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

1 files changed, 29 insertions, 29 deletions

diff --git a/source/know/concept/lorentz-force/index.md b/source/know/concept/lorentz-force/index.md
index d1b216d..03178f6 100644
--- a/source/know/concept/lorentz-force/index.md
+++ b/source/know/concept/lorentz-force/index.md
@@ -10,10 +10,10 @@ layout: "concept"
 ---
 
 The **Lorentz force** is an empirical force used to define
-the [electric field](/know/concept/electric-field/) $\vb{E}$
-and [magnetic field](/know/concept/magnetic-field/) $\vb{B}$.
-For a particle with charge $q$ moving with velocity $\vb{u}$,
-the Lorentz force $\vb{F}$ is given by:
+the [electric field](/know/concept/electric-field/) $$\vb{E}$$
+and [magnetic field](/know/concept/magnetic-field/) $$\vb{B}$$.
+For a particle with charge $$q$$ moving with velocity $$\vb{u}$$,
+the Lorentz force $$\vb{F}$$ is given by:
 
 $$\begin{aligned}
     \boxed{
@@ -25,9 +25,9 @@ $$\begin{aligned}
 
 ## Uniform electric field
 
-Consider the simple case of an electric field $\vb{E}$
+Consider the simple case of an electric field $$\vb{E}$$
 that is uniform in all of space.
-In the absence of a magnetic field $\vb{B} = 0$
+In the absence of a magnetic field $$\vb{B} = 0$$
 and any other forces,
 Newton's second law states:
 
@@ -38,16 +38,16 @@ $$\begin{aligned}
 \end{aligned}$$
 
 This is straightforward to integrate in time,
-for a given initial velocity vector $\vb{u}_0$:
+for a given initial velocity vector $$\vb{u}_0$$:
 
 $$\begin{aligned}
     \vb{u}(t)
     = \frac{q}{m} \vb{E} t + \vb{u}_0
 \end{aligned}$$
 
-And then the particle's position $\vb{x}(t)$
+And then the particle's position $$\vb{x}(t)$$
 is found be integrating once more,
-with $\vb{x}(0) = \vb{x}_0$:
+with $$\vb{x}(0) = \vb{x}_0$$:
 
 $$\begin{aligned}
     \boxed{
@@ -56,16 +56,16 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-In summary, unsurprisingly, a uniform electric field $\vb{E}$
-accelerates the particle with a constant force $\vb{F} = q \vb{E}$.
-Note that the direction depends on the sign of $q$.
+In summary, unsurprisingly, a uniform electric field $$\vb{E}$$
+accelerates the particle with a constant force $$\vb{F} = q \vb{E}$$.
+Note that the direction depends on the sign of $$q$$.
 
 
 ## Uniform magnetic field
 
 Consider the simple case of a uniform magnetic field
-$\vb{B} = (0, 0, B)$ in the $z$-direction,
-without an electric field $\vb{E} = 0$.
+$$\vb{B} = (0, 0, B)$$ in the $$z$$-direction,
+without an electric field $$\vb{E} = 0$$.
 If there are no other forces,
 Newton's second law states:
 
@@ -76,7 +76,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Evaluating the cross product yields
-three coupled equations for the components of $\vb{u}$:
+three coupled equations for the components of $$\vb{u}$$:
 
 $$\begin{aligned}
     \dv{u_x}{t}
@@ -89,15 +89,15 @@ $$\begin{aligned}
     = 0
 \end{aligned}$$
 
-Differentiating the first equation with respect to $t$,
-and substituting $\idv{u_y}{t}$ from the second,
+Differentiating the first equation with respect to $$t$$,
+and substituting $$\idv{u_y}{t}$$ from the second,
 we arrive at the following harmonic oscillator:
 
 $$\begin{aligned}
     \dvn{2}{u_x}{t} = - \omega_c^2 u_x
 \end{aligned}$$
 
-Where we have defined the **cyclotron frequency** $\omega_c$ as follows,
+Where we have defined the **cyclotron frequency** $$\omega_c$$ as follows,
 which may be negative:
 
 $$\begin{aligned}
@@ -108,15 +108,15 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Suppose we choose our initial conditions so that
-the solution for $u_x(t)$ is given by:
+the solution for $$u_x(t)$$ is given by:
 
 $$\begin{aligned}
     u_x(t)
     = u_\perp \cos(\omega_c t)
 \end{aligned}$$
 
-Where $u_\perp \equiv \sqrt{u_x^2 + u_y^2}$ is the constant total transverse velocity.
-Then $u_y(t)$ is found to be:
+Where $$u_\perp \equiv \sqrt{u_x^2 + u_y^2}$$ is the constant total transverse velocity.
+Then $$u_y(t)$$ is found to be:
 
 $$\begin{aligned}
     u_y(t)
@@ -126,10 +126,10 @@ $$\begin{aligned}
 \end{aligned}$$
 
 This means that the particle moves in a circle,
-in a direction determined by the sign of $\omega_c$.
+in a direction determined by the sign of $$\omega_c$$.
 
 Integrating the velocity yields the position,
-where we refer to the integration constants $x_{gc}$ and $y_{gc}$
+where we refer to the integration constants $$x_{gc}$$ and $$y_{gc}$$
 as the **guiding center**, around which the particle orbits or **gyrates**:
 
 $$\begin{aligned}
@@ -140,7 +140,7 @@ $$\begin{aligned}
     = \frac{u_\perp}{\omega_c} \cos(\omega_c t) + y_{gc}
 \end{aligned}$$
 
-The radius of this orbit is known as the **Larmor radius** or **gyroradius** $r_L$, given by:
+The radius of this orbit is known as the **Larmor radius** or **gyroradius** $$r_L$$, given by:
 
 $$\begin{aligned}
     \boxed{
@@ -151,7 +151,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Finally, it is easy to integrate the equation
-for the $z$-axis velocity $u_z$, which is conserved:
+for the $$z$$-axis velocity $$u_z$$, which is conserved:
 
 $$\begin{aligned}
     z(t)
@@ -159,9 +159,9 @@ $$\begin{aligned}
     = u_z t + z_0
 \end{aligned}$$
 
-In conclusion, the particle's motion parallel to $\vb{B}$
+In conclusion, the particle's motion parallel to $$\vb{B}$$
 is not affected by the magnetic field,
-while its motion perpendicular to $\vb{B}$
+while its motion perpendicular to $$\vb{B}$$
 is circular around an imaginary guiding center.
 The end result is that particles follow a helical path
 when moving through a uniform magnetic field:
@@ -177,9 +177,9 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Where $\vb{x}_{gc}(t) \equiv (x_{gc}, y_{gc}, z_{gc})$
+Where $$\vb{x}_{gc}(t) \equiv (x_{gc}, y_{gc}, z_{gc})$$
 is the position of the guiding center.
-For a detailed look at how $\vb{B}$ and $\vb{E}$
+For a detailed look at how $$\vb{B}$$ and $$\vb{E}$$
 can affect the guiding center's motion,
 see [guiding center theory](/know/concept/guiding-center-theory/).