Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Starting with the Boltzmann equation in terms of $f(t,\vec v,\vec x)$ or $f(t,\vec v)$


$$\left(\frac{\partial}{\partial t} + \vec{v} \, \nabla_\vec{x} + \frac{1}{m} \vec{F}\cdot \nabla_\vec{v}\right) f(\vec{x},\vec{v},t) = \left.\frac{\partial f}{\partial t}\right|_\mathrm{collision}\,,$$

how to derive the "two-term approximation", which is expressed in terms of an energy-dependend $f(t,\varepsilon)$?

That one essentially looks like

$$\frac{\partial}{\partial t}f(t,\varepsilon)=\frac{\partial}{\partial \varepsilon}\left(\left(A\ \varepsilon + c\ B\right)\ f(t,\varepsilon)+\frac{\partial}{\partial \varepsilon}\left(B\ \varepsilon\ f(t,\varepsilon)\right)\right)\,.$$

And I think a relatedkeyword is "Fokker-Planck type".

I have here the book "Mathematical theory of transport processes in gases", which discusses the Boltzmann equation in detail, but as far as I can see it never passes to the energy-representation. Searching the web, I found a version of such a $f(t,\varepsilon)$-reformulation described in the software manula for the "Comsol" software, although no derivation of any kind: www.comsol.com/model/download/31731/two_term_boltzmann.pdf

share|improve this question
The Comsol link explains how it can be derived: use the relation $\varepsilon=mv^2/(2e)$ and $f\left(\vec{v}\right)=f_0\left(v\right)+f_1\left(v\right)\cos\theta$. It should be straight-forward from there. –  Kyle Kanos Sep 16 '13 at 13:57
The "two term" approximation refers to the ansatz for $f(\vec{v})$ @KyleKanos mentions. It is the lowest two terms in a systematic expansion in spherical harmonics of the distribution function (assuming axial symmetry). Plug this in the Boltzmann equation and note that once collected together the terms with different angular dependencies must separately vanish. So you get two equations for $f^{(0)},f^{(1)}$ which only depend on $|\vec{v}|$ and not angles. –  Michael Brown Sep 16 '13 at 14:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.