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 $\mu = 0$ (due to the freely varying population):

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

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

\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 $2$ accounts for the photon’s polarization degeneracy. We thus expect that the number of photons $N(E)$ with an energy between $E$ and $E + \dd{E}$ is given by:

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

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

\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 $\nu_{\mathrm{max}}$, which can be found by solving the following equation for $\nu$:

\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 $x \equiv \beta h \nu_{\mathrm{max}}$, this turns into the following transcendental equation:

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

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

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

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

## Stefan-Boltzmann law

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

\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 $\pi^4/15$, leading us to the Stefan-Boltzmann law, which states that the radiated energy is proportional to $T^4$:

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

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

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

## References

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