4. In $\triangle A B C$, it is known that the three angles $A, B, C$ form an arithmetic sequence. Suppose the sides opposite to these angles are $a, b, c$, respectively, and $c-a$ equals the altitude $h$ from vertex $A$ to side $AC$. Then $\sin \frac{C-A}{2}=$ $\qquad$
4. Solution: $\frac{1}{2}$ It is easy to see that $h=c-a=\frac{h}{\sin A}-\frac{h}{\sin C}$, simplifying gives $\sin C-\sin A=\sin A \sin C$.
From the given condition $A+C=120^{\circ}$, hence from the above equation we get
$$
2 \sin \frac{C-A}{2} \cos \frac{120^{\circ}}{2}=\frac{1}{2}[\cos (C-A)-
$$
$\left.\cos 120^{\circ}\right]$,
which simplifies to $\sin ^{2} \frac{C-A}{2}+\sin \frac{C-A}{2}-\frac{3}{4}=0$, solving gives $\sin \frac{C-A}{2}=\frac{1}{2}, \sin \frac{C-A}{2}=-\frac{3}{2}$ (discard).