Expand knowledge base

5 files changed, 540 insertions, 28 deletions

diff --git a/content/know/concept/einstein-coefficients/index.pdc b/content/know/concept/einstein-coefficients/index.pdc
index 37141f2..bd8f76c 100644
--- a/content/know/concept/einstein-coefficients/index.pdc
+++ b/content/know/concept/einstein-coefficients/index.pdc
@@ -5,6 +5,7 @@ publishDate: 2021-07-11
 categories:
 - Physics
 - Optics
+- Electromagnetism
 - Quantum mechanics
 
 date: 2021-07-11T18:22:14+02:00
@@ -105,7 +106,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Since $u(\omega_0)$ represents only black-body radiation,
-our result must agree with Planck's law:
+our result must agree with [Planck's law](/know/concept/plancks-law/):
 
 $$\begin{aligned}
     u(\omega_0)
@@ -143,31 +144,30 @@ Consider the Hamiltonian of an electron with charge $q = - e$:
 
 $$\begin{aligned}
     \hat{H}
-    &= \frac{\vec{P}{}^2}{2 m} - \frac{q}{2 m} (\vec{A} \cdot \vec{P} + \vec{P} \cdot \vec{A}) + \frac{q^2 \vec{A}{}^2}{2m} + q \phi
+    &= \frac{\vec{P}{}^2}{2 m} - \frac{q}{2 m} (\vec{A} \cdot \vec{P} + \vec{P} \cdot \vec{A}) + \frac{q^2 \vec{A}{}^2}{2m} + V
 \end{aligned}$$
 
-With $\vec{A}(\vec{r}, t)$ the magnetic vector potential,
-and $\phi(\vec{r}, t)$ the electric scalar potential.
+With $\vec{A}(\vec{r}, t)$ the electromagnetic vector potential.
 We reduce this by fixing the Coulomb gauge $\nabla \!\cdot\! \vec{A} = 0$,
 such that $\vec{A} \cdot \vec{P} = \vec{P} \cdot \vec{A}$,
-and by assuming that $\vec{A}{}^2$ is negligibly small.
-This leaves us with:
+and by assuming that $\vec{A}{}^2$ is negligible:
 
 $$\begin{aligned}
     \hat{H}
-    &= \frac{\vec{P}{}^2}{2 m} - \frac{q}{m} \vec{P} \cdot \vec{A} + q \phi
+    &= \frac{\vec{P}{}^2}{2 m} - \frac{q}{m} \vec{P} \cdot \vec{A} + V
 \end{aligned}$$
 
 The last term is the Coulomb interaction
 between the electron and the nucleus.
-We can interpret the second term, involving the weak $\vec{A}$, as a perturbation $\hat{H}_1$:
+We can interpret the second term,
+involving the weak $\vec{A}$, as a perturbation $\hat{H}_1$:
 
 $$\begin{aligned}
     \hat{H}
     = \hat{H}_0 + \hat{H}_1
     \qquad \quad
     \hat{H}_0
-    \equiv \frac{\vec{P}{}^2}{2 m} + q \phi
+    \equiv \frac{\vec{P}{}^2}{2 m} + V
     \qquad \quad
     \hat{H}_1
     \equiv - \frac{q}{m} \vec{P} \cdot \vec{A}
@@ -179,7 +179,9 @@ $$\begin{aligned}
     \vec{A}(\vec{r}, t) = \vec{A}_0 \exp\!(i \vec{k} \cdot \vec{r} - i \omega t)
 \end{aligned}$$
 
-The corresponding perturbative electric field $\vec{E}$ points in the same direction:
+The corresponding perturbative
+[electric field](/know/concept/electric-field/) $\vec{E}$
+points in the same direction:
 
 $$\begin{aligned}
     \vec{E}(\vec{r}, t)
@@ -231,6 +233,14 @@ $$\begin{aligned}
 
 Where $\vec{p} \equiv q \vec{r} = - e \vec{r}$ is the electric dipole moment of the electron,
 hence the name *electric dipole approximation*.
+Finally, because electric fields are actually real
+(we made it complex for mathematical convenience),
+we take the real part, yielding:
+
+$$\begin{aligned}
+    \hat{H}_1(t)
+    = - q \vec{r} \cdot \vec{E}_0 \cos\!(- i \omega t)
+\end{aligned}$$
 
