Talk:Gluon

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)

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 <math>r\bar{r}<math>, <math>g\bar{g}<math>, and <math>b\bar{b}<math> 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:

• <math>r\bar{g}<math>
• <math>r\bar{b}<math>
• <math>g\bar{r}<math>
• <math>g\bar{b}<math>
• <math>b\bar{r}<math>
• <math>b\bar{g}<math>

and any two of

• <math>r\bar{r}<math>, <math>g\bar{g}<math>, or <math>b\bar{b}<math>

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)

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?

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)