Why can't a neutron be thought of as a proton plus an electron and neutrino?

[mormonator_rm]

neutrinos, then back to neutron decay

A neutrino and anti-neutrino will have opposite lepton number, which is critical in studying leptonic decay modes. Also, it should be noted that the electron and its anti-neutrino will couple from the W- massive vector boson of the weak force. The positron and electron-neutrino from the W+. Hence the defined direction for neutron decay into proton, electron, and anti-neutrino. It actually conserves lepton number in the process because the electron has L = 1 and the anti-neutrino has L = -1, cancelling to zero. And of course, baryon number is conserved.

I think it may be important to this discussion to mention that the neutron decay proceeds via W- decay as a result of the up quark being changed to a down quark. You could treat it as;

udd -> uud + d(-u)

where the d(-u) term is going to effectively result in the formation of W-. Because the down quark has isospin -1/2 and the anti-up quark also the same isospin, the resulting W- will carry off an isospin of -1, leaving your proton with the opposite isospin from the original neutron. The W- has a very large width, and thus decays very soon after emmission into e- and -(v~e). The full process becomes;

n -> p + W- -> p + e- + -(v~e)

where -(v~e) is the electron anti-neutrino.

 

[mormonator_rm]

Whoops

Um... I didn't realize there were five pages to this thread. My reply was intended for the last message on the first page. Whoops, sorry about that.

Okay, reply to Jeebus' message. I would imagine that such neutron decay is a regular occurance in nuclear reactors, as Cerenkov radiation is very common in reactor pools.

The result of the decay conserves energy. However, in the process you have to make your 940 MeV neutron emit a W- particle with mass 80.41 GeV! Either there is some serious contribution from temporary energy conservation violation (which is permissible for short moments of time: Heisenburg Uncertainty Principle) i.e. vacuum energy, or the W- must be a virtual particle only. In either case, the result is an electron and anti-neutrino, with the electron having considerable momentum (enough to travel faster than light in water, hence the Cerenkov effect being visible).

 

[jcsd]

Originally posted by mormonator_rm
Um... I didn't realize there were five pages to this thread. My reply was intended for the last message on the first page. Whoops, sorry about that.

Okay, reply to Jeebus' message. I would imagine that such neutron decay is a regular occurance in nuclear reactors, as Cerenkov radiation is very common in reactor pools.

The result of the decay conserves energy. However, in the process you have to make your 940 MeV neutron emit a W- particle with mass 80.41 GeV! Either there is some serious contribution from temporary energy conservation violation (which is permissible for short moments of time: Heisenburg Uncertainty Principle) i.e. vacuum energy, or the W- must be a virtual particle only. In either case, the result is an electron and anti-neutrino, with the electron having considerable momentum (enough to travel faster than light in water, hence the Cerenkov effect being visible).

mormonator, wouldn't temporary energy conservation violation mean the W^- has to be a virtual particle.

 

[mormonator_rm]

W- virtual or real

I would imagine that is the case, but I have been informed that the W- in neutron decay is often a real particle, not virtual. Personally, I would think that the violation requires it to be a virtual particle, but this is apparently not always the case. What I think is by no means what is true.

 

[mormonator_rm]

Alright, I went looking for references to real W- decay in n --> p + e- + -(v~e) and found absolutely nothing. I guess I was fed a bunch of BS by some profs and a textbook with misinformation. So yeah, the W- must always be virtual, which made perfect sense all along.

 

[turin]

Originally posted by chroot
There is no way for a composite particle of more than three quarks to be color neutral. There is also no way for a single quark to be color-neutral. Therefore, there are no particles with more than three quarks, or fewer than two. Quarks cannot be found in isolation.

I can't think of a counter example for "quarks cannot be found in isolation," but, what rules out a 5-quark hadron with red, blue, green, and another of these colours with its anti-colour?

How do the protons stay so close to each other in the nucleas inspite of the emmense electrostatic repulsion?

 

[mormonator_rm]

That's one of the latest things going on right now with the finding of X(3872) and other proposals. We are seeing very serious evidence of just the kind of thing you are talking about; hadrons with four or more quarks/antiquarks in combination seem to be turning up suddenly. And yes, they would have to be color neutral.

