# Advanced elasticity/Specific heats of thermoelastic materials

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## Relation between specific heats - 1

For thermoelastic materials, show that the specific heats are related by the relation

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

Proof:

Recall that

${\displaystyle C_{v}:={\frac {\partial {\hat {e}}({\boldsymbol {E}},T)}{\partial T}}=T~{\frac {\partial {\hat {\eta }}}{\partial T}}}$

and

${\displaystyle C_{p}:={\frac {\partial {\tilde {e}}({\boldsymbol {S}},T)}{\partial T}}=T~{\frac {\partial {\tilde {\eta }}}{\partial T}}+{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}~.}$

Therefore,

${\displaystyle C_{p}-C_{v}=T~{\frac {\partial {\tilde {\eta }}}{\partial T}}+{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}-T~{\frac {\partial {\hat {\eta }}}{\partial T}}~.}$

Also recall that

${\displaystyle \eta ={\hat {\eta }}({\boldsymbol {E}},T)={\tilde {\eta }}({\boldsymbol {S}},T)~.}$

Therefore, keeping ${\displaystyle {\boldsymbol {S}}}$ constant while differentiating, we have

${\displaystyle {\frac {\partial {\tilde {\eta }}}{\partial T}}={\frac {\partial {\hat {\eta }}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\frac {\partial {\hat {\eta }}}{\partial T}}~.}$

Noting that ${\displaystyle {\boldsymbol {E}}={\tilde {\boldsymbol {E}}}({\boldsymbol {S}},T)}$, and plugging back into the equation for the difference between the two specific heats, we have

${\displaystyle C_{p}-C_{v}=T~{\frac {\partial {\hat {\eta }}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}+{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}~.}$

Recalling that

${\displaystyle {\frac {\partial {\hat {\eta }}}{\partial {\boldsymbol {E}}}}=-{\cfrac {1}{\rho _{0}}}~{\frac {\partial {\hat {\boldsymbol {S}}}}{\partial T}}}$

we get

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

## Relation between specific heats - 2

For thermoelastic materials, show that the specific heats can also be related by the equations

${\displaystyle C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\frac {\partial {\boldsymbol {E}}}{\partial T}}:\left({\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right)={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\cfrac {T}{\rho _{0}}}~{\frac {\partial {\boldsymbol {E}}}{\partial T}}:\left({\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right)~.}$

We can also write the above as

${\displaystyle C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\boldsymbol {\alpha }}_{E}+{\cfrac {T}{\rho _{0}}}~{\boldsymbol {\alpha }}_{E}:{\boldsymbol {\mathsf {C}}}:{\boldsymbol {\alpha }}_{E}}$

where ${\displaystyle {\boldsymbol {\alpha }}_{E}:={\frac {\partial {\boldsymbol {E}}}{\partial T}}}$ is the thermal expansion tensor and ${\displaystyle {\boldsymbol {\mathsf {C}}}:={\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {E}}}}}$ is the stiffness tensor.

Proof:

Recall that

${\displaystyle {\boldsymbol {S}}=\rho _{0}~{\frac {\partial \psi }{\partial {\boldsymbol {E}}}}=\rho _{0}~{\boldsymbol {f}}({\boldsymbol {E}}({\boldsymbol {S}},T),T)~.}$

Recall the chain rule which states that if

${\displaystyle g(u,t)=f(x(u,t),y(u,t))}$

then, if we keep ${\displaystyle u}$ fixed, the partial derivative of ${\displaystyle g}$ with respect to ${\displaystyle t}$ is given by

${\displaystyle {\frac {\partial g}{\partial t}}={\frac {\partial f}{\partial x}}~{\frac {\partial x}{\partial t}}+{\frac {\partial f}{\partial y}}~{\frac {\partial y}{\partial t}}~.}$

In our case,

${\displaystyle u={\boldsymbol {S}},~~t=T,~~g({\boldsymbol {S}},T)={\boldsymbol {S}},~~x({\boldsymbol {S}},T)={\boldsymbol {E}}({\boldsymbol {S}},T),~~y({\boldsymbol {S}},T)=T,~~{\text{and}}~~f=\rho _{0}~{\boldsymbol {f}}.}$

Hence, we have

${\displaystyle {\boldsymbol {S}}=g({\boldsymbol {S}},T)=f({\boldsymbol {E}}({\boldsymbol {S}},T),T)=\rho _{0}~{\boldsymbol {f}}({\boldsymbol {E}}({\boldsymbol {S}},T),T)~.}$

Taking the derivative with respect to ${\displaystyle T}$ keeping ${\displaystyle {\boldsymbol {S}}}$ constant, we have

${\displaystyle {\frac {\partial g}{\partial T}}={\frac {\partial {\boldsymbol {S}}}{\partial T}}=\rho _{0}~\left[{\frac {\partial {\boldsymbol {f}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\frac {\partial {\boldsymbol {f}}}{\partial T}}~{\frac {\partial T}{\partial T}}\right]}$

or,

${\displaystyle \mathbf {0} ={\frac {\partial {\boldsymbol {f}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\frac {\partial {\boldsymbol {f}}}{\partial T}}~.}$

Now,

${\displaystyle {\boldsymbol {f}}={\frac {\partial \psi }{\partial {\boldsymbol {E}}}}\qquad \implies \qquad {\frac {\partial {\boldsymbol {f}}}{\partial {\boldsymbol {E}}}}={\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}\quad {\text{and}}\quad {\frac {\partial {\boldsymbol {f}}}{\partial T}}={\frac {\partial ^{2}\psi }{\partial T\partial {\boldsymbol {E}}}}~.}$

