Question

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.

(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) Suppose you flip $1,000,000$coins. Would you be surprised to obtain heads and 499,000 tails? Would you be surprised to obtain 510,000 heads and 490,000 tails? Explain.

Short answer

a) The peak of the height within the multiplicity function ${\mathrm{\Omega }}_{max}\approx 2^N$

b) The multiplicity function within the vicinity of the height, $\mathrm{\Omega }\approx {\mathrm{\Omega }}_{max}{e}^{-\frac{2x^2}{N}}$

c) The height within the multiplicity function is $w=\sqrt{2N}$

d) While a results of $501,000$ individuals falls just ahead of the apex, this isn't very astonishing. A conclusion of $510,000$ heads, from the opposite hand , is significantly outside the high and would are unusual.

Step-by-step solution

Microstates Quantity (a)

(a) Assume we get a paramagnet has $N$ magnets, and shut to half them are as in spin-up state, giving in $N\uparrow =$$\frac{1}{2}N\text{,}$, where $N_{\downarrow }$by signifies the energy units. This method provides quantity of microstates:

${\mathrm{\Omega }}_{max}=\left(\begin{array}{c}{N}_{\downarrow }+N_{\uparrow }+1\\ N_{\downarrow }\end{array}\right)=\left(\begin{array}{c}\frac{N}{2}+\frac{N}{2}+1\\ \frac{N}{2}\end{array}\right)$

when N is solely an honest

${\mathrm{\Omega }}_{max}\approx \left(\begin{array}{l}N\\ \frac{N}{2}\end{array}\right)=\frac{N!}{\left(\frac{N}{2}\right)!\left(\frac{N}{2}\right)!}$

For such design variables, we'll employ Stirling's approximation:

$n!\approx \sqrt{2\pi n}{n}^n{e}^{-n}$

$\begin{array}{r}{\mathrm{\Omega }}_{max}\approx \frac{\sqrt{2\pi N}{N}^N{e}^{-N}}{\left(\sqrt{2\pi \left(\frac{N}{2}\right)}{\left(\frac{N}{2}\right)}^{\left(\frac{N}{2}\right)}{e}^{{\left.-\left(\frac{N}{2}\right)\right)}^2}\right.}\\ \end{array}$

${\mathrm{\Omega }}_{max}\approx \frac{\sqrt{2\pi N}{N}^N{e}^{-N}}{\left(2\pi \left(\frac{N}{2}\right){\left(\frac{N}{2}\right)}^N{e}^{-N}\right)}$

$\begin{array}{c}{\mathrm{\Omega }}_{max}\approx \frac{\sqrt{2\pi N}{N}^N}{\pi N{\left(\frac{N}{2}\right)}^N}\\ \end{array}$

$\begin{array}{c}{\mathrm{\Omega }}_{max}\approx \frac{\sqrt{2\pi N}{2}^N}{\pi N}\end{array}$

$\begin{array}{c}{\mathrm{\Omega }}_{max}\approx \frac{2^{N+1}}{\sqrt{2\pi N}}\\ \end{array}$

${\mathrm{\Omega }}_{max}\approx \frac{2^{N+\frac{1}{2}}}{\sqrt{\pi N}}$

${\mathrm{\Omega }}_{max}\approx 2^N$

Vicinity of the height   (b) 

b) We derive its multitude solution within the location of the crest, let:

$x=N_{\uparrow }-\frac{N}{2}\to N_{\uparrow }=\frac{N}{2}+x$

$x=N-N_{\downarrow }-\frac{N}{2}\to N_{\downarrow }=\frac{N}{2}-x$

Therefore, while the subsequent is accurate:

$\mathrm{\Omega }=\left(\begin{array}{c}{N}_{\downarrow }+N_{\uparrow }+1\\ N_{\downarrow }\end{array}\right)=\left(\begin{array}{c}\frac{N}{2}+x+\frac{N}{2}-x+1\\ \frac{N}{2}-x\end{array}\right)\approx \left(\frac{N}{2}-x\right)$

$\mathrm{\Omega }\approx \left(\frac{N}{2}-x\right)=\frac{N!}{\left(\frac{N}{2}-x\right)!\left(\frac{N}{2}+x\right)!}$

As a conclusion, the redundancy think about units of $x$is:

