# Continuum mechanics/Specific heats of thermoelastic materials

Proof:
Recall that
    $C_{v}:={\frac {\partial {\hat {e}}({\boldsymbol {E}},T)}{\partial T}}=T~{\frac {\partial {\hat {\eta }}}{\partial T}}$
and
    $C_{p}:={\frac {\partial {\tilde {e}}({\boldsymbol {S}},T)}{\partial T}}=T~{\frac {\partial {\tilde {\eta }}}{\partial T}}+{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}~.$
Therefore,
    $C_{p}-C_{v}=T~{\frac {\partial {\tilde {\eta }}}{\partial T}}+{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}-T~{\frac {\partial {\hat {\eta }}}{\partial T}}~.$
Also recall that
    $\eta ={\hat {\eta }}({\boldsymbol {E}},T)={\tilde {\eta }}({\boldsymbol {S}},T)~.$
Therefore, keeping ${\boldsymbol {S}}$ constant while differentiating, we have
    ${\frac {\partial {\tilde {\eta }}}{\partial T}}={\frac {\partial {\hat {\eta }}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\frac {\partial {\hat {\eta }}}{\partial T}}~.$
Noting that ${\boldsymbol {E}}={\tilde {\boldsymbol {E}}}({\boldsymbol {S}},T)$, and plugging back into the equation for the difference between the two specific heats, we have
    $C_{p}-C_{v}=T~{\frac {\partial {\hat {\eta }}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}+{\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}~.$
Recalling that
    ${\frac {\partial {\hat {\eta }}}{\partial {\boldsymbol {E}}}}=-{\cfrac {1}{\rho _{0}}}~{\frac {\partial {\hat {\boldsymbol {S}}}}{\partial T}}$
we get
    ${C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}\left({\boldsymbol {S}}-T~{\frac {\partial {\hat {\boldsymbol {S}}}}{\partial T}}\right):{\frac {\partial {\tilde {\boldsymbol {E}}}}{\partial T}}~.}$
Proof:
Recall that
    ${\boldsymbol {S}}=\rho _{0}~{\frac {\partial \psi }{\partial {\boldsymbol {E}}}}=\rho _{0}~{\boldsymbol {f}}({\boldsymbol {E}}({\boldsymbol {S}},T),T)~.$
Recall the chain rule which states that if
    $g(u,t)=f(x(u,t),y(u,t))$
then, if we keep $u$ fixed, the partial derivative of $g$ with respect to $t$ is given by
    ${\frac {\partial g}{\partial t}}={\frac {\partial f}{\partial x}}~{\frac {\partial x}{\partial t}}+{\frac {\partial f}{\partial y}}~{\frac {\partial y}{\partial t}}~.$
In our case,
    $u={\boldsymbol {S}},~~t=T,~~g({\boldsymbol {S}},T)={\boldsymbol {S}},~~x({\boldsymbol {S}},T)={\boldsymbol {E}}({\boldsymbol {S}},T),~~y({\boldsymbol {S}},T)=T,~~{\text{and}}~~f=\rho _{0}~{\boldsymbol {f}}.$
Hence, we have
    ${\boldsymbol {S}}=g({\boldsymbol {S}},T)=f({\boldsymbol {E}}({\boldsymbol {S}},T),T)=\rho _{0}~{\boldsymbol {f}}({\boldsymbol {E}}({\boldsymbol {S}},T),T)~.$
Taking the derivative with respect to $T$ keeping ${\boldsymbol {S}}$ constant, we have
    ${\frac {\partial g}{\partial T}}={\frac {\partial {\boldsymbol {S}}}{\partial T}}=\rho _{0}~\left[{\frac {\partial {\boldsymbol {f}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\frac {\partial {\boldsymbol {f}}}{\partial T}}~{\frac {\partial T}{\partial T}}\right]$
or,
    $\mathbf {0} ={\frac {\partial {\boldsymbol {f}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\frac {\partial {\boldsymbol {f}}}{\partial T}}~.$
Now,
    ${\boldsymbol {f}}={\frac {\partial \psi }{\partial {\boldsymbol {E}}}}\qquad \implies \qquad {\frac {\partial {\boldsymbol {f}}}{\partial {\boldsymbol {E}}}}={\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}\quad {\text{and}}\quad {\frac {\partial {\boldsymbol {f}}}{\partial T}}={\frac {\partial ^{2}\psi }{\partial T\partial {\boldsymbol {E}}}}~.$
Therefore,
    $\mathbf {0} ={\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\frac {\partial ^{2}\psi }{\partial T\partial {\boldsymbol {E}}}}={\frac {\partial }{\partial {\boldsymbol {E}}}}\left({\frac {\partial \psi }{\partial {\boldsymbol {E}}}}\right):{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\frac {\partial }{\partial T}}\left({\frac {\partial \psi }{\partial {\boldsymbol {E}}}}\right)~.$
Again recall that,
    ${\frac {\partial \psi }{\partial {\boldsymbol {E}}}}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}~.$
