Blundell Thermal Physics · Problem 19.4

Equipartition of energy

Physics — Statistical Mechanics · equipartition-mean-square-displacement

Problem

In Example 19.1, show that ⟨xi2⟩=kBT2αi.(19.29)\langle x_i^2 \rangle = \frac{k_\mathrm{B}T}{2\alpha_i}. \tag{19.29}
Intuitive explanation

This problem asks you to find the mean square displacement of a particle in a harmonic potential using the equipartition theorem. The key insight is recognizing that a quadratic term in the energy—like αixi2\alpha_i x_i^2—contributes equally to kinetic and potential energy on average, with each quadratic degree of freedom storing 12kBT\frac{1}{2}k_\mathrm{B}T of energy. By writing the potential energy and applying equipartition directly, you can solve for ⟨xi2⟩\langle x_i^2 \rangle without doing any integrals.

Intuitive explanation

This problem asks you to find the mean square displacement of a particle in a harmonic potential using the equipartition theorem. The key insight is recognizing that a quadratic term in the energy—like αixi2\alpha_i x_i^2—contributes equally to kinetic and potential energy on average, with each quadratic degree of freedom storing 12kBT\frac{1}{2}k_\mathrm{B}T of energy. By writing the potential energy and applying equipartition directly, you can solve for ⟨xi2⟩\langle x_i^2 \rangle without doing any integrals.

Related theory

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