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

1 files changed, 35 insertions, 35 deletions

diff --git a/source/know/concept/lubrication-theory/index.md b/source/know/concept/lubrication-theory/index.md
index d2a615e..4015526 100644
--- a/source/know/concept/lubrication-theory/index.md
+++ b/source/know/concept/lubrication-theory/index.md
@@ -16,15 +16,15 @@ is the study of fluids that are tightly constrained in one dimension,
 especially those in small gaps between moving surfaces.
 
 For simplicity, we limit ourselves to 2D
-by assuming that everything is constant along the $z$-axis.
-Consider a gap of width $d$ (along $y$) and length $L$ (along $x$),
-where $d \ll L$, containing the fluid.
+by assuming that everything is constant along the $$z$$-axis.
+Consider a gap of width $$d$$ (along $$y$$) and length $$L$$ (along $$x$$),
+where $$d \ll L$$, containing the fluid.
 Outside the gap, the lubricant has a
-[Reynolds number](/know/concept/reynolds-number/) $\mathrm{Re} \approx U L / \nu$.
+[Reynolds number](/know/concept/reynolds-number/) $$\mathrm{Re} \approx U L / \nu$$.
 
-Inside the gap, the Reynolds number $\mathrm{Re}_\mathrm{gap}$ is different.
-This is because advection will dominate along the $x$-axis (gap length),
-and viscosity along the $y$-axis (gap width).
+Inside the gap, the Reynolds number $$\mathrm{Re}_\mathrm{gap}$$ is different.
+This is because advection will dominate along the $$x$$-axis (gap length),
+and viscosity along the $$y$$-axis (gap width).
 Therefore:
 
 $$\begin{aligned}
@@ -34,13 +34,13 @@ $$\begin{aligned}
     \approx \frac{d^2}{L^2} \mathrm{Re}
 \end{aligned}$$
 
-If $d$ is small enough compared to $L$,
-then $\mathrm{Re}_\mathrm{gap} \ll 1$.
-More formally, we need $d \ll L / \sqrt{\mathrm{Re}}$,
+If $$d$$ is small enough compared to $$L$$,
+then $$\mathrm{Re}_\mathrm{gap} \ll 1$$.
+More formally, we need $$d \ll L / \sqrt{\mathrm{Re}}$$,
 so we are inside the boundary layer,
 in the realm of the [Prandtl equations](/know/concept/prandtl-equations/).
 
-Let $\mathrm{Re}_\mathrm{gap} \ll 1$.
+Let $$\mathrm{Re}_\mathrm{gap} \ll 1$$.
 We are thus dealing with *Stokes flow*, in which case
 the [Navier-Stokes equations](/know/concept/navier-stokes/equations/)
 can be reduced to the following *Stokes equations*:
@@ -53,16 +53,16 @@ $$\begin{aligned}
     = \eta \: \Big( \pdvn{2}{v_y}{x} + \pdvn{2}{v_y}{y} \Big)
 \end{aligned}$$
 
-Let the $y = 0$ plane be an infinite flat surface,
-sliding in the positive $x$-direction at a constant velocity $U$.
+Let the $$y = 0$$ plane be an infinite flat surface,
+sliding in the positive $$x$$-direction at a constant velocity $$U$$.
 On the other side of the gap,
-an arbitrary surface is described by $h(x)$.
+an arbitrary surface is described by $$h(x)$$.
 
 Since the gap is so narrow,
 and the surfaces' movements cause large shear stresses inside,
-$v_y$ is negligible compared to $v_x$.
+$$v_y$$ is negligible compared to $$v_x$$.
 Furthermore, because the gap is so long,
-we assume that $\ipdv{v_x}{x}$ is negligible compared to $\ipdv{v_x}{y}$.
+we assume that $$\ipdv{v_x}{x}$$ is negligible compared to $$\ipdv{v_x}{y}$$.
 This reduces the Stokes equations to:
 
 $$\begin{aligned}
@@ -74,7 +74,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 This result could also be derived from the Prandtl equations.
-In any case, it tells us that $p$ only depends on $x$,
+In any case, it tells us that $$p$$ only depends on $$x$$,
 allowing us to integrate the former equation:
 
 $$\begin{aligned}
@@ -82,17 +82,17 @@ $$\begin{aligned}
     = \frac{p'}{2 \eta} y^2 + C_1 y + C_2
 \end{aligned}$$
 
