6. Given $\boldsymbol{m}$ is a non-zero vector, $n$ is a unit vector, $\boldsymbol{m} \neq \boldsymbol{n}$, the angle between $\boldsymbol{m}$ and $\boldsymbol{m}-\boldsymbol{n}$ is $60^{\circ}$, $|\boldsymbol{m}| \in(0, a]$, then the minimum value of $a$ is $\qquad$
Given $\overrightarrow{O A}=\boldsymbol{n}, \overrightarrow{O B}=\boldsymbol{m}$, then $\angle O B A=60^{\circ}$
$\Rightarrow$ The trajectory of point $B$ is the two major arcs $\overparen{O B A}$
(the other segment is below the $x$-axis and not shown), circle $C$ is the circumcircle of an equilateral triangle with side length $O A$, the center of the circle $C\left(\frac{1}{2}, \frac{\sqrt{3}}{6}\right)$, radius $r=\frac{\sqrt{3}}{3}$, thus $|\overrightarrow{O B}|_{\max }=\frac{2 \sqrt{3}}{3}$. Therefore, the minimum value of $a$ is $\frac{2 \sqrt{3}}{3}$.