3
$\begingroup$

I am quite confused with a seemly simple question about non-relativistic Boltzmann equation,

The non-relativistic Boltzmann equation is well-explained by the following wiki page

$$\frac{\partial f}{\partial t} + \frac{\mathbf{p}}{m} \cdot \nabla f + \mathbf{F} \cdot\frac{\partial f}{\partial \mathbf{p}} = \left(\frac{\partial f}{\partial t} \right)_\mathrm{coll},$$ where, intuitively, $\mathbf{F}$ is the net force on the massive point $m$.

However, in some references, one talks about the following one-dimensional "Boltzmann equation including accelerations" (see Eq.(48) of the second reference below):

$$\frac{\partial f}{\partial t} + v \frac{\partial}{\partial x}f + \frac{\partial}{\partial v}(a(v)f) = \left(\frac{\partial f}{\partial t} \right)'_\mathrm{coll}.$$

To avoid any mistake from my part, I maintain mostly the original form of the equation as it appears in the literature and put an extra prime on the r.h.s. of the equation. The references are the following

  1. Phys.Rev.E53, 2366 (1996), Gas-Kinetic derivation of Navier-Stocks-like traffic equations Transport.

  2. Transp. Res. 9, 255 (1975), On Boltzmann-like Treatments for Traffic Flow.

Considering $f=f(x,v,t)$, the difference between the two equations comes from the derivative of $a(v)$ with respect to $v$. In the discussions, it seems to me that the collision terms are exactly those as in the first (more common version of the) Boltzmann equation, I am quite confused. What did I miss?

$\endgroup$
  • $\begingroup$ At a guess, this is related to the statement on the Wiki page: "Note that we have used the fact that the phase space volume element $\operatorname{d}^3\mathbf{r} \operatorname{d}^3\mathbf{p}$ is constant, which can be shown using Hamilton's equations (see the discussion under Liouville's theorem). " Try relaxing that, and see if you get the desired derivative structure. The justification for relaxing that is if you're using forces that don't have a Hamilton's equation form (e.g. drag forces). $\endgroup$ – Sean E. Lake Nov 1 '16 at 15:26
  • $\begingroup$ If you do, in the second equation, $a(v)\equiv a$ a constant and do the change of variables to write with $p$ instead of $v$, then you find the first equation. I think first one assumes that the net force $\vec{F}$ is a constant, and in the second one they assume it depends on $v$. $\endgroup$ – VictorSeven Nov 1 '16 at 16:08
  • $\begingroup$ @VitorSeven Thx for the hint. In the first reference, it is written out explicitly that the acceleration $a$ is a function of $v$, $a=(v_0-v)/\tau$, which carries the content of relaxation acceleration, that made me confused. I also looked at the reference therein, by Paveri-Fontana, he did not explain much when the Boltzmann equation was introduced. So I am not sure it is ``well-known" as an alternative form of Boltzmann equation. $\endgroup$ – gamebm Nov 6 '16 at 15:13
1
$\begingroup$

The Boltzmann equation written with velocities follows directly from the Boltzmann equation written with momenta by replacing $\vec p=m\vec v$ by $\vec v$. In the one dimensional equation, you just leave out the other coordinates and velocities. The acceleration $\vec F/m=\vec a$ is already included in the first equation. Thus from the first equation you should obtain $$\frac{\partial f}{\partial t} + v \frac{\partial}{\partial x}f + a\frac{\partial}{\partial v}f = \left(\frac{\partial f}{\partial t} \right)'_\mathrm{coll}.$$ Putting the acceleration in front of the partial derivative $\frac {\partial f} {\partial v}$ like in the second equation is not consistent with the derivation of the Boltzmann equation. Therefore, it is likely that the second equation is not a real Boltzmann transport equation.

Extension of answer. After having had a look at the cited paper (especially the first) it becomes clear that the transport equations there, which correspond to your second equation with the acceleration under the derivative are similar but not the same as the Boltzmann equation. The difference lies in the terms describing the particle number conservation without collisions in phase space which is the continuity equation in phase space $$\frac {\partial f}{\partial t}+∇_{x,v}·\vec V f=0$$ where $\vec V=(dx/dt,dv/dt)$ is the "velocity vector" in phase space. In your second equation this leads to the term $\frac{\partial}{\partial v}(a(v)f)$ with the acceleration $a$ under the derivative. In the derivation of the corresponding terms of the actual Boltzmann equation it is, in essence, assumed that an Hamiltonian exists so that $$\frac {dx}{dt}=\frac{\partial H(x,p)}{\partial p}$$ and $$\frac {dp}{dt}=-\frac{\partial H(x,p)}{\partial x}$$ with $p=mv$. This simplifies the continuity equation in phase space to $$\frac{\partial f}{\partial t} + v \frac{\partial}{\partial x}f + a\frac{\partial}{\partial v}f=0$$ which is the collisionless Boltzmann equation. In the full Boltzmann equation the zero is replaced by the collision term. Thus your second equation (corresponding to ones found in the cited articles) is a transport equation relying on the unsimplified continuity equation in phase space without the Hamiltonian dynamics assumption of the actual Boltzmann equation. The difference is definitely not due to the collision terms which are the same. Various derivations of the Boltzmann equation including the simplifications to the continuity equation in phase space can be found e.g in F. Reif, Fundamentals of statistical and thermal physics, chapter 13.2 Boltzmann equation in the absence of collisions.

$\endgroup$
  • $\begingroup$ Many thanks for the response. But I am afraid you did not clear my doubts. I figured that it was the collision terms which might have a different meaning than in the common Boltzmann equation but cannot get better than this.... if possible, please take a look at the references. $\endgroup$ – gamebm Nov 6 '16 at 15:21
  • $\begingroup$ @gamebm - I had a look at the paper by Helbig and the chapter III. Gas-kinetic (Boltzmann-like) Traffic Models which gives the second equation (with acceleration under the derivative). The collision term is definitely not the reason for the difference to the proper Boltzmann equation. I will explain this in my extended answer above. $\endgroup$ – freecharly Nov 8 '16 at 15:33
  • $\begingroup$ That must be the case, thx! Unfortunately the one I read first was Eq.(48) of the second reference where the first reference is based on. Now I guess it is simply not correct. $\endgroup$ – gamebm Nov 8 '16 at 21:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.