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Requiring the Clausius-Duhem inequality for independent reversible mechanical and thermal variations constrains a constitutive equation. For $U(F,\eta ,\xi )$ it gives $P_{Ii}={\rho }_0{U}_{,F_{iI}}$ and $\theta =U_{,\eta }$. Combining the Lagrangian internal energy balance with the Lagrangian entropy inequality leaves $f_r{\dot{\xi }}_r-q_I{\theta }_{,I}/\theta \ge 0$. The independence assumptions are part of the constitutive argument.

Let $v_i={\dot{x}}_i$ and use a fixed arbitrary material subvolume $B_0$ in the reference configuration, with outward unit normal $N_I$. The reference mass density ${\rho }_0$ is time independent. The global momentum conservation law isThe divergence theorem, differentiation under the integral, and arbitrariness of $B_0$ give

The global conservation of energy law includes both internal energy and kinetic energy:Here $q^0$ is outgoing heat flux per reference area. Localizing givesSubtract the scalar product of the local momentum conservation law with $v$. Since $v_{i,I}={\dot{F}}_{iI}$, this proves the Lagrangian internal energy balance

For positive temperature $\theta$, the global entropy inequality isIts local excess is a nonnegative production density $\gamma$. Expanding the divergence of $q^0/\theta$ proves the Lagrangian entropy inequality in the requested equality form:No constitutive assumption on the heat flux was needed for these balance identities.

For the reference-configuration jump balances, make the orientation convention explicit. Write $[a]=a^{+}-a^{-}$ and let $n^0$ point from the minus side to the plus side. Let $c$ be the actual interface speed in direction $n^0$. A signed level function $\varphi$ with the plus side $\varphi >0$ obeys ${\varphi }_t=-c|{\nabla }_X\varphi |$ on the interface. For $a=a^{-}+[a]H(\varphi )$, the singular time derivative is $-c[a]{\delta }_S$; the singular reference divergence of $b$ is $n_I^0[b_I]{\delta }_S$. Thus a local conservation law $a_t=\mathrm{Div}b+$ bounded sources has the Rankine-Hugoniot conditionThis is also the shrinking moving-pillbox calculation of the global law: finite volume supplies have no surviving surface term.

To match the displayed positive-sign momentum jump in the paper, use its speed parameter as $V=-c$, so the interface travels in direction $-V{n}^0$. Assuming continuous ${\rho }_0$ and no surface force or energy supply, the jump laws are thenand, from ${\partial }_t({\rho }_0\eta )+\mathrm{Div}(q^0/\theta )\ge {\rho }_{0}r/\theta$,The last inequality is nonnegative entropy production concentrated on the interface. Temperature and flux in the entropy term are the traces on their respective sides; one cannot take a common temperature outside the jump unless it is continuous.

If $V$ is instead defined as the interface speed in direction $n^0$, all three $V$ terms above acquire a minus sign. In particular the printed positive-sign momentum formula requires the opposite speed convention; changing just the order of the jump does not remove this convention issue. If reference density is discontinuous, replace ${\rho }_0[a]$ by the jump $[{\rho }_{0}a]$ of the complete conserved density.

An equivalent internal-energy jump follows by subtracting the kinetic-energy jump using $[|v{|}^2/2]=\{v_i\}[v_i]$, where $\{a\}=(a^{+}+a^{-})/2$:The averaged traction appears because $[P_{Ii}{v}_i]=\{P_{Ii}\}[v_i]+[P_{Ii}]\{v_i\}$; using a one-sided traction would generally be incorrect.

Take any fixed material subvolume $B_0$ of the reference configuration, with outward normal vector $N$, current particle velocity $v_i={\dot{x}}_i$, and positive temperature $\theta$. The first law of thermodynamics, including kinetic energy, isThe heat-flux sign is negative because $N_I{q}_I^0$ is outward. The Second law of thermodynamics supplies the corresponding integral Lagrangian entropy inequality:The temperature in each supply or flux term is the local temperature, not a common temperature pulled outside the integrals.

