Starting from (6.51), show that if quarks had spin 0, $F_2(x)$ would still be given by $(9.13)$ but that $F_1(x)=0$ and hence $\sigma_T=0$.
Thus, in contrast to (9.26), spin-0 quarks would yield $\sigma_T / \sigma_L=0$. We can understand this difference by glancing at Fig. 9.4, which shows the head-on collision between the quark and the virtual photon. By conservation of $J_z$ (with $z$ along p), we see that a spin- 0 quark cannot absorb a photon of helicity $\lambda= \pm 1$, so $\sigma_T=0$. Suppose now the quark has spin $\frac{1}{2}$. We recall that its helicity is conserved in a high-energy interaction (see Section 6.6). This can only be achieved by a $\lambda= \pm 1$ photon; hence, $\sigma_L \rightarrow 0$ in this case.