Talk:Gluon

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Help with this template Comments: edit – history – watch – refresh Just as I rated bosons and fermions high priority, I think the same rating is fair here. I gave it a B rating as it is better than your average start article...Snailwalker | talk 01:20, 15 October 2006 (UTC)

Question for specialist: are gluons really massless? Following http://hyperphysics.phy-astr.gsu.edu/hbase/particles/expar.html, this is not true. Also, from my (poor) understanding, it is not simply that gluons "bind" the quarks together; they are themselves, sort of, composed of a quarks pair. -- looxix 10:16 Apr 14, 2003 (UTC)

Yes they are massless, and nowhere in the linked page states they aren't. You may be confusing them with the W and Z bosons. Gluons are not composed of a pair of quarks, however they have two color charges, for instance a red-antigreen gluon.

The fact that gluons themselves have color charge causes somewhat erratic behavior(as gluons are creating and annihilating other gluons as well), including the limited range. The W and Z bosons also have limited range, but in their case it is caused by their mass.

63.205.40.10 04:01, 23 Jan 2004 (UTC)

My copy of the 2002 Review of Particle Physics states that a mass of up to a few MeV may not be precluded, so I added that to the article. -- Schnee 23:59, 26 Jul 2004 (UTC)

Could you add the citation to the article please? HEL 17:54, 1 October 2006 (UTC)

Contents 1 8 gluons 2 gluon exchange 3 Linear algebra and why there are 8 gluons 4 Resolution to the 8 gluon problem 5 Numerology of gluons 6 Copyedit of Properties section 7 Big problem for particle physics pages 8 Flux Tubes

[edit] 8 gluons

I've read several sources that say that due to the subtleties of the theory, there are only 8 gluons, but nobody ever says what they are. Can someone please say what these 8 gluons are, and explain what the subtleties are and why they cause there to be only 8?

Maybe this will help (haven't read it in its entirety myself yet): http://math.ucr.edu/home/baez/physics/ParticleAndNuclear/gluons.html -- Schnee 23:15, 26 Jul 2004 (UTC)

OK, I looked around a bit more and wrote a short section on this. Can someone who actually knows something about QCD look this over? ;) -- Schnee 23:56, 26 Jul 2004 (UTC)

Okay, as I understand it, there's 6 types of charged gluons (red-antigreen, red-antiblue, green-antired, green-antiblue, blue-antired and blue-antigreen) and 2 types of neutral gluons. That doesn't make sense, why the heck are there two types of neutral gluons? What's the diference between the two types of neutral gluons?

Ah, the mysteries of the universe. Who can say? Mashford

You don't seem to get it. This isn't some great unknowable mystery of the universe. We only know about gluons because their existance is inferred by experimental results. If the only difference between tho two types of neutral gluons was something we didn't know about, then we'd think both types were the same thing and assume there were only 7 types of gluons until we figured out the difference, therefore there must be a known difference between the two.

Read the article. The statement "there are 8 types of gluons" is misleading, and shouldn't be taken as being true in the sense that there *really* are only 8 different gluons. There are much more, but they aren't linearly independent, so once you grokked that, it's easier to just say "there are 8 types of gluons", with the unspoken assumption that everyone knows that that's really just a simplification. That being said, colour charge is different from electrical charge etc., anyway - a "red-antired" gluon, for example, wouldn't neutral in the same sense an (electrically) uncharged particle is. Rather, it's "red-neutral" (for lack of a better word); similarly, a "blue-antiblue" gluon would be blue-neutral, and so on. I don't claim to understand all the subtleties of QCD myself (not at all), but I think that it's safe to say that the (natural, from a layman's POV) assumption that colour charge is similar to electrical charge etc. and that there is only one kind of neutrality is wrong. -- Schnee (cheeks clone) 03:19, 17 Jan 2005 (UTC)

Yeah, but if, as you put it, 'red-neutral', 'blue-neutral' and 'green-neutral' were different types of neutral then that would produce nine linearly independant gluon types, instead of 8, and if the three were all the same type of neutral then that would mean only 7 linearly independant gluons. I should probably ask around on some physics messageboard or something.

Okay, quarks can have a charge of r, g or b. Antiquarks can have a charge of -r, -b or -g. Baryons have a net chromatic charge of 0, and are comprised of either 3 quarks with chromatic charges of r, g and b, or three antiquarks with chromatic charges of -r, -g and -b, therefore (r)+(g)+(b)=0, and (-r)+(-g)+(-b)=0. Mesons have a net chromatic charge of 0, and are comprised of one quark with a positive chromitic charge and one antiquark with a corresponding negative chromatic charge, therefore (r)+(-r)=0, (g)+(-g)=0 and (b)+(-b)=0. In other words, colour charges add in a way similar to vectors. Am I getting any of this wrong?

