Question

The argument given above for why ${\mathrm{C}}_{\mathrm{v}}\propto \mathrm{T}$does not depend on the details of the energy levels available to the fermions, so it should also apply to the model considered in Problem 7.16: a gas of fermions trapped in such a way that the energy levels are evenly spaced and non-degenerate.

(a) Show that, in this model, the number of possible system states for a given value of q is equal to the number of distinct ways of writing q as a sum of positive integers. (For example, there are three system states for q = 3, corresponding to the sums 3, 2 + 1, and 1 + 1 + 1. Note that 2 + 1 and 1 + 2 are not counted separately.) This combinatorial function is called the number of unrestricted partitions of q, denoted p(q). For example, p(3) = 3.

(b) By enumerating the partitions explicitly, compute p(7) and p(8).

(c) Make a table of p(q) for values of q up to 100, by either looking up the values in a mathematical reference book, or using a software package that can compute them, or writing your own program to compute them. From this table, compute the entropy, temperature, and heat capacity of this system, using the same methods as in Section 3.3. Plot the heat capacity as a function of temperature, and note that it is approximately linear.

(d) Ramanujan and Hardy (two famous mathematicians) have shown that when q is large, the number of unrestricted partitions of q is given approximately by

$\mathrm{p}\left(\mathrm{q}\right)\approx \frac{{\mathrm{e}}^{\mathrm{\pi }\sqrt{{\displaystyle \raisebox{1ex}{$2\mathrm{q}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}}{4\sqrt{3}\mathrm{q}}$

Check the accuracy of this formula for q = 10 and for q = 100. Working in this approximation, calculate the entropy, temperature, and heat capacity of this system. Express the heat. capacity as a series in decreasing powers of $\mathrm{kT}/\mathrm{\eta }$, assuming that this ratio is large and keeping the two largest terms. Compare to the numerical results you obtained in part (c). Why is the heat capacity of this system independent of N, unlike that of the three dimensional box of fermions discussed in the text?

Short answer

a, explanation has been given.

b. For q=7,

$$\begin{gathered} 7,6+1,5+2,5+1+1,4+3,4+2+1,4+1+1+1,3+1+1+1+1, \\ 2+1+1+1+1+1,1+1+1+1+1+1+1 \end{gathered}$$

For q=8,

$$\begin{gathered} 8,7+1,6+2,6+1+1,5+3,5+2+1,5+1+1+1,4+4,4+3+1,4+2+1+1, \\ 4+1+1+1+1,3+2+1+1+1,3+1+1+1+1+1, \\ 2+1+1+1+1+1+1,1+1+1+1+1+1+1+1 \end{gathered}$$

c. The plot found is using python plot is approximately linear.

d. $$\begin{gathered} \mathrm{p}\left(10\right)=48 \\ \mathrm{p}\left(100\right)=199280893 \end{gathered}$$

$\mathrm{S}=\mathrm{K}\ln \left(\frac{{\mathrm{e}}^{\mathrm{\pi }\sqrt{{\displaystyle \raisebox{1ex}{$2\mathrm{q}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}}{4\sqrt{3}\mathrm{q}}\right)$

$\frac{1}{T}=\frac{K}{\eta }\left(\frac{\mathrm{\pi }}{\sqrt{6\mathrm{q}}}-\frac{1}{\mathrm{q}}\right)$

${\mathrm{C}}_{\mathrm{v}}\approx \mathrm{k}\left(\frac{{\mathrm{\pi }}^2\mathrm{t}}{3}-2\right)$

Step-by-step solution

Part (a) Step 1: Given information

We have given,

${\mathrm{C}}_{\mathrm{v}}\propto \mathrm{T}$

We have to explain the function p in term of q show it is verity.

Simplify

Here the value of q explains the no. of energy levels.

For example as given for $\mathrm{q}=3$, then we can write,

1+1+1

2+1

3

It is explain the how we can distribution the number of q.

Part (b) Step 1: Given information

We have to find$\mathrm{p}\left(7\right),\mathrm{p}\left(8\right)$using above rule.

Simplify

For $\mathrm{q}=7$, we can write

$$\begin{gathered} 7,6+1,5+2,5+1+1,4+3,4+2+1,4+1+1+1,3+1+1+1+1, \\ 2+1+1+1+1+1,1+1+1+1+1+1+1 \end{gathered}$$

For $\mathrm{q}=8$, we can write,

$$\begin{gathered} 8,7+1,6+2,6+1+1,5+3,5+2+1,5+1+1+1,4+4,4+3+1,4+2+1+1, \\ 4+1+1+1+1,3+2+1+1+1,3+1+1+1+1+1, \\ 2+1+1+1+1+1+1,1+1+1+1+1+1+1+1 \end{gathered}$$

part (c) Step 1: Given information,

We have to plot the graph of heat capacity as function of temperature.

