4. Given the dihedral angle $\alpha-l-\beta$ is $60^{\circ}$, moving points $P, Q$ are in planes $\alpha, \beta$ respectively, the distance from $P$ to $\beta$ is $\sqrt{3}$, and the distance from $Q$ to $\alpha$ is $2 \sqrt{3}$, then the minimum distance between points $P, Q$ is $\qquad$ .
Rong's wife $2 \sqrt{3}$.
Analysis: Draw $P B \perp \beta$ intersecting $\beta$ at $B$, and in the plane $\beta$, draw $B D \perp l$. Then $\angle B D P=60^{\circ}$. Knowing that the distance from $P$ to the plane $\beta$ is $\sqrt{3}$, we have $P B=\sqrt{3}$, and thus $P D=2$.

Similarly, the distance from point $Q$ in the plane $\beta$ to the edge $l$ is 4. When the line connecting $P$ and $Q$ is perpendicular to the edge $l$, the distance between $P$ and $Q$ is the shortest (at this time, in the dihedral angle, $P$ and $Q$ should be two points on the edges of the dihedral angle's plane angle). The minimum value is $d^{2}=4+16-2 \times 4 \times 2 \times \cos 60^{\circ}=12$, thus $d=2 \sqrt{3}$.