PX284 - J2 - statistical mechanics of photons

the energy density:

u = \frac{U}{V} = \frac{1}{V} \int_{0}^{\infty} g (ω) ⟨ E (ω) ⟩ d ω

PX284 - H1 - density of states (DOS)

g (k) d k = \frac{V k^{2}}{2 π^{2}} d k \times 2

the factor of $2$ accounts for the two polarisations of light
to convert to $ω$ , the dispersion relation is required
for photons:

ω = c k

g (ω) d ω = \frac{V (ω / c)^{2}}{2 π^{2}} \frac{d ω}{c} \times 2 = \frac{V ω^{2}}{π^{2} c^{2}} d ω

modelling photons has simple harmonic oscillators with:

ε_{n} = (n + \frac{1}{2}) ℏ ω

⟨ ε (ω) ⟩ = U_{S H O} = ℏ ω [\frac{1}{2} + \frac{1}{\exp (β ℏ ω) - 1}]

ignoring the ZPE:

\begin{matrix} (1) & u = \int_{0}^{\infty} \frac{ℏ}{π^{2} c^{3}} \frac{ω^{3}}{\exp (β ℏ ω) - 1} d ω \end{matrix}

using $ω = 2 π ν$ and $ν = c / λ :$

U_{λ} d λ = \frac{8 π h c}{λ^{5}} \frac{1}{\exp (β h / λ) - 1} d λ

this is planck's radiation distribution

consequences

light is quantized into packets of $ℏ ω$ , giving birth to quantum mechanics
stefan-boltzmann law: substituting $x = β ℏ ω$ and $d x = β ℏ d ω$ in equation $(1) :$

u = \frac{ℏ}{π^{2} c^{3}} {(\frac{1}{ℏ β})}^{4} \underset{standard integral = π^{4}}{\underset{⏟}{\int_{0}^{\infty} \frac{x^{3}}{e^{x} - 1} d x}} 15 = \frac{π^{2} k_{B}^{4}}{15 c^{3} ℏ^{3}} T^{4}

∴ u = \frac{4 σ}{c} T^{4}

where, $σ$ is the stefan-boltzmann constant determined from experiments

wien's displacement law: maximising $u_{λ} ⟹ λ_{m a x} T = 2.897$ mm K
planck's constant: planck used the experimentally measured value of $σ$ and wien's law to evaluate $h$ and $k_{B}$