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:

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

  • Apologies in advance if this is a stupid comment! As I understand it, the standard lattice-Boltzmann equation does not conserve energy (only mass and momentum). Generalizing it to an energy-conserving form can be done in several ways, some of which are mentioned on the Wikipedia page. So, it might clarify your question if you could say which version of LB you are considering. – LonelyProf 9 hours ago

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