From 16555851b6514a736c5c9d8e73de7da7fc9b6288 Mon Sep 17 00:00:00 2001 From: Prefetch Date: Thu, 20 Oct 2022 18:25:31 +0200 Subject: Migrate from 'jekyll-katex' to 'kramdown-math-sskatex' --- source/know/concept/step-index-fiber/index.md | 208 +++++++++++++------------- 1 file changed, 104 insertions(+), 104 deletions(-) (limited to 'source/know/concept/step-index-fiber/index.md') 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} -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: -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. -- cgit v1.2.3