Effective collision energy at LHC

Question

The proton is not a fundamental particle, so in high energy proton-proton collision we actually collide proton's sub-constituents: quark-quark and (mostly) gluon-gluon. LHC operates now at 13 TeV center of mass energy but the effective collision energy is less than 13TeV since the sub-constituents carry only a fraction (given by the pdf) of this energy. My question is how would it look the distribution of this effective energy? What is the mean collision energy?

Answer 1

It depends on the final state as can readily by seen by studying the conservation of 4-momentum. Let's consider the simplest case of two particles or two jets in the final state (a $2\to 2$ event as we call it):

$$x_ap_a + x_b p_b = p_1+p_2,$$

where $a$ and $b$ denotes the protons and $1$ and $2$ denotes the final particles, and of course $x$ denotes the fraction of the proton momentum carried by the interacting partons. Then we take the $z$-axis along the beam axis as usual and for each final particles we introduce the transverse energy:

$$E_T^2 = E^2-p_z^2.$$

Then we can express $E$ and $p_z$ in function of the transverse energy $E_T$ and the rapidity $\eta$,

$$\begin{align} E&=E_T\cosh\eta\\ p_z&=E_T\sinh\eta \end{align}$$

The conservation of energy, of momentum along $z$, and of transverse momentum, in this order, reads:

$$\begin{align} x_a \sqrt{S} + x_b\sqrt{S} &= E_{T1}\cosh\eta_1+E_{T2}\cosh\eta_2,\\ x_a \sqrt{S} - x_b\sqrt{S} &= E_{T1}\sinh\eta_1+E_{T2}\sinh\eta_2,\\ 0 &= \vec{p}_{T1}+\vec{p}_{T2}. \end{align}$$

and therefore the two final particles have the same transverse energy $E_T$, and we get the super classic formula,

$$\begin{align} x_a &= \frac{E_T}{2\sqrt{S}}\left(e^{\eta_1}+e^{\eta_2}\right),\\ x_b &= \frac{E_T}{2\sqrt{S}}\left(e^{-\eta_1}-e^{-\eta_2}\right). \end{align}$$

As you can see, the value of $x_a$ and $x_b$ is fixed by the kinematics. This is not just an academic exercise, as this is the kinematic of the Drell-Yann process, which is very useful, and of 2-jet events. The answer to your question is then obtained by looking how $\sqrt{x_ax_bS}$ varies with $\eta_1$ and $\eta_2$. The distribution of $\eta_1$ and $\eta_2$ depends on the process of course but assuming uniform distributions already give an idea.

Further notes: for a $2\to 3$ event, the kinematic constraint loosen a bit, and $x_a$ and $x_b$ are allowed to vary within a range. I am talking about an inclusive cross section here of course, where one of the particle is integrated away. For an exclusive cross section, $x_a$ and $x_b$ are also fixed by the kinematic. The same analysis as above gives the values:

$$\begin{aligned} x_a&=\sum_i \frac{E_{Ti}}{2\sqrt{S}}e^{\eta_i},\\ x_b&=\sum_i \frac{E_{Ti}}{2\sqrt{S}}e^{-\eta_i}, \end{aligned}$$

a formula which is valid for any number of final particles (the sum is over those).

Answer 2

This is a general question which can only be asked for specific interaction, and this is done using monte-carlo simulations which use the parton distribution functions of the proton as far as they have been ascertained by experiments.

See for example this program:

HERAFitter is an open-source package that pro- vides a framework for the determination of the parton distri- bution functions (PDFs) of the proton and for many differ- ent kinds of analyses in Quantum Chromodynamics (QCD). It encodes results from a wide range of experimental measurements in lepton-proton deep inelastic scattering and proton-proton (proton-antiproton) collisions at hadron colliders. These are complemented with a variety of theoretical options for calculating PDF-dependent cross section predictions corresponding to the measurements. The framework covers a large number of the existing methods and schemes used for PDF determination.

Once one has the PDFs one can simulate the center of mass scattering according to the probabilities of interaction and the possible fourvectors, in a monte carlo simulation. This will give the maximum possible energies of parton parton collisions.The mean collision energy will depend on the particular interaction, as pdf's are a function of x and Q , so it is not a simple number derivable from the energy of the beams.

See this talk.