# Ideal gas in the microcanonical ensemble - a subtle point?

## Main Question or Discussion Point

Hi all,

I was brushing up on statistical ensembles, and I found something apparently weird in
microcanonical treatment of the ideal "classical" gas. I'm mainly following K. Huang's
Statistical Mechanics.

So there's a first approach to the problem in which the MC entropy is evaluated
via an integral in the phase space (chapters 6.5, 6.6).
In particular, in chapter 6.6 the Gibbs paradox and the correct Boltzmann
counting are discussed, and the Sackur-Tetrode equation (6.62) is derived
[see eq (C) below].

A different approach is illustrated in chapter 8.5. The system is discretized and
the number of states corresponding to a set of occupation numbers is
evaluated as

$$W\{n_j\} = \prod \frac{g_j^{n_j}}{n_j!}$$

where $$g_j$$ is the degeneracy of the jth level. The entropy is then
calculated by recalling that it should coincide with $$\log W\{\bar n_j\}$$,
where

$$\bar n_j = z g_j e^{-\beta \epsilon_j}$$

is the set of occupation numbers maximizing W.

After some straightforward manipulations one obtains

$$S = \frac{1}{T} (E-\mu N) = \frac{3}{2} N k - N k \log\left[\frac{N}{V} \lambda^{-3}\right] \qquad (A)$$

This is basically eq (8.54), and here's the point that looks weird to me.

From the one hand, this equation seems to provide the correct equation of state
$$PV = NkT$$. However, there's a thermodynamic equation stating

$$E = TS - PV + \mu N \qquad (B)$$

(Gibbs-Duhem equation?). Now if I plug (B) into the first equality in (A) I obtain
$$TS = TS - PV$$ i.e $$PV = 0$$, which disagrees with the equation
of state. In other words, it should be

$$S = \frac{1}{T} (E +PV -\mu N)$$

Furthermore eq. (A), is also referred to as Sackur-Tetrode equation, but differs from
eq. (6.63) which, after some algebra, reads

$$S = \frac{5}{2} N k - N k \log\left[\frac{N}{V} \lambda^{-3}\right] \qquad (C)$$

Now both (A) and (C) are free from the Gibbs paradox, and hence I understand that they
are both Sackur-Tetrode equations.
However (B) has an extra $$NkT=PV$$ term which seems to exclude a clash with the
thermodynamic equation (B).

What does this mean? Should we conclude that the "maximization approach" in section 8.5
does not really work for the classical gas (which, anyway, does not even exist...).
Or perhaps eq. (B) has a more limited scope, and cannot be used in this approach?
For some reason both of these answer sound too dismissive to me...
Also, should eq. (B) be unavailable, I'd have some trouble in solving problem 8.2 (see also https://www.physicsforums.com/showthread.php?t=272585" )

Thanks a lot for any insight and comment.

Franz

PS I was not sure where to post this. There's no statistical physics forum, and after all this is the
"classical gas", so it could fit in classical physics. Apologies in case another forum was more appropriate.

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Well, you know that the entropy is not 1/T (E - mu N), so it goes wrong in Eq. A.

Hi Count Iblis,

I see that (A) has something weird, since I would expect that the thermodynamic equation

$$S = \frac{1}{T} (E +PV -\mu N)\qquad (D)$$

is satisfied.

However (A) is what one gets when pursuing the microcanonical approach in chapter
8.5 of Huang's book. Actually (A) is eq. (8.54) in the book.

Are you suggesting that this calculation is plainly wrong because (A) is not the expected entropy, and hence using it to derive the equations of state makes no sense ?

The book does not discuss this point at all, but simply derives the (correct!) equations of state from the second equality in (A).

That's what's troubling me.

I would have to look into this in more detail. But you would expect that A is related to the logarithm of the Grand Canonical partition function (the total number of particles is not kept constant) and that is proportional to beta PV. So, to get S from Log(W), you would expect to have to subtract an additional term PV and I suspect that this is the source of the missing PV term.

