From 5cacf4ffaf3a9621ab536195f6469f98a420f054 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Sat, 5 Sep 2026 21:55:33 +0200
Subject: Improve knowledge base

---
 source/know/concept/boltzmann-equation/index.md | 20 ++++++++++++--------
 1 file changed, 12 insertions(+), 8 deletions(-)

(limited to 'source/know/concept/boltzmann-equation/index.md')

diff --git a/source/know/concept/boltzmann-equation/index.md b/source/know/concept/boltzmann-equation/index.md
index 5f4add0..3821512 100644
--- a/source/know/concept/boltzmann-equation/index.md
+++ b/source/know/concept/boltzmann-equation/index.md
@@ -67,7 +67,7 @@ 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**:
+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 $$f_0$$:
 
@@ -90,14 +90,15 @@ $$\begin{aligned}
     n(\vb{r}, t) = \int_{-\infty}^\infty f(\vb{r}, \vb{v}, t) \dd{\vb{v}}
 \end{aligned}$$
 
-Consequently, a purely velocity-dependent quantity $$Q(\vb{v})$$ can be averaged like so:
+Consequently, a purely velocity-dependent quantity $$Q(\vb{v})$$
+can be averaged like so:
 
 $$\begin{aligned}
-    \Expval{Q}
-    = \frac{1}{n} \int_{-\infty}^\infty Q(\vb{r}, \vb{v}, t) \: f(\vb{r}, \vb{v}, t) \dd{\vb{v}}
+    \Expval{Q}\!(\vb{r}, t)
+    = \frac{1}{n} \int_{-\infty}^\infty Q(\vb{v}) \: f(\vb{r}, \vb{v}, t) \dd{\vb{v}}
 \end{aligned}$$
 
-With that in mind, we multiply the collisionless BTE equation by $$Q(\vb{v})$$ and integrate,
+With that in mind, we multiply the collisionless BTE by $$Q(\vb{v})$$ and integrate,
 assuming that $$\vb{F}$$ does not depend on $$\vb{v}$$:
 
 $$\begin{aligned}
@@ -136,7 +137,8 @@ $$\begin{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:
+where $$\vb{V} \equiv \Expval{\vb{v}} = n^{-1} \int \vb{v} f \dd{\vb{v}}$$
+is the fluid velocity:
 
 $$\begin{aligned}
     \boxed{
@@ -231,7 +233,8 @@ $$\begin{aligned}
 
 
 {% include proof/start.html id="proof-moment2" -%}
-We insert $$Q = m |\vb{v}|^2 / 2$$ into our prototype and recognize $$\rho$$ wherever possible:
+We insert $$Q = m |\vb{v}|^2 / 2$$ into our prototype
+and recognize $$\rho$$ wherever possible:
 
 $$\begin{aligned}
     0
@@ -244,7 +247,8 @@ $$\begin{aligned}
     - \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:
+We handle these terms one by one.
+Substituting $$\vb{v} = \vb{V} + \vb{w}$$ in the first gives:
 
 $$\begin{aligned}
     \Expval{|\vb{v}|^2}
-- 
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