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At first sight one can say "why not to use only one term, or maybe three or more terms"? Why use two terms? I see that books (see for example good books like Griffiths quantum mechanics or Atkins physical chemistry) use the $\ln(x!)\approx x \ln x - x$ formula, but no one justify this choice. I did calculus using the smaller formula $\ln(x!)\approx x \ln x$ finding, for identical fermions and bosons, the same results that books find using the longer formula (i.e. $N_n = \frac{d_n}{e^{\alpha+\beta e_n} \pm 1}$ where $N_n$ is the number of particles in energy level $E_n$, with degeneracy $d_n$, while $\alpha$ and $\beta$ are the Lagrange multipliers).

The only reason I can speculate to choose $\ln(x!)\approx x \ln x - x$ instead of $\ln(x!)\approx x \ln x$ is that the approximation of the simpler formula is so bad that, despite it works for $x\to\infty$, it doesn't work in the physical context we are playing with. But if this is the reason, the choice should be accompanied by some considerations. A look at this plot suggest that this line of thinking is correct: even for a huge number like Avogadro's one, the difference with the two formulas is about 2\% (the ratio became bigger of 0.99 only for $2.688117141816367\cdot10^{43}$)

enter image description here

But these considerations are not sufficient to justify the use $\ln(x!)\approx x \ln x - x$. I see that $\ln(x!)\approx x \ln x$ is not appropriate for the proof of statistical quantum mechanics, but how can I see that $\ln(x!)\approx x \ln x - x$ is appropriate? The only way seems to be a comparison with the function $\ln(x!)$, but here problems arise.

  1. First. Surely we are interested to a number big and much smaller than Avogadro's number, but this sounds vague... what number should we elect as representative of the typical population of a typical energy level, to test the proper functioning of the approximated formula?
  2. Second. When the argument of $\ln(x!)$ starts becoming big, it put a strain even the powerful computing capabilities of computers

I did this plot

enter image description here

that suggests that this line of thinking too could be correct (for $x=170$, the maximum value hold by the junk of my computer before crashing, the error is about $0,5\%$), but the plot is not decisive and I'd like something stronger, solving satisfactorily the two problems I listed above. A way to proceed could be finding a way to maximize the error done by using $x\ln x - x$ instead of $\ln(x!)$ and then show that this error is not significative for every reasonable value of $x$, but I can't do it.

Adjunct

I'm asked to write down the proof, I'll report the final: it is possible to show under reasonable assumption that the quantum statistics can be found by finding the maximum (with constraints of conservation of particles and energy) of $Q$, where $Q$ for fermions and bosons is respectively $$ Q_{ferm} = \prod_{n} \frac{d_n!}{N_n!(d_n-N_n)!} $$ $$ Q_{bos} = \prod_{n} \frac{(N_n+d_n-1)!}{N_n!(d_n-1)!} $$ and exploiting the hypothesis $1 \ll N_n \ll d_n$. Taking logarithm does not alter the positions of maxima (this step transforms products in sums and will allow us to exploit Stirling). Exploiting Lagrange method we have $$ \frac{\partial}{\partial N_n} \left[ \ln Q + \alpha \left( N - \sum_n N_n \right) + \beta \left( E_n - \sum_n N_n E_n \right) \right]=0 $$ Going on we get to the crucial step, connected with this question: for fermions and bosons we have $$ \frac{\partial }{\partial N_n} \sum_n [ \ln ( d_n! ) - \ln ( N_n ! ) - \ln(d_N - N_n)! ] = \alpha + \beta E_n $$ $$ \frac{\partial }{\partial N_n} \sum_n [ \ln (N_n + d_n - 1)! - \ln ( N_n ! ) - \ln (d_n - 1)! ] = \alpha + \beta E_n $$ from wich you can see that using indifferently $\ln(x!)\approx x \ln x$ or $\ln(x!)\approx x \ln x - x$ you will find $N_n = \frac{d_n}{e^{\alpha+\beta e_n} \pm 1}$. Anyway this is not the real problem that trouble me, the problem is "how can I see that taking two terms in $\ln(x!) \approx \dots$ is reasonably safe?".

