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Our lecturer today mentioned how a piston's head being at equal pressure maximised the multiplicity of states.

enter image description here

He mentioned the following:

If I have a fixed number of particles $N_A$ on left and $N_B$ on the right, and the whole system has a fixed total volume of $M$, we can say the movable piston partitions the total volume into $M_A$ and $M_B$ lattice sites respectively on each side, and therefore

$$M_{total} = M_{A}+M_{B} = constant$$

Hence if we want the volumes of the two sides that maximise the multiplicity function,

$$\Omega(N, M)=\frac{M !}{(M-N) !} \frac{1}{N !}$$

This all made perfect sense. The next step is where I have confusion.

As we are dealing with an ideal gas, the densities are so low that $\frac{N}{M} << 1$ hence

$$\frac{M !}{(M-N)!} \approx M^{N}$$

I understand the approximation as there are fewer molecules N than there are sites M for an ideal gas, but I fail to understand how he managed to reach the conclusion above from the previous equation above that.

I tried to solve this using the approximation:

$$x! = \left(\frac{x}{e}\right)^x$$ but this failed to reach me to the correct working in order to prove the $\approx M^{N}$ approximation.

I managed to get:

$$\frac{M^M}{(M-N)^{M-N}}\times \frac{1}{N^N}$$

but I suspect this is deviating from the main way of reaching the approximation.

How is the approximation achieved?

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  • $\begingroup$ Your multiplicity expression $\Omega$ has a factor $1/N!$ which is missing from the approximation in your title, and in the line you quote after "densities are so low." Is that intentional? $\endgroup$ – rob May 18 at 0:04
  • $\begingroup$ @rob yes the line given in the notes was "We are dealing with gases so low densities N/M<<1" and the approximation was also given as above without the $\frac{1}{N!}$ factor as part of the notes! $\endgroup$ – vik1245 May 18 at 0:10
  • $\begingroup$ @BobSmith By the way: If you accidentally created two accounts, please use the "contact" link at the bottom of any page to request a merge. $\endgroup$ – rob May 18 at 0:50
  • $\begingroup$ @rob have contacted now and just realised! Many thanks. $\endgroup$ – David Smith May 18 at 9:33
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$$ \frac{M !}{\left(M-N\right)!} ~=~ \begin{alignat}{10} M & \times & \left(M - 1\right) & \times & \left(M - 2\right) & \times & ~\cdots~ & \times & \left(M - N + 1\right) & \times & \left(M - N\right) & \times & \left(M - N - 1\right) % & \times & \left(M - N - 2\right) & \times & ~\cdots~ % & \times & 2 & \times & 1 \\[-25px] \hline & & & & & & & & & & \left(M - N\right) & \times & \left(M - N - 1\right) % & \times & \left(M - N - 2\right) & \times & ~\cdots~ % & \times & 2 & \times & 1 \end{alignat} $$

$$ \begin{align} &~=~ M \times \left(M - 1\right) \times \left(M - 2\right) \times ~\cdots~ \times \left(M - N + 1\right) \\[10px] % &~=~ \prod_{i=M-N+1}^{M}{i} \end{align} $$

Since $\frac{N}{M} \ll 1 ,$ then $M \gg N,$ and $M \gg 1,$ so $$ \begin{align} \frac{M !}{\left(M-N\right)!} &~=~ M \times \underbrace{\left(M - 1\right)}_{\approx M} \times \underbrace{\left(M - 2\right)}_{\approx M} \times ~\cdots~ \times \underbrace{\left(M - N + 1\right)}_{\approx M} \\[10px] &~\approx~ \underbrace{M \times \cdots \times M}_{N~\text{times}} \\[10px] &~=~ M^N \,, \end{align}$$ so $$ \frac{M !}{\left(M-N\right)!} ~\approx~ M^N \,.$$

