Download Thermodynamics of the Photon Gas: Equation of State and Black-Body Spectrum - Prof. Luca B and more Study notes Physics in PDF only on Docsity! 12–1 The Photon Gas General Considerations • Goal: Consider the electromagnetic field in a box of volume V , in thermal equilibrium at temperature T . From the point of view of thermodynamics, since the field consists of photons, we have a gas of photons, but the number N is not fixed because photons can be absorbed and emitted by the walls of the box. We would like to obtain: (i) the equation of state, and (ii) the black-body spectrum (Planck’s radiation law). • States: We will start from the quantum partition function. The system consists of photons, spin-1 massless particles whose 1-particle states are specified by the pair (k, λ). In a box of volume V = L1L2L3 with periodic boundary conditions the allowed values of ki are 2πni/Li, for i = 1, 2, 3; λ = ±1 is the polarization. General states in the total Hilbert space, using the Fock representation, are labelled by the occupation numbers for each (k, λ), |nk1,λ1 , nk2,λ2 , ..., nkj ,λj , ...〉, where each nkj ,λj = 0, 1, 2, ... • Hamiltonian: Photons are, to an excellent approximation, non-interacting particles. The single-particle mode (k, λ) has energy Ek = h̄ω, with ω = ck, so Ĥ = ∑ k,λ h̄ω N̂k,λ = ∑ k,λ h̄ω â†k,λâk,λ . • Partition function: Summing over all Fock states, the canonical partition function be- comes (although N is not fixed, we do not use the grand canonical ensemble because N is not an overall conserved quantity in this case) Z = tr e−βĤ = ∑ {nk,λ} e−βΣk,λh̄ωnk,λ = ∏ k,λ ∑ nk,λ e−βh̄ωnk,λ = (∏ k 1 1− e−βh̄ω )2 . 12–2 Thermodynamics • Free energy: From the general expression for F in terms of Z = (∏ k 1 1− e−βh̄ω )2 , if we define x := βh̄ω and integrating by parts in one step, F = −kBT lnZ = 2 kBT ∑ k ln(1− e−βh̄ω) ≈ V (kBT ) 4 π2(h̄c)3 ∫ ∞ 0 dxx2 ln(1− e−x) = V (kBT ) 4 π2(h̄c)3 ( − 1 3 ∫ ∞ 0 dxx3 ex − 1 ) = −V (kBT ) 4 π2(h̄c)3 (2 ζ(4)) = −V (kBT ) 4 π2(h̄c)3 π4 45 = −4σ 3c V T 4 , where σ := π2k4B/60h̄ 3c2 is the Stefan-Boltzmann constant. • Entropy: From the general expression, S = −∂F ∂T ∣∣∣ V = 16σ 3 c V T 3 . • Energy and specific heat: From the general expressions, Ē = F + TS = 4σ c V T 4 , so u = 4σ c T 4 , and cV = T V ∂S ∂T ∣∣∣ V = 16σ c T 3 . • Pressure and equation of state: From the general expression, p = −∂F ∂V ∣∣∣ T = 4σ 3 c T 4 , so Ē = 3 pV , or u = 3 p .
