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Does the equivalence between inertial and gravitational mass imply anything about the Higgs mechanism?

In Higgs mechanism, Higgs field, which likes syrup, slows down particles when they passing through. So it seems Higgs field gives particles inertial mass. But what gives particles gravitational mass? We know, particles can attract each other even when they are static.

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marked as duplicate by Qmechanic, Manishearth Dec 11 '12 at 11:43

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Gravitational mass is a bit of a misnomer, because in General Relativity the spacetime curvature is determined (mostly) by the energy density. Mass is simply treated as equivalent to the amount of energy given by $E = mc^2$, or conversely energy is just treated as the equivalent amount of mass.

So the fact the particles are massless above the electroweak symmetry breaking energy and have a mass below it (acquired through the Higgs mechanism) makes no difference to gravity.

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Do you mean that particles were not affected by Higgs field above electroweak symmetry breaking energy? But if a particle's energy is between electroweak symmetry breaking energy and the GUT energy, then it can also attract others by gravity, i.e. have 'gravitational mass'. –  Popopo Sep 15 '12 at 4:55
Yes, a massless particle will exert a gravitational attraction on objects near it because of its energy density (though for an elementary particle gravity is negligable compared to the other three forces). –  John Rennie Sep 15 '12 at 5:36
So is the mass given by Higgs field different from mass given by the Mass-Energy Formula? –  Popopo Sep 15 '12 at 10:33
This risks getting a bit complicated, but in brief the mass/energy of a particle is the source of the gravitational field of the particle, and it's the total energy density that matters not whether the mass is zero or non-zero. However the way the particle is affected by external gravitational fields does depend on the mass. Zero mass particles are deflected differently from massive particles in an external gravitational field. Zero mass particles always follow null geodesics, and a massive particle cannot follow a null geodesic. –  John Rennie Sep 15 '12 at 15:26
Okay, I see. So does the formula $F_{12}=-G\frac{m_1m_2}{r^2}$ should be rewritten as $F_{12}=-G\frac{\frac{E_1}{c^2} m_2}{r^2}$? –  Popopo Sep 15 '12 at 15:46

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