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diff --git a/content/know/concept/electromagnetic-wave-equation/index.pdc b/content/know/concept/electromagnetic-wave-equation/index.pdc index 59e0125..124bcc6 100644 --- a/content/know/concept/electromagnetic-wave-equation/index.pdc +++ b/content/know/concept/electromagnetic-wave-equation/index.pdc @@ -122,7 +122,7 @@ In fact, thanks to linearity, these **plane waves** can be treated as terms in a Fourier series, meaning that virtually *any* function $f(\vb{k} \cdot \vb{r} - \omega t)$ is a valid solution. -Keep in mind that in reality, $\vb{E}$ and $\vb{B}$ are real, +Keep in mind that in reality $\vb{E}$ and $\vb{B}$ are real, so although it is mathematically convenient to use plane waves, in the end you will need to take the real part. diff --git a/content/know/concept/salt-equation/index.pdc b/content/know/concept/salt-equation/index.pdc new file mode 100644 index 0000000..2f2917b --- /dev/null +++ b/content/know/concept/salt-equation/index.pdc @@ -0,0 +1,286 @@ +--- +title: "SALT equation" +firstLetter: "S" +publishDate: 2022-02-07 +categories: +- Physics +- Optics + +date: 2022-01-20T22:01:48+01:00 +draft: false +markup: pandoc +--- + +# SALT equation + +The **steady-state *ab initio* laser theory** (SALT) is +a theoretical description of lasers, whose mode-centric approach +makes it especially appropriate for microscopically small lasers. + +Consider the [Maxwell-Bloch equations](/know/concept/maxwell-bloch-equations/), +governing the complex polarization +vector $\vb{P}^{+}$ and the scalar population inversion $D$ of a set of +active atoms (or quantum dots) embedded in a passive linear background +material with refractive index $c / v$. +The system is affected by a driving [electric field](/know/concept/electric-field/) +$\vb{E}^{+}(t) = \vb{E}_0^{+} e^{-i \omega t}$, +such that the set of equations is: + +$$\begin{aligned} + - \mu_0 \pdv[2]{\vb{P}^{+}}{t} + &= \nabla \cross \nabla \cross \vb{E}^{+} + \frac{1}{v^2} \pdv[2]{\vb{E}^{+}}{t} + \\ + \pdv{\vb{P}^{+}}{t} + &= - \Big( \gamma_\perp + i \omega_0 \Big) \vb{P}^{+} + - \frac{i}{\hbar} \Big( \vb{p}_0^{-} \cdot \vb{E}^{+} \Big) \vb{p}_0^{+} D + \\ + \pdv{D}{t} + &= \gamma_\parallel (D_0 - D) + \frac{i 2}{\hbar} \Big( \vb{P}^{-} \cdot \vb{E}^{+} - \vb{P}^{+} \cdot \vb{E}^{-} \Big) +\end{aligned}$$ + +Where $\hbar \omega_0$ is the band gap of the active atoms, +and $\gamma_\perp$ and $\gamma_\parallel$ are relaxation rates +of the atoms' polarization and population inversion, respectively. +$D_0$ is the equilibrium inversion, i.e. the value of $D$ if there is no lasing. +Note that $D_0$ also represents the pump, +and both $D_0$ and $v$ depend on position $\vb{x}$. +Finally, the transition dipole matrix elements $\vb{p}_0^{-}$ and $\vb{p}_0^{+}$ are given by: + +$$\begin{aligned} + \vb{p}_0^{-} + \equiv q \matrixel{e}{\vu{x}}{g} + \qquad \qquad + \vb{p}_0^{+} + \equiv q \matrixel{g}{\vu{x}}{e} + = (\vb{p}_0^{-})^* +\end{aligned}$$ + +With $q < 0$ the electron charge, $\vu{x}$ the quantum position operator, +and $\ket{g}$ and $\ket{e}$ respectively +the ground state and first excitation of the active atoms. + +We start by assuming that the cavity has $N$ quasinormal modes $\Psi_n$, +each with a corresponding polarization $\vb{p}_n$ of the active matter. +Note that this ansatz already suggests +that the interactions between the modes are limited: + +$$\begin{aligned} + \vb{E}^{+}(\vb{x}, t) + = \sum_{n = 1}^N \Psi_n(\vb{x}) e^{- i \omega_n t} + \qquad \qquad + \vb{P}^{+}(\vb{x}, t) + = \sum_{n = 1}^N \vb{p}_n(\vb{x}) e^{- i \omega_n t} +\end{aligned}$$ + +Using the modes' linear independence to treat each term of the summation individually, +the first two Maxwell-Bloch equations turn into, respectively: + +$$\begin{aligned} + \mu_0 \omega_n^2 \vb{p}_n + &= \nabla \cross \nabla \cross \Psi_n - \frac{1}{v^2} \omega_n^2 \Psi_n + \\ + i \omega_n \vb{p}_n + &= \big( i \omega_0 + \gamma_\perp \big) \vb{p}_n + + \frac{i}{\hbar} \big(\vb{p}_0^{+} \vb{p}_0^{-}\big) \cdot \Psi_n \: D +\end{aligned}$$ + +With being $\vb{p}_0^{+} \vb{p}_0^{-}$ a dyadic product. +Isolating the latter equation for $\vb{p}_n$ gives us: + +$$\begin{aligned} + \vb{p}_n + &= \frac{\big(\vb{p}_0^{+} \vb{p}_0^{-}\big) \cdot \Psi_n \: D}{\hbar \big((\omega_n - \omega_0) + i \gamma_\perp\big)} + = \frac{\gamma(\omega_n) D}{\hbar \gamma_\perp} \big(\vb{p}_0^{+} \vb{p}_0^{-}\big) \cdot \Psi_n +\end{aligned}$$ + +Where we have defined the Lorentzian gain curve $\gamma(\omega_n)$ as follows, +which represents the laser's preferred frequencies for amplification: + +$$\begin{aligned} + \gamma(\omega_n) + \equiv \frac{\gamma_\perp}{(\omega_n - \omega_0) + i \gamma_\perp} +\end{aligned}$$ + +Inserting this expression for $\vb{p}_n$ +into the first Maxwell-Bloch equation yields +the prototypical form of the SALT equation, +where we still need to replace $D$ with known quantities: + +$$\begin{aligned} + 0 + &= \bigg( \nabla \cross \nabla \cross - \, \omega_n^2 \frac{1}{v^2} + - \omega_n^2 \frac{\mu_0 \gamma(\omega_n) D}{\hbar \gamma_\perp} (\vb{p}_0^{+} \vb{p}_0^{-}) \cdot \bigg) \Psi_n +\end{aligned}$$ + +To rewrite $D$, we turn to its (Maxwell-Bloch) equation of motion, +making the crucial **stationary inversion approximation** $\pdv*{D}{t} = 0$: + +$$\begin{aligned} + D + &= D_0 + \frac{i 2}{\hbar \gamma_\parallel} \Big( \vb{P}^{-} \cdot \vb{E}^{+} - \vb{P}^{+} \cdot \vb{E}^{-} \Big) +\end{aligned}$$ + +This is the most aggressive approximation we will make: +it removes all definite phase relations between modes, +and effectively eliminates time as a variable. +We insert our ansatz for $\vb{E}^{+}$ and $\vb{P}^{+}$, +and point out that only excited lasing modes contribute to $D$: + +$$\begin{aligned} + D + &= D_0 + \frac{i 2}{\hbar \gamma_\parallel} \sum_{\nu, \mu}^\mathrm{active} + \bigg( \vb{p}_\nu^* \cdot \Psi_\mu e^{i (\omega_\nu - \omega_\mu) t} + - \vb{p}_\nu \cdot \Psi_\mu^* e^{i (\omega_\mu - \omega_\nu) t} \bigg) +\end{aligned}$$ + +Here, we make the [rotating wave approximation](/know/concept/rotating-wave-approximation/) +to neglect all terms where $\nu \neq \mu$ +on the basis that they oscillate too quickly, +leaving only $\nu = \mu$: + +$$\begin{aligned} + D + &= D_0 + \frac{i 2}{\hbar \gamma_\parallel} \sum_{\nu}^\mathrm{act.