Therefore,

${\displaystyle \mathbf {0} ={\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\frac {\partial ^{2}\psi }{\partial T\partial {\boldsymbol {E}}}}={\frac {\partial }{\partial {\boldsymbol {E}}}}\left({\frac {\partial \psi }{\partial {\boldsymbol {E}}}}\right):{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\frac {\partial }{\partial T}}\left({\frac {\partial \psi }{\partial {\boldsymbol {E}}}}\right)~.}$

Again recall that,

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

Plugging into the above, we get

${\displaystyle \mathbf {0} ={\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\cfrac {1}{\rho _{0}}}~{\frac {\partial {\boldsymbol {S}}}{\partial T}}={\cfrac {1}{\rho _{0}}}~{\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\cfrac {1}{\rho _{0}}}~{\frac {\partial {\boldsymbol {S}}}{\partial T}}~.}$

Therefore, we get the following relation for ${\displaystyle \partial {\boldsymbol {S}}/\partial T}$:

${\displaystyle {\frac {\partial {\boldsymbol {S}}}{\partial T}}=-\rho _{0}~{\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}=-{\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}~.}$

Recall that

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

Plugging in the expressions for ${\displaystyle \partial {\boldsymbol {S}}/\partial T}$ we get:

${\displaystyle C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}\left({\boldsymbol {S}}+T~\rho _{0}~{\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right):{\frac {\partial {\boldsymbol {E}}}{\partial T}}={\cfrac {1}{\rho _{0}}}\left({\boldsymbol {S}}+T~{\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right):{\frac {\partial {\boldsymbol {E}}}{\partial T}}~.}$

Therefore,

${\displaystyle C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+T~\left({\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right):{\frac {\partial {\boldsymbol {E}}}{\partial T}}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\cfrac {T}{\rho _{0}}}~\left({\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right):{\frac {\partial {\boldsymbol {E}}}{\partial T}}~.}$

Using the identity ${\displaystyle ({\boldsymbol {\mathsf {A}}}:{\boldsymbol {B}}):{\boldsymbol {C}}={\boldsymbol {C}}:({\boldsymbol {\mathsf {A}}}:{\boldsymbol {B}})}$, we have

${\displaystyle {C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+T~{\frac {\partial {\boldsymbol {E}}}{\partial T}}:\left({\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right)={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\cfrac {T}{\rho _{0}}}~{\frac {\partial {\boldsymbol {E}}}{\partial T}}:\left({\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right)~.}}$

## Specific heats of Saint-Venant–Kirchhoff material

Consider an isotropic thermoelastic material that has a constant coefficient of thermal expansion and which follows the Saint-Venant–Kirchhoff model, i.e,

${\displaystyle {\boldsymbol {\alpha }}_{E}=\alpha ~{\boldsymbol {\mathit {1}}}\qquad {\text{and}}\qquad {\boldsymbol {\mathsf {C}}}=\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2\mu ~{\boldsymbol {\mathsf {I}}}}$

where ${\displaystyle \alpha }$ is the coefficient of thermal expansion and ${\displaystyle 3~\lambda =3~K-2~\mu }$ where ${\displaystyle K,\mu }$ are the bulk and shear moduli, respectively.

Show that the specific heats related by the equation

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

Proof:

Recall that,

${\displaystyle C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\boldsymbol {\alpha }}_{E}+{\cfrac {T}{\rho _{0}}}~{\boldsymbol {\alpha }}_{E}:{\boldsymbol {\mathsf {C}}}:{\boldsymbol {\alpha }}_{E}~.}$

Plugging the expressions of ${\displaystyle {\boldsymbol {\alpha }}_{E}}$ and ${\displaystyle {\boldsymbol {\mathsf {C}}}}$ into the above equation, we have

{\displaystyle {\begin{aligned}C_{p}-C_{v}&={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:(\alpha ~{\boldsymbol {\mathit {1}}})+{\cfrac {T}{\rho _{0}}}~(\alpha ~{\boldsymbol {\mathit {1}}}):(\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2\mu ~{\boldsymbol {\mathsf {I}}}):(\alpha ~{\boldsymbol {\mathit {1}}})\\&={\cfrac {\alpha }{\rho _{0}}}~{\text{tr}}{\boldsymbol {S}}+{\cfrac {\alpha ^{2}~T}{\rho _{0}}}~{\boldsymbol {\mathit {1}}}:(\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2\mu ~{\boldsymbol {\mathsf {I}}}):{\boldsymbol {\mathit {1}}}\\&={\cfrac {\alpha }{\rho _{0}}}~{\text{tr}}{\boldsymbol {S}}+{\cfrac {\alpha ^{2}~T}{\rho _{0}}}~{\boldsymbol {\mathit {1}}}:(\lambda ~{\text{tr}}{\boldsymbol {\mathit {1}}}~{\boldsymbol {\mathit {1}}}+2\mu ~{\boldsymbol {\mathit {1}}})\\&={\cfrac {\alpha }{\rho _{0}}}~{\text{tr}}{\boldsymbol {S}}+{\cfrac {\alpha ^{2}~T}{\rho _{0}}}~(3~\lambda ~{\text{tr}}{\boldsymbol {\mathit {1}}}+2\mu ~{\text{tr}}{\boldsymbol {\mathit {1}}})\\&={\cfrac {\alpha }{\rho _{0}}}~{\text{tr}}{\boldsymbol {S}}+{\cfrac {3~\alpha ^{2}~T}{\rho _{0}}}~(3~\lambda +2\mu )\\&={\cfrac {\alpha ~{\text{tr}}{\boldsymbol {S}}}{\rho _{0}}}+{\cfrac {9~\alpha ^{2}~K~T}{\rho _{0}}}~.\end{aligned}}}

Therefore,

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