# Continuum mechanics/Thermoelasticity

## Thermoelastic materials

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.

### Deformation gradient as the strain measure

In thermoelasticity we assume that the fundamental kinematic quantity is the deformation gradient (${\boldsymbol {F}}$ ) which is given by

${\boldsymbol {F}}={\frac {\partial \mathbf {x} }{\partial \mathbf {X} }}={\boldsymbol {\nabla }}_{\circ }\mathbf {x} ~;~~\det {\boldsymbol {F}}>0~.$

A thermoelastic material is one in which the internal energy ($e$ ) is a function only of ${\boldsymbol {F}}$  and the specific entropy ($\eta$ ), that is

$e={\bar {e}}({\boldsymbol {F}},\eta )~.$

For a thermoelastic material, we can show that the entropy inequality can be written as

 ${\rho ~\left({\frac {\partial {\bar {e}}}{\partial \eta }}-T\right)~{\dot {\eta }}+\left(\rho ~{\frac {\partial {\bar {e}}}{\partial {\boldsymbol {F}}}}-{\boldsymbol {\sigma }}\cdot {\boldsymbol {F}}^{-T}\right):{\dot {\boldsymbol {F}}}+{\cfrac {\mathbf {q} \cdot {\boldsymbol {\nabla }}T}{T}}\leq 0~.}$ At this stage, we make the following constitutive assumptions:

1) Like the internal energy, we assume that ${\boldsymbol {\sigma }}$  and $T$  are also functions only of ${\boldsymbol {F}}$  and $\eta$ , i.e.,

${\boldsymbol {\sigma }}={\boldsymbol {\sigma }}({\boldsymbol {F}},\eta )~;~~T=T({\boldsymbol {F}},\eta )~.$

2) The heat flux $\mathbf {q}$  satisfies the thermal conductivity inequality and if $\mathbf {q}$  is independent of ${\dot {\eta }}$  and ${\dot {\boldsymbol {F}}}$ , we have

$\mathbf {q} \cdot {\boldsymbol {\nabla }}T\leq 0\qquad \implies \qquad -({\boldsymbol {\kappa }}\cdot {\boldsymbol {\nabla }}T)\cdot {\boldsymbol {\nabla }}T\leq 0\qquad \implies \qquad {\boldsymbol {\kappa }}\geq \mathbf {0}$

i.e., the thermal conductivity ${\boldsymbol {\kappa }}$  is positive semidefinite.

Therefore, the entropy inequality may be written as

$\rho ~\left({\frac {\partial {\bar {e}}}{\partial \eta }}-T\right)~{\dot {\eta }}+\left(\rho ~{\frac {\partial {\bar {e}}}{\partial {\boldsymbol {F}}}}-{\boldsymbol {\sigma }}\cdot {\boldsymbol {F}}^{-T}\right):{\dot {\boldsymbol {F}}}\leq 0~.$

Since ${\dot {\eta }}$  and ${\dot {\boldsymbol {F}}}$  are arbitrary, the entropy inequality will be satisfied if and only if

${\frac {\partial {\bar {e}}}{\partial \eta }}-T=0\implies T={\frac {\partial {\bar {e}}}{\partial \eta }}\qquad {\text{and}}\qquad \rho ~{\frac {\partial {\bar {e}}}{\partial {\boldsymbol {F}}}}-{\boldsymbol {\sigma }}\cdot {\boldsymbol {F}}^{-T}=\mathbf {0} \implies {\boldsymbol {\sigma }}=\rho ~{\frac {\partial {\bar {e}}}{\partial {\boldsymbol {F}}}}\cdot {\boldsymbol {F}}^{T}~.$

Therefore,

 ${T={\frac {\partial {\bar {e}}}{\partial \eta }}}\qquad {\text{and}}\qquad {{\boldsymbol {\sigma }}=\rho ~{\frac {\partial {\bar {e}}}{\partial {\boldsymbol {F}}}}\cdot {\boldsymbol {F}}^{T}~.}$ Given the above relations, the energy equation may expressed in terms of the specific entropy as

 ${\rho ~T~{\dot {\eta }}=-{\boldsymbol {\nabla }}\cdot \mathbf {q} +\rho ~s~.}$ #### Effect of a rigid body rotation of the internal energy

If a thermoelastic body is subjected to a rigid body rotation ${\boldsymbol {Q}}$ , then its internal energy should not change. After a rotation, the new deformation gradient (${\hat {\boldsymbol {F}}}$ ) is given by

