From 16555851b6514a736c5c9d8e73de7da7fc9b6288 Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Thu, 20 Oct 2022 18:25:31 +0200
Subject: Migrate from 'jekyll-katex' to 'kramdown-math-sskatex'

---
 source/know/concept/material-derivative/index.md | 38 ++++++++++++------------
 1 file changed, 19 insertions(+), 19 deletions(-)

(limited to 'source/know/concept/material-derivative')

diff --git a/source/know/concept/material-derivative/index.md b/source/know/concept/material-derivative/index.md
index 16c2c66..93e8ad0 100644
--- a/source/know/concept/material-derivative/index.md
+++ b/source/know/concept/material-derivative/index.md
@@ -11,22 +11,22 @@ layout: "concept"
 ---
 
 Inside a fluid (or any other continuum), we might be interested in
-the time evolution of a certain intensive quantity $f$,
+the time evolution of a certain intensive quantity $$f$$,
 e.g. the temperature or pressure,
-represented by a scalar field $f(\va{r}, t)$.
+represented by a scalar field $$f(\va{r}, t)$$.
 
-If the fluid is static, the evolution of $f$ is simply $\ipdv{f}{t}$,
+If the fluid is static, the evolution of $$f$$ is simply $$\ipdv{f}{t}$$,
 since each point of the fluid is motionless.
 However, if the fluid is moving, we have a problem:
-the fluid molecules at position $\va{r} = \va{r}_0$ are not necessarily
-the same ones at time $t = t_0$ and $t = t_1$.
-Those molecules take $f$ with them as they move,
+the fluid molecules at position $$\va{r} = \va{r}_0$$ are not necessarily
+the same ones at time $$t = t_0$$ and $$t = t_1$$.
+Those molecules take $$f$$ with them as they move,
 so we need to account for this transport somehow.
 
 To do so, we choose an infinitesimal "blob" or **parcel** of the fluid,
 which always contains the same specific molecules,
-and track its position $\va{r}(t)$ through time as it moves and deforms.
-The value of $f$ for this parcel is then given by:
+and track its position $$\va{r}(t)$$ through time as it moves and deforms.
+The value of $$f$$ for this parcel is then given by:
 
 $$\begin{aligned}
     f(\va{r}, t)
@@ -34,9 +34,9 @@ $$\begin{aligned}
     = f\big(x(t), y(t), z(t), t\big)
 \end{aligned}$$
 
-In effect, we have simply made the coordinate $\va{r}$ dependent on time,
+In effect, we have simply made the coordinate $$\va{r}$$ dependent on time,
 and have specifically chosen the time-dependence to track the parcel.
-The net evolution of $f$ is then its "true" (i.e. non-partial) derivative with respect to $t$,
+The net evolution of $$f$$ is then its "true" (i.e. non-partial) derivative with respect to $$t$$,
 allowing us to apply the chain rule:
 
 $$\begin{aligned}
@@ -46,8 +46,8 @@ $$\begin{aligned}
     &= \pdv{f}{t} + v_x \pdv{f}{x} + v_y \pdv{f}{y} + v_z \pdv{f}{z}
 \end{aligned}$$
 
-Where $v_x$, $v_y$ and $v_z$ are the parcel's velocity components.
-Let $\va{v} = (v_x, v_y, v_z)$ be the velocity vector field,
+Where $$v_x$$, $$v_y$$ and $$v_z$$ are the parcel's velocity components.
+Let $$\va{v} = (v_x, v_y, v_z)$$ be the velocity vector field,
 then we can rewrite this expression like so:
 
 $$\begin{aligned}
@@ -55,12 +55,12 @@ $$\begin{aligned}
     &= \pdv{f}{t} + (\va{v} \cdot \nabla) f
 \end{aligned}$$
 
-Note that $\va{v} = \va{v}(\va{r}, t)$,
-that is, the velocity can change with time ($t$-dependence),
-and depends on which parcel we track ($\va{r}$-dependence).
+Note that $$\va{v} = \va{v}(\va{r}, t)$$,
+that is, the velocity can change with time ($$t$$-dependence),
+and depends on which parcel we track ($$\va{r}$$-dependence).
 
 Of course, the parcel is in our imagination:
-$\va{r}$ does not really depend on $t$;
+$$\va{r}$$ does not really depend on $$t$$;
 after all, we are dealing with a continuum.
 Nevertheless, the right-hand side of the equation is very useful,
 and is known as the **material derivative** or **comoving derivative**:
@@ -75,11 +75,11 @@ $$\begin{aligned}
 The first term is called the **local rate of change**,
 and the second is the **advective rate of change**.
 In effect, the latter moves the frame of reference along with the material,
-so that we can find the evolution of $f$
+so that we can find the evolution of $$f$$
 without needing to worry about the continuum's motion.
 
-That was for a scalar field $f(\va{r}, t)$,
-but in fact the definition also works for vector fields $\va{U}(\va{r}, t)$:
+That was for a scalar field $$f(\va{r}, t)$$,
+but in fact the definition also works for vector fields $$\va{U}(\va{r}, t)$$:
 
 $$\begin{aligned}
     \boxed{
-- 
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