6. In the tetrahedron $O-ABC$, if the three edges meeting at point $O$ are pairwise perpendicular and have a length of $\sqrt{3}$, then the total length of the curve formed by the points on the surface of the tetrahedron that are a distance of 2 from point $A$ is $\qquad$ .
6. $\frac{3 \pi}{2}$.

As shown in Figure 2, points $M$ and $N$ are on edges $AB$ and $AC$, respectively, and $AM = AN = 2$; points $E$ and $F$ are on edges $OB$ and $OC$, respectively, and $OE = OF = 1$.
Thus, $AE = AF = 2$.
Therefore, the curves formed by the points that meet the conditions are:
(1) In the plane $OBC$, an arc with $O$ as the center and a radius of 1
(2) In the plane $AOB$, an arc $\overparen{EM}$ with $A$ as the center and a radius of 2, whose length is $\frac{\pi}{6}$;
(3) In the plane $AOC$, an arc $\overparen{FN}$ with $A$ as the center and a radius of 2, whose length is $\frac{\pi}{6}$;
(4) In the plane $ABC$, an arc $\overparen{MN}$ with $A$ as the center and a radius of 2, whose length is $\frac{2 \pi}{3}$.
Therefore, the sum of the lengths of the curves is
$$
\frac{\pi}{2}+\frac{\pi}{6}+\frac{\pi}{6}+\frac{2 \pi}{3}=\frac{3 \pi}{2} \text {. }
$$