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Why are there 8 gluons?

  1. Mar 31, 2013 #1
    I've only read some popular science books on particle physics so my understanding is pretty basic.



    There are three colors and three anticolors. 3 x 3 will give us 9 gluons.

    Subtracting the red-antired and the other two leaves us with 6 gluons.



    Those science books only explain the first part but they don't explain how we wind up with 8.
     
  2. jcsd
  3. Mar 31, 2013 #2
    because there are 8 generators of the lie algebra of su(3)
     
  4. Apr 1, 2013 #3
    So the popular explanation is off-tangent?
     
  5. Apr 1, 2013 #4
  6. Apr 2, 2013 #5

    tom.stoer

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    The popular explanation misses the color-neutral gluon, which should behave like a photon (and which must be absent b/c we don't see long-range color-electric forces in nature).

    We start with 3 colors for the quarks. We represent these three colors as something like a 3-component vector. Then we act with 3*3 matrices A on these 3-vectors. Now one can represent all 3*3 matrices A as a sum A = ƩaAaTa where we use 9 matrices Ta as a vector space basis. The 9 real numbers Aa represent the gluons.

    Now we have to make a choice regarding the matrices we are talking about. They could be u(3) = u (1) + su(3) matrices, or they could be su(3) matrices w/o the u(1) part. The u(1) corresponds to the T0=1, i.e. to the 3*3 identity matrix. But this would correspond to the photon-like gluon, which we do not see in nature. Therefore we use a=1..8 instand of a=0..8 in the sum Ʃa, i.e. we explicitly exclude the photon-like gluon. That's why we can say that QCD has the gauge symmetry SU(3) instead of U(3).

    Please note that this difference does not show up on the level of the quarks, so any explanation using quarks only will fail to explain the reason for 8 instead of 9 gluons. And it's not math but nature telling us that there are only 8 gluons.
     
    Last edited: Apr 2, 2013
  7. Apr 2, 2013 #6

    Bill_K

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    Why should it behave like a photon? And what does "behave like a photon" mean?
     
  8. Apr 2, 2013 #7
    Both have zero electric charge, zero rest mass and spin 1. A gluon has color charge, a photon has none. (haha that rhymes)


    So the "white" gluon under SU(3) behaves like a photon and would not play a part in QCD. Hence there are 8 left.


    But what are the 8 remaining gluons called?
     
  9. Apr 2, 2013 #8

    tom.stoer

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    It decouples from the other gluons a=1..8 b/c [T0, Ta]=0, i.e. b/c the structure constants f0bc=0. That means that all terms in the Yang-Mills eq. with structure constants are 'abelian' (like Maxwell's eq.) if at least one adjoint index is 0. So for the a=0 gluon there is no coupling to other gluons, which means that it behaves like a color-neutral gauge boson. But this is precisley the case for photons.

    Looking at the Gauß law constraint (DE)aa = 0 which introduces the self-interaction in the color-Coulomb-potential it becomes evident that this constraint reduces to ∂E00 = 0 for a=0, so I expect a standard Coulomb interaction ~ ρ0(x)ρ0(y)/|x-y|, neither with any dynamical gluon-dependent integral operator, nor with non-vanishing Fadeev-Popov determinant.

    In addition the color charge density ρ0 is closely related to the electric charge density of the quark fields. Thefore U(3) ~ U(1) * SU(3) is (up to some algebraic issues) nothing else like QED + QCD.
     
    Last edited: Apr 2, 2013
  10. Apr 2, 2013 #9

    mfb

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    They don't have a special name (unless you count things like "red antiblue" as name), and their identity depends on your choice of a basis anyway.
     
  11. Apr 2, 2013 #10
    If I do that, how would I list out the 8?
     
  12. Apr 2, 2013 #11

    tom.stoer

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    Last edited: Apr 2, 2013
  13. Apr 2, 2013 #12
  14. Apr 2, 2013 #13

    tom.stoer

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    But keep in mind that any other basis T'a = U Ta U-1 will work as well.
     
  15. Apr 3, 2013 #14
    Unfortunately, I'm unable to keep that in mind because I only have a layman understanding :redface:
     
    Last edited: Apr 3, 2013
  16. Apr 3, 2013 #15

    mfb

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    You don't have to use the gluons used there. If ##\psi## and ##\phi## are two gluons, you can replace them by ##\frac{1}{\sqrt{2}}(\psi+\phi)## and ##\frac{1}{\sqrt{2}}(\psi-\phi)##, for example (keep the other 6 gluons), and get another possible set of 8 gluons.
     
  17. Apr 3, 2013 #16

    tom.stoer

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    Think about the eight T-matrices as basis vectors in an eight-dim vector space. You can chose any rotated basis you like, physics remains the same, but you have to change the "names" of the basis vectors (the colors)
     
  18. Apr 3, 2013 #17
    Hmm, thinking about 8-D space? That's a little hard. :smile:

    But I'm sorta getting the picture.
     
  19. Apr 3, 2013 #18

    tom.stoer

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    Think about four dimensions; then forget one dimension b/c of the photon-like gluon; now you can rotate the remaining three axes arbitrarily, respecting pairwise perpendicular dictions
     
    Last edited: Apr 3, 2013
  20. Apr 3, 2013 #19
    So there are nine dimensions but we drop one and it becomes eight?
     
  21. Apr 4, 2013 #20

    tom.stoer

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    yes; but these eight dimensions are not spacetime dimensions but an abstract mathematical concept; but if QCD symmetry were SU(2) instead of SU(3) you could visualize 22-1 = 3 gluons using three mutually orthogonal axes in three-space
     
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