Q. 7.17 - An Introduction to Thermal Physics

Question

In analogy with the previous problem, consider a system of identical spin$‐0bosons$ trapped in a region where the energy levels are evenly spaced. Assume that $N$is a large number, and again let $q$ be the number of energy units.

(a) Draw diagrams representing all allowed system states from $q=0$ up to $q=6$.Instead of using dots as in the previous problem, use numbers to indicate the number of bosons occupying each level.

(b) Compute the occupancy of each energy level, for $q=6$. Draw a graph of the occupancy as a function of the energy at each level.

(c) Estimate values of $\mu$ and $T$ that you would have to plug into the Bose-Einstein distribution to best fit the graph of part(b).

(d) As in part (d) of the previous problem, draw a graph of entropy vs energy and estimate the temperature at $q=6$ from this graph. 

Step-by-Step Solution

Answer

a. The required diagram is given below.

b. The occupancy of each energy level is  $\frac{19}{11} \frac{8}{11} \frac{4}{11} \frac{2}{11} \frac{1}{11} \frac{1}{11} 0 0 ....$

c. The value of $\mu$ and $T$ is ${\overline{n}}_{BE}=\frac{1}{e^{{\displaystyle \raisebox{1ex}{$e$}\!\left/ \!\raisebox{-1ex}{$2.2\eta $}\right.}}-1}$.

d.  The temperature at  $q=6$ from graph is $kT=2.35\eta$.

Part (a) step 1: Given Information

We need to draw the diagram  representing all allowed system states from $q=0$ up to $q=6$Instead of using dots.

Part (a) step 2:Simplifly

Suppose we have a spin-$0 boson$trapped in a region where the energy level is evenly spaced, ,the representation of the states from zero energy units $q=0$to six energy unit $q=6$ is shown unit $q=0$ is shown in the following figure, for $q=1$ there is one state, for $q=1$ there is one state, for  $q=2$ there are two states, for $q=3$ there are three states, for $q=4$ there is five states, for $q=5$ there are seven states and finally for $q=6$ there are eleven states.

Part (b) step 1: Given Information

We need to draw a graph of the occupancy as a function of the energy at each level.

Part (b) step 2: Simplify

To find the occupancy of each level, we add the number in each level and divide it over 11 states, in the lowest level the sum is $19$, in the second-lowest level the sum is $8$, in the third-lowest the sum is $4$, in the fourth-lowest level the sum is $2$, in the fifth-lowest level the sum is$1$, in the sixth-lowest level the sum is $1$,and the sum of the seventh level is zero, so the occupancies  are:

$\frac{19}{11} \frac{8}{11} \frac{4}{11} \frac{2}{11} \frac{1}{11} \frac{1}{11} 0 0 ....$

the energy of the level equals the spacing between the levels $\eta$multiplied by the order of the level, so when we plot a graph between the dimensionless energy and the probability, we draw this between $\raisebox{1ex}{$\epsilon $}\!\left/ \!\raisebox{-1ex}{$\eta $}\right.$ and $\overline{\eta }$, the graph will  looks like this

Part (c) step 1: Given Information

We need to find Estimate values of $\mu$and $T$that you would have to plug into the Bose-Einstein distribution to best fit the graph of part(b).

Part (c) step 2: Simplify

For Bose Einstein distribution, the chemical potential equals the energy level when the occupancy goes to infinity, from the graph we see that the occupancy goes to zero at $\epsilon =0$,so the chemical potential is zero.Bose-Einstein distribution is given by:

${\overline{n}}_{BE}=\frac{1}{e^{{\displaystyle \raisebox{1ex}{$\left(\epsilon -\mu \right)$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}-1}$

set $\mu =0$ then,

${\overline{n}}_{BE}=\frac{1}{e^{{\displaystyle \raisebox{1ex}{$e$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}-1}$

I got a good match when $kT=2.2\eta$, therefore the Bose-Einstein distribution will be

${\overline{n}}_{BE}=\frac{1}{e^{{\displaystyle \raisebox{1ex}{$e$}\!\left/ \!\raisebox{-1ex}{$2.2\eta $}\right.}}-1}$

I plot this function and the points on the same graph,

 

Part (d) step 1:Given Information

We need to draw  draw a graph of entropy vs energy and estimate the temperature at $q=6$ from this graph. 

Part (d) step 2: Simplify

The entropy equals Boltzmann constant multiplies by the natural logarithm of the multiplicity, that is:

$\frac{S}{k}=\ln \left(\Omega \right)$

the multiplicities that correspond each energy units are shown in the following table:

$$\begin{gathered} \frac{q \Omega {\displaystyle \raisebox{1ex}{$s$}\!\left/ \!\raisebox{-1ex}{$k$}\right.}}{0 1 0 } \\ 1 1 0 \\ 2 2 0.693 \\ 3 3 1.098 \\ 4 5 1.609 \\ 5 7 1.946 \\ 6 11 2.397 \end{gathered}$$

a plot between $q and \raisebox{1ex}{$S$}\!\left/ \!\raisebox{-1ex}{$k$}\right.$ looks like this:

Part (d) step 3: Calculation

The slope of the graph is :

$\frac{1∆S}{k∆q}=\frac{2.397-0.693}{6-2}=0.426$

but,

$q=\frac{U}{\eta } ∆q=\frac{∆U}{\eta }$

thus,

$\frac{\eta ∆S}{k∆U}=0.426$

the temperature equals the difference in the energy $U$ over the difference in the entropy that is

$T=\frac{∆U}{∆S}$

thus,

$\frac{\eta }{k}\times \frac{1}{T}=0.426$

$kT=2.35\eta$

and which is a rough agreement with the result of part (c).