Last edited:
Mapes
Homework Helper
Gold Member
I'm not a stat mech expert, but doesn't the microcanonical ensemble assume constant volume? That would preclude the possibility of P-V work and might explain why the corresponding term is missing.

Hi guys,

thank you for your interest! It is nice to talk with other people and hear their point of view.

Count Iblis... I do not know... The whole point of that section of Huang's book is to make use of the microcanonical ensemble, i.e. counting the microstates. The canonical and grand-canonical approaches are fairly clear to me. The only problem I have is that (possibly apparent) inconsistency I am trying to point out.

Mapes. I'm not sure about that. It is true that MC ensemble assumes fixed volume. But I think that that applies during the calculation. At the end you get a volume dependent result, and I think nothing forbids differentiating. Otherwise how could one "derive" thermodynamics? (I'm never sure whether TD is an input or an output... But I'd like that whatever one does is at least consistent with TD).

Actually this is what Huang does in his book. I mean, he derives the "Sackur-Tetrode" equation (A) and says "We straightforwardly find $$PV = NkT$$". And actually if one uses

$$\frac{P}{T} = \frac{\partial S}{\partial V}$$

it works...

Plus, Huang uses the very same MC approach for the quantum statistics, and I see no problem in those cases. The pressure term is there.

I myself have a couple of possible ideas about "my problem":

1) Equation (B) is a bit deceptive, because the variables appearing in are not independent from each other. However, I am not able to back this vague statement with something more detailed.

2) The Boltzmann gas could be inherently pathologic. Actually Huang briefly mentions this about the fact that it does not satisfy the 3rd principle. But I'm a bit surprised there is no mention of the apparent inconsistency I have pointed out. After all, once you spot it, it is hard to ignore it.

I look forward to hearing more from you on this.

F

Count Iblis,

I realize now that I'm not sure about what you denote with A in your recent post. Is that Eq. (A) or the Helmoltz free energy (often called F)?
Anyway wouldn't your attempted explanation create some trouble in the case of Bose-Einstein and Fermi-Dirac statistics? If one follows the very same MC approach, the equivalent of Eq. (A) has an extra term, which turns out to be related to the pressure.

If, as you suggest, one has to introduce another extra term, I expect that things go wrong with the quantum statistics.

Mapes
Homework Helper
Gold Member
Mapes. I'm not sure about that. It is true that MC ensemble assumes fixed volume. But I think that that applies during the calculation. At the end you get a volume dependent result, and I think nothing forbids differentiating. Otherwise how could one "derive" thermodynamics? (I'm never sure whether TD is an input or an output... But I'd like that whatever one does is at least consistent with TD).
This is good point, and all ensembles are equivalent anyway for calculating bulk properties. So scratch that idea.

I took a look at McQuarrie, since it can be useful to compare derivations presented by different authors. He has the equivalent to your equation (C) above, which, as you mentioned, eliminates the problem. If equation (A) was derived without error, then it does seem that that method must make a simplification or assumption that drops the PV term.

(McQuarrie likes to compress terms by adding e inside the log, which is confusing at first because one looks for that variable's definition, or perhaps wonders why electrical charge appears in an ideal gas discussion. For example, he writes your equation (C) as

