I What does the Lattice Boltzmann Equation actually say?

[Swamp Thing]

From this article on the Lattice Boltzmann Method for CFD simulation...
https://feaforall.com/creating-cfd-solver-lattice-boltzmann-method/

The distribution of particles in phase space can be defined by a function f(x,e,t). It corresponds to the probability for a particle at a position [x] to possess a velocity [e] at a time [t].

In the absence of compression or relaxation, the spatial distribution remains essentially unchanged over time. At low Mach and in the absence of collision, the differential of f is conserved. We can thus deduce the following equality:

I am trying to understand what this is saying. For example, let's choose an "e0" at time t0 and position x0, such that the probability maximum with respect to changes in e, occurs at e0. That is, f peaks at e0 when we vary e while keeping x=x0, t=t0. (Assume equilibrium conditions). Now if there are no collisions etc, I would expect that this peak probability would occur at the same value e = e0 even if we look at another instant in time t1 and another point in space x1. But the equation seems to suggest that the peak probability value could shift to e0 + delta_x / delta_t when we move to (x1, t1). But I would expect to be off-peak for this new value of e1 = e0 + delta_x / delta_t.

What am I getting wrong?

 

[div_grad]

@Swamp Thing In (statistical) physics, the phase space of a system generally refers to the space of positions ($x$) and momenta ($p$) of the constituent parts of that system, so that the statistical distribution function $f(x,p,t)$ in the phase space is a function of $x$ and $p$ (and time $t$) rather than the function of positions and velocities $v$. However, if there are no external electromagnetic fields acting on the system, then the relation between momentum and velocity is simple, $p = mv$ ($m$ is the mass), so in this case you can rewrite the statistical result in terms of $v$ (or $e$, as it is called in the article you refer to).

Under conditions of statistical equilibrium one can then show that the distribution function $f(x,p,t)$ is constant in time along the trajectory in the phase space (this is Liouville's theorem),
$$
\frac{\mathrm{d}f}{\mathrm{d}t} = 0 \rm{.}
$$
The above equation is also called the Liouville's equation. Since $f = f(x,p,t)$ is a function of three variables (here we work in 1 dimension, for simplicity), the total time derivative in the Liouville's equation above can be rewritten using the simple chain-rule of differentiation:
$$
\frac{\mathrm{d}f(x,p,t)}{\mathrm{d}t} = \frac{\partial f}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}t} + \frac{\partial f}{\partial p}\frac{\mathrm{d}p}{\mathrm{d}t} + \frac{\partial f}{\partial t} \rm{.}
$$
Now, the term $\frac{\mathrm{d}p}{\mathrm{d}t} \equiv F$ above is just the external force $F$ (this is the known Newton's 2nd law). If there are no external forces ($F=0$) then this term drops out from the Liouville's equation. Noting that the term $\frac{\mathrm{d}x}{\mathrm{d}t} \equiv v$ above is just the velocity $v$, the Liouville's equation can be rewritten as
$$
\frac{\partial f}{\partial t} + v\frac{\partial f}{\partial x} = 0 \rm{,}
$$
which is also given in the article you referred to, albeit there $v \equiv e$ (a difference in notation only).

The Liouville's equation is valid under statistical equilibrium conditions, whence the distribution function $f(x,p,t)$ is constant along the phase trajectory. However, when the system is not in the equilibrium but rather tends to it (processes associated with tending towards the statistical equilibrium conditions are called the kinetic processes, and so the relevant statistical-mechanical equations are often called the kinetic equations) then the right-hand side in the Liouville's equation is not zero, but rather a certain functional of the distribution function $f$:
$$
\frac{\mathrm{d}f}{\mathrm{d}t} = C(f) \rm{.}
$$
This equation is called the Boltzmann equation. The functional $C(f)$ on the right-hand side is called the collision integral and is in general a very complicated integral operator. The kernel of this integral operator can be written in various ways and to a large extent is obtained through phenomenological reasoning. You can read about its simple version for 2-body collisions, e.g., on Wikipedia (https://en.wikipedia.org/wiki/Boltzmann_equation#Two-body_collision_term).

