3
$\begingroup$

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

http://en.wikipedia.org/wiki/Boltzmann_equation

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

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

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ 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. $\endgroup$
    – Kyle Kanos
    Commented Sep 16, 2013 at 13:57
  • $\begingroup$ 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. $\endgroup$
    – Michael
    Commented Sep 16, 2013 at 14:19

1 Answer 1

Reset to default
3
$\begingroup$

G.J.M. Hagelaar and L.C. Pitchford give an elegant derivation of fluid equations in the scope of two-term formulation of the Boltzmann equation. Yours equation above appears in (39) (see "Solving the Boltzmann equation to obtain electron transport coefficients and rate coefficients for fluid models", http://dx.doi.org/10.1088/0963-0252/14/4/011)

Improve this answer
$\endgroup$
1
  • $\begingroup$ Thanks, yeah I know the BOLZIG+ reference - everyone cites it when they use the electron collision cross sections that come with it :) $\endgroup$
    – Nikolaj-K
    Commented Nov 12, 2014 at 9:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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