The blackbody spectrum

Key concepts

1. The electromagnetic fields in an empty cavity can be mapped onto a set of harmonic oscillators, one for each "normal mode" of the cavity. This mapping is a simple exercise in classic electromagnetism. The normal modes of the cavity [assuming periodic boundary conditions in a cubic box of side L] are labeled by two indexes: the wave vector k=m×2π/L [here m is a vector of arbitrary integer components], and the polarization vector ε (restricted to be perpendicular to k by divE=0). The angular frequency of each of these harmonic oscillators is ω = c |k|, where c is the speed of light c=299792458 m/s.
2. Planck's hypothesis: the allowed energy states for an harmonic oscillator are "quantized". A specific oscillator of frequency ω can take only energy E_n = ℏω(n+½), where n is any non-negative integer n=0,1,2,... and ℏ is a "new" constant, introduced "ad hoc" by Planck (ℏ=1.05457168×10^-34 J s). The quanta of the electromagnetic field are called photons, thus n represents the "number of photons".
3. 
   The equilibrium probability distribution of different microstates in any system at thermal equilibrium is determined uniquely by the energy of these states. In particular, at a given temperature T, the probability P(E) of a given state of energy E is proportional to exp(-E/k_BT). Here k_B is known as Boltzmann constant (k_B=1.380657799×10^-23 J/K) and is basically the conversion constant between temperature and energy. The coefficient 1/(k_BT) of the straight line representing Boltzmann's equilibrium probability distribution in a lin-log plot is often indicated as β. This probability distribution is normalized to unity by dividing it by ∑ exp(-E/k_BT), this summation extending over all quantum states of the system. This normalization is known as partition function and it is often indicated as Z = ∑ exp(-E/k_BT). All thermodynamic quantities can be obtained as suitable functions of Z. For example, the mean energy, i.e. the statistical average of the Hamiltonian operator, called the internal energy in thermodynamics, is U₁ = -∂(log Z)/∂β. These basics of statistical mechanics are detailed in this other page.
4. It is a straightforward exercise to obtain the partition function for one quantum oscillator of angular frequency ω:
   Z₁(ℏω) = [2 sinh(ℏω/k_BT)]^-1.
   Application of the internal energy formula to the harmonic oscillator yields
   U₁(ℏω) = ℏω ( [exp(ℏω/k_BT)-1]^-1 +½ ) .
   The ½ ℏω term is a trivial 0-point contribution which may be dropped by a suitable re-definition of the 0 of energy.
   It is natural to interpret [exp(ℏω/k_BT) -1]^-1 as the average number of quanta <n> in that oscillator.
5. Apply these results to the collection of non-interacting quantum oscillators of the e.m. fields. Compute the total internal energy as
   U = ∑_m ∑_ε U₁(ℏω(m))
   where the triple summation over m covers the infinitely many wave vectors k of the oscillators, and the sum over the polarization vector ε has just two terms (perpendicular to k).
6. Call E=ℏω. By introducing the energy-density of oscillators g(E)= E²V/(π² ℏ³ c³), one can replace the four summations with one integration:
   U = V ∫₀^∞ u(E,T) dE ,
   where the energy density per unit volume and per spectral energy interval
   u(E,T) = π^-2 ℏ^-3 c^-3 E³/[exp(E/k_BT) -1] .
   
