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Collisions involving multiple particles

  1. Jan 1, 2013 #1
    Is it ever the case in particle accelerators such as the LHC that a collision occurs involving three or more particles (e.g. two particles collide then some of the debris from that collision impacts a third particle)?

    Does the same phenomenon.also occur in cosmic ray collisions (I would imagine it does since that is how air showers work - particles cascading into other particles) - is that right?
  2. jcsd
  3. Jan 2, 2013 #2
    Yes, most definitely. Remember, the LHC collides proton bunches, not just protons. In addition, protons aren't elementary particles and contain many quarks and gluons. Therefore, it is common to have many collisions. Also, even two particle interactions will involve more particles as spectators or as gluons emitted in initial or final state radiation.
  4. Jan 2, 2013 #3


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    Compared to the total number of collisions, it is extremely rare that the products of a proton-proton collision hit another proton.

    The bunches have some 1011 protons and ~20 inelastic collisions per bunch crossing (both numbers vary a bit, but I don't care about a factor of 2 here). In other words, every proton has a probability of ~2*10-10 to collide in a bunch-crossing. The proton bunches are nearly colliding head-on (with a crossing angle of ~100mrad if I remember correctly) to get as many collisions as possible. Collision products can head in different directions, with a lower probability to collide with anything. On the other hand, they can collide with protons from both beams. As very conservative estimate, I will assume that those effects cancel.
    A proton-proton collisions produces something like ~20 new particles - probably an overestimate for an average collision. Therefore, the probability that collision products collide with a third proton is less than 4*10-9.

    With ~4*108 collisions per second, this corresponds to a "collision chain" every 10 seconds. The second collision does not produce anything special - it is a low-energy collision, the trigger will not detect it (or at least not more frequent than regular collisions). With a trigger rate of ~10-5, this corresponds to a recorded event every ~106 seconds or one per day in stable beams. Again, this is an upper estimate. There should be some recorded collision chains, but I don't think they are visible in all the background.
  5. Jan 2, 2013 #4
    Thanks. Here is a follow up question about heavy ion collisions. I understand that in a heavy ion collision you have a "fireball" that expands and cools, does it ever happen that another ion from elsewhere in the beam goes into the fireball while it still exists? Does anything special happen then? Or does the fireball just dissipate too fast for that to ever happen?
  6. Jan 2, 2013 #5

    Vanadium 50

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    The same argument applies.
  7. Jan 3, 2013 #6


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    It is even worse with the fireball.

    - It is extremely short-living - of the order of femtometers (divided by c). To get 3 nuclei colliding at the same place, you would need two nuclei with a distance of <100fm in the beam direction.
    - It is extremely small - maybe ~10fm across. This gives a total volume of ~104fm3 where a third nucleus has to be, neglecting factors of ~.1 to 10.

    Lead bunches have ~108 nuclei inside with a length of ~0.5 ns (15cm) and a minimal diameter of maybe ~10µm. This gives a volume of 1034fm3. As you can see, even with random motion of nuclei the chance to find two nuclei in the same beam so close together is basically zero.

    Even worse: at a distance of 100fm, you have a potential of ~100 MeV. I would be surprised to see unordered motion with 100 MeV in the beam, but I don't know how much we have there.

    Cross-check: lead-lead collisions have a bunch spacing of ~100ns (? - gives 10MHz bunch crossing rate) and a collision rate of ~5kHz in ALICE if I remember correctly. Therefore, each bunch crossing leads to an average of 0.0005 collisions.
    With the same numbers as above, the maximal collisions per bunch crossings are 108*2*(10fm)2/(10µm)2 = 0.01
    A factor of 20 between those values, looks good.
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