 
 ## Polarized light
@@ -249,19 +259,19 @@ then generally $\ket{1}$ and $\ket{2}$ will be even or odd functions of $z$,
 such that $\matrixel{1}{z}{1} = \matrixel{2}{z}{2} = 0$, leading to:
 
 $$\begin{gathered}
-    \matrixel{1}{H_1}{2} = - q E_0 V
+    \matrixel{1}{H_1}{2} = - q E_0 U
     \qquad
-    \matrixel{2}{H_1}{1} = - q E_0 V^*
+    \matrixel{2}{H_1}{1} = - q E_0 U^*
     \\
     \matrixel{1}{H_1}{1} = \matrixel{2}{H_1}{2} = 0
 \end{gathered}$$
 
-Where $V \equiv \matrixel{1}{z}{2}$ is a constant.
+Where $U \equiv \matrixel{1}{z}{2}$ is a constant.
 The chance of an upward jump (i.e. absorption) is:
 
 $$\begin{aligned}
     P_{12}
-    = \frac{q^2 E_0^2 |V|^2}{\hbar^2} \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2}
+    = \frac{q^2 E_0^2 |U|^2}{\hbar^2} \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2}
 \end{aligned}$$
 
 Meanwhile, the transition probability for stimulated emission is as follows,
@@ -270,7 +280,7 @@ and is therefore symmetric around $\omega_{ba}$:
 
 $$\begin{aligned}
     P_{21}
-    = \frac{q^2 E_0^2 |V|^2}{\hbar^2} \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2}
+    = \frac{q^2 E_0^2 |U|^2}{\hbar^2} \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2}
 \end{aligned}$$
 
 Surprisingly, the probabilities of absorption and stimulated emission are the same!
@@ -295,7 +305,7 @@ Putting this in the previous result gives the following transition probability:
 
 $$\begin{aligned}
     P_{12}
-    = \frac{2 u q^2 |V|^2}{\varepsilon_0 \hbar^2} \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2}
+    = \frac{2 u q^2 |U|^2}{\varepsilon_0 \hbar^2} \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2}
 \end{aligned}$$
 
 For a continuous light spectrum,
@@ -303,7 +313,7 @@ this $u$ turns into the spectral energy density $u(\omega)$:
 
 $$\begin{aligned}
     P_{12}
-    = \frac{2 q^2 |V|^2}{\varepsilon_0 \hbar^2}
+    = \frac{2 q^2 |U|^2}{\varepsilon_0 \hbar^2}
     \int_0^\infty \frac{\sin^2\!\big( (\omega_0 - \omega) t / 2 \big)}{(\omega_0 - \omega)^2} u(\omega) \dd{\omega}
 \end{aligned}$$
 
@@ -319,8 +329,8 @@ which turns out to be $\pi t$:
 
 $$\begin{aligned}
     P_{12}
-    = \frac{q^2 |V|^2}{\varepsilon_0 \hbar^2} u(\omega_0) \int_{-\infty}^\infty \frac{\sin^2\!\big(x t \big)}{x^2} \dd{x}
-    = \frac{\pi q^2 |V|^2}{\varepsilon_0 \hbar^2} u(\omega_0) \:t
+    = \frac{q^2 |U|^2}{\varepsilon_0 \hbar^2} u(\omega_0) \int_{-\infty}^\infty \frac{\sin^2\!\big(x t \big)}{x^2} \dd{x}
+    = \frac{\pi q^2 |U|^2}{\varepsilon_0 \hbar^2} u(\omega_0) \:t
 \end{aligned}$$
 
 From this, the transition rate $R_{12} = B_{12} u(\omega_0)$
@@ -329,7 +339,7 @@ is then calculated as follows:
 $$\begin{aligned}
     R_{12}
     = \pdv{P_{2 \to 1}}{t}
-    = \frac{\pi q^2 |V|^2}{\varepsilon_0 \hbar^2} u(\omega_0)
+    = \frac{\pi q^2 |U|^2}{\varepsilon_0 \hbar^2} u(\omega_0)
 \end{aligned}$$
 