Simplify

The table for p(q) for the values of q up to 100 is given by using the python code as given below.

For this we can use the python code and we will get the plot as shown in figure.

The out put is following

$$\begin{gathered} (1,1,2,3,5,7,11,15,22,30,42,56,77,101,135,176,231,297,385,490,627 \\ 792,1002,1255,1575,1958,2436,3010,3718,4565,5604,6842,8349, \\ 10143,12310,14883,17977,21977,21637,26015,31185,37338,44583, \\ 5269823,614154,715220,831820.............15019136,169229875,190569292) \end{gathered}$$

then entropy will found as

$\mathrm{S}=\mathrm{K}\mathrm{in}\mathrm{\omega }$

then, capacity will be,

${\mathrm{C}}_{\mathrm{V}}=\frac{∆\mathrm{U}}{∆\mathrm{T}}$

Then the code of python to p[lot the curve between the entropy and the temperature is given by,

Then the code of python to p[lot the curve between the entropy and the temperature is given by,

Part (d) Step 1: Given information

We have given,

$\mathrm{p}\left(\mathrm{q}\right)\approx \frac{{\mathrm{e}}^{\mathrm{\pi }\sqrt{{\displaystyle \raisebox{1ex}{$2\mathrm{q}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}}{4\sqrt{3}\mathrm{q}}$

we have to calculate the heat capacity ,temperature and entropy of the system.

Simplify

For $\mathrm{q}=10$, we can write,

$$\begin{gathered} \mathrm{p}\left(\mathrm{q}\right)\approx \frac{{\mathrm{e}}^{\mathrm{\pi }\sqrt{{\displaystyle \raisebox{1ex}{$2\mathrm{q}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}}{4\sqrt{3}\mathrm{q}} \\ \mathrm{p}\left(10\right)=\frac{{\mathrm{e}}^{\mathrm{\pi }\sqrt{{\displaystyle \raisebox{1ex}{$20$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}}{4\sqrt{3}\times 10} \\ \mathrm{p}\left(10\right)=48 \end{gathered}$$

For $q=100$, we can write,

$$\begin{gathered} \mathrm{p}\left(\mathrm{q}\right)\approx \frac{{\mathrm{e}}^{\mathrm{\pi }\sqrt{{\displaystyle \raisebox{1ex}{$2\mathrm{q}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}}{4\sqrt{3}\mathrm{q}} \\ \mathrm{p}\left(100\right)=\frac{{\mathrm{e}}^{\mathrm{\pi }\sqrt{{\displaystyle \raisebox{1ex}{$200$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}}{4\sqrt{3}\times 100} \\ \mathrm{p}\left(10\right)=199280893 \end{gathered}$$

Simplify 

The entropy of the system is given by,

$$\begin{gathered} \mathrm{S}=\mathrm{K}\mathrm{ln\Omega } \\ \mathrm{S}=\mathrm{K}\ln \left(\mathrm{p}\right(\mathrm{q}\left)\right) \\ \mathrm{S}=\mathrm{K}\ln \left(\frac{{\mathrm{e}}^{\mathrm{\pi }\sqrt{{\displaystyle \raisebox{1ex}{$2\mathrm{q}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}}{4\sqrt{3}\mathrm{q}}\right) \end{gathered}$$

Then, the temperature is ,

$$\begin{gathered} \frac{1}{\mathrm{T}}=\frac{1}{\eta }\frac{\partial S}{\partial q} \\ \frac{1}{T}=\frac{K}{\eta }\frac{\partial }{\partial q}\ln \left(\frac{{\mathrm{e}}^{\mathrm{\pi }\sqrt{{\displaystyle \raisebox{1ex}{$2\mathrm{q}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}}{4\sqrt{3}\mathrm{q}}\right) \\ \frac{1}{T}=\frac{K}{\eta }\left(\frac{\mathrm{\pi }}{\sqrt{6\mathrm{q}}}-\frac{1}{\mathrm{q}}\right) \end{gathered}$$

The heat capacity is given by,

$$\begin{gathered} {\mathrm{C}}_{\mathrm{v}}=\mathrm{k}\frac{\partial \mathrm{q}}{\partial \mathrm{t}} \\ {\mathrm{C}}_{\mathrm{v}}\approx \mathrm{k}\left(\frac{{\mathrm{\pi }}^2\mathrm{t}}{3}-2\right) \end{gathered}$$

Here, t is given by,

$\mathrm{t}=\frac{\mathrm{kT}}{\mathrm{\eta }}$