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Which mass of the particle is the source of gravitational field? If we define mass as a pole of the propagator, and calculate loop corrections to the pole we get infinities. Now the way we get rid of these infinities defines what is our renormalized propagator and what is its pole. What is the mass of the particle is dependent on which renormalization scheme we choose. Now, gravity field knows about everything that happens to the particle (as any energy gravitates), all contributions of whatever virtual particles contribute to the mass of particle (up to very short distances).

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If you have a soft graviton emission from a particle, and the particle fluctuates into other stuff, you must also include the soft graviton emmission from the fluctuations, and you get the same soft-graviton emission as from the physical mass, because the energy momentum is conserved at each vertex. The graviton couples to a conserved energy-momentum. So the answer is that gravitons don't care what procedure you use, whatever the physical masses end up being, these will source the overall field. The source of gravity is energy-momentum, not the scalar mass, so the pole position is immaterial. –  Ron Maimon Jan 5 '12 at 17:32
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Classically, the source of the gravitational field is the energy momentum tensor $T_{\mu\nu}$. In the absence of a full quantum theory of gravity, a semiclassical approach is often taken, whereby the source of the gravitational field is the expectation value $\langle T_{\mu\nu}\rangle$. The procedure is quite tricky to carry out though, because $T_{\mu\nu}$ generally contains products of fields, and in the quantum case we would then be evaluating an expectation value of operators at the same spacetime point, with the obvious problem that causes. There are various schemes for handling this, discussed, for example here.

Edit: Here is an online reference talking about the semiclassical approach.

As these references point out, even the definition of a particle is not trivial in curved spacetime.

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The correct answer is "observable mass".

Renormalization can be used to get rid of infinities, but it provides no answer about the values of observable masses. And the problem is not in using different renormalization schemes. One can use the same scheme for electron and muon, and it is known from experiment that the result must be different.

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