# Does the Maxwell-Boltzmann distribution apply to gases only?

The Maxwell-Boltzmann distribution can be used to determine the fraction of particles with sufficient energy to react. I know that the curve applies to gaseous reactants and would like to know whether solids and/or liquids are also described by a similar distribution.

In other words, can I use a Maxwell-Boltzmann distribution to interpret reaction rates at the molecular level in liquids and/or solids?

I'm going to go against the grain here: the Maxwell distribution does describe the distribution of molecule speeds in any (3-D) matter, regardless of phase.

Suppose we have a system of $N$ molecules with masses $m_i$, positions $\vec{r}_i$, and velocities $\vec{v}_i$ (with $i = 1, ..., N$). Assume that the total energy of this system is of the form $$E(\vec{r}_1, \dots, \vec{r}_N; \vec{p}_1, \dots, \vec{p}_N) = U(\vec{r}_1, \dots, \vec{r}_N) + \sum_{i = 1}^N \frac{\vec{p}_i^2}{2 m_i },$$ i.e., a potential energy depending on the molecules' positions and a kinetic energy. According to Boltzmann statistics, the probability of finding this system within a small volume of phase space $d^{3N} \vec{r} \, d^{3N} \vec{p}$ is $$\mathcal{P}(\vec{r}_1, \dots, \vec{r}_N; \vec{p}_1, \dots, \vec{p}_N) d^{3N} \vec{r} \, d^{3N} \vec{p} \propto e^{-E/kT} d^{3N} \vec{r} \, d^{3N} \vec{p} \\= e^{-U(\vec{r}_1, \dots, \vec{r}_N)/kT} \left[ \prod_{i=1}^N e^{-p_i^2/2m_i k T}\right]d^{3N} \vec{r} \, d^{3N} \vec{p}$$ To find the probability distribution of finding molecule #1 with a particular momentum $\vec{p}_1$, we integrate over all other configuration variables (i.e., $\vec{r}_1$ through $\vec{r}_N$ and $\vec{p}_2$ through $\vec{p}_N$). Because the $E$ is the sum of a contribution from the positions and a contribution from the momenta, the Boltzmann factors can be split between these integrals, with the result that $$\mathcal{P}(\vec{p}_1) \, d^3\vec{p}_1 \propto e^{-p_1^2/2 m_1 kT} d^3\vec{p}_1 \left[ \int e^{-U(\vec{r}_1, \dots, \vec{r}_N)/kT} d^{3N}\vec{r}\right] \left\{ \prod_{i = 2}^N \int e^{-p_i^2/2 m_i kT} d^3 \vec{p}_i \right\}$$ These integrals are nasty (especially the one over all $N$ position vectors), but they're just constants with respect to $\vec{p}_1$, which means that the can be folded into the proportionality constant: $$\mathcal{P}(\vec{p}_1) \, d^3\vec{p}_1 \propto e^{-p_1^2/2 m_1 kT} d^3\vec{p}_1.$$ A similar logic applies to every other molecule in my system. In other words, the probability of finding a molecule with a momentum $\vec{p}$ doesn't depend at all on how they interact with each other, assuming that their interaction energy is dependent only on their collective positions. Since these molecules obey the same momentum distribution, they must also have the same velocity distribution, and in particular they obey the Maxwell speed distribution.

So the answer to the question asked in your title, "Does the Maxwell-Boltzmann distribution apply to gases only?" is "no"; it applies to all phases of matter, in the sense that it describes the distribution of particle speeds and energies. However, the question you ask at the bottom of your post, "Can I use a Maxwell-Boltzmann distribution to interpret reaction rates at the molecular level in liquids and/or solids?" may also be "no"; the connection between reaction rate and activation energy is not straightforward if the medium is dense, as was pointed out by @porphyrin in their answer.

• This is fine for a gas but for a condensed phase the coordinates - and therefore the integrals - are not separable. The position and momentum of any particle depends on coordinates of other interacting particles. You can assume a mean-field suffices but that is not the same thing, and it is often not a sufficient approximation. – Buck Thorn Nov 9 '19 at 9:54
• @Thorn: Interesting. I agree that the position coordinates are often not separable, but I wasn’t aware that the same could be true for the momenta. Can you provide a citation discussing non-quadratic momenta so I can edit my answer accordingly? (Note that my derivation can be extended to any system where there are “cross terms” between the momenta of the form $p_1 p_2$, so long as the kinetic energy can be described by a positive definite quadratic form .) – Michael Seifert Nov 9 '19 at 14:36
• Hmm, as sure as I was that I'd spotted something wrong in your line of argument, I am now grasping at straws. A search produced this: scirp.org/journal/PaperInformation.aspx?PaperID=83351 That however used MD with a classical harmonic internuclear potential to confirm a MB distribution in a low T solid. I wonder how particles interacting with a nonlinear potential might behave? – Buck Thorn Nov 9 '19 at 18:19

I'm willing to put myself out there and say that no, a Maxwell-Boltzmann distribution will not be able to make any statement about the ability of liquids or solids to react. Meaning also that you could not study reaction rates with a Maxwell-Boltzmann distribution for the condensed phases.