As I understood the article that Schnee linked to, (with my severley limited memory of linear algebra from my senior year of HS, during which I was not paying attention at all) - it's not that , , and are all mutually equivalent, in which case there would indeed be seven unique gluons, it's that any one of those combinations can be represented in terms of the other two. In other words, the minimal number of unique gluons to describe the theory would be:

• 
• 
• 
• 
• 
• 

and any two of

• , , or

the remaining third being discribable as a linear combination of the other two.

Now, if anybody feeling more prosaic than me wants to incorporate this into the article, Strong interaction and Color charge, I would be much obliged... --Peter Farago 16:36, 16 Apr 2005 (UTC)

This whole digression on the number of gluons doesn't make a lot of sense.

For example: "there actually exists an infinite variety of gluons, each of them being a normalised linear superposition" doesn't make any sense. You could say exactly the same thing about any quantum variable.

And the argument cribbed from Baez is actually an explanation as to why sl(3,R) has 8 generators. Though Baez knows that there's a deep connection between sl(3,R) and su(3) that ensures that su(3) is also 8 dimensional, I doubt it's clear to the vast majority of wiki readers.

Honestly, I don't see why it's necessary to explain the number 8 at all. That's just what it happens to be. If QCD happened to be based on some crazy thing like E8, you wouldn't bother to explain why there are 248 gluons, would you? That's just what the number comes out to be.

-- Xerxes 21:33, 2005 Apr 20 (UTC)

The reason this needs to be explained is that to the casual reader, there appear to be nine possible combinations between each of the three colors and their anticolors, yet in most introductory level sources, the number 8 is casually stated without any explanation. If the math can be reduced to something comprehensible to a normal human, it should be done. I had hoped my understanding and synopsis of Baez was adequate, but if not, I heartily encourage you to explain it rather than simply strike it. --Peter Farago 08:46, 13 Jun 2005 (UTC)

[edit] gluon exchange

just got a question, doesnt the gluons emitted compensate for the new color charge? like blue quark changes to red and emitts antired+blue? in the article its the opposite

and someone tell me whats linear algebra?

[edit] Linear algebra and why there are 8 gluons

All that one needs to know here is that linear algebra involves manipulations of things like vectors: adding them, forming dot products, rotating them, etc.

Now to counting in the usual three dimensional space that we live in. It is clear that there are three components to a vector. In linear algebra one says that there are three basis vectors. One can choose a coordinate system, and write these basis vectors as the unit vectors in the x, y and z directions. Clearly one has an infinite number of vectors, but they are all linear combinations of the three unit vectors.

Saying there are 8 gluons is like saying that there are only 3 vectors. What one means is that there are 8 basis vectors in the space of gluons, although there are actually an infinite variety of gluons. Why is there a space of gluons? Because (I need jargon at this point) quantum states are elements of a Hilbert space of vectors , and quantum theory is a linear algebra on these vectors. Back to english: quantum fields can be added and subtracted like vectors.

It is also important in linear algebra to understand the notion of rotations. If we make an ordinary rotation in 3 space, then that just mixes up the coordinates of a vector: each coordinate becomes a linear combination of the old values of the coordinates. Another way to say this is to say that the x, y and z unit vectors mix under rotations.

Now why 8 basis gluons. There are 3 colours (this means that there are 3 basis quarks and they mix under SU(3) rotations). Each gluon carries one colour and one anticolour label (red-antigreen, for example). This would give nine. Three of these are red-neutral, blue-neutral and green-neutral. Under SU(3) rotations a red-neutral could become a blue-neutral, for example. But there is a colour-singlet (or colour scalar) combinations red-neutral + blue-neutral + green-neutral, which does not mix with anything else under SU(3) rotations. So it plays no part in the force between quarks which is obtained by exchanging gluons (because an SU(3) rotation would always change the colour of the quark). Hence you throw out this combination, finally getting only 8 quarks.

A last bit of jargon: you can get the dimension of the adjoint representation of any SU(N) algebra by repeating this argument. Bambaiah 05:18, Jun 3, 2005 (UTC)

[edit] Resolution to the 8 gluon problem

This is how I understand the situation.

There are in fact two different types of "neutral" QCD particles. There are "colour neutral" particles, which have no net colour charge, where anticolour charges are taken as negative. So red-antired, blue-antiblue, green-antigreen, or any superposition of these, gluons would all be colour-neutral. Also, particles with no colour charge (ie e-) would also be colour neutral. This is the equivalent to QED neutral particles. (ie. hydrogen is neutral because it has 1+ and 1- charge).