Plugging into the above, we get
    $\mathbf {0} ={\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\cfrac {1}{\rho _{0}}}~{\frac {\partial {\boldsymbol {S}}}{\partial T}}={\cfrac {1}{\rho _{0}}}~{\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\cfrac {1}{\rho _{0}}}~{\frac {\partial {\boldsymbol {S}}}{\partial T}}~.$
Therefore, we get the following relation for $\partial {\boldsymbol {S}}/\partial T$:
    ${\frac {\partial {\boldsymbol {S}}}{\partial T}}=-\rho _{0}~{\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}=-{\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}~.$
Recall that
    $C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}\left({\boldsymbol {S}}-T~{\frac {\partial {\boldsymbol {S}}}{\partial T}}\right):{\frac {\partial {\boldsymbol {E}}}{\partial T}}~.$
Plugging in the expressions for $\partial {\boldsymbol {S}}/\partial T$ we get:
    $C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}\left({\boldsymbol {S}}+T~\rho _{0}~{\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right):{\frac {\partial {\boldsymbol {E}}}{\partial T}}={\cfrac {1}{\rho _{0}}}\left({\boldsymbol {S}}+T~{\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right):{\frac {\partial {\boldsymbol {E}}}{\partial T}}~.$
Therefore,
    $C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+T~\left({\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right):{\frac {\partial {\boldsymbol {E}}}{\partial T}}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\cfrac {T}{\rho _{0}}}~\left({\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right):{\frac {\partial {\boldsymbol {E}}}{\partial T}}~.$
Using the identity $({\boldsymbol {\mathsf {A}}}:{\boldsymbol {B}}):{\boldsymbol {C}}={\boldsymbol {C}}:({\boldsymbol {\mathsf {A}}}:{\boldsymbol {B}})$, we have
    ${C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+T~{\frac {\partial {\boldsymbol {E}}}{\partial T}}:\left({\frac {\partial ^{2}\psi }{\partial {\boldsymbol {E}}\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right)={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}+{\cfrac {T}{\rho _{0}}}~{\frac {\partial {\boldsymbol {E}}}{\partial T}}:\left({\frac {\partial {\boldsymbol {S}}}{\partial {\boldsymbol {E}}}}:{\frac {\partial {\boldsymbol {E}}}{\partial T}}\right)~.}$
Proof:
Recall that,
    $C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:{\boldsymbol {\alpha }}_{E}+{\cfrac {T}{\rho _{0}}}~{\boldsymbol {\alpha }}_{E}:{\boldsymbol {\mathsf {C}}}:{\boldsymbol {\alpha }}_{E}~.$
Plugging the expressions of ${\boldsymbol {\alpha }}_{E}$ and ${\boldsymbol {\mathsf {C}}}$ into the above equation, we have
    ${\begin{aligned}C_{p}-C_{v}&={\cfrac {1}{\rho _{0}}}~{\boldsymbol {S}}:(\alpha ~{\boldsymbol {\mathit {1}}})+{\cfrac {T}{\rho _{0}}}~(\alpha ~{\boldsymbol {\mathit {1}}}):(\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2\mu ~{\boldsymbol {\mathsf {I}}}):(\alpha ~{\boldsymbol {\mathit {1}}})\\&={\cfrac {\alpha }{\rho _{0}}}~{\text{tr}}{\boldsymbol {S}}+{\cfrac {\alpha ^{2}~T}{\rho _{0}}}~{\boldsymbol {\mathit {1}}}:(\lambda ~{\boldsymbol {\mathit {1}}}\otimes {\boldsymbol {\mathit {1}}}+2\mu ~{\boldsymbol {\mathsf {I}}}):{\boldsymbol {\mathit {1}}}\\&={\cfrac {\alpha }{\rho _{0}}}~{\text{tr}}{\boldsymbol {S}}+{\cfrac {\alpha ^{2}~T}{\rho _{0}}}~{\boldsymbol {\mathit {1}}}:(\lambda ~{\text{tr}}{\boldsymbol {\mathit {1}}}~{\boldsymbol {\mathit {1}}}+2\mu ~{\boldsymbol {\mathit {1}}})\\&={\cfrac {\alpha }{\rho _{0}}}~{\text{tr}}{\boldsymbol {S}}+{\cfrac {\alpha ^{2}~T}{\rho _{0}}}~(3~\lambda ~{\text{tr}}{\boldsymbol {\mathit {1}}}+2\mu ~{\text{tr}}{\boldsymbol {\mathit {1}}})\\&={\cfrac {\alpha }{\rho _{0}}}~{\text{tr}}{\boldsymbol {S}}+{\cfrac {3~\alpha ^{2}~T}{\rho _{0}}}~(3~\lambda +2\mu )\\&={\cfrac {\alpha ~{\text{tr}}{\boldsymbol {S}}}{\rho _{0}}}+{\cfrac {9~\alpha ^{2}~K~T}{\rho _{0}}}~.\end{aligned}}$
Therefore,
    ${C_{p}-C_{v}={\cfrac {1}{\rho _{0}}}\left[\alpha ~{\text{tr}}{\boldsymbol {S}}+9~\alpha ^{2}~K~T\right]~.}$