More improvements to knowledge base

6 files changed, 107 insertions, 50 deletions

diff --git a/source/know/concept/step-index-fiber/index.md b/source/know/concept/step-index-fiber/index.md
index c0c95d1..210a339 100644
--- a/source/know/concept/step-index-fiber/index.md
+++ b/source/know/concept/step-index-fiber/index.md
@@ -14,17 +14,20 @@ the transverse profile $$F(x,y)$$ of the [electric field](/know/concept/electric
 can be shown to obey the *Helmholtz equation* in 2D:
 
 $$\begin{aligned}
-    \nabla_{\!\perp}^2 F + (n^2 k^2 - \beta^2) F = 0
+    \nabla_{\!\perp}^2 F + (n^2 k^2 - \beta^2) F
+    = 0
 \end{aligned}$$
 
 With $$n$$ being the position-dependent refractive index,
-$$k$$ the vacuum wavenumber $$\omega / c$$,
-and $$\beta$$ the mode's propagation constant, to be determined later.
+$$k = \omega / c$$ the vacuum wavenumber,
+and $$\beta$$ the mode's propagation constant (i.e. wavenumber),
+to be found later.
 In [polar coordinates](/know/concept/cylindrical-polar-coordinates/)
-$$(r,\phi)$$ this equation can be rewritten as follows:
+$$(r,\phi)$$ this can be rewritten as follows:
 
 $$\begin{aligned}
-    \pdvn{2}{F}{r} + \frac{1}{r} \pdv{F}{r} + \frac{1}{r^2} \pdvn{2}{F}{\phi} + \mu F = 0
+    \pdvn{2}{F}{r} + \frac{1}{r} \pdv{F}{r} + \frac{1}{r^2} \pdvn{2}{F}{\phi} + \mu F
+    = 0
 \end{aligned}$$
 
 Where we have defined $$\mu \equiv n^2 k^2 \!-\! \beta^2$$ for brevity.
@@ -37,14 +40,16 @@ such that $$F(r,\phi) = R(r) \, \Phi(\phi)$$.
 Inserting this ansatz:
 
 $$\begin{aligned}
-    R'' \Phi + \frac{1}{r} R' \Phi + \frac{1}{r^2} R \Phi'' + \mu R \Phi = 0
+    R'' \Phi + \frac{1}{r} R' \Phi + \frac{1}{r^2} R \Phi'' + \mu R \Phi
+    = 0
 \end{aligned}$$
 
 We rearrange this such that each side only depends on one variable,
 by dividing by $$R\Phi$$ (ignoring the fact that it may be zero),
 and multiplying by $$r^2$$.
 Since this equation should hold for *all* values of $$r$$ and $$\phi$$,
-this means that both sides must equal a constant $$\ell^2$$:
+this means that both sides must equal a constant,
+which we call $$\ell^2$$:
 
 $$\begin{aligned}
     r^2 \frac{R''}{R} + r \frac{R'}{R} + \mu r^2
@@ -57,11 +62,13 @@ and the well-known *Bessel equation* for $$R$$:
 
 $$\begin{aligned}
     \boxed{
-        \Phi'' + \ell^2 \Phi = 0
+        \Phi'' + \ell^2 \Phi
+        = 0
     }
     \qquad \qquad
     \boxed{
-        r^2 R'' + r R' + (\mu r^2 \!-\! \ell^2) R = 0
+        r^2 R'' + r R' + (\mu r^2 \!-\! \ell^2) R
+        = 0
     }
 \end{aligned}$$
 
@@ -70,9 +77,11 @@ simplest equation. Since the angle $$\phi$$ is limited to $$[0,2\pi]$$,
 $$\Phi$$ must be $$2 \pi$$-periodic, so:
 
 $$\begin{aligned}
-    \Phi(0) = \Phi(2\pi)
+    \Phi(0)
+    = \Phi(2\pi)
     \qquad \qquad
-    \Phi'(0) = \Phi'(2\pi)
+    \Phi'(0)
+    = \Phi'(2\pi)
 \end{aligned}$$
 
 The above equation for $$\Phi$$ with these periodic boundary conditions
@@ -90,13 +99,17 @@ and check if we can then arrive at a non-trivial $$\Phi$$ for each case.
     but the challenge is to satisfy the boundary conditions:
 
     $$\begin{alignedat}{3}
-        \Phi(0) &= \Phi(2 \pi)
+        \Phi(0)
+        &= \Phi(2 \pi)
         \:\quad &&\implies \quad\:\:
-        0 &&= A \sinh(2 \pi \ell) + B \big( \cosh(2 \pi \ell) - 1 \big)
+        0
+        &&= A \sinh(2 \pi \ell) + B \big( \cosh(2 \pi \ell) - 1 \big)
         \\
-        \Phi'(0) &= \Phi'(2 \pi)
+        \Phi'(0)
+        &= \Phi'(2 \pi)
         \: \quad &&\implies \quad \:\:
-        0 &&= A \ell \big( \cosh(2 \pi \ell) - 1 \big) + B \ell \sinh(2 \pi \ell)
+        0
+        &&= A \ell \big( \cosh(2 \pi \ell) - 1 \big) + B \ell \sinh(2 \pi \ell)
     \end{alignedat}$$
 
