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The proton anti-proton annihilation at Wikipedia describes the process as not as simple as the electron-positron annihilation. Especially it states:

Reactions in which proton-antiproton annihilation produces as many as nine mesons have been observed, while production of thirteen mesons is theoretically possible. The generated mesons leave the site of the annihilation at moderate fractions of the speed of light, and decay with whatever lifetime is appropriate for their type of meson.

Now the proton/anti-proton contain a sum of 6 quarks/anti-quarks while 13 mesons contain 26 quarks/anti-quarks.

How does the theory explain the massive increase in quarks/anti-quarks?

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    $\begingroup$ Lots of ways to approach this, but I am surprised at the lack of a easily googlable explanation of "associated production" which is one process that adds two to the quark/anti-quark count (while leaving the baryon number unaffected, of course). Not time to write it just now. $\endgroup$ – dmckee --- ex-moderator kitten Feb 22 '18 at 19:16
  • $\begingroup$ Is it possible you might be less confused if you stated, instead, "3 quarks annihilate with 3 antiquarks to produce 13 quarks and 13 antiquarks"? You recall, in Mev's, very crudely, the masses are 1000+ 1000 > 13 x 140, and the quark masses of 2.3 & 4.8 never mattered, for the light quarks? $\endgroup$ – Cosmas Zachos Feb 22 '18 at 19:48
  • $\begingroup$ Thanks for your answers, the mass conversion now makes sense to me, but I still do not understand the charge. $\endgroup$ – Johannes Maria Frank Feb 23 '18 at 11:59
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In annihilation it is the quantum numbers that add up to zero, baryon number is zero, as strangeness etc. After annihilation, the quarks have to be produced in quark antiquark pairs, which then will combine appropriateley to make hadrons. These can be produced as long as the quantum numbers will be adding up to zero.( for example, if a strange quark is produced an antistrange must accompany it).

It is the energy limitations that will give a limit to the number of hadrons that can be possibly produced.The limits given in the quote in number of possible mesons must come from considering annihilation at rest , or low energies. All the links in the wiki quote state "low energy".

Edit after comment:

The composite particles, protons and antiprotons, proceed in annihilation by the strong interactions that are allowed , and the situation is not simple. Here is a diagram of quark interactions :

quarks

This diagram can also be read time going from left to right, where it depicts the annihilation of a quark antiquark pair to a gluon and the gluon generates a different quark antiquark pair, because only color has to be conserved in the peculiar way of the laws of strong interactions. The other quantum numbers, strangeness, bottomness, topnes, charmness, upness and downness disappear in the interaction. This is in contrast to reading the feynman diagram time from bottom to top.

You might think that the number of pairs is conserved, BUT you are counting without taking into account the nature of strong interactions. Gluons, in contrast to photons, self interact. The intermediary gluon is carrying a lot of energy in proton antiproton annihilations,( about two thousand Mev, to the order of up,down quark masses which are of order of mev) and there is a very large probability that it will radiate away to many gluons which gluons will finally manifest into new quark antiquark pairs. The pair numbers are limited by the available energy and the probabilities of the quantum mechanical calculations.

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  • $\begingroup$ The quark is an elementary particle. How can it be produced out of nothing? Also inside a meson a quark and an anti-quark do not have a problem staying together, at least for several 100 ns. What changes to make the proton anti-proton situation more unstable, or do they also allow for some time to stay together until decay? $\endgroup$ – Johannes Maria Frank Feb 22 '18 at 23:54
  • $\begingroup$ They are not produced out of nothing , they are produced by the summed input energy of the of the masses of proton and antiproton, about, almost 2000Mev. quarks and antiquarks have masses of order of MeV. The input energy has no particle quantum numbers after the annihilation of quarks with antiquarks, ( a gluon at least is produced) and is open for any pair production of quark antiquark . $\endgroup$ – anna v Feb 23 '18 at 5:45
  • $\begingroup$ Thanks, that makes sense now, could you explain the same for the charge? $\endgroup$ – Johannes Maria Frank Feb 23 '18 at 11:58
  • $\begingroup$ The charge is also a quantum number, it has to add up to zero for the whole , input proton antiproton, output sum of charges. The way the charges are distributed on the newly created particles depend on the standard model and follow the various symmetries. Have a look at this answer of mine physics.stackexchange.com/questions/387558/… $\endgroup$ – anna v Feb 23 '18 at 13:07
  • $\begingroup$ Thanks @anna-v for the elaborate answer. Since I already asked too many questions in the comments I posted it as a follow up question physics.stackexchange.com/questions/389172/… $\endgroup$ – Johannes Maria Frank Feb 27 '18 at 23:26
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Have a look at my answer to Conversion of energy to matter? for an explanation of how particles are created in nuclear reactions. You say in a comment to Anna's answer:

The quark is an elementary particle. How can it be produced out of nothing?

But particles are created routinely in particle scattering events. This is exactly how new particles are discovered at particle colliders.

The particles are not created from nothing of course. They are created from energy in accordance with Einstein's famous equation $E=mc^2$. In a collider the energy is supplied from the kinetic energy of the colliding particles. In a proton annihilation the energy comes from the binding energy of the proton.

In a proton the masses of the three valence quarks are around 2MeV for the up quark and 4.8MeV for the down quark, which is only around 1% of the 1GeV mass of the proton. The other 99% of the proton mass comes from the binding energy. Mesons have a much lower binding energy, so the binding energy of the proton can be converted into the creation of the valence quarks of the mesons. If you add up the total energy of the proton and antiproton. i.e. rest mass energy and kinetic energy, and compare it to the total energy of the mesons, photons etc produced in the annihilation then you will discover that the two energies are the same. Hardly surprising since we know energy is conserved.

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  • $\begingroup$ Ah, and every process in physics is reversible, so energy turns into mass. Does something similar also apply to the charge? It is energy that creates the mass, but the sum of the charges is unequal as well. $\endgroup$ – Johannes Maria Frank Feb 23 '18 at 11:57
  • $\begingroup$ @JohannesMariaFrank net charge is always conserved. Charge is not interchangeable with anything in way that mass and energy can be interchanged. Note this is net charge so for example an electron and positron have a net charge of zero, which is why they can annihilate to two uncharged photons. $\endgroup$ – John Rennie Feb 23 '18 at 16:50
  • $\begingroup$ Thanks for the answer John. Mass is easy, it can just be converted from energy so if there is a certain amount of energy there are just enough particles created to make up for that energy. But with charge I am not aware of anything like this. 6 quarks with 2 +2/3 and 4 -1/3 charges create 13 particles, with all kind of different charges. While the net stays the same (0) the sum is a different story. How is this explained? $\endgroup$ – Johannes Maria Frank Feb 23 '18 at 18:19
  • $\begingroup$ @JohannesMariaFrank charge can be created from nothing subject to the restriction that the net charge remains constant. So for example given enough energy you can create electron-positron pairs indefinitely. $\endgroup$ – John Rennie Feb 23 '18 at 18:26
  • $\begingroup$ Thanks @JohnRennie I can't decide which is the better answer. However I have an other question and posted a follow up here: physics.stackexchange.com/questions/389172/… $\endgroup$ – Johannes Maria Frank Feb 27 '18 at 23:28

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