Statistical Mechanics · Problem 6.10

Ideal Bose Systems — R. K. Pathria

Physics — Statistical Mechanics · relativistic-boltzmannian-gas-momentum-distribution

Problem

(a) Show that the momentum distribution of particles in a relativistic Boltzmannian gas, with ε=c(p2+m02c2)1/2\varepsilon = c(p^2 + m_0^2 c^2)^{1/2}, is given by f(p) dp=Ce−βc(p2+m02c2)1/2p2 dp,f(p)\,dp = Ce^{-\beta c(p^2+m_0^2c^2)^{1/2}}p^2\,dp, with the normalization constant C=βm02cK2(βm0c2),C = \frac{\beta}{m_0^2 c K_2(\beta m_0 c^2)}, Kv(z)K_v(z) being a modified Bessel function. (b) Check that in the nonrelativistic limit (kT≪m0c2)(kT \ll m_0 c^2) we recover the Maxwellian distribution, f(p) dp=(β2πm0)3/2e−βp2/2m0(4πp2 dp),f(p)\,dp = \left(\frac{\beta}{2\pi m_0}\right)^{3/2} e^{-\beta p^2/2m_0}(4\pi p^2\,dp), while in the extreme relativistic limit (kT≫m0c2)(kT \gg m_0 c^2) we obtain f(p) dp=(βc)38πe−βpc(4πp2 dp).f(p)\,dp = \frac{(\beta c)^3}{8\pi}e^{-\beta pc}(4\pi p^2\,dp). (c) Verify that, quite generally, ⟨pu⟩=3kT.\langle pu \rangle = 3kT.

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