Still, quarks cannot be found in isolation since the energy to free a single quark would be infinite. The strong potential is approximatley proportional to the distance between the quarks, so it will increase as quarks are brought further apart. Someone pointed this out in another thread I think, that the potential created as quarks are pulled further apart tends to create quark/antiquark pairs, destroying any chance of actually deconfining a quark.

Protons are still attracted to each other through the residual strong force, which is mediated by mesons rather than gluons. Basically, the protons have a "sea" of quarks and anti-quarks in addition to their three commonly known valence quarks (uud). Any quark/antiquark pair from this "sea" of quarks can interact with the other protons, attracting them together even though they are color-neutral in the gluonic strong field regime. Hence it is called the residual strong force, and it is able to hold together color-neutral nucleons of any type (and that includes hyperons in hypernucleii as well).

 

[turin]

Originally posted by mormonator_rm
Someone pointed this out in another thread I think, that the potential created as quarks are pulled further apart tends to create quark/antiquark pairs, destroying any chance of actually deconfining a quark.

This is something that totally confuses me. Why does high potential generate particle/antiparticle pairs? I often hear, in the context of particle physics, that, if something can happen, it will happen (i.e. 1.022 MeV gamma turning into an electron and positron). I don't like that, but I'll deal with it for now if that's the contemporarily acceptable answer.

Originally posted by mormonator_rm
... the protons have a "sea" of quarks and anti-quarks in addition to their three commonly known valence quarks (uud).

How does this get around contributing a huge amount of mass to the baryon?

Originally posted by mormonator_rm
Any quark/antiquark pair from this "sea" of quarks can interact with the other protons, attracting them together even though they are color-neutral in the gluonic strong field regime.

But doesn't the interaction still have to be mediated by bosons (if it's attractive anyway)? I don't yet see how gluons exchange can be ignored in this scheme.

 

[mormonator_rm]

Originally posted by turin
This is something that totally confuses me. Why does high potential generate particle/antiparticle pairs? I often hear, in the context of particle physics, that, if something can happen, it will happen (i.e. 1.022 MeV gamma turning into an electron and positron). I don't like that, but I'll deal with it for now if that's the contemporarily acceptable answer.

Nothing likes to be in a state of higher energy, it will always find a state where it has the least energy. Energy also has a tendency to form massive bodies to carry it off. So as the potential grows, it seeks to bring it back down by creating a quark-antiquark pair. This allows the energy to be released in the form of mass, and also brings the potential back down to where you started.

Originally posted by turin
How does this get around contributing a huge amount of mass to the baryon?

The total energy within the baryon should remain fairly constant as long as it does not decay. Basically, the quark-antiquark "sea" and the potential will trade off, but always add up to the total energy that we see as the baryon's rest mass. More "sea" quarks may be formed and the potential will relax a bit, or more of these "sea" quarks may be released/annihilated and the potential will increase in response.

Originally posted by turin
But doesn't the interaction still have to be mediated by bosons (if it's attractive anyway)? I don't yet see how gluons exchange can be ignored in this scheme.

Mesons are in effect bosons. They have integer spin just like the Gauge Bosons do. This was the foundation of Yukawa's hypothesis on nuclear interactions, when he discovered the pion and measured its mass using the nuclear binding potential. Gluons will not couple to any particle that does not have color charge, so color neutral particles will not couple to gluons, i.e. a gluon can affect a quark within a proton, but it will not effect the whole proton, and because the gluon is colored it is not as likely to bridge the gap between two orbiting protons as a pion would.

 

[turin]

Originally posted by mormonator_rm
Energy also has a tendency to form massive bodies to carry it off.

Is this something like a postulate of particle physics?

Originally posted by mormonator_rm
So as the potential grows, it seeks to bring it back down by creating a quark-antiquark pair. This allows the energy to be released in the form of mass, and also brings the potential back down to where you started.

But the energy is still there, just in a different form (instead of potential energy which I'm assuming to be synonymous with virtual bosons, it is now in the form of fermions). So is mass like the most stable form of stress-energy?

I've also heard that matter and energy are just excitations of a field (I think something like this particle/antiparticle sea). Is there some postulate that says the exitations like to manifest in the form of multiples of 1/2 integer spins?

Originally posted by mormonator_rm
The total energy within the baryon should remain fairly constant as long as it does not decay. Basically, the quark-antiquark "sea" and the potential will trade off, but always add up to the total energy that we see as the baryon's rest mass.

