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

1 files changed, 21 insertions, 21 deletions

diff --git a/source/know/concept/reduced-mass/index.md b/source/know/concept/reduced-mass/index.md
index 6718895..1a05b5c 100644
--- a/source/know/concept/reduced-mass/index.md
+++ b/source/know/concept/reduced-mass/index.md
@@ -8,9 +8,9 @@ layout: "concept"
 ---
 
 Problems with two interacting objects can be simplified
-by combining them into a pseudo-object with **reduced mass** $\mu$,
+by combining them into a pseudo-object with **reduced mass** $$\mu$$,
 whose position equals the relative position of the objects.
-For bodies 1 and 2 with respective masses $m_1$ and $m_2$:
+For bodies 1 and 2 with respective masses $$m_1$$ and $$m_2$$:
 
 $$\begin{aligned}
     \boxed{
@@ -18,11 +18,11 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-If $\va{x}_1$ and $\va{x}_2$ are the objects' respective positions,
+If $$\va{x}_1$$ and $$\va{x}_2$$ are the objects' respective positions,
 then we define
-the relative position $\va{x}_r$,
-the relative velocity $\va{v}_r$,
-and the relative acceleration $\va{a}_r$:
+the relative position $$\va{x}_r$$,
+the relative velocity $$\va{v}_r$$,
+and the relative acceleration $$\va{a}_r$$:
 
 $$\begin{aligned}
     \va{x}_r
@@ -69,9 +69,9 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Meanwhile, Newton's third law states that
-if object 1 experiences a force $\va{F}_1 = m_1 \va{a}_1$ caused by object 2,
-then object 2 experiences an opposite and equal force $\va{F}_2 = - \va{F}_1$.
-In fact, our earlier relation between $\va{a}_1$ and $\va{a}_1$
+if object 1 experiences a force $$\va{F}_1 = m_1 \va{a}_1$$ caused by object 2,
+then object 2 experiences an opposite and equal force $$\va{F}_2 = - \va{F}_1$$.
+In fact, our earlier relation between $$\va{a}_1$$ and $$\va{a}_1$$
 boils down to Newton's third law:
 
 $$\begin{aligned}
@@ -80,7 +80,7 @@ $$\begin{aligned}
     \va{a}_2 = - \frac{m_1}{m_2} \va{a}_1
 \end{aligned}$$
 
-With all that in mind, let us take a closer look at the relative acceleration $\va{a}_r$:
+With all that in mind, let us take a closer look at the relative acceleration $$\va{a}_r$$:
 
 $$\begin{aligned}
     \va{a}_r
@@ -90,15 +90,15 @@ $$\begin{aligned}
     = - \frac{\va{F}_2}{\mu}
 \end{aligned}$$
 
-Where $\mu$ is the reduced mass, as defined above.
-In other words, the relative acceleration $\va{a}_r$
-is just $\va{a}_1 = \va{F}_1 / m_1$ multiplied by $m_1 / \mu$.
+Where $$\mu$$ is the reduced mass, as defined above.
+In other words, the relative acceleration $$\va{a}_r$$
+is just $$\va{a}_1 = \va{F}_1 / m_1$$ multiplied by $$m_1 / \mu$$.
 This can be regarded as focusing on the dynamics of body 1,
 while correcting for the effects of body 2.
 
 This also suggests the following way
-to recover the original positions $\va{x}_1$ and $\va{x}_2$
-from $\va{x}_r$, which you can easily verify for yourself:
+to recover the original positions $$\va{x}_1$$ and $$\va{x}_2$$
+from $$\va{x}_r$$, which you can easily verify for yourself:
 
 $$\begin{aligned}
     \va{x}_1
@@ -110,7 +110,7 @@ $$\begin{aligned}
     = - \frac{m_1}{m_1 + m_2} \va{x}_r
 \end{aligned}$$
 
-With this, we can rewrite the total kinetic energy $T$ in an elegant way:
+With this, we can rewrite the total kinetic energy $$T$$ in an elegant way:
 
 $$\begin{aligned}
     T
@@ -125,11 +125,11 @@ $$\begin{aligned}
     = \frac{1}{2} \mu \va{v}_r^2
 \end{aligned}$$
 
-Then, assuming that the system's potential energy $V$
+Then, assuming that the system's potential energy $$V$$
 only depends on the distance between the two objects,
-i.e. $V = V(|\va{x}_1 - \va{x}_2|) = V(|\va{x}_r|)$,
-we just showed that we can rewrite both $T$ and $V$
-to contain only $\mu$ and relative quantities.
+i.e. $$V = V(|\va{x}_1 - \va{x}_2|) = V(|\va{x}_r|)$$,
+we just showed that we can rewrite both $$T$$ and $$V$$
+to contain only $$\mu$$ and relative quantities.
 This is relevant for both [Lagrangian mechanics](/know/concept/lagrangian-mechanics/)
 and [Hamiltonian mechanics](/know/concept/hamiltonian-mechanics/),
-where $L = T - V$ and $H = T + V$ respectively.
+where $$L = T - V$$ and $$H = T + V$$ respectively.