6. In a regular quadrilateral pyramid $P-A B C D$, $G$ is the centroid of $\triangle P B C$. Then $\frac{V_{G-P A D}}{V_{G-P A B}}=$ $\qquad$
As shown in the figure, $M, N, F$ are the midpoints of the corresponding sides, $M E \perp P N$, $O H \perp P F$. Since $P-A B C D$ is a regular quadrilateral pyramid, we have $\angle M N E=\angle O F H=\alpha$, thus
$$
M E=M N \sin \alpha=2 O F \sin \alpha=2 O H \text {. }
$$

This means the distance from point $M$ to plane $P A D$ is twice the distance from point $M$ to plane $P A B$, hence the distance from point $G$ to plane $P A D$ is also twice the distance from point $G$ to plane $P A B$.
Therefore, $\frac{V_{G-P A D}}{V_{G-P A B}}=2$.