# Derive total energy balance equation from Chapman-Enskog analysis of lattice Boltzmann equation

I'm interested to derive the total energy balance from Chapman-Enskog analysis of lattice Boltzmann equation (LBE). I know, I should go to the second moment of LBE (zeroth moment gives mass conservation and first moment gives momentum conservation). In fact, I derived the main equation for the second moment at convective ($t_{1}$) and diffusive ($t_{2}$) time scales as:

From Chapman-Enskog analysis up to the second order:

Convective time scale: $\frac{\partial f_{i}^{eq}}{\partial t_{1}} + \vec{e_{i}} \cdot \nabla_{1} f_{i}^{eq} = -\frac{f_{i}^{(1)}}{\tau}$

Diffusive time scale: $\frac{\partial f_{i}^{eq}}{\partial t_{2}} + (1-\frac{1}{2 \tau}) \Big [ \frac{\partial f_{i}^{(1)}}{\partial t_{1}} + \vec{e_{i}} \cdot \nabla_{1} f_{i}^{(1)} \Big ] = -\frac{f_{i}^{(2)}}{\tau}$

Where $f_{i}^{eq}$ is the equilibrium distribution function (expanded Maxwell-Boltzmann up to third order):

$f_{i}^{eq} = \rho \omega_{i} (1 + \frac{\vec{e_{i}} \cdot \mathbf{u}}{c_{s}^{2}} + \frac{(\vec{e_{i}} \otimes \vec{e_{i}} - \mathbf{I})\cdot \mathbf{u} \otimes \mathbf{u}}{2 c_{s}^{4}} + \frac{(\vec{e_{i}} \otimes \vec{e_{i}} \otimes \vec{e_{i}} - 3 \vec{e_{i}} \otimes \mathbf{I}) \cdot \mathbf{u} \otimes \mathbf{u} \otimes \mathbf{u}}{6 c_{s}^{6}})$

$f_{i}^{(1)}$ and $f_{i}^{(2)}$ are the first and second order terms of distribution function Taylor's expansion.

We know the total energy and its flux could be derived as:

$\rho e = \rho (\frac{3}{2} U^{2} + \frac{1}{2} |\mathbf{u}|^{2}) = \sum_{i} \frac{1}{2} \vec{e_{i}} \cdot \vec{e_{i}} f_{i}^{eq}$

$\rho e \mathbf{u} = \rho (\frac{3}{2} U^{2} + \frac{1}{2} |\mathbf{u}|^{2}) \mathbf{u} = \sum_{i} \frac{1}{2} \vec{e_{i}} \cdot \vec{e_{i}} \vec{e_{i}} f_{i}^{eq}$

Finally, we have:

Convective time scale :$\frac{\partial}{\partial t_{1}} (\rho e) + \nabla_{1} \cdot (\rho e \mathbf{u}) = -\frac{1}{\tau} \sum_{i} \frac{1}{2} f_{i}^{(1)} \vec{e_{i}} \cdot \vec{e_{i}}$

Diffusive time scale: $\frac{\partial}{\partial t_{2}} (\rho e) + (1-\frac{1}{2 \tau}) \frac{\partial}{\partial t_{1}} (\sum_{i} \frac{1}{2} \vec{e_{i}} \cdot \vec{e_{i}} f_{i}^{(1)}) + (1-\frac{1}{2 \tau}) \nabla_{1} \cdot (\sum_{i} \frac{1}{2} \vec{e_{i}} \cdot \vec{e_{i}} \vec{e_{i}} f_{i}^{(1)}) = -\frac{1}{\tau} \sum_{i} \frac{1}{2} f_{i}^{(2)} \vec{e_{i}} \cdot \vec{e_{i}}$

The terms $\sum_{i} \frac{1}{2} f_{i}^{(1)} \vec{e_{i}} \cdot \vec{e_{i}}$ and $\sum_{i} \frac{1}{2} f_{i}^{(2)} \vec{e_{i}} \cdot \vec{e_{i}}$ are equal to zero because energy is a conserved quantity.

As a result:

Convective time scale :$\frac{\partial}{\partial t_{1}} (\rho e) + \nabla_{1} \cdot (\rho e \mathbf{u}) = 0$

Diffusive time scale: $\frac{\partial}{\partial t_{2}} (\rho e) + (1-\frac{1}{2 \tau}) \nabla_{1} \cdot (\sum_{i} \frac{1}{2} \vec{e_{i}} \cdot \vec{e_{i}} \vec{e_{i}} f_{i}^{(1)}) = 0$

Right now, my question is how can I simplify the term $(1-\frac{1}{2 \tau}) \sum_{i} \frac{1}{2} \vec{e_{i}} \cdot \vec{e_{i}} \vec{e_{i}} f_{i}^{(1)}$ to prove that its divergence is equal to $\nabla \cdot \mathbf{q} - \nabla \cdot (\sigma \cdot \mathbf{u})$, where $\mathbf{q}$ is the heat flux and $\sigma$ is the Cauchy stress tensor:

$\nabla_{1} \cdot ((1-\frac{1}{2 \tau}) \sum_{i} \frac{1}{2} \vec{e_{i}} \cdot \vec{e_{i}} \vec{e_{i}} f_{i}^{(1)}) = \nabla \cdot \mathbf{q} - \nabla \cdot (\sigma \cdot \mathbf{u})$

Finally, we could find the total energy balance equation as:

$\frac{\partial}{\partial t} (\rho e) + \nabla \cdot (\rho e \mathbf{u}) = -\nabla \cdot \mathbf{q} + \nabla \cdot (\sigma \cdot \mathbf{u})$