Fluctuations and Irreversibility Questions and Answers

Answer: B

The mean square displacement is \(N\lambda^{2}\).

Write the total displacement as the sum of N steps, \(\vec{d} = \sum \vec{\lambda}_{i}\), each of length \(\lambda\) in a random direction. Squaring and averaging:

\[ \begin{aligned} \overline{d^{2}} &= \sum_{i}\overline{\lambda_{i}^{2}} + \sum_{i \ne j}\overline{\vec{\lambda}_{i}\cdot\vec{\lambda}_{j}} \\ &= N\lambda^{2} + 0 \end{aligned} \]

The cross terms average to zero because successive directions are independent. So the root mean square distance grows only as \(\sqrt{N}\lambda\). \(N^{2}\lambda^{2}\) would be the result if all steps pointed the same way, and \(N\lambda\) is not even dimensionally a squared length.

Answer: B

The mean square scattering angle is \(N\delta_{0}^{2}\).

Each collision adds an independent random deflection \(\pm\delta_{0}\) with equal probability, so each step has mean zero and mean square \(\delta_{0}^{2}\). For a sum of N independent steps the cross terms average to zero:

\[ \begin{aligned} \overline{\delta^{2}} &= \overline{\left(\sum \delta_{i}\right)^{2}} \\ &= \sum_{i} \overline{\delta_{i}^{2}} \\ &\quad + \sum_{i \ne j} \overline{\delta_{i}\delta_{j}} \\ &= N\delta_{0}^{2} + 0 \end{aligned} \]

This is the random-walk result: the root mean square angle grows as \(\sqrt{N}\). Zero is the mean angle, not the mean square; \(N^{2}\delta_{0}^{2}\) would hold only if all deflections were in the same direction, and \(\delta_{0}^{2}/N\) decreases with N, which is impossible.

Answer: C

The mean displacement after n steps is n/3.

Each step is +1 with probability 2/3 and -1 with probability 1/3, so its mean is

\[ \begin{aligned} \langle s\rangle &= (+1)\tfrac{2}{3} + (-1)\tfrac{1}{3} \\ &= \tfrac{1}{3} \end{aligned} \]

The steps are independent, so the mean displacement after n steps is \(n\langle s\rangle = n/3\). Zero is the mean only for a symmetric walk with equal probabilities. 2n/3 counts the forward steps and forgets to subtract the backward ones, and n/8 has no basis in these probabilities.

Answer: D

The fluctuation is \(\Delta M = \frac{1}{\sqrt{N}\sinh(\mu H/KT)}\). (The paper prints the square of \(\langle M\rangle\) as a trailing 2.)

Each moment points along or against the field with probabilities \(\propto e^{\pm x}\), where \(x = \mu H/KT\). For one moment \(\langle m\rangle = \mu\tanh x\) and \(\langle m^{2}\rangle = \mu^{2}\). For N independent moments the means and variances add:

\[ \begin{aligned} \langle M\rangle &= N\mu\tanh x \\ \langle M^{2}\rangle - \langle M\rangle^{2} &= N\mu^{2}(1 - \tanh^{2}x) = \frac{N\mu^{2}}{\cosh^{2}x} \end{aligned} \]

\[ \begin{aligned} \Delta M &= \frac{\sqrt{N}\mu/\cosh x}{N\mu\tanh x} \\ &= \frac{1}{\sqrt{N}\sinh x} \end{aligned} \]

The cosh and tanh options keep only part of this ratio, and \(1/\sqrt{N}\) drops the field and temperature dependence altogether, although it shows the general rule that relative fluctuations fall as \(1/\sqrt{N}\).

Answer: A

The probability is 1/18.

Each step has 6 equally likely directions, so there are \(6^{3} = 216\) three-step walks. To return in exactly three steps, the walker must go round a small triangle: on a triangular lattice, two steps at 120 degrees to each other followed by the third side.

The first step can be any of 6 directions; for each, there are 2 ways to complete a triangle back to the start (turning left or right). So there are \(6 \times 2 = 12\) returning walks:

\[ P = \frac{12}{216} = \frac{1}{18} \]

1/216 counts just one walk, and 1/36 and 1/12 use the wrong number of closing paths.

Answer: C

The keyed probability is \(\left\{\frac{10!}{4!6!}\left[\frac{1}{2}\right]^{10}\right\}^{2}\).

Each walker takes 10 independent steps, each left or right with probability 1/2. The probability that one walker takes 4 steps one way and 6 the other is binomial:

\[ \begin{aligned} p &= \frac{10!}{4!\,6!}\Big(\frac{1}{2}\Big)^{10} \\ &= \frac{210}{1024} \end{aligned} \]

The walkers are independent, so the probability that both make that same choice, and so stand together on the same site after 10 steps, is the product \(p \times p = p^{2}\).

A single factor \(p\) is the probability for one walker only, and a cube would apply to three walkers. Squaring only the binomial coefficient while squaring the whole bracket again, as in the last option, double-counts the factors.