# Statistical Mechanics -- many copies of a canonical ensemble

1. Jan 29, 2017

### binbagsss

1. The problem statement, all variables and given/known data
Hi

I am looking at the attached extract from David Tong's lecure notes on statistical phsyics

So we have a canonical ensemble system $S$, and the idea is that we take $W>>1$ copies of the system $S$, and the copies of $W$ taken together then can be treated as a microcanonical ensemble with energy $W<E>$.

Each such copy lives in a state $|n>$.

I am stuck on this part
' if $W$ is large enough the number of systems that sit in the state $|n>$ is $p(n)W$' amd therefore we have 'translated probabilities into eventualities'.

MY QUESTIONS

Q1) I don't understand why $W$ is large is needed for $p(n)W$ to describe the number of systems that sit in state $|n>$? Why doesnt this hold for small $n$?
Q2)Also probably a stupid quesiton, but in what ways to the states $|n>$, which physical properties are allowed to differ, since isn't the idea to take a large number $W$ of identical copies of the system $S$, or do they not neeed to be identical?

2. Relevant equations

3. The attempt at a solution

Moderator note:
Moved from homework section to a technical.
The reference it refers to is http://www.damtp.cam.ac.uk/user/tong/statphys/sp.pdf (page 22)
Thx @Stephen Tashi

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2. Feb 5, 2017

### Stephen Tashi

In my opinion, statistical physics uses very cumbersome language because it wishes to use actual frequencies of events instead of dealing with probabilities of events. If we take a small number of systems (e.g. 3 systems ) and there are a large number possible states (e.g. 1000) , then even in the case where all the states have approximately the same probability of being occupied, only at most 3 of them would actually be occupied. So the observed frequency ratios would show many instances of 0/1000 instead of all of them being 1/1000. However, if take a large number of systems then the actual frequency ratios would probably all be close to 1/1000. (Physics wants to proceed on the assumption that the actual frequency ratios are definitely close to 1/1000. So, using the language of ensembles, we speak words that conclude by making this assumption.)

If we take the viewpoint of probability theory, we postulate there is a population of things, each of which is in one and only one of a set of states $\mu_1, \mu_2,...\mu_N$. We assume there exist a set of variables $P_1, P_2,..P_n$ such that each member of the population is associated with one and only set of values of those variables. (i.e. a given member of the population "has" specific numerical values of $P_1,P_2,...P_n$.

Given a specific numerical values of $P_1,P_2,...P_n$, define the subpopulation $W$ to be the subset of the population such that each member of it associated with those specific numerical values. Define $p(\mu_i)$ to be the probability that a randomly selected member of $W$ is in state $\mu_i$.

The members of $W$ are "identical" with respect to the numerical values of $P_1,P_2,..P_n$ that they have, but not identical with respect to which state $\mu_i$ that they are in.

That pattern could be applied to almost any situation, so it's interesting to ask why only particular instances of it are useful. For example we could define the "state" of a person to be his yearly income in dollars. We could define a set of numerical properties associated with a person such as weight, height, length of forearm, etc.

Associated with each specific set of numerical values of the properties, there is a probability distribution for a randomly selected person with those numerical values being in the various states. However, the utility of this viewpoint is dubious because it's unlikely that there are any simplifying formulas that give $p(\mu_i)$ as a function of the numerical values of those properties. Finding $p(\mu_i)$ for a given set of numerical values (e.g. 180 lbs, 6 ft, 23. 9 inches) can be done by sampling or tabulating data from the whole population, but we don't have a simple formula that lets us plug-in the numerical values of the properties and compute $p(\mu_i)$. By contrast, for gases in equilibrium, the properties of temperature, pressure, etc. for gases in equilibrium are related to the $p(\mu_i)$ by mathematical formulae.

3. Feb 5, 2017

### binbagsss

So each system S has the same macroscopic properties, the $P_i$ above you've used.

I'm confused though, are the microstates of the microcanonical system each systems S, or do you also consider the microstates within each system S - so is the number of microstates for the microcananonical system W or NW? where N is the number of microstates of the canonical system S . If it's NW then that's fine. But if it is W, each is defined to be a canonical system itself, and all such systems, these copies of S, then agree macroscopically which has been taken as the microstates... ta

4. Feb 5, 2017

### Stephen Tashi

I'm not saying that each system in the entire population of systems has the same macroscopic properties, but, yes, I am saying that we define a sub-population $W$ so each member of $W$ has the same numerical values of the macroscopic properties. (The adjective "macroscopic" applies to the special situation thermodynamics. In the general scenario, the properties could be any numerical properties that fall short of specifying exactly which state the system is in.)

Each system $S \in W$ is in one and only one (micro-)state $\mu_i$. Different members of $W$ can be in different microstates.

$W$ is a set, not a number. In the above example, the number of microstates is denoted by an uppercase "N". The number of properties is denoted by a lower case "n". The number of elements in $W$ wasn't specified.

The goal of the notes you linked is to have the number of things in microstate $\mu_i$ be $|W| p(\mu_i)$ where "$|W|$" denotes the number of elements in the set $W$.

5. Feb 5, 2017

### binbagsss

But if we consider just one copy of $S$ itself, it is an ensemble and so consists of it's own microstates - the number of degrees of freedom $3N$ if there are no interactions, $N$ particles and we are in 3-d, so if you say each $S \in W$ is in only one microstate, then the only 'microstate' that would make sense in the case that $S$ form the microstates of the ensemble $W$, would be the macrostate of $S$ , rather than picking a random $1/3N$ microstate from a single, each $S$ ?

6. Feb 5, 2017

### Stephen Tashi

I don't understand your definition for "$S$". If are thinking of $S$ as a gas in a container then it is in one and only one microstate. (The fact that its individual particles can have different locations and velocities doesn't alter this fact because the teminology "microstate" doesn't apply to an individual particle.)

If you are thinking of $S$ as set of examples of a gas in a container at given numerical values of temperature and pressure or some other macroscopic variable then different members of $S$ can be in different microstates.

Let's decide how you define $S$ before we say that.

7. Feb 5, 2017

### binbagsss

$S$ is defined to be a canonical ensemble.
Then for the definition of ensemble I have: copies of the same physical system, each in a different microstate, where typically the number of copies tends to infinity.

So because $S$ itself is a canonical ensemble, it is itself made up of the same copies of some physical system ?