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Gluon mass ... yes or no?

  1. Jun 9, 2015 #1
    I know gauge theory predicts gluons are massless. But under spontaneous symmetry breaking do they get a mass? If they are massless why is the strong force such short range? Is it because of quark confinement and asymptotic freedom? It seems like a paradox.
     
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  3. Jun 9, 2015 #2

    Orodruin

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    No, gluons do not acquire a a mass through spontaneous symmetry breaking and as far as we know, the SU(3) colour group is unbroken.

    The range of the strong interaction is limited by confinement. Essentially, it becomes so strong at large scales that it becomes energetically favourable to create new particles rather than allowing free colour states and thys there are no free coulour charges around. The only way for strong interactions to work between hadrons is through residual forces reminiscent of the van der Waals force. Such forces are essentially mediated by virtual pions, which are massive and thus limiting the range of said interactions.
     
  4. Jun 9, 2015 #3

    mfb

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    See also Upper Limits on a Possible Gluon Mass

    It is possible to give gluons a mass in theory. Experimentally you cannot get an exact mass measurement, so a small mass (<1 MeV) cannot be ruled out. There is no deeper theoretical motivation for an additional symmetry breaking, however.
     
  5. Jun 9, 2015 #4
    So I should accept the fact that the gluon may be massless. However, outside of a nucleon I could get a pion (residual strong force) which does have a mass. I assume, this does get its mass from the Higgs.
     
  6. Jun 9, 2015 #5

    Vanadium 50

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    Actually, the pion mass is just about the most complicated mass there is. The quarks get their mass from the Higgs, Is that enough?
     
  7. Jun 10, 2015 #6

    ChrisVer

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    The pion is a Nambu Goldstone boson, associated with the breaking of the U(2) axial symmetry. So it gets its mass from that...
     
  8. Jun 10, 2015 #7
    Not sure I want to go there unless it can be put into much simpler terms.
     
  9. Jun 12, 2015 #8

    vanhees71

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    The pion gets its mass from the explicit breaking of chiral symmetry due to the finite u- and d-quark masses (more precisely their current-quark mass). If these quarks were massless also the pion would be massless as stated in #6.

    Most of the mass of the hadrons made of light quarks (such as proton and neutron) is not due to the Higgs mechanism, which provides just about 2%. The rest is all dynamically generated by the strong interaction. This is validated by lattice QCD which nowadays has a pretty good description of the hadron-mass spectrum with physical current-quark masses.
     
  10. Jun 12, 2015 #9
    Are we talking about a similar mechanism to the way the W and Zs get their mass with the em field explicitly breaking the symmetry? The Goldstone bosons get 'eaten' to form 3 massive pions. Does the rho get its mass in the same way with the different spin accounting for the mass difference?







    /
     
  11. Jun 12, 2015 #10
    No, nothing gets eaten to give the pions mass. If there was an exact symmetry between u and d quarks then the pions would be massless. Since there is no such symmetry the pions have mass. The pions would still have mass even if you turned off electromagnetism because the u and d have different masses.

    I don't think there's any need to explain how the rho "gets its mass." There's no reason to expect the rho to be massless. Even if all the quarks were massless the rho would still be massive. Roughly speaking we can think of the rho's mass as coming from the kinetic energy of its constituent quarks and the energy of the color field between them.
     
  12. Jun 17, 2015 #11
    OK. I have done a lot more reading about this since my last post and realize that we are really talking about 2 different mechanisms here: Spontaneous chiral symmetry breaking and explicit chiral symmetry breaking. Both have to happen to get mass. My interpretation is that the quark condensate (QCD vacuum) provides the spontaneous part and em provides the explicit part. If em were turned off then SSB leads to the Nambu Goldstane bosons which are massless. If em is turned on the symmetry is also explicitly broken and massive pseudo Nambu Goldstone bosons (3 PNGBs are produced for SU(2) and 8 for SU(3) flavor). Because the pion triplet consist of only u and d combinations which are similar and small in mass, the pions have a small mass. The others in the octet contain s quarks which are much heavier.

    The difference between the vector and the pseudoscalar is related to the spin-spin interaction with the chromomagnetic field. There is more energy associated with the J = 1 states than there is with the J = 0 states.

    So it seems to me that both the pseudosclar and vector meson octet components are all PNGBs with the latter case have higher energy because of the different spin orientations.

    Assuming I am right, I am still unclear about how all of this relates to the baryon decuplet. Is it the same mechanism? Also, I have read in some articles that it is the quark mass that is responsible for the explicit symmetry breaking.That seems counter intuitive to me as I thought mass is a consequence and not an instigator.

    How am I doing so far?
     
  13. Jun 17, 2015 #12

    ChrisVer

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  14. Jun 17, 2015 #13
    electromagnetic
     
  15. Jun 17, 2015 #14

    ChrisVer

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    How did you reach to the conclusion that em interactions cause the explicit symmetry breaking? rather than the quark masses?
     
  16. Jun 17, 2015 #15
  17. Jun 18, 2015 #16
    If the masses of u and d were nonzero but exactly equal, would pions still be massless?
     