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

$e^{-N}$with $e^{-\left(\frac{N}{2}+x\right)}{e}^{-\left(\frac{N}{2}-x\right)}$

$\mathrm{\Omega }\approx \frac{\sqrt{2\pi N}{N}^N}{\left(\sqrt{2\pi \left(\frac{N}{2}-x\right)}{\left(\frac{N}{2}-x\right)}^{\left(\frac{N}{2}-x\right)}\right)\left(\sqrt{2\pi \left(\frac{N}{2}+x\right)}{\left(\frac{N}{2}+x\right)}^{\left(\frac{N}{2}+x\right)}\right)}$

$\sqrt{2\pi \left(\frac{N}{2}+x\right)}\times \sqrt{2\pi \left(\frac{N}{2}-x\right)}=2\pi \sqrt{{\left(\frac{N}{2}\right)}^2-x^2}$

Equation

$\mathrm{\Omega }\approx \frac{\sqrt{2\pi N}{N}^N}{2\pi \sqrt{{\left(\frac{N}{2}\right)}^2-x^2}\left({\left(\frac{N}{2}-x\right)}^{\left(\frac{N}{2}-x\right)}\right)\left({\left(\frac{N}{2}+x\right)}^{\left(\frac{N}{2}+x\right)}\right)}$

$\mathrm{\Omega }\approx \frac{\sqrt{2\pi N}{N}^N}{2\pi \sqrt{{\left(\frac{N}{2}\right)}^2-x^2}\left({\left(\frac{N}{2}-x\right)}^{\frac{N}{2}}{\left(\frac{N}{2}-x\right)}^{-x}\right)\left({\left(\frac{N}{2}+x\right)}^{\frac{N}{2}}{\left(\frac{N}{2}+x\right)}^x\right)}$

${\left(\frac{N}{2}+x\right)}^{\frac{N}{2}}{\left(\frac{N}{2}-x\right)}^{\frac{N}{2}}={\left({\left(\frac{N}{2}\right)}^2-x^2\right)}^{\frac{N}{2}}$

$\mathrm{\Omega }\approx \frac{\sqrt{2\pi N}{N}^N}{2\pi \sqrt{{\left(\frac{N}{2}\right)}^2-x^2}{\left({\left(\frac{N}{2}\right)}^2-x^2\right)}^{\frac{N}{2}}{\left(\frac{N}{2}-x\right)}^{-x}{\left(\frac{N}{2}+x\right)}^x}$

$\mathrm{\Omega }\approx \frac{\sqrt{N}{N}^N}{\sqrt{2\pi }{\left({\left(\frac{N}{2}\right)}^2-x^2\right)}^{\frac{N}{2}+\frac{1}{2}}{\left(\frac{N}{2}-x\right)}^{-x}{\left(\frac{N}{2}+x\right)}^x}$

We also may function of notation and handle massive $N$applications without feeling worried concerning square root values. As a basis, here's a conservative calculation:

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

both on sides, have used the expectation andtake care of $\ln \left(ab\right)=\ln \left(a\right)+\ln \left(b\right)\text{and}\ln \left(\frac{a}{b}\right)=\ln \left(a\right)-\ln \left(b\right)$

$\begin{array}{r}\ln \left(\mathrm{\Omega }\right)\approx \ln \left(N^N\right)-\ln {\left({\left(\frac{N}{2}\right)}^2-x^2\right)}^{\frac{\pi }{2}}-\ln {\left(\frac{N}{2}+x\right)}^x-\ln {\left(\frac{N}{2}-x\right)}^{-x}\end{array}$

$\begin{array}{r}\end{array}\begin{array}{r}\ln \left(\mathrm{\Omega }\right)\approx N\ln \left(N\right)-\frac{N}{2}\ln \left({\left(\frac{N}{2}\right)}^2-x^2\right)-x\ln \left(\frac{N}{2}+x\right)+x\ln \left(\frac{N}{2}-x\right)\end{array}$

$\begin{array}{c}\frac{N}{2}\ln \left({\left(\frac{N}{2}\right)}^2-x^2\right)=\frac{N}{2}\ln \left[{\left(\frac{N}{2}\right)}^2\left(1-{\left(\frac{2x}{N}\right)}^2\right)\right]\\ \end{array}$

