4.1.2. DynMaxEnt to Hydrodynamics

Let us now apply the Dynamic MaxEnt reduction so that the conformation tensor **c** relaxes. The energetic representation reversible Equations (32) and irreversible Equations (40) together are

$$\begin{array}{rclclclcl}\frac{\partial\rho}{\partial t} &=& -\partial\_{\bar{i}}\left(\rho u\_{i}^{\dagger}\right),\\ \partial\_{\bar{i}}u\_{i} &=& \left(\begin{array}{c}.\\ \end{array}\right)\_{i} & \begin{array}{c}.\\ \end{array} & \begin{array}{c}.\\ \end{array}\end{array} \tag{41a}$$

$$\begin{split} \frac{\partial \boldsymbol{u}\_{l}}{\partial t} &= -\boldsymbol{\partial}\_{\dot{j}} \left( \boldsymbol{u}\_{l} \boldsymbol{u}\_{\dot{j}}^{\dagger} \right) - \rho \boldsymbol{\partial}\_{l} \rho^{\dagger} - \boldsymbol{u}\_{\dot{j}} \boldsymbol{\partial}\_{l} \boldsymbol{u}\_{\dot{j}}^{\dagger} - s \boldsymbol{\partial}\_{l} \mathbf{s}^{\dagger} - \\ &- c\_{jk} \boldsymbol{\partial}\_{l} \boldsymbol{c}\_{jk}^{\dagger} + \boldsymbol{\partial}\_{k} \left( c\_{kj} \left( \boldsymbol{c}\_{ij}^{\dagger} + \boldsymbol{c}\_{ji}^{\dagger} \right) \right), \end{split} \tag{41b}$$

$$\frac{\partial s}{\partial t}^{\prime} = -\partial\_i \left( s u\_i^{\dagger} \right) + 2 \frac{\Lambda\_c}{(s^{\dagger})^2} c\_{ij}^{\dagger} c\_{ik} c\_{kj\prime}^{\dagger} \tag{41c}$$

$$\frac{\partial \mathbf{c}\_{ij}}{\partial t} = -\partial\_k \left( \mathbf{c}\_{ij} \boldsymbol{u}\_k^\dagger \right) + \mathbf{c}\_{kj} \partial\_k \boldsymbol{u}\_i^\dagger + \mathbf{c}\_{ki} \partial\_k \boldsymbol{u}\_j^\dagger - \frac{2\Lambda\_c}{\mathbf{s}^\dagger} \mathbf{c}\_{ik} \mathbf{c}\_{kj}^\dagger. \tag{41d}$$

Consider now the isothermal incompressible case (The compressible case and the origin of incompressibility were discussed in [19].), i.e., *Se* <sup>=</sup> *<sup>T</sup>* <sup>=</sup> const, *<sup>n</sup>* <sup>=</sup> const and ∇ · **<sup>u</sup>**† <sup>=</sup> 0. Equations (41b) and (41d) at the MaxEnt value of **c** (given by Equation (37)) become

$$\begin{split} \frac{\partial \boldsymbol{u}\_{l}}{\partial t} &= -\boldsymbol{\partial}\_{\boldsymbol{j}} \left( \boldsymbol{u}\_{l} \boldsymbol{u}\_{\boldsymbol{j}}^{\dagger} \right) - \rho \boldsymbol{\partial}\_{l} \boldsymbol{\rho}^{\dagger} - \boldsymbol{u}\_{\boldsymbol{j}} \boldsymbol{\partial}\_{l} \boldsymbol{u}\_{\boldsymbol{j}}^{\dagger} - \boldsymbol{s} \boldsymbol{\partial}\_{l} \boldsymbol{s}^{\dagger} - \\ &- \frac{k\_{B} T n}{H} \boldsymbol{\partial}\_{l} \boldsymbol{\text{Tr} \, \mathbf{c}}^{\dagger} + \frac{k\_{B} T n}{H} \boldsymbol{\partial}\_{k} \left( \boldsymbol{c}\_{\mathrm{ik}}^{\dagger} + \boldsymbol{c}\_{\mathrm{ki}}^{\dagger} \right) \end{split} \tag{42a}$$

$$0 \quad = \ \partial\_j u\_i^\dagger + \partial\_i u\_j^\dagger - 2\frac{\Lambda\_c}{T} c\_{ij}^\dagger. \tag{42b}$$

The last equation has a solution

$$\mathbf{c}^{\dagger} = \frac{T}{2\Lambda\_{c}} \left( \nabla \mathbf{u}^{\dagger} + (\nabla \mathbf{u}^{\dagger})^{T} \right), \quad \text{Tr}\mathbf{c}^{\dagger} = 0. \tag{43}$$

By plugging this solution into the equation for momentum density, we obtain the Navier–Stokes equation for momentum density

$$\begin{split} \frac{\partial u\_i}{\partial t} &= -\eth\_{\dot{\jmath}} \left( u\_i u\_{\dot{\jmath}}^{\dagger} \right) - \rho \eth\_i \rho^{\dagger} - u\_{\dot{\jmath}} \eth\_i u\_{\dot{\jmath}}^{\dagger} - s \eth\_i s^{\dagger} \\ &+ \frac{k\_B T^2 n}{H \Lambda\_{\boldsymbol{\varsigma}}} \eth\_k \left( \eth\_i u\_k^{\dagger} + \eth\_k u\_{\dot{\imath}}^{\dagger} \right), \end{split} \tag{44}$$

where the coefficient *kBT*<sup>2</sup>*n*/*H*Λ*<sup>c</sup>* corresponds to the shear viscosity and **u**† = *E***<sup>u</sup>** = **v** = **u**/*ρ* is the velocity.

The Dynamic MaxEnt reduction of the conformation tensor leads to the Newtonian shear stress tensor.
