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Why are there 8 gluons?

by abbottsys
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abbottsys
#1
Sep9-10, 08:54 PM
P: 13
Why are there exactly 8 gluons? Is this related to the fact that the color force has 3 charges, and if so how? More generally, is there a simple formula that gives the number of gauge bosons for any gauge field?
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Kevin_Axion
#2
Sep9-10, 08:59 PM
P: 920
This should be satisfying although only true understanding of why it is an Octet field comes through mathematical reasoning: http://en.wikipedia.org/wiki/Color_c..._color_charges
abbottsys
#3
Sep9-10, 09:09 PM
P: 13
Thanks, but I was hoping to get a simple explanation without diving into SU(3) mathematics. For example: each charge has 2 values, and number of charges in the color field =3, so then we have 2^3 = 8. Can this be justified by gauge theory? And does the same formula work for other gauge fields, such as the electroweak field?

Atakor
#4
Sep9-10, 09:49 PM
P: 64
Why are there 8 gluons?

Hi,
Quote Quote by abbottsys View Post
Thanks, but I was hoping to get a simple explanation without diving into SU(3) mathematics. For example: each charge has 2 values, and number of charges in the color field =3, so then we have 2^3 = 8. Can this be justified by gauge theory? And does the same formula work for other gauge fields, such as the electroweak field?
The answer is no for both questions.
3 Charges: because the quarks are in the triplet representation of SU(3)
8 Gluons : because the gauge bosons are in the adjoint representation, that has dimension 8.
These two numbers (3 and 8) are not a priori related to each other, as you can put your matter fields in any representation (not only triplet) (but you don't actually do it for physical reasons).
abbottsys
#5
Sep9-10, 09:57 PM
P: 13
But the formula DOES work for the electroweak field. This field has 2 charges (electric, isospin), so the formula says the number of gauge bosons will be 2^2 = 4 and this is correct (photon, W+, W-, Z). So the formula works for 2 gauge fields: color and electroweak. Simple formulas are always nice ;-)
Atakor
#6
Sep9-10, 10:26 PM
P: 64
Quote Quote by abbottsys View Post
But the formula DOES work for the electroweak field. This field has 2 charges (electric, isospin), so the formula says the number of gauge bosons will be 2^2 = 4 and this is correct (photon, W+, W-, Z). So the formula works for 2 gauge fields: color and electroweak. Simple formulas are always nice ;-)
the gauge bosons of SU(2) are W+/- and Z, that is : 3. which is equal to the dimension of SU(2).. the photon comes from the U(1) part of the Standard model.
your "formula" works only for SU(3), because :
N^2-1 = 2^N, for N=3... and ONLY N=3.

ps.
su(2) gives only the isospin charge..
Vanadium 50
#7
Sep9-10, 11:10 PM
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The adjoint representation of SU(N) has N2 - 1 elements. If there were 4 colors, there would be 15 gluions. 10 colors, 99 gluons.

As Akator said, this is equal to 2N only by accident and only when N=3.
tom.stoer
#8
Sep10-10, 12:21 AM
Sci Advisor
P: 5,366
Physically a gluon (living in the adjoint rep.) is something like an object carrying both color-charge and anti-charge at the same time. That's the reason for Nē which is the number of color - anti-color combinations. Now looking at U(N) one finds that it is something like U(1)*SU(N); the U(1) would mean that the gauge boson is color-neutral. As this color-neutral gluon w/o long-range Coulomb-like force does not exist, one has to subtract 1 from Nē.
abbottsys
#9
Sep10-10, 07:47 AM
P: 13
Yes, I understand the Lie Group theoretic architecture.. the adjoint representation of SU(N) has N^2 - 1 elements, color = SU(3), electroweak = U(1)*SU(2) etc. etc. But, if we were to put all that aside, and the only thing we knew was that the color force has 3 charges and the electroweak has 2 charges, then of course the formula works.

I agree it may be an accident, and have no fundamental significance. But it still works. The Gravitational field has 1 charge (mass/energy), so the formula says this gauge field has 2^1 = 2 bosons. The formula has now made a prediction, there are 2 gravitons, not 1. An annoying little formula ;-)
tom.stoer
#10
Sep10-10, 08:11 AM
Sci Advisor
P: 5,366
Unfortunately your counting regarding the graviton is not correct. If you formulate gravity as gauge theory (Palatini / first-order formulation), then you find an underlying SL(2,C) structure which predicts "two charges" in the fundamental rep.; as the graviton lives in the adjoint rep., you will have "three gravitons" according a=1,2,3 where a=1,2,3 referes to the usual Pauli matrices.