   This integration can be performed analytically, to get the total thermal equilibrium energy density of the electromagnetic fields
   U/V = π²(k_BT)⁴/(15ℏ³c³).
7. The radiation power flux crossing any small surface ds in a given direction is c/4×u(E,T) ds dE, if one restricts to photons of energy in the interval (E,E+dE). This power density is generally indicated as R_B(E,T) ds dE. The total power flux R_B(T) = ∫₀^∞ R_B(E,T) dE. The result is of course R_B(T) = c/4×U/V. Using the result above for U/V, this can be rewritten as R_B(T) = σT⁴, with σ = π²k_B⁴/(60ℏ³c²) = 5.67051×10^-8 W m^-2K^-4. This T⁴ dependence of the total radiating power was first established empirically by Stefan (1879), and then derived by thermodynamic arguments by Boltzmann (1884). Here, we re-derived the T⁴ dependence (the Stefan-Boltzmann law) by application of basic concepts of quantum mechanics and statistical thermodynamics. The constant σ is known as Stefan-Boltzmann constant.
8. Where is this "blackbody" name coming from? After all, we have been studying just equilibrium radiation in a cavity: we had no "black bodies" around...
   Any surface absorbs a fraction α and re-emits a fraction γ of the radiant energy it receives. Beside depending on the chemical nature of the surface, both these quantities depend on the frequency ω or photon energy E of the radiation (and on the surface temperature). The way these quantities vary with E expresses the surface color. Energy conservation guarantees that α(E,T) + γ(E,T) = 1. Energy balance between a thermal equilibrium radiation in a cavity and the cavity walls at the same temperature requires that the absorbed power fraction (per unit surface) α(E,T)×R_B(E,T) be entirely re-emitted by the surface. As a consequence, any surface kept at a given temperature T emits [in addition to the reflected/diffused radiation γ(E,T)×(irradiation received)] extra radiation with an equilibrium spectrum multiplied by the surface spectral absorbance. Consequently, a perfectly "black" surface ( α≡1 through the spectrum) kept at a given temperature must emit the spectrum R_B of electromagnetic fields in a cavity to guarantee energy balance. This explains the name blackbody radiation.
9. The explanation at the previous point answers the question: Where is the equilibrium radiation in the cavity coming from?

Problems

1. Given the angular-frequency energy-density distribution u'(ω,T) = π^-2c^-3 ω² ℏ ω/[exp(βℏω) -  1], re-derive the energy distribution u(E,T) discussed at point 5 above.
2. Given the energy-density distribution u(E,T) given at point 5 above, derive the wavelength energy-density distribution u''(λ,T), recalling that λ = h c/E.
   RESULT: u''(λ,T) = 8πch λ^-5 /[exp( βhc/λ) -  1]
3. By multiplying by c/4 all the energy-density distributions of the previous points are converted into suitable radiating power-flux densities. In this way, compute the distributions R_B'(ω,T) and R_B''(λ,T) and check their dimensional correctness.
   NOTE: the distribution R_B''(λ,T) can be studied with the help of this applet.
4. Determine the maximum position λ_max of R_B(λ,T), to discover Wien's displacement law.
   RESULT: λ_max = 0.201405 h c /(k_B T) = 0.00289776 m K / T
5. Determine the maximum position E_max of R_B(E,T), and compare with previous problem.
   RESULT: E_max = 2.82144 k_B T = 0.243133 meV/K×T. Comparison: h c/λ_max = 1.76 E_max. (Understand why!)
6. What fraction of radiant energy in the cavity is below λ_max and what above?
   RESULT: 0.250055 has wavelength below, the rest above.
7. Find the temperature of a cavity having a radiant energy density at 200 nm that is 3.82 times that at 400 nm.
   RESULT: T = 17999.3 K
8. Determine the number of photons per cubic meter having energy between E_max and 1.05  E_max in a black-body radiation field at 300 K, where E_max corresponds to the peak energy density.
   RESULT: Evaluating the integration on the interval (E_max, 1.05 E_max) with the most brutal (rectangular) approximation one finds that the energy density at 300 K in that energy interval is 189 nJ/m³, which corresponds to about 1.62×10¹³ photons m^-3. The accurate expression obtained by carrying out the correct integration of u(E,T)/E is 1.74128 (k_BT/hc)³ = 1.57855×10¹³ photons m^-3.
9. At the Earth surface 1 m² oriented perpendicularly to the Sun rays receives 1353 W of radiating power. Assuming that both the Sun's and the Earth's surfaces are black, estimate:
   a) the temperature of the Sun's surface (Sun radius = 6.95×10⁸ m; average Sun-Earth distance = 1.49×10¹¹ m);
   b) the average Earth temperature, assuming that the Earth is in thermal equilibrium with the solar radiation.
   RESULT: T_Sun = 5755 K; T_Earth = 278 K.

Trivial(?) questions

1. In a dark hot summer night (T=320K) you want to keep as cool as possible to reduce sweating... Being allowed to choose between two walls, a smoke-black one and a shiny white one (both at the same temperature), which one you choose to stand close to?
2. Two stones, a very dark one and a shiny white one are taken out of a hot oven at the same time with the same temperature, and are left to cool far from each other, in a place where all cooling mechanisms but radiation can be neglected. Assuming they have equal shapes and heat capacity, which one cools faster?