4. In $\triangle A B C$, it is known that the three interior angles $\angle A, \angle B, \angle C$ form an arithmetic sequence, with their opposite sides being $a, b, c$ respectively, and $c-a$ equals the height $h$ from vertex $A$ to side $AC$. Then $\sin \frac{C-A}{2}=$
4. $\frac{1}{2}$.

Notice that
$$
\begin{array}{l}
h=c-a=\frac{h}{\sin A}-\frac{h}{\sin C} \\
\Rightarrow \sin C-\sin A=\sin C \cdot \sin A .
\end{array}
$$

Then $2 \cos \frac{C+A}{2} \cdot \sin \frac{C-A}{2}$
$$
=\frac{1}{2}[\cos (C-A)-\cos (C+A)] \text {. }
$$

Since $\angle A, \angle B, \angle C$ form an arithmetic sequence, therefore,
$$
\angle B=60^{\circ}, \angle A+\angle C=120^{\circ} \text {. }
$$

Thus $\sin ^{2} \frac{C-A}{2}+\sin \frac{C-A}{2}-\frac{3}{4}=0$.
Solving gives $\sin \frac{C-A}{2}=\frac{1}{2}$ (negative value is discarded).