Categories: Physics.

Planck’s law

Planck’s law describes the radiation spectrum of a black body: a theoretical object in thermal equilibrium, which absorbs photons, re-radiates them, and then re-absorbs them.

Since the photon population varies with time, this is a grand canonical ensemble, and photons are bosons (see Pauli exclusion principle), this system must obey the Bose-Einstein distribution, with a chemical potential μ=0\mu = 0 (due to the freely varying population):

fB(E)=1exp(βE)1\begin{aligned} f_B(E) = \frac{1}{\exp(\beta E) - 1} \end{aligned}

Each photon has an energy E=ω=ckE = \hbar \omega = \hbar c k, so the density of states is as follows in 3D:

g(E)=2g(k)E(k)=Vk2π2c=VE2π23c3=8πVE2h3c3\begin{aligned} g(E) = 2 \frac{g(k)}{E'(k)} = \frac{V k^2}{\pi^2 \hbar c} = \frac{V E^2}{\pi^2 \hbar^3 c^3} = \frac{8 \pi V E^2}{h^3 c^3} \end{aligned}

Where the factor of 22 accounts for the photon’s polarization degeneracy. We thus expect that the number of photons N(E)N(E) with an energy between EE and E+dEE + \dd{E} is given by:

N(E)dE=fB(E)g(E)dE=8πVh3c3E2exp(βE)1dE\begin{aligned} N(E) \dd{E} = f_B(E) \: g(E) \dd{E} = \frac{8 \pi V}{h^3 c^3} \frac{E^2}{\exp(\beta E) - 1} \dd{E} \end{aligned}

By substituting E=hνE = h \nu, we find that the number of photons N(ν)N(\nu) with a frequency between ν\nu and ν+dν\nu + \dd{\nu} must be as follows:

N(ν)dν=8πVc3ν2exp(βhν)1dν\begin{aligned} N(\nu) \dd{\nu} = \frac{8 \pi V}{c^3} \frac{\nu^2}{\exp(\beta h \nu) - 1} \dd{\nu} \end{aligned}

Multiplying by the energy hνh \nu yields the distribution of the radiated energy, which we divide by the volume VV to get Planck’s law, also called the Plank distribution, describing a black body’s radiated spectral energy density per unit volume:

u(ν)=8πhc3ν3exp(βhν)1\begin{aligned} \boxed{ u(\nu) = \frac{8 \pi h}{c^3} \frac{\nu^3}{\exp(\beta h \nu) - 1} } \end{aligned}

Wien’s displacement law

The Planck distribution peaks at a particular frequency νmax\nu_{\mathrm{max}}, which can be found by solving the following equation for ν\nu:

0=u(ν)    0=3ν2(exp(βhν)1)ν3βhexp(βhν)\begin{aligned} 0 = u'(\nu) \quad \implies \quad 0 = 3 \nu^2 (\exp(\beta h \nu) - 1) - \nu^3 \beta h \exp(\beta h \nu) \end{aligned}

By defining xβhνmaxx \equiv \beta h \nu_{\mathrm{max}}, this turns into the following transcendental equation:

3=(3x)exp(x)\begin{aligned} 3 = (3 - x) \exp(x) \end{aligned}

Whose numerical solution leads to Wien’s displacement law, given by:

hνmaxkBT2.822\begin{aligned} \boxed{ \frac{h \nu_{\mathrm{max}}}{k_B T} \approx 2.822 } \end{aligned}

Which states that the peak frequency νmax\nu_{\mathrm{max}} is proportional to the temperature TT.

Stefan-Boltzmann law

Because u(ν)u(\nu) represents the radiated spectral energy density, we can find the total radiated energy UU per unit volume by integrating over ν\nu:

U=0u(ν)dν=8πhc30ν3exp(βhν)1dν=8πhβ3h3c30(βhν)3exp(βhν)1dν=8πβ4h3c30x3exp(x)1dx\begin{aligned} U &= \int_0^\infty u(\nu) \dd{\nu} = \frac{8 \pi h}{c^3} \int_0^\infty \frac{\nu^3}{\exp(\beta h \nu) - 1} \dd{\nu} \\ &= \frac{8 \pi h}{\beta^3 h^3 c^3} \int_0^\infty \frac{(\beta h \nu)^3}{\exp(\beta h \nu) - 1} \dd{\nu} = \frac{8 \pi}{\beta^4 h^3 c^3} \int_0^\infty \frac{x^3}{\exp(x) - 1} \dd{x} \end{aligned}

This definite integral turns out to be π4/15\pi^4/15, leading us to the Stefan-Boltzmann law, which states that the radiated energy is proportional to T4T^4:

U=4σcT4\begin{aligned} \boxed{ U = \frac{4 \sigma}{c} T^4 } \end{aligned}

Where σ\sigma is the Stefan-Boltzmann constant, which is defined as follows:

σ2π5kB415c2h3\begin{aligned} \sigma \equiv \frac{2 \pi^5 k_B^4}{15 c^2 h^3} \end{aligned}


  1. H. Gould, J. Tobochnik, Statistical and thermal physics, 2nd edition, Princeton.