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Color currents in classical QCD

  1. Nov 23, 2008 #1

    Ben Niehoff

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    I've been playing around with the QCD Lagrangian to get a better understanding of how it works. I can derive some classical, Maxwell-like equations; the inhomogenous ones are

    [tex]\nabla \cdot \vec E^a = -gf^{a}_{bc} \vec A^b \cdot \vec E^c + \rho^a[/tex]

    [tex]\nabla \times \vec B^a - \frac{\partial}{\partial t} \vec E^a = \vec J^a + gf^a_{bc} (\Phi^b \vec E^c - \vec A^b \times \vec B^c)[/tex]

    The problem is that I'm not quite sure how to interpret these equations. The (gluon) color indices {a, b, c} run from 1 to 8. But there are three kinds of color charge. So how do I interpret the sources [itex]\rho^a[/itex] and [itex]\vec J^a[/itex]?

    One thing I attempted was to multiply both equations by the generators [itex]T^a_{ij}[/itex]. This eliminates the gluon color indices {a, b, c} in favor of the quark color indices ij (which then run from 1 to 3). But now there are two indices on everything! One for a color and one for an anti-color. Again, I can't quite figure out how to interpret what it means.
  2. jcsd
  3. Nov 24, 2008 #2

    Ben Niehoff

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    Hmm...no answers. Maybe I didn't explain the question enough. The problem is that these equations are in the adjoint representation, but the quarks are in the fundamental representation, and I wasn't sure how to relate them.

    I think I've figured it out, though. I have to represent some color-charge distribution as the linear combination of adjoint-color matrices (as functions of position and time) that leave that charge distribution invariant. For example, a red point-charge at point [tex]\vec x'[/tex] would be represented by

    [tex]\rho(\vec x) = \left( \frac12 \mathbf \lambda_3 + \frac{\sqrt 3}{2} \mathbf \lambda_8 \right) \delta^{(3)}(\vec x - \vec x')[/tex]

    where [itex]\mathbf \lambda_a[/itex] are the Gell-Mann matrices.

    To anyone more familiar with this stuff: Does this sound right, or have I gone astray?
    Last edited: Nov 24, 2008
  4. Nov 25, 2008 #3

    Ben Niehoff

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    Yes, I had it completely wrong. But now I've figured it out! It comes from the definition of the color currents (of course)

    [tex]j^{\mu}_a = \bar \psi \gamma^{\mu} \frac12 \lambda_a \psi[/tex]


    [tex]\rho_a = j^0_a = \psi^{\dagger} \frac12 \lambda_a \psi = \langle \psi | \; \frac12 \lambda_a \; | \psi \rangle[/tex]

    which for red, green, or blue quarks gives

    [tex]\rho_R = \frac12 \lambda_3 + \frac1{2\sqrt3} \lambda_8[/tex]

    [tex]\rho_G = -\frac12 \lambda_3 + \frac1{2\sqrt3} \lambda_8[/tex]

    [tex]\rho_B = -\frac1{\sqrt3} \lambda_8[/tex]

    which form an equilateral triangle in [itex](\lambda_3, \lambda_8)[/itex] space. Nifty!
  5. Nov 30, 2008 #4


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    you answered your own questions, so that's good. Let me just point out one more thing: "E" and "B" are not describing quarks, but they're describing the gluon field strength. So that is why these equations are in the adjoint and not the fundamental - there are eight types of gluons, and these equations are describing those eight fields, regardless of the form of the source.

    Just a slightly different (wordy) statement of what you wrote down. :wink:
  6. Nov 30, 2008 #5

    Ben Niehoff

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    However, I've reached another point of confusion. These equations are nonlinear, but the nonlinear parts depend upon the structure constants. In particular,

    [tex]f^a_{38} = 0[/tex]

    for all a. So if my quark charges lie in [itex](\lambda_3, \lambda_8)[/itex] space, they will never set up any nonlinear fields! That is, if the vector potentials [itex]\vec A^a[/itex] for a={1, 2, 4, 5, 6, 7} are zero at any time, they are zero for ALL times. The chromoelectric field will behave just as three superimposed Maxwell fields, with completely linear behavior (and hence, no confinement).

    There is an equation which rotates the quark charges:

    [tex]\frac{dQ^a}{dt} = -g f^a_{bc} Q^b \left( \Phi^c - \vec v \cdot \vec A^c \right)[/tex]

    but this too depends on the existence of charges and/or fields that lie outside the [itex](\lambda_3, \lambda_8)[/itex] plane.

    So it appears there are no interesting new dynamics unless one assumes pre-existing gluon fields of the remaining 6 types.

    Is this true, or is there something I'm missing?
  7. Dec 1, 2008 #6

    Vanadium 50

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    I'm not sure what you are doing exactly, but you need the complete gluon octet to make a self-consistent model. Picking 3 of 8 gluons won't work.
  8. Dec 1, 2008 #7

    Ben Niehoff

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    I am using 8 gluon fields. The issue is that it appears that I can set 6 of them to zero without any inconsistency:

    1. The nonlinear parts of the Maxwell-like equations I posted are all multiplied by the structure constants [itex]f^a_{bc}[/itex].

    2. My color currents, as computed above, all lie in the [itex](\lambda_3, \lambda_8)[/itex] plane in the 8-dimensional gauge space.

    3. [itex]f^a_{38} = 0[/itex] for all a in {1, ..., 8}.

    4. Therefore, unless I assume pre-existing gluon fields along one of the {1, 2, 4, 5, 6, 7} axes in gauge space, ALL of the nonlinear terms in the equations are identically zero!

    In other words, there are solutions where the quarks interact with an ordinary, inverse-square force (but with three kinds of charge, instead of just one), and never display any "interesting" dynamics.

    Am I missing something? Is there some principle by which these solutions can be discarded?
  9. Dec 1, 2008 #8
    Gauge invariance prevents you from doing that. Your equations are not SU(3)_c invariant !
  10. Dec 1, 2008 #9

    Ben Niehoff

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    I was hoping it was something like that. But according to Peskin & Schroeder, p. 491, the non-Abelian gauge field is NOT invariant, but transforms as

    [tex]F^a_{\mu \nu} \rightarrow F^a_{\mu \nu} - f^a_{bc} \alpha^b F^c_{\mu \nu}[/tex]

    It's the Lagrangian which should be invariant.

    But maybe there is some subtlety I'm not getting? The quark charges are not gauge-invariant, so maybe I am not allowed to assign them specific values? (Since that would involve fixing an unphysical degree of freedom). Or something like that.
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