# Can Maxwell's speed distribution be given an ensemble picture?

The typically presented view is that for an ideal gas in a container, with total energy being $$E$$, then there is a speed distribution at equilibrium (equilibrium macrostate is the one with the largest phase space volume), say $$f_E$$, which gives the fraction of particles having speeds in the range $$(v, v+dv)$$ as $$f_E(v)dv$$.

But suppose I consider all the possible states (not necessarily the equilibrium ones) that the container containing the ideal gas could be in, with the total energy being $$E$$. Then, what does $$f_E$$ mean for this infinity of states? Does $$f_E(v)dv$$ in this case give the fraction of those states (in which the gas could be found in the container) in which there is at least one gas particle with speed in the range $$(v, v+dv)$$?

Note that this is a mathematically precise question, asking whether the latter probability distribution is exactly the same as the former.

This question arose when I was reading Chapter 1 of Schrödinger's Statistical Thermodynamics, where he writes the following.

... [the problem could be to find] volume ($$=$$ number of states of the assembly) of the 'class' in which one individual member [a gas particle] is in a particular one of its states [speed in range $$(v, v+dv)$$]. Maxwell's law of velocity distribution is the best-known example.

• A few comments to help to find a more precise formulation of the question: I) M-B distribution is not bound to ideal gases. It is the equilibrium velocity distribution even for a solid (within classical statistical mechanics); ii) if you fix the total energy the equilibrium distribution is not the M-B, although it approximates it for large systems; iii) $f_E(v)$ gives the fraction (or the number) of particles as a function of a selected speed. This is not the number of states. By the way, which states? iv) the equilibrium distribution does not maximize the macrostate. That is fixed. Jan 16, 2021 at 13:09
• @GiorgioP Thanks for your comments. I think I don't need to worry about (i) and (ii). Responding to (iii), by states I mean (as I've mentioned in the question), the the state of the entire gas in the container, which is given by specifying the $6N$ position and velocity coordinates of the $N$ gas particles.[...]
– Atom
Jan 16, 2021 at 13:28
• [...] I made the correspondence as I showed in the quote ("an individual member" with "a gas particle" and its "state" being its "velocity") because Schrödinger gives Maxwell's distribution as an example of the former statement. and this naturally leads to my question in boldface.
– Atom
Jan 16, 2021 at 13:32
• @GiorgioP Responding to your comment (iv), I meant that the equilibrium distribution determines the macrostate which has the largest phase space volume among all the possible macrostates. I'll clarify it.
– Atom
Jan 16, 2021 at 13:35

There are different kinds of distributions to be kept in mind in the relation between statistical mechanics and kinetic theory. Maybe the easiest way is to start from statistical mechanics.

We have the probability distribution $$L(x_i,p_i)$$ of the microstates of the system in a given equilibrium macrostate. It's the probability of observing the microstate $$(\bar{x}, \bar{p}) := (x_1,\dotsc,x_N,\ p_1, \dotsc,p_N)$$ in a range $$\mathrm{d}\bar{x}\ \mathrm{d}\bar{p}$$ if a microscopic observation were instantaneously made on the system prepared in the given macrostate. Roughly speaking, this distribution ("ensemble" distribution) is obtained by (1) preparing a copy of the system in the given macrostate, (2) making an instantaneous measurement of its microstate (at an unsystematically chosen time $$t$$), (3) preparing a new copy of the system in the same macrostate, (4) going back to (2) and repeating indefinitely.

For a system described by the canonical ensemble this distribution is the usual $$L(\bar{x}, \bar{p}) \propto \exp[-\beta H(\bar{x}, \bar{p})]$$.

From this probability distribution we can also obtain marginal distributions. In particular the marginal distribution $$L_1(x,p)$$ that one, unsystematically chosen particle of the gas has microstate $$(x,p)$$ in a range $$\mathrm{d}x\ \mathrm{d}p$$. (More precisely $$L_1$$ is the marginal of the symmetrized $$L$$.) Roughly speaking, this marginal distribution is obtained by (1) preparing a copy of the system in the given macrostate, (2) making an instantaneous measurement of the microstate of an unsystematically chosen particle (at an unsystematically chosen time $$t$$), (3) preparing a new copy of the system in the same macrostate, (4) going back to (2) and repeating indefinitely.

