---
title: "Planck's law"
sort_title: "Planck's law"
date: 2021-09-09
categories:
- Physics
layout: "concept"
---

**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](/know/concept/grand-canonical-ensemble/),
and photons are bosons
(see [Pauli exclusion principle](/know/concept/pauli-exclusion-principle/)),
this system must obey the
[Bose-Einstein distribution](/know/concept/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](/know/concept/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.