This is the fundamental assumption of statistical mechanics:

In an isolated system in thermal equilibrium $^1$, all accessible microstates are equally probable.

But why does it mention the condition thermal equilibrium?

As Daniel V. Schroeder mentions:

Any large system in the equilibrium will be found in the macrostate with the greatest entropy.

Equilibrium is the state where there can only be that macrostate which has the greatest multiplicity that is the greatest entropy.

Also, check this pic taken from here; it clearly mentions thermal equilibrium at $q_A =4$; there is only one macrostate corresponding to $q_A= 4$ in thermal equilibrium and not more than one.

enter image description here

But the ergodic hypothesis says otherwise; it says in thermal equilibrium, all microstates are equally probable. Now, among those accessible microstates, there are also many which correspond to $q_A\ne 4$; these microstates define such macrostates which have lesser entropy than the macrostate to that of $q_A=4$ in the example. So, how can it be possible for those microstates of lesser-entropic macrostates $q_A\ne 4$ to exist at thermal equilibrium? For, there can only be one macrostate i.e. $q_A= 4$ at thermal equilibrium.

The famous line goes as:

Entropy is maximum at equilibrium.

You can't expect microstates of lesser-entropic macrostates in thermal equilibrium, isn't it? Equilibrium exists only at the macrostate having the greatest number of microstates; ie. $q_A= 4\;.$

So, how can I say, all microstates are equally likely in thermal equilibrium if microstates other than that of the macrostate having greatest entropy $q_A= 4$ can't exist in thermal equilibrium?

  • 1
    $\begingroup$ A system in thermodynamical equilibrium still has fluctuations that move it out of equilibrium. But the probability of this happening is very low by Poincare's recursion theorem, fluctuations occur at time scales that are exponentially long on the size of the entropy fluctuation . $\endgroup$
    – user83548
    Commented Nov 11, 2015 at 19:39
  • $\begingroup$ @bruce smitherson: So, the state defined at equilibrium is only the most entropic state, isn't it? But the microstates of the other macrostates are also equally likely in thermal equilibrium 7 that implies other macrostste an still occur under thermal equilibrium. This is what I am cofused about why those macrostates which has lesser entropy can be considered in thermal equilibrium. How? Are temperatures same ? $\endgroup$
    – user36790
    Commented Nov 11, 2015 at 19:47
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    $\begingroup$ Yes, if the system moves to a state in which the occupied microstates result in an unlikely macrostate, then you are no longer in thermal equilibrium. However, in most cases you never observe this situation, as it is highly unlikely. I do not know why you claim that "it seems any state, no matter how less the probability be, can be in thermal equilibrium". That is seems to be incorrect. Do you have any reference that state that? $\endgroup$
    – user83548
    Commented Nov 11, 2015 at 20:36
  • $\begingroup$ @bruce smitherson: Yes, I also deem it it incorrect. But see since all the microstates are equally likely, maybe that the system now is in such microstate which is not of the most entropic state, some other state. And as said, the microstates are equally likely, when the system is in thermal equilibrium. That's why I asked how if in thermal equilibrium, some other microstate appeared, how could it be in thermal equilibrium then. Probably that is fluctuation which becomes negligible when the numbers are very large. $\endgroup$
    – user36790
    Commented Nov 12, 2015 at 2:53
  • 1
    $\begingroup$ An isolated system is said to be in thermal equilibrium if you can define a constant temperature across it. The fact that the system can explore all microstates with equal probability does not mean that it is in thermal equilibrium. Thermal equilibrium is actually an average. If the fluctuation is so large (in either space or time) that you can measure it, then the system is momentarily out of thermal equilibrium. $\endgroup$
    – user83548
    Commented Nov 13, 2015 at 18:47

4 Answers 4


A macrostate is a set of microstates. Some microstates are thermal, others are not.

Without the assumption of being in thermal equilibrium you can't assume anything since any possible microstate is possible. And lots of possibilities macrostates could be picked.

Usually you want to group your macrostates according to a state variables such as pressure, volume, total energy or something like that.

And when you break your 30 microstates into three groups: A, B, and C you can ask yourself if each group is classified according to a state variable such as pressure, volume, total energy or something like that.

And even if it is, then all you might know is the state variable, and even that maybe not precisely. For instance the volume isn't precisely known since the exact locations of all the many parts is not known.

Now even when the microstate is a particular microstate, and that microstate is assigned specifically to a particular macrostate, that doesn't tell you how that macrostate assigns probabilities to the microstates in it

And you can assume that each microstate in the macrostate is equally likely. But that is just an assumption. If energy is conserved, then the dynamics will always keep the total system at a configuration with that fixed initial energy, so it doesn't change from any state to just any state.

Doesn't thermal equilibrium mean the macrostate having the greatest multiplicity?

It means so much more. Firstly, it requires that a macrostate is a probability distribution on microstates. Secondly, the macrostate is specified by some (macro) state variables. Thirdly, the particular distribution specified by the hypothesis requires that the space of all microstates be partitioned (partitioned by different values of the state variables) and each partition has an equal probability assigned to every microstate in that part of the partition. I always imagine different floors of a mansion, where your variables constrain you to a different floor and each room in a floor is equally likely.

