Neutron= proton + electron?

[Chi Meson]

Wow. Clean slate! I like the format! And I've got a question.

In elementary physics, it is said that the neutron can "sorta" be thought of as a proton plus an electron together. The mass of the neutron is slightly higher than a proton, by approximately the mass of an electron; in beta decay, a neutron decays into a proton, electron and neutrino; the neutron is neutral, since the two charges cancel.

My question is: I know it is incorrect to say that a neutron is a "Proton plus Electron and Neutrino," but why is it wrong to say this? I'm familiar with the HUP explanation that since the mass of an electron is so small, then it's position is too vague to be contained in the tiny neucleus, but is there some other reason?

[FZ+]

For a start, because neutron decay produces a proton, and electron and an electron ANTI-neutrino?

[mathman]

For starters: A proton consists of 2 up quarks, 1 down quark, and gluons holding it together. A neutron consists of 1 up quark, 2 down quarks, and gluons.

The decay process in simple terms occurs because a down quark changes into an up quark and emits an electron and an anti-neutrino. The important thing to note is that the electron and anti-neutrino weren't there before. They were created at the time this (weak force) reaction took place.

[chroot]

It it correct to say:

Proposition 1: The neutron is composed of two up and one down quark.

but NOT correct to say:

Proposition 2: The neutron is composed of a proton, electron, and anti-electron-neutrino.

Why is this incorrect? It's simple, really. Let's say you put the neutrons in a particle accelerator and bounce electrons off of them. You will see scattering events. If Proposition 1 is correct, you will see by analyzing the scattering events that there are three different scattering centers inside the neutron -- each is one of the quarks. If Proposition 2 is correct, you would instead see only two scatterting centers, one of which is much, much "stronger" than the other, corresponding to the internal electron and proton. You would probably not get enough scattering off of the neutrino to be measurable.

Experiment confirms Proposition 1, and disproves Proposition 2. The neutron is really composed of three quarks.

The electron and anti-electron-neutrino appear when one of the down quarks decays via the weak force into an up quark, and a W- boson. The W- boson then decays into an electron and an anti-electron neutrino, conserving charge and lepton number.

- Warren

[Chi Meson]

Originally posted by FZ+
For a start, because neutron decay produces a proton, and electron and an electron ANTI-neutrino?

yes, okay, for someone as picky as me, I should have caught that.

Thanks to Chroot & Mathman. I was searching my old texbooks, but I could not find a direct answer. Now, a follow up question:

If I were explaining this to advanced high school students, would it be a bad analogy if I were to say that this explanation is similar to: "It is incorrect to say that the uranium atom is composed of a barium plus a krypton atom and a few neutrons" ?

In both situations (beta decay and uranium fission) there is a conservation of a set of quantum numbers, and in both situations there are smaller components that are "shuffled" to create significantly different larger particles. Should I just avoid this analogy?

[selfAdjoint]

Well about "It is incorrect to say that the uranium atom is composed of a barium plus a krypton atom and a few neutrons" it is! Because you could think of many other ways to compose it = uranium is made up of 92 Hydrogen atoms, plus some neutrons. If you mean that Barium and Krypton were found in the first (recognized) fission experiments, be aware that there are many ways fission can happen.

[Chi Meson]

I was about to elaborate further, but I'm seeing now that this analogy is gettingout of hand. I'm leaning to the explanation that the composition of fundamental particles as an arrangement of quarks is NOT analogous to the compsition of nuclei as an arrangement of protons and neutrons.

Would you agree?

[chroot]

In atomic fission, no particles are changing identities. There is no decay or transmutation. A big nucleus splits up into some smaller ones. While I don't particularly care for the explanation that uranium = barium + krypton + neutrons, it's really no more incorrect than a dozen eggs = four eggs + eight eggs.

However, it is entirely incorrect to talk about a neutron = proton + electron + anti-electron-neutrino, since these particles only appear after a down quark decays.

