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Since I studied General Relativity I had this question running on my mind. As I see it (just taking lectures of Quantum Field Theory right now)

"Why you need a gauge boson for gravity when the higgs brings things mass?"

You know, mass is the cause for physicist to study differential geometry so why we need another unproven particle, to exist, in order to describe it?

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You can work out that a theory of gravity based upon a scalar field will not be able to produce gravitational lensing, which is an observed effect. –  Jerry Schirmer Oct 26 '11 at 11:48
    
In the linear approximation, GR predicts gravity waves, which can be described by gravitons. This explanation comes up independent of the Higgs Boson. The need for one doesn't really have any bearing on the other. –  P O'Conbhui Mar 31 '12 at 0:15
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3 Answers 3

The Higgs gives things a rest mass in the standard model, but it is not the only source of rest mass. There is also confining strong interactions, which give hadrons a mass independent of the Higgs.

The graviton mediates gravity, while the Higgs doesn't. The Higgs force is short ranged. The classical graviton is arguably observed already, in the friction of pulsating binaries which matches the prediction of GR for decay due to gravitational waves.

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Uncertainty relation dictates, $m \simeq \dfrac{\hslash}{rv}$, where, $m$ is mass of exchange particle, $r$ is range of force and $v$ velocity of exchange particle.

As, the range of gravitational force in infinite, $r=\infty$, mass of proposed graviton should be zero.

Now, according to Standard model, particles acquire mass by interacting with Higgs field, which pervades everywhere in space. Now, Higgs bosons mediates the action of Higgs field, there mass as predicted is of the order of $ 1 \quad TeV/c^2$.

Hence, as my understanding goes Higgs bosons gives mass, which is the property due to which a force gravity occurs, which is mediated by gravitons which is massless.

They both are needed for different functions.

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One potential source of your confusion is that there are two different notions of mass even in classical physics, inertial mass and gravitational mass that happen to be identified to a high degree of experimental precision. One appears in $F = m_i a$ as a proportionality factor the other in the Newtons law for the gravitational force (I will neglect all factors like $4\pi$ in the following) $$F = \frac{G m_g M }{r^2}.$$ This observation is in fact the key to Einstein's equivalence principle. In general relativity the notion of a gravitational mass for large objects is sort of meaningless, to my knowledge it can only be defined with respect to test particles "at infinity". As you know test objects that are not believed to contribute significantly to the energy momentum tensor $T$, propagate along geodesics with respect to a metric $g$ that is a solution of the Einstein field equations $$R - \alpha\,\text{tr}(R)g = T$$ where $R$ is the Ricci tensor of the Levi-Civita connection of $g$. The Einstein field equations can be seen as a slightly more sophisticated form of the equivalence between gravitational and inertial mass. In fact one way to define the energy momentum tensor of an arbitrary Lagrangian density $L$ is by writing in a coordinate independent form and varying with respect to the metric $$\delta_g L = T \delta g.$$ So if you take $L$ to be the Lagrange density of the standard model, $T$ would determine the energy density of a particular classical solution. Perturbative methods in Quantum field theory always involve an expansion around a classical solution. In the case of the Higgs field in a non-trivial space time this would be the solution to a certain differential equation obtained from the Lagrangian in the usual way, possibly local solutions can only be patched together after gauge transformations. If space time is just flat Minkowski space one gets the usual picture of symmetry breaking.

Perturbative theories of quantum gravity fix a classical solution of the Einstein Field equations (that is a space-time, together with a certain metric $g$ on it), so in the case that $T$ contains expressions involving a Higgs field this would in principle influence the solution, however as soon as you choose $g$ to be anything but the Minkowski metric or the metric of some other symmetric space you will be hard pressed to even write down the propagator of a free field.

Gravitons are the quanta of the field that appear when you quantize the Lagrangian for the linear perturbations $g + \delta h$ around a fixed background metric. Classically those are known as gravitational waves. If you were to write down the whole standard model Lagrangian intrinsically and and compute the perturbation ($g + \delta h$) to determine the couplings, then you would discover that the graviton in fact couples to every field (massless or not) but very weakly. I think Feynman sketches some of those computations in his text on quantum gravity for QED.

So the Higgs field and the graviton appear at conceptually different places in any attempt to perturbatively quantize gravity with matter. If you only consider quantum field theory in a curved background, then there is no graviton. It is just much harder to actually do explicit perturbation theory and there is no well defined notion of mass to begin with (unless you assume that at infinity space time is minkowski space), which in quantum theory is intrinsically tied to the representation theory of the Poincare group.

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