# 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:

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

• 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. – user28077 Aug 18 '18 at 10:14
• @LonelyProf Thank you! In fact, standard LB frameworks (e.g. LBGK) expand the equilibrium distribution ($f_{i}^{eq}$) just up to the second order and the higher order terms (e.g. heat flux) are neglected. As a result, the heat flux (i.e. energy flux) is considered as a ghost moment in conventional LB frameworks. So, in that sense, you are completely right about standard LB equation. But, here I pointed to expansion of equilibrium distribution ($f_{i}^{eq}$) up to third order and it means it should be capable to recover the energy conservation equation. Suggestion or comment is appreciated. – Mehrdad Yousefi Aug 19 '18 at 14:21
• My understanding of LB is fairly basic, so I'm still having difficulty understanding how this would work out. However, the main aim was to seek clarification of the question, and your comment has done that. – user28077 Aug 19 '18 at 14:35
• I have one other suggestion. Most of the posts regarding LB on this site concern its computational implementation, whereas your question seems more related to the fundamental physics. You might have greater success on Physics Stackexchange. If you wanted to transfer it, you can use the "flag" link at the bottom of the question and request that a moderator do it; or alternatively, you could post a copy of the question on that site and delete this one. It's up to you, of course, you might prefer to hold out here for a while longer. – user28077 Aug 19 '18 at 14:40