$$S=\frac{3}{2}Nk+Nk\ln\left[\left(\frac{2\pi m k T}{h^2}\right)^{3/2}\frac{Ve}{N}\right].[/itex] If Huang does the same, you might check for related errors, which could easily account for the missing Nk!) Mapes, this is definitely interesting! Actually I tried to find inspiration in other books, but had no success. I hadn't looked in McQuarrie's book, though, because I didn't know it. I see that the library here has several titles by McQuarrie. I assume you're referring to either "statistical mechanics" or "molecular TD". I'll definitely take look them up. I'll let you know. Thanks again F PS that's a deceptive way of writing eq (C)... I wonder whether the point of all this could be Stirling's approximation to factorials... Count Iblis, I realize now that I'm not sure about what you denote with A in your recent post. Is that Eq. (A) or the Helmoltz free energy (often called F)? Anyway wouldn't your attempted explanation create some trouble in the case of Bose-Einstein and Fermi-Dirac statistics? If one follows the very same MC approach, the equivalent of Eq. (A) has an extra term, which turns out to be related to the pressure. If, as you suggest, one has to introduce another extra term, I expect that things go wrong with the quantum statistics. I meant Eq.A Let'S think a bit more. When you derive: S = 1/T(E - mu N) What you are doing is presumably maximizing Log W and taking into account the constraints that the sum of ni is N and the sum of the enrgies of the single particle states time occupation number is E, via Lagrange multipliers. Then, I think that when you differentiate the expression for the total energy w.r.t. ni and multiply that with the Lagrange multiplier beta you didn't take into account that the energy levels also depend on ni, because you are not keeping the volume constant when you change the nuber of particles, and the energy levels depend on the volume, that volume dependence of the energy levels will yeiled the pressure using the definition of the pressure. Alternatively you can keep the volume constant but then you need another constraint and the Lagrange multiplier for that will yield the pressure term. Let'S think a bit more. When you derive: S = 1/T(E - mu N) What you are doing is presumably maximizing Log W and taking into account the constraints that the sum of ni is N and the sum of the enrgies of the single particle states time occupation number is E, via Lagrange multipliers. Right. This is exactly what I'm doing (or better, what K. Huang is doing). Then, I think that when you differentiate the expression for the total energy w.r.t. ni and multiply that with the Lagrange multiplier beta you didn't take into account that the energy levels also depend on ni, because you are not keeping the volume constant when you change the nuber of particles, and the energy levels depend on the volume, that volume dependence of the energy levels will yeiled the pressure using the definition of the pressure. Alternatively you can keep the volume constant but then you need another constraint and the Lagrange multiplier for that will yield the pressure term. Yes I sort of see what you mean, although the details are not clear to me. You say "when you differentiate the expression for the total energy "... But I am not sure I need to do this. I mean, as far as I understand the whole procedure is: 1. consider a given set of occupation numbers [tex]\{n_j\}$$.
2. count their "degeneracy" $$W\{n_j\}$$
3. find the set $$\{\bar n_j\}$$ maximizing $$W\{n_j\}$$ while meeting the constraint on the number and on the energy.
4. then the entropy is $$S= \ln W\{\bar n_j\}$$

If you write the last equation explicitly, you see it contains two terms, in which you recognize $$E$$ and $$\mu N$$. That is you get the first equality in Eq. (A).

There is no differentiation for the total energy....

The idea of the additional lagrange multiplier is interesting, I'll give it a try.
But if it works I would feel compelled to do the same for the Fermi-Dirac and Bose-Einstein statistics, which apparently work fine without an additional multiplier...

In point 3, when you take into acount the constraint on the total energy, you get a term

$$\beta\epsilon_{i}$$

Beta is the Lagrange multiplier and epsilon_i is the partial derivative of the total energy function w.r.t. n_i. But, in this case, you get an additional term:

$$\beta n_{i }\frac{\partial \epsilon_{i}}{\partial n_{i}}$$

because the partial derivatives w.r.t. the n_i are not at constant volume and the single particle energy levels do depend on the volume (that dependence yields the pressure).

You can write the partial derivative as the derivative of the single particle energy w.r.t. volume times the derivative of the volume w.r.t. n_i, and that last derivative is the same as the derivative of V w.r.t. N.

In point 3, when you take into acount the constraint on the total energy, you get a term

$$\beta\epsilon_{i}$$

Beta is the Lagrange multiplier and epsilon_i is the partial derivative of the total energy function w.r.t. n_i. But, in this case, you get an additional term:

$$\beta n_{i }\frac{\partial \epsilon_{i}}{\partial n_{i}}$$

because the partial derivatives w.r.t. the n_i are not at constant volume and the single particle energy levels do depend on the volume (that dependence yields the pressure).

