1
$\begingroup$

I'm currently reading through a hydrodynamics lecture course and I'm confused at some of the key integrals that are derived. These integrals aid in deriving the moments of the Boltzmann equation.

First, the moments are:

$\int\big(\frac{\partial f}{\partial t}+v_j\frac{\partial f}{\partial x_j}+a_j\frac{\partial f}{\partial v_j})d^3\textbf{v}=0$,

where $f=f(x,v,t),\;d^3\textbf{v}=dv_1dv_2dv_3,\;v=\dot{x},\;a=\dot{v}$ and this equation is calculated three times with different factors: mass, momentum and energy $(m, mv, mv^2/2)$

Using the velocity moments of the the distribution function $f$ I can only get the first part of the integral, where $f$ is only differentiated by time, while the others are a bit more confusing, because I can't get the result they present.

E.g. they suggest that

$\int\frac{\partial f}{\partial v_i}d^3\textbf{v}=0$

if we just integrate by parts and the fact that since the total mass/momentum/energy must be finite then f must follow this condition, which I'm not sure where it comes from:

$\lim_{v \to \infty} v^2f=0$

That is my main problem, the other integrals seem more manageable, but

$\int\frac{\partial f}{\partial v_i}d^3\textbf{v}=0$

is difficult to understand.

The other integrals are:

$\int v_j\frac{\partial f}{\partial v_i}d^3\textbf{v}=-\delta_{ij}\frac{\rho}{m}$

$\frac{1}{2}\int v^2_j\frac{\partial f}{\partial v_i}d^3\textbf{v}=-\frac{\rho}{m}u_i$

I'm less worried about them since I believe that with understanding that one integral that I asked about above the rest would follow logically.

$\endgroup$
1
$\begingroup$

E.g. they suggest that $\int \ \partial f/\partial v_{i} \ d^{3}v = 0$ if we just integrate by parts and the fact that since the total mass/momentum/energy must be finite then f must follow this condition, which I'm not sure where it comes from: $\lim_{v \rightarrow \infty} v^{2} \ f = 0$

If $f\left( \mathbf{v} \right)$ is a velocity distribution function and we know that $f\left( v \right) < \infty$ must be true (i.e., there cannot be an infinite number of particles per unit phase space), then it must follow that: $$ \lim_{v \rightarrow \infty} v^{2} \ f \propto \lim_{v \rightarrow \infty} v^{2} \ e^{-v^{2}} \rightarrow 0 $$ assuming a Maxwellian velocity distribution.

This can be seen when one plots this versus speed (see below).

If you are confused about the integration by parts step (e.g., $\int \ u \ dv = u \ v - \int \ v \ du$), then recall that for a Maxwellian (1D for brevity): $$ \frac{ \partial f }{ \partial v } = f_{o} \frac{ \partial }{ \partial v } \ e^{-v^{2}} = -2 \ v \ f $$ Thus, the $v \ u$ term (expression from Wikipedia link above) goes to $\sim v^{2} \ f$ which is then evaluated at the limits of the original integral (i.e., $\pm \infty$). You know that any terms with odd powers of $v$ will go to zero under integration (i.e., negative values of $v$ result in negative values for the function) so the $\int \ v \ du$ term also goes to zero. v^n-Gaussians vs. Gaussian

Side Note: The velocity moments can be derived in a slightly different manner, where one knows that the units of $f\left( \mathbf{v} \right)$ are number per unit length cubed per unit velocity cubed. Thus, just from dimensional analysis (or physical reasoning), one can argue that the following must be true: $$ n_{s} = \int_{-\infty}^{\infty} \ d^{3}v \ f_{s}\left( \mathbf{v} \right) $$ where $n_{s}$ is the number density (i.e., number per unit volume) of species $s$.

I wrote a more detailed answer and discussion of velocity moments at https://physics.stackexchange.com/a/218643/59023.

$\endgroup$

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.