But isn't the mass of the proton and neutron attributed solely to the masses of the three quarks plus a little bit of binding energy (not even enough extra mass for one more quark)? Or do I have that totally wrong?

Originally posted by mormonator_rm
Mesons are in effect bosons. They have integer spin just like the Gauge Bosons do.

But they are fermionic, are they not (isn't their wavefunction overall antisymmetric)? The s shell electron pair in ground state He is also like a boson in the same respect, right? But the electrons are still fermions, and they only interact with other He electrons by mediating true bosons (and Pauli exclusion?), right?

 

[1100f]

Originally posted by turin


But they are fermionic, are they not (isn't their wavefunction overall antisymmetric)? The s shell electron pair in ground state He is also like a boson in the same respect, right? But the electrons are still fermions, and they only interact with other He electrons by mediating true bosons (and Pauli exclusion?), right?

No, mesons are made of a quark and an anti-quark.
Each one of them has a half spin, sot that total spin can be zero or one. If you add the orbital angular momentum, which is integer, you find that the total spin of a meson is an integer, so they are bosons.

 

[mormonator_rm]

Originally posted by turin
Is this something like a postulate of particle physics?

Pretty close. This is simply the tendency of energy released from one form or another; it must cannot simply exist as energy independently, but must be redistributed as mass and momentum energies.

Originally posted by turin
But the energy is still there, just in a different form (instead of potential energy which I'm assuming to be synonymous with virtual bosons, it is now in the form of fermions). So is mass like the most stable form of stress-energy?

Yes, the energy is still there, but I would not say that mass is neccesarily a more stable form of energy than any other.

Originally posted by turin
I've also heard that matter and energy are just excitations of a field (I think something like this particle/antiparticle sea). Is there some postulate that says the exitations like to manifest in the form of multiples of 1/2 integer spins?

I do not believe there is such a postulate. The only excitations I know of are when the radial quantum number N is increased, such as in the case of the pion and the first "radially excited pion" or pi(1300). The pion occupies the isotriplet position of the 1(1)S0 multiplet, and pi(1300) occupies the isotriplet position of the 2(1)S0 multiplet (they have the same quantum numbers IG(JPC) but differ only by the radial number N).

Originally posted by turin
But isn't the mass of the proton and neutron attributed solely to the masses of the three quarks plus a little bit of binding energy (not even enough extra mass for one more quark)? Or do I have that totally wrong?

Actually, the quarks in the proton and neutron make up so little of the total energy of these nucleons. The binding energy is by far the largest contributor. The masses of the three quarks come out to less than 20 MeV for both nucleons, while nucleon total masses are 938.27 MeV (proton) and 939.57 MeV (neutron). If charge assymetry did not exist here, then they would actually be totally degenerate in mass (as would be the up and down quark masses). Also, an important item here is that the gluons that bind these quarks together will produce a cloud of quark-antiquark pairs evrywhere they go, leading to "seas" of quarks and antiquarks in hadrons of all types. If a quark-antiquark pair (a meson) is lost from a baryon, the potential energy increases in response; this may sound strange, but it is the direct result of the strong force acting like a harmonic oscillator potential. Alot like a spring. If you break a stretched spring at its center, you form two new nodes, and the two springs both relax. This is the same effect as when a quark-antiquark pair is produced within the "sea" of a hadron. So the potential changes and the quark pair production/annihilation rates will always balance.

Originally posted by turin
But they are fermionic, are they not (isn't their wavefunction overall antisymmetric)? The s shell electron pair in ground state He is also like a boson in the same respect, right? But the electrons are still fermions, and they only interact with other He electrons by mediating true bosons (and Pauli exclusion?), right?

On the contrary, their wave functions are symmetric. This very behavior is manifested in the spectroscopy of mesons; their rest masses appear to progress through the quantum states as a harmonic oscillator function. The comment by 1100f is totally correct. The mesons do not obey the Pauli Exclusion Principle; rather, they obey Bose-Einstein statistics.

 

[turin]

Originally posted by mormonator_rm
Pretty close. This is simply the tendency of energy released from one form or another; it must cannot simply exist as energy independently, but must be redistributed as mass and momentum energies.

I think I understand/accept that. Gluons have no mass energy, only momentum energy? Why is the energy so "unhappy" as a gluon, or so much "happier" as a meson? Is there any theory/explanation, or is this just something that we observe and therefore accept (for now)?

