Migrate from 'jekyll-katex' to 'kramdown-math-sskatex'

1 files changed, 44 insertions, 41 deletions

diff --git a/source/know/concept/boltzmann-equation/index.md b/source/know/concept/boltzmann-equation/index.md
index 20399df..9ed2fd2 100644
--- a/source/know/concept/boltzmann-equation/index.md
+++ b/source/know/concept/boltzmann-equation/index.md
@@ -10,17 +10,17 @@ layout: "concept"
 ---
 
 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:
+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:
+no longer have a net effect, so $$f$$ is fixed:
 
 $$\begin{aligned}
     \dv{f}{t}
@@ -35,7 +35,7 @@ $$\begin{aligned}
     = \bigg(\! \pdv{f}{t} \!\bigg)_\mathrm{\!col}
 \end{aligned}$$
 
-Where the right-hand side simply means "all changes in $f$ due to collisions".
+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:
 
 $$\begin{aligned}
@@ -49,8 +49,8 @@ $$\begin{aligned}
     &= \pdv{f}{t} + \vb{v} \cdot \nabla f + \vb{a} \cdot \pdv{f}{\vb{v}}
 \end{aligned}$$
 
-Where we have introduced the shorthand $\ipdv{f}{\vb{v}}$.
-Inserting Newton's second law $\vb{F} = m \vb{a}$
+Where we have introduced the shorthand $$\ipdv{f}{\vb{v}}$$.
+Inserting Newton's second law $$\vb{F} = m \vb{a}$$
 leads us to the **Boltzmann equation** or
 **Boltzmann transport equation** (BTE):
 
@@ -64,41 +64,41 @@ $$\begin{aligned}
 But what about the collision term?
 Expressions for it exist, which are almost exact in many cases,
 but unfortunately also quite difficult to work with.
-In addition, $f$ is a 7-dimensional function,
+In addition, $$f$$ is a 7-dimensional function,
 so the BTE is already hard to solve without collisions.
 We only present the simplest case,
 known as the **Bhatnagar-Gross-Krook approximation**:
-if the equilibrium state $f_0(\vb{r}, \vb{v})$ is known,
-then each collision brings the system closer to $f_0$:
+if the equilibrium state $$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.
+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 $n$:
+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:
+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 $\vb{v}$:
+With that in mind, we multiply the collisionless BTE equation by $$Q(\vb{v})$$ and integrate,
+assuming that $$\vb{F}$$ does not depend on $$\vb{v}$$:
 
 $$\begin{aligned}
     0
@@ -110,10 +110,10 @@ $$\begin{aligned}
     + \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 $\vb{v} \to \pm\infty$,
+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:
 
 $$\begin{aligned}
@@ -134,9 +134,9 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-If we set $Q = m$, then the mass density $\rho = n \Expval{Q}$,
+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 $$\vb{V} \equiv \Expval{\vb{v}} = \int \vb{v} f \dd{\vb{v}}$$ is the fluid velocity:
 
 $$\begin{aligned}
     \boxed{
@@ -150,8 +150,8 @@ $$\begin{aligned}
 <label for="proof-moment0">Proof</label>
 <div class="hidden" markdown="1">
 <label for="proof-moment0">Proof.</label>
-We insert $Q = m$ into our prototype,
-and since $m$ is constant, the rest is trivial:
+We insert $$Q = m$$ into our prototype,
+and since $$m$$ is constant, the rest is trivial:
 
 $$\begin{aligned}
     0
@@ -159,12 +159,13 @@ $$\begin{aligned}
     \\
     &= \pdv{\rho}{t} + \nabla \cdot \big(\rho \Expval{\vb{v}}\big) - 0
 \end{aligned}$$
+
 </div>
 </div>
 
-If we instead choose the momentum $Q = m \vb{v}$,
+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/):
+where $$\hat{P}$$ is the [Cauchy stress tensor](/know/concept/cauchy-stress-tensor/):
 
 $$\begin{aligned}
     \boxed{
@@ -178,7 +179,7 @@ $$\begin{aligned}
 <label for="proof-moment1">Proof</label>
 <div class="hidden" markdown="1">
 <label for="proof-moment1">Proof.</label>
-We insert $Q = m \vb{v}$ into our prototype and recognize $\rho$ wherever possible:
+We insert $$Q = m \vb{v}$$ into our prototype and recognize $$\rho$$ wherever possible:
 
 $$\begin{aligned}
     0
@@ -189,11 +190,11 @@ $$\begin{aligned}
     - \vb{F} \cdot \bigg( n \Expval{\pdv{\vb{v}}{\vb{v}}} \bigg)
 \end{aligned}$$
 
-With $\vb{v} \vb{v}$ being a dyadic product.
+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 $\vb{V}$:
+we split $$\vb{v} = \vb{V} \!+\! \vb{w}$$,
+where $$\vb{V}$$ is the average velocity vector,
+and $$\vb{w}$$ is the local deviation from $$\vb{V}$$:
 
 $$\begin{aligned}
     \Expval{\vb{v} \vb{v}}
@@ -202,7 +203,7 @@ $$\begin{aligned}
     = \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, $\Expval{\vb{w}} = 0$.
+Since $$\vb{w}$$ represents a deviation from the mean, $$\Expval{\vb{w}} = 0$$.
 We define the pressure tensor:
 
 $$\begin{aligned}
@@ -212,19 +213,20 @@ $$\begin{aligned}
 \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:
+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 $Q = m |\vb{v}|^2 / 2$,
+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:
+where $$U$$ is the thermal energy density and $$\vb{J}$$ is the heat flux:
 
 $$\begin{aligned}
     \boxed{
@@ -240,7 +242,7 @@ $$\begin{aligned}
 <label for="proof-moment2">Proof</label>
 <div class="hidden" markdown="1">
 <label for="proof-moment2">Proof.</label>
-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:
 
 $$\begin{aligned}
     0
@@ -253,7 +255,7 @@ $$\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:
 
 $$\begin{aligned}
     \Expval{|\vb{v}|^2}
@@ -265,7 +267,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 And likewise for the second term,
-where we recognize the stress tensor $\Expval{\vb{w} \vb{w}}$:
+where we recognize the stress tensor $$\Expval{\vb{w} \vb{w}}$$:
 
 $$\begin{aligned}
     \Expval{|\vb{v}|^2 \vb{v}}
@@ -294,7 +296,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 To clarify the physical interpretation,
-we define $U$, $\vb{J}$ and $\hat{P}$ as follows:
+we define $$U$$, $$\vb{J}$$ and $$\hat{P}$$ as follows:
 
 $$\begin{aligned}
     U
@@ -347,6 +349,7 @@ $$\begin{aligned}
     \end{bmatrix}
     = \sum_{i=1}^{3} \sum_{j=1}^{3} \pdv{P_{ij}}{x_j} V_i
 \end{aligned}$$
+
 </div>
 </div>