4.2.1. Non-Equilibrium Thermodynamics of Reynolds Stress

Let the state variables be **x** = (*ρ*, **u**,*s*, **R**). The Poisson bracket expressing kinematics of **x** is

$$\begin{split} \left\{ A, B \right\}^{(\mathbf{R})} &= -\left\{ A, B \right\}^{(\mathbf{F} \mathbf{M})} + \int \mathrm{d} \mathbf{r} \mathrm{R}\_{ij} \left( \partial\_{\mathbf{k}} A\_{R\_{ij}} B\_{\mathbf{u}\_{\mathbf{k}}} - \partial\_{\mathbf{k}} B\_{R\_{ij}} A\_{\mathbf{u}\_{\mathbf{k}}} \right) - \\ &- \int \mathrm{d} \mathbf{r} \mathrm{R}\_{ij} \left( \left( A\_{R\_{ij}} + A\_{R\_{jk}} \right) \partial\_{\mathbf{k}} B\_{\mathbf{u}\_{\mathbf{i}}} - \left( B\_{R\_{ij}} + B\_{R\_{jk}} \right) \partial\_{\mathbf{k}} A\_{\mathbf{u}\_{\mathbf{i}}} \right), \end{split} \tag{45}$$

see, e.g., [14,30]. The reversible evolution equations are then

$$\frac{\partial \rho}{\partial t}\_{i} = -\partial\_{i} \left(\rho E\_{u\_{i}}\right),\tag{46a}$$

$$\begin{split} \frac{\partial \boldsymbol{u}\_{i}}{\partial t} &= -\boldsymbol{\partial}\_{\dot{j}} \left( \boldsymbol{u}\_{i} \boldsymbol{E}\_{\boldsymbol{u}\_{j}} \right) - \rho \boldsymbol{\partial}\_{i} \boldsymbol{E}\_{\rho} - \boldsymbol{u}\_{j} \boldsymbol{\partial}\_{i} \boldsymbol{E}\_{\boldsymbol{u}\_{j}} - s \boldsymbol{\partial}\_{i} \boldsymbol{E}\_{\boldsymbol{s}} - \\ &- R\_{kj} \boldsymbol{\partial}\_{i} \boldsymbol{E}\_{R\_{kj}} - \boldsymbol{\partial}\_{k} \left( R\_{ij} \left( \boldsymbol{E}\_{R\_{kj}} + \boldsymbol{E}\_{R\_{jk}} \right) \right), \end{split} \tag{46b}$$

$$\frac{\partial s}{\partial t}\_{\alpha} = -\partial\_i \left( s E\_{\mathbb{U}\_i} \right), \tag{46c}$$

$$\frac{\partial R\_{ij}}{\partial t} = -\partial\_k \left( R\_{ij} E\_{u\_k} \right) - R\_{kj} \partial\_i E\_{u\_k} - R\_{ki} \partial\_j E\_{u\_k}.\tag{46d}$$

Let entropy be given by

$$S^{(\mathbf{R})} = \int \mathrm{d}\mathbf{r} \left[ s \left( \rho\_\prime \varepsilon - \frac{\mathbf{u}^2}{2\rho} - \frac{1}{2m} \mathrm{Tr} \mathbf{R} \right) + \frac{1}{2} k\_B \frac{\rho}{m} \ln \det \frac{\mathbf{R}}{Q\_R} \right],\tag{47}$$

where *QR* is an appropriately chosen constant. The reason for this entropy is analogical to the entropy (35). A derivative of entropy (47) with respect to **R** is

$$R\_{ij}^{\*} = \frac{\partial S^{(\mathbf{R})}}{\partial R\_{ij}} = -\frac{1}{2m} S\_{\mathbf{c}} \delta\_{ij} + \frac{k\_B \rho}{2m} (\mathbf{R}^{-1})\_{ij\prime} \tag{48}$$

