Physics 4311: Thermal Physics – HW Solution 1

Problem 1: Joint probabilities (13 points)

The random variables $x$ and $y$ are jointly distributed. $x$ can take values 1, 2, or 3, whereas $y$ can take the values 7 or 8. The joint probabilities are given by $p_{\mathit{𝑥𝑦}}(1,7)=3∕8$, $p_{\mathit{𝑥𝑦}}(2,7)=1∕8$, $p_{\mathit{𝑥𝑦}}(3,7)=1∕4$, $p_{\mathit{𝑥𝑦}}(1,8)=1∕8$, $p_{\mathit{𝑥𝑦}}(2,8)=1∕24$, $p_{\mathit{𝑥𝑦}}(3,8)=1∕12$.

a)

Normalization:

$$\begin{array}{rcll}& & p_{\mathit{𝑥𝑦}}(1,7)+p_{\mathit{𝑥𝑦}}(2,7)+p_{\mathit{𝑥𝑦}}(3,7)+p_{\mathit{𝑥𝑦}}(1,8)+p_{\mathit{𝑥𝑦}}(2,8)+p_{\mathit{𝑥𝑦}}(3,8)=& \\ & & =3∕8+1∕8+1∕4+1∕8+1∕24+1∕12=1& \end{array}$$

.

b)

Reduced probabilities:

$$\begin{array}{rcll}{p}_x(1)& =& p_{\mathit{𝑥𝑦}}(1,7)+p_{\mathit{𝑥𝑦}}(1,8)=3∕8+1∕8=1∕2& \\ p_x(2)& =& p_{\mathit{𝑥𝑦}}(2,7)+p_{\mathit{𝑥𝑦}}(2,8)=1∕8+1∕24=1∕6& \\ p_x(3)& =& p_{\mathit{𝑥𝑦}}(3,7)+p_{\mathit{𝑥𝑦}}(3,8)=1∕4+1∕12=1∕3& \end{array}$$

c)

Reduced probabilities

$$\begin{array}{rcll}{p}_y(7)& =& p_{\mathit{𝑥𝑦}}(1,7)+p_{\mathit{𝑥𝑦}}(2,7)+p_{\mathit{𝑥𝑦}}(3,7)=3∕8+1∕8+1∕4=3∕4& \\ p_y(8)& =& p_{\mathit{𝑥𝑦}}(1,8)+p_{\mathit{𝑥𝑦}}(2,8)+p_{\mathit{𝑥𝑦}}(3,8)=1∕8+1∕24+1∕12=1∕4& \end{array}$$

d)

Conditional probabilities:

$$\begin{array}{rcll}{p}_x(2|y=7)=\frac{p_{\mathit{𝑥𝑦}}(2,7)}{p_y(7)}=\frac{1∕8}{3∕4}=1∕6& & & \\ p_x(2|y=8)=\frac{p_{\mathit{𝑥𝑦}}(2,8)}{p_y(8)}=\frac{1∕24}{1∕4}=1∕6& & & \end{array}$$

e)

The variables $x$ and $y$ are statistically independent because $p_{\mathit{𝑥𝑦}}(x,y)=p_x(x)p_y(y)$.

Problem 2: Gaussian distribution (15 points)

a)

$${\int \; <!--\; nolimits\; -->}_{-\infty }^{\infty }\mathit{𝑑𝑥𝑃}(x)=1\Rightarrow C=\frac{1}{\sqrt{2\mathit{𝜋𝐴}}}$$

b)

$$\begin{array}{rcll}& & ⟨x⟩={\int \; <!--\; nolimits\; -->}_{-\infty }^{\infty }\mathit{𝑑𝑥𝑃}(x)x=x_0& \\ & & {\int \; <!--\; nolimits\; -->}_{-\infty }^{x_M}\mathit{𝑑𝑥𝑃}(x)=1∕2\Rightarrow x_M=x_0& \\ & & {\frac{d}{\mathit{𝑑𝑥}}P(x)|}_{x=x_p}=0\Rightarrow x_p=x_0& \\ & & & \end{array}$$

c)

Second moment and variance

$$\begin{array}{rcll}& & ⟨x^2⟩={\int \; <!--\; nolimits\; -->}_{-\infty }^{\infty }\mathit{𝑑𝑥𝑃}(x)x^2=x_0^2+A& \\ & & {\sigma }_x^2=⟨x^2⟩={⟨x⟩}^2=A& \end{array}$$

Problem 3: Probability of a 10% density fluctuation (12 points)

Consider two identical boxes, A and B.

a)

20 particles are distributed over the two identical boxes A and B at random.

$$\begin{array}{rcll}& & P(10)=\frac{20!}{10!10!}{\left(\frac{1}{2}\right)}^{10}{\left(\frac{1}{2}\right)}^{10}=0.17620& \\ & & P(11)=\frac{20!}{11!9!}{\left(\frac{1}{2}\right)}^{11}{\left(\frac{1}{2}\right)}^9=0.16018& \\ & & P(11)∕P(10)=0.9091& \end{array}$$

b)

Repeat the calculations for $200$ particles.

$$\begin{array}{rcll}& & P(100)=0.05635& \\ & & P(110)=0.02080& \\ & & P(110)∕P(100)=0.3691& \end{array}$$

c)

Repeat the calculations for $2000$ particles.

$$\begin{array}{rcll}& & P(1000)=0.01784& \\ & & P(1100)=8.005\times 10^{-7}& \\ & & P(1100)∕P(1000)=4.49\times 10^{-5}& \end{array}$$