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

1 files changed, 104 insertions, 104 deletions

diff --git a/source/know/concept/step-index-fiber/index.md b/source/know/concept/step-index-fiber/index.md
index e0b26af..dd83334 100644
--- a/source/know/concept/step-index-fiber/index.md
+++ b/source/know/concept/step-index-fiber/index.md
@@ -9,31 +9,31 @@ categories:
 layout: "concept"
 ---
 
-As light propagates in the $z$-direction through an optical fiber,
-the transverse profile $F(x,y)$ of the [electric field](/know/concept/electric-field/)
+As light propagates in the $$z$$-direction through an optical fiber,
+the transverse profile $$F(x,y)$$ of the [electric field](/know/concept/electric-field/)
 can be shown to obey the *Helmholtz equation* in 2D:
 
 $$\begin{aligned}
     \nabla_{\!\perp}^2 F + (n^2 k^2 - \beta^2) F = 0
 \end{aligned}$$
 
-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.
+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.
 In [polar coordinates](/know/concept/cylindrical-polar-coordinates/)
-$(r,\phi)$ this equation can be rewritten as follows:
+$$(r,\phi)$$ this equation can be rewritten as follows:
 
 $$\begin{aligned}
     \pdvn{2}{F}{r} + \frac{1}{r} \pdv{F}{r} + \frac{1}{r^2} \pdvn{2}{F}{\phi} + \mu F = 0
 \end{aligned}$$
 
-Where we have defined $\mu \equiv n^2 k^2 \!-\! \beta^2$ for brevity.
-From now on, we only consider choices of $\mu$ that do not depend on $\phi$ or $z$,
-but may vary with $r$.
+Where we have defined $$\mu \equiv n^2 k^2 \!-\! \beta^2$$ for brevity.
+From now on, we only consider choices of $$\mu$$ that do not depend on $$\phi$$ or $$z$$,
+but may vary with $$r$$.
 
 This Helmholtz equation can be solved by *separation of variables*:
-we assume that there exist two functions $R(r)$ and $\Phi(\phi)$
-such that $F(r,\phi) = R(r) \, \Phi(\phi)$.
+we assume that there exist two functions $$R(r)$$ and $$\Phi(\phi)$$
+such that $$F(r,\phi) = R(r) \, \Phi(\phi)$$.
 Inserting this ansatz:
 
 $$\begin{aligned}
@@ -41,10 +41,10 @@ $$\begin{aligned}
 \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 $\ell^2$:
+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 $$\ell^2$$:
 
 $$\begin{aligned}
     r^2 \frac{R''}{R} + r \frac{R'}{R} + \mu r^2
@@ -52,8 +52,8 @@ $$\begin{aligned}
     = \ell^2
 \end{aligned}$$
 
-This gives an eigenvalue problem for $\Phi$,
-and the well-known *Bessel equation* for $R$:
+This gives an eigenvalue problem for $$\Phi$$,
+and the well-known *Bessel equation* for $$R$$:
 
 $$\begin{aligned}
     \boxed{
@@ -65,9 +65,9 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-We will return to $R$ later; we start with $\Phi$, because it has the
-simplest equation. Since the angle $\phi$ is limited to $[0,2\pi]$,
-$\Phi$ must be $2 \pi$-periodic, so:
+We will return to $$R$$ later; we start with $$\Phi$$, because it has the
+simplest equation. Since the angle $$\phi$$ is limited to $$[0,2\pi]$$,
+$$\Phi$$ must be $$2 \pi$$-periodic, so:
 
 $$\begin{aligned}
     \Phi(0) = \Phi(2\pi)
@@ -75,18 +75,18 @@ $$\begin{aligned}
     \Phi'(0) = \Phi'(2\pi)
 \end{aligned}$$
 
-The above equation for $\Phi$ with these periodic boundary conditions
+The above equation for $$\Phi$$ with these periodic boundary conditions
 is a [Sturm-Liouville problem](/know/concept/sturm-liouville-theory/).
-Consequently, there are infinitely many allowed values of $\ell^2$,
+Consequently, there are infinitely many allowed values of $$\ell^2$$,
 all real, and one of them is lowest, known as the *ground state*.
 
