4. In $\triangle A B C$, the sides opposite to $\angle A$, $\angle B$, and $\angle C$ are denoted as $a$, $b$, and $c$ respectively. If $b^{2}=a^{2}+c^{2}-a c$, and $c-a$ equals the height $h$ from vertex $A$ to side $A C$, then $\sin \frac{C-A}{2}=$ $\qquad$
4. $\frac{1}{2}$.

From $b^{2}=a^{2}+c^{2}-a c$, using the cosine rule, we know $\angle B=60^{\circ} \Rightarrow \angle A+\angle C=120^{\circ}$.

Also, $h=c-a=\frac{h}{\sin A}-\frac{h}{\sin C}$, so
$$
\begin{array}{l}
\sin C-\sin A=\sin C \cdot \sin A \\
\Rightarrow 2 \cos \frac{C+A}{2} \cdot \sin \frac{C-A}{2} \\
\quad=-\frac{1}{2}(\cos (C+A)-\cos (C-A)) .
\end{array}
$$

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