Again, the left-hand side of the above Boltzmann equation is rewritten using the chain-rule and if there are no external forces $F$ acting on the system, the relevant Boltzmann equation takes the simplified form
$$
\frac{\partial f}{\partial t} + v\frac{\partial f}{\partial x} = C(f) \rm{,}
$$
which is also given in the article you referred to (of course, the author uses a different notation $C(t) \equiv (\partial_t f)_{\text{coll}}$ for the collision integral).

 

[Swamp Thing]

Thank you for the detailed answer. I will be reading it and trying to pick up the prerequisites from the links you provide. Meanwhile, referring to the equality in my original post, is this what that is saying?

Edit: No, that's probably not correct. I would expect the two curves to be the same if the temperature is constant over the fluid and if the macroscopic velocities are the same in the region we are looking at. So in terms of this graph, what would the visual/graphical/intuitive explanation look like?

 

[div_grad]

@Swamp Thing In the plot you show, what is on the horizontal axis and what is on the vertical axis? How did you obtain these blue and orange curves?

 

[Swamp Thing]

Sorry for not labeling the axes. The plots show f(x,e,t) as a function of e for some fixed x, t. They are not to scale but just represent the general shape of the Maxwell-Boltzmann distribution.
In Mathematica:

Plot[Evaluate[Table[(x)^2 Exp[-((x)/a)^2], {a, {5, 5.5}}]], {x, 0, 15},
  PlotRange -> All]

It's based on this from the Wikipedia article on the Maxwell-Boltzmann Distribution:

The tutorial I referred to in my original post shows similar-looking plots and says they are based on this distribution.

 

[div_grad]

@Swamp Thing In this case the statistical distribution function $f_m(v)$ depends solely on the speed $v$ (I added the subscript "$m$" in $f_m$, which stands for "Maxwellian", to distinguish it from your "full" $f(x,e,t)$). This means that the probability $\mathrm{d}w$ to find the speed of a random particle between $v$ and $v+\mathrm{d}v$ is given by $\mathrm{d}w = f_m(v)\mathrm{d}v$. Now, if you introduce the most probable speed $\tilde{v} = \sqrt{2k_{\rm{B}}T / m}$ and use it to define the dimensionless ratio $\alpha \equiv v/\tilde{v}$, then said probability can be rewritten as $\mathrm{d}w = f_m(\alpha)\mathrm{d}\alpha$, where
$$
f_m(\alpha) = \frac{4}{\sqrt{\pi}} \alpha^2 e^{-\alpha^2} \rm{.}
$$
This is just the Wikipedia formula you quoted, but rewritten using a dimensionless variable $\alpha$, which may be more convenient, e.g., for plotting and using it in computer simulations (in contrast, note that in your Mathematica prompt you additionally defined the parameter "$a$").

Now, changing the speed by some $\delta v = \delta x / \delta t$ is the same as changing the dimensionless ratio $\alpha$ by some $\delta \alpha$. If you now set up the horizontal grid for $\alpha$'s and then use it to plot the Maxwellian functions: $f_m(\alpha)$ and $f_m(\alpha + \delta\alpha$), then you will find that the "height-of-the-peak" of both distributions is the same and only their positions are shifted (by the amount $\delta\alpha$).

Here is the appropriate plot I made in Python (here, $\delta\alpha = 0.2$):

Does this clarify some of your issues?

 

[Swamp Thing]

Now, changing the speed by some δv=δx/δt is the same as changing the dimensionless ratio α by some δα.

The LaTex editor isn't working properly here so I will have to write it like this:

I'm having trouble with delta_v = delta_x / delta_t. I can understand that delta_v = delta (dx/dt) but not the former.

I will read the other stuff in detail and get back if I have more questions.

Again, thanks for the help!

 

[div_grad]

I'm having trouble with delta_v = delta_x / delta_t. I can understand that delta_v = delta (dx/dt) but not the former.

OK, writing $\delta v = \delta x / \delta t$ is a slight abuse of notation on my part. But then again, the same abuse of notation occurs in the article you refer to, when the author writes (now I will use the notation of the article explicitly) $f(x + \delta x, e + \frac{\delta x}{\delta t}, t + \delta t)$, instead of writing $f(x + \delta x, e + \delta e, t + \delta t)$. The latter of these two forms is correct.

 

[Swamp Thing]

the same abuse of notation occurs in the article you refer to,

OK, I think that was the source of most of my confusion. And the rest of your explanations are also helping me get my head around the idea. Thanks again.