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In many books on nonlinear fiber optics, the Taylor series expansion of the mode-propagation constant $\beta$ is performed about a frequency $\omega_0$ at which a pulse's spectrum is to be centered.

$\beta(\omega) = n(\omega)\frac{\omega}{c} = \beta_0 + \beta_1(\omega - \omega_0) + \frac{1}{2}\beta_2(\omega - \omega_0)^2 + ...$

where

$\beta_m = (\frac{d^m\beta}{d\omega^m})_{\omega = \omega_0} (m = 0,1,2,...)$

Now the authors say that the parameters $\beta_1$ and $\beta_2$ are related to the refractive index and its derivatives through

$\beta_1 = \frac{1}{v_g} = \frac{n_g}{c} = \frac{1}{c}(n + \omega\frac{dn}{d\omega})\\$

$\beta_2 = \frac{1}{c}(2\frac{dn}{d\omega} + \omega\frac{d^2n}{d\omega^2})$

Can someone explain how these relations can be found?

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let the wavenumber be $$ \beta(\omega)= n(\omega)\omega/c. $$ Then by Leibnitz rule for differentiating products we have $$ \frac{\partial \beta}{\partial \omega}= \frac {n(\omega)}{c}+ \frac{\omega}{c} \frac{\partial n}{\partial \omega}. $$ Differentiating again
$$ \frac{\partial^2 \beta}{\partial \omega^2}=\frac{\partial }{\partial \omega}\left(\frac {n(\omega)}{c}\right)+ \frac{\partial }{\partial \omega}\left( \frac{\omega}{c} \frac{\partial n}{\partial \omega}\right) \\= \frac 2 c \frac{\partial n}{\partial \omega}+ \frac{\omega}{c} \frac{\partial^2 n}{\partial \omega^2}. $$ So the Taylor expansion about $\omega_0$
$$ \beta(\omega)= \beta(\omega_0)+ (\omega-\omega_0) \frac{\partial \beta}{\partial \omega}\Big|_{\omega_0}+\frac 12 (\omega-\omega_0)^2 \frac{\partial^2 \beta}{\partial \omega^2}\Big|_{\omega_0}+ \ldots $$ gives what you want. Note that the group velocity is $$ v_g= \frac{\partial \omega}{\partial k} $$ and the wavenumber $k$ is here denoted by "$\beta$" so $\beta_1$ is the reciprocal of $v_g$.

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  • $\begingroup$ Why is $\beta$ represented by $k$? Is it because $\alpha$ is small enough to be neglected in optical fibers? $\endgroup$ – Paddy Jul 17 '20 at 22:40
  • $\begingroup$ Is not $\beta$ just the fibre-optics engineers' name for $k$? Your def in terms of the refractive index seems to suggest this. $n(\omega)$ can be complex if there is absorbtion, so $k$ can have an imaginary part if needed. Obvously in practical application the wavenumber is of the order of inverse nanometers while any imaginary part is of the order of inverse kilometers. $\endgroup$ – mike stone Jul 17 '20 at 23:34
  • $\begingroup$ Honestly, I am not sure. I am very new to the subject and I am not very clear with terminology. $\endgroup$ – Paddy Jul 17 '20 at 23:36
  • $\begingroup$ Okay, a document I found says that $\beta$ is the component of $k$ in the direction of propagation. brunel.ac.uk/~eestprh/EE5514/lesson1_new.pdf $\endgroup$ – Paddy Jul 17 '20 at 23:40

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