In the standard model gravitron is a particle but Higgs boson is not?

  1. In the standard model gravitron is a particle but Higgs boson is not. The Higgs field causes particles to acquire mass. Mass generates a gravitation field. Is gravitron is in any way related to Higgs boson? Is the Higgs field any way related to gravitation field?
     
  2. jcsd
  3. Matterwave

    Matterwave 3,860
    Science Advisor
    Gold Member

    What makes you say the Higgs boson not a particle? Wouldn't that be the most confusing name to give it, since a boson is a type of particle?

    The graviton is not part of the Standard Model. The standard model does not give a description of the gravitational force. The gravitational field has not been successfully quantized as of yet, and that is a field of active research.
     
  4. Thank you for the response. In any reference regarding standard model 1. Photon 2. Gluon 3. W+,W-,W0 4. Graviton are only mentioned but not Higgs boson.
     
  5. Matterwave

    Matterwave 3,860
    Science Advisor
    Gold Member

    Those are the force mediating particles. The graviton is only a theoretical one. It's not been discovered, and there is as yet no coherent and renormalizable theory developed with it that reproduces the correct phenomena.

    There are many other particles, the quarks and the leptons. The Higgs is the last one that rounds out the bunch.

    Here is the wikipedia article: http://en.wikipedia.org/wiki/Standard_model

    Check here for the full particle content: http://en.wikipedia.org/wiki/Standard_model#Full_particle_count
     
  6. Thank you matterwave I got it!
     
  7. mfb

    Staff: Mentor

    Where? It is completely wrong.
    "W0" should be Z, I guess. It is a different particle.
     
  8. Yes it is z. it was my mistake in typing
     
  9. ChrisVer

    ChrisVer 2,328
    Gold Member

    Well in the particle references they don't have the Higgs boson as a force mediator (or gauge boson), because it's not a mediator.

    I don't really like the name "gauge boson" for the set of particles containing Z, because Z appears after the symmetry breaking (so what's the gauge???).
     
  10. Bill_K

    Bill_K 4,157
    Science Advisor

    The Higgs boson participates in a Yukawa interaction with the fermions. One could consider this as the mediation of a weak, short-range "force."

    The Z exists beforehand. All the symmetry breaking does is give it a mass, making it a "massive gauge boson". The gauge is the (unnamed) transformation in SU(2)L x U(1)Y that's orthogonal to the electromagnetic gauge.
     
  11. ChrisVer

    ChrisVer 2,328
    Gold Member

    what force is it then? EM? WI? SI? G? non of them... the Higgs boson doesn't exist in the Yukawa term- it is the Higgs field that does and which gets a vev. The Higgs boson is just the direct confirmation of the existence of that field.
    Well when I hear the term "gauge boson" I think of the particles existing in the adjoint representation of the gauge group. Eg SU(2) contains 3 gauge bosons or SU(3) contains 8.
    However after the SSB there is no SU(2), while you get the massive bosons W,Z and the massless photon.
    So Z is not a gauge boson- in particular it's a combination of the two chargeless+massless gauge bosons existing before the SB. So the gauge bosons before SB are [itex]W^{0, \pm}_{\mu}, B_{\mu}[/itex] and after you only have the [itex] A_{\mu}[/itex]. Maybe that's a misconception....
     
  12. Bill_K

    Bill_K 4,157
    Science Advisor

    The full interaction term is (v + h)(¯eLeR + ¯eReL). The field part v¯ee produces a mass, while the boson part h¯ee produces a Yukawa interaction.

    The point of the Higgs Mechanism is to break the symmetry while not destroying it. The SU(2) x U(1) symmetry is still there, underlying everything. The Lagrangian still respects the symmetry, the gauge bosons are still there, the particle multiplets are there...

    Any linear combination of the massless gauge bosons W0 and B0 is still a perfectly good massless gauge boson. After symmetry breaking you focus on the mass eigenstates A and Z.
     
  13. Thank you all could you please help me in clearing one more doubt !
    gamma ray is a photon as such massless. If it is made to pass through bubble chamber, it is resolved as an electron & positron.
    Now the electron has mass. Does the Higgs mechanism (with generation of particle with mass 128 Gev as that took place at LHC/ATLAS ) take place for the electron to acquire mass.
     
  14. mfb

    Staff: Mentor

    It is not "resolved". It can get converted into two other particles.

    The Higgs field is (more precise: can be) responsible for the electron mass. The Higgs mechanism with the Higgs boson is not. This has nothing to do with the production process.
     
  15. ChrisVer

    ChrisVer 2,328
    Gold Member

    Also for the photon... for the production of the electron+positron you need two photons.... one photon cannot be converted in two particles because for those two particles you can always find a Center of Mass reference frame where their total momentum is zero, while you can't do that for a single photon.
    (I need someone to reconfirm this statement of mine)

    Plus I don't see how the SU(2) exists after the SSB.... :(
    the mass term breaks the SU(2)... and I think in general the scheme is:
    [itex] SU(2)_{L} \times U(1)_{y} \rightarrow U(1)_{em}[/itex]
    so you remain with only the U(1)
     
    Last edited: Jun 21, 2014
  16. Nugatory

    Staff: Mentor

    Single-photon pair production can and does happen as long as there is a suitable mass (usually an atomic nucleus) nearby to soak up the excess momentum. A photon and a nucleus go into the interaction, an electron and a positron and the nucleus come out.
     
  17. PAllen

    PAllen 5,797
    Science Advisor
    Gold Member

    I think the one and only requirement for these other particle(s) is that they interact with photons. Thus, if not another photon, there must be electric charge. An isolated neutron should work (theoretically, with very low probability) due to charge of its quarks. Of course, near earth, mediation by a nucleus must account for nearly 100% of pair production.
     
  18. ChrisVer

    ChrisVer 2,328
    Gold Member

    PAllen I think that's for if the photon is energetic enough to see the quarks in the neutron (should be of GeV energy I guess?)....
    Otherwise the neutron is totally chargeless for the photon. It's the same for example in astroparticle physics, when we say that when Hydrogen (neutral) atoms formed - or when the photon Temperature fell ~1eV - the universe became transparent for the photons.
     
  19. PAllen

    PAllen 5,797
    Science Advisor
    Gold Member

    Transparent is a matter of probabilities. For that matter, it is not clear that photon-photon interactions have ever been observed due to very low probability, but they are theoretically allowed. Maybe the probability is a a cascade of 10-100, but I am not seeing an argument that it is exactly zero. Put another way, I would think you could say (per the SM) that a Z particle cannot facilitate pair production, while a neutron has some vanishing probability of doing so.
     
  20. mfb

    Staff: Mentor

    The Z couples to quarks, and quarks have charge, so on loop-level, you get photon/Z interactions as well (not sure how/if they might cancel out).

    Pair production at a Z would be... let's call it exotic :D.

    For the neutron, if the energy is low the quark charges are practically invisible because they cancel out nearly exactly then.
    Multi-GeV photons don't care about the hadronic structure and scatter at the partons.
     
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