However, QCD is more complex in that it also has "colourless" particles. A colourless particle is one that has equal amounts of RGB and equal amounts of antiR antiG antiB. This is the inspiration for the colour terminology since if you add red green and blue light together you get white (colourless) light. So, a proton is definately colourless (only colourles particles are stable) since it has 3 quarks, one of each colour. However, it is not colour neutral. It's colour charge is RGB. This is what leads to the residual QCD force that holds the colourless protons and neutrons together (as opposed to the fundamental QCD force which holds non-colourless quarks together inside the nucleons). The electron is also colourless since it has equal amounts of each colour (namly none!). It is thus colourless and colour-neutral.

A colourless gluon would look like:

• 

(note that this is also colour-neutral)

Ok....so now that we have the two types of QCD neutral particles straight the 8 gluon problem is easy. Colour-neutral gluons are allowed, it's the colourless gluons that are forbid. This is because colourless particles are the only stable particles and if a colourless gluon existed it should have been detected long ago. The colour-neutral gluons, however, are not stable and can't exist outside the nucleous...thus evading our perception except at high energy.

So, there are nine combos of colour-anticolour gluons that are theoretically allowed (including the 3 colour-neutral gluons). However, the requirement of no colourless gluons imposes a further restriction upon the gluon colours. This reduces the number or degrees of freedom by 1. Thus there are only 8. These could be take to be (for example):

• 
• 
• 
• 
• 
• 

and

• 
• 

These are all linearly independent and additionally there is no way of forming a colourless gluon through superposition of these gluons. Thus, they form a basis for all experimentally allowed gluons.

If nobody has any objections about this I will update the main page to reflect this.

Danlaycock 20:45, Aug 13, 2005 (UTC)

Please don't. This is a solution in search of a problem. There is a simple formula for deriving the number of gluons: N²−1. That's really the end of the story. -- Xerxes 01:56, 2005 August 14 (UTC)

I disagree. The fact that there is a simple formula to derive the number of gluons is all the more reason why it should be explained. If there was a more complicated relation then it would be easier to justify glossing over it.

If there had of been 9 gluons it wouldn't have needed any further consideration, but since the number doesn't agree with what one would expect given the number of colour-anticolour combinations it should be explained. Anyone with some basic statistical knowledge can calculate that there should be 9, which leads to much confusion when the number 8 is quoted. This isn't like trying to explain the number of quarks, which as far as I understand is a purely emperical result. There is nothing special about 6. The number of gluons, however, is directly a function of the number of colours in the theory. And since it is relatively easy to show qualitatively why we think there are only 8 I think that it should be explained.

Obviously there are a significant number of people just on this talk page who are interested and/or confused about this topic. I know it's something that bothered me for quite a while.

I agree with you in that there is no point trying to explain why it is that there is no colourless gluon...that's just the way nature is. However, how this leads to only 8 gluons should be explained in my opinion. Obviously the fact that their aren't colourless gluons is incredibly important to the universe that we live in. If there was one, the strong force could have been a long range force and thus had macroscopic effects. Thus this fact does have very significant implications.

If you disagree with my argument feel free to add or comment on it but I really feel that this issue should be addressed. Afterall, colour content is essentially the most important characteristic of a gluon and barely any mention of it is made in the article.

Danlaycock 04:04, Aug 14, 2005 (UTC)

As someone who came to this page trying to find out why there are only 8 gluons and found Danlaycock's explanation sorted it out for me, I'd say this Xerxes guy needs to think about who this thing is written for... people like me who would like an answer, not a load of "this is a solution in search of a problem" rubbish. So thank you Danlaycock for explaining it.

Using the explanation given here, can you now determine the number of force-carrying bosons in an SO(10) gauge theory? If you thought it was 99, you're wrong; the correct answer is 45. How about E₆? Is it 35? No, it's 78. That's because the hand-wavy explanation given here is not the right way to solve this problem.

For SU(N), the adjoint happens to have dimensionality N²−1. And it happens that there's a deep fundamental connection between SU(N) and the number of linearly-independent traceless real N×N matrices. And it happens that you can make up a little story about the N colors and N anti-colors that matches well with those matrices (you take out the totally-symmetric combination to make the matrices traceless). But at the end of that story, you haven't learned anything! Specifically, you still have no idea why it works and when it doesn't. It's bad intuition that will just have to get unlearned later. Or worse, never will.

The right way to find the number of gluons is to know that it's the dimensionality of the adjoint representation of the appropriate group, which may or may not bear any simple relation to the dimensionality of the fundamental group (number of "colors"). -- Xerxes 15:39, 2005 August 14 (UTC)

I haven't got a mathematical background, so that "explanation" makes no sense to me whatsoever - I don't even know what an SO(10) gauge theory or E₆ *is*. And even if it turns out that I have heard of them I can't understand that notation. So it might not be 'correct' but I'll be sticking to the kiddy-simplified version that isn't flying way over my head, ta very much...

