I am reading a book about the boltzmann equation, the author gives the expression of the fluid density $\rho$ as follows: $$\rho(\mathbf r,t) = \int {M\,f(\mathbf r,\mathbf c,t) \, \mathrm d\mathbf c}$$ where:

  • $\rho$ is the fluid density.
  • $\mathbf r$ is the position of particles.
  • $\mathbf c$ is the velocity of particles.
  • $t$ is time.
  • $M$ is the molecular mass.
  • $f$ is the Boltzmann distribution function which gives the number of molecules at time $t$ positioned between $\mathbf r $ and $ \mathbf r + \mathrm d \mathbf r$ which have velocities between $\mathbf c$ and $ \mathbf c + \mathrm d \mathbf c$

Can you, please, help me to derive this expression of fluid density?

Thank you

  • $\begingroup$ This is basically a self-sufficient definition. What do you want to derive? Do you understand the notion of distribution function? $\endgroup$ – Andrii Magalich Jun 27 '16 at 20:54
  • $\begingroup$ @AndriiMagalich: I do not really understand why $\rho$ is given by that expression. $\endgroup$ – Syntax_ErrorX00 Jun 27 '16 at 20:58
  • $\begingroup$ hmmm... you can probably 'derive' this from the Vlasov /Boltzmann Equation, but I don't know if you will find that insightful. I tend to assume there is some probability density \rho(x,t) that can be expressed as an integral representation over the velocity distribution. $\endgroup$ – anon01 Jun 27 '16 at 21:33
  • $\begingroup$ Just saw the comments above. The expression you have posted is a useful definition (provided you know f), for answering questions like: what is the density of particles below velocity c_0? Then you could just integrate out to c_0 to get an idea of what slower particles are doing. $\endgroup$ – anon01 Jun 27 '16 at 21:39

Density is the average amount of mass per unit volume $\rho(\vec{r},t) = \frac{M}{V_r}$.

Distribution function is defined as a number of particles per unit phase space volume $ f(\vec{r}, \vec{c},t) = \frac{N}{V_r V_c} $ (which is space volume times velocity volume). Each particle has mass $M$, so to get the total mass density, we need to sum distribution function multiplied by particle mass over all velocities in the given space volume.

Then, summing over small unit volumes in the velocity space $d^3 c$

$$\rho(\vec{r},t) = \sum_\vec{c} M f(\vec{r}, \vec{c},t) \Delta V_c \approx M \int f(\vec{r}, \vec{c},t) d\vec{c} $$

  • $\begingroup$ Why did you ignore the space volume? I do expect the density (based on your definition) to be given by: $$\rho ( \vec r, t) = \int {M f( \vec r, \vec c, t )d \vec c d \vec r}$$ $\endgroup$ – Syntax_ErrorX00 Jun 27 '16 at 22:12
  • $\begingroup$ This will give you the total mass in that volume, not the density $\endgroup$ – Andrii Magalich Jun 27 '16 at 22:13
  • $\begingroup$ @AndriiMagadish: Thanks I got it, that was so helpful $\endgroup$ – Syntax_ErrorX00 Jun 27 '16 at 22:15

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.