     This only has non-trivial solutions
@@ -121,13 +134,17 @@ and check if we can then arrive at a non-trivial $$\Phi$$ for each case.
     Putting this in the boundary conditions:
 
     $$\begin{alignedat}{3}
-        \Phi(0) &= \Phi(2 \pi)
+        \Phi(0)
+        &= \Phi(2 \pi)
         \qquad &&\implies \qquad
-        A &&= 0
+        A
+        &&= 0
         \\
-        \Phi'(0) &= \Phi'(2 \pi)
+        \Phi'(0)
+        &= \Phi'(2 \pi)
         \qquad &&\implies \qquad
-        B &&= B
+        B
+        &&= B
     \end{alignedat}$$
 
     $$B$$ can be nonzero, so this a valid solution.
@@ -137,13 +154,17 @@ and check if we can then arrive at a non-trivial $$\Phi$$ for each case.
     $$\Phi(\phi) = A \sin(\phi \ell) + B \cos(\phi \ell)$$, therefore:
 
     $$\begin{alignedat}{3}
-        \Phi(0) &= \Phi(2 \pi)
+        \Phi(0)
+        &= \Phi(2 \pi)
         \quad &&\implies \quad
-        0 &&= A \sin(2 \pi \ell) + B \big(\cos(2\pi \ell) - 1\big)
+        0
+        &&= A \sin(2 \pi \ell) + B \big(\cos(2\pi \ell) - 1\big)
         \\
-        \Phi'(0) &= \Phi'(2 \pi)
+        \Phi'(0)
+        &= \Phi'(2 \pi)
         \quad &&\implies \quad
-        0 &&= A \big(\cos(2 \pi \ell) - 1\big) - B \sin(2 \pi \ell)
+        0
+        &&= A \big(\cos(2 \pi \ell) - 1\big) - B \sin(2 \pi \ell)
     \end{alignedat}$$
 
     This system only has nontrivial solutions
@@ -163,13 +184,17 @@ and check if we can then arrive at a non-trivial $$\Phi$$ for each case.
     We revisit the boundary conditions and indeed see:
 
     $$\begin{alignedat}{3}
-        0 &= A \sin(2 \pi \ell) + B \big(\cos(2 \pi \ell) - 1\big)
+        0
+        &= A \sin(2 \pi \ell) + B \big(\cos(2 \pi \ell) - 1\big)
         \qquad &&\implies \qquad
-        0 &&= 0
+        0
+        &&= 0
         \\
-        0 &= A \big(\cos(2 \pi \ell) - 1\big) - B \sin(2 \pi \ell)
+        0
+        &= A \big(\cos(2 \pi \ell) - 1\big) - B \sin(2 \pi \ell)
         \qquad &&\implies \qquad
-        0 &&= 0
+        0
+        &&= 0
     \end{alignedat}$$
 
     So $$A$$ and $$B$$ are *both* unconstrained,
@@ -185,7 +210,8 @@ we get the following for $$\ell = 0, 1, 2, ...$$:
 
 $$\begin{aligned}
     \boxed{
-        \Phi_\ell(\phi) = A \cos(\phi \ell)
+        \Phi_\ell(\phi)
+        = A \cos(\phi \ell)
     }
 \end{aligned}$$
 
@@ -198,7 +224,8 @@ Let us now revisit the Bessel equation for the radial function $$R(r)$$,
 which should be continuous and differentiable throughout the fiber:
 
 $$\begin{aligned}
-    r^2 R'' + r R' + \mu r^2 R - \ell^2 R = 0
+    r^2 R'' + r R' + \mu r^2 R - \ell^2 R
+    = 0
 \end{aligned}$$
 
