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/meniscus/index.md | 68 +++++++++++++++++------------------
 1 file changed, 34 insertions(+), 34 deletions(-)

(limited to 'source/know/concept/meniscus')

diff --git a/source/know/concept/meniscus/index.md b/source/know/concept/meniscus/index.md
index e032a4e..a53b5e9 100644
--- a/source/know/concept/meniscus/index.md
+++ b/source/know/concept/meniscus/index.md
@@ -15,19 +15,19 @@ touches a flat solid wall, it will curve to meet it.
 This small rise or fall is called a **meniscus**,
 and is caused by surface tension and gravity.
 
-In 2D, let the vertical $y$-axis be a flat wall,
-and the fluid tend to $y = 0$ when $x \to \infty$.
-Close to the wall, i.e. for small $x$, the liquid curves up or down
-to touch the wall at a height $y = d$.
+In 2D, let the vertical $$y$$-axis be a flat wall,
+and the fluid tend to $$y = 0$$ when $$x \to \infty$$.
+Close to the wall, i.e. for small $$x$$, the liquid curves up or down
+to touch the wall at a height $$y = d$$.
 
 Three forces are at work here:
-the first two are the surface tension $\alpha$ of the fluid surface,
-and the counter-pull $\alpha \sin\phi$ of the wall against the tension,
-where $\phi$ is the contact angle.
+the first two are the surface tension $$\alpha$$ of the fluid surface,
+and the counter-pull $$\alpha \sin\phi$$ of the wall against the tension,
+where $$\phi$$ is the contact angle.
 The third is the [hydrostatic pressure](/know/concept/hydrostatic-pressure/) gradient
 inside the small portion of the fluid above/below the ambient level,
 which exerts a total force on the wall given by
-(for $\phi < \pi/2$ so that $d > 0$):
+(for $$\phi < \pi/2$$ so that $$d > 0$$):
 
 $$\begin{aligned}
     \int_0^d \rho g y \dd{y}
@@ -35,7 +35,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 If you were wondering about the units,
-keep in mind that there is an implicit $z$-direction here too.
+keep in mind that there is an implicit $$z$$-direction here too.
 This results in the following balance equation for the forces at the wall:
 
 $$\begin{aligned}
@@ -43,7 +43,7 @@ $$\begin{aligned}
     = \alpha \sin\phi + \frac{1}{2} \rho g d^2
 \end{aligned}$$
 
-We isolate this relation for $d$
+We isolate this relation for $$d$$
 and use some trigonometric magic to rewrite it:
 
 $$\begin{aligned}
@@ -52,10 +52,10 @@ $$\begin{aligned}
     = \sqrt{\frac{\alpha}{\rho g}} \sqrt{4 \sin^2\!\Big(\frac{\pi}{4} - \frac{\phi}{2}\Big)}
 \end{aligned}$$
 
-Here, we recognize the definition of the capillary length $L_c = \sqrt{\alpha / (\rho g)}$,
-yielding an expression for $d$
-that is valid both for $\phi < \pi/2$ (where $d > 0$)
-and $\phi > \pi/2$ (where $d < 0$):
+Here, we recognize the definition of the capillary length $$L_c = \sqrt{\alpha / (\rho g)}$$,
+yielding an expression for $$d$$
+that is valid both for $$\phi < \pi/2$$ (where $$d > 0$$)
+and $$\phi > \pi/2$$ (where $$d < 0$$):
 
 $$\begin{aligned}
     \boxed{
@@ -66,9 +66,9 @@ $$\begin{aligned}
 
 Next, we would like to know the exact shape of the meniscus.
 To do this, we need to describe the liquid surface differently,
-using the elevation angle $\theta$ relative to the $y = 0$ plane.
-The curve $\theta(s)$ is a function of the arc length $s$,
-where $\dd{s}^2 = \dd{x}^2 + \dd{y}^2$,
+using the elevation angle $$\theta$$ relative to the $$y = 0$$ plane.
+The curve $$\theta(s)$$ is a function of the arc length $$s$$,
+where $$\dd{s}^2 = \dd{x}^2 + \dd{y}^2$$,
 and is governed by:
 
 $$\begin{aligned}
@@ -82,23 +82,23 @@ $$\begin{aligned}
     = \frac{1}{R}
 \end{aligned}$$
 
-The last equation describes the curvature radius $R$
-of the surface along the $x$-axis.
+The last equation describes the curvature radius $$R$$
+of the surface along the $$x$$-axis.
 Since we are considering a flat wall,
 there is no curvature in the orthogonal principal direction.
 
