---
title: "Planck's law"
firstLetter: "P"
publishDate: 2021-09-09
categories:
- Physics

date: 2021-09-09T08:12:14+02:00
draft: false
markup: pandoc
---

# 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](/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.