  18. Jun 18, 2015 #17

    ChrisVer

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    http://www.physics.umd.edu/courses/Phys741/xji/chapter5.pdf
    Have a look at section 5.3

    Yes because after the mathematics of the whole thing you obtain that the pion mass should be:
    [itex] m_{\pi0}^2 = \frac{(m_u + m_d)4u_{vev}}{f_\pi^2}[/itex]
    and for the charged pions you also get an electromagnetic correction ([itex]\Delta[/itex]):
    [itex] m_{\pi \pm}^2 = \frac{(m_u + m_d)4u_{vev}}{f_\pi^2}+ \Delta[/itex]
    For the derivation you could have a look in Weinberg's QFT Vol2 book (sec.19.7)
     
  19. Jun 20, 2015 #18

    vanhees71

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    Let's sort the foundations a bit.

    The fundamental theory describing the strong interactions is quantum chromodynamics (QCD). It's fundamental symmetry is the local color-charge group ##\mathrm{SU}(3)_{\text{col}}##. The fields in the Lagrangian are the 6 quark flavors, which are represented by Dirac fields and the color-gauge fields, the gluons. The remarkable feature of QCD is its asymptotic freedom, i.e., the running coupling in renormalized perturbation theory gets small for the large energy-momentum scales, i.e., for high-energy scattering events.

    At low energies, we cannot observe quarks and gluons as asymptotic free states. The reason is called confinement. The asymptotic free states observed are the colorless hadrons (mesons made of a quark and an antiquark and baryons made of three quarks; there are also hints for "exotic" configurations like tetra quark states). Confinement cannot be described in the realm of perturbation thery, but there are very convincing results from lattice QCD, where QCD is evaluated non-perturbatively on a space-time grid on a computer, and the observed mass spectrum of hadrons is reproduced very well nowadays.

    So at low energies the relevant degrees of freedom are hadrons, and thus one has to use effective field theories for them. Now the question is, how to model these interactions, and there nature is kind to us in providing more symmetries of the QCD Lagrangian than the fundamental local color symmetry. These are global ("ungauged") symmetries. The most important one in the light-quark sector (u- and d-quarks) is chiral symmetry. Compared to typical scales of the light hadrons (around 1 GeV) the current-quark masses of the light quarks with only some MeV (inferred from deep inelastic scattering, where perturbative QCD is applicable) are small. So you can first consider the limit, where you set these quark masses to 0. Then the u and d quark interaction in QCD becomes symmetric under chiral transformations, i.e., you can rotate the left- and the right-handed part of the corresponding Dirac spinors with independent SU(2) rotations in flavor space. This is the ##\mathrm{SU}(2)_L \otimes \mathrm{SU}(2)_R## symmetry.

    Now the strong interaction leads to spontaneous breaking of this symmetry, i.e., the ground state is not symmetric under this group. The reason is that the vacuum expectation value (vev) of the quark condensate ##\langle \overline{\psi} \psi \rangle \neq 0##. This breaks the chiral symmetry such that only the part of the symmetry which keeps the quark-condensate vev invariant, and that's the vector part of this group, the ##\mathrm{SU}(2)_V##, the isospin rotations. So from the originally 6-dimensional group parameters you are left with only 3. Goldstone's theorem now tells you that thus there must be 3 massless pseudoscalar states, and these are identified with the pions, because those are very light (##m_{\pi} \simeq 140 \; \mathrm{MeV}##).

    In this chiral limit, thus you can argue as follows: We build a Lagrangian for scalar bosons with chiral symmetry and break this symmetry spontaneously to the ##\mathrm{SU}(2)_V## such that you get three massless pseudoscalar bosons which you identify with the pions. The most simple example is the linear ##\sigma## model, invented by Gell-Mann and Levy in 1960. You realize the chiral symmetry as an ##\mathrm{SO}(4)## acting on a quartet of real fields ##\vec{\phi}## and write a mass term with the wrong sign and a symmetric four-field coupling
    $$\mathcal{L}=\frac{1}{2} (\partial_{\mu} \vec{\phi}) \cdot (\partial_{\mu} \vec{\phi} ) + \frac{\mu^2}{2} \vec{\phi}^2 - \frac{\lambda}{8} (\vec{\phi} \cdot \vec{\phi})^4.$$
    The ground state is not at ##\vec{\phi}=0##, because the potential has no minimum there. You can now choose the direction of the thus needed vev for the fields ##\vec{\phi}## arbitrarily (the ground state for a spontaneously broken global symmetry is always degenerate!), e.g., ##\langle \vec{\phi} \rangle=(\sigma_0,0,0,0)##. Then the rotations of the last three components leave the vacuum unchanged and rotations in a plane containing a component in the first direction don't. Thus rewriting the Lagrangian by setting ##\vec{\phi}=(\sigma_0+\sigma, \vec{\pi})##, you'll find that the three ##\vec{\pi}## field degrees of freedom represent massless particles, the Goldstone bosons of the spontaneously broken symmetry, and you are left with another scalar boson, represented by ##\sigma##, it's known as the ##\sigma## meson or in the more modern naming scheme as the ##f_0##. It is a very broad resonance with a mass between 400-550 MeV and a width of 400-700 MeV.