Originally posted by mormonator_rm
Yes, the energy is still there, but I would not say that mass is neccesarily a more stable form of energy than any other.

Then does the meson has an equally likely probability to become a gluon? Does the particle indefinitely oscillate between the gluon and meson state?

Originally posted by mormonator_rm
The only excitations I know of are when the radial quantum number N is increased, such as in the case of the pion and the first "radially excited pion" or pi(1300). The pion occupies the isotriplet position of the 1(1)S0 multiplet, and pi(1300) occupies the isotriplet position of the 2(1)S0 multiplet (they have the same quantum numbers IG(JPC) but differ only by the radial number N).

I don't have any clue what you just said to me.

Originally posted by mormonator_rm
Actually, the quarks in the proton and neutron make up so little of the total energy of these nucleons. The binding energy is by far the largest contributor. The masses of the three quarks come out to less than 20 MeV for both nucleons, while nucleon total masses are 938.27 MeV (proton) and 939.57 MeV (neutron).

I'll take your word for it.

Originally posted by mormonator_rm
If charge assymetry did not exist here, then they would actually be totally degenerate in mass (as would be the up and down quark masses).

I'll take your word for it. Does this mean that there would not be any definite mass but some probability distribution of mass?

Originally posted by mormonator_rm
Also, an important item here is that the gluons that bind these quarks together will produce a cloud of quark-antiquark pairs evrywhere they go, leading to "seas" of quarks and antiquarks in hadrons of all types.

I don't understand why a gluon would do this.

Originally posted by mormonator_rm
If a quark-antiquark pair (a meson) is lost from a baryon, the potential energy increases in response; this may sound strange, but it is the direct result of the strong force acting like a harmonic oscillator potential. Alot like a spring. If you break a stretched spring at its center, you form two new nodes, and the two springs both relax. This is the same effect as when a quark-antiquark pair is produced within the "sea" of a hadron. So the potential changes and the quark pair production/annihilation rates will always balance.

I don't see this analogy to the spring. The strong interaction is an interaction with colour, right? A meson is colour neutral, right? So why does the strong force care what a meson does, as long as it stays a meson? Is the meson a colour dipole? If anything, it seems like the meson should take energy away from the baryon (in terms of mass and whatever amount of momentum), not add to it. If the baryon and emitted meson are considered as a system, why do they interact (strong, weak, and/or electromagnetic)?

Originally posted by mormonator_rm
On the contrary, their wave functions are symmetric. This very behavior is manifested in the spectroscopy of mesons; their rest masses appear to progress through the quantum states as a harmonic oscillator function.

What's a harmonic oscillator function? I don't understand what this means at all.

Originally posted by mormonator_rm
The comment by 1100f is totally correct. The mesons do not obey the Pauli Exclusion Principle; rather, they obey Bose-Einstein statistics.

That is strange to me. I guess I'll have to take your words for it.

 

[mormonator_rm]

Originally posted by turin
I think I understand/accept that. Gluons have no mass energy, only momentum energy? Why is the energy so "unhappy" as a gluon, or so much "happier" as a meson? Is there any theory/explanation, or is this just something that we observe and therefore accept (for now)?

No "happiness" involved. It just has to do with the allowed decay paths. Mesons typically just decay into other lighter mesons, or radiatively (via photon emmission, especially the case in systems of heavy quarkonium). Mesons don't appear to have a strong tendency to decay into gluons, although there appear to be some mesons that are actually "glueballs", bound gluon states that act like mesons. It all has to do with the allowed decay paths by quantum numbers, parity, charge-conjugation, G-parity, etc...

Originally posted by turin
Then does the meson has an equally likely probability to become a gluon? Does the particle indefinitely oscillate between the gluon and meson state?

There is probably a small oscillation, but not so much that you can't tell its still a meson. There are some mesons that do contain a gluonic flux tube, but they always have exotic quantum numbers like 1-+ or 2+-, etc., numbers that are unnallowed for regular mesons.

The quarks themselves are what oscillate/mix. You will find that quark-composite states such as mesons, especially those with no isospin, will mix their states. Thus, we have examples like the eta mesons and the neutral pion in the 1(1)S0 multiplet, which are all mixtures of the three pure states;

(u + -u) - (d + -d)
(s + -s)
((u + -u) + (d + -d))/2^1/2

The other pure states, which have isospins, are;

(u + -d) and (d + -u)
(s + -d) and (d + -s)
(s + -u) and (u + -s)

don't appear to mix as much as the isospin=0 states.