 To continue, we need to specify the refractive index $$n(r)$$, contained in $$\mu(r)$$.
@@ -231,11 +258,13 @@ Let $$\pm$$ be the sign of $$\mu$$:
 $$\begin{aligned}
     \begin{cases}
         \displaystyle
-        0 = \rho^2 \pdvn{2}{R}{\rho} + \rho \pdv{R}{\rho} \pm \rho^2 R - \ell^2 R
+        0
+        = \rho^2 \pdvn{2}{R}{\rho} + \rho \pdv{R}{\rho} \pm \rho^2 R - \ell^2 R
         & \mathrm{for}\; \mu \neq 0
         \\
         \displaystyle
-        0 = r^2 \pdvn{2}{R}{r} + r \pdv{R}{r} - \ell^2 R
+        0
+        = r^2 \pdvn{2}{R}{r} + r \pdv{R}{r} - \ell^2 R
         & \mathrm{for}\; \mu = 0
     \end{cases}
 \end{aligned}$$
@@ -249,7 +278,8 @@ Of the remaining candidates, $$\ln(r)$$, $$r^\ell$$ and $$I_\ell(\rho)$$ do not
 leading to the following $$R_o$$:
 
 $$\begin{aligned}
-    R_{o,\ell}(r) =
+    R_{o,\ell}(r)
+    =
     \begin{cases}
         r^{-\ell}
         & \mathrm{for}\; \mu = 0 \;\mathrm{and}\; \ell = 1,2,3,...
@@ -267,8 +297,9 @@ at the boundary $$r = a$$, so they can never be continuous with $$R_o'$$.
 This leaves $$J_\ell(\rho)$$ for $$\mu > 0$$:
 
 $$\begin{aligned}
-    R_{i,\ell}(r) =
-    J_\ell(\rho) = J_\ell(r \sqrt{\mu})
+    R_{i,\ell}(r)
+    = J_\ell(\rho)
+    = J_\ell(r \sqrt{\mu})
     \qquad \mathrm{for}\; \mu > 0 \;\mathrm{and}\; \ell = 0,1,2,...
 \end{aligned}$$
 
@@ -295,7 +326,8 @@ Since $$\mu \equiv n^2 k^2 \!-\! \beta^2$$ by definition,
 this discovery places a constraint on the propagation constant $$\beta$$:
 
 $$\begin{aligned}
-    n_i^2 k^2 > \beta^2 \ge n_o^2 k^2
+    n_i^2 k^2 > \beta^2
+    \ge n_o^2 k^2
 \end{aligned}$$
 
 Therefore, $$n_i > n_o$$ in a step-index fiber,
@@ -304,7 +336,7 @@ the fiber is not able to guide the light outside this range.
 
 However, not all $$\beta$$ in this range are created equal for all $$k$$.
 To investigate further, let us define the quantities
-$$\xi_\mathrm{core}$$ and $$\xi_\mathrm{clad}$$ like so,
+$$\xi_i$$ and $$\xi_o$$ like so,
 assuming $$n_i$$ and $$n_o$$ do not depend on $$k$$:
 
 $$\begin{aligned}
@@ -318,7 +350,8 @@ $$\begin{aligned}
 It is important to note that the sum of their squares is constant with respect to $$\beta$$:
 
 $$\begin{aligned}
-    \xi_i^2 + \xi_o^2 = (\mathrm{NA})^2 k^2
+    \xi_i^2 + \xi_o^2
+    = (\mathrm{NA})^2 k^2
 \end{aligned}$$
 
 Where $$\mathrm{NA}$$ is the so-called **numerical aperture**,
@@ -333,7 +366,7 @@ $$\begin{aligned}
 \end{aligned}$$
 
 From this, we define a new fiber parameter: the $$V$$-**number**,
-which is extremely useful:
+which is very useful:
 
 $$\begin{aligned}
     \boxed{
@@ -360,16 +393,19 @@ meanwhile defining $$X \equiv a \xi_i$$ and $$Y \equiv a \xi_o$$
 for convenience, such that $$X^2 + Y^2 = V^2$$:
 
 $$\begin{aligned}
-    X \frac{J_\ell'(X)}{J_\ell(X)} = Y \frac{K_\ell'(Y)}{K_\ell(Y)}
+    X \frac{J_\ell'(X)}{J_\ell(X)}
+    = Y \frac{K_\ell'(Y)}{K_\ell(Y)}
 \end{aligned}$$
 
 We can turn this result into something a bit nicer
 by using the following identities:
 
 $$\begin{aligned}
-    J_\ell'(x) = -J_{\ell+1}(x) + \ell \frac{J_\ell(x)}{x}
-    \qquad \quad
-    K_\ell'(x) = -K_{\ell+1}(x) + \ell \frac{K_\ell(x)}{x}
+    J_\ell'(x)
+    = - \ell \frac{J_\ell(x)}{x} + J_{\ell-1}(x)
+    \qquad \qquad
+    K_\ell'(x)
+    = - \ell \frac{K_\ell(x)}{x} - K_{\ell-1}(x)
 \end{aligned}$$
 
 With this, the transcendental equation for $$\beta$$
@@ -377,13 +413,14 @@ takes this convenient form:
 
 $$\begin{aligned}
     \boxed{
-        X \frac{J_{\ell+1}(X)}{J_\ell(X)} = Y \frac{K_{\ell+1}(Y)}{K_\ell(Y)}
+        X \frac{J_{\ell-1}(X)}{J_\ell(X)}
+        = - Y \frac{K_{\ell-1}(Y)}{K_\ell(Y)}
     }
 \end{aligned}$$
 