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  • $\begingroup$ Can you discuss which derivations in statmech you are talking about, and give examples where the other approximation also works? This is likely more valuable than plots comparing the approximation. $\endgroup$ – Norbert Schuch Jul 27 '18 at 21:45
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – David Z Jul 28 '18 at 9:15
  • $\begingroup$ I have found it easier to understand some derivations that do not use entropy and logarithms. For example, it is easier to demonstrate that the fractional change of multiplicity with energy is a measure of temperature, and then proceed that this is the same as ${\rm d} \ln \Omega/{\rm d}E$ by the chain rule. $\endgroup$ – Pieter Jul 30 '18 at 11:44
  • $\begingroup$ @Pieter where can I find your proof? $\endgroup$ – Fausto Vezzaro Jul 30 '18 at 21:48
  • $\begingroup$ @FaustoVezzaro Not a rigid proof, but I wrote here: physics.stackexchange.com/questions/404797/… And then one can proceed to get the Boltzmann factor: physics.stackexchange.com/questions/412215/… $\endgroup$ – Pieter Jul 31 '18 at 7:56
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The essential point is: it depends on the necessary accuracy. However generally in the thermodynamic limit we can ignore terms subleading $N$ (see edit).

An example: the Sackur-Tetrode formula

Say we're trying to obtain the Sackur-Tetrode formula for entropy. Starting from the parition function for a $3D$ ideal gas, $$Z=\frac{V^N}{\lambda^{3N}N!}$$ and using the formula for entropy $S = \frac{\partial}{\partial T}(kT \log{Z})$ one gets exactly $$S =N k\log{\frac{V}{\lambda^{3}}}+\frac{3}{2}Nk-k\log{N!}$$ Now we can make the approximation $$\log{N!} = N\log{N}+\mathcal{O}(N)$$ and we get: $$S =N k\log{\frac{V}{\lambda^{3}N}}+\mathcal{O}(N)\tag{1}$$ which is the correct formula to $\mathcal{O}(N)$ because $V$ scales like $N$. If instead we took one further term: $$\log{N!} = N\log{N}-N+\mathcal{O}(1)$$ The approximation would be more precise: $$S =N k\left[\log{\frac{V}{\lambda^{3}N}}+\frac{5}{2}\right]+\mathcal{O}(1)\tag{2}$$ Equation $(2)$ is correct to one further order, but depending on what you're looking at $(1)$ may be just as good. Note that it's pointless to take further orders (say $\mathcal{O}(1/N)$) because we're in the large $N$ regime.

Another example: deriving the quantum statistics

I assume the derivation goes like in this question. You're trying to maximise an entropy of the form: $$S = \log{\frac{N!}{n_i ! (N-n_i)!}}$$ subject to certain constraints. In order to take derivatives of $S$ we must apply Stirling's approximation: $$S = \log{N!} - n_i \log{n_i}-(N-n_i) \log{(N-n_i)}+\mathcal{O}(n_i)$$ or to one further order after cancellation: $$S = \log{N!} - n_i \log{n_i}-(N-n_i) \log{(N-n_i)}+N+\mathcal{O}(1)$$ In this case the correction is actually independent of $n_i$, so when taking derivatives both approximations yield the same result. However you do not know beforehand whether this is case, so you should use the $\mathcal{O}(1)$ approximation.

EDIT: When doing statistical mechanics we almost always work in the so-called thermodynamic limit: $N,V \to \infty$ while keeping $N/V$ constant. This means that extensive quantities (for instance the entropy), which scale like $N$, will go to infinity. In order to extract useful results, it is useful to define stuff like "entropy per particle" $s = S/N$. This remains finite in the thermodynamic limit. It is clear at this point that the only terms which contribute in the thermodynamic limit are at least $\mathcal{O}(N)$, and subleading terms are irrelevant.

