4. Let the base $A B C D$ of the right square prism $A B C D-A_{1} B_{1} C_{1} D_{1}$ be a unit square. If the dihedral angle $A_{1}-B D-C_{1}$ is $\frac{\pi}{3}$, then $A A_{1}=$ $\qquad$
Answer: $\frac{\sqrt{6}}{2}$. Solution: Take the midpoint $\mathrm{O}$ of $\mathrm{BD}$, and connect $\mathrm{OA}, \mathrm{OA}_{1}, \mathrm{OC}_{1}$.
Then $\angle \mathrm{A}_{1} \mathrm{OC}_{1}$ is the plane angle of the dihedral angle $\mathrm{A}_{1}-\mathrm{BD}-\mathrm{C}_{1}$, so $\angle \mathrm{A}_{1} \mathrm{OC}_{1}=\frac{\pi}{3}$, and $\triangle \mathrm{OA}_{1} \mathrm{C}_{1}$ is an equilateral triangle. Therefore, $\mathrm{A}_{1} \mathrm{O}=\mathrm{A}_{1} \mathrm{C}_{1}=\sqrt{2}$, so $A A_{1}=\sqrt{A_{1} O^{2}-A O^{2}}=\sqrt{(\sqrt{2})^{2}-\left(\frac{\sqrt{2}}{2}\right)^{2}}=\frac{\sqrt{6}}{2}$.