# Are elements of statistical ensemble fixed or dynamic?

I will use the simplest example I can think of to explain what I am trying to understand. Consider a system with ~$10^{23}$ particles in an equilibrium with fixed values of pressure, volume and temperature. Let's say it's a ideal gas in the piston. The statistical ensamble of this system is a set of elements where each element represents one possible microstate.

1. Are these elements fixed in time? In other words, is each microstate from the ensemble a static "snapshot", as if we took the picture of it (that contains info about value of $(p,q)$ of each particle at that particular moment) and put it in a set (ensemble)? Or are they dynamic, that is, are they allowed to evolve in time (all particles are moving and so their $(p,q)$ are changing) with the initial conditions taken at the moment of snapshot?

2. If we change pressure of a piston (say reversibly and by an infinitesimal amount) and put the system in a new equilibrium state, does the previously created ensamble still applies? Or do we have to create a new ensamble for each different equilibrium? In other words, does the ensamble of a system contains all possible microstates of all possible equilibriums of the system (like there is one ultimate ensamble), or each equilibrium state has it's own private ensamble associated with it and independent from others?

• my understanding: 1. for certain condition, a microstate is fixed. 2. if you change condition, it will be differetn microstate. Jul 27, 2016 at 3:39

Are these elements fixed in time? In other words, is each microstate from the ensemble a static "snapshot", as if we took the picture of it (that contains info about value of $(p,q)$ of each particle at that particular moment) and put it in a set (ensemble)?

Yes. Quoting from Wikipedia:

In mathematical physics, especially as introduced into statistical mechanics and thermodynamics by J. Willard Gibbs in 1902, an ensemble (also statistical ensemble) is an idealization consisting of a large number of virtual copies (sometimes infinitely many) of a system, considered all at once, each of which represents a possible state that the real system might be in. In other words, a statistical ensemble is a probability distribution for the state of the system.

From a macroscopic point of view, a thermodynamical system can be described by a set of thermodynamic variables: for example $(N,V,E)$ (microcanonical ensemble) or $(N,V,T)$ (canonical ensemble).

From a microscopic point of view, the system is described by the $6N$ canonical variables $\mathbf q_1 \dots \mathbf q_N$ and $\mathbf p_1 \dots \mathbf p_N$. Every possible state of the system is thus a point in a $6N$-dimensional phase space. Of those points, only a certain subset corresponds to our macroscopic variables. This subset is given by the partition function:

$$Z =\int d^{N} \mathbf p \int_{D(V)} d^N \mathbf q \ \rho(\{\mathbf p, \mathbf q \})$$

where $D(V)$ is the spacial domain defined by the containing volume and $\rho$ is an appropriate (normalized) probability density. Such probability density will depend on $E$ in the microcanonical ensemble, on $T$ in the canonical ensemble etc.

So the microstates are already every possible state your system can attain: when you consider an ensemble, you are already considering every possible microscopic dynamical state of your system.

If we change pressure of a piston (say reversibly and by an infinitesimal amount) and put the system in a new equilibrium state, does the previously created ensamble still applies? Or do we have to create a new ensamble for each different equilibrium?

The ensemble will change. If you change the pressure of your system, you are changing $V$ or $T$ or both. So, if we take the canonical ensemble to have a concrete example, you will have one of the following

$$Z(N,V,T) \to Z(N,V',T')$$ $$Z(N,V,T) \to Z(N,V,T')$$ $$Z(N,V,T) \to Z(N,V',T)$$

So the ensemble has changed.

Last but not least: remember that all these considerations are only valid at thermodynamic equilibrium, i.e. the probability distribution $\rho$ describes an equilibrium state.

• +1 for mentioning that Gibbs's definition only applies at equilibrium. Sep 19, 2016 at 12:56

An ensemble is a collection of possible microstates of a system which are consistent with the macroscopic properties of the system. So for example, in the canonical ensemble, the number of particles is fixed but the energy is allowed to change. In the microcanonical ensemble both particle number and energy are fixed. Suppose you fix a number of macroscopic variables, say $(N, E, p, V)$, then there are only a limited amount of microstates possible which are consistent with these macroscopic constraints.

In equilibrium statistical physics, dynamics plays no or almost no role. Hence thinking of an ensemble as a collection of snapshots of a gas of particles specifying each $(p_i, q_i)$ would make sense. The key point is that in statistical mechanics, with $10^{23}$ particles it would be nearly impossible and not very insightful to study the time evolution of all these particles. Hence we limit ourselves to macroscopic observables which arise from the collective behaviour.

