Could there exist a “locality” field? [closed]

What I mean is (and I'm a layperson on the subject), can there exist a field that pervades the universe - like the Higgs field - that interacts with particles to give them "distance" or "space" between one another, in a similar way that the Higgs field give particles their mass?

And could an excitation or de-excitation of this "locality" field affect the space in between two particles in space? If such a field could even be possible, that is.

According to General Relativity, there is a dynamic field, called the “metric field”, that pervades the universe and determines its geometry. The spacetime distance between events is determined by this field. It determines which events can causally influence other events. Its influence on particles is what we know as gravity.

Ripples in this field are called gravitational waves. They were first detected in 2015.

Quantum excitations in this field are called gravitons. They are theoretical and have not been observed. We may never be able to detect them because they are so weak.

• +1: This is an interesting way to think about the metric field. I am not sure how relevant this point is but one doesn't need a dynamical metric field to determine the geometry of spacetime. One can have a fully Minkowskian non-dynamic metric field in a universe absent of gravity which can equally determine the geometry of spacetime. – Dvij D.C. Jul 14 '19 at 18:24
• @FeynmansOutforGrumpyCat Not really. Spacetime is a Fourier conjugate of matter (energy-momentum). They cannot exist without or independently of each other just like the spectrum of light cannot exist without light. – safesphere Jul 15 '19 at 1:19

Would like to point out the similarity and dissimilarity between "Higgs field" and the "locality field" (a.k.a. metric):

• Like the Higgs field, the "locality field" also acquires a non-zero vacuum expectation value (Minkowski metric) that breaks the local Lorentz gauge symmetry and diffeomorphism invariance. There is nothing, I mean NOTHING, in the general relativity that tells you that the vacuum should be Minkowskian. Within the framework of general relativity, it's perfect fine that the vacuum metric is $$g_{\mu\nu}=0$$, which preserves both the local Lorentz gauge symmetry and diffeomorphism invariance. The non-zero Minkowskian metric $$g_{\mu\nu}=\eta_{\mu\nu}$$ is accidental, in the sense that it happens to be determined by the evolutionary history of our Universe.

• Gravitons are Nambu-Goldston bosons, which are "excitation or de-excitation" from the symmetry breaking Minkowskian VEV. Whereas, for the Higgs field, the Nambu-Goldston boson is "eaten" by the Higgs mechanism. The tricky part of the story is that metric field is a tensor while Higgs is a scalar, which makes the NG boson analogy only partially correct.

• The gauge field acquires mass via the kinetic term $$(D\phi)^2$$ of the Higgs field. For the "locality field", the situation is a bit different: the covariant derivative $$De$$ of the tetrad $$e$$ (in lieu of the "locality field" $$g_{\mu\nu}$$) vanishes, since the zero torsion condition enforces that $$De = de + \omega e = 0$$, where $$\omega$$ is the spin connection (which is the Lorentz "gauge field" in gravity).
• In my understanding, the graviton field (i.e. the dynamical metric field) should break the global Lorentz invariance but neither the gauged local Lorentz invariance or the diff invariance. If the graviton field were to break a local gauge invariance, gravitons should acquire mass, right? In other words, a Nambu-Goldstone boson should correspond to the SSB of a global symmetry and not a local one. Otherwise, it would be a Higgs-like massive boson. Let me know where I am going wrong. – Dvij D.C. Jul 14 '19 at 19:09
• @FeynmansOutforGrumpyCat, it's the non-zero Minkowski metric (not the deviation from it, i.e. the graviton) that breaks both the local Lorentz gauge symmetry and diffeomorphism invariance. The tricky part of the story is that metric field is a tensor while Higgs is a scalar, which makes the NG boson analogy only partially correct. – MadMax Jul 14 '19 at 19:22
• Ah, I see. Thanks. To confirm, it breaks the diff invariance spontaneously in the sense that the Einstein equations are manifestly diff invariant but a particular non-Minkowskian metric isn't, correct? – Dvij D.C. Jul 14 '19 at 19:27
• @FeynmansOutforGrumpyCat, the standard model is electroweak invariant, however, once the Higgs acquires the non-zero VEV, the symmetry is said to be spontaneously broken. The same situation goes for the "locality field", when it acquires the Minkowskian metric, the local Lorentz symmetry is lost. There is nothing, I mean NOTHING, in the general relativity that tells you that the vacuum should be Minkowskian: It's happens to be an observational fact, like the Higg VEV, which is a free parameter set by the initial condition of our Universe. – MadMax Jul 14 '19 at 19:45
• @FeynmansOutforGrumpyCat, within the framework of general relativity, it's perfect fine that the vacuum metric is $g_{\mu\nu} = 0$, which preserves both the local Lorentz gauge symmetry and diffeomorphism invariance. The non-zero Minkowskian metric $g_{\mu\nu} = \eta_{\mu\nu}$ is accidental, that happens to be determined by the evolutionary history of our Universe. – MadMax Jul 14 '19 at 19:55