${\hat {\boldsymbol {F}}}={\boldsymbol {Q}}\cdot {\boldsymbol {F}}~.$

Since the internal energy does not change, we must have

$e={\bar {e}}({\hat {\boldsymbol {F}}},\eta )={\bar {e}}({\boldsymbol {F}},\eta )~.$

Now, from the polar decomposition theorem, ${\boldsymbol {F}}={\boldsymbol {R}}\cdot {\boldsymbol {U}}$  where ${\boldsymbol {R}}$  is the orthogonal rotation tensor (i.e., ${\boldsymbol {R}}\cdot {\boldsymbol {R}}^{T}={\boldsymbol {R}}^{T}\cdot {\boldsymbol {R}}={\boldsymbol {\mathit {1}}}$ ) and ${\boldsymbol {U}}$  is the symmetric right stretch tensor. Therefore,

${\bar {e}}({\boldsymbol {Q}}\cdot {\boldsymbol {R}}\cdot {\boldsymbol {U}},\eta )={\bar {e}}({\boldsymbol {F}},\eta )~.$

We can choose any rotation ${\boldsymbol {Q}}$ . In particular, if we choose ${\boldsymbol {Q}}={\boldsymbol {R}}^{T}$ , we have

${\bar {e}}({\boldsymbol {R}}^{T}\cdot {\boldsymbol {R}}\cdot {\boldsymbol {U}},\eta )={\bar {e}}({\boldsymbol {\mathit {1}}}\cdot {\boldsymbol {U}},\eta )={\tilde {e}}({\boldsymbol {U}},\eta )~.$

Therefore,

${\bar {e}}({\boldsymbol {U}},\eta )={\bar {e}}({\boldsymbol {F}},\eta )~.$

This means that the internal energy depends only on the stretch ${\boldsymbol {U}}$  and not on the orientation of the body.

### Other strain and stress measures

The internal energy depends on ${\boldsymbol {F}}$  only through the stretch ${\boldsymbol {U}}$ . A strain measure that reflects this fact and also vanishes in the reference configuration is the Green strain

 ${{\boldsymbol {E}}={\frac {1}{2}}({\boldsymbol {F}}^{T}\cdot {\boldsymbol {F}}-{\boldsymbol {\mathit {1}}})={\frac {1}{2}}({\boldsymbol {U}}^{2}-{\boldsymbol {\mathit {1}}})~.}$ Recall that the Cauchy stress is given by

${\boldsymbol {\sigma }}=\rho ~{\frac {\partial {\bar {e}}}{\partial {\boldsymbol {F}}}}\cdot {\boldsymbol {F}}^{T}~.$

We can show that the Cauchy stress can be expressed in terms of the Green strain as

 ${{\boldsymbol {\sigma }}=\rho ~{\boldsymbol {F}}\cdot {\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}\cdot {\boldsymbol {F}}^{T}~.}$ Also, recall that the first Piola-Kirchhoff stress tensor is defined as

${\boldsymbol {P}}=J~({\boldsymbol {\sigma }}\cdot {\boldsymbol {F}}^{-T})~{\text{where}}~J=\det {\boldsymbol {F}}$

Alternatively, we may use the nominal stress tensor

${\boldsymbol {N}}=J~({\boldsymbol {F}}^{-1}\cdot {\boldsymbol {\sigma }})$

From the conservation of mass, we have $\rho _{0}=\rho ~\det {\boldsymbol {F}}$ . Hence,

 ${{\boldsymbol {P}}={\cfrac {\rho _{0}}{\rho }}~{\boldsymbol {\sigma }}\cdot {\boldsymbol {F}}^{-T}~~{\text{and}}~~{\boldsymbol {N}}={\cfrac {\rho _{0}}{\rho }}~{\boldsymbol {F}}^{-1}\cdot {\boldsymbol {\sigma }}}$ 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 (${\boldsymbol {S}}$ ):

 ${{\boldsymbol {S}}:={\boldsymbol {F}}^{-1}\cdot {\boldsymbol {P}}={\boldsymbol {N}}\cdot {\boldsymbol {F}}^{-T}={\cfrac {\rho _{0}}{\rho }}~{\boldsymbol {F}}^{-1}\cdot {\boldsymbol {\sigma }}\cdot {\boldsymbol {F}}^{-T}~.}$ In terms of the derivatives of the internal energy, we have