Originally posted by turin
I'll take your word for it.

You don't have to take my word for it. Just check out the particle listings of the Physical Review at http://pdg.lbl.gov. There is a section on quarks, giving all the known info on their quantum numbers, charges, and masses; there is even additional commentary for those who are interested.

Originally posted by turin
Does this mean that there would not be any definite mass but some probability distribution of mass?

Not exactly. Degeneracy just means that they become equal in mass because there is nothing to cause them to be different in mass. But you have hit a key concept right on the head! The fact of particle physics is that nothing has a totally definite mass! When you look at the particle listings, they do give you the mass of the particle (with the associated errors), but they also give you a quantity called the "width" of the particle. The "width" is, for all intensive purposes, the uncertainty in mass caused by the fact that the particle doesn't actually exist in the same state for very long. The "narrower" a particle is, the longer its lifetime, and therefore the more certain its mass is. The mass presented as the mass of the particle is actually derived from taking the center point of the distribution of its mass signature (another discussion, perhaps).

Originally posted by turin
I don't understand why a gluon would do this.

Because gluons are colored, and also because they exist within a very strong field. Photons will couple into cascades of electron-positron pairs in an intense electromagnetic field or in passing through barrier. Gluons are a lot like photons in this way. Quarks are continuously emmitting gluons; electrons can do this as well, but with photons, such as in the Cerenkov effect.

Originally posted by turin
I don't see this analogy to the spring. The strong interaction is an interaction with colour, right? A meson is colour neutral, right? So why does the strong force care what a meson does, as long as it stays a meson? Is the meson a colour dipole? If anything, it seems like the meson should take energy away from the baryon (in terms of mass and whatever amount of momentum), not add to it. If the baryon and emitted meson are considered as a system, why do they interact (strong, weak, and/or electromagnetic)?

The analogy to the spring was an attempt to illustrate the shape of the potential curve over distance. The strong force operates on the principle of "asymptotic freedom"; i.e. when two quarks are at a distance of zero they experience no attraction, yet quarks can never be deconfined because the binding energy between them increases with distance, unlike the static potentials in electric and gravitational forces which are infinite at close range and drop off to zero over distance.

When quarks are removed from the "sea" of a baryon, this increases the net distance between the remaining quarks, increasing the potential within the baryon.

The interactions between baryons are the reason that mesons are transferred between them, ultimately. Like the interactions between charged particles cause a photon to transfer between them, the interactions between two quark-composite particles causes a meson to be transferred.

Originally posted by turin
What's a harmonic oscillator function? I don't understand what this means at all.

Basically it means that the energy levels of mesons will increase in even steps, as opposed to the quickly decreasing size of steps to continuum (as in the Fermi potential). The quarks will never be free, so adding energy just gives you new mesons.

Originally posted by turin
That is strange to me. I guess I'll have to take your words for it.

There is an excersize you can do to convince yourself of it, but it takes a lot of time. If you make models of the SU(3) meson octects and put them together in order of ascending quantum number, you will find that you could do this forever in just one energy level. On the other hand, make models of the SU(3) baryon octets and decuplets, and you will find that there are only two multipltets allowed in the ground state. I tried it, and now I have this really cool set of multiplet models (in addition to seeing the effect for myself). If you have the time and interest, I would reccomend it (using the Physical Review section on the Quark Model for a reference).

 

[turin]

Originally posted by mormonator_rm
It all has to do with the allowed decay paths by quantum numbers, parity, charge-conjugation, G-parity, etc...


There are some mesons that do contain a gluonic flux tube, but they always have exotic quantum numbers like 1-+ or 2+-, etc., numbers that are unnallowed for regular mesons.


You will find that quark-composite states such as mesons, especially those with no isospin, will mix their states. Thus, we have examples like the eta mesons and the neutral pion in the 1(1)S0 multiplet, which are all mixtures of the three pure states;

(u + -u) - (d + -d)
(s + -s)
((u + -u) + (d + -d))/2^1/2

The other pure states, which have isospins, are;

(u + -d) and (d + -u)
(s + -d) and (d + -s)
(s + -u) and (u + -s)

don't appear to mix as much as the isospin=0 states.


Just check out the particle listings of the Physical Review at http://pdg.lbl.gov.


