4. A cylinder is intersected by a plane, and the shape of the intersection is an ellipse. The eccentricity of this ellipse is $\frac{\sqrt{5}}{3}$. The acute angle formed between this intersection and the base of the cylinder is . $\qquad$
4. $\arccos \frac{2}{3}$ Detailed Explanation: Let the acute angle formed by this section and the base of the cylinder be $\theta$, the length of the major axis of the ellipse be $2 a$, and the length of the minor axis be $2 b$ (i.e., the diameter of the cylinder's base). Since $e=\frac{\sqrt{5}}{3}$, therefore $\frac{b}{a}=\frac{2}{3}$, thus $\cos \theta=\frac{2 b}{2 a}=\frac{2}{3}$, therefore $\theta=\arccos \frac{2}{3}$.