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Classical Limit and the Maxwell-Boltzmann Distribution

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(εμ)1e^{\beta(\varepsilon - \mu)} \gg 1

For all relevant energies. This is equivalent to nλth31n\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)=eβ(εμ)f_{\mathrm{FD}(\varepsilon) \approx f_{\mathrm{BE}(\varepsilon) \approx f_{\mathrm{MB}(\varepsilon) = e^{-\beta(\varepsilon - \mu)}}}}

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

1eβ(εμ)±11eβ(εμ)=eβ(εμ)\frac{1}{e^{\beta(\varepsilon - \mu)} \pm 1} \approx \frac{1}{e^{\beta(\varepsilon - \mu)}} = e^{-\beta(\varepsilon - \mu)}

\blacksquare

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

f(v)dv=4π(m2πkBT)3/2v2emv2/(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}}.

For a system of NN indistinguishable non-interacting particles, the canonical partition function factorises:

ZN=1N!Z1NZ_N = \frac{1}{N!} Z_1^N

where Z1Z_1 is the single-particle partition function. For a classical ideal gas in three dimensions:

Z1=V(2πmkBTh2)3/2=Vλth3Z_1 = V \left(\frac{2\pi m k_B T}{h^2}\right)^{3/2} = \frac{V}{\lambda_{\mathrm{th}}^3}

with thermal de Broglie wavelength λth=h/2πmkBT\lambda_{\mathrm{th}} = h/\sqrt{2\pi m k_B T}.

The Helmholtz free energy is F=kBTlnZNF = -k_B T \ln Z_N, from which all thermodynamic quantities follow:

P=FV=NkBTVP = -\frac{\partial F}{\partial V} = \frac{N k_B T}{V}

S=FT=NkB[ln ⁣(VNλth3)+52]S = -\frac{\partial F}{\partial T} = Nk_B\left[\ln\!\left(\frac{V}{N\lambda_{\mathrm{th}}^3}\right) + \frac{5}{2}\right]

Theorem 7.2 (Equipartition). For a classical system in thermal equilibrium at temperature TT, each quadratic degree of freedom in the Hamiltonian contributes 12kBT\frac{1}{2}k_B T to the mean energy.

For a monatomic ideal gas with 3 translational degrees of freedom: E=32NkBT\langle E \rangle = \frac{3}{2}Nk_B T. For a diatomic gas with additional rotational degrees of freedom (at sufficiently high TT): E=52NkBT\langle E \rangle = \frac{5}{2}Nk_B T.

The equipartition theorem fails at low temperatures when quantum effects freeze out degrees of freedom (the equipartition theorem is a classical result valid only in the high-temperature limit).

The Maxwell-Boltzmann distribution can be derived by maximising the Boltzmann entropy S=kBipilnpiS = -k_B \sum_i p_i \ln p_i subject to constraints ipi=1\sum_i p_i = 1 and ipiεi=E\sum_i p_i \varepsilon_i = \langle E \rangle:

δ[kBipilnpiα(ipi1)β(ipiεiE)]=0\delta\left[-k_B \sum_i p_i \ln p_i - \alpha\left(\sum_i p_i - 1\right) - \beta\left(\sum_i p_i \varepsilon_i - \langle E \rangle\right)\right] = 0

This yields pi=eα1eβεip_i = e^{-\alpha - 1} e^{-\beta \varepsilon_i}, and normalisation gives:

pi=eβεijeβεj=eβεiZ1p_i = \frac{e^{-\beta\varepsilon_i}}{\sum_j e^{-\beta\varepsilon_j}} = \frac{e^{-\beta\varepsilon_i}}{Z_1}

With β=1/(kBT)\beta = 1/(k_B T), this is the Maxwell-Boltzmann distribution for discrete energy states.

Problem. Find the density of an ideal gas at height zz in a uniform gravitational field, assuming constant temperature TT. This is the barometric formula.

Solution

The gravitational potential energy of a molecule at height zz is mgzmgz. In equilibrium, the number density follows the Maxwell-Boltzmann distribution:

n(z)=n0exp ⁣(mgzkBT)n(z) = n_0 \exp\!\left(-\frac{mgz}{k_B T}\right)

where n0n_0 is the density at z=0z = 0. The pressure is P(z)=n(z)kBT=P0emgz/(kBT)P(z) = n(z) k_B T = P_0 e^{-mgz/(k_B T)}.

The scale height H=kBT/(mg)H = k_B T/(mg) characterises the exponential decay. For Earth’s atmosphere at T=288T = 288 K: H8.5H \approx 8.5 km.

\blacksquare

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 vf(v)dvv \cdot f(v)\,dv (faster molecules hit the hole more frequently). The effusion distribution is:

feff(v)dvvv2emv2/(2kBT)dv=v3emv2/(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)2v3emv2/(2kBT)f_{\mathrm{eff}(v) = \frac{1}{2(k_BT/m)^2}\,v^3\,e^{-mv^2/(2k_BT)}}

The mean kinetic energy:

εeff=12mv2eff=12m0v5emv2/(2kBT)dv0v3emv2/(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 0vneav2dv=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):

v2eff=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}}

εeff=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

Problem. Estimate the mean free path of nitrogen molecules in air at STP (T=273T = 273 K, P=1P = 1 atm). The molecular diameter of N2_2 is approximately 0.370.37 nm.

Solution

The mean free path λ\lambda is the average distance a molecule travels between collisions:

λ=12πd2n\lambda = \frac{1}{\sqrt{2}\,\pi d^2 n}

where nn is the number density and dd is the molecular diameter. From the ideal gas law:

n=PkBT=1.013×1051.381×1023×2732.69×1025 m3n = \frac{P}{k_B T} = \frac{1.013 \times 10^5}{1.381 \times 10^{-23} \times 273} \approx 2.69 \times 10^{25}\ \text{m}^{-3}

λ=12π(3.7×1010)2×2.69×10256.8×108 m68 nm\lambda = \frac{1}{\sqrt{2}\,\pi (3.7 \times 10^{-10})^2 \times 2.69 \times 10^{25}} \approx 6.8 \times 10^{-8}\ \text{m} \approx 68\ \text{nm}

This is about 200 times the molecular diameter, confirming the diluteness of the gas and the validity of the classical limit.

\blacksquare

The Maxwell-Boltzmann distribution fails when quantum effects become significant:

  • Degenerate Fermi gases (high density, low temperature): Fermi-Dirac statistics must be used; the Pauli exclusion principle prevents multiple occupancy of quantum states.
  • Bose-Einstein condensation occurs when nλth32.612n\lambda_{\mathrm{th}}^3 \gtrsim 2.612; the classical approximation breaks down as bosons accumulate in the ground state.
  • Equipartition failure at low temperatures: rotational and vibrational degrees of freedom freeze out when kBTωk_B T \ll \hbar\omega, violating the classical prediction.