 Just below the liquid surface in the meniscus,
 we expect the hydrostatic pressure
 and the [Young-Laplace law](/know/concept/young-laplace-law/)
-to agree about the pressure $p$,
-where $p_0$ is the external air pressure:
+to agree about the pressure $$p$$,
+where $$p_0$$ is the external air pressure:
 
 $$\begin{aligned}
     p_0 - \rho g y
     = p_0 - \frac{\alpha}{R}
 \end{aligned}$$
 
-Rearranging this yields that $R = L_c^2 / y$.
+Rearranging this yields that $$R = L_c^2 / y$$.
 Inserting this into the curvature equation gives us:
 
 $$\begin{aligned}
@@ -106,8 +106,8 @@ $$\begin{aligned}
     = \frac{y}{L_c^2}
 \end{aligned}$$
 
-By differentiating this equation with respect to $s$
-and using $\idv{y}{s} = \sin\theta$, we arrive at:
+By differentiating this equation with respect to $$s$$
+and using $$\idv{y}{s} = \sin\theta$$, we arrive at:
 
 $$\begin{aligned}
     \boxed{
@@ -115,7 +115,7 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-To solve this equation, we multiply it by $\idv{\theta}{s}$,
+To solve this equation, we multiply it by $$\idv{\theta}{s}$$,
 which is nonzero close to the wall:
 
 $$\begin{aligned}
@@ -123,16 +123,16 @@ $$\begin{aligned}
     = \dv{\theta}{s} \sin\theta
 \end{aligned}$$
 
-We integrate both sides with respect to $s$
-and set the integration constant to $1$,
-such that we get zero when $\theta \to 0$ away from the wall:
+We integrate both sides with respect to $$s$$
+and set the integration constant to $$1$$,
+such that we get zero when $$\theta \to 0$$ away from the wall:
 
 $$\begin{aligned}
     \frac{L_c^2}{2} \Big( \dv{\theta}{s} \Big)^2
     = 1 - \cos\theta
 \end{aligned}$$
 
-Isolating this for $\idv{\theta}{s}$ and using a trigonometric identity then yields:
+Isolating this for $$\idv{\theta}{s}$$ and using a trigonometric identity then yields:
 
 $$\begin{aligned}
     \dv{\theta}{s}
@@ -142,7 +142,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 We use trigonometric relations on the equations
-for $\idv{x}{s}$ and $\idv{y}{s}$ to get $\theta$-derivatives:
+for $$\idv{x}{s}$$ and $$\idv{y}{s}$$ to get $$\theta$$-derivatives:
 
 $$\begin{aligned}
     \dv{x}{\theta}
@@ -156,7 +156,7 @@ $$\begin{aligned}
     = - L_c \cos\!\Big( \frac{\theta}{2} \Big)
 \end{aligned}$$
 
-Let $\theta_0 = \phi - \pi/2$ be the initial elevation angle $\theta(0)$ at the wall.
+Let $$\theta_0 = \phi - \pi/2$$ be the initial elevation angle $$\theta(0)$$ at the wall.
 Then, by integrating the above equations, we get the following solutions:
 
 $$\begin{gathered}
@@ -172,9 +172,9 @@ $$\begin{gathered}
     }
 \end{gathered}$$
 
-Where the integration constant has been chosen such that $y \to 0$ for $\theta \to 0$ away from the wall,
-and $x = 0$ for $\theta = \theta_0$.
-This result is consistent with our earlier expression for $d$:
+Where the integration constant has been chosen such that $$y \to 0$$ for $$\theta \to 0$$ away from the wall,
+and $$x = 0$$ for $$\theta = \theta_0$$.
+This result is consistent with our earlier expression for $$d$$:
 
 $$\begin{aligned}
     d
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