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The Maxwell-Boltzmann velocity distribution arises from non-reactive elastic collisions of particles and is usually discussed in the context of the kinetic theory (for gases). There are various offhand remarks, for example here (slide 5), that state without reference that particles observe a similar velocity distribution in liquid. Is that true? References?

The main reason I'm curious is that it seems as though the mean free path would be extremely short in liquid vs. gas. I'm actually most curious about the nature of the collisions in liquid vs. gas, i.e., are collisions in liquid still (on average) elastic?

EDIT: The linked post on particle velocity in liquids is definitely interesting and weighs in on this question, and I appreciate the distinction made between local position fluctuation vs. long range movement. Still, let me frame this question in a few different ways.

  1. Gillespie proposed a stochastic framework for simulating chemical reactions which is formulated on the idea that non-reactive elastic collisions serve to 'uniformize' particle position so that the assumption of well-mixedness is always satisfied (see page 409 in the linked version). This is formulated from kinetic theory. A corollary to this is that a non-reactive collision between two molecules that are able to react does not induce local correlation, i.e., two particles able to react with each other that just collided, but didn't react are no more likely to react with each other in the next dt than any other particle pair in the volume. Gillespie's algorithm is commonly used in biology where biochemical species are modeled in the aqueous environment of cells. Is this valid, and if so why?

  2. On a microscopic scale, suppose we are interested in two 'A' particles diffusing in one dimension in an aqueous environment. The two A particles collide but don't react. What is their behavior immediately after the collision? Is it a 'reflection' which conserves velocity and might correspond to a Neumann BC? In a gas that approach seems natural, but in liquid the short mean-free path makes me think that diffusive forces would rapidly dissipate any momentum from the A-A collision, which might imply the A particles collide and 'stop'. How should I be thinking about this?

Just to bring it back to the original question, I think both (1) and (2) depend on the statistical velocity behavior of particles in liquid.

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    $\begingroup$ I'm curious about it too. This question arised recently in my work, and I haven't found a satisfatory answer yet. $\endgroup$
    – Hydro Guy
    Commented May 20, 2014 at 21:46
  • $\begingroup$ I think your edits 1 and 2 would be better posted as a new question and possibly as two separate questions. $\endgroup$ Commented May 21, 2014 at 15:59
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    $\begingroup$ @vector07, I do not think that your question was well answered here or in the "duplicate" question suggested by the people below. I think Maxwell-Boltzmann distribution should be valid for molecules in liquid too, at least according to classical statistical physics, because the factor $e^{-\beta p^2/2m}$ in the Gibbs-Boltzmann probability density does not depend on potential energy and is the same whether the molecule is in gas, or a liquid. I do not know if there is a measurement supporting this theoretical result. $\endgroup$ Commented May 21, 2014 at 19:34
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    $\begingroup$ @JánLalinský If Maxwell-Boltzmann distribution is valid in liquids, then the equation of state of ideal gas is also valid for liquid. And apparently, this is not the case. $\endgroup$ Commented Aug 24, 2016 at 17:05
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    $\begingroup$ @Physicist137, it may seem like the derivation is general, but when you look at it closely, it assumes that the particles do not interact with each other. When repulsive interaction is admitted, the particles exert pressure on the wall even if they stand still. Therefore pressure can be higher than the value formula gives. Pressure much higher can easily be produced in liquids. $\endgroup$ Commented Aug 30, 2016 at 21:14

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Generally speaking, it depends on the nature of the liquid. The assumptions behind the Maxwell-Boltzmann distribution are fairly simple: molecules can be approximated as point-masses and their only interactions are through collisions that exchange momentum and energy.

So if you have liquids where this is true, then yes, the distribution will be correct. However, if you have liquids (or gases for that matter) that have large molecules (so that the point-mass assumption is invalid) or that have long-range interaction forces (like water for example), then the distribution will not be Maxwell-Boltzmann.

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    $\begingroup$ Another very important example of fluids with long-range interaction forces are plasmas. $\endgroup$ Commented May 21, 2014 at 15:52
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    $\begingroup$ But don't thermalized plasmas follow MB distribution too in spite of their long range interaction forces? $\endgroup$
    – alex
    Commented Nov 19, 2017 at 9:48
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Classical particles must follow the Maxwell-Boltzmann velocity distribution, and this is a consequence of separability of momenta and position in the partition function, and that the Boltzmann factor weighing the probability of each state is $\propto \exp(-\beta\mathcal{H})$. This is not, however, to say that this is the distribution one would always obtain experimentally. I am not aware of an experimental procedure where one can directly sample velocities, but rather you observe displacements between times $0$ and $t$, say, and then you divide the size of the displacement by the time $t$ to get an effective velocity.