(this is the same as the three vector bosons in SU(2), namely W+- and Z; of course the three bosons W+- and Z have nothing to do with the two polarizations)

But of course we also know that physically we have only two polarizations as expected for massless spin-two particles (in gravity the relation between the "charges" and the polarization is rather indirect - but it exists).

Lessons learned? counting physical degrees of freedom will not work purely algebraically as the number of physical degrees of freedom (#p.d.o.f.) and the number of algebraic degrees of freedom (#a.d.o.f.) are related via the number of constraints (#o.c.) somehow as follows

#p.d.o.f. = #a.d.o.f. - #o.c.

But the number of constraints follows from a Lagrangian = from the dynamics, not from an algebraic structure.

I think the relations you are talking about are nothing else but contingencies for small numbers
abbottsys
#11
Sep10-10, 09:22 AM
P: 13
Hold on a second. Pulling Palatini into this is a bit unfair. f(R) gravity is not an established theory, it's an active field of research, and it has known problems, not least of which is that it cannot be reconciled with the Standard Model. It cannot in any way be called a quantum theory of gravity. The bottom line is that #gravitons is still very much an open issue.

But your comments on dof are well taken.

btw: in the formula 2^N for the number of bosons of a gauge field with N charges I'm surprised that nobody has commented on the 2. Where does it come from. It's NOT a random choice just to fit the data ;-)
tom.stoer
#12
Sep10-10, 09:37 AM
Sci Advisor
P: 5,366
Palatini has a priori nothing to to with f(R) - even it may be used in that context.

The Palatini formulation allowes you to rewrite ART or any other f(R) action as a gauge theory with a local Lorentz gauge symmetry. Doing that you find that the connection becomes the gauge field and that you have something like a gauge theory equipped with an extra diffeomorphism constraint which further restricts you local degrees of freedom.
xepma
#13
Sep10-10, 09:41 AM
P: 527
Here's one way to look at it. There are three colors in SU(3) gauge theory. That just means you have three types of charges, usually referred to as red (r), green (g) and blue (b). The gluon carries two charges, meaning there are 3*3 = 9 possible combinations:
rr
gg
bb
rg
rb

etc. Now, more technically one of these charges is actually labeled by the "anti-color", which I denote with an overbar. The number of charges is still the same, the labeling is now:

[tex]r\bar{r}[/tex]
[tex]g\bar{g}[/tex]
[tex]b\bar{b}[/tex]
[tex]r\bar{b}[/tex]

etc. So again, the gluons carries a "charge" and an "anticharge" and this would imply that there are 9 different kind of gluons. But this is not the whole story..

What do gluons do? Well, they interact -- with each other and with quarks. A quark carries only one type of charge, so it is either charged with red, green or blue. When a gluon interacts with a quark the following happens: the "anti-charge" of the gluon annihilates the color charge of the quark (they have to be compatible!), and the quark will now carry the remaining charge of the gluon. So a [tex]r\bar{b}[/tex] gluons interacts with a blue quark results in a single red colored quark. That's easy right?

Now, a new element comes into play: instead of being restricted to the type of gluons I (partially) listed above, you are also allowed to take linear combinations to form new gluons. So the object:

[tex]r\bar{r} + r\bar{b}[/tex]

is also a gluon! When it interacts with a blue-charged quark the first part ([tex]r\bar{r}[\tex]) does nothing, while the second part ([tex] r\bar{b}[/tex]) replaced the blue charge of the quark by a red one. Clear enough right? So you can take linear combinations of charges and construct new types of gluons that way.

This still gives 9 types of gluons. The reason is that we still need 9 "basis" gluons to form all other linear combinations with. So why are there 8 gluons? Consider the following, very specific gluon:

[tex]r\bar{r} + b\bar{b} + g\bar{g}[/tex]

What happens when this gluon interacts with a quark? The answer is: nothing. No matter what the charge of the quark is, the interaction with this type of gluon does not result in a change of charge. In fact, you can even go as far in saying that this gluon is simply invisible to the quarks and also other type of gluons. The gluon is "color neutral" -- it therefore does not contribute to the action of the strong force. Simply put: we should ignore this type of gluon.

When we get rid of this type of gluon we end up with one less "basis vector" in our space of "possible linear combinations of gluons". So the dimensionality of this space is simply reduced by one: 9-1 = 8. Eight different type of (Linear independent) gluons.

