= Solution
{c}
Place the <optical lens> at $z=0$. For a sufficiently broad incident <Gaussian beam>, the <beam waist> positions approach the <focal lengths> frequency by frequency:
$$
z_\omega\simeq f_0+\beta(\omega-\omega_0).
$$
Add the <group delay> at the <optical lens> to propagation through <vacuum>:
$$
\boxed{t_\omega=2\eta(\omega-\omega_0)+\frac{z_\omega}{c}
=\frac{f_0}{c}+\left(2\eta+\frac\beta c\right)(\omega-\omega_0).}
$$
Here an arbitrary common time origin has been omitted. Differentiating both equations with respect to <angular frequency> yields
$$
\boxed{\frac{v_f}{c}=\frac{\beta}{\beta+2\eta c}
=\left(1+\frac{2\eta c}{\beta}\right)^{-1}.}
$$
For $\beta>0$, define $u=2\eta c/\beta$. The velocity graph in the preceding figure has horizontal asymptote zero and a vertical asymptote at $u=-1$. For $\eta>0$, $0<v_f<c$: the higher-<frequency> components both enter later and travel farther. At $\eta=0$, $v_f=c$. For $-\beta/(2c)<\eta<0$, $v_f>c$: the farther-focusing components leave earlier, partly compensating their extra travel. At $\eta=-\beta/(2c)$ every component focuses at the same time, so the trajectory is not a finite-velocity function of time. For $\eta<-\beta/(2c)$ the farther positions focus first, giving $v_f<0$; it tends to zero from below as $\eta\to-\infty$.
A <flying focus> is a moving pattern assembled from different portions of an <optical pulse>. Each portion propagates causally at $c$ in this model. The superluminal or backward motion of the focus therefore does not describe a material object or information overtaking <light>. The calculation assumes negligible additional frequency-dependent <group delay> from the <optical lens>; any such delay must be included in $\psi$.
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