I What is the difference between charge and force carriers?

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What is the difference between charge and force carriers?
I'm a bit confused about what difference there is (if any) between charge and force carriers. I imagine charge as a field around a particle that pushes or attracts other particles. But we're also told that bosons mediate the forces by exchanging. So there seems to be a redundancy unless charge is itself the exchange of bosons, in which case they are the same thing or two aspects of the same thing. Please elaborate. I'll understand answers with QM and a bit of QFT.
 
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JohnH said:
I'm a bit confused about what difference there is (if any) between charge and force carriers.
Force carriers, as you note, are bosons that mediate interactions by being exchanged. That means, heuristically, that the interaction involves fermions emitting and absorbing "force carrier" bosons. For example, in quantum electrodynamics, electrons (fermions) emit and absorb photons (the "force carrier" boson for the electromagnetic interaction).

"Charge" is the coupling constant that governs the emission and absorption of force carrier bosons by fermions. For example, the charge of the electron is the coupling constant that governs its emission and absorption of photons.
 
Appreciate the answer.
 
I think, you should forget about thinking in terms of "force carriers". The particles and interactions are described by relativistic quantum field theory, and the interaction is due to the fields, and usually it is evaluated using perturbation theory and Feynman diagrams, which are only on a superficial point of view depicting scattering processes being caused by the "exchange of virtual particles" (the internal lines of a Feynman diagram). They are rather symbols with a clear mathematical meaning, telling you how to calculate transition-probabilities between asymptotic free incoming and asymptotic free outgoing single-particle states, which can be detected. That's why the corresponding "external lines" symbolize in fact something that's really observable with a detector (say, e.g., an electron with some momentum and spin component). The internal lines stand for certain functions, the socalled time-ordered propagators and cannot so easily directly interpreted as "particles".

This becomes the more evident if you think about that not only the gauge bosons of a standard model are "force carriers" in this loose sense, because in a process like Compton scattering at lowest order perturbation theory the internal line is an electron line, i.e., in this case you'd think the electron is the "charge carrier".

Charge is indeed a property of the particles, and charged particles indeed always carry a corresponding field around them. A "bare electron" without its electromagnetic field is just an approximation to begin with the perturbation theory, where you just neglect all interactions between particles and thus also the fields each particle carries around are neglected. If you start to calculate interactions with perturbation theory, particularly for the case of QED you realize that this picture leads to some trouble. E.g., when calculating cross sections for Bremsstrahlung, i.e., the emission of an additional photon when scattering charged particles at each other, you get a divergent result for the total cross section due to socalled infrared and collinear divergences. The resolution is that you have to consider the particle together with its own electromagnetic field around it and that there are arbitrarily soft photons which might get undetected when measuring bremsstrahlung, i.e., at lowest order you have to take into account the possibility that there is also a scattering process without the extra photon but at higher-order contribution to the elastic scattering process (then instead of an observable photon there's an internal photon line leading to a closed loop in the Feynman diagram, i.e., a "radiative correction"). In other cases you have to even resum infinitely many diagrams. E.g., to find the description of an electron around a atomic nucleus by just treating the atomic nucleus as providing a Coulomb potential for the electron from quantum field theory you have to resum all "soft-photon ladder diagrams", which due to the singularities of the photon propagator all contribute at the same formal order in the power of the coupling constant. That's due to the masslessness of the electromagnetic field/photon, which results in the long-ranged nature of the electromagnetic interaction. From another point of view this implies that the "true asymptotic states" of charged particles are not the plane waves of the bare non-interacting particles but "Coulomb distorted" ones.
 
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