 Using the relations from earlier with $g_1 = g_2$,
@@ -338,9 +348,9 @@ for a polarized incoming light spectrum:
 
 $$\begin{aligned}
     \boxed{
-        B_{21} = B_{12} = \frac{\pi q^2 |V|^2}{\varepsilon_0 \hbar^2}
+        B_{21} = B_{12} = \frac{\pi q^2 |U|^2}{\varepsilon_0 \hbar^2}
         \qquad
-        A_{21} = \frac{\omega_0^3 q^2 |V|^2}{\pi \varepsilon \hbar c^3}
+        A_{21} = \frac{\omega_0^3 q^2 |U|^2}{\pi \varepsilon \hbar c^3}
     }
 \end{aligned}$$
 
@@ -375,9 +385,9 @@ Evaluating the integrals yields:
 
 $$\begin{aligned}
     \expval{|W|^2}
-    = \frac{2 \pi}{4 \pi} |V|^2 \int_0^\pi \cos^2(\theta) \sin\!(\theta) \dd{\theta}
-    = \frac{|V|^2}{2} \Big[ \!-\! \frac{\cos^3(\theta)}{3} \Big]_0^\pi
-    = \frac{|V|^2}{3}
+    = \frac{2 \pi}{4 \pi} |U|^2 \int_0^\pi \cos^2(\theta) \sin\!(\theta) \dd{\theta}
+    = \frac{|U|^2}{2} \Big[ \!-\! \frac{\cos^3(\theta)}{3} \Big]_0^\pi
+    = \frac{|U|^2}{3}
 \end{aligned}$$
 
 With this additional constant factor $1/3$,
@@ -385,16 +395,16 @@ the transition rate $R_{12}$ is modified to:
 
 $$\begin{aligned}
     R_{12}
-    = \frac{\pi q^2 |V|^2}{3 \varepsilon_0 \hbar^2} u(\omega_0)
+    = \frac{\pi q^2 |U|^2}{3 \varepsilon_0 \hbar^2} u(\omega_0)
 \end{aligned}$$
 
 From which it follows that the Einstein coefficients for unpolarized light are given by:
 
 $$\begin{aligned}
     \boxed{
-        B_{21} = B_{12} = \frac{\pi q^2 |V|^2}{3 \varepsilon_0 \hbar^2}
+        B_{21} = B_{12} = \frac{\pi q^2 |U|^2}{3 \varepsilon_0 \hbar^2}
         \qquad
-        A_{21} = \frac{\omega_0^3 q^2 |V|^2}{3 \pi \varepsilon \hbar c^3}
+        A_{21} = \frac{\omega_0^3 q^2 |U|^2}{3 \pi \varepsilon \hbar c^3}
     }
 \end{aligned}$$
 
