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

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1. Jun 16, 2014

### tlnarasimham

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. Jun 16, 2014

### Matterwave

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.

3. Jun 17, 2014

### tlnarasimham

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.

4. Jun 17, 2014

### Matterwave

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

5. Jun 17, 2014

### tlnarasimham

Thank you matterwave I got it!

6. Jun 17, 2014

### Staff: Mentor

Where? It is completely wrong.
"W0" should be Z, I guess. It is a different particle.

7. Jun 17, 2014

### tlnarasimham

Yes it is z. it was my mistake in typing

8. Jun 19, 2014

### ChrisVer

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???).

9. Jun 19, 2014

### Bill_K

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.

10. Jun 19, 2014

### ChrisVer

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 $W^{0, \pm}_{\mu}, B_{\mu}$ and after you only have the $A_{\mu}$. Maybe that's a misconception....

11. Jun 19, 2014

### Bill_K

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.

12. Jun 21, 2014

### tlnarasimham

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.

13. Jun 21, 2014

### 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.

14. Jun 21, 2014

### ChrisVer

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:
$SU(2)_{L} \times U(1)_{y} \rightarrow U(1)_{em}$
so you remain with only the U(1)

Last edited: Jun 21, 2014
15. Jun 21, 2014

### 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.

16. Jun 21, 2014

### PAllen

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.

17. Jun 21, 2014

### ChrisVer

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.

18. Jun 21, 2014

### PAllen

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.

19. Jun 21, 2014

### 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.