Fluctuations and Irreversibility Questions and Answers

Q: A gas molecule moves equal distances \(\lambda\) between successive collisions with equal probability in any direction. After a total of N such collisions, the mean square displacement of the molecule \(\overline{d}^{2}\) will be :

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.

Q: A particle passes through a medium and is deflected by a small angle \(\pm \delta_{0}\) upon each collision within the medium with equal probability in any direction. The mean square angular scattering after N such collision, \(\overline{\delta^{2}}\) is :

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.

A gas molecule moves equal distances \(\lambda\) between successive collisions with equal probability in any direction. After a total of N such collisions, the mean square displacement of the molecule \(\overline{d}^{2}\) will be :

A.\(N\lambda\)
B.\(N\lambda^{2}\)
C.\(N^{2}\lambda^{2}\)
D.\(N^{2}\lambda\)

Exam Year(s)

MH SET - 2019, SET Exam - 2019,

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.

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A particle passes through a medium and is deflected by a small angle \(\pm \delta_{0}\) upon each collision within the medium with equal probability in any direction. The mean square angular scattering after N such collision, \(\overline{\delta^{2}}\) is :

A.0
B.\(N\delta_{0}^{2}\)
C.\(\frac{\delta_{0}^{2}}{N}\)
D.\(N^{2}\delta_{0}^{2}\)

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

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.

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