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Bhatnagar, Gross and Krook (BGK) proposed a relaxation term for the collision integral $ Q$ as follows $$J = \frac{1}{\tau} (f^{eq} - f)$$ where $f^{eq}$ is the distribution at equilibrium. $Q$ has following property: $$\int Q \psi_k d\mathbf{v} = 0$$ where $\psi_0 = 1,\hspace{.5cm} (\psi_1, \psi_2,\psi_3) = \mathbf{v}, \hspace{.5cm} \psi_4 = \mathbf{v}^2 $

Now we say that this term should have the following constraints:

a) It should have a tendency towards Maxwellian distribution

What is meant by this expression? My understanding is that f should be close to Maxwell distribution at all times. Is this correct? if not what is the correct interpretation of this constraint?

b) J Should conserve collision invariants of Q

i.e. $\int J \psi_k d\mathbf{v}d\mathbf{x} = 0 $ where $\psi_k$ is one of the collision invariant.

Note that now integral is performed over both velocity and spatial space instead of velocity alone as in case of $Q$. $\int J \psi_k d\mathbf{v} = 0 \implies \int J \psi_k d\mathbf{v}d\mathbf{x} = 0$ but $\int J \psi_k d\mathbf{v}d\mathbf{x} = 0 \not \implies \int J \psi_k d\mathbf{v} = 0$. So why don't we put condition on J as $\int J \psi_k d\mathbf{v} = 0$? By adding extra spatial space in the integral do we put less restrictions on J?

if I substitute the first expression for J into the above constraint b), then it implies $\int f \psi_k d\mathbf{v}d\mathbf{x}$ = 0 but this implies that f should be already Maxwellian for this relation to be true! What am I missing here?

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  • $\begingroup$ Do you have a reference for the constraint "a"? For the second one, you are right for momentum ($\psi=m\vec v$): it is not conserved. However, it conserves the number of particles ($\psi=1$). Thus, the integral of $J$ is not zero for all $\psi$. $\endgroup$ – fffred Jun 3 '13 at 22:40
  • $\begingroup$ In fact, maybe the constraint (a) is simply saying that the term $J$ relaxes the distribution towards an equilibrium $f^{eq}$, not a priori maxwellian. It may not be a constraint but an interpretation. $\endgroup$ – fffred Jun 3 '13 at 22:45
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    $\begingroup$ I guess a) means that the collision integral should lead to an evolution of the system towards the equilibrium, where the Maxwell distribution holds. Could b) just mean, that the nonequilibrium distribution f has to be isotropic? Darn, I have the feeling that I should know that, I'll have to check. $\endgroup$ – Dilaton Jun 3 '13 at 22:55
  • $\begingroup$ @fffred, reference for "a" and "b" is the Textbook :Lattice-gas Cellular Automata and Lattice Boltzmann Models by Dieter A.Wolf-Gladrow $\endgroup$ – alekhine Jun 3 '13 at 23:01
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    $\begingroup$ I just read parts of the original BGK paper. It clearly states that the number of particles is instantaneously conserved because $\int Jd\mathbf{v}=0$ and the momenta and energy are NOT instantaneously conserved, i.e. $\int J\;m\mathbf{v}d\mathbf{v}\neq0$ and $\int J\;m\mathbf{v}^2d\mathbf{v}\neq0$. $\endgroup$ – fffred Jun 3 '13 at 23:09
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For (a), as stated in the comments, it is not a constraint by simply an interpretation of the term $J$ as a relaxation towards equilibrium.

For (b), $\int J \psi d\mathbf v$ is not always zero. BGK find that reducing $f^{eq}$ such as $$J=\frac{1}{\tau}\left(f^{eq}(\mathbf{v})n(\mathbf{x},t)-f\right)$$ where $n(\mathbf{x},t)=\int fd\mathbf{v}$, makes the integral of the first moment ($\psi=1$) zero, meaning that the number of particles is conserved. They show the higher moments ($\mathbf v$ and $\mathbf v^2$) to have a non-zero integral over velocities. They interpret this as a non-conservation of momentum and energy locally, but seem to indicate that they are conserved on the average over a cycle. My interpretation of this is that the total momentum and energy can only be imposed on the full system, which corresponds to the integrals on both velocity and space that you cited. You cannot impose, with this model, local conservation.

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