Higgs field

  • #1
Why dont the photon and the neutrinos couple to the higgs field????
Is there any associated charge which determines which particles couple to the Higgs field????
 

Answers and Replies

  • #2
33
0
Photons do not couple to Higgs field by an experimental fact as we now that the photon is massless. It is not a mathematical reason but an experimental datum.

Instead, neutrinos have mass and this somewhat changed the way standard model should be formulated.
 
  • #3
But is there no underlying reason????
are you implying that the photon lacks a kind of charge which prevents it from coupling to the Higgs field?????
What is this charge called as by the physics community???
eg, electric charge,weak charge,colour charge etccc
 
  • #4
blechman
Science Advisor
779
8
But is there no underlying reason????
are you implying that the photon lacks a kind of charge which prevents it from coupling to the Higgs field?????
What is this charge called as by the physics community???
eg, electric charge,weak charge,colour charge etccc

it's called electric charge! the Higgs boson is neutral, and hence cannot couple to the photon (to leading order). It DOES couple at higher order in perturbation theory. for example, the decay higgs -> 2 photons is a potentially very interesting decay channel, but it has to go through loops (higgs -> top+antitop, which then annihilate into two photons).

The neutrino is much more subtle, as Lester says. Depending on HOW the neutrino gets a mass (is there a right-handed neutrino?!) will determine how it couples to the Higgs. This is still an open and hotly debated question.
 
  • #5
I've got an idea. Photons are the left overs of the interaction of Higgs bosons with Black Holes. I know that this idea may sound crazy, but it won't hurt to speculate. In General Relativity, we know that gravity drags escaping Photons making them to lose energy (redshift). In the same way picture Higgs bosons escaping the tremendous gravitational pull of a Black Hole, but instead of redshifting, they get stripped of mass and become massless photons while its mass decays into several other subatomic particlas such as quarks or leptons. This statement is oversimplified, but it may be the start of something in the understanding of the Higgs mechanism.
 
  • #6
4,254
1
it's called electric charge! the Higgs boson is neutral, and hence cannot couple to the photon (to leading order). It DOES couple at higher order in perturbation theory. for example, the decay higgs -> 2 photons is a potentially very interesting decay channel, but it has to go through loops (higgs -> top+antitop, which then annihilate into two photons).

The neutrino is much more subtle, as Lester says. Depending on HOW the neutrino gets a mass (is there a right-handed neutrino?!) will determine how it couples to the Higgs. This is still an open and hotly debated question.

Blechman, I'm somewhat confused. If the Higgs couple to electric charge, how can the weak Z and the neutrino have mass, or do you mean there should be additional mechanisms?
 
  • #7
blechman
Science Advisor
779
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Blechman, I'm somewhat confused. If the Higgs couple to electric charge, how can the weak Z and the neutrino have mass, or do you mean there should be additional mechanisms?

Not the Higgs, the PHOTON - the photon can only couple (at leading order) to objects with charge. Since the Higgs boson is electrically neutral, it cannot couple to photons except through higher-order processes (through virtual top quarks, for instance). The Z boson doesn't couple to charge, but to a linear combination of charge and weak isospin. Since the Higgs has weak isospin (1/2) it can couple to the Z, but not the photon.

Hope that helps.
 
  • #8
blechman
Science Advisor
779
8
I've got an idea. Photons are the left overs of the interaction of Higgs bosons with Black Holes. I know that this idea may sound crazy, but it won't hurt to speculate. In General Relativity, we know that gravity drags escaping Photons making them to lose energy (redshift). In the same way picture Higgs bosons escaping the tremendous gravitational pull of a Black Hole, but instead of redshifting, they get stripped of mass and become massless photons while its mass decays into several other subatomic particlas such as quarks or leptons. This statement is oversimplified, but it may be the start of something in the understanding of the Higgs mechanism.

And where does electromagetism come from? Photons and Higgs bosons share Bose-Einstein statistics, but that's about all they share! They have VERY different interactions; they have different spin (photon is spin-1 while Higgs is spin-0); they have different charges (photon is neutral while Higgs has weak isospin and weak hypercharge); photons are gauge bosons that transform nonlinearly under gauge transformations, while Higgs bosons transform linearly (in fact, trivially)... The list goes on. Gravity cannot account for these fundamental differences. Black holes may be weird, but they're not THAT weird!

This program won't work; but you should keep studying and trying out ideas...
 
  • #9
4,254
1
Not the Higgs, the PHOTON - the photon can only couple (at leading order) to objects with charge.

Hope that helps.

No, sorry. One of us misunderstood the question posed by quantumfireball. I think the man was asking what charge is conserved on interaction with the Higgs, as though the Higgs bosons were force carriers.
 
  • #10
No, sorry. One of us misunderstood the question posed by quantumfireball. I think the man was asking what charge is conserved on interaction with the Higgs, as though the Higgs bosons were force carriers.

Well i meant whatever i asked and Mr Belchmann has answered my question well.
However since i have not learnt about weak interactions i am not familiar with weak isospin.
However its clear to me now why the electromagnetic field does not interact with the higgs field,since th em field only couples to charged fields.That is by imposing local U(1) invariance on the charged field one has to compulsory introduce a four vector A,and field lagrangian is invariant together under U(1) and A->A+del(A).
But i was wondering what would happen to neutral fields under u(1)
 

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