Even if we can't have single quarks in nature because of the charge colour, what would be the result of the collision of two down quarks at high velocities (0,99% c) at high energies, like the ones of LHC proton's collision..?


1 Answer 1


99% of the speed of light generates a Lorentz factor of only

$$ \gamma = \left[ 1 - (.99)^2 \right]^{-1/2} \approx 7 $$

which means that you have only about 14 times the mass of a down-quark to make additional particles. The PDG puts the bare mass of the down quark in the neighborhood of 5 MeV, so $14 \times 5\,\mathrm{MeV} = 70\,\mathrm{MeV}$ isn't enough energy to create any pair except an electron--positrons.

So, here are some possible outcomes (taking into account that you must be doing baryon collisions and just looking at cases where you have collisions between constituent down quarks

  • Elastic collision. Your baryons go in and come out with different moment but otherwise unchanged.
  • Electron-positron pair creation, but the baryons still come out unchanged.
  • (If you have at least on nucleus rather than a bare nucleon in the input state) nuclear excitation.

In other words: nothing much. That is a result of your specifying a very low energy regime.

Some things you don't have enough energy to do:

  • Muon--antimuon creation.
  • Meson creation (not even a $\pi^0$).
  • Nucleon excitation (assuming a nucleon or nuclear beam).
  • Most meson excitations (assuming a meson beam for at least one of the quarks)

BTW--Particle physicists rarely talk about speed in this kind of context, because most speeds approach $c$ (often very closely). Instead we generally talk about energy and/or momentum. The exception when we do talk about speeds is when we are using Cerenkov detectors.

  • $\begingroup$ I were wanting to say at high energies, like the ones of LHC proton's collision.. $\endgroup$
    – Hakim
    Jul 14, 2013 at 21:33
  • $\begingroup$ p=mv and that's why I referred to speed... $\endgroup$
    – Hakim
    Jul 15, 2013 at 0:53
  • 1
    $\begingroup$ Momentum is not linearly related to velocity in a relativistic regime. You can write $p = \gamma m v$ if you insist on that form, but as I said particle physicist don't reach for velocity first to understand kinematics; simply because it is more algebraically complicated to use velocity than it is to simply rely on the four-momentum. $\endgroup$ Jul 15, 2013 at 0:59
  • $\begingroup$ So what would be the result of the collision at high energies..? $\endgroup$
    – Hakim
    Jul 15, 2013 at 13:53
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    $\begingroup$ That is one of many (very many) possible outcomes if there is enough energy. Moreover because of the neutrino we know that you are looking at a weak scattering event at low energy (far below the $Z$ mass), so it is not a very likely one. $\endgroup$ Aug 12, 2013 at 1:05

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