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  • $\begingroup$ Apparently MathJax complains about nesting align blocks? $\endgroup$ – Nat May 18 at 0:18
  • $\begingroup$ More like the line width got overflow. Nice typesetting by the way. $\endgroup$ – acarturk May 18 at 0:28
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$$M! = M \times (M-1) \cdots \times (M-N+1) \times (M-N)!$$ $$\frac{M!}{(M-N)!}=\frac{M \times (M-1) \cdots \times (M-N+1) \times (M-N)!}{(M-N)!}$$ The terms $(M-N)!$ cancel, $$\frac{M!}{(M-N)!}=\prod\limits_{i=0}^{N-1}\,(M-i)$$ We know $i < N$. Assuming $N \ll M$, we get $i \ll M$ and we can write $(M-i)$ as $M$, then $$\frac{M!}{(M-N)!}=\prod\limits_{i=0}^{N-1}\,M = M^N.$$

Sorry, my initial response was wrong.

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  • $\begingroup$ But surely if you approximate $(M-N) \approx M$ then $\frac{M!}{(M-N)!} \ approx \frac{M!}{M!} = 1$ how can that work? or am I thinking about this matter incorrectly? $\endgroup$ – vik1245 May 18 at 0:00
  • $\begingroup$ @BobSmith I realised that I did some mistakes. Can you look again?. $\endgroup$ – acarturk May 18 at 0:02
  • $\begingroup$ I understand the first part correctly but my only confusion comes from how you managed to go from $\frac{M!}{(M-N)!}$ to $\prod\limits_{i=0}^{N-1}\,M$ although I understand why you write $(M-i)$ as $M$ using the approximation - once I understand that then I can accept your answer! The final step makes sense too - just the middle part of the final step! $\endgroup$ – vik1245 May 18 at 0:05
  • $\begingroup$ @BobSmith The assumption $N<<M$ also means that the dummy variable $\forall i: i<<M$. The trick is to open the biggest $N$ terms of $M!$, and see that the rest, i.e. $(M-N)!$ cancel with the denominator. After that? $M>>i$ therefore the first degree terms should consist only of $M$. Since this assumption gives us the correct result, I need not factor in $i$. $\endgroup$ – acarturk May 18 at 0:09
  • $\begingroup$ Minor note: the spacing is nicer if you use \ll and \gg rather than << and >>. Compare $\ll$ to $<<$. $\endgroup$ – rob May 18 at 0:15
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Just to see how to do it using Stirling's formula: \begin{align} \frac{M!}{(M- N)!} &= \frac{M^M e^{-M}}{(M-N)^{M-N} e^{-(M-N)}} \\ &= \frac{M^N}{e^N} \left( \frac{M}{M- N} \right)^{M-N} \\ &= \frac{M^N}{e^N} \left( 1 - \frac{N}{M-N} \right)^{-(M-N)} \end{align} But since $N \gg M$, we have $x \equiv (M-N)/N \gg 1$ and so
$$ \left( 1 - \frac{1}{x} \right)^{-Nx} = \left[ \left( 1 - \frac{1}{x} \right)^{x}\right]^{-N} \approx e^{N} $$ using the approximation $(1 - 1/x)^x \approx e^{-1}$ for $x \gg 1$. Thus, $$ \frac{M!}{(M- N)!} \approx M^N. $$

More generally, you can use Stirling's formula in the form $\ln n! \approx n \ln n - n$ to calculate corrections to this approximation. If I've done my derivation correctly, for example, it turns out that $$ \ln \left[ \frac{M!}{(M- N)!} \right] \approx N \ln M + \mathcal{O} \left(\frac{N}{M}\right)^2. $$

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When density is low one could in principle follow the particles for a while. Assume the particles are distinguishable. Suppose there is one particle and that there are six sites: Ω = 6. Now two particles, a red one and a blue one - Ω = 36. Etcetera.

Interesting to note though that this approximation fails if there is a small hole in the partition. What final distribution would maximize Ω?

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