Expand knowledge base

4 files changed, 397 insertions, 22 deletions

diff --git a/content/know/concept/binomial-distribution/index.pdc b/content/know/concept/binomial-distribution/index.pdc
index e644164..183e7e9 100644
--- a/content/know/concept/binomial-distribution/index.pdc
+++ b/content/know/concept/binomial-distribution/index.pdc
@@ -115,7 +115,8 @@ By inserting $q = 1 - p$, we arrive at the desired expression.
 </div>
 
 As $N \to \infty$, the binomial distribution
-turns into the continuous normal distribution:
+turns into the continuous normal distribution,
+a fact that is sometimes called the **de Moivre-Laplace theorem**:
 
 $$\begin{aligned}
     \boxed{
@@ -142,9 +143,9 @@ We use Stirling's approximation to calculate the factorials in $D_m$:
 
 $$\begin{aligned}
     \ln\!\big(P_N(n)\big)
-    &= \ln(N!) - \ln(n!) - \ln\!\big((N - n)!\big) + n \ln(p) + (N - n) \ln(q)
+    &= \ln\!(N!) - \ln\!(n!) - \ln\!\big((N - n)!\big) + n \ln\!(p) + (N - n) \ln\!(q)
     \\
-    &\approx \ln(N!) - n \big( \ln(n)\!-\!\ln(p)\!-\!1 \big) - (N\!-\!n) \big( \ln(N\!-\!n)\!-\!\ln(q)\!-\!1 \big)
+    &\approx \ln\!(N!) - n \big( \ln\!(n)\!-\!\ln\!(p)\!-\!1 \big) - (N\!-\!n) \big( \ln\!(N\!-\!n)\!-\!\ln\!(q)\!-\!1 \big)
 \end{aligned}$$
 
 For $D_0(\mu)$, we need to use a stronger version of Stirling's approximation
@@ -152,16 +153,16 @@ to get a non-zero result. We take advantage of $N - N p = N q$:
 
 $$\begin{aligned}
     D_0(\mu)
-    &= \ln(N!) - \ln\!\big((N p)!\big) - \ln\!\big((N q)!\big) + N p \ln(p) + N q \ln(q)
+    &= \ln\!(N!) - \ln\!\big((N p)!\big) - \ln\!\big((N q)!\big) + N p \ln\!(p) + N q \ln\!(q)
     \\
-    &= \Big( N \ln(N) - N + \frac{1}{2} \ln(2\pi N) \Big)
-    - \Big( N p \ln(N p) - N p + \frac{1}{2} \ln(2\pi N p) \Big) \\
-    &\qquad - \Big( N q \ln(N q) - N q + \frac{1}{2} \ln(2\pi N q) \Big)
-    + N p \ln(p) + N q \ln(q)
+    &= \Big( N \ln\!(N) - N + \frac{1}{2} \ln\!(2\pi N) \Big)
+    - \Big( N p \ln\!(N p) - N p + \frac{1}{2} \ln\!(2\pi N p) \Big) \\
+    &\qquad - \Big( N q \ln\!(N q) - N q + \frac{1}{2} \ln\!(2\pi N q) \Big)
+    + N p \ln\!(p) + N q \ln\!(q)
     \\
-    &= N \ln(N) - N (p + q) \ln(N) + N (p + q) - N - \frac{1}{2} \ln(2\pi N p q)
+    &= N \ln\!(N) - N (p + q) \ln\!(N) + N (p + q) - N - \frac{1}{2} \ln\!(2\pi N p q)
     \\
-    &= - \frac{1}{2} \ln(2\pi N p q)
+    &= - \frac{1}{2} \ln\!(2\pi N p q)
     = \ln\!\Big( \frac{1}{\sqrt{2\pi \sigma^2}} \Big)
 \end{aligned}$$
 