-To find the eigenvalues $\ell^2$ and their corresponding $\Phi$,
-we in turn assume that $\ell^2 < 0$, $\ell^2 = 0$, or $\ell^2 > 0$,
-and check if we can then arrive at a non-trivial $\Phi$ for each case.
+To find the eigenvalues $$\ell^2$$ and their corresponding $$\Phi$$,
+we in turn assume that $$\ell^2 < 0$$, $$\ell^2 = 0$$, or $$\ell^2 > 0$$,
+and check if we can then arrive at a non-trivial $$\Phi$$ for each case.
 
-* For $\ell^2 < 0$, solutions have the form $\Phi(\phi) = A \sinh(\phi \ell) + B \cosh(\phi \ell)$,
-    where $A$ and $B$ are unknown linearity constants.
-    At least one of these constants must be nonzero for $\Phi$ to be non-trivial,
+* For $$\ell^2 < 0$$, solutions have the form $$\Phi(\phi) = A \sinh(\phi \ell) + B \cosh(\phi \ell)$$,
+    where $$A$$ and $$B$$ are unknown linearity constants.
+    At least one of these constants must be nonzero for $$\Phi$$ to be non-trivial,
     but the challenge is to satisfy the boundary conditions:
 
     $$\begin{alignedat}{3}
@@ -112,12 +112,12 @@ and check if we can then arrive at a non-trivial $\Phi$ for each case.
         = 2 \big( \cosh(2 \pi \ell) - 1 \big)
     \end{aligned}$$
 
-    This can only be zero if $\ell = 0$,
-    which contradicts the premise that $\ell^2 < 0$,
-    so we conclude that $\ell^2$ cannot be negative,
+    This can only be zero if $$\ell = 0$$,
+    which contradicts the premise that $$\ell^2 < 0$$,
+    so we conclude that $$\ell^2$$ cannot be negative,
     because no non-trivial solutions exist here.
 
-* For $\ell^2 = 0$, the solution is $\Phi(\phi) = A \phi + B$.
+* For $$\ell^2 = 0$$, the solution is $$\Phi(\phi) = A \phi + B$$.
     Putting this in the boundary conditions:
 
     $$\begin{alignedat}{3}
@@ -130,11 +130,11 @@ and check if we can then arrive at a non-trivial $\Phi$ for each case.
         B &&= B
     \end{alignedat}$$
 
-    $B$ can be nonzero, so this a valid solution.
-    We conclude that $\ell^2 = 0$ is the ground state.
+    $$B$$ can be nonzero, so this a valid solution.
+    We conclude that $$\ell^2 = 0$$ is the ground state.
 
-* For $\ell^2 > 0$, all solutions have the form
-    $\Phi(\phi) = A \sin(\phi \ell) + B \cos(\phi \ell)$, therefore:
+* For $$\ell^2 > 0$$, all solutions have the form
+    $$\Phi(\phi) = A \sin(\phi \ell) + B \cos(\phi \ell)$$, therefore:
 
     $$\begin{alignedat}{3}
         \Phi(0) &= \Phi(2 \pi)
@@ -159,7 +159,7 @@ and check if we can then arrive at a non-trivial $\Phi$ for each case.
         = 2 \big(\cos(2 \pi \ell) - 1\big)
     \end{aligned}$$
 
-    Meaning that $\ell$ must be an integer.
+    Meaning that $$\ell$$ must be an integer.
     We revisit the boundary conditions and indeed see:
 
     $$\begin{alignedat}{3}
@@ -172,16 +172,16 @@ and check if we can then arrive at a non-trivial $\Phi$ for each case.
         0 &&= 0
     \end{alignedat}$$
 
-    So $A$ and $B$ are *both* unconstrained,
-    and each integer $\ell$ is a doubly-degenerate eigenvalue.
+    So $$A$$ and $$B$$ are *both* unconstrained,
+    and each integer $$\ell$$ is a doubly-degenerate eigenvalue.
     The two linearly independent solutions,
-    $\sin(\phi \ell)$ and $\cos(\phi \ell)$,
+    $$\sin(\phi \ell)$$ and $$\cos(\phi \ell)$$,
     represent the polarization of light in the mode.
     For simplicity, we assume that all light is in a single polarization,
-    so only $\cos(\phi \ell)$ will be considered from now on.
+    so only $$\cos(\phi \ell)$$ will be considered from now on.
 
