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1 files changed, 35 insertions, 23 deletions
diff --git a/source/know/concept/nonlinear-schrodinger-equation/index.md b/source/know/concept/nonlinear-schrodinger-equation/index.md index 2ea1b23..820b361 100644 --- a/source/know/concept/nonlinear-schrodinger-equation/index.md +++ b/source/know/concept/nonlinear-schrodinger-equation/index.md @@ -212,20 +212,20 @@ $$\begin{aligned}  \end{aligned}$$  Next, we take the [Fourier transform](/know/concept/fourier-transform/) -$$t \to (\omega\!-\!\omega_0)$$ of the wave equation, -once again treating $$|E|^2$$ (inside $$\varepsilon_r$$) as a constant. +$$t \to \omega$$ of the wave equation, +again treating $$|E|^2$$ (inside $$\varepsilon_r$$) as a constant.  The constant $$s = \pm 1$$ is included here  to deal with the fact that different authors use different sign conventions:  $$\begin{aligned}      0 -    &= \hat{\mathcal{F}}\bigg\{ \nabla^2 E - \frac{\varepsilon_r}{c^2} \pdvn{2}{E}{t} \bigg\} +    &= \hat{\mathcal{F}}\bigg\{ \bigg( \nabla^2 E - \frac{\varepsilon_r}{c^2} \pdvn{2}{E}{t} \bigg) e^{-i \omega_0 t} \bigg\}      \\      &= \int_{-\infty}^\infty      \bigg( \nabla^2 E - \frac{\varepsilon_r}{c^2} \pdvn{2}{E}{t} \bigg)      e^{i s (\omega - \omega_0) t} \dd{t}      \\ -    &= \nabla^2 E + s^2 \frac{\varepsilon_r}{c^2} (\omega - \omega_0)^2 E +    &= \nabla^2 E + s^2 (\omega - \omega_0)^2 \frac{\varepsilon_r}{c^2} E  \end{aligned}$$  We use $$s^2 = 1$$ and define $$\Omega \equiv \omega - \omega_0$$ @@ -392,8 +392,8 @@ with all the arguments shown for clarity:  $$\begin{aligned}      \boxed{          \Delta{\beta}(\omega) -        = \frac{\omega}{c \mathcal{A}_\mathrm{mode}(\omega)} -        \iint_{-\infty}^\infty \Delta{n}(x, y, \omega) \: |F(x, y, \omega)|^2 \dd{x} \dd{y} +        = \frac{\omega}{c \mathcal{A}_\mathrm{mode}} +        \iint_{-\infty}^\infty \Delta{n}(x, y, \omega) \: |F(x, y)|^2 \dd{x} \dd{y}      }  \end{aligned}$$ @@ -403,8 +403,8 @@ $$F$$ must be dimensionless,  and consequently $$A$$ has (SI) units of an electric field.  $$\begin{aligned} -    \mathcal{A}_\mathrm{mode}(\omega) -    \equiv \iint_{-\infty}^\infty |F(x, y, \omega)|^2 \dd{x} \dd{y} +    \mathcal{A}_\mathrm{mode} +    \equiv \iint_{-\infty}^\infty |F|^2 \dd{x} \dd{y}  \end{aligned}$$  Now we finally turn our attention to the equation for $$A$$. @@ -442,7 +442,7 @@ Recall that earlier, in order to treat $$\chi^{(3)}$$ as instantaneous,  we already assumed a temporally broad  (spectrally narrow) pulse.  Hence, for simplicity, we can cut off this Taylor series at $$\beta_2$$, -which is good enough for many cases. +which is good enough in many cases.  Inserting the expansion into $$A$$'s equation:  $$\begin{aligned} @@ -450,10 +450,11 @@ $$\begin{aligned}      &= i \pdv{A}{z} + i \frac{\beta_1}{s} (-i s \Omega) A - \frac{\beta_2}{2 s^2} (- i s \Omega)^2 A + \Delta{\beta}_0 A  \end{aligned}$$ -Which we have rewritten as preparation for taking the inverse Fourier transform, +Which we have rewritten in preparation for taking the inverse Fourier transform,  by introducing $$s$$ and by replacing $$\Delta{\beta}(\omega)$$  with $$\Delta{\beta_0} \equiv \Delta{\beta}(\omega_0)$$ -in order to remove all explicit dependence on $$\omega$$. +in order to remove all explicit dependence on $$\omega$$, +i.e. we only keep the first term of $$\Delta{\beta}$$'s Taylor expansion.  