There is an excersize you can do to convince yourself of it, but it takes a lot of time. If you make models of the SU(3) meson octects and put them together in order of ascending quantum number, you will find that you could do this forever in just one energy level. On the other hand, make models of the SU(3) baryon octets and decuplets, and you will find that there are only two multipltets allowed in the ground state. I tried it, and now I have this really cool set of multiplet models (in addition to seeing the effect for myself). If you have the time and interest, I would reccomend it (using the Physical Review section on the Quark Model for a reference).

I didn't really follow any of this. I think all of this stuff is out of my league. I never took a class or read a book on QFT, and have only recently become aware of this theory. Does understanding of this stuff rely on an understanding of QFT? How (what method do you recommend) does one learn QFT: class at university, book, internet ...?

Originally posted by mormonator_rm
Because gluons are colored, and also because they exist within a very strong field. Photons will couple into cascades of electron-positron pairs in an intense electromagnetic field or in passing through barrier. Gluons are a lot like photons in this way. Quarks are continuously emmitting gluons; electrons can do this as well, but with photons, ...


The interactions between baryons are the reason that mesons are transferred between them, ultimately. Like the interactions between charged particles cause a photon to transfer between them, the interactions between two quark-composite particles causes a meson to be transferred.

Why do you say "... in an intense electromagnetic field ...?" Is this a catalyst or a requirement or what?

I don't have it clear why charged particles emit photons (what motivates the emition, symmetry or something?)

Originally posted by mormonator_rm
when two quarks are at a distance of zero they experience no attraction, yet quarks can never be deconfined because the binding energy between them increases with distance, unlike the static potentials in electric and gravitational forces which are infinite at close range and drop off to zero over distance.


When quarks are removed from the "sea" of a baryon, this increases the net distance between the remaining quarks, ...


Basically it means that the energy levels of mesons will increase in even steps, as opposed to the quickly decreasing size of steps to continuum (as in the Fermi potential). The quarks will never be free, so adding energy just gives you new mesons.

OK, so the strong potential goes as x², like the quarks are connected by rubberbands or something? Then, around each baryon, is there a "meson cloud" that can couple to the "meson cloud" of another baryon? It still just seems that the one baryon's business is it's own. I don't see where the other baryon comes in. Does another baryon come up with a "knife" and cut the meson loose, thus lowering the energy of the other baryon that had the stray meson?

 

[mormonator_rm]

Originally posted by turin
Does understanding of this stuff rely on an understanding of QFT? How (what method do you recommend) does one learn QFT: class at university, book, internet ...?

Yeah, a knowledge of quantum flavor dynamics is the basis for dealing with these things in general. Classes are good, but personally I enjoy reading books on my own much better. I would actually recommend a book called Introduction to Quarks and Partons; it was published in the early 1980's, so its a bit of a throwback to the days when charmonium physics had just arrived on the scene. It does an excellent job of laying out the transformations and mixing for SU(3) flavor states of hadron multiplets. SU(3) only takes into account the three lightest quarks, so the mixing between the states is prevalent, whereas the heavier quarks beyond SU(3) hardly mix at all.

Originally posted by turin
Why do you say "... in an intense electromagnetic field ...?" Is this a catalyst or a requirement or what?

I don't have it clear why charged particles emit photons (what motivates the emition, symmetry or something?)

This is defined as Bremstrahlung. Look up the process on the internet or in a text.

Photons naturally couple to and from charged particles via the electromagnetic term of the electro-weak Lagrangian;

QeA

where Q is the electric charge (in units of the positron charge), A is the massless photon field, and e is the coupling (equal to the positron charge, related to the natural coupling g by the Weinberg angle).

Originally posted by turin
OK, so the strong potential goes as x², like the quarks are connected by rubberbands or something? Then, around each baryon, is there a "meson cloud" that can couple to the "meson cloud" of another baryon? It still just seems that the one baryon's business is it's own. I don't see where the other baryon comes in. Does another baryon come up with a "knife" and cut the meson loose, thus lowering the energy of the other baryon that had the stray meson?

Yes, that is a fairly good approximation. Just remember that due to wave-functions and such, no baryon (or meson, or anything for that matter) can be treated as a totally isolated system. Just like electrons do not need to collide in order to interact, baryons do not need to collide to interact, either.

