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/young-laplace-law/index.md | 32 +++++++++++++-------------
 1 file changed, 16 insertions(+), 16 deletions(-)

(limited to 'source/know/concept/young-laplace-law')

diff --git a/source/know/concept/young-laplace-law/index.md b/source/know/concept/young-laplace-law/index.md
index 28c3026..f890beb 100644
--- a/source/know/concept/young-laplace-law/index.md
+++ b/source/know/concept/young-laplace-law/index.md
@@ -13,21 +13,21 @@ In liquids, the **Young-Laplace law** governs surface tension:
 it describes the tension forces on a surface
 as a pressure difference between the two sides of the liquid.
 
-Consider a small rectangle on the surface with sides $\dd{\ell_1}$ and $\dd{\ell_2}$,
+Consider a small rectangle on the surface with sides $$\dd{\ell_1}$$ and $$\dd{\ell_2}$$,
 orientated such that the sides are parallel to the (orthogonal)
 principal directions of the surface' [curvature](/know/concept/curvature/).
 
 Surface tension then pulls at the sides with a force
-of magnitude $\alpha \dd{\ell_2}$ and $\alpha \dd{\ell_2}$,
-where $\alpha$ is the energy cost per unit of area,
+of magnitude $$\alpha \dd{\ell_2}$$ and $$\alpha \dd{\ell_2}$$,
+where $$\alpha$$ is the energy cost per unit of area,
 which is the same as the force per unit of distance.
 However, due to the surface' curvature,
 those forces are not quite in the same plane as the rectangle.
 
 Along both principal directions,
 if we treat this portion of the surface as a small arc of a circle
-with a radius equal to the principal radius of curvature $R_1$ or $R_2$,
-then the tension forces are at angles $\theta_1$ and $\theta_2$
+with a radius equal to the principal radius of curvature $$R_1$$ or $$R_2$$,
+then the tension forces are at angles $$\theta_1$$ and $$\theta_2$$
 calculated from the arc length:
 
 $$\begin{aligned}
@@ -38,14 +38,14 @@ $$\begin{aligned}
     = \frac{1}{2} \dd{\ell_1}
 \end{aligned}$$
 
-Pay attention to the indices $1$ and $2$:
-to get the angle of the force pulling at $\dd{\ell_1}$,
-we need to treat $\dd{\ell_2} / 2$ as an arc,
+Pay attention to the indices $$1$$ and $$2$$:
+to get the angle of the force pulling at $$\dd{\ell_1}$$,
+we need to treat $$\dd{\ell_2} / 2$$ as an arc,
 and vice versa.
 
 Since the forces are not quite in the plane,
 they have a small component acting *perpendicular* to the surface,
-with the following magnitudes $\dd{F_1}$ and $\dd{F_2}$
+with the following magnitudes $$\dd{F_1}$$ and $$\dd{F_2}$$
 along the principal axes:
 
 $$\begin{aligned}
@@ -62,13 +62,13 @@ $$\begin{aligned}
     = \frac{\alpha}{R_2} \dd{A}
 \end{aligned}$$
 
-The initial factor of $2$ is there since
+The initial factor of $$2$$ is there since
 the same force is pulling at opposide sides of the rectangle.
-We end up with $\alpha / R_{1,2}$ multiplied by
-the surface area $\dd{A} = \dd{\ell_1} \dd{\ell_2}$ of the rectangle.
+We end up with $$\alpha / R_{1,2}$$ multiplied by
+the surface area $$\dd{A} = \dd{\ell_1} \dd{\ell_2}$$ of the rectangle.
 
-Adding together $\dd{F_1}$ and $\dd{F_2}$ and
-dividing out $\dd{A}$ gives us the force-per-area (i.e. the pressure)
+Adding together $$\dd{F_1}$$ and $$\dd{F_2}$$ and
+dividing out $$\dd{A}$$ gives us the force-per-area (i.e. the pressure)
 added by surface tension,
 which is given by the **Young-Laplace law**:
 
@@ -79,9 +79,9 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-The total excess pressure $\Delta p$ is called the **Laplace pressure**,
+The total excess pressure $$\Delta p$$ is called the **Laplace pressure**,
 and fully determines the effects of surface tension:
-a certain interface shape leads to a certain $\Delta p$,
+a certain interface shape leads to a certain $$\Delta p$$,
 and the liquid will flow (i.e. the surface will move)
 to try to reach an equilibrium.
 
-- 
cgit v1.3