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I am trying to understand Pauling's derivation of the residual entropy of ice (link).

First let me paraphrase Pauling's argument.

He supposes that every oxygen atom (let's call one of them $A$) is "joined" to four other oxygen atoms (call them $B_1,B_2,B_3,B_4$) in a wurtzite structure. By "joined" we mean there is a single hydrogen atom between them, and the hydrogen atom can lie either closer to $A$ or to one of the adjacent $B_i$, so that there are $2^{2n}$ different configurations of hydrogen atoms if we have $n$ oxygen atoms.

He then imposes the restriction that every oxygen atom must have two close hydrogen atoms and then asks what fraction of the original $2^{2n}$ configurations remain.

He then analyzes a single oxygen atom and finds that of the 16 possible configurations of the four surrounding hydrogen atoms only 6 satisfy the restriction. He then concludes that of the original $2^{2n}$ configurations only a fraction $\left(\frac{6}{16}\right)^n = \left( \frac{3}{8} \right)^n$ of them satisfy the constraint.

The last step is what I'm struggling to understand. He appears to be arguing the following:

Suppose we take all the $\Omega_m$ configurations that satisfy the restriction for $m < n$ of the oxygen atoms and consider an additional oxygen atom. Since each configuration of its surrounding four hydrogen atoms occurs equally often, and only $\frac{3}{8}$ of them satisfy the restriction, we get that $\Omega_{m+1} = \frac{3}{8}\Omega_m$ so that $\Omega_n = 2^{2n} \left( \frac{3}{8} \right)^n = \left(\frac{3}{2}\right)^n$.

Am I correct that this is indeed his reasoning? If so I do not see how to prove the italicized statement.

I would start by partitioning the set $\tilde{\Omega}_m$ of configurations satisfying the restriction (i.e. $|\tilde{\Omega}_m|=\Omega_m$) for the first $m$ oxygen atoms and partitioning it into sets $\tilde{\Omega}_m^c$ of configurations that have the same configuration $c$ for the four hydrogen atoms surrounding the $m+1^{\text{th}}$ oxygen atom (i.e. $c=1,\ldots,16$). Then there should be one-to-one mappings between the different $\tilde{\Omega}_m^c$. I can see how, for instance, if $c$ is related to $c'$ by toggling each of the surrounding hydrogen atoms, we can show that $|\tilde{\Omega}^c_m|=|\tilde{\Omega}^{c'}_m|$ by the mapping which toggles all $2n$ hydrogen atoms. But for two general $c$ and $c'$ it is not obvious to me that $|\tilde{\Omega}^c_m|=|\tilde{\Omega}^{c'}_m|$.


Note: If $c$ and $c'$ differ by a single hydrogen atom, it seems possible to make a mapping by toggling the hydrogen atom and then moving outwards from the oxygen atom towards the surface of the ice, toggling hydrogen atoms as necessary so that the resulting configuration still satisfies the restriction. It seems plausible that this mapping is invertible and that any $c$ and $c'$ could be brought into one-to-one correspondance by composition of such maps. The strategy seems pretty convoluted though.

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    $\begingroup$ He seems to be pretending the constraints are all independent, and therefore saying that (3/8)^N of the possible configurations satisfy all constraints. This can't be exactly true - because the number of states you end up with, (3/2)^N, isn't an integer! I am having trouble thinking of a way to figure out how good an approximation it is, however. $\endgroup$
    – Kuma
    Commented Sep 26, 2019 at 22:36
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    $\begingroup$ He is indeed using an approximation of independent constraints. Checking the accuracy of his formula is nontrivial. See e.g. the history section of this article: en.wikipedia.org/wiki/Ice-type_model . $\endgroup$ Commented Sep 27, 2019 at 7:43
  • $\begingroup$ @Kuma and Zarathustra: Interesting thanks. It is odd how matter of factly he presents the proof considered its fundamental invalidity. $\endgroup$ Commented Sep 27, 2019 at 15:37
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    $\begingroup$ Thanks for the great link, @Zarathustra. $\endgroup$
    – Kuma
    Commented Sep 27, 2019 at 15:52

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