How to simplify using stirling approximation?

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The discussion focuses on the Stirling approximation in the context of statistical mechanics, specifically in deriving the multiplicity function \(\Omega(N, q) = \frac{(N-1+q)!}{q!(N-1)!}\) from Daniel Schroeder's "Introduction to Thermal Physics." Participants clarify that while both \(N\) and \(q\) are large, the factorial growth rates dictate which terms can be neglected. It is established that neglecting \(N-1\) in the numerator is incorrect due to its significant contribution when combined with \(q\) in the factorial, as demonstrated with numerical examples showing the vast difference in factorial sizes.

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eprparadox
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Hey!

So we're deriving something in Daniel Schroeder's Introduction to Thermal Physics and it starts with this:

[tex] \Omega \left( N,q\right) =\dfrac {\left( N-1+q\right) !} {q!\left( N-1\right) !}[/tex]

Both N and q are large numbers and q >> N.

The derivation is in the book, but I am always confused with when I can throw away terms.

For example, in this case intuitively, I would want to say that since q >> N and we have N - 1 + q, I would just keep the q and throw away the N - 1. I know this is wrong, but I don't know why.

In the book, they begin by throwing away the "- 1" term since both q and N are large numbers. I would have thrown away the N - 1. Why can't I do this and when can I/can't I throw away terms?

Thanks!
 
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Because N is involved in the numerator's factorial. Since factorials grow very fast, then a larger factorial will contribute quite a lot, even though the added size isn't quite that large in comparison.
Example, if N=10 and q=1000, then q! is on the order of 102567 and while (q+N)! is on the order of 102597 and may not seem much larger, it's in fact 1030 (a million trillion trillion) times larger.
[tex]q!=1\times2\times3\times...\times999\times1000 \approx 10^{2567}[/tex]
[tex]N! = 1\times2\times...\times10 \approx 10^6[/tex]
[tex](q+N)! = 1\times2\times3\times...\times999\times1000\times 1001\times 1002 \times ... \times 1010[/tex]
[tex]= q! \times 1001 \times 1002 \times... \times 1010 \approx q! \times 10^{30}[/tex]
N! is only 106 on its own, but when included with q in the factorial, it multiplies by 1030. If you neglect a factor of this size, it'll definitely make a noticeable difference in your approximation.
 
Last edited:
Hey, got it. Thanks so much!
 

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