@@ -170,17 +171,17 @@ This is indeed the case:
 
 $$\begin{aligned}
     D_1(n)
-    &= - \big( \ln(n)\!-\!\ln(p)\!-\!1 \big) + \big( \ln(N\!-\!n)\!-\!\ln(q)\!-\!1 \big) - 1 + 1
+    &= - \big( \ln\!(n)\!-\!\ln\!(p)\!-\!1 \big) + \big( \ln\!(N\!-\!n)\!-\!\ln\!(q)\!-\!1 \big) - 1 + 1
     \\
-    &= - \ln(n) + \ln(N - n) + \ln(p) - \ln(q)
+    &= - \ln\!(n) + \ln\!(N - n) + \ln\!(p) - \ln\!(q)
     \\
     D_1(\mu)
-    &= \ln(N q) - \ln(N p) + \ln(p) - \ln(q)
-    = \ln(N p q) - \ln(N p q) 
+    &= \ln\!(N q) - \ln\!(N p) + \ln\!(p) - \ln\!(q)
+    = \ln\!(N p q) - \ln\!(N p q)
     = 0
 \end{aligned}$$
 
-For the same reason, we expect that $D_2(\mu)$ is negative
+For the same reason, we expect that $D_2(\mu)$ is negative.
 We find the following expression:
 
