Question 5.15: Calculation of the Gluonic Contribution to FL with the Cross......
Calculation of the Gluonic Contribution to F_{L} with the Cross-Section Method
Calculate the contribution of photon-gluon scattering to the longitudinal structure function. Use the cross-section method as in the previous example 5.14.
The necessary diagrams we encounter are shown in the Fig. 5.28.
Let us first collect the necessary definitions from Examples 5.13 and 5.14. The gluonic structure function is given by a convolution
F_{L}^{g}=\int\limits_{x}^{1} \frac{\mathrm{d} z}{z} \mathcal{F}_{L}\left(z, Q^{2}\right) F_{2}^{(0) g}\left(\frac{x}{z}, Q^{2}\right) (1)
with a function \mathcal{F}_{L}\left(z, Q^{2}\right) calculable in perturbation theory and the gluon density F_{2}^{(0) g}\left(x, Q^{2}\right)=x G\left(x, Q^{2}\right). We calculate \mathcal{F}_{L} by projecting onto the partonic scattering tensor W_{\mu \nu} (see (6) and (29) of Example 5.14):
\mathcal{F}_{L}\left(z, Q^{2}\right)=\frac{4 z^{2}}{Q^{2}} p^{\mu} p^{v} W_{\mu \nu}, (2)
where the partonic scattering tensor to the order we are working is given by
W_{\mu \nu}=\frac{1}{(4 \pi)} \frac{1}{(8 \pi)} \int\limits_{0}^{1} \mathrm{~d} y M_{\mu}(2) M_{\nu}^{*}(2), (3)
where we have used the representation (2) and (19) from Example 5.14 for the two-particle phase space (l=2). The kinematics are the same as previously; i.e. for the reaction
\mathrm{g}(p)+\gamma^{*}(p) \rightarrow \overline{\mathrm{q}}(k)+\mathrm{q}\left(p^{\prime}\right) (4)
we choose
\begin{aligned} & p=(|p|, 0,0,|p|), \\ & k=(|k|, 0,|k| \sin \theta,|k| \cos \Theta), \\ & p^{\prime}=(|k|, 0,-|k| \sin \theta,-|k| \cos \Theta), & (5) \end{aligned}
and define the Mandelstam variables as
\begin{aligned} & s=(p+q)^{2}=\frac{Q^{2}}{z}(1-z), \\ & t=(p-k)^{2}=\frac{-Q^{2}}{z}(1-y), \\ & u=\left(p-p^{\prime}\right)^{2}=\frac{-Q^{2}}{z} y, & (6) \end{aligned}
with y=\frac{1}{2}(1+\cos \Theta). Remember also that p^{\prime 2}=p^{2}=0=k^{2}, q^{2}=-Q^{2} and that the Bjorken variable with respect to the parton momentum p is defined as z=Q^{2} / 2 p \cdot q. With these definitions the calculation is straightforward. The amplitude of the process reads
\begin{aligned} M_{\mu}(2)=g & {\left[\bar{u}_{a}\left(p^{\prime}\right) \gamma_{\mu} \frac{\mathrm{i}}{\not p-\not k} \frac{\lambda_{a b}^{A}}{2} \gamma_{\alpha} u_{b}(k)\right.} \left.+\bar{u}_{\alpha}\left(p^{\prime}\right) \gamma_{\alpha} \frac{\lambda_{a b}^{A}}{2} \frac{\mathrm{i}}{\not q-\not k} \gamma_{\mu} u_{b}(k)\right] \cdot \varepsilon^{\alpha} . \end{aligned} (7)
When squaring we find for the color sum
\frac{1}{N^{2}-1} \sum\limits_{A, B, a, b} \frac{\lambda_{a b}^{A}}{2} \frac{\lambda_{b a}^{A}}{2}=\frac{1}{N^{2}-1} \sum\limits_{A, B} \operatorname{tr} \frac{\lambda^{A}}{2} \frac{\lambda^{A}}{2}=\frac{1}{N^{2}-1} \sum\limits_{A, B} \frac{1}{2} \delta^{A B}=\frac{1}{2} . (8)
Together with the sum over quark flavors this gives a factor T_{\mathrm{f}}=\frac{1}{2} N_{\mathrm{f}}. With p^{\prime}-p=q-k the squared amplitude becomes
