Sterling approximation, multiplicity, and two-state paramagnet

[Ecoi]

Homework Statement

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

(a) Use Stirling's approximation to estimate the height of the peak in the multiplicity function. (I am fairly confident in my answer here)

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

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

(d) I should be okay on this part.

(My professor said to not do b, but c seems impossible to me without having to do b first.)

Book: Thermal Physics, Danial Schroeder - Chapter 2, section 4.

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Homework Equations

Multiplicity: $$\Omega = \frac{N!}{N_{\uparrow}! N_{\downarrow}!}$$

Displaying wrong?
(or... omega = N! / [(Nup)! (Ndown)!]

Displaying wrong?
Sterling approximation: $$N! = {N^N}{e^(-N)}{\sqrt{2N \pi} }$$

(or... N! = (N^N)(e^-N)(sqrt 2piN)***I had a heck of a time getting the formulas on here. I would type something and it would pop up as "$$\Omega = \frac {N_{\uparrow}!}{N_{\downarrow}!}$$" whatever I typed in. Sorry for not all of them being in latex format. Maybe I'm doing something wrong.

*Edit: If the formulas display incorrectly, I typed them out explicitly what they should be... If someone knows what I'm doing incorrectly on the latex code, it would help me to see what (I am not that used to it yet).

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The Attempt at a Solution

For part (a), I got:

$$N_{\uparrow} = N/2$$

so... using the sterling approximation I obtain

Multiplicity (1/2) = $$\frac{2^{(2N + 1)/2}}{\sqrt{\pi N}}$$

(or... multiplicity (1/2) = 2 ^ [(2N + 1) / 2] / (sqrt pi N)

I am fairly confident about this, I was wondering whether or not to neglect the 1 in the 2N + 1 or not. I figured I would see if it was required later and drop it if it is convenient to later.For part (c) (since part b wasn't required), I tried to think of approximating it using a rectangle, but that would be a very rough approximation. I don't really see how we are suppose to do part (c) without (b).I'm also a little confused on part (b) if I do have to do it to do part (c). What I have to do is derive a formula for the multiplicity near the peak. This would seem fairly straightforward, but I really don't see any formulas close to what was done in the book. The book derived the expression for two large Einstein solids. However, the author assumed q << N. In this example, however, I can't say N(up) or N(down) << N.

There's a formula for N(down) << N that is: multiplicity = [ Ne / N(down)] ^ N(down), but I see no reason to assume N(down) << N. These are really the only formulas I see for a two-state paramagnet.

If I don't need to do part (b) for (c), then help on that would be great. But it seems to me that I need part (b) for (c). I am pretty sure I could do part (c) if I knew how to do part (b) though. (Because it should be the maximum multiplicity and have some term multiplied by it.)

 

[mark726]

For part (b), you can use the binomial theorem to expand the multiplicity function in terms of x = N_up - (N/2).

\Omega = \frac{N!}{N_{\uparrow}! N_{\downarrow}!} = \frac{N!}{(N/2 + x)!(N/2 - x)!}

Using Stirling's approximation for N!, we have:

\Omega \approx \frac{\left(\frac{N}{2}\right)^N e^{-N}}{\left(\frac{N}{2}+x\right)^{\frac{N}{2}+x} e^{-(\frac{N}{2}+x)}\left(\frac{N}{2}-x\right)^{\frac{N}{2}-x} e^{-(\frac{N}{2}-x)}}

Simplifying, we get:

\Omega \approx \frac{2^N e^{-N}}{\left(\frac{N}{2}\right)^N e^{-N} (1+\frac{2x}{N})^{\frac{N}{2}+x} (1-\frac{2x}{N})^{\frac{N}{2}-x}}

Using the fact that (1+x)^n \approx 1+nx for small values of x, we can further simplify:

\Omega \approx \frac{2^N e^{-N}}{\left(\frac{N}{2}\right)^N e^{-N} (1+\frac{2}{N}(N/2)x) (1-\frac{2}{N}(N/2)x)}

\Omega \approx \frac{2^N e^{-N}}{\left(\frac{N}{2}\right)^N e^{-N} (1+x)(1-x)}

\Omega \approx \frac{2^N e^{-N}}{\left(\frac{N}{2}\right)^N e^{-N} (1-x^2)}

\Omega \approx \frac{2^N e^{-N}}{\left(\frac{N}{2}\right)^N e^{-N}} \frac{1}{(1-x^2)}

\Omega \approx 2^N \frac{1}{(1-x^2)}

This is the formula for the multiplicity function near the peak, in terms of x. When x =