 $$\begin{aligned}
diff --git a/content/know/concept/boltzmann-equation/index.pdc b/content/know/concept/boltzmann-equation/index.pdc
new file mode 100644
index 0000000..84e75cd
--- /dev/null
+++ b/content/know/concept/boltzmann-equation/index.pdc
@@ -0,0 +1,363 @@
+---
+title: "Boltzmann equation"
+firstLetter: "B"
+publishDate: 2022-10-02
+categories:
+- Physics
+- Thermodynamics
+- Fluid mechanics
+
+date: 2022-09-25T17:32:30+02:00
+draft: false
+markup: pandoc
+---
+
+# Boltzmann equation
+
+Consider a collection of particles,
+each with its own position $\vb{r}$ and velocity $\vb{v}$.
+We can thus define a probability density function $f(\vb{r}, \vb{v}, t)$
+describing the expected number of particles at $(\vb{r}, \vb{v})$ at time $t$.
+Let the total number of particles $N$ be conserved, then clearly:
+
+$$\begin{aligned}
+    N = \iint_{-\infty}^\infty f(\vb{r}, \vb{v}, t) \dd{\vb{r}} \dd{\vb{v}}
+\end{aligned}$$
+
+At equilibrium, all processes affecting the particles
+no longer have a net effect, so $f$ is fixed:
+
+$$\begin{aligned}
+    \dv{f}{t}
+    = 0
+\end{aligned}$$
+
+If each particle's momentum only changes due to collisions,
+then a non-equilibrium state can be described as follows, very generally:
+
+$$\begin{aligned}
+    \dv{f}{t}
+    = \bigg(\! \pdv{f}{t} \!\bigg)_\mathrm{\!col}
+\end{aligned}$$
+
+Where the right-hand side simply means "all changes in $f$ due to collisions".
+Applying the chain rule to the left-hand side then yields:
+
+$$\begin{aligned}
+    \bigg(\! \pdv{f}{t} \!\bigg)_\mathrm{\!col}
+    &= \pdv{f}{t} + \bigg( \pdv{f}{x} \dv{x}{t} \!+\! \pdv{f}{y} \dv{y}{t} \!+\! \pdv{f}{z} \dv{z}{t} \bigg)
+    + \bigg( \pdv{f}{v_x} \dv{v_x}{t} \!+\! \pdv{f}{v_y} \dv{v_y}{t} \!+\! \pdv{f}{v_z} \dv{v_z}{t} \bigg)
+    \\
+    &= \pdv{f}{t} + \bigg( v_x \pdv{f}{x} \!+\! v_y \pdv{f}{y} \!+\! v_z \pdv{f}{z} \bigg)
+    + \bigg( a_x \pdv{f}{v_x} \!+\! a_y \pdv{f}{v_y} \!+\! a_z \pdv{f}{v_z} \bigg)
+    \\
+    &= \pdv{f}{t} + \vb{v} \cdot \nabla f + \vb{a} \cdot \pdv{f}{\vb{v}}
+\end{aligned}$$
+
+Where we have introduced the shorthand $\pdv*{f}{\vb{v}}$.
+Inserting Newton's second law $\vb{F} = m \vb{a}$
+leads us to the **Boltzmann equation** or
+**Boltzmann transport equation** (BTE):
+
+$$\begin{aligned}
+    \boxed{
+        \pdv{f}{t} + \vb{v} \cdot \nabla f + \frac{\vb{F}}{m} \cdot \pdv{f}{\vb{v}}
+        = \bigg(\! \pdv{f}{t} \!\bigg)_\mathrm{\!col}
+    }
+\end{aligned}$$
+
+But what about the collision term?
+Expressions for it exist, which are almost exact in many cases,
+but unfortunately also quite difficult to work with.
+In addition, $f$ is a 7-dimensional function,
+so the BTE is already hard to solve without collisions.
+We only present the simplest case,
+known as the **Bhatnagar-Gross-Krook approximation**:
+if the equilibrium state $f_0(\vb{r}, \vb{v})$ is known,
+then each collision brings the system closer to $f_0$:
+
+$$\begin{aligned}
+    \pdv{f}{t} + \vb{v} \cdot \nabla f + \frac{\vb{F}}{m} \cdot \pdv{f}{\vb{v}}
+    = \frac{f_0 - f}{\tau}
+\end{aligned}$$
+
+Where $\tau$ is the average collision period.
+The right-hand side is called the **Krook term**.
+
+
+
+## Moment equations
+
+From the definition of $f$,
+we see that integrating over all $\vb{v}$ yields the particle density $n$:
+
+$$\begin{aligned}
+    n(\vb{r}, t) = \int_{-\infty}^\infty f(\vb{r}, \vb{v}, t) \dd{\vb{v}}
+\end{aligned}$$
