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As a preamble, I have several questions and I know that is preferable to ask one question per post, but they are really linked together. This is why I chose to ask them together.

I am considering a set of particles at a given position $x$ in a fluid at rest, and I am interested by their repartition in the phase-space. Let us define $f(x,v)$ their distribution in the phase-space. Therefore, $f(x,v)$ is the number of particles at the position $x$ having the velocity $v$. Since the fluid is at rest, the average velocity is zero:

$$\int_{\mathbb{R}}vf(x,v)dv=0.$$

Moreover, the temperature is given by : $$\int_{\mathbb{R}}\frac{1}{2}mv^2f(x,v)dv=\frac{3}{2}k_B T.$$

To me, we usually take a normal distribution for $f$, where the standard deviation is linked to the temperature and as a normal distribution, $f$, is symmetric and isotropic.

Thus, my questions are :

  • Is it possible to have an asymmetric $f$ distribution ? ie there is more particles going in the direction $x^+$, but particles going in the direction $x^-$ can go faster, such as the average velocity remains zero.

  • Is it possible to have an distribution for $f$ that is not isotropic ? ie the standard deviation is greater over the x-axis than the standard deviation over the y-axis.

  • Is the definition of the temperature would change in such cases?

  • What could the possible causes and/or consequences of such cases?

R. N
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  • Some previous posts of relevance: https://physics.stackexchange.com/q/216819/59023 and https://physics.stackexchange.com/a/218643/59023. Note that 2nd velocity moment is not the temperature but the pressure tensor (i.e., a rank-2 tensor). It can reduce to a temperature with a proper equation of state. – honeste_vivere Aug 08 '18 at 14:17

1 Answers1

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There are an infinite number of different distribution functions that can be used to describe any situation of interest. You can have asymmetry, bumps, or even point-like delta functions in your distribution. In most realistic situations, collisions in the fluid will act to increase the entropy of the system by randomizing the particle velocities, and the normal distribution is the particular distribution that has maximum entropy. Thus, distributions in thermal equilibrium are isotropic normal distributions.

Away from equilibrium, a good way to quantify the asymmetry of the temperature is to define the pressure tensor:

$$ \overset{\text{$\leftrightarrow$}}{P}(x) = \int d^3v\; m\vec v \vec v f(x,v), $$

which can carry a different temperature in each spatial direction, as well as off-diagonal elements that define the shear-stress. Often this quantity is split as

$$ \overset{\text{$\leftrightarrow$}}{P}(x) = p(x)\overset{\text{$\leftrightarrow$}}{I} + \overset{\text{$\leftrightarrow$}}{\pi}(x), $$

where $p$ is taken to be the isotropic pressure and $\overset{\text{$\leftrightarrow$}}{\pi}$ is termed the stress tensor which measures the deviation of the distribution from spherical symmetry.

One can also define the heat-flux vector as the third-moment of the distribution, $$ \vec q(x) = \int d^3v\; {m \over 2}\;(\vec v-\langle \vec v\rangle)^2\; \vec v \;f(x,v), $$ which defines the flux-density of heat-energy.

To gain even further insight into the dynamics of this process, it is best to describe things in terms of kinetic theory where the distribution function is evolved in time according to the equation $$ {\partial f \over \partial t} + \vec v \cdot \vec \nabla f + {\vec F \over m} \cdot {\partial f \over \partial \vec v} = \mathcal{C}[f]. $$ The collision term on the r.h.s. of the above equation is responsible for the randomization that leads to entropy production. This term forces the distribution towards a normal distribution on a time-scale on the order of the inter-particle collision-time. When one considers times much faster than this or situations where collisions can be neglected (i.e., a high-temperature plasma or a low-density fluid), the terms on the l.h.s. can dominate and the distribution can stray far from being normal.

We can also perturb the kinetic equation to determine the stability of a given distribution. If we perturb the given distribution and our perturbation damps out in time, we can conclude the distribution is stable. However, if our perturbation grows in time, then this implies the given distribution is unstable.

Including the self-consistent effects of electromagnetism/collisions complicates things further and are actively researched fields. These are probably out of the scope of this answer. But kinetic theory is definitely a worthwhile endeavor.

Hope this helps answer your questions.

Matt P.
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  • Your answer seems very good, but before to accept it and to be sure that I understood well, I would like some precisions. So, viscous effects (shear-stress) and thermal conduction (heat-flux) can create anisotropic and asymmetric distributions of particles ? – R. N Jul 30 '18 at 13:32
  • Not quite, shear-stress and heat-flux are measures of the deviation away from an isotropic equilibrium distribution. If we are given a distribution, we can calculate the heat-flux and shear-stress. Often what is done to close the kinetic hierarchy of equations is to equate the heat-flux with the product of the thermal conduction coefficient and the gradient of the temperature (usually called Fick's law). The thermal conduction coefficient is called a transport coefficient, which is calculated from a perturbative expansion of the collision term in the kinetic equation. – Matt P. Jul 30 '18 at 13:49
  • I did not very well expressed myself, but your explications are going in the direction of what I understood. So thank you. I validated your answer. – R. N Jul 30 '18 at 13:57