Now, remember that the diffusion constant is defined as (in 3D) $$D = \lim_{t\to\infty}\frac{1}{6t}\langle (\mathbf{r}(t) - \mathbf{r}(0))^2\rangle$$

Note that here we assume that the displacements are taken an infinite time apart. Because the collisions cause random motion, and in stochastic Ito calculus the time is proportional to the square of displacements, this indeed evaluates as a finite constant in most cases. And in most cases the distribution of displacements (and therefore of effective velocities) at infinity is Gaussian.

Now there are important exceptions to this rule. Fractional Brownian motion, for example, does not yield a regular diffusion constant, but undergoes something called anomalous diffusion (a hot topic in current physics research, anything dealing with anomalous diffusion, or diffusion seemingly anomalous often gets published in top venues). Anomalous diffusion looks like the following: $$D = \lim_{t\to\infty}\frac{1}{6t^\alpha}\langle (\mathbf{r}(t) - \mathbf{r}(0))^2\rangle$$

where the limit makes sense only for some $\alpha$. If $\alpha$ is larger than $1$, one calls the motion superdiffusive, and when it is smaller, subdiffusive.

How is this related to velocity distributions? Well, to get an idea through the diffusion constant of the underlying velocity distribution, one wants the displacements to be a very short time apart. $$D = \lim_{t\to0}\frac{1}{6t^\alpha}\langle (\mathbf{r}(t) - \mathbf{r}(0))^2\rangle$$

Note the change of limit. Now what is $\alpha$ going to be? Obviously $2$, because there are no collisions in very short time scales and one is then dealing with ballistic motion, where displacements scale linearly with time. Shortly after (if the limit in $t$ were to be in the picosecond scale) one enters the subdiffusive regime caused by viscoelastic behaviour of the material, and finally, when the displacements are measured long enough apart, to the normal, diffusive regime (in most liquids, that is, but there are exceptions, of course, like fractional Brownian motion).

Very few experimental methods can access the femtosecond, ballistic, regime, and it is only here that the displacement distribution should follow Maxwell-Boltzmann. For longer time scales, as you typically observe and are interested in, a Gaussian distribution might be a better approximation, but this does depend on the type of liquid.

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I think Maxwell-Boltzmann distribution should be valid for molecules in liquid too, at least according to classical statistical physics, because the factor $e ^{−\beta p^2/2m}$ in the Gibbs-Boltzmann probability density does not depend on potential energy and is the same whether the molecule is in gas, or a liquid. I do not know if there is a measurement supporting this theoretical result.

are collisions in liquid still (on average) elastic?

Elastic collision means that appreciable change in the kinetic and potential energy of two bodies happens to them only during short time interval and the energy long after that is the same as the energy long before that - the interaction of the two molecules is thought of as a scattering process. In liquids the interaction of the molecules may not be idealizable in this way, as the molecules are believed to be in incessant complicated motion constantly influencing each other (Brownian motion...) This does not seem to be a reason to abandon the Boltzmann statistics, however.

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The velocity distribution in a liquid does follow MB. This is routinelly seen in molecular dynamics (MD) simulations in liquids such as water (and in more complex systems!!, proteins in liquids, membranes , everything). In MD we do not consider simplified collision rules (which is a useless concept in liquids), rather we model complex interatomic potentials in dense systems. The reason for MB validity is, as said before, in the fact that the hamiltonian separates into two different parts, one containing the coordinates (the potential energy, independent of velocities) and another one containing the velocities (the kinetic energy). A good reference could be the first chapter in "Understanding Molecular Simulations" by Frenkel and Smit.

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Yes.

For reference, please see the derivation of the MB distribution by Landau and Lifshitz.

Their derivation is very general. It depends only on the Gibbs distribution and the usual classical assumption about energy being the sum of kinetic energy and potential energy.

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    $\begingroup$ But in an harmonic oscillator, momentum and coordinates are not independent. If I think about an Einstein model of a solid at classical temperatures, I would guess that near-zero velocities (at the classical turning points) are more likely than in the MB distribution. (A bit too lazy to do the calculation now.) $\endgroup$
    – user137289
    Commented Feb 22, 2021 at 20:30
  • $\begingroup$ Nice answer, see also this (MB distribution at equilibrium arises from detailed balance) physics.stackexchange.com/a/745726/226902 $\endgroup$
    – Quillo
    Commented Jun 17 at 8:22

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