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+---
+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.