The reason why I am fairly confident this is true is because the Maxwell-Boltzmann distribution makes the assumption that it is working with an ideal gas. Making that assumption is both powerful and very limiting. The ideal gas approximation is quite good in a lot of cases, but it could not even come close to describing the behavior of a condensed phase.

If some such distribution were used to study condensed phase reaction rates, it simply wouldn't be a Maxwell-Boltzmann distribution.

I would also point out that the Maxwell-Boltzmann distribution doesn't describe all gas-phase reactions all that well. This can be inferred quite easily from the fact that the M-B distribution treats all gases as ideal, so the only possible difference between two systems is the mass of the particle and the temperature. That means also that any gas phase reaction which depends strongly on the orientation of the collision would likely be assumed by M-B to happen much more frequently than it really does.

All that to say, M-B is deeply rooted in the gas phase, and even there it falls short for certain systems, so it probably isn't possible to apply to the condensed phases.

On the other hand, it is possible that a partition function for a liquid or a solid phase could be used to study reactions in some way, and the partition function for a generic gas particle can be used to derive the M-B distribution so . . . There's that.

The problem with that is that the partition function requires you to sum the Boltzmann Factor over all states, and for a condensed phase system . . . That's basically gonna be an infinite number of states.

Anytime you approach stat. mech. or the name Boltzmann, it's wise to keep this quote by David Goodstein in mind:

“Ludwig Boltzmann, who spent much of his life studying statistical mechanics, died in 1906, by his own hand. Paul Ehrenfest, carrying on the work, died similarly in 1933. Now it is our turn to study statistical mechanics. Perhaps it will be wise to approach the subject cautiously.”

In liquids the Boltzmann distribution limits reaction in what are called 'activation limited' reactions. In these reactions the two species collide but do not have sufficient energy to react because the activation energy is high, so they diffuse apart again. The number of particles with sufficient energy to react is given by the Boltzmann distribution. This falls with increase in energy so rapidly that reaction becomes a rate event, perhaps once in 100 million collisions.

Reactions with a low activation energy can react at first approach, by the Boltzmann distribution there is enough energy to surmount the activation energy to reaction. However, in these cases the reaction rate is limited by how fast the species can diffuse together, so reaction rate becomes a property of the solvent.

I agree with the answer of Michael Seifert. I just wanted to point out, that this applies to classical systems only (even if they are non-ideal).

For example systems in which relativistic effects play a role (Maxwell–Jüttner distribution), or for systems in which quantum effects play a role, the Maxwell-Boltzmann distribution will be violated.

No. Maxwell-Boltzman statistics are not followed. The Equipartition Theorem is followed (at equilibrium) but the distribution tends to be more like a Planck Distribution giving different equations for $$v_{avg}$$. Consider that solids arguably don't have an atomic $$v_{avg}$$ (or rather it is zero) and that liquids differ from gases because they have additional degrees of freedom. (That is, they have Potential Energy modes due to their molecule-to-molecule interactions which Ideal Gases do not have by definition. The assumption of non-correlated (random) movement can't be made (at sufficiently low densities)).

Some notes:

This area is difficult to get a handle on. Here's two references I found useful (along with the Wikipedia articles on Phonons,the Equipartion Theorem, Diffusion, and Molecular Diffusion):

The 2nd demonstrates that the way in which liquids move (are correctly modeled (hopefully!)) depends on their "modes" or potential energy degrees of freedom (potential function). There are both internal degrees of freedom (rotations, twistings, vibrations,...) and external (Van Der Waals forces, etc.) not to mention steric constraints (is there a size above which the concept of average molecular velocity has no meaning? (or does the molecular velocity concept transition into the macroscopic particle concept as size increases? Does it matter if "size" is significantly nonisotropic? (e.g. linear, planar, or spherical molecules?))

• Perhaps you mean: " The assumption of non-correlated (random) movement can't be made (at sufficiently high densities))" – Buck Thorn Nov 9 '19 at 9:50

The distributions of energy in liquids and solids fall off faster in liquids and solids than the Maxwell-Boltzmann distribution - if a molecule in a liquid or solid has high energy, it is going to transfer it to other molecules much sooner than in a gas.

• I think the difference in distributions is due to the number of degrees of freedom due to interparticle interactions, and not due to faster interactions rate. Otherwise you could make the argument that a hotter gas is less likely to observe the MB distribution. – Buck Thorn Nov 9 '19 at 10:01