Question

Use a computer to plot formula $2.22$directly, as follows. Define $z$=$q\_\left\{A\right\}/q$, so that $(1-z)$=$q\_\left\{B\right\}/q$. Then, aside from an overall constant that we'll ignore, the multiplicity function is $\left[4z\right(1-z){]}^N$, where $z$ranges from $0$to$1$and the factor of $4$ensures that the height of the peak is equal to $1$for any $N$. Plot this function for$N$=$1,10,100,1000$, and $10,000$. Observe how the width of the peak decreases as$N$increases.

Short answer

The width of the peak decreases as $N$increases.

Step-by-step solution

High temperature 

For large systems at high temperatures, so that $q>>N$we have the approximate formula:

$\Omega \approx {\left(\frac{qe}{N}\right)}^N$

If we now have two such solids and allow them to interact, the number of microstates for the combined system for any given macrostate (that is, a given division of the total energy $q$=$q\_\left\{A\right\}+q\_\left\{B\right\}$ between the two solids) is just the product of the numbers for the two separate solids:

$\Omega \approx {\left(\frac{q_{A}e}{N_A}\right)}^{N_A}{\left(\frac{q_{B}e}{N_B}\right)}^{N_B}$

We've already seen that the most probable state for a pair of interacting solids is the state in which the energy quanta are distributed evenly between the two systems, so that $\frac{q_A}{q_B}=\frac{N_A}{N_B}$ Our goal is to investigate how likely it is that the distribution of energy will deviate significantly from this most probable state. To make things simpler, we'll take $N\_\left\{A\right\}$=$N_B$=$N$so that both solids are the same size. Then:

$\Omega \approx {\left(\frac{e{q}_A}{N}\right)}^N{\left(\frac{e{q}_B}{N}\right)}^N={\left(\frac{e}{2N}\right)}^{2N}{\left(q_A{q}_B\right)}^N$

The shape of this curve

but, $q\_\left\{B\right\}$=$q-q\_\left\{A\right\}$, so:

$\Omega \approx {\left(\frac{e}{2N}\right)}^{2N}{\left(q_A\left(q-q_A\right)\right)}^N$

With $N$held constant, the shape of this curve is determined by the ${\left(q_A\left(q-q_A\right)\right)}^N$, factor. If we pull out a factor of $q^2$, we get:

${\left(q_A\left(q-q_A\right)\right)}^N={\left(q^2\left[\frac{q_A}{q}\right]\left[\left(1-\frac{q_A}{q}\right)\right]\right)}^N$${\left(q_A\left(q-q_A\right)\right)}^N=q^{2N}{\left(\left[\frac{q_A}{q}\right]\left[\left(1-\frac{q_A}{q}\right)\right]\right)}^N$

${\left(q_A\left(q-q_A\right)\right)}^N={\left(\left[q\frac{q_A}{q}\right]\left[q\left(1-\frac{q_A}{q}\right)\right]\right)}^N$

let, $\mathit{z}=\frac{{\mathit{q}}_A}{\mathit{q}}$, we get:

${\left(q_A\left(q-q_A\right)\right)}^N=q^{2N}\left[z\right(1-z){]}^N$

We can get a feel for how the curve's shape changes as we increase $N$by plotting

$\left[z\right(1-z){]}^N$

for several values of$N$

Graph of vertical change

$z$itself ranges from $0$to $1$and $z(1-z)$has a maximum value of $0.25$a$0.25$t $z=0.5$, so we can scale the graph to a vertical range of $0$to $1$by inserting a factor of $4$inside the parentheses. That is, we plot:

$\left(4z\right(1-z){)}^N$

I used python to draw the graph, the code is illustrated in following picture

Shape of diagram

Curves of violet

The curves are for$N=1$$N=10$ (blue), (orange), $N=100$(green), $N=1000$(red) and $N=10000$(violet)