-By combining our result for $\ell^2 = 0$ and $\ell^2 > 0$,
-we get the following for $\ell = 0, 1, 2, ...$:
+By combining our result for $$\ell^2 = 0$$ and $$\ell^2 > 0$$,
+we get the following for $$\ell = 0, 1, 2, ...$$:
 
 $$\begin{aligned}
     \boxed{
@@ -189,28 +189,28 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Here, $\ell$ is called the **primary mode index**.
-We exclude $\ell < 0$ because $\cos(x) \propto \cos(-x)$
-and $\sin(x) \propto \sin(-x)$,
-and because $A$ is free to choose thanks to linearity.
+Here, $$\ell$$ is called the **primary mode index**.
+We exclude $$\ell < 0$$ because $$\cos(x) \propto \cos(-x)$$
+and $$\sin(x) \propto \sin(-x)$$,
+and because $$A$$ is free to choose thanks to linearity.
 
-Let us now revisit the Bessel equation for the radial function $R(r)$,
+Let us now revisit the Bessel equation for the radial function $$R(r)$$,
 which should be continuous and differentiable throughout the fiber:
 
 $$\begin{aligned}
     r^2 R'' + r R' + \mu r^2 R - \ell^2 R = 0
 \end{aligned}$$
 
-To continue, we need to specify the refractive index $n(r)$, contained in $\mu(r)$.
+To continue, we need to specify the refractive index $$n(r)$$, contained in $$\mu(r)$$.
 We choose a **step-index fiber**,
-whose cross-section consists of a **core** with radius $a$,
-surrounded by a **cladding** that extends to infinity $r \to \infty$.
-In the core $r < a$, the index $n$ is a constant $n_i$,
-while in the cladding $r > a$ it is another constant $n_o$.
+whose cross-section consists of a **core** with radius $$a$$,
+surrounded by a **cladding** that extends to infinity $$r \to \infty$$.
+In the core $$r < a$$, the index $$n$$ is a constant $$n_i$$,
+while in the cladding $$r > a$$ it is another constant $$n_o$$.
 
-Since $\mu$ is different in the core and cladding,
-we will get different solutions $R_i$ and $R_o$ there,
-so we must demand that the field is continuous at the boundary $r = a$:
+Since $$\mu$$ is different in the core and cladding,
+we will get different solutions $$R_i$$ and $$R_o$$ there,
+so we must demand that the field is continuous at the boundary $$r = a$$:
 
 $$\begin{aligned}
     R_i(a) = R_o(a)
@@ -219,14 +219,14 @@ $$\begin{aligned}
 \end{aligned}$$
 
 Furthermore, for a physically plausible solution,
-we require that $R_i$ is finite
-and that $R_o$ decays monotonically to zero when $r \to \infty$.
-These constraints will turn out to restrict $\mu$.
+we require that $$R_i$$ is finite
+and that $$R_o$$ decays monotonically to zero when $$r \to \infty$$.
+These constraints will turn out to restrict $$\mu$$.
 
