0
$\begingroup$

Let's take into consideration the classical Boltzmann Equation for gas kinetics (i'm writing down the one obtained in Huang K. - Statistical mechanics, equation (3.36)):

$$\left(\frac{\partial}{\partial t} +\frac{\mathbf p_1}{m}\nabla_r+\mathbf F\nabla_{p_1} \right)f_1=\int d^3p_2 d^3p'_1 d^3p'_2 \delta^4(P_f-P_i)|T_{fi}|^2(f'_2f'_1-f_1f_2)$$

which is an integro-differential nonlinear (due to the $f'_2f'_1$ and $f_1f_2$ terms) equation in the distribution function $f_1$.

In a textbook I'm reading it is said that another feature of such equation is that

the nonlinear terms $f'_2f'_1$ and $f_1f_2$ are calculated for different values of the momentum.

What could be the meaning of this statement? Could "the momentum" be the $\mathbf p_1$ in $f_1(\mathbf x_1, \mathbf p_1, t$)?

$\endgroup$
  • 1
    $\begingroup$ I think it means that $f_1$ is a shortcut notation for $f(\vec{x}, \vec{p}_1,t) $; then $f'_1 = f(\vec{x},\vec{p}'_1,t) $ etc... $\endgroup$ – Matteo Feb 4 '18 at 17:34
0
$\begingroup$

The meaning is that in the scattering term on the RHS of the Boltzmann transport equation you have the products of the distribution functions $f_1$ and $f_2$ for different momenta $p_1$ and $p_2$.

$\endgroup$
  • $\begingroup$ Thank you. But in the case of $f'_2f'_1$ we take into consideration every value of $p'_1$ and $p'_2$, due to integration... $\endgroup$ – Lo Scrondo Feb 4 '18 at 21:54

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.