The Boltzmann equation without collision operator $\Omega$ is as follows: $$\dfrac{\partial f}{\partial t} + \mathbf v \cdot \nabla f = 0 \tag{1}$$ Where $\mathbf v$ is the velocity, and $f$ is the distribution function which is a function of position, velocity, and time.

The Maxwell-Boltzmann distribution function is: $$f(\mathbf v) = 4\pi v^2\left( \dfrac{m}{2k\pi T}\right)^{3/2}e^{-\dfrac{mv^2}{2kT}}\tag{2}$$ My question is: what is the relation between the equation $(1)$ and the function $(2)$? is this latter function a solution to the equation $(1)$? if it is, I do not see explicitly the time and the position in the function $f$


Your equation (2) is trivially a solution of (1), because $v$ and $T$ are constant. This is a disappointing answer, because it leaves unanswered the question what makes the Boltzmann distribution unique. The answer is that you only wrote down the collision-less Boltzmann equation, but in the real world collisions are always present (and indeed, systems cannot equilibrate without them). The Boltzmann distribution is unique in the sense that the collision term $C[f]$ vanishes. This follows easily for a two-body collision term that conserves enrgy $E\sim v^2$. If $$ C[f_1]\sim \int d\Gamma_{234} \,(f_1f_2-f_3f_4)\, w(12;34) $$ where $f_i=f(v_i)$, $d\Gamma_i=d^3p_i/(2\pi)^3$ and $w(12;34)$ is the collision rate for $1+2\to 3+4$ scattering. Then $$ C[f_1^{eq}] \sim \int d\Gamma_{234}\, \exp(-(E_1+E_2-E_3-E_4)/T)\, w(12;34) \sim 0. $$

You can generalize the Boltzmann distribution to include a potential, $f\sim \exp(-V(x)/T)$. Then the left hand side of the Boltzmann equation becomes more interesting, gradients of $f$ have to cancel against the force term in the Boltzmann equation (which you also do not write down).

Finally, we can make things even more interesting by allowing $T$ and $v$ to depend on position and time. Then the Boltzmann distribution is an (approximate) solution of the Boltzmann equation provided, $T(x,t)$ and $v(x,t)$ satisfy the Euler (Navier-Stokes) equation of fluid dynamics.

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    $\begingroup$ but if any function for example $f(v) = v^2$ is also a solution based on your answer. $\endgroup$ – Syntax_ErrorX00 Jun 28 '16 at 22:01
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    $\begingroup$ That's my point. You have to include the collision term. Then, only $f\sim \exp(-v^2)$ is a solution. $\endgroup$ – Thomas Jun 28 '16 at 22:22
  • $\begingroup$ the collision operator is a function of $f$,so the equation will be an-integro-differential which is difficult to solve. therfore I need to use the BGKW approximation of the collision operator as follows: $$\Omega = \dfrac 1\tau \left(f^{eq}-f \right)$$ $\endgroup$ – Syntax_ErrorX00 Jun 28 '16 at 22:28
  • $\begingroup$ You don't have to solve the Boltzmann equation to show that $C[f^{eq}]=0$. I added a short proof to my answer. For the BGK kernel the collision term obviously vanishes by definition, $C_{BGK}[f^{eq}]=(f^{eq}-f^{eq})/\tau=0$. $\endgroup$ – Thomas Jun 28 '16 at 23:30
  • $\begingroup$ @Thomas this is a good answer. Maybe you could explain some of the variables in your collision operator? $\endgroup$ – anon01 Jun 28 '16 at 23:35

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