Newtonian stress and Fourier conductive closure

Use a Cartesian inertial frame in Newtonian spacetime. At an event let $u$ be the fluid velocity, $T>0$ its temperature, $\rho >0$ its mass density, and $\xi =(\rho ,T)$ its local state in a specified open set $\mathcal{S}\subset (0,\mathrm{\infty }{)}^2$. The velocity gradient has components $L_{ij}={\mathrm{\partial }}_{x_j}{u}_i$; define the symmetric rate matrix $D=(L+L^T)\mathrm{/}2$, its trace $\theta =\mathrm{tr}D$, and its trace-free part $D_0=D-\theta I_3\mathrm{/}3$, where $I_3$ is the $3\times 3$ identity matrix. For $A,B\in {\mathbb{R}}^{3\times 3}$, use the Frobenius contraction $A:B={\sum }_{i,j=1}^3{A}_{ij}{B}_{ij}$.

Adopt specified scalar state laws $\mu ,\zeta :\mathcal{S}\to [0,\mathrm{\infty })$ and a specified conductivity matrix law $K:\mathcal{S}\to {\mathbb{R}}^{3\times 3}$, with $x^T\mathrm{sym}K(\xi )x\ge 0(x\in {\mathbb{R}}^3),\mathrm{sym}K:=(K+K^T)\mathrm{/}2.$ The conductivity law is tensor-valued and may be anisotropic; no restriction $K=k{I}_3$ is imposed. Define $\mathrm{Sym}(3):=\{D\in {\mathbb{R}}^{3\times 3}:D^T=D\}$ and $\mathrm{SO}(3):=\{Q\in {\mathbb{R}}^{3\times 3}:Q^{T}Q=I_3,\det Q=1\}$. At a fixed state, let ${\mathcal{A}}_{\xi }:\mathrm{Sym}(3)\to \mathrm{Sym}(3)$ assign the viscous stress to the symmetric rate matrix. Assume it is real-linear, meaning ${\mathcal{A}}_{\xi }(aD+bE)=a{\mathcal{A}}_{\xi }(D)+b{\mathcal{A}}_{\xi }(E)$ for all $a,b\in \mathbb{R}$ and $D,E\in \mathrm{Sym}(3)$, and equivariant in the sense ${\mathcal{A}}_{\xi }(QD{Q}^T)=Q{\mathcal{A}}_{\xi }(D)Q^T$ for every $Q\in \mathrm{SO}(3)$. Adopt $\tau ={\mathcal{A}}_{\xi }(D)=2\mu (\xi )D_0+\zeta (\xi )\theta I_3,\sigma =-p{I}_3+\tau ,q=-K(\xi )\mathrm{\nabla }T\mathrm{.}$ Here $p$ is the scalar pressure field in the tension-positive convention: the traction on an outward unit normal $n$ is $\sigma n$. Under a rigid observer rotation $Q\in \mathrm{SO}(3)$, tensor fields transform by $S^{\ast }=QS{Q}^T$; $\mu ,\zeta$ are scalar state functions and $K^{\ast }=QK{Q}^T$. The viscous response is isotropic in this proper-rotation sense, while the conductivity tensor transforms covariantly and may be anisotropic. The mathematical linear-isotropic classification identifies this stress form from the stated domain, codomain, linearity and equivariance hypotheses; it does not establish that any material adopts them.

In SI units, $D,\theta ,D_0$ are in ${\mathrm{s}}^{-1}$, $\mu ,\zeta$ in $\mathrm{P}\mathrm{a}\mathrm{s}$, $K$ in $\mathrm{W}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}$, $q$ in $\mathrm{W}{\mathrm{m}}^{-2}$, and $\sigma ,\tau ,p$ in $\mathrm{P}\mathrm{a}$. If the same stress is written $2\mu D+\lambda \theta I_3$, then $\lambda =\zeta -2\mu \mathrm{/}3$; no nonnegativity of $\lambda$ is assumed. These specified constitutive functions are adopted model inputs, not measured coefficient values or consequences of balance laws.

David Tong, Fluid Mechanics, Part II lecture notes, §§3.1–3.1.1, printed pp. 58–61

An Introduction to Continuum Mechanics, Second Edition, Chapter 6, §§6.5.1–6.5.3 and §6.7.2, printed pp. 242–245 and 251–252