What happens to free quarks in particle accelerators?

  • Thread starter Thread starter Uchuujin_7
  • Start date Start date
  • Tags Tags
    Particle Quarks
Uchuujin_7
Messages
4
Reaction score
0
I'm new to quantium physics but taking an interesting and trying to read more every day.

I have a bit of a silly questions - in particle accelerators when collisions with protons or neutrons free quarks, what is the ultimate fate of the free quarks? It's my understanding that a quark isn't content to just exist alone and there isn't really an abundance of free quarks for it to join with. Does it decay into other elementry particles like photons?
 
Physics news on Phys.org
They undergo a process termed hadronisation. Basically, you're not going to get a singular quark produced, as this would defy basic conservation laws. If we look at a general case where a quark and an anti-quark pair are moving away from each other (in the center of mass frame), due to the strong force increasing with distance, a potential exists between the pair of quarks. We can think of this 'flux tube' as consisting of lots of gluons. When the potential energy of the flux tube (gained from the kinetic energy of the moving particles) is great enough to create another pair of particles, this happens, the orginal quark pairing up with an antiquark, and the original antiquark pairing up with a quark.

This is a simplified description, but I hope it makes sense.
 
Thanks for the answer. What you are saying makes some sense to me.

When particles collide and free quarks go flying - are they quarks from the anti-/protons or neutrons that collided, or were they created in the energies of the collision?

One of the reasons I'm asking about this is I recently saw some bubble chamber images then went and read about how old bubble chambers worked and I realized that the particles tended to leave a trail and vanish at some point (when their energies decreased enough to not leave a trail anymore.) I wondered what became of them at this point.

Bubble chambers are old though, I'm not sure if at that time they were working with energies high enough for quarks or what the particles were creating the bubble trails.
 
When you talk about particle collisions creating quarks that's completely different to neutron research...

Anyway, the quarks which then hadronise and whose products we see are created 'out of' the energy from the initial matter-antimatter collisions.

Coming to your bubble chamber question, all hadrons are made of quarks, so when we look at tracks due to hadrons interacting / decaying we are looking at particles made of quarks. i.e. http://rd11.web.cern.ch/RD11/rkb/PH14pp/node17.html Shows 8 incoming particles (K mesons), one of which strikes a proton and decays, as can be seen by the branching of the tracks in a number of places.

You are made of up and down quarks, just because you're not accelerated to speeds close to the speed of light doesn't mean there aren't quarks there...

When particles stop leaving a trail, that doesn't mean stuff still isn't going to happen to them. Particles will continue to decay until they reach a final stable state; just about everything ends up as photons (most things decay to pions, which then usually decay to photon pairs).

What you should realize is that there is no such thing as a realisable free quark - they've never been observed at the energies we can produce.
 
Last edited by a moderator:
Not to mention fundamentally experimentally impossible. They would react with nuclei in the detector/vacuum long before it could be ever detected, even if you could wrench them apart for long enough (say with a Planckian accelerator the size of our galaxy).
 
Haelfix said:
(say with a Planckian accelerator the size of our galaxy).

Now that is an accelerator I'd like to see...
 
Thread 'Why is there such a difference between the total cross-section data? (simulation vs. experiment)'
Well, I'm simulating a neutron-proton scattering phase shift. The equation that I solve numerically is the Phase function method and is $$ \frac{d}{dr}[\delta_{i+1}] = \frac{2\mu}{\hbar^2}\frac{V(r)}{k^2}\sin(kr + \delta_i)$$ ##\delta_i## is the phase shift for triplet and singlet state, ##\mu## is the reduced mass for neutron-proton, ##k=\sqrt{2\mu E_{cm}/\hbar^2}## is the wave number and ##V(r)## is the potential of interaction like Yukawa, Wood-Saxon, Square well potential, etc. I first...
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...
Back
Top