One important point to note is that in general we cannot recover the full distribution $$L(\bar{x}, \bar{p})$$ from the one-particle marginal $$L_1(x,p)$$. The only exception if for non-interacting particles, for which $$L(\bar{x}, \bar{p}) = \prod_{i=1}^N L_1(x,p)$$.

Another probability distribution that we can consider is that of the number of particles $$n$$ that have microstate $$(x,p)$$ in a range $$\mathrm{d}x\ \mathrm{d}p$$. Since the number of particles is finite, this is a somewhat complicated distribution. For example, the probability that we find only one particle in the state $$(x,p)$$ is given by $$F(n\!=\!1 | x,p) = \sum_{i=1}^N \int\delta(x_i-x)\,\delta(p_i-p)\ %\prod_{j}^{j\ne i}\delta(x_j-x)\,\delta(p_i-p)\ L(\bar{x}, \bar{p}) \ \mathrm{d}\bar{x}\ \mathrm{d}\bar{p}\ ,$$ and for two $$F(n\!=\!2 | x,p) = \sum_{i\ne j}^{\text{all combos}} \int \delta(x_i-x)\,\delta(p_i-p)\ \delta(x_j-x)\,\delta(p_j-p)\ %\prod_{j}^{j\ne i}\delta(x_j-x)\,\delta(p_i-p)\ L(\bar{x}, \bar{p}) \ \mathrm{d}\bar{x}\ \mathrm{d}\bar{p}\ ,$$ and so on. It's important to note that this is not a probability distribution over values of $$(x,p)$$, because these are in the conditional of the probability. Again roughly speaking this probability distribution is obtained by (1) fixing a microstate $$(x,p)$$ and range $$\mathrm{d}x\ \mathrm{d}p$$, (2) preparing a copy of the system in the given macrostate, (3) making an instantaneous measurement of the microstate (at an unsystematically chosen time $$t$$) and counting how many particles have state $$(x,p)$$, (3) preparing a new copy of the system in the same macrostate, (4) going back to (3) and repeating indefinitely, (5) going back to (1) choosing a different fixed state $$(x,p)$$ and repeating indefinitely.

Of this probability distribution we can calculate the expected value, $$\mathrm{E}(n | x,p) = \sum_{n=0}^N n F(n | x,p)$$ This is the molecular density of kinetic theory.

When $$N$$ is extremely large, it turns out that $$\mathrm{E}(n | x,p) \approx N\ L_1(x,p)$$. We can also restrict our attention to momenta (or velocities) only, obtaining the marginal $$\mathrm{E}(n | p) \approx N\ L_1(p)$$, which is the (unnormalized) Maxwell-Boltzmann distribution.

So we can say, at least approximately, that what you write as $$f_E$$ is the expectation of a marginal distribution obtained from a canonical ensemble distribution.

The relation and difference between these distributions is well explained in a paper by Jaynes:

The definition of the molecular density (and the fact that it's an expectation) can be found for example in Chapter III of

Short answer: No, they are not the same. The most probable speed distribution is not the same as the weighted average of all possible distributions. But they are very very close to each other, and this is a major point in Schrödinger's book.

The way I see it (and this view was my takeaway from Schrödinger's book that you mention) is that you shouldn't view a single container of gas as an ensemble. Instead, you should imagine an ensemble to be a large number of containers (I think he said this was one of Gibbs' insights). Each can be in its own (micro)state, and the macrostate is just the distribution of microstates (number of containers in each microstate). If the containers don't exchange energy or particles, like in your case, you have the microcanonical ensemble, where the macrostate is that each microstate of the gas, with the correct amount of energy, is equally likely.

