path: root/source/know/concept/step-index-fiber/index.md

---
title: "Step-index fiber"
sort_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,
$$k = \omega / c$$ 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 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,
which we call $$\ell^2$$:

$$\begin{aligned}
    r^2 \frac{R''}{R} + r \frac{R'}{R} + \mu r^2
    = -\frac{\Phi''}{\Phi}
    = \ell^2
\end{aligned}$$

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 is 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 $$\mu$$:

$$\begin{aligned}
    \begin{cases}
        \displaystyle
        0
        = \rho^2 \pdvn{2}{R}{\rho} + \rho \pdv{R}{\rho} \pm \rho^2 R - \ell^2 R
        & \mathrm{for}\; \mu \neq 0
        \\
        \displaystyle
        0
        = r^2 \pdvn{2}{R}{r} + r \pdv{R}{r} - \ell^2 R
        & \mathrm{for}\; \mu = 0
    \end{cases}
\end{aligned}$$

{% include image.html file="bessel-full.png" width="100%"
    alt="First few solutions to Bessel's equation" %}

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
$$\xi_i$$ and $$\xi_o$$ 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 very 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:

$$\begin{aligned}
    J_\ell'(x)
    = - \ell \frac{J_\ell(x)}{x} + J_{\ell-1}(x)
    \qquad \qquad
    K_\ell'(x)
    = - \ell \frac{K_\ell(x)}{x} - K_{\ell-1}(x)
\end{aligned}$$

With this, the transcendental equation for $$\beta$$
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**,
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 $$\ell$$.

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:

{% include image.html file="transcendental-full.png" version="2" width="100%"
    alt="Graphical solution of transcendental equation" %}

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.
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.