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I was reading an answer in this exchange trying to explain Hawking Radiation to non-physicists (like myself) An explanation of Hawking Radiation .

It informed me that the whole "virtual particle explanation" is just an analogy that shouldn't be taken literally. That makes me wonder if virtual particles in general are just an analogy to explain processes in quantum fields that are too mathematically involved for the layman to understand?

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I think that is one useful way of thinking of it. You can say that Feynman found this to be a neat way of organizing the quantum field calculations, but that it is just a computational convenience.

Another way of thinking about it is to say that while those virtual particles are not real in any sense, the whole Feynman diagram approach does come close to an important pattern in nature or that it allows us to use our intuitions to think well about these things. That is, there "is something there", but it is not so much virtual particles or graph vertices as an overall pattern of relations (this can shade over into the first view).

No doubt there is some physicist or philosopher who insists that virtual particles are actual things mostly to be contrary, but I think most people fall somewhere on this convenience/pattern continuum.

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  • $\begingroup$ Virtual particles are by definition not observable (by the HUP), so how could anyone rightly claim that such particles do exist? There's no way to know! $\endgroup$ – N. Steinle Aug 29 '18 at 22:00
  • $\begingroup$ @N.Steinle: there are lots of things that aren't observable. Nobody has ever seen the molten iron core at the center of the Earth; we just infer that it is there by looking at its effects. So how can you rightly claim that it actually exists? (Indeed it does, while virtual particles are clearly in some sense less real, even though they have real effects. But the argument you make in your comment just doesn't hold water.) $\endgroup$ – Peter Shor Sep 22 '18 at 22:45
  • $\begingroup$ @AndersSandberg Fair enough. It's just that virtual particles are far more "unrealistic" than anything else - especially than something like a planetary core - because not even in principle can we ever directly measure a virtual particle, but we still hold out that some day there might be a way to directly measure the core of a planet. $\endgroup$ – N. Steinle Sep 22 '18 at 23:37
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Virtual particles exist in the mathematics of Feynman diagrams, and are just labeled by the particle name because they have the quantum numbers of the particle, so that incoming and outgoing balances of quantum numbers can be checked, but not the mass, which is off mass shell, and variable under the integral described by the feynman diagram. An example:

enter image description here

This is electron electron scattering into an electron electron, by the exchange of a virtual photon. The diagram represents an integral where the input and output particles are on mass shell, i.e. the length of their four vectors are the mass of the particles. The intermediate photon has a varying four vector with no fixed mass.

For Hawking radiation see my recent answer and also here.

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TLDR: Thinking of virtual particles as an "analogy" is a trick that makes our brains hurt less when we think about quantum mechanics (and thus is very useful), but it in no way corresponds to an actual real-life distinction between two different kinds of particles. They are two ends of a phenomenon that stretches continuously between the two. So if you say that virtual particles are just an analogy, philosophically you would have to conclude that real particles are also just an analogy. (But I wouldn't say that this is an inaccurate statement.)

Detailed answer:

There is actually no sharp distinction between "virtual" and "real" particles—they are two ends of a continuum. Looking at the virtual particles page in Wikipedia, it describes a number of ways in which virtual particles differ from real particles. Let me go through these.

A virtual particle does not precisely obey the energy–momentum relation $m^2c^4 = E^2 − p^2c^2.$ Its kinetic energy may not have the usual relationship to velocity–indeed, it can be negative. This is expressed by the phrase off mass shell.

In fact, "real" particles do not obey this relationship exactly, either ... the longer the real particles exist, the closer they need to come to obeying the relationship, so any particle that is around for a long time nearly satisfies it, and thus assuming that it satisfies it exactly is a very good approximation. But unless the particle lives forever, it is only an approximation and not exact.

Virtual particles are also viewed as excitations of the underlying fields, but appear only as forces, not as detectable particles. They are "temporary" in the sense that they appear in calculations, but are not detected as single particles. Thus, in mathematical terms, they never appear as indices to the scattering matrix, which is to say, they never appear as the observable inputs and outputs of the physical process being modelled.

What this says is that in a scattering matrix, any particle that you start out with or end up with is "real", but any particle that is created and then destroyed in the interaction is "virtual". But in real life, a particle coming out of one interaction is going to end up being destroyed in a different one. If it lasts a long time between the two interactions, you call it "real", and if it lasts only a short time, you call it "virtual". But these are two ends of a continuum, for which no exact dividing line exists.

So if you say that virtual particles are just an analogy, then philosophically, wouldn't you also have to conclude that all particles are also an analogy?

It actually is very useful to differentiate between "virtual" particles and "real" particles when thinking about high energy physics, and I don't want to discourage anybody from making this distinction. But this distinction doesn't correspond to any actual sharp distinction in real life.

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  • $\begingroup$ Thanks for clearing things up for me. It's too bad I can check only one answer here. $\endgroup$ – Jack R. Woods Sep 10 '18 at 17:31

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