which is equal to zero if and only if

$$R\_{ij} = k\_B T \rho \delta\_{ij}.\tag{49}$$

This is the MaxEnt value of **R**, at which the Reynolds stress is proportional to the unit matrix. Similarly as in the preceding section, we choose dissipation potential

$$\Xi^{(\mathbf{R})} = \int d\mathbf{r} \Lambda\_R \mathcal{R}\_{ij}^\* \mathcal{R}\_{jk} \mathcal{R}\_{ki'}^\* \tag{50}$$

the derivative of which is

$$
\Xi\_{R\_{ij}^{\ast}}^{(\mathbf{R})} = 2\Lambda\_{\mathbf{R}} R\_{jk} R\_{ki}^{\ast}.\tag{51}
$$

Note that *Rij* was identified with *Rij* and similarly for **R**<sup>∗</sup> for simplicity of notation. Evolution equation of **R** then gains an irreversible term

$$\frac{\partial \Xi^{(\mathbf{R})}}{\partial \mathcal{R}\_{ij}^{\*}}\Big|\_{\mathbf{R}^{\*} = \mathcal{S}\_{\mathbf{R}}} = -\frac{\Lambda\_{\mathcal{R}}}{Tm} \left(\mathbf{R} - k\_{\mathcal{B}} T \rho \mathbf{I}\right). \tag{52}$$

By combining the reversible evolution (46) and irreversible (52), we obtain

$$\begin{array}{rcl}\frac{\partial\rho}{\partial t} &=& -\partial\_i \left(\rho E\_{u\_i}\right),\\\frac{\partial}{\partial t}\end{array} \tag{53a}$$

$$\frac{\partial \boldsymbol{u}\_{i}}{\partial t} = -\partial\_{\boldsymbol{j}} \left( \boldsymbol{u}\_{i} \boldsymbol{E}\_{\boldsymbol{u}\_{j}} \right) - \rho \partial\_{\boldsymbol{i}} \boldsymbol{E}\_{\rho} - \boldsymbol{u}\_{j} \partial\_{\boldsymbol{i}} \boldsymbol{E}\_{\boldsymbol{u}\_{j}} - \boldsymbol{s} \partial\_{\boldsymbol{i}} \boldsymbol{E}\_{\boldsymbol{s}} - \boldsymbol{s} \partial\_{\boldsymbol{i}} \boldsymbol{E}\_{\boldsymbol{s}} - \boldsymbol{s} \partial\_{\boldsymbol{i}} \boldsymbol{E}\_{\boldsymbol{s}}$$

$$-R\_{kj}\partial\_i E\_{R\_{kj}} - \partial\_k \left( R\_{ij} \left( E\_{R\_{kj}} + E\_{R\_{jk}} \right) \right) \\ \text{ } \tag{53b}$$

$$\frac{\partial \mathbf{s}}{\partial t} = -\partial\_i \left( s E\_{\mu\_i} \right) + 2 \frac{\Lambda\_R}{(E\_s)^2} E\_{R\_{ij}} R\_{jk} E\_{R\_{ki}} \tag{53c}$$

$$\begin{aligned} \frac{\partial R\_{ij}}{\partial t} &= -\partial\_k \left( R\_{ij} E u\_k \right) - R\_{kj} \partial\_i E u\_k - R\_{ki} \partial\_j E u\_k \\ &- 2 \frac{\Lambda\_R}{E\_s} R\_{jk} E\_{R\_{ki}} \end{aligned} \tag{53d}$$

which are the GENERIC evolution equations for fluid mechanics with Reynolds stress. For instance, the last equation can be rewritten as

$$\frac{\partial R\_{ij}}{\partial t} = -\partial\_k \left( R\_{ij} v\_k \right) - R\_{kj} \partial\_l v\_k - R\_{ki} \partial\_j v\_k$$

$$-\frac{\Lambda\_R}{Tm} \left( \mathbf{R} - k\_B T \rho \mathbf{I} \right), \tag{54}$$

from which the tendency to relaxation of the Reynolds stress tensor to the respective MaxEnt value is obvious.
