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Kinematics and Cross Section for a Two-body Process

In this section we begin the description of kinematics selection and cross-section calculation. The example is for the case of a $2 \to 2$ process, with final-state masses assumed to be vanishing. Later on we will expand to finite fixed masses, and to resonances.

Consider two incoming beam particles in their c.m. frame, each with energy $E_{\mathrm{beam}}$. The total squared c.m. energy is then $s = 4 E_{\mathrm{beam}}^2$. The two partons that enter the hard interaction do not carry the total beam momentum, but only fractions $x_1$ and $x_2$, respectively, i.e. they have four-momenta

$\displaystyle p_1$ $\textstyle =$ $\displaystyle E_{\mathrm{beam}}(x_1; 0, 0, x_1) ~,$  
$\displaystyle p_2$ $\textstyle =$ $\displaystyle E_{\mathrm{beam}}(x_2; 0, 0, -x_2) ~.$ (67)

There is no reason to put the incoming partons on the mass shell, i.e. to have time-like incoming four-vectors, since partons inside a particle are always virtual and thus space-like. These space-like virtualities are introduced as part of the initial-state parton-shower description, see section [*], but do not affect the formalism of this section, wherefore massless incoming partons is a sensible ansatz. The one example where it would be appropriate to put a parton on the mass shell is for an incoming lepton beam, but even here the massless kinematics description is adequate as long as the c.m. energy is correctly calculated with masses.

The squared invariant mass of the two partons is defined as

\hat{s} = (p_1 + p_2)^2 = x_1 \, x_2 \, s ~.
\end{displaymath} (68)

Instead of $x_1$ and $x_2$, it is often customary to use $\tau$ and either $y$ or $x_{\mathrm{F}}$:
$\displaystyle \tau$ $\textstyle =$ $\displaystyle x_1 x_2 = \frac{\hat{s}}{s} ~;$ (69)
$\displaystyle y$ $\textstyle =$ $\displaystyle \frac{1}{2} \ln \frac{x_1}{x_2} ~;$ (70)
$\displaystyle x_{\mathrm{F}}$ $\textstyle =$ $\displaystyle x_1 - x_2 ~.$ (71)

In addition to $x_1$ and $x_2$, two additional variables are needed to describe the kinematics of a scattering $1 + 2 \to 3 + 4$. One corresponds to the azimuthal angle $\varphi$ of the scattering plane around the beam axis. This angle is always isotropically distributed for unpolarized incoming beam particles, and so need not be considered further. The other variable can be picked as $\hat{\theta}$, the polar angle of parton 3 in the c.m. frame of the hard scattering. The conventional choice is to use the variable

\hat{t} = (p_1 - p_3)^2 = (p_2 - p_4)^2 =
- \frac{\hat{s}}{2} (1 - \cos \hat{\theta}) ~,
\end{displaymath} (72)

with $\hat{\theta}$ defined as above. In the following, we will make use of both $\hat{t}$ and $\hat{\theta}$. It is also customary to define $\hat{u}$,
\hat{u} = (p_1 - p_4)^2 = (p_2 - p_3)^2 =
- \frac{\hat{s}}{2} (1 + \cos \hat{\theta}) ~,
\end{displaymath} (73)

but $\hat{u}$ is not an independent variable since
\hat{s} + \hat{t} + \hat{u} = 0 ~.
\end{displaymath} (74)

If the two outgoing particles have masses $m_3$ and $m_4$, respectively, then the four-momenta in the c.m. frame of the hard interaction are given by

\hat{p}_{3,4} = \left(
\frac{\hat{s} \pm (m_3^2-m_4^2)}{2\sq...
...{\sqrt{\hat{s}}}{2} \, \beta_{34} \cos \hat{\theta}
\right) ~,
\end{displaymath} (75)

\beta_{34} = \sqrt{ \left( 1 - \frac{m_3^2}{\hat{s}} -
...)^2 - 4 \, \frac{m_3^2}{\hat{s}} \,
\frac{m_4^2}{\hat{s}} } ~.
\end{displaymath} (76)

Then $\hat{t}$ and $\hat{u}$ are modified to
\hat{t}, \hat{u} = - \frac{1}{2} \left\{ ( \hat{s} - m_3^2 - m_4^2 )
\mp \hat{s} \, \beta_{34} \cos \hat{\theta} \right\} ~,
\end{displaymath} (77)

\hat{s} + \hat{t} + \hat{u} = m_3^2 + m_4^2 ~.
\end{displaymath} (78)

The cross section for the process $1 + 2 \to 3 + 4$ may be written as

$\displaystyle \sigma$ $\textstyle =$ $\displaystyle \int \int \int \d x_1 \, \d x_2 \, \d\hat{t} \,
f_1(x_1, Q^2) \, f_2(x_2, Q^2) \, \frac{\d\hat{\sigma}}{\d\hat{t}}$  
  $\textstyle =$ $\displaystyle \int \int \int \frac{\d\tau}{\tau} \, \d y \, \d\hat{t} \,
x_1 f_1(x_1, Q^2) \, x_2 f_2(x_2, Q^2) \,
\frac{\d\hat{\sigma}}{\d\hat{t}} ~.$ (79)

The choice of $Q^2$ scale is ambiguous, and several alternatives are available in the program. For massless outgoing particles the default is the squared transverse momentum

Q^2 = \hat{p}_{\perp}^2 = \frac{\hat{s}}{4} \sin^2\hat{\theta} =
\frac{\hat{t}\hat{u}}{\hat{s}} ~,
\end{displaymath} (80)

which is modified to
Q^2 = \frac{1}{2} (m_{\perp 3}^2 + m_{\perp 4}^2) =
...m_3^2 + m_4^2) +
\frac{\hat{t} \hat{u} - m_3^2 m_4^2}{\hat{s}}
\end{displaymath} (81)

when masses are introduced in the final state. The mass term is selected such that, for $m_3 = m_4 = m$, the expression reduces to the squared transverse mass, $Q^2 = \hat{m}_{\perp}^2 = m^2 + \hat{p}_{\perp}^2$. For cases with space-like virtual incoming photons, of virtuality $Q_i^2 = - m_i^2 = \vert p_i^2\vert$, a further generalization to
Q^2 = \frac{1}{2} (Q_1^2 + Q_2^2 + m_3^2 + m_4^2) + \hat{p}_{\perp}^2
\end{displaymath} (82)

is offered.

The $\d\hat{\sigma}/\d\hat{t}$ expresses the differential cross section for a scattering, as a function of the kinematical quantities $\hat{s}$, $\hat{t}$ and $\hat{u}$, and of the relevant masses. It is in this function that the physics of a given process resides.

The performance of a machine is measured in terms of its luminosity $\cal L$, which is directly proportional to the number of particles in each bunch and to the bunch crossing frequency, and inversely proportional to the area of the bunches at the collision point. For a process with a $\sigma$ as given by eq. ([*]), the differential event rate is given by $\sigma {\cal L}$, and the number of events collected over a given period of time

N = \sigma \, \int {\cal L} \, \d t ~.
\end{displaymath} (83)

The program does not calculate the number of events, but only the integrated cross sections.

next up previous contents
Next: Resonance Production Up: Process Generation Previous: Equivalent photon flux in   Contents
Stephen Mrenna 2007-10-30