4.2.2. DynMaxEnt to Hydrodynamics

As in the case of the conformation tensor in Section 4.1, let us now show how the Reynolds stress relaxes. The MaxEnt value (49) can be plugged into Equations (53). Assuming again isothermal incompressible flow, *<sup>ρ</sup>* <sup>=</sup> const, ∇ · **<sup>u</sup>**† <sup>=</sup> 0 and *Se* <sup>=</sup> *<sup>T</sup>* <sup>=</sup> const, the equations become

$$\frac{\partial \boldsymbol{u}\_{i}}{\partial t} = -\boldsymbol{\partial}\_{\dot{j}} \left( \boldsymbol{u}\_{i} \boldsymbol{u}\_{j}^{\dagger} \right) - \rho \boldsymbol{\partial}\_{i} \rho^{\dagger} - \boldsymbol{u}\_{j} \boldsymbol{\partial}\_{i} \boldsymbol{u}\_{j}^{\dagger} - \boldsymbol{s} \boldsymbol{\partial}\_{i} \mathbf{s}^{\dagger} - $$
 
$$ -k\_{B} T \rho \boldsymbol{\partial}\_{i} \boldsymbol{R}\_{kk}^{\dagger} - k\_{B} T \rho \boldsymbol{\partial}\_{k} \left( \boldsymbol{R}\_{ki}^{\dagger} + \boldsymbol{R}\_{ik}^{\dagger} \right), \tag{55a} $$

$$0 \quad = \quad -k\_B T \rho \left(\partial\_i u\_j^\dagger + \partial\_j u\_i^\dagger\right) - 2\Lambda\_R k\_B \rho \mathcal{R}\_{ji}^\dagger. \tag{55b}$$

The last equation has a solution

$$\ddot{\mathbf{R}}^{\dagger} = -\frac{T}{2\Lambda\_R} \left( \nabla \mathbf{u}^{\dagger} + (\nabla \mathbf{u}^{\dagger})^T \right) \qquad \text{and} \qquad \text{Tr}\tilde{\mathbf{R}} = 0. \tag{56}$$

Plugging this solution back into the equation for **u** leads to

$$\begin{split} \frac{\partial u\_{l}}{\partial t} &= -\eth\_{\dot{\jmath}} \left( u\_{l} u\_{\dot{\jmath}}^{\dagger} \right) - \rho \eth\_{l} \rho^{\dagger} - u\_{\dot{\jmath}} \eth\_{l} u\_{\dot{\jmath}}^{\dagger} - s \eth\_{l} s^{\dagger} \\ &+ \frac{k\_{B} T^{2} \rho}{\Delta\_{R}} \eth\_{k} \left( \eth\_{k} u\_{l}^{\dagger} + \eth\_{l} u\_{k}^{\dagger} \right), \end{split} \tag{57}$$

which is again the Navier–Stokes equation with shear viscosity. Relaxation of Reynolds stress thus leads to extra (also called turbulent) viscosity by means of the Dynamic MaxEnt reduction.
