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I'm a computational chemist trying to understand more deeply the concepts of statistical mechanics. I followed Susskind's lectures. He starts by dividing a system in N subsystems; for my brain to like this approach and understand what comes next, I want to think about the subsystems as the actual particles (atoms/molecules) that make up the system. Each particle will be in state $i$, and if we ignore degeneracy the state will have a unique energy $E_i$. The occupation number $n_i$ identifies the number of particles in state $i$ and the probability of finding a particle in state $i$ is by definition $P_i = \frac{n_i}{N}$. Just one more piece of terminology (please correct me if I use anything incorrectly): I call microstate the collective state in which the particles are at any given time.

Then, each possible microstate corresponds to a particular set of occupation numbers; on the other hand, a set of occupation numbers might describe more than one microstate. This allows us to conclude that the most probable set of occupation numbers will be the one with the maximum number of ways of redistributing the particles among the energy states (while keeping the set ${n_{i}}$ fixed), that is the occupation numbers corresponding to the greatest number of equivalent microstates. The "number of rearrangements" is given by:

$$C = \frac{N!}{\prod\limits_{i} n_{i}!}$$

and by maximizing this quantity (this Wikipedia page goes through the same derivation) we arrive at a definition of the partition function as:

$$Z = \sum_{i} e^{-\beta E_{i}}$$

where, importantly, the $E_i$ are the energy states of the particles. The energy:

$$\sum_{i} P_{i} E_{i} = E$$

will be the average energy of a particle. In many other places, I've seen written that the $E_i$ are the values of the total energy of the system in a particular microstate. How are these things related? The total energy would be the sum of the individual particle energies (assuming ideal gas) and the total partition function would be the product of the individual partition functions . . . still, the derivation seems to make sense if I'm considering the energy states for each particle, rather than the total microstate energy.

Susskind goes on to state that $\beta$ is the inverse of the temperature, and he derives an equation for the Helmholtz free energy as a function of $Z$:

$$E - TS = -T\log Z$$ $$F = -T\log Z$$

. . . again, if we have defined $E$ to be the average energy of a particle, does this equation still make sense?

If you have taken the time to read all this and plan on taking more to answer, first of all thank you, but secondly please answer in a way that doesn't use complex mathematical constructions that a chemist might not be familiar with.

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2 Answers 2

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Your initial definitions ect. are ok. Let me supplement a few things.

Your macro states s are described by a vector $(n_1,n_2, ..,?)$ having $\sum n_i = N$

The partition function is in general derived from the number microstates $\Omega(s)$ and the entropy $S(s) = k ln \Omega(s)$.

You can use the partition function to calculate the probability: $P_i = e^{\beta E_i}/Z, e^{\beta E_i}$ is the famous Boltzmann factor.

$\beta$ is just $1/kT$, for people who is feed up with writing $1/kT$ all the time - $k$ is the Boltzmann constant.

Your $E_i$ is the energy of a particle in the i'th state lets stick with that. The average energy formular is correct.

The total energy $E_T$ for the macrostate s is simply given by $E_T = \sum_{i} n_i E_i$ many microstates has this total energy ("number of rearrangements").

The equations for the Helmholtz free energy are designed for and valid for the $E_T$, I don't see the meaning of it for an average energy E.

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  • $\begingroup$ I understand everything you wrote, though my doubt still stands. With regards to the Helmholtz free energy, Susskind does define it using the average energy E in the lecture I was mentioning. If you have the patience it starts here and in about 5 mins the revelant part is over. youtu.be/2BJYXuZZK3c?t=802 (these lectures need to be watched at 2x speed :D) $\endgroup$ Commented Jan 5, 2022 at 17:51
  • $\begingroup$ In supplement to the above answer, I think that the i's of the video is a simple indexing of the macro states, arriving at E the total average energy, in other terms the probable total energy of the System. So the meaning of the states and indexing is misaligned with the video. $\endgroup$
    – Karsten B
    Commented Jan 5, 2022 at 20:55
  • $\begingroup$ Yes, so maybe this is how the mess gets untangled. You are right, there is a slight misalignment between his derivation and my understanding: he imagines having a system and dividing it in N weakly-interacting subsystems that act as a heath bath for any given subsystem. Then, the i's are indexing the states occupied by a subsystem. Why can't my particles be the "subsystems"? It seems that by treating the system as an ideal gas this should be possible. Everything makes so much more sense if I think of particles instead of "subsystems", so I'm trying to reconcile the two things. $\endgroup$ Commented Jan 5, 2022 at 21:21
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for my brain to like this approach and understand what comes next, I want to think about the subsystems as the actual particles (atoms/molecules) that make up the system

No, this is not how it works. The subsystems are macroscopic in size with huge numbers of particles and the $n_i$ count subsystems, not particles. Perhaps this cartoon will help:

enter image description here

Here is what is going on:

  • The original system is $(E_\text{tot},V_\text{tot},N_\text{tot})$.

  • We divide it into $K$ macroscopic systems all with the same volume $V=V_\text{tot}/K$ and number of particles $N=N_\text{tot}/K$, but with their energy unspecified as long as the total energy of the $K$ subsystems is $E_\text{tot}$. This makes the average energy per subsystem $$\bar E = E_\text{tot} /K$$

  • Now look at the subsystems and count how many are in the same microstate. For example, A and E have their particles in the same positions, and if the atoms also have identical energies, call this "microstate 1" and set $n_1$ equal to the number of times this microstate appears in the collection of subsystems. This is how we construct distribution $\{n_i\} = (n_1,n_2\cdots)$. This distribution satisfies $$\sum n_i = K,\quad \sum n_i E_i = E_\text{tot} = K \bar E$$

This construction models the internal microstate of the big system $(E_\text{tot}, V_\text{tot}, N_\text{tot})$ by allowing its internal parts to exchange any amount of energy. It is important in this construction to recognize that the system is partitioned into macroscopic subsystems, not to its constituent particles.

In this thought experiment we begin with an extremely large system so that we can partition it into an extremely large number of macroscopic partitions, large enough that each possible microstate will appear a large number of times. It is a thought experiment so all these assumptions are OK.

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