diff --git a/content/know/concept/hermite-polynomials/index.pdc b/content/know/concept/hermite-polynomials/index.pdc
new file mode 100644
index 0000000..7f59945
--- /dev/null
+++ b/content/know/concept/hermite-polynomials/index.pdc
@@ -0,0 +1,100 @@
+---
+title: "Hermite polynomials"
+firstLetter: "H"
+publishDate: 2021-09-08
+categories:
+- Mathematics
+- Statistics
+
+date: 2021-09-08T17:00:42+02:00
+draft: false
+markup: pandoc
+---
+
+# Hermite polynomials
+
+The **Hermite polynomials** are a set of functions
+that appear in physics and statistics,
+although slightly different definitions are used in those fields.
+
+
+## Physicists' definition
+
+The **Hermite equation** is an eigenvalue problem for $n$,
+and the Hermite polynomials $H_n(x)$ are its eigenfunctions $u(x)$,
+subject to the boundary condition that $u$ grows at most polynomially,
+in which case the eigenvalues $n$ are non-negative integers:
+
+$$\begin{aligned}
+    \boxed{
+        u'' - 2 x u' + 2 n u = 0
+    }
+\end{aligned}$$
+
+The $n$th-order Hermite polynomial $H_n(x)$
+is therefore as follows, according to physicists:
+
+$$\begin{aligned}
+    H_n(x)
+    &= (-1)^n \exp\!(x^2) \dv[n]{x} \exp\!(- x^2)
+    \\
+    &= \Big( 2 x - \dv{x} \Big)^n 1
+\end{aligned}$$
+
+This form is known as a *Rodrigues' formula*.
+The first handful of Hermite polynomials are:
+
+$$\begin{gathered}
+    H_0(x) = 1
+    \qquad \quad
+    H_1(x) = 2 x
+    \qquad \quad
+    H_2(x) = 4 x^2 - 2
+    \\
+    H_3(x) = 8 x^3 - 12 x
+    \qquad \quad
+    H_4(x) = 16 x^4 - 48 x^2 + 12
+\end{gathered}$$
+
+And then more $H_n$ can be computed quickly
+using the following recurrence relation:
+
+$$\begin{aligned}
+    \boxed{
+        H_{n + 1}(x) = 2 x H_n(x) - 2n H_{n-1}(x)
+    }
+\end{aligned}$$
+
+They (almost) form an *Appell sequence*,
+meaning their derivatives are like so:
+
+$$\begin{aligned}
+    \boxed{
+        \dv[k]{x} H_n(x)
+        = 2^k \frac{n!}{(n - k)!} H_{n - k}(x)
+    }
+\end{aligned}$$
+
+Importantly, all $H_n$ are orthogonal with respect to the weight function $w(x) \equiv \exp\!(- x^2)$:
+
+$$\begin{aligned}
+    \boxed{
+        \braket{H_n}{w H_m}
+        \equiv \int_{-\infty}^\infty H_n(x) \: H_m(x) \: w(x) \dd{x}
+        = \sqrt{\pi} 2^n n! \: \delta_{nm}
+    }
+\end{aligned}$$
+
+Where $\delta_{nm}$ is the Kronecker delta.
+Finally, they form a basis in the [Hilbert space](/know/concept/hilbert-space/)
+of all functions $f(x)$ for which $\braket{f}{w f}$ is finite.
+This means that every such $f$ can be expanded in $H_n$:
+
+$$\begin{aligned}
+    \boxed{
+        f(x)
+        = \sum_{n = 0}^\infty a_n H_n(x)
+        = \sum_{n = 0}^\infty \frac{\braket{H_n}{w f}}{\braket{H_n}{w H_n}} H_n(x)
+    }
+\end{aligned}$$
+
diff --git a/content/know/concept/laguerre-polynomials/index.pdc b/content/know/concept/laguerre-polynomials/index.pdc
new file mode 100644
index 0000000..a4be1ff
--- /dev/null
+++ b/content/know/concept/laguerre-polynomials/index.pdc
@@ -0,0 +1,131 @@
+---
+title: "Laguerre polynomials"
+firstLetter: "L"
+publishDate: 2021-09-08
+categories:
+- Mathematics
+
+date: 2021-09-08T17:00:48+02:00
+draft: false
+markup: pandoc
+---
+
+# Laguerre polynomials
+
+The **Laguerre polynomials** are a set of useful functions that arise in physics.
+They are the non-singular eigenfunctions $u(x)$ of **Laguerre's equation**,
+with the corresponding eigenvalues $n$ being non-negative integers:
+
+$$\begin{aligned}
+    \boxed{
+        x u'' + (1 - x) u' + n u = 0
+    }
+\end{aligned}$$
+
+The $n$th-order Laguerre polynomial $L_n(x)$