${\boldsymbol {S}}={\cfrac {\rho _{0}}{\rho }}~{\boldsymbol {F}}^{-1}\cdot \left(\rho ~{\boldsymbol {F}}\cdot {\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}\cdot {\boldsymbol {F}}^{T}\right)\cdot {\boldsymbol {F}}^{-T}=\rho _{0}~{\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}$

Therefore,

${\boldsymbol {P}}=\rho _{0}~{\boldsymbol {F}}\cdot {\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}~.$

and

${\boldsymbol {N}}=\rho _{0}~{\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}\cdot {\boldsymbol {F}}^{T}~.$

That is,

 ${{\boldsymbol {S}}=\rho _{0}~{\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}}~;~~{{\boldsymbol {P}}=\rho _{0}~{\boldsymbol {F}}\cdot {\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}}~;~~{\boldsymbol {N}}=\rho _{0}~{\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}\cdot {\boldsymbol {F}}^{T}$ ### Stress Power

The stress power per unit volume is given by ${\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v}$ . In terms of the stress measures in the reference configuration, we have

${\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} =\left(\rho ~{\boldsymbol {F}}\cdot {\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}\cdot {\boldsymbol {F}}^{T}\right):({\dot {\boldsymbol {F}}}\cdot {\boldsymbol {F}}^{-1})~.$

Using the identity ${\boldsymbol {A}}:({\boldsymbol {B}}\cdot {\boldsymbol {C}})=({\boldsymbol {A}}\cdot {\boldsymbol {C}}^{T}):{\boldsymbol {B}}$ , we have

${\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} =\left[\left(\rho ~{\boldsymbol {F}}\cdot {\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}\cdot {\boldsymbol {F}}^{T}\right)\cdot {\boldsymbol {F}}^{-T}\right]:{\dot {\boldsymbol {F}}}=\rho ~\left({\boldsymbol {F}}\cdot {\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}\right):{\dot {\boldsymbol {F}}}={\cfrac {\rho }{\rho _{0}}}~{\boldsymbol {P}}:{\dot {\boldsymbol {F}}}={\cfrac {\rho }{\rho _{0}}}~{\boldsymbol {N}}^{T}:{\dot {\boldsymbol {F}}}~.$

We can alternatively express the stress power in terms of ${\boldsymbol {S}}$  and ${\dot {\boldsymbol {E}}}$ . Taking the material time derivative of ${\boldsymbol {E}}$  we have

${\dot {\boldsymbol {E}}}={\frac {1}{2}}({\dot {{\boldsymbol {F}}^{T}}}\cdot {\boldsymbol {F}}+{\boldsymbol {F}}^{T}\cdot {\dot {\boldsymbol {F}}})~.$

Therefore,

${\boldsymbol {S}}:{\dot {\boldsymbol {E}}}={\frac {1}{2}}[{\boldsymbol {S}}:({\dot {{\boldsymbol {F}}^{T}}}\cdot {\boldsymbol {F}})+{\boldsymbol {S}}:({\boldsymbol {F}}^{T}\cdot {\dot {\boldsymbol {F}}})]~.$

Using the identities ${\boldsymbol {A}}:({\boldsymbol {B}}\cdot {\boldsymbol {C}})=({\boldsymbol {A}}\cdot {\boldsymbol {C}}^{T}):{\boldsymbol {B}}=({\boldsymbol {B}}^{T}\cdot {\boldsymbol {A}}):{\boldsymbol {C}}$  and ${\boldsymbol {A}}:{\boldsymbol {B}}={\boldsymbol {A}}^{T}:{\boldsymbol {B}}^{T}$  and using the symmetry of ${\boldsymbol {S}}$ , we have

${\boldsymbol {S}}:{\dot {\boldsymbol {E}}}={\frac {1}{2}}[({\boldsymbol {S}}\cdot {\boldsymbol {F}}^{T}):{\dot {\boldsymbol {F}}}^{T}+({\boldsymbol {F}}\cdot {\boldsymbol {S}}):{\dot {\boldsymbol {F}}}]={\frac {1}{2}}[({\boldsymbol {F}}\cdot {\boldsymbol {S}}^{T}):{\dot {\boldsymbol {F}}}+({\boldsymbol {F}}\cdot {\boldsymbol {S}}):{\dot {\boldsymbol {F}}}]=({\boldsymbol {F}}\cdot {\boldsymbol {S}}):{\dot {\boldsymbol {F}}}~.$