-Where $C_1$ and $C_2$ are integration constants.
-At $y = 0$, the viscous *no-slip* condition demands that $v_x = U$, so $C_2 = U$.
-Likewise, at $y = h(x)$, we need $v_x = 0$, leading us to:
+Where $$C_1$$ and $$C_2$$ are integration constants.
+At $$y = 0$$, the viscous *no-slip* condition demands that $$v_x = U$$, so $$C_2 = U$$.
+Likewise, at $$y = h(x)$$, we need $$v_x = 0$$, leading us to:
 
 $$\begin{aligned}
     v_x
     = \frac{p'}{2 \eta} y^2 - \Big( \frac{p'}{2 \eta} h + \frac{U}{h} \Big) y + U
 \end{aligned}$$
 
-The moving bottom surface drags fluid in the $x$-direction
-at a volumetric rate $Q$, given by:
+The moving bottom surface drags fluid in the $$x$$-direction
+at a volumetric rate $$Q$$, given by:
 
 $$\begin{aligned}
     Q
@@ -103,9 +103,9 @@ $$\begin{aligned}
 
 Assuming that the lubricant is incompressible,
 meaning that the same volume of fluid must be leaving a point as is entering it.
-In other words, $Q$ is independent of $x$,
-which allows us to write $p'(x)$ in terms of
-measurable constants and the known function $h(x)$:
+In other words, $$Q$$ is independent of $$x$$,
+which allows us to write $$p'(x)$$ in terms of
+measurable constants and the known function $$h(x)$$:
 
 $$\begin{aligned}
     \boxed{
@@ -114,7 +114,7 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Then we insert this into our earlier expression for $v_x$, yielding:
+Then we insert this into our earlier expression for $$v_x$$, yielding:
 
 $$\begin{aligned}
     v_x
@@ -130,7 +130,7 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-With this, we can find $v_y$ by exploiting incompressibility,
+With this, we can find $$v_y$$ by exploiting incompressibility,
 i.e. the continuity equation states:
 
 $$\begin{aligned}
@@ -139,7 +139,7 @@ $$\begin{aligned}
     = - 2 h' \frac{U h - 3 Q}{h^4} \big( 2 h y - 3 y^2 \big)
 \end{aligned}$$
 
-Integrating with respect to $y$ thus leads to the following transverse velocity $v_y$:
+Integrating with respect to $$y$$ thus leads to the following transverse velocity $$v_y$$:
 
 $$\begin{aligned}
     \boxed{
@@ -150,7 +150,7 @@ $$\begin{aligned}
 
 Typically, the lubricant is not in a preexisting pressure differential,
 i.e it is not getting pumped through the system.
-Although the pressure gradient $p'$ need not be zero,
+Although the pressure gradient $$p'$$ need not be zero,
 we therefore expect that its integral vanishes:
 
 $$\begin{aligned}
@@ -159,7 +159,7 @@ $$\begin{aligned}
     = 6 \eta U \int_L \frac{1}{h(x)^2} \dd{x} - 12 \eta Q \int_L \frac{1}{h(x)^3} \dd{x}
 \end{aligned}$$
 
-Isolating this for $Q$, and defining $q$ as below, yields a simple equation:
+Isolating this for $$Q$$, and defining $$q$$ as below, yields a simple equation:
 
 $$\begin{aligned}
     Q
@@ -169,7 +169,7 @@ $$\begin{aligned}
     \equiv \frac{\int_L h^{-2} \dd{x}}{\int_L h^{-3} \dd{x}}
 \end{aligned}$$
 
-We substitute this into $v_x$ and rearrange to get an interesting expression:
+We substitute this into $$v_x$$ and rearrange to get an interesting expression:
 
 $$\begin{aligned}
     v_x
@@ -180,7 +180,7 @@ $$\begin{aligned}
 
 The first factor is always positive,
 but the second can be negative,
-if for some $y$-values:
+if for some $$y$$-values:
 
 $$\begin{aligned}
     h^2 < 3 y (h - q)
@@ -188,8 +188,8 @@ $$\begin{aligned}
     y > \frac{h^2}{3 (h - q)}
 \end{aligned}$$
 
-Since $h > y$, such $y$-values will only exist
-if $h$ is larger than some threshold:
+Since $$h > y$$, such $$y$$-values will only exist
+if $$h$$ is larger than some threshold:
 
 $$\begin{aligned}
     3 (h - q) > h
@@ -201,7 +201,7 @@ If this condition is satisfied,
 there will be some flow reversal:
 rather than just getting dragged by the shearing motion,
 the lubricant instead "rolls" inside the gap.
-This is confirmed by $v_y$:
+This is confirmed by $$v_y$$:
 
 $$\begin{aligned}
     v_y