A baryon does not contain additional mesons; rather it emits mesons on a regular basis, some real and some virtual, from the "sea" of quarks and antiquarks. A baryon does not simply come up to another baryon and collide with it to "cut off" a meson. The meson is emitted, not removed forcibly. Through the interaction of the baryons with the emitted mesons, baryons can be confined within nuclear bonds. The Yukawa hypothesis identified mesons, in particular the pion, as the force carrying particles that enabled protons and neutrons to be confined in the nucleus of an atom

Now, collisions of baryons do occur in the lab, and they are excellent tools for probing the constituents of baryons (as in the pp collision Drell-Yan process, elastic scattering scaling behaviors, etc.).

 

[NEOclassic]

Page two of string revisited

Originally posted by jcsd
That's if neutrinos aren't their own antiparticles which is still a matter of conteion among particle physicists.

Hi jc,
As I recall, the spin-a-half that is intrinsically associated with the electron (and was labeled "neutrino" by Fermi's students and colleagues in recognition of Fermi's nationality whereby the "-ino" suffix implies "tiny" as in "bambino"). It is really not a particle but rather a property without which, quantum orbital Pauli pairing would not be allowed. The direction of rotation of the electron is opposite that of the positron [the inertial spins of the electron and positron during pair production are parallel while the magnetic spins are opposed]. In the case of free neutron decay to an extremely energetic electron (that retains its own spin character) carries off with it, the spin character of the retained positron - so-called anti-neutrino - a positron-anti-neutrino rather than an electron-anti- neutrino. When this spin is stripped off, its directon of rotation is such that it is called an anti-neutrino; but since it no longer influences either of the dipoles, magnetism and torque, it looks like a regular neutrino from its backside. I believe this was what jc was alluding to. Cheers, Jim

 

[turin]

Originally posted by mormonator_rm
Yeah, a knowledge of quantum flavor dynamics is the basis for dealing with these things in general.

I thought "QFT" stood for "Quantum Field Theory." I've never heard of "Flavor Dynamics." Thanks for the feedback.

Originally posted by mormonator_rm
A baryon does not contain additional mesons; rather it emits mesons on a regular basis, some real and some virtual, from the "sea" of quarks and antiquarks. A baryon does not simply come up to another baryon and collide with it to "cut off" a meson. The meson is emitted, not removed forcibly.

How can a meson escape the baryon while it is attracted to the baryon by a x² potential? Is the answer in that book you recommended?

 

[jcsd]

Neoclassic, the neutrino is a particle, it can exist on it's own, it's just that some particle physicists think that it may be a self-conjugate particle (like the photon), though in the standard model this isn't the case.

turin, QFT does stand for quantum field theory and quantum flavourdynamics is a quantum field theory, but it's more commonly known as the electroweak theory.

 

[turin]

Thanks, jcsd, that clears that up a bit.

Mormonator (or anyone else),
Is the x² potential between baryons and NOT between the baryon and the meson? Like the 1/x potential is between the electron and the proton and NOT the electron and the photon? Then, the exchange of photons somehow cause a 1/x potential (something I don't understand) and the mesons somehow cause an x² potential analogously?

I was thinking that the meson was being attracted as x², but if it is playing the same role between the baryons as the photon plays between charged particles, then I will have reached the next level in understanding.

 

[mormonator_rm]

First of all, reply to turin;

A meson and a baryon will each be color-white, and hence no direct strong force attraction between them if they happen to separate slightly. The moment the meson is emitted, its constituent quarks fall outside of the range of free gluons from the baryon. Only the gluons between the constituents of the meson will affect the structure of the meson at that point, as far as strong potential is concerned.
The actual emmission of the meson is most likely just the reaction of a baryon to its close proximity to another baryon (like an electron emiting a photon in response to the proximity of another electron), or if decay is involved it is linked to the weak force rather than the strong force.
Thankyou to jcsd for elaborating on QFD as a QFT.

*I just saw your new message post, and you are totally correct about your last comment; the mesons do act very much in the same way between baryons as photons act between charged particles. I think you are beginning to get a clearer idea of the concept. Just one thing to keep in mind; the x^2 potential is between quarks within composite baryons and mesons, not between baryons and other baryons or mesons (and by the way, x^2 is only an educated approximation at this time, we do not know for sure the exact form).

