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

1 files changed, 33 insertions, 33 deletions

diff --git a/source/know/concept/hydrostatic-pressure/index.md b/source/know/concept/hydrostatic-pressure/index.md
index 2e55246..020fc75 100644
--- a/source/know/concept/hydrostatic-pressure/index.md
+++ b/source/know/concept/hydrostatic-pressure/index.md
@@ -9,7 +9,7 @@ categories:
 layout: "concept"
 ---
 
-The pressure $p$ inside a fluid at rest,
+The pressure $$p$$ inside a fluid at rest,
 the so-called **hydrostatic pressure**,
 is an important quantity.
 Here we will properly define it,
@@ -20,8 +20,8 @@ both with and without an arbitrary gravity field.
 ## Without gravity
 
 Inside the fluid, we can imagine small arbitrary partition surfaces,
-with normal vector $\vu{n}$ and area $\dd{S}$,
-yielding the following vector element $\dd{\va{S}}$:
+with normal vector $$\vu{n}$$ and area $$\dd{S}$$,
+yielding the following vector element $$\dd{\va{S}}$$:
 
 $$\begin{aligned}
     \dd{\va{S}}
@@ -29,7 +29,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 The orientation of these surfaces does not matter.
-The **pressure** $p(\va{r})$ is defined as the force-per-area
+The **pressure** $$p(\va{r})$$ is defined as the force-per-area
 of these tiny surface elements:
 
 $$\begin{aligned}
@@ -38,9 +38,9 @@ $$\begin{aligned}
 \end{aligned}$$
 
 The negative sign is there because a positive pressure is conventionally defined
-to push from the positive (normal) side of $\dd{\va{S}}$ to the negative side.
-The total force $\va{F}$ on a larger surface inside the fluid is
-then given by the surface integral over many adjacent $\dd{\va{S}}$:
+to push from the positive (normal) side of $$\dd{\va{S}}$$ to the negative side.
+The total force $$\va{F}$$ on a larger surface inside the fluid is
+then given by the surface integral over many adjacent $$\dd{\va{S}}$$:
 
 $$\begin{aligned}
     \va{F}
@@ -57,8 +57,8 @@ $$\begin{aligned}
     = - \int_V \nabla p \dd{V}
 \end{aligned}$$
 
-Since the total force on the blob is simply the sum of the forces $\dd{\va{F}}$
-on all its constituent volume elements $\dd{V}$,
+Since the total force on the blob is simply the sum of the forces $$\dd{\va{F}}$$
+on all its constituent volume elements $$\dd{V}$$,
 we arrive at the following relation:
 
 $$\begin{aligned}
@@ -71,8 +71,8 @@ $$\begin{aligned}
 If the fluid is at rest, then all forces on the blob cancel out
 (otherwise it would move).
 Since we are currently neglecting all forces other than pressure,
-this is equivalent to demanding that $\dd{\va{F}} = 0$,
-which implies that $\nabla p = 0$, i.e. the pressure is constant.
+this is equivalent to demanding that $$\dd{\va{F}} = 0$$,
+which implies that $$\nabla p = 0$$, i.e. the pressure is constant.
 
 $$\begin{aligned}
     \boxed{
@@ -84,17 +84,17 @@ $$\begin{aligned}
 ## With gravity
 
 If we include gravity, then,
-in addition to the pressure's *contact force* $\va{F}_p$ from earlier,
-there is also a *body force* $\va{F}_g$ acting on
-the arbitrary blob $V$ of fluid enclosed by $S$:
+in addition to the pressure's *contact force* $$\va{F}_p$$ from earlier,
+there is also a *body force* $$\va{F}_g$$ acting on
+the arbitrary blob $$V$$ of fluid enclosed by $$S$$:
 
 $$\begin{aligned}
     \va{F}_g
     = \int_V \rho \va{g} \dd{V}
 \end{aligned}$$
 
-Where $\rho$ is the fluid's density (which need not be constant)
-and $\va{g}$ is the gravity field given in units of force-per-mass.
+Where $$\rho$$ is the fluid's density (which need not be constant)
+and $$\va{g}$$ is the gravity field given in units of force-per-mass.
 For a fluid at rest, these forces must cancel out:
 
 $$\begin{aligned}
@@ -116,7 +116,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 On Earth (or another body with strong gravity),
-it is reasonable to treat $\va{g}$ as only pointing in the downward $z$-direction,
+it is reasonable to treat $$\va{g}$$ as only pointing in the downward $$z$$-direction,
 in which case the above condition turns into:
 
 $$\begin{aligned}
@@ -124,9 +124,9 @@ $$\begin{aligned}
     = \rho g_0 z
 \end{aligned}$$
 
-Where $g_0$ is the magnitude of the $z$-component of $\va{g}$.
+Where $$g_0$$ is the magnitude of the $$z$$-component of $$\va{g}$$.
 We can generalize the equilibrium condition by treating
-the gravity field as the gradient of the gravitational potential $\Phi$:
+the gravity field as the gradient of the gravitational potential $$\Phi$$:
 
 $$\begin{aligned}
     \va{g}(\va{r})
@@ -142,13 +142,13 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-In practice, the density $\rho$ of the fluid
-may be a function of the pressure $p$ (compressibility)
-and/or temperature $T$ (thermal expansion).
+In practice, the density $$\rho$$ of the fluid
+may be a function of the pressure $$p$$ (compressibility)
+and/or temperature $$T$$ (thermal expansion).
 We will tackle the first complication, but neglect the second,
 i.e. we assume that the temperature is equal across the fluid.
 
-We then define the **pressure potential** $w(p)$ as
+We then define the **pressure potential** $$w(p)$$ as
 the indefinite integral of the density:
 
 $$\begin{aligned}
@@ -166,7 +166,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 From this, let us now define the
-**effective gravitational potential** $\Phi^*$ as follows:
+**effective gravitational potential** $$\Phi^*$$ as follows:
 
 $$\begin{aligned}
     \Phi^* \equiv \Phi + w(p)
@@ -181,7 +181,7 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-At every point in the fluid, despite $p$ being variable,
+At every point in the fluid, despite $$p$$ being variable,
 the force that is applied by the pressure must have the same magnitude in all directions at that point.
 This statement is known as **Pascal's law**,
 and is due to the fact that all forces must cancel out
@@ -193,15 +193,15 @@ $$\begin{aligned}
     = 0
 \end{aligned}$$
 
-Let the blob be a cube with side $a$.
-Now, $\va{F}_p$ is a contact force,
-meaning it acts on the surface, and is thus proportional to $a^2$,
-however, $\va{F}_g$ is a body force,
-meaning it acts on the volume, and is thus proportional to $a^3$.
+Let the blob be a cube with side $$a$$.
+Now, $$\va{F}_p$$ is a contact force,
+meaning it acts on the surface, and is thus proportional to $$a^2$$,
+however, $$\va{F}_g$$ is a body force,
+meaning it acts on the volume, and is thus proportional to $$a^3$$.
 Since we are considering a *point* in the fluid,
-$a$ is infinitesimally small,
-so that $\va{F}_p$ dominates $\va{F}_g$.
-Consequently, at equilibrium, $\va{F}_p$ must cancel out by itself,
+$$a$$ is infinitesimally small,
+so that $$\va{F}_p$$ dominates $$\va{F}_g$$.
+Consequently, at equilibrium, $$\va{F}_p$$ must cancel out by itself,
 which means that the pressure is the same in all directions.