 All $$\beta$$ that satisfy this indicate the existence
-of a **linearly polarized** mode.
-These modes are called $$\mathrm{LP}_{\ell m}$$,
+of a **linearly polarized mode**,
+each labelled $$\mathrm{LP}_{\ell m}$$,
 where $$\ell$$ is the primary (azimuthal) mode index,
 and $$m$$ the secondary (radial) mode index,
 which is needed because multiple $$\beta$$ may exist for a single $$\ell$$.
@@ -394,12 +431,32 @@ where red and blue respectively denote the left and right-hand side:
 
 {% include image.html file="transcendental-full.png" width="100%" alt="Graphical solution of transcendental equation" %}
 
-This shows that each $$\mathrm{LP}_{\ell m}$$ has an associated cut-off $$V_{\ell m}$$,
-so that if $$V > V_{\ell m}$$ then $$\mathrm{LP}_{lm}$$ exists,
+For the ground state the light is well-confined in the core,
+but for higher modes it increasingly leaks into the cladding,
+thereby reducing the wavenumber $$\beta$$ (because $$n_i > n_o$$)
+until the fiber can no longer guide the light $$\beta < n_o k$$,
+and the modes thus stop existing.
+Therefore, there is a mode cutoff when $$\beta = n_o k$$
+or equivalently when $$\xi_o = 0$$.
+With this is mind, consider a slightly rearranged version
+of the above transcendental equation:
+
+$$\begin{aligned}
+    X J_{\ell-1}(X) K_\ell(Y)
+    = - Y K_{\ell-1}(Y) J_\ell(X)
+\end{aligned}$$
+
+Because $$Y \equiv a \xi_o$$ and $$X^2 = V^2 - Y^2$$,
+if $$\xi_o = 0$$ then $$Y = 0$$ and $$X = V$$,
+and the right-hand side is zero;
+for it to be satisfiable (i.e. for the mode to exist),
+the other side should also vanish.
+Therefore, all $$\mathrm{LP}_{\ell m}$$ have cutoffs $$V_{\ell m}$$
+equal to the $$m$$th roots of $$J_{\ell-1}(V_{\ell m}) = 0$$,
+so if $$V > V_{\ell m}$$ then $$\mathrm{LP}_{\ell m}$$ exists,
 as long as $$\beta$$ stays in the allowed range.
-The cut-offs of the secondary modes for a given $$\ell$$
-are found as the $$m$$th roots of $$J_{\ell-1}(V_{\ell m}) = 0$$.
-In the above figure, they are $$V_{01} = 0$$, $$V_{11} = 2.405$$, and $$V_{02} = V_{21} = 3.832$$.
+In the above figure, they are $$V_{01} = 0$$,
+$$V_{11} = 2.405$$, and $$V_{02} = V_{21} = 3.832$$.
 
 All differential equations have been linear,
 so a linear combination of these solutions is also valid.
diff --git a/source/know/concept/step-index-fiber/transcendental-full.png b/source/know/concept/step-index-fiber/transcendental-full.png
index 294d66a..0506cbf 100644
--- a/source/know/concept/step-index-fiber/transcendental-full.png
+++ b/source/know/concept/step-index-fiber/transcendental-full.png
diff --git a/source/know/concept/step-index-fiber/transcendental-half.avif b/source/know/concept/step-index-fiber/transcendental-half.avif
index fd69e2b..064b1b4 100644
--- a/source/know/concept/step-index-fiber/transcendental-half.avif
+++ b/source/know/concept/step-index-fiber/transcendental-half.avif
diff --git a/source/know/concept/step-index-fiber/transcendental-half.jpg b/source/know/concept/step-index-fiber/transcendental-half.jpg
index d3862e6..e08bba4 100644
--- a/source/know/concept/step-index-fiber/transcendental-half.jpg
+++ b/source/know/concept/step-index-fiber/transcendental-half.jpg
diff --git a/source/know/concept/step-index-fiber/transcendental-half.png b/source/know/concept/step-index-fiber/transcendental-half.png
index dda7620..fab9a27 100644
--- a/source/know/concept/step-index-fiber/transcendental-half.png
+++ b/source/know/concept/step-index-fiber/transcendental-half.png
diff --git a/source/know/concept/step-index-fiber/transcendental-half.webp b/source/know/concept/step-index-fiber/transcendental-half.webp
index b669cc5..ff1082c 100644
--- a/source/know/concept/step-index-fiber/transcendental-half.webp
+++ b/source/know/concept/step-index-fiber/transcendental-half.webp