# Proton-proton collisions

I have a question about proton-proton collisions at the LHC. Firstly, the 4-momentum $p^\mu=(E/c,\vec{p})$ can be represented as $p^\mu =(m_T \cosh \Psi, p_T \cos \phi , p_T \sin \phi, m_T c \sinh \Psi)$ where $\Psi$ is the rapidity, $\phi$ is the azimuthal angle and $p_T=\sqrt{p_x^2+p_y^2}$ the magnitude of the momentum in the transverse plane. Then there is $m_T$ which if you use the conservation of 4-momentum $m_T^2=m_0^2+p_T^2/c^2$.

The question is about a collision between two quarks inside the protons which create a new particle with mass $m$ and rapidity $\Psi$, known. The protons in the beam have 4-momentum $p^\mu_1=(E_{\text{beam}}/c,0,0,E_{\text{beam}}/c)$ and $p^\mu_2=(E_{\text{beam}}/c,0,0,-E_{\text{beam}}/c)$ (i.e. travelling in opposite directions) quarks inside the protons carry the fractions $\xi_1, \xi_2$ of their respective proton's momentum.

Can I assume here that the form of the 4-momentum for a given quark is $p^\mu_{\text{q1}}=\xi_1 p_1^\mu$? It seems that there could be a way that the quark could have transverse momentum which cancels with the other quarks within the system giving a net transverse momentum of zero?

If I proceed as above (no transverse momentum) the invariant mass squared of the the quark-quark system before the collision is $m^2=4E_{\text{beam}}\xi_1\xi_2/c^2$ if this gives rise to a new particle with mass and rapidity known, can I constrain either of the $\xi$s? If the quarks before the collision had $p_T=0$ is this necessarily true for the new particle? Would this mean that $4E_{\text{beam}}\xi_1\xi_2/c^2=m_T^2c^2-p_T^2=P^2$?

• Some reactions (notably Drell-Yan to muons) can be used to measure the distributions of the $\xi$s with relative little background. This leads to Parton Distribution Functions and generalizations thereof. But is should be clear that there are ambiguities in any given measurement (if nothing else which parton should be labeled 1 and which 2). Apr 22, 2013 at 17:45
• In terms of the dynamics, the protons are initially approaching each other with opposite 3-momentum, would it be reasonable to assume that the quark inside the proton, if it has that fraction of the 4-momentum, to actually have 4-momentum $p_{\text{q}}^\mu=\xi p^\mu$? I can see the reasoning behind the 3-momentum being proportional in this way since for high energy beams the transverse 3-momentum must be negligibly small in comparison to the z-direction. Apr 22, 2013 at 18:11
• Yes but the only information you get---no matter how good your detector---relates to the combined properties of two partons. The sum of their transverse momenta. The difference of the momentum fraction. The sum of their strangness or charm or... And so on. Even in a result with no neutrinos (or neutral exotica) you have no way of know which parton carried any transverse momentum you see. That's the labeling problem. Apr 22, 2013 at 18:17
• I am trying to show that $\xi_1=\exp(y) mc^2/2E_{\text{beam}}$ is this the right direction to be going? It doesn't seem to me that I will be able to eliminate the $\xi_2$ because, like you said, the 4-momentum of the combined system of quarks before the collision has them $p^\mu = E_{\text{beam}}/c(\xi_1+\xi_2,0,0,\xi_1-\xi_2)$, I just have the sums of the energies and the difference of the 3-momenta. It looks like I can get cancellation by taking the Minkowski scalar product with some 4-vector Apr 22, 2013 at 18:33

(2) Can one ever reconstruct an LHC event to the extent that one can determine/constrain the momentum of each parton in the collision? I'm not sure - could one measure the final state energy to find $\xi_1+\xi_2$ and longitudinal momentum to find $\xi_1-\xi_2$? I don't know if the required precision would be possible.
(3) What is the relation between the invariant mass of the initial and final system? $$p^2 \approx 4 E^2\xi_1\xi_2 = (\sum_i p^\mu_i)^2$$ where the sum runs over all visible and invisible products, rather than just the new particle, which won't be produced on its own. The products (including the new particle) needn't each have $p_T=0$, but $\sum_i p_T^i=0$.
NB your $m_T$ is more commonly called $E_T$, the transverse energy, and $m_T$ is reserved for transverse mass.