To derive the entropy of an ideal gas via the ergodic hypothesis, we first find the density of states function: $$g(E)=\frac{V^{N}}{h^{3N}}\frac{(2\pi m E)^{\frac{3N}{2}}}{\left(\frac{3N}{2}-1\right)!}\frac{1}{E}$$

The multiplicity is then give by $\Omega=\int_{E-\frac{\Delta}{2}}^{E+\frac{\Delta}{2}}g(E')dE'\approx \frac{V^{N}}{h^{3N}}\frac{(2\pi m E)^{\frac{3N}{2}}}{\left(\frac{3N}{2}-1\right)!}\frac{\Delta}{E}$, for small $\Delta$.

Then using the definition of entropy $S=k_{B} \ln(\Omega)$, and applying Stirling's approximation, as well as assuming that $N \gg 1$, I obtained:

$$S=Nk_{B}\left(\frac{5}{2}+\ln\left(\frac{V}{N}\left(\frac{4\pi m E}{3Nh^{2}}\right)^{\frac{3N}{2}}\right)\right)+k_{B}\ln\left(\frac{\Delta}{E}\right)$$

The first term is the Sackur Tetrode equation which is known to be correct, but the second term seems wrong.

Most resources seem to appeal to the fact that $\Delta$ is arbitrarily small, and set $k_{B}\ln\left(\frac{\Delta}{E}\right)=0$, but if $\Delta \ll E$, then $\frac{\Delta}{E}\ll1$, which would mean $k_{B}\ln\left(\frac{\Delta}{E}\right)\ll 0$, i.e. the extra term is approaching $-\infty$, rather than $0$, and is thus not negligible. In short, I can't find a way to jsutify throwing away this tem, as it seems to blow up rather than vanish for small $\Delta$, and an esplanation would be much appreciated.

  • $\begingroup$ Since $\frac{\Delta}{E}\sim\frac{1}{\sqrt{N}}$ the second term is negligibly small in the thermodynamic limit ($N\rightarrow\infty$) $\endgroup$ Mar 4, 2018 at 15:54
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    $\begingroup$ That makes sense, but I dont understand why $\frac{\Delta}{E}\sim\frac{1}{\sqrt{N}}$. I also realised I had left out a factor of $\Delta$ in the multiplicity, but that is fixed now. Thanks for your help! $\endgroup$
    – CoffeeCrow
    Mar 6, 2018 at 2:07
  • $\begingroup$ @CoffeeCrow, $\Delta$ is a peak energy width, $g(E)=d \Gamma/dE $ by definition so it can not include $\Delta$ $\endgroup$ Mar 6, 2018 at 6:05
  • $\begingroup$ Sorry, I accidentaly changed it on both the multiplicity and density of states instead of just the multiplicity. Its fixed now. $\endgroup$
    – CoffeeCrow
    Mar 6, 2018 at 8:45
  • $\begingroup$ You should ask yourself why the integral is in a band of size $\Delta$ around $E$. What you really want is $S(E)$ so why integrate at all. The physical motivation is that if you are looking at a subsystem (so open) with average energy $E$ then being extensive $E \sim N$ and $Var(E) \sim N$. The second quantity is by definition $\Delta^2$. This comes from independence of fluctuations of individual molecules. $\endgroup$ Mar 6, 2018 at 9:07

1 Answer 1


You seem to have calculated $\ln \Omega$ assuming permutations are distinguishable and assuming energy is distributed in range $E-\Delta/2..E+\Delta/2$. This is OK mathematically, but leads to:

  • exponent $3N/2$ which means the result is not a homogeneous function of first order of variables $E,V,N$ like thermodynamic entropy is. One consequence of this is that this "entropy" does not obey usual thermodynamic relations like $T=1/(\partial S/\partial E)$. For the concept of homogeneity in thermodynamics, check e.g. https://faculty.uca.edu/saddison/GordonConference/GordonConferencePoster.pdf ;

  • even if correcting factor is introduced to $\Omega$ so the resulting expression is homogeneous, the result depends on the width $\Delta$, which to some extent is arbitrary.

In expressions of function $S(E,V,N)$ (for thermodynamic entropy), magnitude of some term does not matter much; a term may be installed or written off even if it is very big, provided it does not depend on $N$ and $E$. So magnitude is not the reason some terms are written off.

The way you did the calculation - using some arbitrary $\Delta$ - leads to result that depends on $\Delta$, but only very weakly. Sometimes such non-complying term is written off, if it can be shown that the term has negligible consequences on the measurable physical quantities.

What matters are magnitudes of derivatives of entropy with respect to $N$ and $E$. Let us assume we have expression that is Sackur-Tetrode plus your problematic term:

$$ S = Nk_{B}\left(\frac{5}{2}+\ln\left(\frac{V}{N}\left(\frac{4\pi m E}{3Nh^{2}}\right)^{\frac{3}{2}}\right)\right)+k_{B}\ln\left(\frac{\Delta}{E}\right) $$

Calculating chemical potential defined as

$$ - T \frac{\partial S}{\partial N}_{E,V=const} $$

the arbitrary term gives no contribution, as it does not depend on $N$.

Calculating inverse of temperature defined as

$$ 1/T = \frac{\partial S}{\partial E}_{V,N=const} $$

the Sackur-Tetrode expression gives contribution $\frac{3}{2}N k_B / E$, while the arbitrary term gives contribution $-k_B/E$. This second one is much smaller and makes only a small difference in the resulting temperature.

Since the arbitrary term does not depend on any other physical quantity, those are all derivatives that matter. So the arbitrary term can be neglected.

  • $\begingroup$ I may be wrong here, but I think that the quantity that should go into $S=k_B \ln (...)$ is not OP's $\Omega$, but $g(E)$, which is positive and independent from $\Delta$...See my answer and the references from Tuckerman. $\endgroup$
    – valerio
    Mar 7, 2018 at 13:57
  • $\begingroup$ @valerio, it is true that $S=k_B\ln (d\Omega/dE)$ is a more elegant definition for statistical entropy, as there is no arbitrary energy width involved. But I think the real moral of the story of OP's problem is that the presence or magnitude of $\Delta$ does not matter, at least not for thermodynamics; all variants lead to same results in practice, where $N$ is very large number. $\endgroup$ Mar 7, 2018 at 19:48
  • $\begingroup$ You are right, it shouldn't matter in the limit of large $N$. $\endgroup$
    – valerio
    Mar 7, 2018 at 22:12

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