4. Let the tetrahedron $P-ABC$ satisfy $PA=PB=3, AB=BC=CA=2$, then the maximum volume of the tetrahedron is $\qquad$ .
Let the midpoint of $AB$ be $D$, then $PD=\sqrt{3^{2}-1}=2 \sqrt{2}$. Therefore, $V_{P-ABC} \leqslant \frac{1}{3} S_{\triangle ABC} \cdot PD=\frac{1}{3} \cdot \sqrt{3} \cdot 2 \sqrt{2}=\frac{2 \sqrt{6}}{3}$, equality holds when $PD \perp$ the base $ABC$.