} + \bigg( \vb{p}_\nu^* \cdot \Psi_\nu - \vb{p}_\nu \cdot \Psi_\nu^* \bigg) +\end{aligned}$$ + +Inserting our earlier equation for $\vb{p}_n$ +and using the fact that $\vb{p}_0^{+} = (\vb{p}_0^{-})^*$ leads us to: + +$$\begin{aligned} + D + &= D_0 + \frac{i 2 D}{\hbar^2 \gamma_\parallel \gamma_\perp} \sum_{\nu}^\mathrm{act.} + \bigg( \gamma^*(\omega_\nu) \big(\vb{p}_0^{+} \vb{p}_0^{-}\big)^* \!\cdot\! \Psi_\nu^* \cdot \Psi_\nu + - \gamma(\omega_\nu) \big(\vb{p}_0^{+} \vb{p}_0^{-}\big) \!\cdot\! \Psi_\nu \cdot \Psi_\nu^* \bigg) + \\ + &= D_0 + \frac{i 2 D}{\hbar^2 \gamma_\parallel \gamma_\perp} \sum_{\nu}^\mathrm{act.} + \bigg( \gamma^*(\omega_\nu) \big(\vb{p}_0^{+} \cdot \Psi_\nu^*\big) \vb{p}_0^{-} \cdot \Psi_\nu + - \gamma(\omega_\nu) \big(\vb{p}_0^{-} \cdot \Psi_\nu\big) \vb{p}_0^{+} \cdot \Psi_\nu^* \bigg) + \\ + &= D_0 + \frac{i 2 D}{\hbar^2 \gamma_\parallel \gamma_\perp} \sum_{\nu}^\mathrm{act.} + \Big( \gamma^*(\omega_\nu) - \gamma(\omega_\nu) \Big) \big|\vb{p}_0^{-} \cdot \Psi_\nu\big|^2 +\end{aligned}$$ + +By putting the terms on a common denominator, it is easily shown that: + +$$\begin{aligned} + \gamma^*(\omega_\nu) - \gamma(\omega_\nu) + &= \frac{\gamma_\perp ((\omega_\nu - \omega_0) + i \gamma_\perp)}{(\omega_\nu - \omega_0)^2 + \gamma_\perp^2} + - \frac{\gamma_\perp ((\omega_\nu - \omega_0) - i \gamma_\perp)}{(\omega_\nu - \omega_0)^2 + \gamma_\perp^2} + \\ + &= \frac{\gamma_\perp (i \gamma_\perp + i \gamma_\perp)}{(\omega_\nu - \omega_0)^2 + \gamma_\perp^2} + = i 2 \big|\gamma(\omega_\nu)\big|^2 +\end{aligned}$$ + +Inserting this into our equation for $D$ gives the following expression: + +$$\begin{aligned} + D + &= D_0 - \frac{4 D}{\hbar^2 \gamma_\parallel \gamma_\perp} \sum_{\nu}^\mathrm{act.} + \Big|\gamma(\omega_\nu) \vb{p}_0^{-} \cdot \Psi_\nu\Big|^2 +\end{aligned}$$ + +We then properly isolate this for $D$ to get its final form, namely: + +$$\begin{aligned} + D + &= D_0 \bigg( 1 + \frac{4}{\hbar^2 \gamma_\parallel \gamma_\perp} \sum_{\nu}^\mathrm{act.} + \Big|\gamma(\omega_\nu) \vb{p}_0^{-} \cdot \Psi_\nu\Big|^2 \bigg)^{-1} +\end{aligned}$$ + +Substituting this into the prototypical SALT equation from earlier +yields the most general form of the **SALT equation**, +upon which the theory is built: + +$$\begin{aligned} + \boxed{ + 0 + = \bigg( \nabla \cross \nabla \cross + -\,\omega_n^2 \bigg[ \frac{1}{v^2(\vb{x})} + \frac{\mu_0 \gamma(\omega_n)}{\hbar \gamma_\perp} + \frac{D_0(\vb{x})}{1 + h(\vb{x})} (\vb{p}_0^{+} \vb{p}_0^{-}) \cdot \bigg] \bigg) \Psi_n(\vb{x}) + } +\end{aligned}$$ + +Where we have defined **spatial hole burning** function $h(\vb{x})$ like so, +representing the depletion of the supply of charge +carriers as they are consumed by the active lasing modes: + +$$\begin{aligned} + \boxed{ + h(\vb{x}) + \equiv \frac{4}{\hbar^2 \gamma_\parallel \gamma_\perp} \sum_{\nu}^\mathrm{act.