---
title: "Self-steepening"
sort_title: "Self-steepening"
date: 2021-02-26
categories:
- Physics
- Optics
- Fiber optics
- Nonlinear optics
layout: "concept"
---

For a laser pulse travelling through an optical fiber,
its intensity is highest at its peak, so the Kerr effect will be strongest there.
This means that the peak travels slightly slower
than the rest of the pulse, leading to **self-steepening** of its trailing edge.
Mathematically, this is described by adding a new term to the
nonlinear Schrödinger equation:

$$\begin{aligned}
    0
    = i\pdv{A}{z} - \frac{\beta_2}{2} \pdvn{2}{A}{t} + \gamma \Big(1 + \frac{i}{\omega_0} \pdv{}{t} \Big) \big(|A|^2 A\big)
\end{aligned}$$

Where $$\omega_0$$ is the angular frequency of the pump.
We will use the following ansatz,
consisting of an arbitrary power profile $$P$$ with a phase $$\phi$$:

$$\begin{aligned}
    A(z,t) = \sqrt{P(z,t)} \, \exp\!\big(i \phi(z,t)\big)
\end{aligned}$$

For a long pulse travelling over a short distance, it is reasonable to
neglect dispersion ($$\beta_2 = 0$$).
Inserting the ansatz then gives the following, where $$\varepsilon = \gamma / \omega_0$$:

$$\begin{aligned}
    0 &= i \frac{1}{2} \frac{P_z}{\sqrt{P}} - \sqrt{P} \phi_z + \gamma P \sqrt{P} + i \varepsilon \frac{3}{2} P_t \sqrt{P} - \varepsilon P \sqrt{P} \phi_t
\end{aligned}$$

This results in two equations, respectively corresponding to the real
and imaginary parts:

$$\begin{aligned}
    0 &= - \phi_z - \varepsilon P \phi_t + \gamma P
    \\
    0 &= P_z + \varepsilon 3 P_t P
\end{aligned}$$

The phase $$\phi$$ is not so interesting, so we focus on the latter equation for $$P$$.
As it turns out, it has a general solution of the form below, which shows that
more intense parts of the pulse will tend to lag behind compared to the rest:

$$\begin{aligned}
    P(z,t) = f(t - 3 \varepsilon z P)
\end{aligned}$$

Where $$f$$ is the initial power profile: $$f(t) = P(0,t)$$.
The derivatives $$P_t$$ and $$P_z$$ are then given by:

$$\begin{aligned}
    P_t
    &= (1 - 3 \varepsilon z P_t) \: f'
    \qquad \quad \implies \quad
    P_t
    = \frac{f'}{1 + 3 \varepsilon z f'}
    \\
    P_z
    &= (-3 \varepsilon P - 3 \varepsilon z P_z) \: f'
    \quad \implies \quad
    P_z
    = \frac{- 3 \varepsilon P f'}{1 + 3 \varepsilon z f'}
\end{aligned}$$

These derivatives both go to infinity when their denominator is zero,
which, since $$\varepsilon$$ is positive, will happen earliest where $$f'$$
has its most negative value, called $$f_\mathrm{min}'$$,
which is located on the trailing edge of the pulse.
At the propagation distance where this occurs, $$L_\mathrm{shock}$$,
the pulse will "tip over", creating a discontinuous shock:

$$\begin{aligned}
    \boxed{
        L_\mathrm{shock} = -\frac{1}{3 \varepsilon f_\mathrm{min}'}
    }
\end{aligned}$$

In practice, however, this will never actually happen, because by the time
$$L_\mathrm{shock}$$ is reached, the pulse spectrum will have become so
broad that dispersion can no longer be neglected.

A simulation of self-steepening without dispersion is illustrated below
for the following Gaussian initial power distribution,
with $$T_0 = 25\:\mathrm{fs}$$, $$P_0 = 3\:\mathrm{kW}$$,
$$\beta_2 = 0$$ and $$\gamma = 0.1/\mathrm{W}/\mathrm{m}$$:

$$\begin{aligned}
    f(t) = P(0,t) = P_0 \exp\!\Big(\! -\!\frac{t^2}{T_0^2} \Big)
\end{aligned}$$


Its steepest points are found to be at $$2 t^2 = T_0^2$$, so
$$f_\mathrm{min}'$$ and $$L_\mathrm{shock}$$ are given by:

$$\begin{aligned}
    f_\mathrm{min}' = - \frac{\sqrt{2} P_0}{T_0} \exp\!\Big(\!-\!\frac{1}{2}\Big)
    \quad \implies \quad
    L_\mathrm{shock} = \frac{T_0}{3 \sqrt{2} \varepsilon P_0} \exp\!\Big(\frac{1}{2}\Big)
\end{aligned}$$

This example Gaussian pulse therefore has a theoretical
$$L_\mathrm{shock} = 0.847\,\mathrm{m}$$,
which turns out to be accurate,
although the simulation breaks down due to insufficient resolution:

<a href="pheno-steep.jpg">
<img src="pheno-steep-small.jpg" style="width:100%">
</a>

Unfortunately, self-steepening cannot be simulated perfectly: as the
pulse approaches $$L_\mathrm{shock}$$, its spectrum broadens to infinite
frequencies to represent the singularity in its slope.
The simulation thus collapses into chaos when the edge of the frequency window is reached.
Nevertheless, the general trends are nicely visible:
the trailing slope becomes extremely steep, and the spectrum
broadens so much that dispersion cannot be neglected anymore.

When self-steepening is added to the nonlinear Schrödinger equation,
it no longer conserves the total pulse energy $$\int |A|^2 \dd{t}$$.
Fortunately, the photon number $$N_\mathrm{ph}$$ is still
conserved, which for the physical envelope $$A(z,t)$$ is defined as:

$$\begin{aligned}
    \boxed{
        N_\mathrm{ph}(z) = \int_0^\infty \frac{|\tilde{A}(z,\omega)|^2}{\omega} \dd{\omega}
    }
\end{aligned}$$


## References
1. B.R. Suydam, [Self-steepening of optical pulses](https://doi.org/10.1007/0-387-25097-2_6), 2006, Springer.