As for your second question, the ensemble in a new equilibrium state does change, provided that the associated macroscopic variables are different. Say you go from fixed energy $E_1$ to a different energy $E_2$, then the possible positions and momenta of the particles are different because their total energy has to add up to different constants.

• +1 for mentioning that Gibbs's definition only applies at equilibrium. Sep 19, 2016 at 12:56

In general statistical mechanics (either in/out of equilibrium), yes the ensemble is dynamic. Note there are two equivalent ways to imagine ensembles:

1. A bunch of virtual copies of the same system, all having a slightly different state. Each copy represents a possible state of the system, and each evolves in time according to regular mechanics, i.e., Hamilton's equations in the classical case.
2. A probability distribution over state-space (phase space), a space which has $6^N$ dimensions for an ideal gas of $N$ particles. This probability distribution evolves according to the Liouville equation.

Note that way #1 can also be seen as a cloud of points in phase space, each point representing one virtual copy of the system. Way #2 is just saying that in the limit of an infinite number of virtual copies, we can smear these points into a continuous distribution.

The definition of equilibrium in statistical mechanics is that the probability distribution is not changing in time. Indeed all the virtual copies are evolving in time, however there are other virtual copies which are slightly earlier / slightly later in time, and so "move in" to replace each other. In equilibrium, any state has equal probability to all of its future and past states, and so the probability distribution does not change.

To answer your second question, yes, each different thermodynamic state is associated with a different ensemble (probability distribution). Note that these different ensembles can largely overlap with each other in phase space, but it is the average values that really determine the macroscopic observables.

Corresponding to a macroscopic equilibrium state of a system, there exists a set of microscopic states, such that at any time the system can be in any of those microscopic states with equal probability. In other words, the system passes through every micro-state in that set, without preference for one micro-state rather than the other. This $\textit{set}$ is fixed in time i.e. elements in this set do not change with time; in fact this set is what characterizes an equilibrium state.

Therefore if the system moves to a new equilibrium state then the set of micro-states corresponding to it will be different from the old one. There is no one ultimate ensemble, because there is no one ultimate equilibrium state.

• Nope. The states in which a particle is close to the above of the volume cannot exist anymore when the piston has moves downwards. Sep 19, 2016 at 12:55

I like to think of what happens in thermodynamics as follows:

In principle, one might be able to precisely measure the state of a system, and then use quantum or classical mechanics to see how it evolves in time. This turns out to be exremely hard if the system is macroscopic (i.e. has $\sim 10^{23}$ particles), because you cannot practically measure the state of so many particles at the same time. However, such large systems tend to be ergodic: their state explores large chunks of the sphase space if one waits long enough. This means that one can try to instead characterise what percentage of time the system spends in each portion of phase space. This is what thermodynamics (more properly statistical physics) tries to do.

This leads us to the notion of micro and macro state. The microstate is the actual point in phase space that the system obeys at a precise moment in time. The microstate is not constant, but goes all over the place by ergodicity. The macrostate is the probability distribution for finding the system in a particular microstate at a specific time.

As luck has it, the probability distributions defining a macrostate have to be of a very specific form (a Gibbs ensemble) because of physical reasons, and we can uniquely describe them by range of observable numbers called thermodynamic variables. These numbers are things like entropy, pressure, etc. You cannot vary all of them independently, by knowing the right subset, you can construct the macrostate, and the value for all other variables; the thermodynamic identity tells you how to do this.

So the microstate is not constant, but as long as you don't change the thermodynamic variables, the macrostate is. However, by changing the pressure of a piston, as you describe, you also change the thermodynamic variables, in this case volume and pressure and possibly temperature. Consequently, the macrostate of the system also changes. This means that the probability of finding particular microstates upon measurement also varies, and in particular, you might see microstates that you would never encounter before changing the pressure.

1) I'm not sure what you're asking exactly but QM says that all possible combinations of locations are a possibility, there are an infinite number of microstates as long as we consider space to be a continuum (ie. there are an infinite # of locations the particle can exist at as opposed to a finite amount). Each microstate has a different probability associated with it.

That being said, when talking in probabilities, the state that we capture the system at doesn't tell us anything about the function that determines the ensemble of possibilities at any point in time.

2) Disregarding any possible changes in total energy the change in space the system operates in also changes the ensemble of possibilities.