8. 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 height $h$ from vertex $A$ to side $AC$, then $\sin \frac{C-A}{2}=$ $\qquad$ .
$$
\begin{array}{l}
A, B, C \text{ form an arithmetic sequence } \Rightarrow C+A=120^{\circ}, \\
c-a=h=c \sin A \Rightarrow \sin C-\sin A=\sin C \sin A \\
\Rightarrow 2 \cos \frac{C+A}{2} \sin \frac{C-A}{2}=-\frac{1}{2}[\cos (C+A)-\cos (C-A)] \\
\Rightarrow \sin \frac{C-A}{2}=-\frac{1}{2}\left[-\frac{1}{2}-\left(1-2 \sin ^{2} \frac{C-A}{2}\right)\right]=\frac{3}{4}-\sin ^{2} \frac{C-A}{2} \\
\Rightarrow 4 \sin ^{2} \frac{C-A}{2}+4 \sin \frac{C-A}{2}-3=0 \Rightarrow \sin \frac{C-A}{2}=\frac{1}{2} .
\end{array}
$$