But wait, surely it was you who said further up that the formula was just N²−1, but it's using that on the SO(10) and E₆ that comes up with the answers you just said were wrong... So how do you know which is the appropriate group or the fundamental group or whatever? *confused*

And just one more thing... would a red quark turning into a blue quark emit a red-blue, red-antiblue, or antired-blue gluon? I've seen different things in different explanations and I don't know which is right (where does this "anti" thing come from and what does it mean?)

I'm jumping in here, but here's my comment: maybe there is no simple explanation. The simplest place I know of to look would be in David Griffiths' particle physics book, which says something about the 9 quark colors being rearranged into an octet and a singlet, and the singlet not existing in nature. I assume that's a watered-down version of what Xerxes said, which I do not yet have the education to understand either. To answer your last question, it's red-antiblue. That way there remains a total of 1 unit of red, and 0 units of blue. -- SCZenz 17:06, 1 September 2005 (UTC)

Ooh I think I get it, clever! So does that mean that the antiparticle of a red-antiblue gluon is the antired-blue gluon? And this *magic number* of 8 gluons includes those antiparticles? Or am I barking up the wrong tree... I was reading a book by Murray Gell-Mann and he seems to have conveniently ignored the "anti" bits.

The gluon is its own antiparticle. But yes, if you include color, then flipping CP (which produces antiparticles, at least as long as you're just dealing with the strong force) will turn a red-antiblue into a blue-antired. That won't help resolve the counting issue, though, as they're already included in the naive count of 9. (Red-antired, blue-antiblue, and green-antigreen would all stay the same under a CP flip, which is why you can have an odd number.) -- SCZenz 18:02, 1 September 2005 (UTC)

Eeeexcellent - simplistic though my understanding of this may be, 'tis still enough to supergeek my A level physics coursework... mwahahaha :D Thankies!

[edit] Numerology of gluons

Again with the silly 3×3 and take out the singlet. Unless you're prepared to explain why red-antired, blue-antiblue and green-antigreen form nonsinglet linear combinations, your argument does not follow. And once you've gotten into that level of detail, you might as well use an actually valid explanation of traceless hermitian matrices. The argument as it stood was misleading and wrong.

I still don't see why we should try to convince a novice reader that 3×3 is right and then convince him it is wrong (but only by 1). A novice reader does not (and should not) have an intuition for how many force-carrying bosons are associated with a particular gauge theory. It's a complicated topic. These hand-wavy explanations do not do it service. -- Xerxes 18:42, 8 December 2005 (UTC)

It should be acknowledged that it appears there should be nine gluons. Once that's out of the way, it must be explained why the appearance is nonetheless wrong, which shouldn't need an especially complex argument. Whatever form the argument should take, it can't rely upon SU(3) to demonstrate there should be eight, because that's circular reasoning. ‣ᓛᖁᑐ 19:58, 8 December 2005 (UTC)

I would not be opposed to a discussion of why we think QCD is an SU(3) gauge theory and not a U(3) gauge theory. But then a U(3) gauge theory would just decompose into SU(3)×U(1), so you'd really just get 8 gluons and 1 sorta-like-a-photon. -- Xerxes 03:18, 9 December 2005 (UTC)

[edit] Copyedit of Properties section

The Properties section has some garbled sentences:

The gluon is a vector boson like the photon; it has a spin state of 1. Vector particles usually have three spin states, but gauge invariance reduces the number of spin states of a gluon to two; negative intrinsic parity, and zero isospin. In quantum field theory, unbroken gauge invariance requires that gauge bosons have zero mass (experiment limits the gluon's mass to less than a few MeV). The gluon is its own antiparticle.

I will try to sort this out a bit. HEL 17:46, 1 October 2006 (UTC)

[edit] Big problem for particle physics pages

The discussion pages associated with many particle physics articles are full of questions/discussion/nonsense from people who have no idea about the physics, instead of discussions about improving the pages. I propose we delete all the above rubbish about "why are there only 8 gluons", and anything else like it, so we can focus on actually improving page content. An encyclopaedia entry is not the place for a thorough pedagogical exposition; people can visit discussion groups etc. if they want to learn the stuff properly. Any objections? Shambolic Entity 07:41, 1 November 2006 (UTC)

[edit] Flux Tubes

"Flux tubes" are mentioned throughout this article, but not much is said about them. Who discovered (or imagined) them, when were they integrated into the particle theory, Etc. These explanations could be given either in the Gluon page, or on a separate entry. Thanks. Hugo Dufort 19:57, 15 November 2006 (UTC)