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  • $\begingroup$ Well, in some sense (2) is more correct than (1). (1) pretends that the $V/\lambda^3$ is meaningful, while in fact it isn't, since it is just another $O(N)$ term, so you should correctly write $S=-Nk\log N + O(N)$ (which seems odd, so I guess you want to use that $V$ scales in some way with $N$). $\endgroup$ – Norbert Schuch Jul 28 '18 at 15:21
  • $\begingroup$ @Norbert Schuch Yes I was thinking about extensivity. Perhaps a better statement would be that it is as good depending on what you're doing. I will update later $\endgroup$ – John Donne Jul 28 '18 at 16:03
  • $\begingroup$ @JohnDonne In the proof I wrote above (you can find more details in Griffiths) there is no explicit mention of entropy and the logarithm only serves to break production in summation and to exploit Stirling approximation (even if the maximization of entropy is certainly a possible angle from which see this problem). In the proof we are searching the maximum with constraints of a $\mathbb{R}^\infty \to \mathbb{R}$ function (each point in the domain is a configuration). I find difficult what you wrote: shouldn't $N$ have a subscript $N_n$? And with respect to derive? Probably is the same proof? $\endgroup$ – Fausto Vezzaro Jul 30 '18 at 21:46
  • $\begingroup$ @Fausto Vezzaro My proof is the same as yours. Note that the constant $N$ in my proof corresponds to the $d_n$ in yours, which are given and nowhere one differentiates with respect to them. Generally you should take as many terms as needed. As explained in my answer there's no need to take terms $\mathcal{O}(1/N)$ and since we're differentiating it's also pointless to take the constant term. The only terms left are those of the form $x\log{x}-x$ $\endgroup$ – John Donne Jul 30 '18 at 22:09
  • $\begingroup$ I think it is useful to emphasize that while (1) is correct, it is rather meaningless: the first term is itself $\mathcal{O}(N)$, so it's of the same order as the neglected term. Only (2) yields a sensible approximation for large $N$ (indeed, 5/2 might be much larger than the other term, depending on the values of $V/N$ and $\lambda$). $\endgroup$ – Ori Aug 8 '18 at 14:28
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Let's start by writing two more terms in Stirling's approximation: $$\log N! = N \log N - N +\frac{1}{2}\log N + \frac{1}{2}\log 2\pi + \mathcal{O}(\frac{1}{N}).$$

On an intuitive level, the main point is that $\log$ increases very slowly, and thus for the typical numbers of statistical mechanics, i.e., Avogadro's number, $\log 10^{24} = 24 \log 10$ which is about 24, i.e., much much less than $10^{24}$ and not much more than 1. Thus, $\log N$ can be safely neglected compared to $N$, but if $\mathcal{O}(\log N)$ terms are kept it is often prudent to also keep $\mathcal{O}(1)$ terms.

In fact, statistical mechanics is almost always accurate only "to leading order in the exponent" (before taking logs, i.e., when calculating partition functions or the number of states). So in the thermodynamic variables, terms of order $\log N$ are always neglected (think, e.g., of the difference between the Boltzmann and Gibbs entropies, or the difference between the log of the partition function and the Legendre transform of the entropy). The more mathematically sophisticated (or at least jargon-heavy) way of saying this uses the language of the theory of large deviations: thermodynamic potentials such as entropies and free energies are rate functions, defined (at least implicitly) by limits of the form $S(E) = \lim_{N\to\infty} \frac{1}{N} \log \Omega(E)$ (where $\Omega(E)$ is the number of states with energy $E$). In these limits, the higher order terms in Sterling's approximation do not contribute.

A further comment on why both $N \log N$ and $N$ are kept and not only the first of these: usually, thermodynamic quantities are either extensive ($\mathcal{O}(N)$) or intensive ($\mathcal{O}(1)$). So the $N \log N$ terms usually cancels out in the calculation and the leading term is $\mathcal{O}(N)$. This is what happens, e.g., in the Sackur-Tetrode example given by John Donne, or when approximating bynomial coefficients: $\log {{N}\choose{xN}}\simeq N[x \log x + (1-x)\log(1-x)]$ (where $0<x<1$). (In fact, when the $N\log N$ term does not cancel, people often subtract it by hand, e.g. by choosing to work with relative entropies rather than entropies).

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  • $\begingroup$ I upvoted your answer. Observing how the development of ln(x!) continues is interesting: if N is some thousand (or bigger until Avogadro's number too) the first two terms are approximately of the same order of magnitude while other terms are negligible. This is not a proof, but it is a convincing justification: it is safer to use both terms. $\endgroup$ – Fausto Vezzaro Aug 28 '18 at 19:33

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