+
+Consequently, a purely velocity-dependent quantity $Q(\vb{v})$ can be averaged like so:
+
+$$\begin{aligned}
+    \expval{Q}
+    = \frac{1}{n} \int_{-\infty}^\infty Q(\vb{r}, \vb{v}, t) \: f(\vb{r}, \vb{v}, t) \dd{\vb{v}}
+\end{aligned}$$
+
+With that in mind, we multiply the collisionless BTE equation by $Q(\vb{v})$ and integrate,
+assuming that $\vb{F}$ does not depend on $\vb{v}$:
+
+$$\begin{aligned}
+    0
+    &= \int_{-\infty}^\infty Q \bigg( \pdv{f}{t} + \vb{v} \cdot \nabla f + \frac{\vb{F}}{m} \cdot \pdv{f}{\vb{v}} \bigg) \dd{\vb{v}}
+    \\
+    &= \int Q \pdv{f}{t} \dd{\vb{v}} + \int (\vb{v} \cdot \nabla f) \: Q \dd{\vb{v}} + \frac{\vb{F}}{m} \cdot \int Q \pdv{f}{\vb{v}} \dd{\vb{v}}
+    \\
+    &= \pdv{t} \int Q f \dd{\vb{v}} + \int \Big( \nabla \cdot (\vb{v} f) - f (\nabla \cdot \vb{v}) \Big) Q \dd{\vb{v}}
+    + \frac{\vb{F}}{m} \cdot \int \bigg( \pdv{\vb{v}} (Q f) - f \pdv{Q}{\vb{v}} \bigg) \dd{\vb{v}}
+\end{aligned}$$
+
+The first integral is simply $n \expval{Q}$.
+In the second integral, note that $\vb{v}$ is a coordinate
+and hence not dependent on $\vb{r}$, so $\nabla \cdot \vb{v} = 0$.
+Since $f$ is a probability density, $f \to 0$ for $\vb{v} \to \pm\infty$,
+so the first term in the third integral vanishes after it is integrated:
+
+$$\begin{aligned}
+    0
+    &= \pdv{t} \big(n \expval{Q}\big) + \int \nabla \cdot (\vb{v} f) \: Q \dd{\vb{v}}
+    + \frac{\vb{F}}{m} \cdot \bigg( \Big[ Q f \Big]_{-\infty}^\infty - \int f \pdv{Q}{\vb{v}} \dd{\vb{v}} \bigg)
+    \\
+    &= \pdv{t} \big(n \expval{Q}\big) + \nabla \cdot \int Q \vb{v} f \dd{\vb{v}}
+    - \frac{\vb{F}}{m} \cdot \int f \pdv{Q}{\vb{v}} \dd{\vb{v}}
+\end{aligned}$$
+
+We thus arrive at the prototype of the BTE's so-called **moment equations**:
+
+$$\begin{aligned}
+    \boxed{
+        0
+        = \pdv{t} \big(n \expval{Q}\big) + \nabla \cdot \big(n \expval{Q \vb{v}}\big) - \frac{\vb{F}}{m} \cdot \bigg( n \expval{\pdv{Q}{\vb{v}}} \bigg)
+    }
+\end{aligned}$$
+
+If we set $Q = m$, then the mass density $\rho = n \expval{Q}$,
+and we find that the **zeroth moment** of the BTE describes conservation of mass,
+where $\vb{V} \equiv \expval{\vb{v}} = \int \vb{v} f \dd{\vb{v}}$ is the fluid velocity:
+
+$$\begin{aligned}
+    \boxed{
+        0
+        = \pdv{\rho}{t} + \nabla \cdot \big(\rho \vb{V}\big)
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-moment0"/>
+<label for="proof-moment0">Proof</label>
+<div class="hidden">
+<label for="proof-moment0">Proof.</label>
+We insert $Q = m$ into our prototype,
+and since $m$ is constant, the rest is trivial:
+
+$$\begin{aligned}
+    0
+    &= \pdv{t} \big(n \expval{m}\big) + \nabla \cdot \big(n \expval{m \vb{v}}\big) - \frac{\vb{F}}{m} \cdot \bigg( n \expval{\pdv{m}{\vb{v}}} \bigg)
+    \\
+    &= \pdv{\rho}{t} + \nabla \cdot \big(\rho \expval{\vb{v}}\big) - 0
+\end{aligned}$$
+</div>
+</div>
+
+If we instead choose the momentum $Q = m \vb{v}$,
+we find that the **first moment** of the BTE describes conservation of momentum,
+where $\hat{P}$ is the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/):
+
+$$\begin{aligned}
+    \boxed{
+        0
+        = \pdv{t} \big(\rho \vb{V}\big) + \rho \vb{V} (\nabla \cdot \vb{V}) + \nabla \cdot \hat{P} - n \vb{F}
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-moment1"/>
+<label for="proof-moment1">Proof</label>
+<div class="hidden">