The Boltzmann distribution, and by extension the Maxwell-Boltzmann, comes from the canonical ensemble. Here, you take a large number of containers but allow them to exchange energy with each other. Then, you take a large number of these ensembles and put them into a microcanonical ensemble, which is equivalent to saying that each state of the composite system of containers (= set of microstates of all the containers) is equally likely. The macrostate is the number of containers in each microstate (= number of microstates in the macrostate, = $$e^\mathrm{entropy}$$, = phase space volume). So actually, there is a probability distribution over different possible macrostates, or a "distribution of distributions". But normally, you make an approximation and say that you only care about the most probable macrostate, the "most probable distribution" (i.e. what you define as the equilibrium macrostate), and this gives you the Boltzmann distribution. I think Schrödinger expends quite a bit of effort to prove that this approximation is very accurate in the limit of a large ensemble.

So to answer your question, no, the distributions are not the same. Including also "non-equilibrium" macrostate, i.e. less probable ones, in principle alters the Boltzmann distribution, but only a little bit, and this effect is smaller the larger ensemble you have (i.e. the bigger the environment is).

I hope these ramblings answer some part of your question. It is a confusing topic, because there are many layers: you collect particles into a gas system, then collect systems into a canonical ensemble, then collect canonical ensembles into a huge microcanonical one.

(Okay, you actually asked about if the container's energy $$E$$ is fixed (= no environment), while I am now talking about fixed temperature. But even for fixed energy, you make this same approximation in choosing the most probable speed distribution, rather than averaging over all of them.)

• Just a comment: A macrostate is a state of the system where the macroscopic observables (pressure, volume, temperature, etc) have definite values. Multiple microstates can correspond to the same macrostate, but a macrostate itself is not a distribution. I like to think of a macrostate as being an equivalence class of microstates that have the same macroscopic observables. Meanwhile, ensembles are probability distributions over microstates. In the microcanonical ensemble, all states in the distribution have the same energy and number of particles. In the canonical ensemble, the energy can vary. Jan 17, 2021 at 1:26
• @Andrew But macroscopic observables are generally not defined on individual microstates. For example, you can't look at a microstate and say what temperature it has. Temperature is a property of the distribution of microstates, so is pressure. I'd argue that the set of macroscopic variables can be thought of as nothing but a convenient representation of the probability distribution over microstates. For example, knowing $T$ and $V$ for an ideal gas (with known composition) in the canonical ensemble is equivalent to knowing the full Boltzmann distribution. Jan 17, 2021 at 19:10
• Fair enough, I've just never heard it described this way. For example, I tend to think of the probability that the system will be in a given macrostate as being the number of microstates corresponding to that macrostate, divided by the total number of accessible microstates. I don't see how to square this with your definition. Note that a microstate does define the set of macroscopic observables; knowing the microstate of a gas you could calculate the internal energy, number of particles, and volume, for instance. (The pressure can be computed from the momenta of particles near the boundaries) Jan 17, 2021 at 21:39
• @Andrew I have to (mostly) disagree about the observables. Each microstate has an energy, but I've never seen internal energy defined as anything else than the average energy of a distribution of microstates ($U = -\frac{\partial Z}{\partial\beta}$ in the canonical ensemble), not all with the same energy. Pressure is defined in terms of internal energy ($p = -\frac{\partial U}{\partial V}$). Entropy of a macrostate is defined in terms of the number of microstates in it, and temperature as $T = \frac{\partial U}{\partial S}$. Particle number and volume are interesting cases, though. Jan 17, 2021 at 22:29
• @Andrew In my understanding, particle number $N$ is like energy: Every microstate has a particle number $n$, but when you talk about $N$, you either mean a constant (all microstates have the same $n$, microcanonical or canonical ensemble), or you mean the average $n$ (grand canonical ensemble). The same for volume: each microstate has a volume, but either all have the same volume $V$ (constant-$V$ ensemble) or $V$ means the average volume (like in a constant-$p$ ensemble). Jan 17, 2021 at 22:35