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
$$
4 k k^{\prime}=-\hat{t}-\hat{u}=\hat{s}+Q^{2}
$$
The interesting quantity is the transverse momentum of the outgoing quark, $p_{T}=k^{\prime} \sin \theta$. Show that
$$
p_{T}^{2}=\frac{\operatorname{sit} 2}{\left(s+Q^{2}\right)^{2}}
$$
or, in the limit of small-angle scattering. $-\hat{t} \approx \hat{s}$, that
$$
p_{T}^{2}=\frac{s(-\hat{\imath})}{s+Q^{2}} .
$$
Further, show that for small scattering angles $(\cos \theta \simeq 1)$,
$$
d \Omega=\frac{4 \pi}{s} d p_{T}^{2}
$$
$\underset{\rightarrow q_{2} g .}{\text { Figy } 10.7 \text { Center-of-mass frame for } \gamma^{*} q_{1}}$
10.4 The Gluon Emission Cross Section
213
It is clear from $(10.17)$ that at high energy ( $\vec{s}$ large), the $\gamma^{*} \mathrm{q} \rightarrow \mathrm{qg}$ cross section peaks as $-\hat{i} \rightarrow 0$. Referring back to Fig. $4.8$ and the related discussion, we see that this is due to quark exchange in the $t$ channel. We can therefore approximate the cross section by its forward peak. For forward scattering, we find, from (10.26) and (4.35), that
$$
\frac{d \dot{\sigma}}{d p_{T}^{2}}=\frac{1}{16 \pi s^{2}} \overline{|S M|^{2}}
$$