Derivatives of the propagation constant in optical fibers

Question

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?

Answer 1

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 $v_g$.