4. In Rt $\triangle A B C$, $\angle A=90^{\circ}, A D \perp B C$ at $D$, the angle bisector of $\angle B$ intersects $A D$ at $M$, and intersects $A C$ at $N$, and $A B-A M=\sqrt{3}-1, B M \cdot B N=2$. Then $\frac{B C-A B}{C N}=$ $\qquad$
4. $\frac{\sqrt{3}}{2}$.

As shown in Figure 6, with point $A$ as the center and $AN$ as the radius, a circle is drawn intersecting $AB$ at point $F$.
$$
\begin{array}{c}
\because \angle 2+\angle 3=\angle 4+\angle 5 \\
=90^{\circ}, \text { and } \angle 3=\angle 4, \\
\therefore \angle 5=\angle 2 . \\
\text { Also, } \because \angle 1=\angle 5,
\end{array}
$$
$\therefore \angle 1=\angle 2$. Therefore, circle $\odot A$ passes through point $M$.
By the secant theorem, we have
$$
\begin{array}{l}
B M \cdot B N=B F \cdot B E=(A B+A E)(A B-A F) \\
=(A B+A N)(A B-A N)=A B^{2}-A N^{2}=2 . \\
\text { Also, } \because A B-A M=A B-A N=\sqrt{3}-1, \\
\therefore A B+A N=\frac{2}{A B-A N}=\sqrt{3}+1 .
\end{array}
$$

Thus, $A B=\sqrt{3}, A N=1$.
In the right triangle $\triangle NAB$,
$$
\because \tan \angle 2=\frac{A B}{A N}=\sqrt{3}, \therefore \angle 2=60^{\circ} \text {. }
$$

Draw $NT \perp BC$, with the foot of the perpendicular at $T$. Then $BT=AB$. Therefore,
$$
\frac{B C-A B}{C N}=\frac{B C-B T}{C N}=\frac{C T}{C N}=\sin \angle 6=\sin 60^{\circ}=\frac{\sqrt{3}}{2} .
$$