} + \Big|\gamma(\omega_\nu) \vb{p}_0^{-} \cdot \Psi_\nu(\vb{x})\Big|^2 + } +\end{aligned}$$ + +Many authors assume that $\vb{p}_0^- \parallel \Psi_n$, +so that only its amplitude $|g|^2 \equiv \vb{p}_0^{+} \cdot \vb{p}_0^{-}$ matters. +In that case, they often non-dimensionalize $D$ and $\Psi_n$ +by dividing out the units $d_c$ and $e_c$: + +$$\begin{aligned} + \tilde{\Psi}_n + \equiv \frac{\Psi_n}{e_c} + \qquad + e_c + \equiv \frac{\hbar \sqrt{\gamma_\parallel \gamma_\perp}}{2 |g|} + \qquad \qquad + \tilde{D} + \equiv \frac{D}{d_c} + \qquad + d_c + \equiv \frac{\varepsilon_0 \hbar \gamma_\perp}{|g|^2} +\end{aligned}$$ + +And then the SALT equation and hole burning function $h$ are reduced to the following, +where the vacuum wavenumber $k_n = \omega_n / c$: + +$$\begin{aligned} + 0 + = \bigg( \nabla \cross \nabla \cross -\,k_n^2 \bigg[ \varepsilon_r + + \gamma(c k_n) \frac{\tilde{D}_0}{1 + h} \bigg] \bigg) \tilde{\Psi}_n + \qquad + h(\vb{x}) + = \sum_{\nu}^\mathrm{act.} \Big|\gamma(c k_\nu) \tilde{\Psi}_\nu(\vb{x})\Big|^2 +\end{aligned}$$ + + +In addition, some papers only consider 1D or 2D *transverse magnetic* (TM) modes, +in which case the fields are scalars. Using the vector identity + +$$\begin{aligned} + \nabla \cross \nabla \cross \Psi + = \nabla (\nabla \cdot \Psi) - \nabla^2 \Psi +\end{aligned}$$ + +Where $\nabla \cdot \Psi = 0$ thanks to [Gauss' law](/know/concept/maxwells-equations/), +so we get an even further simplified SALT equation: + +$$\begin{aligned} + 0 + = \bigg( \nabla^2 +\,k_n^2 \bigg[ \varepsilon_r + + \gamma(c k_n) \frac{\tilde{D}_0}{1 + h} \bigg] \bigg) \tilde{\Psi}_n +\end{aligned}$$ + +The challenge is to solve this equation for a given $\varepsilon_r(\vb{x})$ and $D_0(\vb{x})$, +with the boundary condition that $\Psi_n$ is a plane wave at infinity, +i.e. that there is light leaving the cavity. + +If $k_n$ has a negative imaginary part, then that mode is behaving as an LED. +Gradually increasing the pump $D_0$ in a chosen region +causes the $k_n$'s imaginary parts become less negative, +until one of them hits the real axis, at which point that mode starts lasing. +After that, $D_0$ can be increased even further until some other $k_n$ become real. + +Below threshold (i.e. before any mode is lasing), the problem is linear in $\Psi_n$, +but above threshold it is nonlinear, and the amplitude of $\Psi_n$ is adjusted +such that the corresponding $k_n$ never leaves the real axis. +When any mode is lasing, hole burning makes it harder for other modes to activate, +since it effectively reduces the pump $D_0$. + + +## References +1. L. Ge, Y.D. Chong, A.D. Stone, + [Steady-state *ab initio* laser theory: generalizations and analytic results](http://dx.doi.org/10.1103/PhysRevA.82.063824), + 2010, American Physical Society. + diff --git a/content/know/concept/step-index-fiber/bessel.jpg b/content/know/concept/step-index-fiber/bessel.jpg Binary files differnew file mode 100644 index 0000000..464a1e7 --- /dev/null +++ b/content/know/concept/step-index-fiber/bessel.jpg diff --git a/content/know/concept/step-index-fiber/index.pdc b/content/know/concept/step-index-fiber/index.pdc new file mode 100644 index 0000000..8847fff --- /dev/null +++ b/content/know/concept/step-index-fiber/index.pdc @@ -0,0 +1,427 @@ +--- +title: "Step-index fiber" +firstLetter: "S" +publishDate: 2022-02-11 +categories: +- Physics +- Optics +- Fiber optics + +date: 2022-01-31T19:29:33+01:00 +draft: false +markup: pandoc +--- + +# Step-index fiber + +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. +In [polar coordinates](/know/concept/cylindrical-polar-coordinates/) +$(r,\phi)$ this equation can be rewritten as follows: + +$$\begin{aligned} + \pdv[2]{F}{r} + \frac{1}{r} \pdv{F}{r} + \frac{1}{r^2} \pdv[2]{F}{\phi} + \mu F = 0 +\end{aligned}$$ + +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 $r$. + +This Helmholtz equation can be solved by *separation of variables*: +we assume that there exist two functions $R(r)$ and $\Phi(\phi)$ +such that $F(r,\phi) = R(r) \, \Phi(\phi)$. +Inserting this ansatz: + +$$\begin{aligned} + R'' \Phi + \frac{1}{r} R' \Phi + \frac{1}{r^2} R \Phi'' + \mu R \Phi = 0 +\end{aligned}$$ + +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 $\ell^2$: + +$$\begin{aligned} + r^2 \frac{R''}{R} + r \frac{R'}{R} + \mu r^2 + = -\frac{\Phi''}{\Phi} + = \ell^2 +\end{aligned}$$ + +This gives an eigenvalue problem for $\Phi$, +and the well-known *Bessel equation* for $R$: + +$$\begin{aligned} + \boxed{ + \Phi'' + \ell^2 \Phi = 0 + } + \qquad \qquad + \boxed{ + r^2 R'' + r R' + (\mu r^2 \!-\! \ell^2) R = 0 + } +\end{aligned}$$ + +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: + +$$\begin{aligned} + \Phi(0) = \Phi(2\pi) + \qquad \qquad + \Phi'(0) = \Phi'(2\pi) +\end{aligned}$$ + +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 $\ell^2$, +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. + +* 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: + + $$\begin{alignedat}{3} + \Phi(0) &= \Phi(2 \pi) + \:\quad &&\implies \quad\:\: + 0 &&= A \sinh\!(2 \pi \ell) + B \big( \cosh\!(2 \pi \ell) - 1 \big) + \\ + \Phi'(0) &= \Phi'(2 \pi) + \: \quad &&\implies \quad \:\: + 0 &&= A \ell \big( \cosh\!(2 \pi \ell) - 1 \big) + B \ell \sinh\!(2 \pi \ell) + \end{alignedat}$$ + + This only has non-trivial solutions + if the determinant of the system matrix is zero: + + $$\begin{aligned} + 0 + &= \mathrm{det} + \begin{bmatrix} + \sinh\!(2 \pi \ell) & \cosh\!(2 \pi \ell) - 1 \\ + \cosh\!(2 \pi \ell) - 1 & \sinh\!(2 \pi \ell) + \end{bmatrix} + = 2 \big( \cosh\!(2 \pi \ell) - 1 \big) + \end{aligned}$$ + + 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, + because no non-trivial solutions exist here. + +* For $\ell^2 = 0$, the solution is $\Phi(\phi) = A \phi + B$. + Putting this in the boundary conditions: + + $$\begin{alignedat}{3} + \Phi(0) &= \Phi(2 \pi) + \qquad &&\implies \qquad + A &&= 0 + \\ + \Phi'(0) &= \Phi'(2 \pi) + \qquad &&\implies \qquad + B &&= B + \end{alignedat}$$ + + $B$ 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: + + $$\begin{alignedat}{3} + \Phi(0) &= \Phi(2 \pi) + \quad &&\implies \quad + 0 &&= A \sin\!