Physics 4311: Thermal Physics – HW Solution 11

due date: Wednesday, April 22, 2026

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Problem 1: System of 3-level atoms (12 points)

a)

Z_{1} = 2 + e^{- 𝛽𝜖}, Z_{N} = Z_{1}^{N} = {(2 + e^{- 𝛽𝜖})}^{N} (if atoms are distinguishable)

b)

F = - k_{B} 𝑇𝑙𝑛 Z_{N} = - N k_{B} T ln Z_{1} = - N k_{B} T ln (2 + e^{- 𝛽𝜖})

U = - \frac{\partial}{∂𝛽} 𝑙𝑛 Z_{N} = N \frac{𝜖 e^{- 𝛽𝜖}}{2 + e^{- 𝛽𝜖}} = \frac{𝑁𝜖}{2 e^{𝛽𝜖} + 1}

c)

S = \frac{1}{T} (U - F) = \frac{𝑁𝜖}{T (2 e^{𝛽𝜖} + 1)} + N k_{B} 𝑙𝑛 (2 + e^{- 𝛽𝜖})

first term vanishes for $T \to 0$ as $e^{𝛽𝜖}$ diverges; second term contributes $N k_{B} 𝑙𝑛 (2)$

S \to N k_{B} 𝑙𝑛 (2) for T \to 0

each atom contributes $k_{B} 𝑙𝑛 (2)$ because of doubly degenerate ground state

Problem 2: Generalized equipartition theorem (8 points)

Z = \int_{- \infty}^{\infty} 𝑑𝑞 e^{- 𝛽𝐴 | q |^{n} ∕ 2} = 2 \int_{0}^{\infty} 𝑑𝑞 e^{- 𝛽𝐴 q^{n} ∕ 2}

Z = \frac{2}{n} Γ (1 ∕ n) {(\frac{𝛽𝐴}{2})}^{- 1 ∕ n}

contribution to internal energy:

⟨ E ⟩ = - \frac{\partial}{∂𝛽} 𝑙𝑛 Z = - \frac{\partial}{∂𝛽} ln (β^{- 1 ∕ n}) = k_{B} T ∕ n

Problem 3: Polymer under tension (20 points)

a)

Z_{1} = \int_{0}^{π} 𝑑𝛩 \int_{0}^{2 π} 𝑑𝜙 sin Θ e^{𝛽𝑀𝑔ℓ cos Θ} = 2 π \int_{- 1}^{1} 𝑑𝜂 e^{𝛽𝑀𝑔ℓ𝜂} = \frac{4 π}{𝛽𝑀𝑔ℓ} sinh (𝛽𝑀𝑔ℓ)

b)

Z_{N} = Z_{1}^{N} = {[\frac{4 π}{𝛽𝑀𝑔ℓ} 𝑠𝑖𝑛h (𝛽𝑀𝑔ℓ)]}^{N}

F = - k_{B} 𝑇𝑙𝑛 Z_{N} = - N k_{B} T ln [\frac{4 π}{𝛽𝑀𝑔ℓ} sinh (𝛽𝑀𝑔ℓ)]

c)

L_{z} = 𝑁ℓ ⟨ 𝑐𝑜𝑠 Θ ⟩ = 𝑁ℓ \frac{\partial}{\partial (𝛽𝑀𝑔ℓ)} ln Z_{1}

L_{z} = 𝑁ℓ [- \frac{1}{𝛽𝑀𝑔ℓ} + 𝑐𝑜𝑡h (𝛽𝑀𝑔ℓ)]

d)

For $T \to 0$ ( $β \to \infty$ ), use $𝑐𝑜𝑡h (\infty) = 1$ $\Rightarrow L_{z} = 𝑁ℓ$ (fully stretched)
For $T \to \infty$ ( $β \to 0$ ), use Taylor expansion $𝑐𝑜𝑡h (x) = 1 ∕ x + x ∕ 3 + \dots$
$\Rightarrow L_{z} = 𝑁ℓ (𝛽𝑀𝑔ℓ ∕ 3) \to 0$ (rods equally likely to point in all directions)

e)

From above: $L_{z} = 𝛽𝑁 ℓ^{2} 𝑀𝑔 ∕ 3$ , compare with Hooke’s law $F = 𝑘𝑥$
spring constant $k = 3 k_{B} T ∕ (N ℓ^{2})$ proportional to $T$ (entropic force)