- Warren

[bdkeenan00]

what is the charge of an anti-electron-neutrino,anti-muon-neutrino,and the anti-tau-neutrino?

[chroot]

The neutrinos and and their antiparticles all have no electrical charge.

- Warren

[Chi Meson]

Thank you chroot. That clears it up.

[bdkeenan00]

what's the difference between the neutrinos and their anti particles.

[chroot]

Their interaction with the weak force.

- Warren

[bdkeenan00]

Okay,thank you very much.

[jcsd]

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

[garytse86]

hey mathman, you said that the elctron and the electron-antineutrino were created when the weak force decay took place. How can matter be actually created? Mustn't there be some sort of energy being converted in the particles?

[chroot]

The neutron's rest mass is equal to the sum of the kinetic energy and rest masses of the decay products.

- Warren

[garytse86]

no I mean where does the electron come from?

can't have been released from the neutron because the neutron is composed of 2 down quarks and 1 up quark. therefore it is not from the neutron.

[Nereid]

The electron is a lepton, and the anti-neutrino is an anti-lepton, so lepton number is conserved. Baryon number is also conserved. Energy is conserved. Since no conservation laws are broken, the neutron must decay (ergodic principle; what is not forbidden is compulsory).

A down quark is transformed into an up quark, via the weak interaction (force carrier is the W- particle).

So, yes, it did 'come from' the neutron; but it didn't exist until the decay happened!

[chroot]

Originally posted by garytse86
no I mean where does the electron come from?

can't have been released from the neutron because the neutron is composed of 2 down quarks and 1 up quark. therefore it is not from the neutron.
Did you not read this thread? The beta decay is a result of a down quark decaying into an up quark.

- Warren

[FZ+]

Originally posted by garytse86
no I mean where does the electron come from?

Promotion of virtual particles, IIRC.

[garytse86]

yes, but I didn't know where that lepton (electron) came from.

[1100f]

Originally posted by bdkeenan00
what's the difference between the neutrinos and their anti particles.

The neutrino has a property called helicity, which is like the spin for massless particles. The neutrino's helicity is the opposite of the antineutrino's.
One is called left-handed, the other is right-handed

[jcsd]

As I said before though, it's not 100% certain that neutrinos aren't their own anti-particles.

[theriddler876]

how can these particles have no mass? even light has mass

[theriddler876]

one more thing if the charge of an up quark is 2/3, and one down is -1/3 then a couln't a neutron be any combination that cancels them out, IE 2 up and 4 down, or any 1:2 ratio that stays within the confines of the mass of a neutron

[chroot]

Originally posted by theriddler876
how can these particles have no mass? even light has mass
This is incorrect. Light has no mass.

- Warren

[chroot]

Originally posted by theriddler876
one more thing if the charge of an up quark is 2/3, and one down is -1/3 then a couln't a neutron be any combination that cancels them out, IE 2 up and 4 down, or any 1:2 ratio that stays within the confines of the mass of a neutron
The quarks interact not only electromagnetically, but also via the strong force. The strong force requires all hadrons (particles composed of quarks) to be color-neutral. Therefore, there are only two kinds of hadrons:

1) The mesons, composed of a quark-antiquark color-neutral pair. The two quarks must be a combination like red + anti-red, making the composite particle color-neutral. The colors are transmuted by strong force interactions -- when quarks exchange gluons, they change color -- and thus (red, anti-red), (blue, anti-blue) and (green, anti-green) are all indistinguishable.

2) The baryons, composed of three quarks or three antiquarks, also color-neutral. Baryons must be made of one each red, green, and blue quark or anti-red, anti-green, and anti-blue, and are therefore also color neutral.

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.

- Warren

[theriddler876]

Originally posted by chroot
This is incorrect. Light has no mass.

- Warren

You're reffering to rest mass

[Nereid]

chroot: The strong force requires all hadrons (particles composed of quarks) to be color-neutral.
Can anyone recommend a good, non-technical overview of QCD, its origins, how the ideas developed, and the landmark experiments which validated it?