(2 \pi \ell) + B \big(\cos\!(2\pi \ell) - 1\big) + \\ + \Phi'(0) &= \Phi'(2 \pi) + \quad &&\implies \quad + 0 &&= A \big(\cos\!(2 \pi \ell) - 1\big) - B \sin\!(2 \pi \ell) + \end{alignedat}$$ + + This system only has nontrivial solutions + if the determinant of its matrix is zero: + + $$\begin{aligned} + 0 + &= \mathrm{det} + \begin{bmatrix} + \sin\!(2 \pi \ell) & \cos\!(2 \pi \ell) - 1 \\ + \cos\!(2 \pi \ell) - 1 & -\sin\!(2 \pi \ell) + \end{bmatrix} + = 2 \big(\cos\!(2 \pi \ell) - 1\big) + \end{aligned}$$ + + Meaning that $\ell$ must be an integer. + We revisit the boundary conditions and indeed see: + + $$\begin{alignedat}{3} + 0 &= A \sin\!(2 \pi \ell) + B \big(\cos\!(2 \pi \ell) - 1\big) + \qquad &&\implies \qquad + 0 &&= 0 + \\ + 0 &= A \big(\cos\!(2 \pi \ell) - 1\big) - B \sin\!(2 \pi \ell) + \qquad &&\implies \qquad + 0 &&= 0 + \end{alignedat}$$ + + 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 $\cos\!(\phi \ell)$, + 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. + +By combining our result for $\ell^2 = 0$ and $\ell^2 > 0$, +we get the following for $\ell = 0, 1, 2, ...$: + +$$\begin{aligned} + \boxed{ + \Phi_\ell(\phi) = A \cos(\phi \ell) + } +\end{aligned}$$ + +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. + +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 $\mu(r)$. +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 $n_o$. + +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 $r = a$: + +$$\begin{aligned} + R_i(a) = R_o(a) + \qquad \qquad + R_i'(a) = R_o'(a) +\end{aligned}$$ + +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 $\mu$. + +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 $\mu$: + +$$\begin{aligned} + \begin{cases} + \displaystyle + 0 = \rho^2 \pdv[2]{R}{\rho} + \rho \pdv{R}{\rho} \pm \rho^2 R - \ell^2 R + & \mathrm{for}\; \mu \neq 0 + \\ + \displaystyle + 0 = r^2 \pdv[2]{R}{r} + r \pdv{R}{r} - \ell^2 R + & \mathrm{for}\; \mu = 0 + \end{cases} +\end{aligned}$$ + +<a href="bessel.jpg"> +<img src="bessel.jpg" style="width:90%;display:block;margin:auto;"> +</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 $R_o$: + +$$\begin{aligned} + R_{o,\ell}(r) = + \begin{cases} + r^{-\ell} + & \mathrm{for}\; \mu = 0 \;\mathrm{and}\; \ell = 1,2,3,... + \\ + K_\ell(\rho) = K_\ell(r \sqrt{-\mu}) + & \mathrm{for}\; \mu < 0 \;\mathrm{and}\; \ell = 0,1,2,... + \end{cases} +\end{aligned}$$ + +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 $\mu > 0$: + +$$\begin{aligned} + R_{i,\ell}(r) = + J_\ell(\rho) = J_\ell(r \sqrt{\mu}) + \qquad \mathrm{for}\; \mu > 0 \;\mathrm{and}\; \ell = 0,1,2,... +\end{aligned}$$ + +Putting this all together, we now know what the full solution for $F$ should look like: + +$$\begin{aligned} + F_\ell(r, \phi) + = R_\ell(r) \, \Phi_\ell(\phi) + = + \begin{cases} + A_\ell \: R_{i,\ell}(r) \, \cos\!(\phi \ell) + & \mathrm{for}\; r \le a + \\ + B_\ell \: R_{o,\ell}(r) \, \cos\!