Pions - how they interact with Baryons

[JPC]

hey, i was told that in a Atom's Nucleus, Nucleons keep changing from Proton state to Neutron state , and vice versa , by the means of Pions exchange.

But how does this works

I know :
p+ : u u d
n : u d d

Positive pion (that turns neutron into proton ?) : (d) u
u d d + (d) u = u u d ??

Negative pion (that turns proton into neutron ?) : (u) d
u u d + (u) d = u d d ??

///

Also : how can a Baryon produce a pion to be able to exchange it ??
i mean how does it produce the anti quark ?

 

[arivero]

I am amazed about how a source of information can tell you about quarks and at the same time it is not telling you the rest of the history. At least the beta decay think you asked in the parallel thread. This one you are asking is more complicated. In principle nucleons exchange pions, yes, and they do it by creating and destroying pairs of quark-antiquark yes. Moreover, the mass of the pion determines the reach of the interaction. But the total interaction needs to consider spin and then not only exchange of pions, but also other mesons, and there it is not easy to calculate the total force.

 

[JPC]

Ok
ill surely get to know this latter in my studies

But, if in a Nucleus, protons keep changing into neutons, and neutrons into protons, would this mean that Number of protons we usually use in class is actually the average. Because like if they keep changing, there must be a very short interval I of time where The Proton number is Z-1 or Z+1, ect ??

And , what is a nucleon that has just emited a pion, but not received any yet ?

Thus, is there something that determines the emition of a pion ? Would a Hydrogen with 0 neutron emmit any pions ?

If we have a hydrogen with 0 neutron, then would it always be a proton, or will it also keep changing from proton to neutron also ?

 

[blechman]

But, if in a Nucleus, protons keep changing into neutons, and neutrons into protons, would this mean that Number of protons we usually use in class is actually the average. Because like if they keep changing, there must be a very short interval I of time where The Proton number is Z-1 or Z+1, ect ??

It's not so simple. In fact, all of this is a little misleading, since what "really" happens is that the quarks are exchanging things called gluons, and these things sort of turn into pions and other things. There's a lot of stuff going on here!

But protons and neutrons are not switching back and forth macroscopically, so this effect you're talking about doesn't really happen (if you want a technical explanation, it's because of the isospin violation from the mass difference between the proton and neutron). Anyway, this is deep stuff...

And , what is a nucleon that has just emited a pion, but not received any yet ?

This question doesn't make sense. These are all quantum mechanical things, so you can't talk about things like that thanks to Heisenberg uncertainty. This gets into things called "virtual particles"; you can look those things up if you're interested.

Thus, is there something that determines the emition of a pion ? Would a Hydrogen with 0 neutron emmit any pions ?

If we have a hydrogen with 0 neutron, then would it always be a proton, or will it also keep changing from proton to neutron also ?

The only things that can emit gluons (notice that I don't talk about pions!) are things with "color charge" - that is, quarks and other gluons. Again, as I said above, it's not that a proton spontaneously turns into a neutron, so no, a H atom would not do this. However, you might think of a proton and a neutron inside a deuterium nucleus are exchanging pions among themselves.

 

[malawi_glenn]

Pion exchange (strong nuclear force) is the analogy in QCD with wan.der-vaals "force" in QED. In a very crude approximate picture :)

 

[blechman]

sure, you're right. I don't personally do nuclear physics, so I don't like to think in terms of "pion exchange" since I know it's really gluons. The old theories of Feynman, etc that talk about pion exchange are very model-dependent and approximate, as you rightly point out. That's why I don't like to talk about pions as force-mediators (it goes back to the days before people knew of QCD).

 

[malawi_glenn]

sure, you're right. I don't personally do nuclear physics, so I don't like to think in terms of "pion exchange" since I know it's really gluons. The old theories of Feynman, etc that talk about pion exchange are very model-dependent and approximate, as you rightly point out. That's why I don't like to talk about pions as force-mediators (it goes back to the days before people knew of QCD).

Well I don't know what and how the strong nuclear force is mediated. All I know is that it is in low energy region and doing pure QCD is very very very complicated. And I also do final year undergraduate course in Nuclear physics now and Meson exchange potentials are disscussed and derived and also how this gives the nuclear shell model potential and how nice it fits with experimental data. But also our teacher who is Professor in Hadron-physics has told us that we should see the Strong Nuclear force as the analogy with Van Der Vaals residual force in electrodynamics. So you "have to" start in the right end sort of say, it is "only" a model, and JPC asked a question about that model, so therefore i think we shall try to answer his quesion THEN point out that it is just a approximate model and better theories are in progress.