Now, thermal equilibrium doesn't mean having the greatest multiplicity. You could have $N$ particles of some gas at a certain pressure $P_0$ and volume $V$ and that could be one macrostate, and there might be macrostates with a larger multiplicity with the same $N$ and $V$ and larger $P_+\gt P_0$ but there isn't enough energy for those $N$ particles to have that pressure, there just isn't enough kinetic energy to spread around to get them to the $P_+\gt P_0$ macrostate variable.

If you want to go to the mansion example. Imagine that you have two mansions and one person can go up a floor if (and only if) the other person goes down a floor. When they are both by some stairs then one can go up and the other can go down. But if that places one of them into a floor with trillions more rooms available than the other one had, then they are way way more likely to stay in the configuration with the one stuck in the floor with way more rooms.

So energy can be exchanged between the two people, but the additional energy spends most of it's time with one of them having more energy if they can exchange it. Eventually they could get to a level where one gains as many rooms as the other one loses. And that joint collection roughly is the macrostate of the combined system.

And when that happens we say they are thermal equilibrium. And the thing they have in common, temperature, is how many additional rooms they gain per bit of energy. Maybe one has stairs that are longer, so going up/down one flight for it is going up/down 5 flights for the other. But maybe the rate at which the floor have more rooms changes with floors at a different rate. There could still be a $\textrm dS/\textrm dE$ in common.

Can thermal equilibrium have fluctuation?

The macrostate could, in principle change from thermal equilibrium and more to have to two subsystems be at different temperature instead of the same temperature. But that would require bouncing around until you are by some stairs going down instead of towards any of the many more options on staying on the same floor, and then continuing to do the improbable floor after floor until the temperature of the two subsystems are very out of line.

And even if it happened, it could just go back to equilibrium. The idea is that for a large enough system, the time to wait for such a thing to happen is just really really long.

  • $\begingroup$ Edited my question. Please check that. $\endgroup$
    – user36790
    Commented Feb 3, 2016 at 5:33

I think there is some confusion of terminology here..

A system is not at thermal equilibrium if you lock it into the most probable state. A single state is not an equilibrium, it is the complete evolution of the system which has to be observed (or known) to say if it's in equilibrium.

If we take your A,B,C example, the system will be in thermal equilibrium if (over time) there is a passage between the A,B and C macrostates as related to their respective differences in free energy (which is directly related to their phase-space size of microstates).

So if you keep observing your system, it will jump between A, B and C, with C being most probable. This is indeed what the definition says as well (all microstates being equally probable) and needs the ergodic hypothesis to work.

A system out of thermal equilibrium is for example if you have some external force that keeps the system in C or similar.

  • $\begingroup$ Firstly, thanks for the answer. So, could you tell what should be the definition of thermal equilibrium then? $\endgroup$
    – user36790
    Commented Nov 12, 2015 at 17:47
  • $\begingroup$ The definition you started your post with is good. You just have to realize that equilibrium is a temporal process and not a statement of which single state is "in most equilibrium".. $\endgroup$
    – BjornW
    Commented Nov 12, 2015 at 17:52
  • $\begingroup$ No, I just wanted to get a little more elaboration of A system is not at thermal equilibrium if you lock it into the most probable state. What does that mean? It implies just by remaining at $A$ is not thermal equilibrium. Does it switch to $B,C$ even a thermal equilibrium? I would be grateful if you, please, elaborate the second para. $\endgroup$
    – user36790
    Commented Nov 12, 2015 at 18:42
  • $\begingroup$ Yes - the system can explore the possible configurations of the system, without bias. Thermal equilibrium really just means that the temperature has become homogenous throughout the system, not that the system has stopped moving. Of course at zero temperature, the state won't change at all, but at any non-zero temperature, there is a chance that the system will move between states (more easily if more thermal energy is available). This "state-hopping" can be quantified by statistical mechanics. Maybe have a look at this: en.wikipedia.org/wiki/Thermal_fluctuations $\endgroup$
    – BjornW
    Commented Nov 12, 2015 at 23:15

In statistical mechanics, a microstate is a specific microscopic configuration of a thermodynamic system that the system may occupy with a certain probability in the course of its thermal fluctuations. In contrast, the macrostate of a system refers to its macroscopic properties, such as its temperature, pressure, volume and density. Treatments on statistical mechanics, define a macrostate as follows. A particular set of values of energy, number of particles and volume of an isolated thermodynamic system is said to specify a particular macrostate of it. In this description, microstates appear as different possible ways the system can achieve a particular macrostate.

Now lets come to your example:

Suppose there are three macrostates A,B,C ; A contains 3 microstates; B contains 6 microstates & C contains 10 microstates.

To be in thermal equilibrium, they have to belong to one and the same isolated system. Three isolated systems from each other do not define one macrostate, but three, in contrast to what you assume . These cannot come into thermodynamic equilibrium because by your construction they have to be isolated.