    Now the chiral symmetry is only approximate, because the u- and d-quarks are not exactly massless but have a small mass of some MeV. Thus the pion becomes massive, and one has to add a small symmetry-breaking term in the effective Lagrangian. Since it's small it can be treated as a perturbation ("chiral perturbation theory"). If the u- and the d-quark masses were equal the isospin symmetry would be still exact, but it's also already violated by the difference of the quark masses, but it's still a pretty good approximate theory.

    Now, if you add the electromagnetic interaction to the game, the isospin symmetry is also broken by this, but again it's only a small breaking, because the em. interaction is so weak.

    Phenomenologically the socalled vector-meson dominance model is quite successful to describe the electromagnetic interactions of hadrons. It is based on the assumption that the electromagnetic current is proportional to the (charge neutral) light-vector meson fields (##\rho##, ##\omega## and if you add strangeness also ##\phi##). These are massive vector fields. If you take only the charge-neutral vector mesons you can make a abelian gauge model with massive gauge bosons. There's nothing forbidding naively added mass terms to the gauge fields for an Abelian U(1) gauge group (Stueckelberg formalism). The theory is still gauge invariant (and in minimal-coupling form even renormalizable). It's like QED with a massive photon, but now describing the strong interaction among vector mesons and pions.

    Of course, one is also interested in describing all the vector mesons, not only the uncharged. The ##\rho## meson, e.g., is a isovector meson and comes thus as three such mesons, one neutral and one positively and one negatively charged. Then you'd like to describe this as a SU(3) gauge theory, which is non-abelian. This has indeed been done by Bando, Kugo at al in the mid 1980ies: You take an additional "hidden" gauge group (one does NOT gauge the chiral symmetry, because then the pions would get eaten up as the third degree of freedom of the then massive ##\rho## mesons when Higgsing it).

    As in the standard model of the electroweak interaction you'd end up with an additional massive Higgs boson in the physical particle spectrum. Since this is unwanted, one integrates this out by realizing the hidden-local symmetry non-linearly, i.e., without a Higgs boson. This is then of course no longer renormalizable, but that doesn't matter, because one sees it anyway as a low-energy representation, i.e., an effective field theory, which has to be expanded in powers of momentum and thus carries infinitely many low-energy constants. The only constraint are the underlying symmetries. The same is true of the chiral perturbation theory describing hadrons.
     
  20. Jun 21, 2015 #19
    What is the contribution of uponium and downonium to neutral pion?
     
  21. Jun 21, 2015 #20

    vanhees71

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    According to the standard picture of mesons the pion is a pseudo-scalar bound state of a specific mixture of up-antiup and down-antidown states, such that it belongs correctly to the isospin-1 triplet of pseudoscalar mesons. The details are as follows.

    [This introductory paragraph is corrected from a wrong previous version (see the discussion in the following two postings).]

    From this the most simple way to figure the quark content out is to write down the corresponding current. From a Dirac spinor you can build with help of the Clifford algebra generated by the ##\gamma## matrices these currents. For a single spinor like the ##u##-quark spinor a pseudo scalar is given by ##\overline{u} \mathrm{i} \gamma_5 u##.

    Now in the SU(2) quark model the u and d quark build a isospin-1/2 doublet
    $$\psi=\begin{pmatrix} u \\ d \end{pmatrix}.$$
    With help of the generators of the isospin rotations ##\vec{\tau}=\vec{\sigma}/2##, where ##\vec{\sigma}## are the usual Pauli matrices, you can then build the pseudoscalar isospin triplet
    $$\vec{\pi}=\psi \mathrm{i} \gamma_5 \vec{\tau} \psi.$$
    The electric charge operator is encoded in the charge matrix, which is related by isospin and hypercharge, but here we can just take the charge matrix for the quarks
    $$\hat{Q}_q=\mathrm{diag}(2/3,-1/3).$$
    For the anti-antiquarks, of course we have ##\hat{Q}_{\bar{q}}=-\hat{Q}_q##. This matrices act on the flavor doublet

    Now we write out the bilinear forms for the pion fields
    $$\pi_1 \propto \overline{u} \gamma_5 d+\overline{d} \gamma_5 u,$$
    $$\pi_2 \propto -\mathrm{i} \overline{u} \gamma_5 d+ \mathrm{i} \overline{d} \gamma_5 u,$$
    $$\pi_3 \propto \overline{u} \gamma_5 u + \overline{d} \gamma_5 d.$$
    Since in the bilinear form the charge is given by the sum of the charges of its constituents, obviously only ##\pi_3## has a definite charge, namely 0. So this combination already gives the neutral pion.

    The charge eigenstates made out of the quarks are obviously
    $$\pi^{\pm} \propto \pi_1 \mp \mathrm{i} \pi_2.$$
    So the valence quarks for a ##\pi_+## are an up and and anti-down quark for a ##\pi_-## an anti-up and a down quark.
     
    Last edited: Jun 21, 2015
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