You can write the partial derivative as the derivative of the single particle energy w.r.t. volume times the derivative of the volume w.r.t. n_i, and that last derivative is the same as the derivative of V w.r.t. N.
I might be wrong but your argument above does not convince me.
I see that the energy levels depend on the volume, but the link with the total number is not clear to me. Plus, here you are not deriving w.r.t. the total number. You are considering different arrangements of the same number of particles, isnt'it? How can the energy levels depend on the way you distribute the particles on them, or even on the total N? They are single particle levels. And aren't N and V assumed independent of each other?
Also, I think that whathever differentiation you are making during this calculation is at fixed volume by definition of MC ensemble. Of course the final result depends on the volume, and you can differentiate w.r.t V. But only when you're done with the calculation, I think.

As I say, if your argument makes sense one should apply it also in the quantum cases. Everything is the same, except the W's. But as far as I see the quantum cases work well with no need of an additional term. Is there a reason for introducing the additional term you are suggesting only for the Boltzmann gas?

Anyways thanks for your help. It's nice to hear different points of view, and I'll give more thought on your suggestion.

Looking at the previous points in more detail I think I have spotted a further term.

$$W\{n_j\}= \prod_j \frac{g_j^{n_j}}{n_j!}$$

$$\ln W\{n_j\}= \sum_j \ln \frac{g_j^{n_j}}{n_j!} = \sum_j n_j \ln \frac{g_j}{n_j} + \sum_j \ln n_j$$

Where I used the Stirling's approximation $$\ln n_j! = n_j \ln n_j - n_j$$

The book by Huang actually gives only the first term in the last equality.
However the additional sum does not look very easy to handle and right now it's not clear to me how it can result in the missing $$NkT$$ term.

Hi guys,

I think I've solved it. With hindsight it was fairly trivial.

My feeling that it had to do with the Stirling approximation was
correct after all, only I've made a really silly mistake in my last post.
I have written the missing term as $$\ln n_j$$ but it is simply $$n_j$$.
It occurred to me last night, when I decided I would make an attempt
to handle the "difficult term".
The correct calculation goes like this (I'm discarding the overbar
in the most probable distribution)

$$\frac{S}{k}= \ln W\{\bar n_j\} = \sum_j \left n_j [\ln g_j - \ln n_j + 1 \right] = \sum_j n_j \ln \frac{g_j}{n_j} \sum_j n_j = \beta(E - \mu N) + N$$

Thus, the sum inexplicably ignored by Huang appears to provide the
missing term, since in the case of an ideal classical gas $$N = PV/kT$$.

Both the thermodynamic equation (B) and the "real" Sackur-Tetrode
equation (C) are satisfied.

It remains totally unclear to me the reason why K. Huang carried out
the entire calculation without the above term, ending up with the doubly
defective "Sackur-Tetrode" equation (8.54).

I had a doubt that it was because one ignores such a term also in the
derivation of the maximal distribution. But, if I'm not making a blunder,
it appears not to be the case. Looking at section 4.3, where the calculation
is explicitly carried out (with a good deal of typos..) , one recognized that
term as the 1 appearing at exponent in the fugacity: $$z = e^{-\alpha - 1}$$,
where $$\alpha$$ is the Lagrange multiplier for the total number.

The only other thing I can think of is the assumption just before Eqs. (8.47) that
1 is negligible with respect to $$g_i$$.
But ignoring the missing term would be assuming that 1 is negligible with respect to
$$\ln g_i/n_i$$, which is not the same. And, obviously, the missing term is clearly
not negligible w.r.t. to the energy term, being 2/3 of it.

F

Hi, Franz and everybody interested in the entropy of gases!

It might be interesting for you to know that the Sackur-Tetrode equation is valid also for the translational part of the entropy of real and molecular gases with changing of the 5/2RT on the translational part of the enthalpy: H - Eint, where Eint means zero pressure limit of internal movements of molecules (rotations and vibrations). In this case the density of real gas in the S-T equation should be substituted by the monomer fraction density.
It was confirmed on a wide experimental basis and described in my publication:
Monomer fraction in real gases, Int. J. of Thermodynamics, Vol 11 (N1), March 2008, pp 1-9.

So, your question about correct form of the Sackur-Tetrode equation has answer - (C), with enthalpy of ideal gas, equal to 5/2RT.

Best regards,

Boris Sedunov