# Why do we get the same result using different ensembles?

There are different kinds of ensembles: microcanonical, canonical, grand-canonical... But for a particular system, no matter whether the system is isolated or closed, they just give the same result of $E$ and $N$.

For example, to calculate energy for gas in a box, the number of particles $N$ is fixed. However, if we use grand-canonical ensemble, just replace $\bar{N}$ with $N$ in the final result, we get the correct answer, which is the same as the result using canonical ensemble.

For example, to calculate Maxwell-Boltzmann distribution, use canonical ensemble and grand-canonical ensemble will give the same result.

I'm confused with it. Since the physical condition is different, why these ensembles give just the same answer?

• They give the same result because you take the thermodynamic limit, that is, the case of large numbers (particles, volume, and other extensive quantities). In this limit, the bulk properties do not depend on the nature of the walls of the system, i.e whether they are conducting/insulating, allowing particles through or not. – Abhinav Dec 12 '13 at 14:04
• But as long as thermodynamic quantity makes sense, we must take thermodynamic limit? – Andy Dec 12 '13 at 14:38
• Yes, strictly speaking, for 'actual' systems, the ensemble matters. But the number of particles is so large (~ upwards of 10^22 in everyday life) that this difference is very tiny and not detectable in any way. – Abhinav Dec 12 '13 at 14:42
• So I gauss, for small systems, we can still define thermodynamics quantity, but the fluctuations is usually very large, so we must take them into account. And in those systems, different ensembles do have different result. Is that right? – Andy Dec 12 '13 at 15:47
• Yes, that is how I understand it. Since most systems we apply thermodynamics to are fairly large, this difference can be safely ignored. There's a slight correction, though- for small systems, it is not the fluctuation from the equilibrium quantity that is large, but the fractional deviation from the equilibrium. – Abhinav Dec 12 '13 at 16:08

Here's a close analogy which (besides having has its own practical use) can shed some light, if you're looking at the problem from the probabilistic-mathematical point of view.

Imagine we play the following game ($$A$$) : we throw $$N$$ balls into $$K$$ boxes, with uniform probability. Let call the result (or configuration) $$X=(x_1,x_2 ...x_k)$$, where $$x_i$$ is the number of balls inside box $$i$$).

It's clear that the probability of each configuration is given by a multinomial distribution. It's also clear that $$E(x_i)=N/K$$.

We'll then be interested in computing some function of each result (configuration), averaged over several trials. Typically some $$E(g(X))$$ or some probabilty asociated with $$X$$ (eg: which is the variance of each $$x_i$$? which is the probability that the first half of the boxes have more than $$N/2+\delta$$ balls? etc) This is feasible, but no very easy, because it's a little cumbersome to integrate (sum) over a multinomial distribution.

Now, imagine a different game ($$B$$) : we throw inside each box a random number of balls, independently, following a Poisson distribution, with mean $$\lambda = N/K$$ In this case, the configurations $$Y=(y_1,y_2 ...y_k)$$ will follow a joint Poisson distribution. This is clearly not the same as the above; for one thing, now we have that the total number of balls $$n_y=\sum y_i$$ is a random variable, while in game $$A$$ it was fixed ($$n_x = \sum x_i=N$$). However, they resemble at least in the average : $$E(n_y)=E(n_x)$$ and the same is true for the marginals $$E(x_i)=E(y_i)$$. But we can say much more: we can check that our Poisson conditioned to the sum being equal $$N$$, ie $$P(Y|n_Y = N_y)$$, is the same multinomial as above (that's why we chose a Poisson).

Then, game $$B$$ is not equivalent to $$A$$... but it is when given the event $$n_y = N$$. And, by continuity, we expect them to be "almost equivalent" when the ideal event "almost" happens" ($$n_y \approx N$$). Furthermore, $$E(n_y)=N$$, and, if $$N$$ is large, by the law of large numbers, we expect $$n_y \to N$$; i.e., with high probability we'll be at or near the conditioning event. Hence, the conditioning will eventually turn irrelevant (informally: we'll be conditioning over an almost-sure event region), and the games will be asympotically statistically equivalent.

Now, consider that game $$B$$, though it appears more complex than the original $$A$$ (the total number of balls is variable instead of fixed), it's actually much more mathematically tractable, because the variables $$y_i$$ are independent. Then, instead of computing our $$E(g(X))$$ we compute $$E(g(Y))$$, and we are justified in assume that as $$N\to \infty$$ the results will be the same.

Notice also that game $$A$$ has two fixed paramenters: $$N,K$$. In game $$B$$, instead $$N$$ is not fixed, only its average; but, on the other side, we now have a new fixed parameter $$\lambda$$. This is totally analogous to what happens when transforming between ensembles. And it is (IMO) also totally analogous in spirit: instead of the original "natural" physic model (fixed number of particles, or kinetic energy), we devise an alternative model where the old parameter is not fixed but variable, and we add a new "artificial" fixed parameter (fugacity, temperature) that guarantees that asymptotically the models are equivalent. Surely, we can give physical justifications for the alternative models (reservoirs), but the main reason to prefer them is that, in spite of looking more complicated, actually they are mathematically simpler.

• That's very nice of you to give such an impressive demonstrating. I understand the example of game A & B and understand the spirit behind. But I just wonder how can this game be related to the ensembles. It seems that grand-canonical ensemble is somehow related to Poisson distribution? – Andy Dec 15 '13 at 15:24
• The game A is related to micro-canonical enseble and the game B to canonical. – Nogueira Apr 17 '15 at 1:11