Question

This problem gives an alternative approach to estimating the width of the peak of the multiplicity function for a system of two large Einstein solids.

(a) Consider two identical Einstein solids, each with $\mathrm{N}$oscillators, in thermal contact with each other. Suppose that the total number of energy units in the combined system is exactly $2\mathrm{N}$. How many different macrostates (that is, possible values for the total energy in the first solid) are there for this combined system?

(b) Use the result of Problem$2.18$to find an approximate expression for the total number of microstates for the combined system. (Hint: Treat the combined system as a single Einstein solid. Do not throw away factors of "large" numbers, since you will eventually be dividing two "very large" numbers that are nearly equal. Answer: $2^{4\mathrm{N}}/\sqrt{8\mathrm{\pi N}}$.)

(c) The most likely macrostate for this system is (of course) the one in which the energy is shared equally between the two solids. Use the result of Problem $2.18$to find an approximate expression for the multiplicity of this macrostate. (Answer:$2^{4\mathrm{N}}/\left(4\mathrm{\pi N}\right)$ .)

(d) You can get a rough idea of the "sharpness" of the multiplicity function by comparing your answers to parts (b) and (c). Part (c) tells you the height of the peak, while part (b) tells you the total area under the entire graph. As a very crude approximation, pretend that the peak's shape is rectangular. In this case, how wide would it be? Out of all the macrostates, what fraction have reasonably large probabilities? Evaluate this fraction numerically for the case $\mathrm{N}={10}^{23}$.

Short answer

a. Different macrostates is$2\mathrm{N}+1$.

b. Total no of microstates is $\frac{2^{4\mathrm{N}}}{\sqrt{2\mathrm{\pi }4\mathrm{N}}}$.

c. Multiplicity of macrostate is$\frac{2^{4\mathrm{N}}}{4\mathrm{\pi N}}$.

d. The fraction is$3.963\times {10}^{-13}$.

Step-by-step solution

Calculation for macrostates (part a)

a.

Consider the following simple system, in which each solid has $\mathrm{N}$oscillators.

And the total number of quanta is $\mathrm{q}=2\mathrm{N}$.

If we start counting macrostates from zero,

we have:

$0,1,2,3,\dots 2\mathrm{N}$macrostates

As a result, the number of macro-stats is,

$2\mathrm{N}+1$

Expression for total no of microstates (part b)

b.

The number of microstates in a solid with quanta$\mathrm{q}$and $\mathrm{n}$oscillators is,

$\mathrm{\Omega }=\sqrt{\frac{\mathrm{n}}{2\mathrm{\pi q}(\mathrm{q}+\mathrm{n})}}{\left(\frac{\mathrm{q}+\mathrm{n}}{\mathrm{n}}\right)}^{\mathrm{n}}{\left(\frac{\mathrm{q}+\mathrm{n}}{\mathrm{q}}\right)}^{\mathrm{q}}$

Total no of microstates:

For $\mathrm{q}=2\mathrm{N}$,$\mathrm{n}=2\mathrm{N}$

${\mathrm{\Omega }}_{\text{total}}=\sqrt{\frac{2\mathrm{N}}{2\mathrm{\pi }\left(2\mathrm{N}\right)(2\mathrm{N}+2\mathrm{N})}}{\left(\frac{2\mathrm{N}+2\mathrm{N}}{2\mathrm{N}}\right)}^2\mathrm{N}{\left(\frac{2\mathrm{N}+2\mathrm{N}}{2\mathrm{N}}\right)}^{2\mathrm{N}}$

${\mathrm{\Omega }}_{\text{total}}=\frac{2^{4\mathrm{N}}}{\sqrt{2\mathrm{\pi }4\mathrm{N}}}$

Expression for multiplicity of macrostate (part c)

(c) .

For ${\mathrm{q}}_{\mathrm{A}}={\mathrm{q}}_{\mathrm{B}}=\frac{\mathrm{q}}{2}=\mathrm{N}$.

The total number of microstates is,

${\mathrm{\Omega }}_{\mathrm{mp}}={\left[\sqrt{\frac{\mathrm{N}}{2\mathrm{\pi N}(\mathrm{N}+\mathrm{N})}}{\left(\frac{\mathrm{N}+\mathrm{N}}{\mathrm{N}}\right)}^{\mathrm{N}}{\left(\frac{\mathrm{N}+\mathrm{N}}{\mathrm{N}}\right)}^{\mathrm{N}}\right]}^2$

${\mathrm{\Omega }}_{\mathrm{mp}}={\left[\sqrt{\frac{1}{4\mathrm{\pi N}}}{2}^{2\mathrm{N}}\right]}^2$

For $\mathrm{n}=\mathrm{N}$oscillators and ${\mathrm{q}}_{\mathrm{A}}=\mathrm{N}$energy quanta,

${\mathrm{\Omega }}_{\mathrm{mp}}=\frac{2^{4\mathrm{N}}}{4\mathrm{\pi N}}$

Explanation  (part d)

d.

The width is,

$\mathrm{w}=\frac{{\mathrm{\Omega }}_{\text{total}}}{{\mathrm{\Omega }}_{\mathrm{mp}}}$

$=\frac{2^{4\mathrm{N}}}{\sqrt{2\mathrm{\pi }4\mathrm{N}}}\times \frac{4\mathrm{\pi N}}{2^{4\mathrm{N}}}$

$\mathrm{w}=\sqrt{2\mathrm{\pi N}}$

$\approx 2.5\sqrt{\mathrm{N}}$

The Gaussian width is $2\sqrt{\mathrm{N}}$.

As a result, the rectangular approximation isn't all that awful.

Large probabilities is:

$\frac{\sqrt{2\mathrm{\pi N}}}{2\mathrm{N}}=\sqrt{\frac{\mathrm{\pi }}{2\mathrm{N}}}$

For a macroscopic solid with $\mathrm{N}={10}^{23}$,

Fraction is,

$\sqrt{\frac{\mathrm{\pi }}{2\left({10}^{23}\right)}}=3.963\times {10}^{-13}$