(\phi l) + & \mathrm{for}\; r \ge a + \end{cases} +\end{aligned}$$ + +Where $A_\ell$ and $B_\ell$ are constants to be chosen +based on the light's intensity, and to satisfy the continuity condition at $r = a$. + +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; +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, +assuming $n_i$ and $n_o$ do not depend on $k$: + +$$\begin{aligned} + \xi_i(k) + \equiv \sqrt{ n_i^2 k^2 - \beta^2(k) } + \qquad \qquad + \xi_o(k) + \equiv \sqrt{ \beta^2(k) - n_o^2 k^2 } +\end{aligned}$$ + +It is important to note that the sum of their squares is constant with respect to $\beta$: + +$$\begin{aligned} + \xi_i^2 + \xi_o^2 = (\mathrm{NA})^2 k^2 +\end{aligned}$$ + +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: + +$$\begin{aligned} + \boxed{ + \mathrm{NA} + \equiv \sqrt{n_i^2 - n_o^2} + } +\end{aligned}$$ + +From this, we define a new fiber parameter: the $V$-**number**, +which is extremely useful: + +$$\begin{aligned} + \boxed{ + V + \equiv a \sqrt{\xi_i^2 + \xi_o^2} + = a k \: \mathrm{NA} + } +\end{aligned}$$ + +Now, the allowed values of $\beta$ are found +by fulfilling the boundary conditions (for $\mu \neq 0$): + +$$\begin{aligned} + A_\ell J_\ell(a \xi_i) + &= B_\ell K_\ell(a \xi_o) + \\ + A_\ell \xi_i J_\ell'(a \xi_i) + &= B_\ell \xi_o K_\ell'(a \xi_o) +\end{aligned}$$ + +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 $X^2 + Y^2 = V^2$: + +$$\begin{aligned} + X \frac{J_\ell'(X)}{J_\ell(X)} = Y \frac{K_\ell'(Y)}{K_\ell(Y)} +\end{aligned}$$ + +We can turn this result into something a bit nicer +by using the following identities: + +$$\begin{aligned} + J_\ell'(x) = -J_{\ell+1}(x) + \ell \frac{J_\ell(x)}{x} + \qquad \quad + K_\ell'(x) = -K_{\ell+1}(x) + \ell \frac{K_\ell(x)}{x} +\end{aligned}$$ + +With this, the transcendental equation for $\beta$ +takes this convenient form: + +$$\begin{aligned} + \boxed{ + X \frac{J_{\ell+1}(X)}{J_\ell(X)} = Y \frac{K_{\ell+1}(Y)}{K_\ell(Y)} + } +\end{aligned}$$ + +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 $\ell$. + +An example graphical solution of the transcendental equation +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.jpg" style="width:90%;display:block;margin:auto;"> +</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 $V_{02} = V_{21} = 3.832$. + +All differential equations have been linear, +so a linear combination of these solutions is also valid. +Therefore, the fiber modes represent independent "channels" of light. +However, in practice, they can interact nonlinearly, +and light can scatter between them, and between polarizations. + + + +## References +1. O. Bang, + *Applied mathematics for physicists: lecture notes*, 2019, + unpublished. +2. B.E.A. Saleh, M.C. Teich, + *Fundamentals of photonics*, 1st edition, 1991, + Wiley. diff --git a/content/know/concept/step-index-fiber/modes.jpg b/content/know/concept/step-index-fiber/modes.jpg Binary files differnew file mode 100644 index 0000000..85682d7 --- /dev/null +++ b/content/know/concept/step-index-fiber/modes.jpg |