[jcsd]

Originally posted by theriddler876
You're reffering to rest mass

mass = rest mass

[theriddler876]

do these flavours have a charge, and if so, do they have a charge to mass ratio?

[jcsd]

No, neutrinoes are electrically neutral and possibly massless (if you look at a chart of elementary particles their mass will usually be given as zero, however a couple of experiments have suggested they may have a small but non-zero mass so the issue is still up in the air at the momen. It seems to me though that more people are coming down on the side that they have mass which is useful as they are a candidate for non-baryonic dark matter). They are generally cofined to weak interactions.

[Nereid]

Originally posted by jcsd
No, neutrinoes are electrically neutral and possibly massless (if you look at a chart of elementary particles their mass will usually be given as zero, however a couple of experiments have suggested they may have a small but non-zero mass so the issue is still up in the air at the momen. It seems to me though that more people are coming down on the side that they have mass which is useful as they are a candidate for non-baryonic dark matter). They are generally cofined to weak interactions.
At least one variety (mixture) must have mass ("rest mass"), otherwise there'd be no neutrino oscillations; these are now firmly established from observation. There's furious work going on to re-do core-collapse supernova models, to incorporate these oscillations.

Neutrinos are not likely to be more than a small fraction of non-baryonic dark matter as they're too hot; dark matter seems to be quite cold.

Here's an image of the Sun, in 'neutrino light'; it's fuzzy (image is >20o cf <1o actual) because we can't (yet) focus neutrinos as we can light:
http://antwrp.gsfc.nasa.gov/apod/ap980605.html

[jcsd]

It's still not 100% sure but IIRC neutrinoes cannot have both a well-defined flavour and a well-defined mass.

[Nereid]

This paper is a good, non-technical overview of neutrino oscillations, incorporating much of the most recent experimental results (thanks to SelfAdjoint):
http://arxiv.org/abs/physics/0303116

[theriddler876]

at some point someting must have come from nothing,

I believe that nothing, fundamentally cannot exsist, we just can't detect it, take the neutrinos' for example, which "make up" sub atomic particles, well what makes them up?, you could keep asking that question, so would you ever truly have nothing?

[Floyd Flanigan]

Excellent point. It lends it's self to the theory of infinite halves. I like it.

[Nereid]

Originally posted by theriddler876
at some point something must have come from nothing,

I believe that nothing, fundamentally cannot exist, we just can't detect it, take the neutrinos' for example, which "make up" sub atomic particles, well what makes them up?, you could keep asking that question, so would you ever truly have nothing?
The Gordon Kane article, "The Dawn of Physics Beyond the Standard Model", in the June 2003 issue of Scientific American, is a good review of the current state of play - the successes of the Standard Model, and 'ten mysteries' which it cannot address.

BTW, neutrinos do not "make up" sub-atomic particles.

[Tom Mattson]

I took the cacophany on photon mass and split it off into a seperate thread, which you can find here:

http://www.physicsforums.com/showthread.php?threadid=6616

A "real physicist" would know that--since a photon isn't even emitted in the beta decay of neutrons--that the subject of photon mass is, among other things, off topic.

Originally posted by chroot
Take your pseudoscientific blather somewhere else. Here at pf.com, we have a forum called "philosophy" just for this kind of discussion.


Not Philosophy, but Theory Development. We take our philosophy seriously here.

[PRodQuanta]

Floyd Flanigan,

I think by now, you have noticed chroot is this type of guy,"If it looks like an orange, smells like an orange, and there is mathematics that proves it is an orange; then it is an orange. That's all there is to it, no if, ands, or buts about it.