Now, ${\boldsymbol {S}}={\boldsymbol {F}}^{-1}\cdot {\boldsymbol {P}}$ . Therefore, ${\boldsymbol {S}}:{\dot {\boldsymbol {E}}}={\boldsymbol {N}}^{T}:{\dot {\boldsymbol {F}}}$ . Hence, the stress power can be expressed as

 ${{\frac {\rho _{0}}{\rho }}~{\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} ={\boldsymbol {P}}:{\dot {\boldsymbol {F}}}={\boldsymbol {N^{T}}}:{\dot {\boldsymbol {F}}}={\boldsymbol {S}}:{\dot {\boldsymbol {E}}}~.}$ If we split the velocity gradient into symmetric and skew parts using

${\boldsymbol {\nabla }}\mathbf {v} ={\boldsymbol {l}}=\mathbf {d} +{\boldsymbol {w}}$

where $\mathbf {d}$  is the rate of deformation tensor and ${\boldsymbol {w}}$  is the spin tensor, we have

${\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} ={\boldsymbol {\sigma }}:\mathbf {d} +{\boldsymbol {\sigma }}:{\boldsymbol {w}}={\text{tr}}~({\boldsymbol {\sigma }}^{T}\cdot \mathbf {d} )+{\text{tr}}~({\boldsymbol {\sigma }}^{T}\cdot {\boldsymbol {w}})={\text{tr}}~({\boldsymbol {\sigma }}\cdot \mathbf {d} )+{\text{tr}}~({\boldsymbol {\sigma }}\cdot {\boldsymbol {w}})~.$

Since ${\boldsymbol {\sigma }}$  is symmetric and ${\boldsymbol {w}}$  is skew, we have ${\text{tr}}~({\boldsymbol {\sigma }}\cdot {\boldsymbol {w}})=0$ . Therefore, ${\boldsymbol {\sigma }}:{\boldsymbol {\nabla }}\mathbf {v} ={\text{tr}}~({\boldsymbol {\sigma }}\cdot \mathbf {d} )$ . Hence, we may also express the stress power as

 ${{\frac {\rho _{0}}{\rho }}~{\text{tr}}~({\boldsymbol {\sigma }}\cdot \mathbf {d} )={\text{tr}}~({\boldsymbol {P}}^{T}\cdot {\dot {\boldsymbol {F}}})={\text{tr}}~({\boldsymbol {N}}\cdot {\dot {\boldsymbol {F}}})={\text{tr}}~({\boldsymbol {S}}\cdot {\dot {\boldsymbol {E}}})~.}$ ### Helmholtz and Gibbs free energy

Recall that

${\boldsymbol {S}}=\rho _{0}~{\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}~.$

Therefore,

${\frac {\partial {\bar {e}}}{\partial {\boldsymbol {E}}}}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}~.$

Also recall that

${\frac {\partial {\bar {e}}}{\partial \eta }}=T~.$

Now, the internal energy $e={\bar {e}}({\boldsymbol {E}},\eta )$  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

${\boldsymbol {E}}={\tilde {\boldsymbol {E}}}({\boldsymbol {S}},T)\qquad {\text{and}}\qquad \eta ={\tilde {\eta }}({\boldsymbol {S}},T)~.$

Then the specific internal energy, the specific entropy, and the stress can also be expressed as functions of ${\boldsymbol {S}}$  and $T$ , or ${\boldsymbol {E}}$  and $T$ , i.e.,

$e={\bar {e}}({\boldsymbol {E}},\eta )={\tilde {e}}({\boldsymbol {S}},T)={\hat {e}}({\boldsymbol {E}},T)~;\qquad \eta ={\tilde {\eta }}({\boldsymbol {S}},T)={\hat {\eta }}({\boldsymbol {E}},T)~;\qquad {\text{and}}\qquad {\boldsymbol {S}}={\hat {\boldsymbol {S}}}({\boldsymbol {E}},T)$

We can show that

${\cfrac {d}{dt}}(e-T~\eta )=-{\dot {T}}~\eta +{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\dot {\boldsymbol {E}}}\qquad {\text{or}}\qquad {\cfrac {d\psi }{dt}}=-{\dot {T}}~\eta +{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\dot {\boldsymbol {E}}}~.$

and

${\cfrac {d}{dt}}(e-T~\eta -{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\boldsymbol {E}})=-{\dot {T}}~\eta -{\cfrac {1}{\rho _{0}}}~{\dot {\boldsymbol {S}}}:{\boldsymbol {E}}\qquad {\text{or}}\qquad {\cfrac {dg}{dt}}={\dot {T}}~\eta +{\cfrac {1}{\rho _{0}}}~{\dot {\boldsymbol {S}}}:{\boldsymbol {E}}~.$

