When I learned statistical physics, and ensembles and such things, the term "macrostate" was introduced as some vague thing that "described the state of the system, and is defined completely by a few variables". I couldn't wrap my head around what this was supposed to mean.

Upon learning that one can define a probability distribution $\rho$ over set of all microstates $\Gamma$ of a system, and that an equilibrium state is a state where the entropy

\begin{align} S = -\int_{\Gamma} d\Omega~\rho(\Gamma) \ln \rho (\Gamma) \end{align}

is at a maximum (under appropriate boundary conditions, like a mean energy value, or a mean particle number), I simply identified a macrostate with a certain probability distribution described by the given variables.

This identification worked very well for me. In equilibrium state (and when the ensemble is fixed), the distribution is defined by few variables, and it fully describes the state of the system, in the sense that you can calculate any statistical observable you'd like to see.

I didn't have any problem with handling it like that so far - but however I don't know if it's actually true: So the question is at first:

Can I always map a macrostate to a probability distribution over a set of the microstates of the system?

If so - Is this map injective?

If so, and I constrain the space of probability distributions to the ones where the equilibrium condition holds, is this map bijective?

My own answers to those questions would be "yes", but I don't know if I'm missing something. For example, the proposed map only holds if you specify an ensemble (microcanonic, canonic, grand-canonical and so on ...).


1 Answer 1


Can I always map a macrostate to a probability distribution over a set of the microstates of the system?

Basically, yes.

The logic is that the actual state of the system is some microstate. We don't know (and don't really care about) what specific microstate the system is in, but we can (at least in principle) compute the probability distribution over microstates. Given the probability distribution over microstates, we can evaluate expectation values of observables like temperature, pressure, and volume. The term macrostate refers to a "complete" set of these expectation values. The word "complete" here is a little wishy washy, but basically means the full set of stable (not varying on microscopic timescales) quantities we can measure from a macroscopic system in equilibrium.

The probability distribution is given by the one which maximizes the entropy, subject to various constraints (eg: fixed total energy and number of particles, or fixed number of particles but varying energy, or varying energy and particles). The combination of a given set of constraints and the corresponding probability distribution is a thermodynamic ensemble. For example, the canonical ensemble refers to a probability distribution over microstates that maximizes the entropy given that the energy can flow in and out of the system, but with a fixed number of particles.

If so - Is this map injective?

I think I can rephrase this question as: "if I have two systems which have the same set of possible microstates are observed to have different macroscopic observables, do they have different probability distributions over microstates?" In which case the answer is yes. A simple example would be to consider the ideal gas at two different temperatures. Since the temperature is proportional to the average kinetic energy, the distribution of kinetic energies must be different. In general, the macroscopic observables are computed as averages over the microstates with respect to some probability distribution, and so if the average quantities come out different and the space of microstates is the same, then the distribution must be different.

If so, and I constrain the space of probability distributions to the ones where the equilibrium condition holds, is this map bijective?

If I understand the question, I think your phrasing is a bit too vague for a sharp answer, unfortunately. Mathematically you can define probability distributions over all kinds of bizarre spaces without a clear physical interpretation, you can compute the entropy for these distributions, and you can even maximize the entropy over a class of weird and unphysical distributions. Given that, it's certainly not the case that any probability distribution that can be described as a maximum entropy distribution from a mathematical point of view, represents a physical system.

But, I don't know exactly what you mean by "constrain[ing] the space of probability distributions to the ones where the equilibrium condition holds." You could make this statement vacuously true if you define this phrase to mean that you only want to consider probability distributions which do have a physical interpretation, in which case your question is logically something like, "Assuming A, is A true?"

So, to summarize, I'd say the answers to your questions are: yes, yes, and no.

  • $\begingroup$ Thank you for your answer. I tried to better define what I meant by constraining the probability distributions - but basically I almost always end up with a tautology. The real question behind it is that I want to know if I can treat macrostate and prob. distribution with as the same thing, because there is something like a canonical bijection between them, or alternatively, wether a distribution like the ones described in the beginning of my question do contain all the information that is needed to fully describe the macro state. $\endgroup$ Commented Apr 10, 2021 at 6:20
  • $\begingroup$ @Quantumwhisp Sorry I am just responding now. I would say that they are different but closely related. Given a distribution the macrostate is completely defined. But, there are some distributions that don't correspond to what you would normally call a macro state. For example, if you have a gas in a box and the distribution of particle positions is a delta functional on the state where all the particles are in the corner of the box with some momentum, there's not really a well defined pressure or temperature since the system is not in equilibrium. $\endgroup$
    – Andrew
    Commented Apr 14, 2021 at 0:46

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