# I Higgs and the Graviton

1. May 27, 2016

### putongren

What's the difference between the Higgs boson and the hypothetical graviton particle? Do they both have to do with the mass of matter? I guess what I'm trying to ask is how are the Higgs particle and the graviton related.

Last edited: May 27, 2016
2. May 28, 2016

### Orodruin

Staff Emeritus
They have absolutely nothing to do with each other.

Gravity couples to the stress-energy tensor, not to mass.

3. May 28, 2016

The graviton is the "gauge boson" which mediates the gravitational force (couples to the stress energy tensor) between objects in the QFT framework. I put this in quotes because although the graviton transforms under diffeomorphisms, gravity is not a typical gauge theory. For example, you have the Weinberg Witten theorem which says a theory with a poincare and gauge invariant conserved stress energy tensor in 3+1 d cannot admit a particle with spin greater than 1, and you usually you have a conserved/covariantly conserved current in a gauge theory which the photon for example couples to. However, this may be resolved somewhat in the AdS/CFT correspondence. Using the AdS/CFT "dictionary" the boundary stress tensor corresponds to the graviton.

The Higg's boson as a fundamental particle gives the bosons mediating the electroweak force mass. If you have a mass term in the Langrangian, it is violates gauge invariance. Adding the Higg's field resolves this. The Higg's mechanism also happens in superconductors, which I think is a good first example. With global U(1) spontaneous symmetry breaking you get a gapless Goldstone boson like the phonon in a superfluid. However this does not happen for a gauge theory. Adding the Higg's boson causes the Goldston boson to be eaten and just generates a massive mode for the photon. This is what prevents a magnetic field from penetrating a SC and causes the Meissner effect.

Another interesting thing is that in AdS/CFT, a way to break translation symmetry on the boundary is to "Higgs" the graviton. This is something I am interested in since this is often done to calculate conductivities.

4. May 30, 2016

### putongren

Radium, I had a hard time understanding your post. I just want to check if my understanding is correct. The Higgs boson "causes" inertial mass in certain sub atomic particles, while the graviton carries the gravitational force. Is that correct?

5. May 30, 2016

### ftr

you are correct on the first one.The gravitons carry the energy away in the form of "gravitational wave"(the recent LIGO experiment as an example), however, the "virtual gravitons" are responsible for the gravitational force.

6. Nov 28, 2016

### putongren

If Einstein said that inertial and gravitational masses are equivalent due to the Equivalence principle, wouldn't the Higgs boson and the graviton be equivalent also?

7. Nov 28, 2016

### Orodruin

Staff Emeritus
No. This is simply not how the world or physics works. You cannot take some words that look nice together and base your theory on it.

And that is not what the equivalence principle says. It says it is always possible to find a local inertial frame. It explains the observation from classical physics that inertial and gravitational masses are equal.

8. Nov 29, 2016

### haushofer

Isn't this reversed reasoning? As I see it, you use the equivalence of inertial and grav.mass to reason that gravity is geometrical. But the reason why they are both equivalent, we don't know.

9. Nov 29, 2016

### haushofer

One confusion people sometimes have, is that because the Higgs particle is a boson, it should carry a force. This is not the case.

10. Nov 29, 2016

### Orodruin

Staff Emeritus
This might have been the historical path, but I have always seen it the other way around. Because gravity is geometrical, "gravitational mass" is just the inertia in an accelerating frame.

To me, the question is "how does the concept of gravitational = inertial mass emerge in classical mechanics limit of GR?" The equivalence is a postdiction of GR. You lift the question of why they are equal to "why is GR true?"

11. Nov 29, 2016

### Staff: Mentor

It does, sort of, it leads to a Yukawa interaction between all particles with mass, with coupling constants given by their mass. Jester discussed it in more detail.

12. Nov 29, 2016

### vanhees71

Of course, the Higgs field (as any other field in QFT) both represents particles (in a very specific meaning as asymptotic free Fock states) and provides interactions between fields/particles (including itself). In addition the Higgs field has a non-vanishing vacuum expectation value, which leads to masses for the quarks and leptons via the mentioned Yukawa couplings of the Higgs field to the fermion fields.

13. Dec 1, 2016

### ChrisVer

This is wrong...
The Higgs Boson has nothing to do with the mass of the particles (except for its couplings)... It's connected to masses just as much as an undertaker is connected to a murder: it's an observable leftover of the theory we had to incorporate masses into the Standard Model.

14. Dec 1, 2016

### vanhees71

The vacuum expectation value of the Higgs field leads to the masses of the leptons and quarks. The "eating-up" of the would-be Goldstone modes of the Higgs fields by the gauge fields (in the unitary gauge) leads to the mass of the corresponding vector mesons. You are "higgsing" the local gauge group of quantum flavor dynamics $\mathrm{SU}(2)_{L,\text{flavor}} \times \mathrm{U}(1)_{Y}$ down to $\mathrm{U}(1)_{\text{em}}$, implying that you have 3 massive gauge fields (the W and Z boson fields) and one massless gauge field (the photon), the quarks and leptons as well as one Higgs boson in the physical spectrum.

15. Dec 3, 2016

Yes that is what I was referring to. If the Higgs wasn't there, the gauge bosons could not have a mass since that would violate gauge symmetry.

Consider the U(1) case. If the photon were to have a mass, the little group would be SO(3) instead of ISO(2) since you can choose a rest frame for a massive particle. It is easy to see however that the longitudinal polarization is forbidden by the Ward identity when you look at Lorentz transformations (which is a way to derive the Ward identity) of matrix elements. Having a mass would also make the theory look nonrenormalizable from the large k limit of the propagator.

The abelian Higgs mechanism for example basically comes from expanding about a new choice of vacuum in the Mexican hat potential when you have a negative mass term for the complex scalar field and hence a nonzero vacuum expectation value. Without a gauge field, you have massless goldstone bosons (from the phase of the scalar field) are along the bottom of the hat and the massive mode is up the sides. (The amplitude fluctuations).

Usually when you have a massive photon for example in let's say just pure maxwell theory,it clearly violates gauge invariance. But when you consider coupling to a gauge field in the system I mentioned above a mass of the gauge field is generated from expanding about the solution with the higgs, and the goldstone boson becomes the longitudinal polarization of the photon. So you now you have the Higgs, 2 transverse polarizations of the photon and one longitudinal (1+2+1), whereas before you had a complex scalar and transverse polarizations of the photon (2+2).

So here when I say the gauge boson has a mass, I am referring to a particle with a longitudinal polarization and the fact that in a superconductor, the B field has a penetration depth which looks like what you would get if you solved for the photon progator including a mass.

16. Dec 5, 2016

### vanhees71

That's only partially right. For the Abelian case, i.e., U(1) gauge symmetry you can have a massive gauge boson without an additional Higgs boson via the Stueckelberg mechanism. The upshot of the analysis is that you can just write down a naive mass term for the gauge field. As long as you minimally couple to a conserved current, you have a gauge theory, and it's renormalizable like QED. Only the "photon" becomes massive.