4. The axial section of a cone with vertex $P$ is an isosceles right triangle. $A$ is a point on the circumference of the base circle, $B$ is a point inside the base circle, $O$ is the center of the base circle, $A B \perp O B$, with the foot of the perpendicular at $B$, $O H \perp H B$, with the foot of the perpendicular at $H$, and $P A=4$, $C$ is the midpoint of $P A$. Then, when the volume of the tetrahedron $O-H P C$ is maximized, the length of $O B$ is $\qquad$.
Rong's wife $\frac{2 \sqrt{6}}{3}$.
Analysis $A B \perp O B \Rightarrow P B \perp A B \Rightarrow A B \perp$ plane $P O B \Rightarrow$ plane $P A B \perp$ plane $P O B$;
$O H \perp P B \Rightarrow O H \perp$ plane $P A B \Rightarrow O H \perp H C, O H \perp P C$;
Also $P C \perp O C \Rightarrow P C \perp$ plane $O C H \Rightarrow P C$ is the height of the tetrahedron $P-O C H$, $P C=O C=2$, and the area of $\triangle O C H$ reaches its maximum value when $O H=H C=\sqrt{2}$ (a right triangle with hypotenuse $=2$).
When $O H=\sqrt{2}$, knowing $P O=2 \sqrt{2}$, we find $\angle O P B=30^{\circ}, O B=P O \tan 30^{\circ}=\frac{2 \sqrt{6}}{3}$.