+<label for="proof-moment1">Proof.</label>
+We insert $Q = m \vb{v}$ into our prototype and recognize $\rho$ wherever possible:
+
+$$\begin{aligned}
+    0
+    &= \pdv{t} \big(n \expval{m \vb{v}}\big) + \nabla \cdot \big(n \expval{m \vb{v} \vb{v}}\big)
+    - \frac{\vb{F}}{m} \cdot \bigg( n \expval{\pdv{(m \vb{v})}{\vb{v}}} \bigg)
+    \\
+    &= \pdv{t} \big(\rho \expval{\vb{v}}\big) + \nabla \cdot \big(\rho \expval{\vb{v} \vb{v}}\big)
+    - \vb{F} \cdot \bigg( n \expval{\pdv{\vb{v}}{\vb{v}}} \bigg)
+\end{aligned}$$
+
+With $\vb{v} \vb{v}$ being a dyadic product.
+To give it a physical interpretation,
+we split $\vb{v} = \vb{V} \!+\! \vb{w}$,
+where $\vb{V}$ is the average velocity vector,
+and $\vb{w}$ is the local deviation from $\vb{V}$:
+
+$$\begin{aligned}
+    \expval{\vb{v} \vb{v}}
+    &= \expval{(\vb{V} \!+\! \vb{w}) (\vb{V} \!+\! \vb{w})}
+    = \expval{\vb{V} \vb{V} + 2 \vb{V} \vb{w} + \vb{w} \vb{w}}
+    = \vb{V} \vb{V} + 2 \vb{V} \expval{\vb{w}} + \expval{\vb{w} \vb{w}}
+\end{aligned}$$
+
+Since $\vb{w}$ represents a deviation from the mean, $\expval{\vb{w}} = 0$.
+We define the pressure tensor:
+
+$$\begin{aligned}
+    \hat{P}
+    \equiv \rho \expval{\vb{w} \vb{w}}
+    = \rho \expval{(\vb{v} \!-\! \vb{V}) (\vb{v} \!-\! \vb{V})}
+\end{aligned}$$
+
+This leads to the expected result,
+where $\nabla \cdot (\rho \vb{V}\vb{V})$ represents the fluid momentum,
+and $\nabla \cdot \hat{P}$ the viscous/pressure momentum:
+
+$$\begin{aligned}
+    0
+    &= \pdv{t} \big(\rho \vb{V}\big) + \nabla \cdot \big(\rho \vb{V} \vb{V} + \hat{P}\big) - n \vb{F}
+\end{aligned}$$
+</div>
+</div>
+
+Finally, if we choose the kinetic energy $Q = m |\vb{v}|^2 / 2$,
+we find that the **second moment** gives conservation of energy,
+where $U$ is the thermal energy density and $\vb{J}$ is the heat flux:
+
+$$\begin{aligned}
+    \boxed{
+        0
+        = \pdv{t} \bigg(\frac{\rho}{2} |\vb{V}|^2 + U \bigg)
+        + \nabla \cdot \bigg(\frac{\rho}{2} |\vb{V}|^2 \vb{V} + \vb{V} \cdot \hat{P} + U \vb{V} + \vb{J} \bigg)
+        - \vb{F} \cdot \big( n \vb{V} \big)
+    }
+\end{aligned}$$
+
+<div class="accordion">
+<input type="checkbox" id="proof-moment2"/>
+<label for="proof-moment2">Proof</label>
+<div class="hidden">
+<label for="proof-moment2">Proof.</label>
+We insert $Q = m |\vb{v}|^2 / 2$ into our prototype and recognize $\rho$ wherever possible:
+
+$$\begin{aligned}
+    0
+    &= \pdv{t} \bigg(n \expval{\frac{m |\vb{v}|^2}{2}}\bigg)
+    + \nabla \cdot \bigg(n \expval{\frac{m |\vb{v}|^2}{2} \vb{v}}\bigg)
+    - \frac{\vb{F}}{m} \cdot \bigg( n \expval{\pdv{\vb{v}} \frac{m |\vb{v}|^2}{2}} \bigg)
+    \\
+    &= \pdv{t} \bigg(\frac{\rho}{2} \expval{|\vb{v}|^2}\bigg)
+    + \nabla \cdot \bigg(\frac{\rho}{2} \expval{|\vb{v}|^2 \vb{v}}\bigg)
+    - \frac{\vb{F}}{2} \cdot \bigg( n \expval{\pdv{|\vb{v}|^2}{\vb{v}}} \bigg)
+\end{aligned}$$
+
+We handle these terms one by one. Substituting $\vb{v} = \vb{V} + \vb{w}$ in the first gives:
+
+$$\begin{aligned}
+    \expval{|\vb{v}|^2}
+    &= \expval{(\vb{V} \!+\! \vb{w}) \cdot (\vb{V} \!+\! \vb{w})}
+    = \expval{|\vb{V}|^2 + 2 \vb{V} \cdot \vb{w} + |\vb{w}|^2}
+    \\
+    &= |\vb{V}|^2 + 2 \vb{V} \cdot \expval{\vb{w}} + \expval{|\vb{w}|^2}
+    = |\vb{V}|^2 + \expval{|\vb{w}|^2}
+\end{aligned}$$
+
+And likewise for the second term,
+where we recognize the stress tensor $\expval{\vb{w} \vb{w}}$:
+
+$$\begin{aligned}
