(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

For a single large two-state paramagnet, the multiplicity function is very sharply peaked about [tex]N_{\uparrow} = N/2[/tex]

a) Use Stirling's approximation to estimate the height of the peak in the multiplicity function.

b) Use the methods in this section to derive a formula for the multiplicity function in the vicinity of the peak, in terms of [tex]x \equiv N_{\uparrow} - (N/2)[/tex] Check that your formula agrees with your answer to part (a) when x = 0.

c) How wide is the peak in the multiplicity function?

2. Relevant equations

Stirling's approximation: [tex]ln(N!) = N ln(N) - N [/tex]

# of microstates = [tex] \frac{N!}{N_{\uparrow}!N_{\downarrow}!}[/tex]

3. The attempt at a solution

The first question seems simple enough; I'll just plug in N/2:

[tex] ln \Omega = ln(N!) - ln((N/2)!) - ln((N/2)!) [/tex]

[tex] ln \Omega \approx Nln(N) - N - 2((N/2)ln(N/2)-N/2) = Nln(N)-Nln(N/2) = Nln(2)[/tex]

So the answer to (a) is 2^N. This seems problematic though because I think 2^N should be the total number of microstates for all macrostates, not just the macrostate where half the dipoles are "up"...Do I need to include the [tex]\sqrt{2{\pi}N}[/tex] term in Stirling's approximation?

For (b), I managed to get the formula:

[tex]\Omega = N^N[(N/2+x)(N/2-x)]^{-N/2}(\frac{N/2-x}{N/2+x})^x[/tex]

which does give me 2^N where x=0, but I expected to get a gaussian curve, especially considering part (c). But I can't see how I can get an exponential in the final answer because the exponentials seem to cancel out whether or not the square root term is in there.

What am I missing here? Am I on the right track at all?

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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# Homework Help: Stirling's approximation/Multiplicity

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