LHCb « Benjamin Cantabile in Europe

12/01/2012

: Image via Wikipedia

At the LHC, protons in the beam 1 have the four momentum

$\left(E_{1},p_{1}\right)$

and protons in the beam 2 have the four momentum

$\left(E_{2},p_{2}\right).$

They collide at the collison point at the center of the detectors (ATLAS, CMS, LHCb, ALICE). Since the velocity is very very near to the light velocity $c=1$ , we can safely approximate $v=1$ . This is equivalent to neglect the effect of the rest mass of the proton ( $0.94\left[\mathrm{GeV/c^{2}}\right]$ ) in high energy collisions.

With this approximation,E holds. In the collision center-of-mass frame, the vectorial sum of the three-momenta of the colliding particles is zero

$p_{1}+p_{2} =0(p_{1}\mathrm{\ and\ }p_{2}\mathrm{\ are\ 3-vectors})$

At the LHC, protons of both beams have the same and opposite momentum in the laboratory frame. It may be called asymmetric collider. (A good example of asymmetric collider is KEKB or PEP2 B-factory colliders).

The four momentum (energy and three-momentum) of the colliding two-proton system is

$E_{1}+E_{2} = E + E = 2E$
$p_{1}+p_{2} = E - E = 0$

in the laboratory frame, and we call the total energy 2E as the collision energy.

It is often expressed as

$2E =\sqrt{s}$

by using the Lorentz invariant

$s= (E_1+E_2)^2 - (p_1 + p_2)^2$

Let parton $a$ (quark, anti-quark or gluon) inside the proton 1 have $x_{1}$ ( $0<x_{1}<1$ ) times the energy of the proton 1 ( $E=\sqrt{s}/2$ in the laboratory frame) in the proton 1 momentum direction, and let parton b have $x_{2}$ ( $0<x_{2}<1$ ) times the energy of the proton 2 in the proton 2 momentum direction.