PX156 - A3c - boltzmann distribution

P (E) \sim \exp (- β E)

where, $β = \frac{1}{k_{B} T}$

P (n h f) = \frac{\exp (- β n h f)}{\sum_{n}^{\infty} \exp (- β n h f)}

for $y \to 0 : (1 - y)^{- 1} = 1 + y + y^{2} + y^{3} + \dots :$

(1 - e^{- x})^{- 1} = 1 + e^{- x} + e^{- 2 x} + e^{- 3 x} + \dots = \sum_{n = 0}^{\infty} e^{- n x}

P (n h f) = e^{- β n h f} (1 - e^{- β h f})

⟨ E ⟩ = \sum_{n}^{\infty} P (E_{n}) E_{n} = \sum_{n = 0}^{\infty} n h f e^{- β n h f} (1 - e^{- β h f})

\begin{aligned} \sum_{n = 0}^{\infty} \frac{d}{d β} e^{- β n h f} & = \sum_{m = 0}^{\infty} - n h f e^{- β n h f} \\ = \frac{d}{d β} \frac{1}{1 - e^{- β h f}} \\ = \frac{- e^{- β h f} h f}{(1 - e^{- β h f})^{2}} \\ = \frac{- e^{- β h f} h f}{(1 - e^{- β h f})^{2}} \end{aligned}

⟨ E ⟩ = \frac{h f}{e^{β h f} - 1}

high $T$ limit: $k_{B} T >> h f ⟹ β h f << 1 :$

e^{β h f} - 1 \approx β h f + \frac{(β h f)^{2}}{2} + \dots ⟹ ⟨ E ⟩ = \frac{h f}{β h f} = k_{B} T

low $T$ limit $k_{B} T << h f ⟹ β h f >> 1 :$

⟨ E ⟩ = \frac{h f}{β h f} = h f e^{- \frac{h f}{k_{B} T}} ⟹ s m a l l

I (λ) = \frac{2 π c}{λ^{4}} k_{B} T = \frac{2 π c}{λ^{4}} \frac{h f}{e^{\frac{h f}{k_{B} T}} - 1} = \frac{2 π c^{2} h}{λ^{5}} \frac{1}{e^{\frac{h f}{k_{B} T}} - 1}

taking $T = 10^{4} K$ , $λ = 5 \times 10^{- 7} m$ :

\exp (\frac{h c}{λ k_{B} T}) \approx 18

comparable to

I (λ, T) = \frac{a \exp (\frac{- b}{λ T})}{λ^{5}}

to find $λ_{m a x} : \frac{d I}{d λ} = 0$
to find $I$ , integrate $I (λ) :$

I = \frac{2 π^{5} k_{B}^{4}}{15 h^{3} c^{2}} T^{4}