28
$\begingroup$

When two like magnetic poles are brought together, there's a repulsive force felt that's inversely proportional to their separation. In the standard model, the answer to "What is transmitting this repulsive force through empty space between the two magnets?" is described as virtual photons.

If I want to measure 15 newtons of force between two north poles of adjacent magnets, I can position my magnets accordingly and measure the force directly. I'll never see photons involved because of their virtual nature, but the force they're delivering is very real, and easy to measure.

If I want to produce the same amount of force on my magnet by directly bombarding it with real photons, however, it would take an enormous amount of energy.

It seems strange to think that the same particles responsible for producing a force strong enough to keep two massive objects apart, are barely capable of moving a light sail in microgravity.

Why are real photons so much less efficient in carrying momentum than virtual photons?

I have to believe virtual particles are the topic for a sizeable portion of questions on SE; if this is a duplicate please feel free to close, but from my review I haven't seen this addressed directly.

Improve this question
$\endgroup$
5
  • 8
    $\begingroup$ FWIW, I think that this question is just as well posed within classical E&M ("Why do static electric or magnetic fields generate strong forces so much more easily than EM waves?"), so perhaps there is nothing really "quantum" about it. $\endgroup$
    – Rococo
    Commented Mar 22, 2022 at 0:19
  • 2
    $\begingroup$ It sounds like you are saying you need "more" real photons than virtual photons, without ever asking the question 'How "many" virtual photons are needed to produce 15 Newtons of force?' $\endgroup$
    – chepner
    Commented Mar 22, 2022 at 13:14
  • $\begingroup$ @chepner I think closer to my intent is asking why the magnetic repulsion transmitted by these virtual photons is so robust in the absence of other effects, whereas to produce that much force with real photons you'd be able to detect many other effects due to the large amount of energy involved to impart that much momentum. Whether it's done using one high energy photon or 10^30 photons, to transfer the equivalent momentum would be a tall order. $\endgroup$
    – JPattarini
    Commented Mar 22, 2022 at 15:48
  • 1
    $\begingroup$ @Rococo That's a great point... and I don't have a good handle on the classical answer to that either $\endgroup$
    – JPattarini
    Commented Mar 22, 2022 at 15:50
  • $\begingroup$ @JPattarini I'm inclined to say that ACuriousMind's explanation works just as well classically- EM waves are constrained to a particular dispersion relation that does not apply to static fields or near-field radiation. $\endgroup$
    – Rococo
    Commented Mar 24, 2022 at 19:22

2 Answers 2

Reset to default
40
$\begingroup$

Within the usual handwavy accounts of virtual particles, the answer is "simple": Virtual particles aren't required to obey on-shell mass-energy relations (in this case $E=pc$), so there can be virtual particles with large momenta but very small energies.

However and as usual, I would advise not to think in terms of virtual particles at all - they are artifacts of drawing perturbation theory as Feynman diagrams and you cannot even say non-perturbatively what they are supposed to be. The reason "virtual photons" act so differently from actual photons is that the term "virtual photon" doesn't describe a quantum state that would resemble a free, real photon at all, it describes a certain computation in an interacting quantum field theory. There isn't really any reason except for the name to expect this to have anything to do with the behaviour of actual photons.

See this answer of mine for a lengthier discussion on why it is misleading to think of "virtual particles" as particles or as actual intermediate states at all.

Improve this answer
$\endgroup$
1
  • 8
    $\begingroup$ I would add for clarity that virtual lines in Feynman diagrams have the names of particles because it is a useful mnemonic for quantum number conservation between vertices. There are not only virtual photons but virtual particles of the whole standard model table of particles, for particle physics calculations. $\endgroup$
    – anna v
    Commented Mar 22, 2022 at 5:02
-6
$\begingroup$

The electric fields of charges of the same sign behave like elastic bodies (where there is a body, there cannot be a second one). If two charges of the same name are approached, they recede in the free state, as elastic bodies do.

Opposite charges, however, attract each other and it is observable that they emit EM radiation. In order to loosen a connection again - i.e. to remove an electron from the atom, photons are sufficient (as sole and at the same time necessary condition) to restore the fields around the opposite charges and to consider the particles isolated from each other again.

To get a capacitor with separated charges, you have to separate the electrons from the rest atom. This can be done by a solar cell. Photons hit electrons, these absorb the photons, are accelerated, penetrate an isolation layer and must take the way back over a consumer. So for example our capacitor.

A magnet, on the other hand, is only a mediator. If you move a magnet near an electrical conductor, the magnet is not consumed. Its magnetic field causes a Lorentz force, i.e. the displacement of charges. Thereby, the movement of the magnet is opposed by a resistance, but its magnetic field remains unchanged.

What enables us to turn the concept of virtual photons into a science of interaction between fields is the modeling of the structure of these fields. Because why fields with the same sign behave like elastic bodies, there is currently no model-like idea for it.

Virtual photons are an embarrassment, which will dissolve as soon as the scientific interest in the investigation of the inner structure of fields leads to new model conceptions.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Re "Virtual photons are an embarrassment": Because they are not a physical reality (whatever that means)? Couldn't the same be said about forces in general? Or the wave function? They can be very useful (say, predictive power). $\endgroup$
    – Peter Mortensen
    Commented Mar 23, 2022 at 16:46
  • $\begingroup$ I don't think they are on the same ground, you could carry out calculations without virtual phonons and obtain the same prediction, which is a measurable result. This is not the case for forces in classical mechanics or the wave function in quantum mechanics, they are "real" in the sense that they can be measured (somewhat directly), while this can not be done for virtual phonons, as far as I know. I agree that the answer is a bit excessive, but predictive power and physical reality (in the sense of measurement) I would say are quite distinct. $\endgroup$
    – Another User
    Commented Mar 26, 2022 at 10:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.