More improvements to knowledge base

1 files changed, 20 insertions, 9 deletions

diff --git a/source/know/concept/self-steepening/index.md b/source/know/concept/self-steepening/index.md
index e06b0b5..fd48e0f 100644
--- a/source/know/concept/self-steepening/index.md
+++ b/source/know/concept/self-steepening/index.md
@@ -27,7 +27,8 @@ We will use the following ansatz,
 consisting of an arbitrary power profile $$P$$ with a phase $$\phi$$:
 
 $$\begin{aligned}
-    A(z,t) = \sqrt{P(z,t)} \, \exp\!\big(i \phi(z,t)\big)
+    A(z,t)
+    = \sqrt{P(z,t)} \, \exp\!\big(i \phi(z,t)\big)
 \end{aligned}$$
 
 For a long pulse travelling over a short distance, it is reasonable to
@@ -35,16 +36,19 @@ neglect dispersion ($$\beta_2 = 0$$).
 Inserting the ansatz then gives the following, where $$\varepsilon = \gamma / \omega_0$$:
 
 $$\begin{aligned}
-    0 &= i \frac{1}{2} \frac{P_z}{\sqrt{P}} - \sqrt{P} \phi_z + \gamma P \sqrt{P} + i \varepsilon \frac{3}{2} P_t \sqrt{P} - \varepsilon P \sqrt{P} \phi_t
+    0
+    &= i \frac{1}{2} \frac{P_z}{\sqrt{P}} - \sqrt{P} \phi_z + \gamma P \sqrt{P} + i \varepsilon \frac{3}{2} P_t \sqrt{P} - \varepsilon P \sqrt{P} \phi_t
 \end{aligned}$$
 
 This results in two equations, respectively corresponding to the real
 and imaginary parts:
 
 $$\begin{aligned}
-    0 &= - \phi_z - \varepsilon P \phi_t + \gamma P
+    0
+    &= - \phi_z - \varepsilon P \phi_t + \gamma P
     \\
-    0 &= P_z + \varepsilon 3 P_t P
+    0
+    &= P_z + \varepsilon 3 P_t P
 \end{aligned}$$
 
 The phase $$\phi$$ is not so interesting, so we focus on the latter equation for $$P$$.
@@ -53,7 +57,8 @@ which shows that more intense parts of the pulse
 will lag behind compared to the rest:
 
 $$\begin{aligned}
-    P(z,t) = f(t - 3 \varepsilon z P)
+    P(z,t)
+    = f(t - 3 \varepsilon z P)
 \end{aligned}$$
 
 Where $$f$$ is the initial power profile: $$f(t) = P(0,t)$$.
@@ -85,7 +90,8 @@ $$\begin{aligned}
     = 1 + 3 \varepsilon z f_\mathrm{min}'
     \qquad \implies \qquad
     \boxed{
-        L_\mathrm{shock} \equiv -\frac{1}{3 \varepsilon f_\mathrm{min}'}
+        L_\mathrm{shock}
+        \equiv -\frac{1}{3 \varepsilon f_\mathrm{min}'}
     }
 \end{aligned}$$
 
@@ -99,7 +105,8 @@ with $$T_0 = 25\:\mathrm{fs}$$, $$P_0 = 3\:\mathrm{kW}$$,
 $$\beta_2 = 0$$ and $$\gamma = 0.1/\mathrm{W}/\mathrm{m}$$:
 
 $$\begin{aligned}
-    f(t) = P(0,t) = P_0 \exp\!\Big(\! -\!\frac{t^2}{T_0^2} \Big)
+    f(t)
+    = P(0,t) = P_0 \exp\!\Big(\! -\!\frac{t^2}{T_0^2} \Big)
 \end{aligned}$$
 
 
@@ -107,9 +114,11 @@ Its steepest points are found to be at $$2 t^2 = T_0^2$$, so
 $$f_\mathrm{min}'$$ and $$L_\mathrm{shock}$$ are given by:
 
 $$\begin{aligned}
-    f_\mathrm{min}' = - \frac{\sqrt{2} P_0}{T_0} \exp\!\Big(\!-\!\frac{1}{2}\Big)
+    f_\mathrm{min}'
+    = - \frac{\sqrt{2} P_0}{T_0} \exp\!\Big(\!-\!\frac{1}{2}\Big)
     \quad \implies \quad
-    L_\mathrm{shock} = \frac{T_0}{3 \sqrt{2} \varepsilon P_0} \exp\!\Big(\frac{1}{2}\Big)
+    L_\mathrm{shock}
+    = \frac{T_0}{3 \sqrt{2} \varepsilon P_0} \exp\!\Big(\frac{1}{2}\Big)
 \end{aligned}$$
 
 This example Gaussian pulse therefore has a theoretical
@@ -127,6 +136,7 @@ Nevertheless, the general trends are nicely visible:
 the trailing slope becomes extremely steep, and the spectrum
 broadens so much that dispersion cannot be neglected anymore.
 
+{% comment %}
 When self-steepening is added to the nonlinear Schrödinger equation,
 it no longer conserves the total pulse energy $$\int |A|^2 \dd{t}$$.
 Fortunately, the photon number $$N_\mathrm{ph}$$ is still
@@ -137,6 +147,7 @@ $$\begin{aligned}
         N_\mathrm{ph}(z) = \int_0^\infty \frac{|\tilde{A}(z,\omega)|^2}{\omega} \dd{\omega}
     }
 \end{aligned}$$
+{% endcomment %}