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/newtons-bucket/index.md | 28 ++++++++++++++--------------
 1 file changed, 14 insertions(+), 14 deletions(-)

(limited to 'source/know/concept/newtons-bucket/index.md')

diff --git a/source/know/concept/newtons-bucket/index.md b/source/know/concept/newtons-bucket/index.md
index 757262d..21740b9 100644
--- a/source/know/concept/newtons-bucket/index.md
+++ b/source/know/concept/newtons-bucket/index.md
@@ -10,23 +10,23 @@ layout: "concept"
 ---
 
 **Newton's bucket** is a cylindrical bucket
-that rotates at angular velocity $\omega$.
+that rotates at angular velocity $$\omega$$.
 Due to [viscosity](/know/concept/viscosity/),
 any liquid in the bucket is affected by the rotation,
-eventually achieving the exact same $\omega$.
+eventually achieving the exact same $$\omega$$.
 
 However, once in equilibrium, the liquid's surface is not flat,
 but curved upwards from the center.
-This is due to the centrifugal force $\va{F}_\mathrm{f} = m \va{f}$ on a molecule with mass $m$:
+This is due to the centrifugal force $$\va{F}_\mathrm{f} = m \va{f}$$ on a molecule with mass $$m$$:
 
 $$\begin{aligned}
     \va{f}
     = \omega^2 \va{r}
 \end{aligned}$$
 
-Where $\va{r}$ is the molecule's position relative to the axis of rotation.
+Where $$\va{r}$$ is the molecule's position relative to the axis of rotation.
 This (fictitious) force can be written as the gradient
-of a potential $\Phi_\mathrm{f}$, such that $\va{f} = - \nabla \Phi_\mathrm{f}$:
+of a potential $$\Phi_\mathrm{f}$$, such that $$\va{f} = - \nabla \Phi_\mathrm{f}$$:
 
 $$\begin{aligned}
     \Phi_\mathrm{f}
@@ -34,8 +34,8 @@ $$\begin{aligned}
     = - \frac{\omega^2}{2} (x^2 + y^2)
 \end{aligned}$$
 
-In addition, each molecule feels a gravitational force $\va{F}_\mathrm{g} = m \va{g}$,
-where $\va{g} = - \nabla \Phi_\mathrm{g}$:
+In addition, each molecule feels a gravitational force $$\va{F}_\mathrm{g} = m \va{g}$$,
+where $$\va{g} = - \nabla \Phi_\mathrm{g}$$:
 
 $$\begin{aligned}
     \Phi_\mathrm{g}
@@ -43,7 +43,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Overall, the molecule therefore feels an "effective" force
-with a potential $\Phi$ given by:
+with a potential $$\Phi$$ given by:
 
 $$\begin{aligned}
     \Phi
@@ -51,7 +51,7 @@ $$\begin{aligned}
     = \mathrm{g} z - \frac{\omega^2}{2} (x^2 + y^2)
 \end{aligned}$$
 
-At equilibrium, the [hydrostatic pressure](/know/concept/hydrostatic-pressure/) $p$
+At equilibrium, the [hydrostatic pressure](/know/concept/hydrostatic-pressure/) $$p$$
 in the liquid is the one that satisfies:
 
 $$\begin{aligned}
@@ -59,7 +59,7 @@ $$\begin{aligned}
     = - \nabla \Phi
 \end{aligned}$$
 
-Removing the gradients gives integration constants $p_0$ and $\Phi_0$,
+Removing the gradients gives integration constants $$p_0$$ and $$\Phi_0$$,
 so the equilibrium equation is:
 
 $$\begin{aligned}
@@ -67,16 +67,16 @@ $$\begin{aligned}
     = - \rho (\Phi - \Phi_0)
 \end{aligned}$$
 
-We isolate this for $p$ and rewrite $\Phi_0 = \mathrm{g} z_0$,
-where $z_0$ is the liquid height at the center:
+We isolate this for $$p$$ and rewrite $$\Phi_0 = \mathrm{g} z_0$$,
+where $$z_0$$ is the liquid height at the center:
 
 $$\begin{aligned}
     p
     = p_0 - \rho \mathrm{g} (z - z_0) + \frac{\omega^2}{2} \rho (x^2 + y^2)
 \end{aligned}$$
 
-At the surface, we demand that $p = p_0$, where $p_0$ is the air pressure.
-The $z$-coordinate at which this is satisfied is as follows,
+At the surface, we demand that $$p = p_0$$, where $$p_0$$ is the air pressure.
+The $$z$$-coordinate at which this is satisfied is as follows,
 telling us that the surface is parabolic:
 
 $$\begin{aligned}
-- 
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