From 6ce0bb9a8f9fd7d169cbb414a9537d68c5290aae Mon Sep 17 00:00:00 2001 From: Prefetch Date: Fri, 14 Oct 2022 23:25:28 +0200 Subject: Initial commit after migration from Hugo --- source/know/concept/plancks-law/index.md | 140 +++++++++++++++++++++++++++++++ 1 file changed, 140 insertions(+) create mode 100644 source/know/concept/plancks-law/index.md (limited to 'source/know/concept/plancks-law') diff --git a/source/know/concept/plancks-law/index.md b/source/know/concept/plancks-law/index.md new file mode 100644 index 0000000..7ea06f5 --- /dev/null +++ b/source/know/concept/plancks-law/index.md @@ -0,0 +1,140 @@ +--- +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. -- cgit v1.2.3