We define the Helmholtz free energy as

 ${\psi ={\hat {\psi }}({\boldsymbol {E}},T):=e-T~\eta ~.}$ We define the Gibbs free energy as

 ${g={\tilde {g}}({\boldsymbol {S}},T):=-e+T~\eta +{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\boldsymbol {E}}~.}$ The functions ${\hat {\psi }}({\boldsymbol {E}},T)$  and ${\tilde {g}}({\boldsymbol {S}},T)$  are unique. Using these definitions it can be shown that

${\frac {\partial {\hat {\psi }}}{\partial {\boldsymbol {E}}}}={\cfrac {1}{\rho _{0}}}~{\hat {\boldsymbol {S}}}({\boldsymbol {E}},T)~;~~{\frac {\partial {\hat {\psi }}}{\partial T}}=-{\hat {\eta }}({\boldsymbol {E}},T)~;~~{\frac {\partial {\tilde {g}}}{\partial {\boldsymbol {S}}}}={\cfrac {1}{\rho _{0}}}~{\tilde {\boldsymbol {E}}}({\boldsymbol {S}},T)~;~~{\frac {\partial {\tilde {g}}}{\partial T}}={\tilde {\eta }}({\boldsymbol {S}},T)$

and

${\frac {\partial {\hat {\boldsymbol {S}}}}{\partial T}}=-\rho _{0}~{\frac {\partial {\hat {\eta }}}{\partial {\boldsymbol {E}}}}\qquad {\text{and}}\qquad {\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}=\rho _{0}~{\frac {\partial {\tilde {\eta }}}{\partial {\boldsymbol {S}}}}~.$

### Specific Heats

The specific heat at constant strain (or constant volume) is defined as

 ${C_{v}:={\frac {\partial {\hat {e}}({\boldsymbol {E}},T)}{\partial T}}~.}$ The specific heat at constant stress (or constant pressure) is defined as

 ${C_{p}:={\frac {\partial {\tilde {e}}({\boldsymbol {S}},T)}{\partial T}}~.}$ We can show that

$C_{v}=T~{\frac {\partial {\hat {\eta }}}{\partial T}}=-T~{\frac {\partial ^{2}{\hat {\psi }}}{\partial T^{2}}}$

and

$C_{p}=T~{\frac {\partial {\tilde {\eta }}}{\partial T}}+{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}=T~{\frac {\partial ^{2}{\tilde {g}}}{\partial T^{2}}}+{\boldsymbol {S}}:{\frac {\partial ^{2}{\tilde {g}}}{\partial {\boldsymbol {S}}\partial T}}~.$

Also the equation for the balance of energy can be expressed in terms of the specific heats as

 {\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

${\boldsymbol {\beta }}_{S}:={\frac {\partial {\hat {\boldsymbol {S}}}}{\partial T}}\qquad {\text{and}}\qquad {\boldsymbol {\alpha }}_{E}:={\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}~.$

The quantity ${\boldsymbol {\beta }}_{S}$  is called the coefficient of thermal stress and the quantity ${\boldsymbol {\alpha }}_{E}$  is called the coefficient of thermal expansion.

The difference between $C_{p}$  and $C_{v}$  can be expressed as

$C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}\left({\boldsymbol {S}}-T~{\frac {\partial {\hat {\boldsymbol {S}}}}{\partial T}}\right):{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}~.$

However, it is more common to express the above relation in terms of the elastic modulus tensor as

 ${C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\boldsymbol {\alpha }}_{E}+{\cfrac {T}{\rho _{0}}}~{\boldsymbol {\alpha }}_{E}:{\boldsymbol {\mathsf {C}}}:{\boldsymbol {\alpha }}_{E}}$ where the fourth-order tensor of elastic moduli is defined as

${\boldsymbol {\mathsf {C}}}:={\frac {\partial {\hat {\boldsymbol {S}}}}{\partial {\tilde {\boldsymbol {E}}}}}=\rho _{0}~{\frac {\partial ^{2}{\hat {\psi }}}{\partial {\tilde {\boldsymbol {E}}}\partial {\tilde {\boldsymbol {E}}}}}~.$

For isotropic materials with a constant coefficient of thermal expansion that follow the St. Venant-Kirchhoff material model, we can show that

$C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}\left[\alpha ~{\text{tr}}{\boldsymbol {S}}+9~\alpha ^{2}~K~T\right]~.$