Same thing holds for the guy in the Beta-decay thread, he asked a quesion in the realm of standard model, so the answer should first of all be in that realm too; then we can start discussing things outside the standard model. In fact, there is a subforum for those issues =)

 

[JPC]

ok

and would the equations be like that :

with Proton+
energy from received pion + uud -> (d)u + udd
energy from received pion + uud -> (u)d + uuu

with Neutron
energy from received pion + udd -> (d)u + ddd
energy from received pion + udd -> (u)d + uud

with ddd
energy from received pion + ddd -> (u)d + udd

with uuu
cannot emmit any pion

////////

 

[mormonator_rm]

ok

and would the equations be like that :

with Proton+
energy from received pion + uud -> (d)u + udd
energy from received pion + uud -> (u)d + uuu

with Neutron
energy from received pion + udd -> (d)u + ddd
energy from received pion + udd -> (u)d + uud

with ddd
energy from received pion + ddd -> (u)d + udd

with uuu
cannot emmit any pion

////////

ok, but please understand there are no uuu or ddd baryons in the ground state (i.e. nowhere near nucleon mass, and not Jp=1/2+). uuu and ddd occur in the 3/2- multiplet at about 1232 MeV.

and don't forget uuu = (d)u + uud

so please restate your thought

 

[JPC]

Oh yes, i must have made a mistake trying to find the pions that uuu could produce

But, is it the 3quark of a baryon that produce/recieve pions, or the antiquark-quark pairs and gluons in the sea around ? (Q1)

IF answer (Q1) : 3quark then ->
to release a pion the 3 quarks must first receive a pion ?? (Q2)

IF answer (Q2) : Yes then -> Well how can this work , because if to release a pion they must first receive a pion, then how was the first pion emmited ? by Big bang energy ? (Q3)

 

[mormonator_rm]

Oh yes, i must have made a mistake trying to find the pions that uuu could produce

But, is it the 3quark of a baryon that produce/recieve pions, or the antiquark-quark pairs and gluons in the sea around ? (Q1)

IF answer (Q1) : 3quark then ->
to release a pion the 3 quarks must first receive a pion ?? (Q2)

IF answer (Q2) : Yes then -> Well how can this work , because if to release a pion they must first receive a pion, then how was the first pion emmited ? by Big bang energy ? (Q3)

You could think of the nucleons as baryons (three "valence" quarks) in a "boiling cauldron" of spontaneously produced quark-antiquark pairs ("sea" quarks and antiquarks) that annihilate quickly. This is the picture that is painted for us by Drell-Yan processes. In other words, if a "sea" antiquark and one "valence" quark escape from the baryon to become a pion, the not only has a pion been emitted, but a further "sea" quark (now unable to annihilate because one "sea" antiquark is missing) will now take the "valence" spot vacated by the departing "valence" quark. So however it happens, there is always a way to quickly, even if not instantly, fill the ranks of the "valence" quarks when changes take place. Similarly, when two protons collide at high-energy, and produce a particle-antiparticle pair during recoil, one may ask "how did an antiparticle come out when there are no antiquarks in the protons to begin with?" The "sea" of quark-antiquark pairs that surround the "valence" quarks is a likely answer: a "valence" quark from one proton may have annihilated with a "sea" antiquark in the other proton, thus allowing a simple source for inelastic energy loss into new particle-antiparticle pairs following the initial collision. And the lost "valence" quarks will always be restored from the ranks of teaming "sea" quarks.

 

[JPC]

f a "sea" antiquark and one "valence" quark escape from the baryon to become a pion, the not only has a pion been emitted, but a further "sea" quark (now unable to annihilate because one "sea" antiquark is missing) will now take the "valence" spot vacated by the departing "valence" quark

So if we have for example , as "valence" : uud
we could have : uud + (d)d = u(d) + udd ?
which we could simplify to uud = u(d) + udd , knowing that this equation oviously implies a (d)d ?

/////

Very clear explanation , thanks

 

[JPC]

And also :

in this quark-antiquark sea , is there only (d)d , (u)u, or can there be for example (c)c ?

 

[blechman]

And also :

in this quark-antiquark sea , is there only (d)d , (u)u, or can there be for example (c)c ?

Yes, the sea can have any quark-antiquark pairs, including c, b, and t.

 

[mormonator_rm]

Yes, the sea can have any quark-antiquark pairs, including c, b, and t.

Yes this is very true, and hypernucleic reactions are an excellent example of interactions that include s quarks and antiquarks. However, remember that the heavier quarks are less likely to occur (as their production is suppressed according to mass squared), and generally do not exist for long enough to interact (since the "sea quarks" are generally considered to be virtual, and hence only live for a time proportional to Planck's Constant and inversely proportional to their rest mass) with the valence quarks. Furthurmore, there must be sufficient energy to support the transition of the virtual heavy quark pair into a real pair, since they will be separated between the initial baryon and the emitted meson. The heavier baryons only emerge in high-energy reactions, which are observed only in man-made processes here on earth; and it is highly, highly unlikely to see a neutron transform into a charmed baryon, much less something even heavier like a bottom baryon...