Question

Consider a two-state paramagnet with ${10}^{23}$elementary dipoles, with the total energy fixed at zero so that exactly half the dipoles point up and half point down.

(a) How many microstates are "accessible" to this system?

(b) Suppose that the microstate of this system changes a billion times per second. How many microstates will it explore in ten billion years (the age of the universe)?

(c) Is it correct to say that, if you wait long enough, a system will eventually be found in every "accessible" microstate? Explain your answer, and discuss the meaning of the word "accessible."

Short answer

(a) The accessible microstates to the present system is $\mathrm{\Omega }\approx 2^N=2^{N^{23}}$

(b) There are $3.15576\times {10}^{26}$microstates will it explore in ten billion years.

(c) For all this to develop, we'd to require almost a vast quantity of days (at least here on scope of the planet's present era).

Step-by-step solution

Microstates (a)

(a) Assume we get a paramagnet has $N={10}^{23}$ magnets, and shut to half them are as in spin-up state, giving in $N\uparrow =$$N\downarrow =\frac{1}{2}N\text{,}$, whereby $N\downarrow$ signifies the energy units..This technique provides quantity of microstates:

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

when N is simply a good number

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

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

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

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

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

localid="1650343548696" $\begin{array}{r}\mathrm{\Omega }\approx \frac{\sqrt{2\pi N}{N}^N}{\pi N{\left(\frac{N}{2}\right)}^N}\\ \end{array}$

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

$\mathrm{\Omega }\approx 2^N=2^{N^{23}}$

where every other line probably applies for $N$amounts about ${10}^{23}$, although this makes $2N$ a really vast concentration, but dividing by the denominator won't make a big difference.

Independent States (b) and (c)

(b) Above a ten-billion-year interval, if the machine was during a changing microstate a billion views a second, it'd examine:

${10}^9\times {10}^{10}\times 365.25\times 24\times 3600=3.15576\times {10}^{26}$microstates

The variability of microstates allowed is:

$2^{{10}^{23}}\approx {10}^{3\times {10}^{22}}$

As both a result, the quantity of independent states accessed over 10 billion centuriesis simply alittle fraction among those possible.