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Wyatt's Notes — University

Classical Limit and the Maxwell-Boltzmann Distribution

7.1 Derivation from Quantum Statistics

In the classical (dilute) limit, both Fermi-Dirac and Bose-Einstein distributions reduce to the Maxwell-Boltzmann distribution. The condition for the classical limit is

eβ(ε−μ)≫1e^{\beta(\varepsilon - \mu)} \gg 1

For all relevant energies. This is equivalent to nλth3≪1n\lambda_{\mathrm{th}^3 \ll 1} (the thermal de Broglie wavelength is much smaller than the inter-particle spacing).

Theorem 7.1. In the classical limit:

fFD(ε)≈fBE(ε)≈fMB(ε)=e−β(ε−μ)f_{\mathrm{FD}(\varepsilon) \approx f_{\mathrm{BE}(\varepsilon) \approx f_{\mathrm{MB}(\varepsilon) = e^{-\beta(\varepsilon - \mu)}}}}

Proof. When eβ(ε−μ)≫1e^{\beta(\varepsilon - \mu)} \gg 1The +1+1 or −1-1 in the denominator is negligible:

1eβ(ε−μ)±1≈1eβ(ε−μ)=e−β(ε−μ)\frac{1}{e^{\beta(\varepsilon - \mu)} \pm 1} \approx \frac{1}{e^{\beta(\varepsilon - \mu)}} = e^{-\beta(\varepsilon - \mu)}

■\blacksquare

7.2 Maxwell-Boltzmann Speed Distribution

For a classical ideal gas, the probability distribution of molecular speeds is

f(v) dv=4π(m2πkBT)3/2v2e−mv2/(2kBT) dvf(v)\,dv = 4\pi\left(\frac{m}{2\pi k_BT}\right)^{3/2} v^2 e^{-mv^2/(2k_BT)}\,dv

Characteristic speeds:

  • Most probable: vp=2kBT/mv_p = \sqrt{2k_BT/m}
  • Mean: ⟨v⟩=8kBT/(πm)\langle v \rangle = \sqrt{8k_BT/(\pi m)}
  • RMS: vrms=3kBT/mv_{\mathrm{rms} = \sqrt{3k_BT/m}}

The ordering is vp<⟨v⟩<vrmsv_p < \langle v \rangle < v_{\mathrm{rms}}.

7.3 Worked Example: Effusion

Problem. A gas of molecular mass mm at temperature TT effuses through a small hole. Find the distribution of speeds of the effusing molecules and the mean kinetic energy per effusing molecule.

Solution

The effusion rate for molecules with speed between vv and v+dvv + dv is proportional to v⋅f(v) dvv \cdot f(v)\,dv (faster molecules hit the hole more frequently). The effusion distribution is:

feff(v) dv∝v⋅v2e−mv2/(2kBT) dv=v3e−mv2/(2kBT) dvf_{\mathrm{eff}(v)\,dv \propto v \cdot v^2 e^{-mv^2/(2k_BT)}\,dv = v^3 e^{-mv^2/(2k_BT)}\,dv}

Normalising:

feff(v)=12(kBT/m)2 v3 e−mv2/(2kBT)f_{\mathrm{eff}(v) = \frac{1}{2(k_BT/m)^2}\,v^3\,e^{-mv^2/(2k_BT)}}

The mean kinetic energy:

⟨ε⟩eff=12m⟨v2⟩eff=12m⋅∫0∞v5e−mv2/(2kBT) dv∫0∞v3e−mv2/(2kBT) dv\langle \varepsilon \rangle_{\mathrm{eff} = \frac{1}{2}m\langle v^2 \rangle_{\mathrm{eff} = \frac{1}{2}m \cdot \frac{\int_0^\infty v^5 e^{-mv^2/(2k_BT)}\,dv}{\int_0^\infty v^3 e^{-mv^2/(2k_BT)}\,dv}}}

Using ∫0∞vne−av2 dv=12a(n+1)/2Γ ⁣(n+12)\int_0^\infty v^n e^{-av^2}\,dv = \frac{1}{2a^{(n+1)/2}}\Gamma\!\left(\frac{n+1}{2}\right):

⟨v2⟩eff=Γ(3)/(2a3)Γ(2)/(2a2)=2a=4kBTm\langle v^2 \rangle_{\mathrm{eff} = \frac{\Gamma(3)/(2a^3)}{\Gamma(2)/(2a^2)} = \frac{2}{a} = \frac{4k_BT}{m}}

⟨ε⟩eff=2kBT\langle \varepsilon \rangle_{\mathrm{eff} = 2k_BT}

This is 4/34/3 times the bulk average 32kBT\frac{3}{2}k_BT --- effusing molecules are “hotter” because faster molecules escape preferentially. ■\blacksquare