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Why don't we call the fermions in the standard model force carriers? Because the Standard Model is what it is. By the way, I think it's far less complete than people make it out to be, and that it comes with some unfortunate baggage. Consider for example an electron and a positron interacting. People say they interact via virtual-photon force carriers: ...


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If we identify a force as a scattering process, i.e. with a mediator of some interactions, then this need not be a vector boson of course. One can speak of "Higgs" force for instance if the process under consideration is mediated by the the Higgs (which is a scalar). There are also numerous cases where the interaction is mediated by a fermion. Therefore, ...


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Maybe there IS an effect similar to what you describe. In photoelectric absorption, the permissible departing electron states in a material determine the probability of absorption (the rate). This is easily seen in X-ray absorption near a photoelectric excitation threshold (like a K-edge), and is the basis of XAFS (X-ray absorption fine structure) ...


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It might help if you provided context or references for your issue. As it stands, for non vanishing m, which is necessary if you are talking about spin and not helicity, the first equation is a definition of the auxiliary field function $$ \psi^\mu \equiv \frac{\hat{p}^\mu}{mc} ~\phi $$ which makes the second equation the K-G equation for φ. I'm not sure ...


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This is a very tough question, since you do not make any further assumptions about the force - most importantly its strength, sign, variation with distance and objects it acts upon. In general, yes, the universe would probably look absolutely different and we would not be here to ask this question. Somebody or something absolutely else might. Our ...


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The situation is not symmetric at all: This diagram describes a force between two fermions, but a diagram such as just doesn't exist (in the Standard Model). Fermions can in fact mediate a force between bosons, like in: Such diagrams are highly suppressed loop diagrams though, and the one above would after renormalization be seen as just one ...


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I think that this is more about the historical construction of the theory than about the actual interactions. In a lagrangian, two fields A, B interact when there is a product term of both such as AB. So, I see no real fundamental distinction there, even with more complicated expressions. But when one introduces the interaction bosons, it's by the mean of ...



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