This is kind of a a follow up question to my last question. If I know the microstates a system can be in, and if I know the probability for each of those microstates, I can calculate average observables (like all the extensive Variables, $U$, $V$, $N$) and the entropy $S$.

If I don't restrict to probability distributions that satisfy an equilibrium condition of maximum entropy, then there is in general no functional dependence between the observables that I can calculcate. As well as that I can't calculate observables that involve derivates of the macroscopic variables (like $T$, $p$ or $\mu$).

Question is: Can I still call the state (which isn't necessarily an equilibrium state), described by the probability distribution, a macrostate?

If not, what word is there to describe non-equilibrium states? And is the probability distribution even enough to define a non-equilibrium state?


2 Answers 2


From Stanford lectures (similar treatment in Chandler's textbook):

"A macrostate is defined by specifying the value of every macroscopic variable. There may be a huge number of microstates all corresponding to the same macrostate.",

where the macroscopic variables are defined as "the total energy of the system, the total number of gas molecules, the volume of space it takes up, etc.".

Specifying a macroscopic state for a classical system is equivalent to specifying a lower-dimensional subspace of the phase space.

If you define your microscopic state not as a point in the phase space, but as a probability distribution on the phase space (which is not a usual definition), then specifying the average value of energy, number of particles and etc. will specify the subspace of the space of all possible probability distributions similar to the definition of the regular macroscopic state.

PS. I changed my response to reflect on the comments

  • 1
    $\begingroup$ This is not the standard terminology, nor is it the terminology given in the linked wikipedia page. A microstate is a specific microscopic configuration (for instance, the values taken by all the spins in an Ising system). A macrostate is the relevant set of thermodynamical parameters (for instance, temperature and magnetic field). A macrostate is often identified with the corresponding probability measure on microstates. $\endgroup$ Commented Apr 14, 2021 at 5:41
  • $\begingroup$ I object to the same points as Yvan - I was always under the impression hat a microstate is a specific configuration of the physical system - a point in the 6N-dimensional phase space in the classical case, for example. $\endgroup$ Commented Apr 14, 2021 at 6:33
  • $\begingroup$ Hm... You are right, Wikipedia defines microstate differently. The treatment there though implies that the state of the system at hand is described by a point in the phase space $(p, q)$. I, on the other hand, consider the state to be a function on the phase space $P(p ,q)$ (i.e. the probability distribution is the state). Such consideration makes it possible to include entropy as another macroscopic parameter, and have the thermal state as an equilibrium state of the probability distribution (which is convenient for justifying laws of thermodynamics). But you are right, it in not standard $\endgroup$
    – Pavlo. B.
    Commented Apr 14, 2021 at 6:37
  • $\begingroup$ The confusion in the OP arises precisely because the definition of macrostate is vague in wiki (and many other sources). Volume, number of particles and energy are good quantities, because they are defined for any state of the system. But temperature should not be mentioned on the equal footing with the others, because it is defined only for equilibrium distributions. $\endgroup$
    – Pavlo. B.
    Commented Apr 14, 2021 at 6:47

I will rephrase your question:

If the probability distribution that maximizes entropy represents the equilibrium state, what do the other distributions in the feasible set represent?

The answer is, they represent non equilibrium states.

Here is a simple example. Start with an $(E,V,N)$ system, which we imagine as rigid box, and divide it into two parts with volumes $V_1$ and $V_2=V-V_1$, energies $E_1$ and $E_2=E-E_1$ and numbers of particles $N_2=N-N_1$. Allow each part to reach internal equilibrium but prevent all transfers between the two parts. We now have a non equilibrium distribution for the entire box: Μicrostates that satisfy the conditions $(E_1,V_1,N_1)$ and $(E_2,V_2,N_2)$ have equal probability. All other microstates have zero probability. Mathematically $$ P'(\text{microstate}) = \begin{cases} \displaystyle \frac{1}{\Omega(E_1,V_1,N_1)\Omega(E_2,V_2,N_2)} & \text{if permissible}\\ 0& \text{otherwise} \end{cases} $$ where permissible is a microstate $(E,V,N)$ that satisfies the partitioning into $(E_1,V_1,N_1)$ and $(E_2,V_2,N_2)$. The equilibrium distribution is $$ P^*(\text{microstate}) = \frac{1}{\Omega(E_1+E_2,V_1+V_2,N_1+N_2)} $$

If we remove the barrier that prevents transfers between the compartments, the probability of microstate at time $t=0$ is given by $P'$ and at $t=\infty$ by $P^*$.

In this example $P'$ is a feasible distribution with lower entropy than the equilibrium distribution and can be realized as the probability distribution of a non equilibrium macrostate.


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