Question 5.4: The Maximum Transverse Momentum (a) Determine the maximum tr......
The Maximum Transverse Momentum
(a) Determine the maximum transverse momentum of a parton occurring in the final state if the photon momentum before the collision is given in the Breit system (see Exercise 3.6) by
q_{\mu}=(0 ; 0,0,-Q) (1)
and the initial parton momentum in the Breit system is
p_{\mu}=(p ; 0,0, p). (2)
(b) Verify the relation \nu_{\max }=s / 2 and investigate the kinematical region accessible at the HERA collider.
(a) We first assume that two partons escape after the collision. These partons carry momenta
\begin{aligned} p_{1}^{\prime} & =\left(\frac{p}{2} ; {p}_{\perp}, \frac{p-Q}{2}\right) \\ p_{2}^{\prime} & =\left(\frac{p}{2} ;-{p}_{\perp}, \frac{p-Q}{2}\right) & (3) \end{aligned}
Obviously the four-momentum is conserved:
q+p=p_{1}^{\prime}+p_{2}^{\prime}. (4)
Furthermore squaring the parton momentum p_{1}^{\prime} yields
\begin{aligned} \left(p_{1}^{\prime}\right)^{2} & =0=\left(\frac{p}{2}\right)^{2}-\left({p}_{\perp}\right)^{2}-\left(\frac{p-Q}{2}\right)^{2} \\ & =\left(\frac{p}{2}\right)^{2}-\left({p}_{\perp}\right)^{2}-\left(\frac{p}{2}\right)^{2}+\frac{2 p \cdot Q}{4}-\left(\frac{Q}{2}\right)^{2} \\ & =-\left({p}_{\perp}\right)^{2}+\frac{p \cdot Q}{2}-\left(\frac{Q}{2}\right)^{2}, & (5) \end{aligned}
because it is assumed to be of rest mass zero, i.e. p_{1}^{\prime 2}=0. Introducing the momentum fraction x=Q^{2} /(2 p \cdot Q), one gets
-\left({p}_{\perp}\right)^{2}-\left(\frac{Q}{2}\right)^{2}+\frac{Q^{2}}{4 x}=0 (6)
or
\left|{p}_{\perp}\right|=\frac{Q}{2} \sqrt{\frac{1-x}{x}}. (7)
For more than two partons in the final state this is the maximum transverse momentum that can be achieved.
(b) According to the definition (5.11), \nu=q \cdot p_{p}, we have in the rest system of the nucleon
\begin{aligned} \nu & \stackrel{\text { R.S. }}{=}\left(E_{e}-E_{e}^{\prime}\right) M_{p}, \\ \nu_{\max } & =E_{e} M_{p} & (8) \end{aligned}
For the center-of-mass energy
\begin{aligned} & s=\left(p_{p}+p_{e}\right)^{2}=M_{p}^{2}+2 p_{p} \cdot p_{e}+m_{e}^{2} \sim 2 p_{p} \cdot p_{e} \\ & \stackrel{\text { R.S. }}{=} 2 E_{e} M_{p}=2 \nu_{\max }. & (9) \end{aligned}
From this we get for x_{\min }
x_{\min }=\frac{Q^{2}}{2 \nu_{\max }} \simeq \frac{Q^{2}}{s}. (10)
For HERA kinematics E_{e}=30 \mathrm{GeV}, E_{p}=820 \mathrm{GeV}, so that
s \simeq e E_{e} E_{p} \simeq 10^{5} \mathrm{GeV}^{2}. (11)
Then we get for Q^{2} \sim 5 \mathrm{GeV} a minimal value x_{\min } \sim 10^{-4}.