Apply the divergence theorem and use the reference momentum conservation equation ${\rho }_0{\dot{v}}_i=P_{Ii,I}+{\rho }_0{g}_i$. In the total-energy equation, the kinetic derivative cancels the body-force work and the $v_i{P}_{Ii,I}$ term. Since $v_{i,I}={\dot{F}}_{iI}$, the remaining local Lagrangian internal energy balance and Lagrangian entropy inequality areThey hold pointwise because $B_0$ was arbitrary. In particular,

The specific Helmholtz free energy is $\psi =u-\theta \eta$. Substitution yields its balanceand elimination of the heat supply yields the Clausius-Duhem inequalityFor $\psi =\psi (F,\theta ,{\xi }_r)$, expand the derivative to obtainThe usual Coleman-Noll procedure admits independent reversible choices of $\dot{F}$ and $\dot{\theta }$ at a fixed state. Their coefficients must vanish, since either sign and arbitrary magnitude would otherwise violate the inequality. ThusInternal-variable evolution and heat conduction must satisfy this remaining inequality. Substituting the constitutive relations back into the energy balance givesThe last printed equation is missing a factor ${\rho }_0$ on its internal-variable term. The PDF defines $f_r=-{\psi }_{,{\xi }_r}$ with $\psi$ per unit mass, so the expression above is the dimensionally consistent consequence. Equivalently define the volumetric thermodynamic force conjugate to an internal variable ${\hat{f}}_r=-{\rho }_0{\psi }_{,{\xi }_r}$; the term is then ${\hat{f}}_r{\dot{\xi }}_r$ with no extra density. The two conventions must not be mixed.

For thermoelasticity with a temperature-dependent constraint, admissible rates obeyThe reversible coefficient functional need now vanish only on this tangent hyperplane. Hence it can be a scalar multiple $q$ of the constraint normal:ThereforeThe reaction terms cancel on admissible rates, leaving the same internal-variable/conduction inequality and entropy heating equation. The multiplier $q$ is determined by equilibrium and boundary conditions rather than by the free energy alone.

For linear thermoelasticity with prescribed thermal volume, write $F=I+H$, $H_{ij}=u_{i,j}$, and $\vartheta =\theta -{\theta }_0$. Hold the internal variables fixed in this reversible linearization, and evaluate all derivatives below at $(I,{\theta }_0)$ and zero reaction. The stress-free reference state has ${\psi }_{,F}=0$. Since $\partial \det F/\partial F=I$ there, first-order nominal and Cauchy stress tensors agree. Their constitutive laws becomewhere equality of mixed free-energy derivatives givesThe linearized constraint is $u_{i,i}=h^{\prime }({\theta }_0)\vartheta$. Thus volume is thermally prescribed; strict temperature-independent incompressibility corresponds to $h^{\prime }=0$. The coefficient $C_e$ is an entropy slope; the corresponding volumetric heat capacity at fixed deformation is ${\theta }_0{C}_e$.

Together with ${\rho }_0{\ddot{u}}_i={\sigma }_{ji,j}+{\rho }_0{g}_i$, the first-order thermal balance is ${\theta }_0(C_e\dot{\vartheta }-{\beta }_{ji}{\dot{u}}_{i,j}+h^{\prime }({\theta }_0)\dot{q})={\rho }_{0}r-q_{I,I}^0$. A heat-flux constitutive law completes the evolution problem. The paired signs of the coupling tensor follow from the same free energy, rather than from separately postulated mechanical and thermal equations.

Use the material-first nominal stress tensor convention $P_{Ii}$, with $F_{iI}$ the deformation gradient. For any fixed reference material subvolume $V_0$ with outward unit normal $N$, conservation of energy isHere $b$ is body force per mass, $r$ is heat supply per mass and $q$ is nominal heat flux. Using momentum conservation to remove the kinetic energy balance gives the integral Lagrangian internal energy balance:The integral Lagrangian entropy inequality iswhere $\theta >0$ is temperature and $q/\theta$ is the thermodynamic entropy flux.

Since the subvolume is arbitrary, the divergence theorem givesExpand $\dot{U}=U_{,F_{iI}}{\dot{F}}_{iI}+U_{,\eta }\dot{\eta }+U_{,{\xi }_r}{\dot{\xi }}_r$ and eliminate the heat supply between these relations. The Clausius-Duhem inequality becomesUnder the usual local constitutive independence assumptions, the reversible mechanical and thermal variations can have either sign. The Coleman-Noll procedure therefore imposesDefine the thermodynamic force conjugate to an internal variable by $f_r=-{\rho }_0{U}_{,{\xi }_r}$. The energy balance and the remaining entropy constraint now give, respectively,