Now this is a topic I have argued before, that the photon has a mass. And if you ask me, I think it is very possible it does, like expressed in this book here: Photon :
Old Problems in Light of New Ideas
(A Volume in Contemporary Fundamental Physics - Valerie V. Dvoeglazov - Editor)
Nova Science Publishers, Inc.; ISBN 1560728108

What chroot is saying, is that according to the mathematics, the fact that a photon has a mass isn't very probable. Which brings me to one of my favorite sayings:
"It's possible, but not probable."
Paden Roder

[jcsd]

It's possible because you cannot detrimine the mass of light exactly by experiment, so the best any experiment, even an idealized one can do is place an upper limit on it.

But if light has mass then much of quantum physics and the standard modle go straight out of the window.

[chroot]

Originally posted by jcsd
But if light has mass then much of quantum physics and the standard modle go straight out of the window.
That's the ticket, jcsd.

What you don't realize, PRodQuanta, is that an enormous amount of theory and experiment depends upon the mass of light being zero. Were it not zero, none of our theories would agree with any of the experiments done in the last, oh, 50 years or so. QED is the most successful scientific theory of all time, and has predicted experimental results to the precision of, IIRC, parts per 100 billion. It would be hard to swallow that all this agreement between theory and experiment is nothing but coincidence!

- Warren

[Nereid]

Originally posted by chroot
That's the ticket, jcsd.

What you don't realize, PRodQuanta, is that an enormous amount of theory and experiment depends upon the mass of light being zero. Were it not zero, none of our theories would agree with any of the experiments done in the last, oh, 50 years or so. QED is the most successful scientific theory of all time, and has predicted experimental results to the precision of, IIRC, parts per 100 billion. It would be hard to swallow that all this agreement between theory and experiment is nothing but coincidence!

- Warren In other words, alternative theories have a lot of heavy lifting to do just to get to the starting line. If you have a pet theory which proposes something as radical as a non-zero (rest) mass for the photon, you have to show that your theory can explain (is consistent with) all the experiments which have been, up till now, explained with a zero mass photon. Not a task for the faint-hearted.

[neutroncount]

I must mention that the engineering behind some of what we take for granted today are built from the pure theory and prediction of quantum mechainics. If the theory doesn't hold as you say, most of that research wouldn't have produced squat for products.

[theriddler876]

Originally posted by chroot
That's the ticket, jcsd.

What you don't realize, PRodQuanta, is that an enormous amount of theory and experiment depends upon the mass of light being zero. Were it not zero, none of our theories would agree with any of the experiments done in the last, oh, 50 years or so.
- Warren

Perhaps all experiments have a slight margin of error, Perhaps the mass of light is so small that the values attained by the experiments do and will continue to work, we have limited technology, it would be trying to get the exact measurement of a rock, by weighing it with a balance measuring only in whole Kg

[jcsd]

As I said you cannot verify the mass of light exactly even in an idealized experiment only place upper limits on it.

The thing is though the fact that light has no mass is axiomatic in theories like Q.E.D., it's a key assumption and one which if fals invalidates the whole theory.

[bdkeenan00]

Wouldn't the electromagnetic force be reduced to a short range force if the photon had a mass according to QED?

[chroot]

bdkeenan00,

Indeed. We would no longer be able to see those quasars....

- Warren

[Jeebus]

These are all mechanical scenarios for the reaction occurence, right?

http://xxx.arXiv.org/abs/physics/0205057

"Small perturbations of averaged ideal turbulence reproduce the electromagnetic field. A hollow cavity models the neutron. The cavity stabilized via a perturbation of the turbulence energy serves as a model of the proton. An isle of quiescent fluid models a localized electron. The antineutrino corresponds to a positive disturbance of the turbulence energy needed in order to compensate the difference in perturbations of the energy produced by the electron and proton." --arXiv

The Neutron mass exceeds the sum of the Proton and Electron masses. Thus the decay can conserve energy, right?

Isn't this decay commonly involved in fission of radioactive nuclei in the nuclear reactors that are used to generate electricity for the detonation of nuclear weapons?

[mormonator_rm]

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]

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]

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 alot 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 alot 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 alot 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 alot 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 reccommend 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]

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]

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