acfb)
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We define the Gibbs free energy as
![{\displaystyle {g={\tilde {g}}({\boldsymbol {S}},T):=-e+T~\eta +{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\boldsymbol {E}}~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/150a8528b674e724e91ad6a5365bc577a49e5ea3)
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The functions
and
are unique. Using
these definitions it can be shown that
![{\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)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/49a234ce9d7f8db16c8d282dfa0c5e84d76ac051)
and
![{\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}}}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9f88dcf04f04a72144232cf8a12a832c25cdaf43)
The specific heat at constant strain (or constant volume) is defined as
![{\displaystyle {C_{v}:={\frac {\partial {\hat {e}}({\boldsymbol {E}},T)}{\partial T}}~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4e3c7aa3ce0eb65163454e9cf59630f20cf24f81)
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The specific heat at constant stress (or constant pressure) is
defined as
![{\displaystyle {C_{p}:={\frac {\partial {\tilde {e}}({\boldsymbol {S}},T)}{\partial T}}~.}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bd0404e2064bf907f28f585390bd43bff820236a)
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We can show that
![{\displaystyle C_{v}=T~{\frac {\partial {\hat {\eta }}}{\partial T}}=-T~{\frac {\partial ^{2}{\hat {\psi }}}{\partial T^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6822e3f2cf472b540820c639e7bc9b6d9579b386)
and
![{\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}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/74b84168ee9d75967e51cfa8e3601471fa1167d5)
Also the equation for the balance of energy can be expressed in terms
of the specific heats as
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7fd660552c42cafc874d7db7a5bb048e5dc6df1f)
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where
![{\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}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/123ee75202719e13613ae9840dd55b30e1d34080)
The quantity
is called the coefficient of thermal stress and the quantity
is called the coefficient of thermal expansion.
The difference between
and
can be expressed as
![{\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}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5db092d1b29240a683d54c3a6014ee499a67bc75)
However, it is more common to express the above relation in terms of the elastic modulus tensor as
![{\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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ecaaf69f4156b100a553d692d241443edee86afc)
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where the fourth-order tensor of elastic moduli is defined as
![{\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}}}}}~.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c40bf362b595bcddb2f001ee82eb4e128b685d63)
For isotropic materials with a constant coefficient of thermal expansion that follow the St. Venant-Kirchhoff material model, we can show that
![{\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)
- 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.