1 files changed, 20 insertions, 20 deletions

diff --git a/source/know/concept/lawson-criterion/index.md b/source/know/concept/lawson-criterion/index.md
index ff9594b..f2f2fe0 100644
--- a/source/know/concept/lawson-criterion/index.md
+++ b/source/know/concept/lawson-criterion/index.md
@@ -13,9 +13,9 @@ the **Lawson criterion** must be met,
 from which some required properties
 of the plasma and the reactor chamber can be deduced.
 
-Suppose that a reactor generates a given power $P_\mathrm{fus}$ by nuclear fusion,
-but that it leaks energy at a rate $P_\mathrm{loss}$ in an unusable way.
-If an auxiliary input power $P_\mathrm{aux}$ sustains the fusion reaction,
+Suppose that a reactor generates a given power $$P_\mathrm{fus}$$ by nuclear fusion,
+but that it leaks energy at a rate $$P_\mathrm{loss}$$ in an unusable way.
+If an auxiliary input power $$P_\mathrm{aux}$$ sustains the fusion reaction,
 then the following inequality must be satisfied
 in order to have harvestable energy:
 
@@ -24,8 +24,8 @@ $$\begin{aligned}
     \le P_\mathrm{fus} + P_\mathrm{aux}
 \end{aligned}$$
 
-We can rewrite $P_\mathrm{aux}$ using the definition
-of the **energy gain factor** $Q$,
+We can rewrite $$P_\mathrm{aux}$$ using the definition
+of the **energy gain factor** $$Q$$,
 which is the ratio of the output and input powers of the fusion reaction:
 
 $$\begin{aligned}
@@ -45,12 +45,12 @@ $$\begin{aligned}
     = P_\mathrm{fus} \Big( \frac{Q + 1}{Q} \Big)
 \end{aligned}$$
 
-We assume that the plasma has equal species densities $n_i = n_e$,
-so its total density $n = 2 n_i$.
-Then $P_\mathrm{fus}$ is as follows,
-where $f_{ii}$ is the frequency
+We assume that the plasma has equal species densities $$n_i = n_e$$,
+so its total density $$n = 2 n_i$$.
+Then $$P_\mathrm{fus}$$ is as follows,
+where $$f_{ii}$$ is the frequency
 with which a given ion collides with other ions,
-and $E_\mathrm{fus}$ is the energy released by a single fusion reaction:
+and $$E_\mathrm{fus}$$ is the energy released by a single fusion reaction:
 
 $$\begin{aligned}
     P_\mathrm{fus}
@@ -59,11 +59,11 @@ $$\begin{aligned}
     = \frac{n^2}{4} \Expval{\sigma v} E_\mathrm{fus}
 \end{aligned}$$
 
-Where $\Expval{\sigma v}$ is the mean product
-of the velocity $v$ and the collision cross-section $\sigma$.
+Where $$\Expval{\sigma v}$$ is the mean product
+of the velocity $$v$$ and the collision cross-section $$\sigma$$.
 
-Furthermore, assuming that both species have the same temperature $T_i = T_e = T$,
-the total energy density $W$ of the plasma is given by:
+Furthermore, assuming that both species have the same temperature $$T_i = T_e = T$$,
+the total energy density $$W$$ of the plasma is given by:
 
 $$\begin{aligned}
     W
@@ -71,8 +71,8 @@ $$\begin{aligned}
     = 3 k_B T n
 \end{aligned}$$
 
-Where $k_B$ is Boltzmann's constant.
-From this, we can define the **confinement time** $\tau_E$
+Where $$k_B$$ is Boltzmann's constant.
+From this, we can define the **confinement time** $$\tau_E$$
 as the characteristic lifetime of energy in the reactor, before leakage.
 Therefore:
 
@@ -84,7 +84,7 @@ $$\begin{aligned}
     = \frac{3 n k_B T}{\tau_E}
 \end{aligned}$$
 
-Inserting these new expressions for $P_\mathrm{fus}$ and $P_\mathrm{loss}$
+Inserting these new expressions for $$P_\mathrm{fus}$$ and $$P_\mathrm{loss}$$
 into the inequality, we arrive at:
 
 $$\begin{aligned}
@@ -101,8 +101,8 @@ $$\begin{aligned}
 \end{aligned}$$
 
 However, it turns out that the highest fusion power density
-is reached when $T$ is at the minimum of $T^2 / \Expval{\sigma v}$.
-Therefore, we multiply by $T$ to get the Lawson triple product:
+is reached when $$T$$ is at the minimum of $$T^2 / \Expval{\sigma v}$$.
+Therefore, we multiply by $$T$$ to get the Lawson triple product:
 
 $$\begin{aligned}
     \boxed{
@@ -112,7 +112,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 For some reason,
-it is often assumed that the fusion is infinitely profitable $Q \to \infty$,
+it is often assumed that the fusion is infinitely profitable $$Q \to \infty$$,
 in which case the criterion reduces to:
 
 $$\begin{aligned}