 Solution

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:Add the group delay at the optical lens to propagation through vacuum:Here an arbitrary common time origin has been omitted. Differentiating both equations with respect to angular frequency yieldsFor $\beta >0$, 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 $\psi$.