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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.