You could say "take an isolated system and cut the volume in three, A, B, C" and continue with your example. The subsets do not define a macrostate of an isolated system, only a particular microstate of a single isolated system ,with a very small probability of ever manifesting, similar as an example to "what is the probability for all the molecules in the room to be up at the ceiling" .

The inclusion of "thermal equilibrium" is in place so that a fixed temperature can be defined for the system. Temperature is a measure of the macrostate and statistical mechanics needs to connect to thermodynamic quantities in a one to one way with statistical quantities in the state. Otherwise the identification of temperature with average kinetic energy would not be one to one.

  • $\begingroup$ I thought equilibrium is a macrostate which has the maximum multiplicities. C has the maximum multiplicity; that must be the macrostate which is in equilibrium; why then would anyone say entropy is maximum in equilibrium? Equilibrium is a specific macrostate; the fluctuations comprise of another macrostate. All the molecules confined in a corner is a macrostate comprising very small multiplicity- if any of these microstates happen to occur, you can't say it happened in equilibrium which is simply another macrostate. $\endgroup$
    – user36790
    Commented Jan 30, 2016 at 11:26
  • $\begingroup$ You are using the definition of "macrostate " wrongly. A macrostate is an "isolated system", one.. It is not comprised of systems. Thermal equilibrium means the temperature is uniform throughout. This ensures that the average kinetic energy is uniform throughout $\endgroup$
    – anna v
    Commented Jan 30, 2016 at 12:37
  • $\begingroup$ So, what should you say when it is said equilibrium has the greatest entropy? $\endgroup$
    – user36790
    Commented Jan 30, 2016 at 12:45
  • $\begingroup$ It is a comparison in time for one macrostate. When in equilibrium it is in the greatest entropy for the system. It supposes no more microstates are created , which of course is true in a classical neutral gas but wrong when one considers black body radiation. ( there is no isolation when radiation is considered, unless one counts the leaving photons) $\endgroup$
    – anna v
    Commented Jan 30, 2016 at 14:23
  • $\begingroup$ So, does that mean thermal equilibrium is not a macrostate? $\endgroup$
    – user36790
    Commented Jan 31, 2016 at 20:46

Your question does not make a lot of sense, which is due to the words "microstate" and "macrostate" floating around without being given a precise meaning.

A microstate is a point in the phase space of the system - it refers to a single, unambiguous configuration of the system. For a system of $N$ freely moving particles (an ideal gas) the phase space is typically $\mathbb{R}^{6N}$.

A macrostate is a probability distribution $\rho(x,p)$ on the phase space - it is not only a "set of microstates", but it also assigns a probability to every subset of microstates, which is the probability with which the system is found in that particular microstate. In the case of finitely many microstates, it assigns a non-zero probability to each single microstate.

A macrostate (or mixed state, as this concept is called outside the history-laden field of thermodynamics) is in equilibrium if it does not change in time.

The ergodic hypothesis now asserts that the macrostates with $\partial_t \rho = 0$ are states where $\rho$ is uniform on the set where it is non-zero. This is compatible with

There is often a notion that a macrostate is given by specifying thermodynamic properties like pressure or temperature, or even defined to do so. With the definitions in this post, different macrostates can realize the same sets of such state variables, that is, such a thermodynamic state does not correspond to a unique macrostate. If you start from the notions of thermodynamic state and microstate, then the ergodic hypothesis says that a thermodynamic state that is in equilibrium is uniquely realized by the uniform macrostate, i.e. no other macrostate produces the same state variables and is in equilibrium.

  • $\begingroup$ Thanks for answering; I've edited my question and tried to present my argument through an example. Please check that. Meanwhile, I have to go to bed for in India, it is 4am; I was waiting for you the whole night to talk with you at the chat : P Good night : ) $\endgroup$
    – user36790
    Commented Feb 2, 2016 at 22:28
  • $\begingroup$ @user36790: The uniform distribution on a set is the maximum entropy distribution among all continuous distributions. I do not understand where you see the contradiction between equilibrium being the maximally entropic macrostate and it being the macrostate in which all microstates with non-zero probability are equiprobabe. $\endgroup$
    – ACuriousMind
    Commented Feb 2, 2016 at 22:31
  • $\begingroup$ the contradiction between equilibrium being the maximally entropic macrostate and it being the macrostate in which all microstates with non-zero probability are equiprobable- let me go step by step: at equilibrium, the macrostate prevailing corresponds to $q_A= 4$ as seen in the pic, right? $\endgroup$
    – user36790
    Commented Feb 3, 2016 at 14:37
  • $\begingroup$ At equilibrium, there are microstates corresponding to $q_A= 4$ and not any other microstates corresponding to $q_A\ne 4$, right? $\endgroup$
    – user36790
    Commented Feb 3, 2016 at 14:39
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    $\begingroup$ @user36790: "Accessible" means "compatible with the assigned state variables". Since the state variable $q_A$ has been assigned the value 4, no microstate with $q_A\neq 4$ is "accessible" - realizing such a microstate would change the state variable $q_A$, hence the thermodynamic state, so it is not a microstate compatible or "accessible" from that thermodynamic state. $\endgroup$
    – ACuriousMind
    Commented Feb 3, 2016 at 15:00

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