In general relativity, energy, including potential energy, creates mass. That accounts for things like the mass defect in atomic nuclei. But that potential mass-energy must also generate a gravitational force (curve space-time). My question is: where is that mass? Distributed between the 2 interacting particles? Distributed uniformly along the line connecting them? In the center of mass?
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$\begingroup$ In the classical case, you can simply check for the stress-energy associated with the EM field. $\endgroup$– SlereahCommented Feb 20, 2018 at 8:23
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Classically we need to consider the energy of the field associated with the interaction. For example if you consider a hydrogen atom it has a mass 13.6eV lower than an isolated proton and electron. The stress energy tensor for a Bohr hydrogen atom (this is classical remember) would include the masses of the proton and electron, the momentum of the electron and the stress energy of the electrostatic field. The mass deficit is associated with the change in the combined electrostatic fields as the electron and proton approach each other, so in this sense it would be spread out across the electrostatic field.
However life gets more difficult when we consider anything mediated by the strong and weak forces because these don't have a classical field. I think in that case we'd have to shrug and admit we need a theory of quantum gravity for a full description.
For nuclei I suppose we could treat the strong nuclear force as a classical field, though it is not a fundamental force. We'd have to associate a stress energy tensor with this effective field. I imagine that would work, though it would be an approximation at best.
answered Feb 20, 2018 at 9:22
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$\begingroup$ You don't need a theory of quantum gravity, you just need to compute the expectation value of the stress-energy tensor $\langle \psi | \hat{T}_{\mu\nu}(x) | \psi \rangle$ $\endgroup$– SlereahCommented Feb 20, 2018 at 10:37
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$\begingroup$ @Slereah doesn't that give you problems when dealing with superpositions? $\endgroup$ Commented Feb 20, 2018 at 10:39
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$\begingroup$ That it does, but it is expected to work in the limit of not having weird superpositions of states with disjoint support and $$\langle T_{\mu\nu}(p) T_{\rho \sigma}(q) \rangle \approx \langle T_{\mu\nu}(p) \rangle \langle T_{\rho \sigma}(q) \rangle$$ $\endgroup$– SlereahCommented Feb 20, 2018 at 10:40