+is given in the form of a *Rodrigues' formula* by:
+
+$$\begin{aligned}
+    L_n(x)
+    &= \frac{1}{n!} \exp\!(x) \dv[n]{x} \big(x^n \exp\!(-x)\big)
+    \\
+    &= \frac{1}{n!} \Big( \dv{x} - 1 \Big)^n x^n
+\end{aligned}$$
+
+The first couple of Laguerre polynomials $L_n(x)$ are therefore as follows:
+
+$$\begin{gathered}
+    L_0(x) = 1
+    \qquad \quad
+    L_1(x) = 1 - x
+    \qquad \quad
+    L_2(x) = \frac{1}{2} (x^2 - 4 x + 2)
+\end{gathered}$$
+
+Based on Laguerre's equation,
+**Laguerre's generalized equation** is as follows,
+with an arbitrary real (but usually integer) parameter $\alpha$,
+and $n$ still a non-negative integer:
+
+$$\begin{aligned}
+    \boxed{
+        x u'' + (\alpha + 1 - x) u' + n u = 0
+    }
+\end{aligned}$$
+
+Its solutions, denoted by $L_n^\alpha(x)$,
+are the **generalized** or **associated Laguerre polynomials**,
+which also have a Rodrigues' formula.
+Note that if $\alpha = 0$ then $L_n^\alpha = L_n$:
+
+$$\begin{aligned}
+    L_n^\alpha(x)
+    &= \frac{1}{n!} x^{-\alpha} \exp\!(x) \dv[n]{x} \big( x^{n + \alpha} \exp\!(-x) \big)
+    \\
+    &= \frac{x^{-\alpha}}{n!} \Big( \dv{x} - 1 \Big)^n x^{n + \alpha}
+\end{aligned}$$
+
+The first couple of associated Laguerre polynomials $L_n^\alpha(x)$ are therefore as follows:
+
+$$\begin{aligned}
+    L_0^\alpha(x) = 1
+    \qquad
+    L_1^\alpha(x) = \alpha + 1 - x
+    \qquad
+    L_2^\alpha(x) = \frac{1}{2} (x^2 - 2 \alpha x - 4 x + \alpha^2 + 3 \alpha + 2)
+\end{aligned}$$
+
+And then more $L_n^\alpha$ can be computed quickly
+using the following recurrence relation:
+
+$$\begin{aligned}
+    \boxed{
+        L_{n + 1}^\alpha(x)
+        = \frac{(\alpha + 2 n + 1 - x) L_n^\alpha(x) - (\alpha + n) L_{n - 1}^\alpha(x)}{n + 1}
+    }
+\end{aligned}$$
+
+The derivatives are also straightforward to calculate
+using the following relation:
+
+$$\begin{aligned}
+    \boxed{
+        \dv[k]{x} L_n^\alpha(x)
+        = (-1)^k L_{n - k}^{\alpha + k}(x)
+    }
+\end{aligned}$$
+
+Noteworthy is that these polynomials (both normal and associated)
+are all mutually orthogonal for $x \in [0, \infty[$,
+with respect to the weight function $w(x) \equiv x^\alpha \exp\!(-x)$:
+
+$$\begin{aligned}
+    \boxed{
+        \braket{L_m^\alpha}{w L_n^\alpha}
+        = \int_0^\infty L_m^\alpha(x) \: L_n^\alpha(x) \: w(x) \dd{x}
+        = \frac{\Gamma(n + \alpha + 1)}{n!} \delta_{nm}
+    }
+\end{aligned}$$
+
+Where $\delta_{nm}$ is the Kronecker delta.
+Moreover, they form a basis in
+the [Hilbert space](/know/concept/hilbert-space/)
+of all functions $f(x)$ for which $\braket{f}{w f}$ is finite.
+Any such $f$ can thus be expanded as follows:
+
+$$\begin{aligned}
+    \boxed{
+        f(x)
+        = \sum_{n = 0}^\infty a_n L_n^\alpha(x)
+        = \sum_{n = 0}^\infty \frac{\braket{L_n}{w f}}{\braket{L_n}{w L_n}} L_n^\alpha(x)
+    }
+\end{aligned}$$
+
+Finally, the $L_n^\alpha(x)$ are related to
+the [Hermite polynomials](/know/concept/hermite-polynomials/) $H_n(x)$ like so:
+
+$$\begin{aligned}
+    H_{2n(x)} &= (-1)^n 2^{2n} n! \: L_n^{-1/2}(x^2)
+    \\
+    H_{2n + 1(x)} &= (-1)^n 2^{2n + 1} n! \: L_n^{1/2}(x^2)
+\end{aligned}$$
diff --git a/content/know/concept/legendre-polynomials/index.pdc b/content/know/concept/legendre-polynomials/index.pdc
new file mode 100644
index 0000000..d21f263
--- /dev/null
+++ b/content/know/concept/legendre-polynomials/index.pdc
@@ -0,0 +1,125 @@
+---
+title: "Legendre polynomials"
+firstLetter: "L"
+publishDate: 2021-09-08
+categories:
+- Mathematics
+
+date: 2021-09-08T17:00:53+02:00
+draft: false
+markup: pandoc
+---
+