Now, to NEOclassic;

jcsd is correct about the SM view of neutrinos. For one, if neutrinos and antineutrinos were self-conjugates, like photons, there would be a violation of lepton number conservation in many well-known interactions that follow the principle; that is, unless the lepton number of a neutrino can be arbitrarily assigned, and thus a neutrino of L = -1 and a neutrino of L = 1 can be produced as a pair(hence a reason for calling some neutrinos and some antineutrinos). This definition, however, would intrinsically define them as non-self-conjugate antiparticles, as electrons and positrons have opposite lepton numbers as well (in addition to opposite charge). If we consider electrons to have an isospin of -1/2 and a lepton number of -1, then their charge could be described as;

Q = I~3 + L/2

which equals one for the positron, -1 for the electron, if we are able to use the convention where the sign of L is the same as the sign of Q in charged leptons. The neutrinos, to conserve quantum numbers, must then be assigned the opposite isospins, so that Q = 0 for both neutrino and antineutrino. This puts neutrinos in isospin doublets with their associated leptons. This doublet form for the leptons is exactly how they are represented in the SM.
In conclusion, I would say that the opposite lepton numbers required for neutrinos and antineutrinos is sufficient to define them as not being self-conjugates, if the current SM is accurate with regard to the isospin doublet structure of quarks and leptons.

 

[arivero]

Spin, lads, spin

Long long time ago it was said that a nucleus was composed of protons and electrons. This was ruled out on spin arguments. Take for instance a 208 Pb 82, nucleus. It should have 208 protons and 126 electrons. The charge fits, but the angular momentum fails then.

Same, but more subtle, for the neutron=proton+electron+(anti)neutrino. In this view the freedom for neutron total angular momentum differs from the one of the proton, because the proton should be considered a single particle while the neutron should be seen as a sum of three.

With quarks, both proton and neutron are in equal footing.

 

[NEOclassic]

Originally posted by arivero
Long long time ago it was said that a nucleus was composed of protons and electrons. This was ruled out on spin arguments. Take for instance a 208 Pb 82, nucleus. It should have 208 protons and 126 electrons. The charge fits, but the angular momentum fails then.

Same, but more subtle, for the neutron=proton+electron+(anti)neutrino. In this view the freedom for neutron total angular momentum differs from the one of the proton, because the proton should be considered a single particle while the neutron should be seen as a sum of three.

With quarks, both proton and neutron are in equal footing.

Hi Arivero,
It seems to me that each paragraph of your post contains an inconsistency.
1. J.J. Thompson's plum-pudding applied to a whole atom - not a nucleus.
2. The total angular momentum (spin = 1/2) obtains for each nucleon.
3. Magnetically speaking, the proton and the neutron will never be on equal footing - neutron = -1.91 nuclear magnetons vs proton = +2.79 nm. Thanks for your audience. Cheers

 

[mormonator_rm]

Before the neutrino was conjectured by Fermi, neutrons were often treated as protons with an added electron (a totally incorrect concept, but nobody knew better). If isospin symmetry was an exact symmetry, and if electric charge dissappeared, the proton and neutron would be on equal footing in every way; they would even become degenerate in mass. That is, the strong force could be taken to be an exact symmetry of nature, and up and down quarks would become degenerate. Remember that the nucleons fall in the ground state of baryonic resonances, thus the total spin j = 1/2 is due completely to spin momentum and not angular momentum (because N = 0, l = 0, according to Fermi-Dirac statistics). They also have an isospin I = 1/2, where the neutron and proton have opposite magnitudes of isospin.

 

[arivero]

Originally posted by NEOclassic
Hi Arivero,

Hi neo,

1. J.J. Thompson's plum-pudding

Did I named Thomson? I was referring, as the other poster has already remarked, to a theory of the nucleus.

2. The total angular momentum (spin = 1/2) obtains for each nucleon.

Ok I have not been clear here. For the neutron=electron+proton, the spin issue fails completely. If you add the neutrino, it must be bound to the inside of the neutron, and you must assume that it is in the ground state, only with both conditions you can get the 1/2 of the neutron.

3. Magnetically speaking, the proton and the neutron will never be on equal footing

It is obvious I was referring to spin issues. I know that the proton has charge +1, too... was this the next argument?

The point is that in current theory both nucleons are composed of three spin 1/2 particles. In the theory proposed in this thread, the neutron would be composed of three spin 1/2 particles but the proton would be composed of a single particle.

Cheers

Alejandro