+    \expval{|\vb{v}|^2 \vb{v}}
+    &= \expval{(\vb{V} \!+\! \vb{w}) \cdot (\vb{V} \!+\! \vb{w}) (\vb{V} \!+\! \vb{w})}
+    = \expval{(|\vb{V}|^2 + 2 \vb{V} \cdot \vb{w} + |\vb{w}|^2) (\vb{V} \!+\! \vb{w})}
+    \\
+    &= \expval{|\vb{V}|^2 \vb{V} + |\vb{V}|^2 \vb{w}
+    + 2 (\vb{V} \cdot \vb{w}) \vb{V} + 2 (\vb{V} \cdot \vb{w}) \vb{w}
+    + |\vb{w}|^2 \vb{V} + |\vb{w}|^2 \vb{w}}
+    \\
+    &= |\vb{V}|^2 \vb{V} + |\vb{V}|^2 \expval{\vb{w}}
+    + 2 (\vb{V} \cdot \expval{\vb{w}}) \vb{V} + 2 \expval{(\vb{V} \cdot \vb{w}) \vb{w}}
+    + \expval{|\vb{w}|^2} \vb{V} + \expval{|\vb{w}|^2 \vb{w}}
+    \\
+    &= |\vb{V}|^2 \vb{V} + 0 + 0 + 2 \vb{V} \cdot \expval{\vb{w} \vb{w}}
+    + \expval{|\vb{w}|^2} \vb{V} + \expval{|\vb{w}|^2 \vb{w}}
+\end{aligned}$$
+
+The third term is fairly obvious, but we calculate it rigorously just to be safe:
+
+$$\begin{aligned}
+    \pdv{|\vb{v}|^2}{\vb{v}}
+    &= \pdv{\vb{v}} \big( v_x^2 + v_y^2 + v_z^2 \big)
+    = \vu{e}_x \pdv{v_x^2}{v_x} + \vu{e}_y \pdv{v_y^2}{v_y} + \vu{e}_z \pdv{v_z^2}{v_z}
+    = 2 \vb{v}
+\end{aligned}$$
+
+To clarify the physical interpretation,
+we define $U$, $\vb{J}$ and $\hat{P}$ as follows:
+
+$$\begin{aligned}
+    U
+    &\equiv \frac{\rho}{2} \expval{|\vb{w}|^2}
+    = \frac{\rho}{2} \expval{(\vb{v} \!-\! \vb{V}) \cdot (\vb{v} \!-\! \vb{V})}
+    \\
+    \vb{J}
+    &\equiv \frac{\rho}{2} \expval{|\vb{w}|^2 \vb{w}}
+    = \frac{\rho}{2} \expval{(\vb{v} \!-\! \vb{V}) \cdot (\vb{v} \!-\! \vb{V})(\vb{v} \!-\! \vb{V})}
+    \\
+    \hat{P}
+    &\equiv \rho \expval{\vb{w} \vb{w}}
+    = \rho \expval{(\vb{v} \!-\! \vb{V}) (\vb{v} \!-\! \vb{V})}
+\end{aligned}$$
+
+Putting it all together, we arrive at the expected result, namely:
+
+$$\begin{aligned}
+    0
+    &= \pdv{t} \bigg(\frac{\rho}{2} |\vb{V}|^2 + U \bigg)
+    + \nabla \cdot \bigg(\frac{\rho}{2} |\vb{V}|^2 \vb{V} + \vb{V} \cdot \hat{P} + U \vb{V} + \vb{J} \bigg)
+    - \vb{F} \cdot \big( n \vb{V} \big)
+\end{aligned}$$
+
+For the sake of clarity, we write out the pressure term, including the outer divergence:
+
+$$\begin{aligned}
+    \nabla \cdot (\vb{V} \cdot \hat{P})
+    &= (\nabla \cdot \hat{P}{}^{\mathrm{T}}) \cdot \vb{V}
+    = \nabla \cdot
+    \begin{bmatrix}
+        P_{xx} & P_{xy} & P_{xz} \\
+        P_{yx} & P_{yy} & P_{yz} \\
+        P_{zx} & P_{zy} & P_{zz}
+    \end{bmatrix}
+    \cdot
+    \begin{bmatrix}
+        V_x \\ V_y \\ V_z
+    \end{bmatrix}
+    \\
+    &=
+    \begin{bmatrix}
+        \displaystyle \pdv{P_{xx}}{x} + \pdv{P_{xy}}{y} + \pdv{P_{xz}}{z} \\
+        \displaystyle \pdv{P_{yx}}{x} + \pdv{P_{yy}}{y} + \pdv{P_{yz}}{z} \\
+        \displaystyle \pdv{P_{zx}}{x} + \pdv{P_{zy}}{y} + \pdv{P_{zz}}{z}
+    \end{bmatrix}^{\mathrm{T}}
+    \cdot
+    \begin{bmatrix}
+        V_x \\ V_y \\ V_z
+    \end{bmatrix}
+    = \sum_{i=1}^{3} \sum_{j=1}^{3} \pdv{P_{ij}}{x_j} V_i
+\end{aligned}$$
+</div>
+</div>
+
+
+
+## References
+1.  M. Salewski, A.H. Nielsen,
+    *Plasma physics: lecture notes*,
+    2021, unpublished.
diff --git a/content/know/concept/central-limit-theorem/index.pdc b/content/know/concept/central-limit-theorem/index.pdc
index a957736..518c224 100644
--- a/content/know/concept/central-limit-theorem/index.pdc
+++ b/content/know/concept/central-limit-theorem/index.pdc
@@ -53,7 +53,7 @@ $$\begin{aligned}
     \phi(k) = \int_{-\infty}^\infty p(x) \exp(i k x) \dd{x}
 \end{aligned}$$
 
