In the center-of-mass frame of the parton process $\gamma^* \mathrm{q}_1 \rightarrow \mathrm{q}_2 \mathrm{~g}$ of Fig. 10.7, show that
$$
\begin{aligned}
& \hat{s}=2 k^2+2 k q_0-Q^2=4 k^{\prime 2}, \\
& \hat{t}=-Q^2-2 k^{\prime} q_0+2 k k^{\prime} \cos \theta=-2 k k^{\prime}(1-\cos \theta), \\
& \hat{u}=-2 k k^{\prime}(1+\cos \theta),
\end{aligned}
$$
where $k, k^{\prime}$ are the magnitudes of the center-of-mass momenta $\mathbf{k}, \mathbf{k}^{\prime}$. Note that for the virtual photon, $q_0^2=k^2-Q^2$. A useful result is
$$
4 k k^{\prime}=-\hat{t}-\hat{u}=\hat{s}+Q^2 \text {. }
$$
The interesting quantity is the transverse momentum of the outgoing quark, $p_T=k^{\prime} \sin \theta$. Show that
$$
p_T^2=\frac{\hat{s} \hat{t} \hat{u}}{\left(\hat{s}+Q^2\right)^2}
$$
or, in the limit of small-angle scattering, $-\hat{i} \ll \hat{s}$, that
$$
p_T^2=\frac{\hat{s}(-\hat{t})}{\hat{s}+Q^2} .
$$
Further, show that for small scattering angles ( $\cos \theta \simeq 1$ ),
$$
d \Omega=\frac{4 \pi}{s} d p_T^2 .
$$