-Introducing a new coordinate $\rho \equiv r \sqrt{|\mu|}$
+Introducing a new coordinate $$\rho \equiv r \sqrt{|\mu|}$$
 gives the Bessel equation's standard form,
 which has well-known solutions called *Bessel functions*, shown below.
-Let $\pm$ be the sign of $\mu$:
+Let $$\pm$$ be the sign of $$\mu$$:
 
 $$\begin{aligned}
     \begin{cases}
@@ -244,11 +244,11 @@ $$\begin{aligned}
 <img src="bessel-small.jpg" style="width:100%">
 </a>
 
-Looking at these solutions with our constraints for $R_o$ in mind,
-we see that for $\mu > 0$ none of the solutions decay
-*monotonically* to zero, so we must have $\mu \le 0$ in the cladding.
-Of the remaining candidates, $\ln\!(r)$, $r^\ell$ and $I_\ell(\rho)$ do not decay at all,
-leading to the following $R_o$:
+Looking at these solutions with our constraints for $$R_o$$ in mind,
+we see that for $$\mu > 0$$ none of the solutions decay
+*monotonically* to zero, so we must have $$\mu \le 0$$ in the cladding.
+Of the remaining candidates, $$\ln\!(r)$$, $$r^\ell$$ and $$I_\ell(\rho)$$ do not decay at all,
+leading to the following $$R_o$$:
 
 $$\begin{aligned}
     R_{o,\ell}(r) =
@@ -261,12 +261,12 @@ $$\begin{aligned}
     \end{cases}
 \end{aligned}$$
 
-Next, for $R_i$, we see that when $\mu < 0$ all solutions are invalid
-since they diverge at $r = 0$,
-and so do $\ln\!(r)$, $r^{-\ell}$ and $Y_\ell(\rho)$.
-Of the remaining candidates, $r^0$ and $r^\ell$ have a non-negative slope
-at the boundary $r = a$, so they can never be continuous with $R_o'$.
-This leaves $J_\ell(\rho)$ for $\mu > 0$:
+Next, for $$R_i$$, we see that when $$\mu < 0$$ all solutions are invalid
+since they diverge at $$r = 0$$,
+and so do $$\ln\!(r)$$, $$r^{-\ell}$$ and $$Y_\ell(\rho)$$.
+Of the remaining candidates, $$r^0$$ and $$r^\ell$$ have a non-negative slope
+at the boundary $$r = a$$, so they can never be continuous with $$R_o'$$.
+This leaves $$J_\ell(\rho)$$ for $$\mu > 0$$:
 
 $$\begin{aligned}
     R_{i,\ell}(r) =
@@ -274,7 +274,7 @@ $$\begin{aligned}
     \qquad \mathrm{for}\; \mu > 0 \;\mathrm{and}\; \ell = 0,1,2,...
 \end{aligned}$$
 
-Putting this all together, we now know what the full solution for $F$ should look like:
+Putting this all together, we now know what the full solution for $$F$$ should look like:
 
 $$\begin{aligned}
     F_\ell(r, \phi)
@@ -289,25 +289,25 @@ $$\begin{aligned}
     \end{cases}
 \end{aligned}$$
 
-Where $A_\ell$ and $B_\ell$ are constants to be chosen
-based on the light's intensity, and to satisfy the continuity condition at $r = a$.
+Where $$A_\ell$$ and $$B_\ell$$ are constants to be chosen
+based on the light's intensity, and to satisfy the continuity condition at $$r = a$$.
 
-We found that $\mu \le 0$ in the cladding and $\mu > 0$ in the core.
-Since $\mu \equiv n^2 k^2 \!-\! \beta^2$ by definition,
-this discovery places a constraint on the propagation constant $\beta$:
+We found that $$\mu \le 0$$ in the cladding and $$\mu > 0$$ in the core.
+Since $$\mu \equiv n^2 k^2 \!-\! \beta^2$$ by definition,
+this discovery places a constraint on the propagation constant $$\beta$$:
 
 $$\begin{aligned}
     n_i^2 k^2 > \beta^2 \ge n_o^2 k^2
 \end{aligned}$$
 
-Therefore, $n_i > n_o$ in a step-index fiber,
-and there is only a limited range of allowed $\beta$-values;
+Therefore, $$n_i > n_o$$ in a step-index fiber,
+and there is only a limited range of allowed $$\beta$$-values;
 the fiber is not able to guide the light outside this range.
 
