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The wikipedia article describes how in QM exchanges of gauge bosons carry force, and describes how the graviton may also be a gauge boson.

If the observable universe is made of around 10^80 hydrogen atoms*, this implies each atom should be exchanging gauge bosons with 10^80 other atoms. So each atom should launch 10^80 gravitons, and a similar number of photons, gluons etc.

Is this really how it works? Does one proton launch 10^80 bosons when it comes into existance?


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Bosons are quanta of energy in a quantum field, and there is an uncertainty principle relating field intensity and particle number (see "coherent state" vs "number state"). The static gravitational field of a proton would probably be represented as a coherent state, meaning that it was a superposition of number states for every possible number of gravitons. – Mitchell Porter Jan 11 '13 at 3:56
These concepts of quantum field theory are a little hard to grasp at first encounter, but perhaps the simplest answer to your question is that the quantum gravitational interaction between the proton and the rest of the universe should involve arbitrarily large numbers of gravitons, and not just one for each other particle. – Mitchell Porter Jan 11 '13 at 3:59
@MitchellPorter is correct. Also, in the language of particle exchange (which is by no means the only way to talk about this), static forces are generated by "virtual" particles. These are particles which don't obey the usual relationship between energy and momentum (really this is because they are an artifact of a certain approximation method in quantum field theory - the fundamental thing is the field, not the particles). The virtual particles behind a static force have zero energy, so there is no obstruction to making an infinite number of them. – Michael Brown Jan 11 '13 at 4:35

2 Answers 2

It is meaningless to talk about the precise number of virtual particles corresponding to the interaction of a certain particle with other particles. It is simply not well-defined, neither in the mathematical framework of the path integral formalism of quantum field theory, nor if one uses them as a heuristic picture of reality. Virtual particles appear in a context where one needs properly defined in- and out-states, then one can formulate a path integral and draw Feynman diagrams, and in principle there are infinitely many of them. As virtual particles correspond to internal lines in Feynman diagrams, this circumstance makes counting them quite hard. Heuristically, one could say that we only count those diagrams that contribute the most to a given process and say that it is mediated by the corresponding virtual particles. However, one cannot formulate the process of the interaction of a proton with "all other atoms" in such a way that is is possible to acquire a meaningful answer.

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Since the commenters are not turning their comments to an answer, I will summarize:

If the observable universe is made of around 10^80 hydrogen atoms*, this implies each atom should be exchanging gauge bosons with 10^80 other atoms.

So far so good. The field theoretical view of interactions says that these happen by the exchange of virtual particles between "real particles", real means that they are on their mass shell.

This Feynman diagram

enter image description here

Shows the on shell particles: The kaon which decays to three pions, which have definite mass, and off shell ones, the quarks and the gluon and the W which are off mass shell and virtual and thus do not obey energy and momentum conservation laws. This is most striking in the case of the W exchange, where the mass of the W on shell (real) is orders of magnitude larger than the mass of the real Kaon. They are just quantum number representations similar to the correspondingly named on shell particles ( gluons and quarks are never free, so never on shell, but that is another story).

So each atom should launch 10^80 gravitons, and a similar number of photons, gluons etc.t

No. Each atom theoretically can interact with every other atom through virtual particle exchanges, though these interactions will be extremely extremely weak the larger the distance between the on shell particles. These off shell particles may be one or many, from each hydrogen atom, though in the calculation usually a few exchanges are dominant.

Launch implies on shell manifestation, in my vocabulary.

What you should keep is that there is no conservation on the number of off the mass shell virtual particles other than the diminishing probability of multiple exchanges, diminishing at least for weak, electromagnetic and gravitational interaction. Gluons are another type of animal and another story too.

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Though I do hesitate to say that the off-shell particles are "there" at all. The feynman diagram above is, after all, ultimately an artifact of perturbation theory, which depends on our approximation scheme, not on physics. – Jerry Schirmer Jul 24 '14 at 20:08
@JerrySchirmer Sure. But the very concept of a virtual particle comes from such diagrams , i.e. the format of calculating crossections for a process. It is the successful fitting of data with calculations that gives them "existence" , as poles, with the quantum numbers of the "real" particles. – anna v Jul 25 '14 at 4:12

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