# Approximation of multiplicity when Ideal gas low density is applied $\frac{M !}{(M-N)!} \approx M^{N}$

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

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?

• 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? – rob May 18 '19 at 0:04
• @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! – vik1245 May 18 '19 at 0:10
• @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. – rob May 18 '19 at 0:50
• @rob have contacted now and just realised! Many thanks. – David Smith May 18 '19 at 9:33

\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 $$\frac{M !}{\left(M-N\right)!} ~\approx~ M^N \,.$$

• Apparently MathJax complains about nesting align blocks? – Nat May 18 '19 at 0:18
• More like the line width got overflow. Nice typesetting by the way. – acarturk May 18 '19 at 0:28

$$M! = M \times (M-1) \cdots \times (M-N+1) \times (M-N)!$$ $$\frac{M!}{(M-N)!}=\frac{M \times (M-1) \cdots \times (M-N+1) \times (M-N)!}{(M-N)!}$$ 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.

• 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? – vik1245 May 18 '19 at 0:00
• @BobSmith I realised that I did some mistakes. Can you look again?. – acarturk May 18 '19 at 0:02
• 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! – vik1245 May 18 '19 at 0:05
• @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 $i$. – acarturk May 18 '19 at 0:09
• Minor note: the spacing is nicer if you use \ll and \gg rather than << and >>. Compare $\ll$ to $<<$. – rob May 18 '19 at 0:15

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 $$\ln \left[ \frac{M!}{(M- N)!} \right] \approx N \ln M + \mathcal{O} \left(\frac{N}{M}\right)^2.$$

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?