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Topic 2 of 5

Boltzmann Distribution

When a system is in thermal contact with a heat reservoir at temperature T, the probability of finding it in a microstate of energy ε is proportional to e^{-ε/k_BT}. This Boltzmann factor is the cornerstone of statistical mechanics.

Key Concepts

  • Boltzmann factor: P(ε) ∝ e^{-ε/k_BT}
  • Partition function: Z = Σᵢ e^{-εᵢ/k_BT}
  • Average energy: ⟨E⟩ = -∂ ln Z/∂β, β=1/k_BT
  • Helmholtz free energy: F = -k_BT ln Z
  • Equipartition theorem: each quadratic degree of freedom contributes k_BT/2

Key Equations

Boltzmann probability
Pi=e−εi/kBTZP_i = \frac{e^{-\varepsilon_i/k_BT}}{Z}
Partition function
Z=∑ie−εi/kBT=∑ie−βεiZ = \sum_i e^{-\varepsilon_i/k_BT} = \sum_i e^{-\beta\varepsilon_i}
Mean energy
⟨E⟩=−∂ln⁡Z∂β\langle E \rangle = -\frac{\partial \ln Z}{\partial \beta}
Free energy
F=−kBTln⁡ZF = -k_BT \ln Z
Worked Example

Example Problem

Problem

A two-level system has ε₁=0, ε₂=0.1 eV at T=300 K. Find P₂/P₁.

Solution

β = 1/(k_BT) = 1/(0.02585 eV). P₂/P₁ = e^{-βε₂} = e^{-0.1/0.02585} = e^{-3.868} = 0.0209.

Practice

Exercises

7 problems
1 of 7

A two-level system has ε₁=0, ε₂=0.05 eV at T=300 K (k_BT=0.02585 eV). Find the ratio P₂/P₁.

2 of 7

For the same system, find the partition function Z = 1 + e^{-βε₂}.

3 of 7

For the two-level system (ε₁=0, ε₂=0.05 eV, T=300 K), find ⟨E⟩ in eV.

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4 of 7

Monatomic ideal gas: equipartition gives ⟨E⟩ = f×k_BT/2 per molecule where f=3 (translational DOF). Find ⟨E⟩ per molecule at T=300 K in eV.

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5 of 7

A harmonic oscillator at T=300 K has ℏω=0.01 eV. Using the quantum partition function Z = 1/(1-e^{-βℏω}), find Z.

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6 of 7

For a particle with 3 equally-spaced levels (ε=0, ε₀, 2ε₀) with ε₀=k_BT, find Z at temperature T.

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7 of 7

For a classical ideal gas, the translational partition function per particle is Z₁ = V(2πmk_BT/h²)^{3/2}. At T=300 K, m=4.65×10⁻²⁶ kg (N₂), V=1 L. Find ln Z₁.

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Key Takeaways

  • The Boltzmann factor e^{-ε/k_BT} is the weight of each microstate in thermal equilibrium
  • The partition function Z normalizes probabilities and encodes all thermodynamic information
  • Average energy follows from -∂lnZ/∂β
  • The equipartition theorem assigns k_BT/2 to each quadratic degree of freedom
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