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+---
+title: "Step-index fiber"
+date: 2022-02-11
+categories:
+- Physics
+- Optics
+- Fiber optics
+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/)
+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:
+
+$$\begin{aligned}
+    \pdvn{2}{F}{r} + \frac{1}{r} \pdv{F}{r} + \frac{1}{r^2} \pdvn{2}{F}{\phi} + \mu F = 0
+\end{aligned}$$
+
+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 $r$.
+
+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 $\mu$:
+
+$$\begin{aligned}
+    \begin{cases}
+        \displaystyle
+        0 = \rho^2 \pdvn{2}{R}{\rho} + \rho \pdv{R}{\rho} \pm \rho^2 R - \ell^2 R
+        & \mathrm{for}\; \mu \neq 0
+        \\
+        \displaystyle
+        0 = r^2 \pdvn{2}{R}{r} + r \pdv{R}{r} - \ell^2 R
+        & \mathrm{for}\; \mu = 0
+    \end{cases}
+\end{aligned}$$
+
+<a href="bessel.jpg">
+<img src="bessel.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 $R_o$:
+
+$$\begin{aligned}
+    R_{o,\ell}(r) =
+    \begin{cases}
+        r^{-\ell}
+        & \mathrm{for}\; \mu = 0 \;\mathrm{and}\; \ell = 1,2,3,...
+        \\
+        K_\ell(\rho) = K_\ell(r \sqrt{-\mu})
+        & \mathrm{for}\; \mu < 0 \;\mathrm{and}\; \ell = 0,1,2,...
+    \end{cases}
+\end{aligned}$$
+
+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: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 $V_{02} = V_{21} = 3.832$.
+
+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.
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