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+---
+title: "Self-steepening"
+firstLetter: "S"
+publishDate: 2021-02-26
+categories:
+- Physics
+- Optics
+- Fiber optics
+- Nonlinear dynamics
+
+date: 2021-02-26T15:18:25+01:00
+draft: false
+markup: pandoc
+---
+
+# Self-steepening
+
+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:
+
+$$\begin{aligned}
+    0
+    = i\pdv{A}{z} - \frac{\beta_2}{2} \pdv[2]{A}{t} + \gamma \Big(1 + \frac{i}{\omega_0} \pdv{t}\Big) \big(|A|^2 A\big)
+\end{aligned}$$
+
+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:
+
+<img src="pheno-steep.jpg">
+
+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 Media.