Problem:

Inside a blackbody cavity, the energy density per unit frequency interval, ρ(ν), is given by Planck's formula

ρ(ν) = (8πν²/c³)hν/(exp(hν/(kT)) - 1).

(a)  Derive an expression for the intensity per unit frequency interval, I(ν), of the radiation emitted by the blackbody.
(b)  Derive the Stefan-Boltzmann law.
(c)  Derive Wien's displacement law.

∫₀^∞x³dx/(e^x - 1) = π⁴/15.

Solution:

• Concepts:
  Blackbody radiation
  Reasoning:
  We are asked to derive the  Stefan-Boltzmann law and the Wien law.
  Experimental realization of a black body:
  Use the inside of a large box (oven) held at a constant temperature T. A small hole is cut in one side. Any radiation that enters through the hole bounces around inside and has little chance of ever getting out again. Eventually it gets absorbed. Radiation coming out the hole is as good a representation of the radiation from a perfect emitter.
  How do you calculate the intensity of the radiation coming out of the small hole?
• Details of the calculation:
  (a)  Consider an area dA and the energy per unit frequency interval in a ring at a distance between r and r + dr from dA.
  

  The probability that this energy will reach the area dA is dA cosθ/(4πr²). 
  The amount of energy per unit frequency interval reaching dA from the ring is
  ρ(ν)2πr²sinθ dr dθ dA cosθ/(4πr²) = ρ(ν)sinθ dr dθ dA cosθ/2.
  The total energy per unit frequency interval reaching dA in a time interval dt comes from rings with radii between r = 0 and r = cdt.  It is therefore given by
  I(ν) dA dt = ∫₀^π/2 dθ ρ(ν) dA cdt sinθ cosθ/2 = ¼ ρ(ν) dA cdt.
  I(ν) = ¼ ρ(ν) c.
  
  (b)  To find the total intensity, integrate over all frequencies.
  ∫₀^∞(2π/c²)hν³ dν/(exp(hν/(kT)) - 1) = (2πk⁴T⁴/(h³c²))∫₀^∞ x³ dx/(exp(x) - 1)
  = (2πk⁴T⁴/(h³c²) π⁴/15 = (2π⁵k⁴T⁴/(15h³c²) = σT⁴.
  Here σ is the Stefan-Boltzmann constant.

  (c)  The Wien Law gives the wavelength of the peak of the radiation distribution.
  |I(ν)dν| = |I(λ)dλ|,  I(λ) = I(ν)|dν/dλ| = (c/λ²)I(ν).
  I(λ,T) = [2hc²/λ⁵][1/(exp(hc/(λkT)) - 1)].

  dI(λ)/dλ = 0 --> (d/dλ)(λ⁵(exp(hc/λkT - 1))^-1 = 0.
  [hc/(λkT)][[exp(hc/(λkT))/(exp(hc/(λkT)) - 1)²] - 5 = 0,
  xe^x/(e^x - 1) - 5 = 0.

  Use a calculator to find x ~ 4.96.
  λ_max = hc/(xkT)  ~ (2.9*10^-3 mK/T).