\begin{aligned} p^{\mu} p^{v} \sum_{\text {spins }} M_{\mu}(2) M_{v}^{*}(2) & \\ =T_{\mathrm{f}} g^{2}\{ & \operatorname{tr}\left[\not p(\not p-\not k) \gamma_{\alpha} \not k \gamma_{\beta}(\not p-\not k) \not p \not p^{\prime}\right] \cdot \frac{1}{(p-k)^{4}} \\ + & \operatorname{tr}\left[\gamma_{\alpha}\left(\not p^{\prime}-\not p\right) \not p \not k \not p\left(\not p^{\prime}-\not p\right) \gamma_{\beta} \not p^{\prime}\right] \cdot \frac{1}{\left(p-p^{\prime}\right)^{4}} & (9) \\ + & \operatorname{tr}\left[\not p(\not p-\not k) \gamma_{\alpha} \not k \not p\left(\not p^{\prime}-\not p\right) \gamma_{\beta} \not p^{\prime}\right] \frac{1}{\left(p-p^{\prime}\right)^{2}(p-k)^{2}} \\ + & \left.\operatorname{tr}\left[\gamma_{\alpha}\left(\not p^{\prime}-\not p\right) \not p \not k \gamma_{\beta}(\not p-\not k) \not p \not p^{\prime}\right] \frac{1}{\left(p-p^{\prime}\right)^{2}(p-k)^{2}}\right\}\left(-g^{\alpha \beta}\right) \\ =T_{\mathrm{f}} g^{2} 2 & \left\{\operatorname{tr}\left[\not p \not k \not k \not k \not p \not p^{\prime}\right] \cdot \frac{1}{t^{2}} \operatorname{tr}\left[\not p^{\prime} \not k \not p \not p^{\prime} \not p^{\prime}\right] \cdot \frac{1}{u^{2}}\right. \\ & \left.-\operatorname{tr}\left[\not p k \not p^{\prime} \not p k \not p^{\prime}\right] \cdot \frac{1}{u t}-\operatorname{tr}\left[\not k \not p \not p^{\prime} \not k \not p \not p^{\prime}\right] \cdot \frac{1}{u t}\right\}. & (10)\end{aligned}
The first two traces vanish due to k^{2}=p^{2}=0 and the second two are identical because \operatorname{tr}\left(\gamma_{\alpha} \gamma_{\beta} \ldots \gamma_{\varrho} \gamma_{\sigma}\right)=\operatorname{tr}\left(\gamma_{\sigma} \gamma_{\varrho} \ldots \gamma_{\beta} \gamma_{\alpha}\right). The remaining two traces give
\begin{aligned} -\operatorname{tr}\left[\not p \not k \not p^{\prime} \not p \not k \not p^{\prime}\right] & =\operatorname{tr}\left[\not p \not k \not p^{\prime} \not k \not p \not p^{\prime}\right] \\ & =\left(2 p \cdot p^{\prime}\right)\left(2 p^{\prime} \cdot k\right) \cdot \operatorname{tr}[\not p \not k] \\ & =2 u \cdot t\left(2 p^{\prime} \cdot k\right)=2 u \cdot t \cdot s \cdot & (11) \end{aligned}
What remains is therefore
p^{\mu} p^{\nu} \sum\limits_{s p i n s} M_{\mu}(2) M_{v}^{*}(2)=T_{\mathrm{f}} g^{2} \cdot 8 s=8 \frac{Q^{2}}{z}(1-z) T_{\mathrm{f}} g^{2}. (12)
Inserting this into (2) we finally find
\begin{aligned} \mathcal{F}_{L}\left(x, Q^{2}\right) & =\frac{g^{2}}{\pi} T_{\mathrm{f}} z(1-z) \\ & =16\left(\frac{\alpha_{\mathrm{s}}}{4 \pi}\right) T_{\mathrm{f}} z(1-z), \qquad (13) \end{aligned}
which inserted into (1) gives the final answer
\begin{aligned} F_{L}^{g}\left(x, Q^{2}\right) & =16\left(\frac{\alpha_{s}}{4 \pi}\right) T_{\mathrm{f}} \int\limits_{x}^{1} \frac{\mathrm{d} z}{z}[z(1-z)] F_{2}^{(0) g}\left(\frac{x}{z}, Q^{2}\right) \\ & =16\left(\frac{\alpha_{\mathrm{s}}}{4 \pi}\right) T_{\mathrm{f}} \int\limits_{x}^{1} \frac{\mathrm{d} y}{y}\left(\frac{x}{y}\left(1-\frac{x}{y}\right)\right) F_{2}^{(0) g}\left(y, Q^{2}\right), & (14) \end{aligned}
which is completely in agreement with our previous result in (5.282).
F_L^g\left(x, Q^2\right)=T_{\mathrm{f}}\left(\frac{\alpha_s}{4 \pi}\right) 16 \int\limits_x^1 \frac{\mathrm{d} y}{y}\left(\frac{x}{y}\right)\left(1-\frac{x}{y}\right) F_2^g(y), (5.282)