In group theoretical language we have started with two types of charges ([tex]\mathbf{3}[/tex] and [tex]\mathbf{\bar{3}}[/tex]) and formed all possible objects that are combinations of these charges ( [tex]\mathbf{3} \otimes \mathbf{\bar{3}}[/tex] ). From there we see that there are nine types of gluons we can form. But one of these gluons is neutral in color and should be disregarded. This is usually written as: [tex]\mathbf{3} \otimes \mathbf{\bar{3}} = \mathbf{8} \oplus \mathbf{1}[/tex]. The [tex]\mathbf{1}[/tex] represents the neutral colored gluon. What's important is that these numbers represent vector spaces, and the space associated to [tex]\mathbf{1}[/tex] is completely trivial and can be thrown away.
granpa
#14
Sep10-10, 09:50 AM
P: 2,258
like a meson?
tom.stoer
#15
Sep10-10, 09:51 AM
Sci Advisor
P: 5,366
Quote Quote by xepma View Post
Here's one way to look at it. There are three colors in SU(3) gauge theory. That just means you have three types of charges, usually referred to as red (r), green (g) and blue (b). The gluon carries two charges, meaning there are 3*3 = 9 possible combinations:
rr
gg
bb
rg
rb

etc.
This is intuitively clear but you always run into trouble how to explain the "minus one". My explanation is not better, it just avoids to talk too much about the "minus one" :-)
xepma
#16
Sep10-10, 09:58 AM
P: 527
Quote Quote by tom.stoer View Post
This is intuitively clear but you always run into trouble how to explain the "minus one". My explanation is not better, it just avoids to talk too much about the "minus one" :-)
Wasn't finished. I hit the wrong button.
xepma
#17
Sep10-10, 10:01 AM
P: 527
Quote Quote by granpa View Post
like a meson?
Yes, the coloring of a meson is neutral as well: the quarks carry color, the antiquarks carry anti-color. But in this case you want the mesons to be color neutral, so you are only allowed to look at the neutral colored combinations!

However, with mesons you can also choose different types of quarks (6 in total) so the number of possible mesons is very large because of this.
Dickfore
#18
Sep10-10, 11:24 AM
P: 3,014
Quote Quote by abbottsys View Post
Why are there exactly 8 gluons? Is this related to the fact that the color force has 3 charges, and if so how? More generally, is there a simple formula that gives the number of gauge bosons for any gauge field?
Yes, it is. The number of possible colors is the dimension of the SU gauge symmetry group. The gauge fields that mediate this interaction correspond to the generators of the symmetry.

In general, any unitary matrix ([itex]\hat{U} \cdot \hat{U}^{\dagger} = \hat{U}^{\dagger} \cdot \hat{U} = \hat{1}[/itex]) may be written as:

[tex]
\hat{U} = \exp{\left[i \, \hat{A}\right]}, \; \hat{A}^{\dagger} = \hat{A}
[/tex]

An [itex]N \times N[/itex] hermitian (self-adjoint) matrix has N real numbers along the main diagonal and the elements in the lower triangle are just the complex conjugate of the elements in the upper triangle. The number of elements in the upper triangle is equal to the number of pairs (for the indices) that you can make without repetition (they are not on the diagonal) and without the ordering being important (if one pair is in the upper triangle, then the other one is in the lower triangle). Therefore, their number is:

[tex]
C^{2}_{N} = \left(\stackrel{N}{2}\right) = \frac{N (N - 1)}{2}
[/tex]

But, for each of these elements we need 2 real numbers (since they are complex numbers) and so, we need [itex]N(N - 1)[/itex] off-diagonal parameters.

The condition of the group being 'special' means that:

[tex]
\mathrm{det}\left(\hat{U}\right) = \exp{\left[i \, \mathrm{Tr}(\hat{A})\right]} = 1 \Rightarrow \mathrm{Tr}{A} = 0
[/tex]

This gives one linear constraint among the N real diagonal elements. Thus, we actually have [itex]N - 1[/itex] real parameters along the main diagonal. Finally, the total number of real parameters is:

[tex]
N (N - 1) (N - 1) = (N - 1)(N + 1) = N^{2} - 1
[/tex]

The number of real parameters which determine a special unitary [itex]N \times N[/itex] matrix is equal to the number of generators of the SU(N) group.

For [itex]N = 3[/itex] you get [itex]3^{2} - 1 = 9 - 1 = 8[/itex] generators. This is the number of gluons.

Questions for you:

What if [itex]N = 2[/itex]? How many 'charges' are there? How many gauge bosons are there? Do you know of such a theory?

What if [itex]N = 1[/itex]? Do you really need the specialty condition? Do you know of such a theory?


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