-Note that $\phi(k)$ can be interpreted as the average of $\exp(i k x)$.
+Note that $\phi(k)$ can be interpreted as the average of $\exp\!(i k x)$.
 We take its Taylor expansion in two separate ways,
 where an overline denotes the mean:
 
@@ -62,7 +62,7 @@ $$\begin{aligned}
     = \sum_{n = 0}^\infty \frac{k^n}{n!} \: \phi^{(n)}(0)
     \qquad
     \phi(k)
-    = \overline{\exp(i k x)} = \sum_{n = 0}^\infty \frac{(ik)^n}{n!} \overline{x^n}
+    = \overline{\exp\!(i k x)} = \sum_{n = 0}^\infty \frac{(ik)^n}{n!} \overline{x^n}
 \end{aligned}$$
 
 By comparing the coefficients of these two power series,
@@ -88,12 +88,12 @@ using our earlier relation:
 $$\begin{aligned}
     C^{(1)}
     &= - i \dv{k} \Big(\ln\!\big(\phi(k)\big)\Big) \big|_{k = 0}
-    = - i \frac{\phi'(0)}{\exp(0)}
+    = - i \frac{\phi'(0)}{\exp\!(0)}
     = \overline{x}
     \\
     C^{(2)}
     &= - \dv[2]{k} \Big(\ln\!\big(\phi(k)\big)\Big) \big|_{k = 0}
-    = \frac{\big(\phi'(0)\big)^2}{\exp(0)^2} - \frac{\phi''(0)}{\exp(0)}
+    = \frac{\big(\phi'(0)\big)^2}{\exp\!(0)^2} - \frac{\phi''(0)}{\exp\!(0)}
     = - \overline{x}^2 + \overline{x^2} = \sigma^2
 \end{aligned}$$
 