After transforming and using $$s^2 = 1$$,  we get the following equation for $$A(z, t)$$: @@ -468,11 +469,11 @@ according to which effects we want to include.  Earlier, we approximated $$\varepsilon_r \approx n^2$$,  so if we instead say that $$\varepsilon_r = (n \!+\! \Delta{n})^2$$,  then $$\Delta{n}$$ should include absorption and nonlinearity. -A simple and commonly used form for $$\Delta{n}$$ is therefore: +The most commonly used form for $$\Delta{n}$$ is therefore:  $$\begin{aligned} -    \Delta{n} -    = n_2 I + i \frac{\alpha c}{2 \omega} +    \Delta{n}(x, y, \omega) +    = n_2(\omega) \: I(x, y, \omega) + i \frac{c \alpha(\omega)}{2 \omega}  \end{aligned}$$  Where $$I$$ is the intensity (i.e. power per unit area) of the light, @@ -491,12 +492,13 @@ $$\begin{aligned}      + \frac{3 \omega \Imag\{\chi^{(3)}_{xxxx}\}}{2 \varepsilon_0 c^2 n^2} I      \qquad      I -    = \frac{\varepsilon_0 c n}{2} |E|^2 +    = \frac{\varepsilon_0 c n}{2} |F|^2 |A|^2  \end{aligned}$$ -For simplicity, we set $$\Imag\{\chi^{(3)}_{xxxx}\} = 0$$, -which is a good approximation for fibers made of silica. -Inserting this form of $$\Delta{n}$$ into $$\Delta{\beta_0}$$ then yields: +For simplicity we set $$\Imag\{\chi^{(3)}_{xxxx}\} = 0$$, +which is a good approximation for silica fibers. +Inserting this form of $$\Delta{n}$$ into $$\Delta{\beta_0}$$ +and neglecting the $$(x, y)$$-dependence of $$\Delta{n}$$ yields:  $$\begin{aligned}      \Delta{\beta}_0 @@ -507,24 +509,31 @@ $$\begin{aligned}      + \gamma_0 \frac{\varepsilon_0 c n}{2} \mathcal{A}_\mathrm{mode} |A|^2  \end{aligned}$$ -Where we have defined the nonlinear parameter $$\gamma_0$$ like so, +Where we have defined the parameter $$\gamma_0 \equiv \gamma(\omega_0)$$ like so,  involving the **effective mode area** $$\mathcal{A}_\mathrm{eff}$$,  which contains all information about $$F$$ needed for solving $$A$$'s equation:  $$\begin{aligned}      \boxed{ -        \gamma_0 -        = \gamma(\omega_0) -        \equiv \frac{\omega_0 n_2}{c \mathcal{A}_\mathrm{eff}} +        \gamma(\omega) +        \equiv \frac{\omega n_2(\omega)}{c \mathcal{A}_\mathrm{eff}(\omega)}      }      \qquad \qquad      \boxed{ -        \mathcal{A}_\mathrm{eff}(\omega_0) +        \mathcal{A}_\mathrm{eff}(\omega)          \equiv \frac{\displaystyle \bigg( \iint_{-\infty}^\infty |F|^2 \dd{x} \dd{y} \bigg)^2}          {\displaystyle \iint_{-\infty}^\infty |F|^4 \dd{x} \dd{y}}      }  \end{aligned}$$ +Note the $$\omega$$-dependence of $$A_\mathrm{eff}$$: +so far we have conveniently ignored that $$F$$ also depends on $$\omega$$, +because it is a parameter in its eigenvalue equation. +This is valid for spectrally narrow pulses, so we will stick with it. +Just beware that some people make the ad-hoc generalization +$$\gamma_0 \to \gamma(\omega)$$, which is not correct in general +(this is an advanced topic, see Lægsgaard). +  Substituting $$\Delta{\beta_0}$$ into the main problem  yields a prototype of the NLS equation: @@ -694,3 +703,6 @@ so many authors only show that case.  2.  O. Bang,      *Nonlinear mathematical physics: lecture notes*,      2020, unpublished. +3.  J. Lægsgaard, +    [Mode profile dispersion in the generalized nonlinear Schrödinger equation](https://doi.org/10.1364/OE.15.016110), +    2007, Optica.  | 
