= Which wavelength enters the derivative formula
For the \i[wavelength in the medium], $\lambda_m=2\pi/k$, differentiate $\omega=kv_p(k)$:
$$
v_g=v_p+k\frac{dv_p}{dk}
=v_p+k\frac{dv_p}{d\lambda_m}\left(-\frac{\lambda_m}{k}\right),\qquad
\boxed{v_g=v_p-\lambda_m\frac{dv_p}{d\lambda_m}}.
$$
The distinction in <group velocity in material and vacuum wavelength coordinates> is essential for the numerical data. For incident \i[vacuum] wavelength $\lambda_0=2\pi c/\omega$, differentiation of $k=\omega/v_p(\lambda_0)$ instead gives
$$
\boxed{v_g=\frac{v_p}{1+(\lambda_0/v_p)\,dv_p/d\lambda_0}
=\frac{c}{n-\lambda_0\,dn/d\lambda_0}}.
$$
Thus the printed derivative identity is valid, but its wavelength coordinate cannot be carried unchanged into data specified before refraction.
Back to article page
