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Equilibrium statistical mechanics is (amongst other things) about deriving the equations of state of thermodynamic systems (in equilibrium) from a microscopic basis (i.e. starting with a microscopic Hamiltonian).

In order to do that, we observe the system over a very long time, which means taking the limit of time average and variance of a phase space function. For quasi-ergodic systems, this is equivalent to the (appropriate) ensemble average/variance. We get a very sharp peaked average value which is constant in time and reproduces the thermodynamic e.o.s for a system in equilibrium.

So far, so good.

How can one now define a 'system in equilibrium' in terms of statistical mechanics? Would it be convenient to define a subset of phase space in which the macroscopic variables (like total energy,...) differ only by a small value (eg. the variance) from the ensemble average and call all points of this subset equilibrium-states (and the other ones non-equilibrium states) of the system? Or is there another definition?

EDIT: Maybe this thought experiment will help to clearify my question. Let's assume that we have a small container filled with an (ideal) gas. The container itself is placed within another but much larger isolated container with no other gas in it. At time T1 we open the small container and simultanously measure the full microstate of the gas. Then we wait "long enough" and at T2 we measure the full microstate again. Intuitively, one would say that the system was out of equilibrium at T1 and in equilibrium at T2. Yet, both microstates are part of the microcanonical ensemble. If we would measure a macroscopic phase space function (where no particle is somehow favoured) at T1 we would propably get a different result compared to a measurement at T2 or the ensemble average of the function. Furthermore, because of the recurrence theorem, a state like at T1 will come again at some point in the future. So, how could one define equilibrium with this experiment in mind?

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    $\begingroup$ From Wikipedia's Statistical ensemble (mathematical physics):"the ensemble will not evolve if it contains all past and future phases of the system. Such a statistical ensemble, one that does not change over time, is called stationary and can said to be in statistical equilibrium." Also, "that state of a closed statistical system in which the average values of all the physical quantities characterizing the state are independent of time." $\endgroup$ – Conifold Jan 28 '17 at 0:02
  • $\begingroup$ Statistical mechanics aims at doing much more than merely deriving the equations of states of thermodynamic systems. Indeed, it allows you to address a multitude of questions that cannot be addressed with thermodynamics. Moreover, ergodicity (in its usual form) is irrelevant to statistical mechanics (this is discussed in many places on this site). $\endgroup$ – Yvan Velenik Jan 28 '17 at 8:50
  • $\begingroup$ @Conifold What does "the ensemble can said to be in statistical equilibrium" mean? That the system is in equilibrium? $\endgroup$ – user2224350 Jan 28 '17 at 11:52
  • $\begingroup$ @YvanVelenik You're right. I just left this out, yet $\endgroup$ – user2224350 Jan 28 '17 at 12:13
  • $\begingroup$ This is a typical turn of phrase in definitions, "can be said to be X" is that we can use X as a name for what is defined. In this case, "in statistical equilibrium" can be used as synonymous to the ensemble being stationary. $\endgroup$ – Conifold Jan 30 '17 at 18:39
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Let $\rho(p,q,t)$ be your ensemble probability density. Notice that I am assuming that $\rho$ can have an explicit time dependence.

The average of some quantity $Q(p,q)$ will then be calculated as

$$\langle Q(p,q) \rangle = \int_{\Omega} \rho(p,q,t) Q(p,q) dp dq \tag{1}\label{1}$$

where $\Omega$ is the phase space. Notice that in general this average will depend on time.

For an Hamiltonian system $\rho$ must satisfy Liouville's equation:

$$\frac{d\rho(p,q,t)}{dt}=\partial_t \rho (p,q,t) +\{\rho(p,q,t),H(p,q,t)\} = 0\tag{2}\label{2}$$

where $H$ is the Hamiltonian and $\{\cdot\}$ are Poisson's brakets.

Now, in thermodynamic equilibrium you want phase space averages as \ref{1} to be time-independent. This is realized if $\rho$ has no explicit time dependence:

$$\partial_t \rho = 0 \tag{3}\label{3}$$

In this case, Liouville's equaiton \ref{2} becomes

$$\{\rho,H\} = 0 \tag{4}\label{4}$$

The general solution of \ref{4} is any function of the Hamiltonian

$$\rho(p,q) = f(H) \tag{5}\label{5}$$

The specific form of $f$ depends on the constraints required by the ensemble (ex. microcanonical = fixed $N,V,E$).

[too long, didn't read]: An ensemble (set of systems) is at equilibrium when the probability density $\rho$ has no explicit time dependence: $\partial_t \rho=0$.

References: M. E. Tuckerman, Statistical Mechanics: Theory and Molecular Simulation


Update (after comment discussion)

Actually, $\partial_t \rho$ is a necessary condition for thermodynamic equilibrium, meaning that

$$\text{Equilibrium} \ \Rightarrow \partial_t \rho=0$$

However, I don't know if it is also a sufficient condition, i.e. if

$$\text{Equilibrium} \ \Leftarrow \partial_t \rho=0$$

is also true. I will leave the answer anyway, because I believe that it could be useful.

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    $\begingroup$ I do not agree with this answer. There is a difference in between a stationary state and a thermal state. $\endgroup$ – Steven Mathey Jan 30 '18 at 10:19
  • $\begingroup$ @StevenMathey What would this difference be? $\endgroup$ – valerio Jan 30 '18 at 10:27
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    $\begingroup$ It's easy to construct a counter example. Pick for example $f(H) = (\delta(H-h_1) + \delta(H-h_2))/2$. It's normalised and represents a stationary state, but is not thermal. $\endgroup$ – Steven Mathey Jan 30 '18 at 12:33
  • $\begingroup$ @StevenMathey I am afraid I don't understand what do you mean by "thermal". $\endgroup$ – valerio Jan 30 '18 at 12:34
  • $\begingroup$ I explain it in this answer. $\endgroup$ – Steven Mathey Jan 30 '18 at 12:39
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When probabilities dont change with time.

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  • $\begingroup$ There would however be fluctuations about the mean. $\endgroup$ – SAKhan Jan 30 '18 at 16:23

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