-However, not all $\beta$ in this range are created equal for all $k$.
+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,
-assuming $n_i$ and $n_o$ do not depend on $k$:
+$$\xi_\mathrm{core}$$ and $$\xi_\mathrm{clad}$$ like so,
+assuming $$n_i$$ and $$n_o$$ do not depend on $$k$$:
 
 $$\begin{aligned}
     \xi_i(k)
@@ -317,13 +317,13 @@ $$\begin{aligned}
     \equiv \sqrt{ \beta^2(k) - n_o^2 k^2 }
 \end{aligned}$$
 
-It is important to note that the sum of their squares is constant with respect to $\beta$:
+It is important to note that the sum of their squares is constant with respect to $$\beta$$:
 
 $$\begin{aligned}
     \xi_i^2 + \xi_o^2 = (\mathrm{NA})^2 k^2
 \end{aligned}$$
 
-Where $\mathrm{NA}$ is the so-called **numerical aperture**,
+Where $$\mathrm{NA}$$ is the so-called **numerical aperture**,
 often mentioned in papers and datasheets as one of a fiber's key parameters.
 It is defined as:
 
@@ -334,7 +334,7 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-From this, we define a new fiber parameter: the $V$-**number**,
+From this, we define a new fiber parameter: the $$V$$-**number**,
 which is extremely useful:
 
 $$\begin{aligned}
@@ -345,8 +345,8 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-Now, the allowed values of $\beta$ are found
-by fulfilling the boundary conditions (for $\mu \neq 0$):
+Now, the allowed values of $$\beta$$ are found
+by fulfilling the boundary conditions (for $$\mu \neq 0$$):
 
 $$\begin{aligned}
     A_\ell J_\ell(a \xi_i)
@@ -356,10 +356,10 @@ $$\begin{aligned}
     &= B_\ell \xi_o K_\ell'(a \xi_o)
 \end{aligned}$$
 
-To remove $A_\ell$ and $B_\ell$,
+To remove $$A_\ell$$ and $$B_\ell$$,
 we divide the latter equation by the former,
-meanwhile defining $X \equiv a \xi_i$ and $Y \equiv a \xi_o$
-for convenience, such that $X^2 + Y^2 = V^2$:
+meanwhile defining $$X \equiv a \xi_i$$ and $$Y \equiv a \xi_o$$
+for convenience, such that $$X^2 + Y^2 = V^2$$:
 
 $$\begin{aligned}
     X \frac{J_\ell'(X)}{J_\ell(X)} = Y \frac{K_\ell'(Y)}{K_\ell(Y)}
@@ -374,7 +374,7 @@ $$\begin{aligned}
     K_\ell'(x) = -K_{\ell+1}(x) + \ell \frac{K_\ell(x)}{x}
 \end{aligned}$$
 
-With this, the transcendental equation for $\beta$
+With this, the transcendental equation for $$\beta$$
 takes this convenient form:
 
 $$\begin{aligned}
@@ -383,27 +383,27 @@ $$\begin{aligned}
     }
 \end{aligned}$$
 
-All $\beta$ that satisfy this indicate the existence
+All $$\beta$$ that satisfy this indicate the existence
 of a **linearly polarized** mode.
-These modes are called $\mathrm{LP}_{\ell m}$,
-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 $\ell$.
+These modes are called $$\mathrm{LP}_{\ell m}$$,
+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 $$\ell$$.
 
 An example graphical solution of the transcendental equation
-is illustrated below for a fiber with $V = 5$,
+is illustrated below for a fiber with $$V = 5$$,
 where red and blue respectively denote the left and right-hand side:
 
 <a href="modes.jpg">
 <img src="modes-small.jpg" style="width:100%">
 </a>
 
-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,
-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 $V_{02} = V_{21} = 3.832$.
+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,
+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 $$V_{02} = V_{21} = 3.832$$.
 
 All differential equations have been linear,
 so a linear combination of these solutions is also valid.