+# Legendre polynomials
+
+The **Legendre polynomials** are a set of functions that sometimes arise in physics.
+They are the eigenfunctions $u(x)$ of **Legendre's differential equation**,
+which is a ([Sturm-Liouville](/know/concept/sturm-liouville-theory/))
+eigenvalue problem for $\ell (\ell + 1)$,
+where $\ell$ turns out to be a non-negative integer:
+
+$$\begin{aligned}
+    \boxed{
+        (1 - x^2) u'' - 2 x u' + \ell (\ell + 1) u = 0
+    }
+\end{aligned}$$
+
+The $\ell$th-degree Legendre polynomial $P_\ell(x)$
+is given in the form of a *Rodrigues' formula* by:
+
+$$\begin{aligned}
+    P_\ell(x)
+    &= \frac{1}{2^\ell \ell!} \dv[\ell]{x} (x^2 - 1)^\ell
+\end{aligned}$$
+
+The first handful of Legendre polynomials $P_\ell(x)$ are therefore as follows:
+
+$$\begin{gathered}
+    P_0(x) = 1
+    \qquad \quad
+    P_1(x) = x
+    \qquad \quad
+    P_2(x) = \frac{1}{2} (3 x^2 - 1)
+    \\
+    P_3(x) = \frac{1}{2} (5 x^3 - 3 x)
+    \qquad \quad
+    P_4(x) = \frac{1}{8} (35 x^4 - 30 x^2 + 3)
+\end{gathered}$$
+
+And then more $P_\ell$ can be computed quickly
+using **Bonnet's recursion formula**:
+
+$$\begin{aligned}
+    \boxed{
+        (\ell + 1) P_{\ell + 1}(x) = (2 \ell + 1) x P_\ell(x) - \ell P_{\ell - 1}(x)
+    }
+\end{aligned}$$
+
+The derivative of a given $P_\ell$ can be calculated recursively
+using the following relation:
+
+$$\begin{aligned}
+    \boxed{
+        \dv{x} P_{\ell + 1}
+        = (\ell + 1) P_\ell(x) + x \dv{x} P_\ell(x)
+    }
+\end{aligned}$$
+
+Noteworthy is that the Legendre polynomials
+are mutually orthogonal for $x \in [-1, 1]$:
+
+$$\begin{aligned}
+    \boxed{
+        \braket{P_m}{P_n}
+        = \int_{-1}^{1} P_m(x) \: P_n(x) \dd{x}
+        = \frac{2}{2 n + 1} \delta_{nm}
+    }
+\end{aligned}$$
+
+As was to be expected from Sturm-Liouville theory.
+Likewise, they form a complete basis in the
+[Hilbert space](/know/concept/hilbert-space/)
+of piecewise continuous functions $f(x)$ on $x \in [-1, 1]$,
+meaning:
+
+$$\begin{aligned}
+    \boxed{
+        f(x)
+        = \sum_{\ell = 0}^\infty a_\ell P_\ell(x)
+        = \sum_{\ell = 0}^\infty \frac{\braket{P_\ell}{f}}{\braket{P_\ell}{P_\ell}} P_\ell(x)
+    }
+\end{aligned}$$
+
+Each Legendre polynomial $P_\ell$ comes with
+a set of **associated Legendre polynomials** $P_\ell^m(x)$
+of order $m$ and degree $\ell$.
+These are the non-singular solutions of the **general Legendre equation**,
+where $m$ and $\ell$ are integers satisfying $-\ell \le m \le \ell$:
+
+$$\begin{aligned}
+    \boxed{
+        (1 - x^2) u'' - 2 x u' + \Big( \ell (\ell + 1) - \frac{m^2}{1 - x^2} \Big) u = 0
+    }
+\end{aligned}$$
+
+The $\ell$th-degree $m$th-order associated Legendre polynomial $P_\ell^m$
+is as follows for $m \ge 0$:
+
+$$\begin{aligned}
+    P_\ell^m(x)
+    = (-1)^m (1 - x^2)^{m/2} \dv[m]{x} P_\ell(x)
+\end{aligned}$$
+
+Here, the $(-1)^m$ in front is called the **Condon-Shortley phase**,
+and is omitted by some authors.
+For negative orders $m$,
+an additional constant factor is necessary:
+
+$$\begin{aligned}
+    P_\ell^{-m}(x) = (-1)^m \frac{(\ell - m)!}{(\ell + m)!} P_\ell^m(x)
+\end{aligned}$$
+
+Beware, the name is misleading:
+if $m$ is odd, then $P_\ell^m$ is actually not a polynomial.
+Moreover, not all $P_\ell^m$ are mutually orthogonal
+(but some are).
diff --git a/content/know/concept/plancks-law/index.pdc b/content/know/concept/plancks-law/index.pdc
new file mode 100644
index 0000000..6e01b9b
--- /dev/null
+++ b/content/know/concept/plancks-law/index.pdc
@@ -0,0 +1,146 @@
+---
+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.