Derive (11.19). Also show that the angle $\theta$ between the $\mathrm{q}$ and $\bar{q}$ directions in Fig. 11.7 is determined by the relation
$$
x_{\bar{q}}=\frac{2\left(1-x_q\right)}{2-x_q-x_q \cos \theta} .
$$
Let us now compute the cross section corresponding to Fig. 11.7. For this particular graph, the $\overline{\mathrm{q}}$ emits a softer gluon, so that
$$
x_q \geq x_{\bar{q}} \geq x_g \text {. }
$$
The most obvious experimental signature of gluon emission is that the $\mathrm{q}$ and $\overline{\mathrm{q}}$ are no longer produced back to back. The $\overline{\mathrm{q}}$ is produced with a transverse momentum fraction $x_T$ relative to the direction of the quark. The relevant observable quantity is therefore $d \sigma / d x_T^2$. This cross section can readily be obtained using the Altarelli-Parisi-Weizsäcker-Williams technique of Section 10.9. Referring to Fig. 11.8, we obtain
$$
\frac{d \sigma}{d x_{\bar{q}} d p_T^2}=\sigma\left(\mathrm{e}^{-} \mathrm{e}^{+} \rightarrow \mathrm{q} \bar{q}\right) \gamma_{\tilde{q} \tilde{q}}\left(x_{\bar{q}}, p_T^2\right),
$$
see (10.57), where $\sigma$ gives the probability for producing a $q \bar{q}$ pair and $\gamma_{\tilde{q} \bar{q}}$ is the probability that the $\bar{q}$ subsequently emits a gluon with a fraction $\left(1-x_{\bar{q}}\right)$ of its momentum and a transverse momentum $\left|p_T\right|$. From (11.3) and (10.58), we have
$$
\begin{gathered}
\sigma\left(\mathrm{e}^{-} \mathrm{e}^{+} \rightarrow \mathrm{q} \bar{q}\right)=\frac{4 \pi \alpha^2}{Q^2} e_q^2, \\
\gamma_{\bar{q} \bar{q}}\left(x_{\bar{q}}, p_T^2\right)=\gamma_{q q}\left(x_{\bar{q}}, p_T^2\right)=\frac{\alpha_s}{2 \pi} \frac{1}{p_T^2} P_{q q}\left(x_{\bar{q}}\right) .
\end{gathered}
$$
On substitution of (11.23) into (11.22), we find
$$
\frac{1}{\sigma} \frac{d \sigma}{d x_{\tilde{q}} d x_T^2}=\frac{\alpha_s}{2 \pi} \frac{1}{x_T^2} P_{q q}\left(x_{\tilde{q}}\right) .
$$
To calculate $d \sigma / d x_T^2$, it remains to integrate over all possible $\overline{\mathrm{q}}$ energy fractions $x_{\bar{q}}$. Using (10.31) for $P_{q q}$, we obtain
$$
\frac{1}{\sigma} \frac{d \sigma}{d x_T^2}=2 \frac{\alpha_s}{2 \pi} \frac{1}{x_T^2} \int_{\left(x_{\bar{q}}\right)_{\min }}^{\left(x_{\bar{q}}\right)_{\max }} d x \frac{4}{3}\left(\frac{1+x^2}{1-x}\right) .
$$
An extra factor of 2 is included to allow for the equally probable diagram with $\mathrm{q} \leftrightarrow \overline{\mathrm{q}}$.
The integrand in (11.25) diverges when $x_{\bar{q}} \rightarrow 1$. The kinematic situation where $x_{\bar{q}}$ reaches its maximum value is therefore of special interest. From (11.21), we see that the largest value allowed for $x_{\bar{q}}$ is
$$
x_{\bar{q}}=x_q .
$$
This value can be approached if we make the emitted gluon as soft as possible, which, remembering that $x_T$ is fixed, occurs when
$$
x_g=x_T .
$$
This kinematic configuration is shown in Fig. 11.9. Thus, from (11.17), we have
$$
\left(x_q\right)_{\min }=\left(x_{\bar{q}}\right)_{\max } \simeq 1-\frac{x_T}{2} .
$$
Momentum is conserved for $\theta$ or $x_T$ not too large, but $x_g$ is only exactly equal to $x_T$ when $\theta \rightarrow 0$. Such approximations are implicit in the Altarelli-Parisi calculation of (11.25), see Chapter 10. Using (11.28), (11.25) becomes
$$
\frac{1}{\sigma} \frac{d \sigma}{d x_T^2}=\frac{8 \alpha_s}{3 \pi} \frac{1}{x_T^2} \int_{\left(x_{\bar{q}}\right)_{\min }}^{1-\frac{1}{2} x_T} \frac{d x}{1-x},
$$
where we have approximated $1+x^2$ by 2 . Finally, omitting all but the leading logarithmic term, we obtain
$$
\frac{1}{\sigma} \frac{d \sigma}{d x_T^2} \simeq \frac{4 \alpha_s}{3 \pi} \frac{1}{x_T^2} \log \left(\frac{1}{x_T^2}\right) .
$$