$\frac{N}{2}\ln \left({\left(\frac{N}{2}\right)}^2-x^2\right)=\frac{N}{2}\left[2\ln \left(\frac{N}{2}\right)+\ln \left(1-{\left(\frac{2x}{N}\right)}^2\right)\right]$

Remaining Equation

$\ln (1+x)=x\text{for}\left|x\right|<<1$

$\begin{array}{c}\frac{N}{2}\ln \left({\left(\frac{N}{2}\right)}^2-x^2\right)=\frac{N}{2}\left[2\ln \left(\frac{N}{2}\right)-{\left(\frac{2x}{N}\right)}^2\right]\end{array}$

$\begin{array}{c}\frac{N}{2}\ln \left({\left(\frac{N}{2}\right)}^2-x^2\right)=N\ln \left(\frac{N}{2}\right)-\frac{2x^2}{N}\\ \end{array}$

$\frac{N}{2}\ln \left({\left(\frac{N}{2}\right)}^2-x^2\right)=N\ln N-N\ln 2-\frac{2x^2}{N}$

$\begin{array}{c}x\ln \left(\frac{N}{2}+x\right)=x\ln \left(\frac{N}{2}\right)+x\ln \left(1+\frac{2x}{N}\right)\\ \end{array}$

$x\ln \left(\frac{N}{2}+x\right)=x\ln \left(\frac{N}{2}\right)+\frac{2x^2}{N}$

Fourth Term of Equation

$\begin{array}{c}x\ln \left(\frac{N}{2}-x\right)=x\ln \left(\frac{N}{2}\right)+x\ln \left(1-\frac{2x}{N}\right)\\ \end{array}$

$x\ln \left(\frac{N}{2}-x\right)=x\ln \left(\frac{N}{2}\right)-\frac{2x^2}{N}$

$\begin{array}{c}\ln \left(\mathrm{\Omega }\right)\approx N\ln \left(N\right)-N\ln N+N\ln 2+\frac{2x^2}{N}-x\ln \left(\frac{N}{2}\right)-\frac{2x^2}{N}+x\ln \left(\frac{N}{2}\right)-\frac{2x^2}{N}\end{array}$

$\begin{array}{c}\ln \left(\mathrm{\Omega }\right)\approx N\ln 2+\frac{2x^2}{N}-\frac{2x^2}{N}-\frac{2x^2}{N}\\ \end{array}$

$\ln \left(\mathrm{\Omega }\right)\approx \ln \left(2^N\right)-\frac{2x^2}{N}$

$\begin{array}{c}\mathrm{\Omega }\approx 2^N{e}^{-\frac{2x^2}{N}}\end{array}$

$\begin{array}{c}\to \mathrm{\Omega }\approx {\mathrm{\Omega }}_{max}{e}^{-\frac{2x^2}{N}}\\ \end{array}$

$\mathrm{\Omega }\approx {\mathrm{\Omega }}_{max}=2^N$

Peak value (c)

c) Oncethe top declines to $\frac{1}{e}$of its previous peak, the length of the crest is:

$\frac{{\Omega }_{\max }}{e}={\Omega }_{\max }{e}^{-\frac{2x^2}{N}}$

$\to e^{-1}=e^{-\frac{2x^2}{N}}$

$\to \frac{2x^2}{N}=1$

As a response, the peak's girth is increased.

Coins  (d)

d) If $1000000$ currencies were spun, the probability of getting 501000 heads is (let $N_{\uparrow }$be the head):

Substitution giving in:

$\begin{array}{c}x=N_{\uparrow }-\frac{N}{2}\\ \end{array}$

$=501000-\frac{1000000}{2}=1000$

$\begin{array}{c}\\ P=\frac{\mathrm{\Omega }}{{\mathrm{\Omega }}_{max}}\end{array}$

$P=e^{-\frac{2x^2}{N}}$

$x=1000$and $N=1000000$are changed, giving in:

$P=e^{-\frac{2(1000{)}^2}{1000000}}=0.1353$

It wouldn't be a very controversial outcome, given the big likelihood.

Obtaining a conclusion of $510000$heads seems to own a likelihood of:

$x=N_{\uparrow }-\frac{N}{2}$

$=510000-\frac{1000000}{2}$

$=10000$

$P=e^{-\frac{2(10000{)}^2}{1000000}}$

$=1.384\times {10}^{-87}$