@@ -153,7 +153,7 @@ with $\sqrt{N}$ appearing in the arguments of $\phi_n$:
 $$\begin{aligned}
     \phi_z(k)
     &= \idotsint
-    \Big( \prod_{n = 1}^N p(x_n) \Big) \: \delta\Big( z - \frac{1}{\sqrt{N}} \sum_{n = 1}^N x_n \Big) \exp(i k z)
+    \Big( \prod_{n = 1}^N p(x_n) \Big) \: \delta\Big( z - \frac{1}{\sqrt{N}} \sum_{n = 1}^N x_n \Big) \exp\!(i k z)
     \dd{x_1} \cdots \dd{x_N}
     \\
     &= \idotsint
@@ -184,7 +184,7 @@ $$\begin{aligned}
     = i k \overline{z} - \frac{k^2}{2} \sigma_z^2
     \\
     \phi_z(k)
-    &\approx \exp(i k \overline{z}) \exp\!(- k^2 \sigma_z^2 / 2)
+    &\approx \exp\!(i k \overline{z}) \exp\!(- k^2 \sigma_z^2 / 2)
 \end{aligned}$$
 
 We take its inverse Fourier transform to get the density $p(z)$,
@@ -194,7 +194,7 @@ which is even already normalized:
 $$\begin{aligned}
     p(z)
     = \hat{\mathcal{F}}^{-1} \{\phi_z(k)\}
-    &= \frac{1}{2 \pi} \int_{-\infty}^\infty \exp\!\big(\!-\! i k (z - \overline{z})\big) \exp(- k^2 \sigma_z^2 / 2) \dd{k}
+    &= \frac{1}{2 \pi} \int_{-\infty}^\infty \exp\!\big(\!-\! i k (z - \overline{z})\big) \exp\!(- k^2 \sigma_z^2 / 2) \dd{k}
     \\
     &= \frac{1}{\sqrt{2 \pi \sigma_z^2}} \exp\!\Big(\!-\! \frac{(z - \overline{z})^2}{2 \sigma_z^2} \Big)
 \end{aligned}$$
diff --git a/content/know/concept/ritz-method/index.pdc b/content/know/concept/ritz-method/index.pdc
index 320771a..1fe0d31 100644
--- a/content/know/concept/ritz-method/index.pdc
+++ b/content/know/concept/ritz-method/index.pdc
@@ -229,6 +229,10 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
+In the context of quantum mechanics, this is not surprising,
+since any superposition of multiple states
+is guaranteed to have a higher energy than the ground state.
+
 Note that the convergence to $\lambda_0$ goes as $|c_n|^2$,
 while $u$ converges to $u_0$ as $|c_n|$ by definition,
 so even a fairly bad guess $u$ will give a decent estimate for $\lambda_0$.
@@ -348,10 +352,17 @@ in a limited basis would yield a matrix $\overline{H}$ giving rough eigenvalues.
 The point of this discussion is to rigorously show
 the validity of this approach.
 
+If we only care about the ground state,
+then we already know $\lambda$ from $R[u]$,
+so all we need to do is solve the above matrix equation for $a_n$.
+Keep in mind that $\overline{M}$ is singular,
+and $a_n$ are only defined up to a constant factor.
+
 Nowadays, there exist many other methods to calculate eigenvalues
 of complicated operators $\hat{H}$,
 but an attractive feature of the Ritz method is that it is single-step,
 whereas its competitors tend to be iterative.
+That said, the Ritz method cannot recover from a poorly chosen basis.