4. Let the edge length of a regular octahedron be 1, then the distance between its two parallel faces is $\qquad$
Answer
As shown in the figure, the regular octahedron $P-ABCD-Q$, it is easy to know that $PQ=\sqrt{2}$. $E, F$ are the midpoints of $CD$ and $AB$ respectively. In $\triangle PEF$, $PE=PF=\frac{\sqrt{3}}{2}, EF=1$, then $\frac{\sqrt{3}}{2} \cdot EG=1 \cdot \frac{\sqrt{2}}{2} \Rightarrow EG=\frac{\sqrt{6}}{3}$.
Therefore, the distance between two parallel faces of the regular octahedron is $\frac{\sqrt{6}}{3}$.