1
$\begingroup$

In the Standard Model of QM, all forces are mediated or carried by particles (for want of a better word) called bosons. The photon is an example of a force-carrying gauge boson, and mediates the electromagnetic force.

The electrostatic and magnetostatic forces are not mediated by real photons, and the answer here explains that they are not mediated by virtual photons either, while the answer here goes way over my head.

Put simply, what particles are these forces carried by - or aren't they after all?

As an example, consider a charged fragment of paper levitated above a charged metal plate. Nothing appears to be moving or changing its momentum or anything; no work is being done, no energy is going anywhere. The electrostatic levitating force is counterbalanced not by another electromagnetic force but by gravity. How does that work, then?

$\endgroup$
9
  • $\begingroup$ The answer you last cited is actually also over my head. But I think it sounds pretty correct if I say: static fields do not exist in nature. Anytime a charge is accelerated in the apparently static field of another charge, this acceleration violates the assumption of "static-ness" because it means that (classically) EM waves get radiated by the mutually accelerated charges. In the quantum field theoretical description these actual waves correspond to real photons. $\endgroup$
    – oliver
    Mar 25 at 17:55
  • $\begingroup$ Then, if you go to the limit of infinitely slow processes, the exchanged photons make up the changes in the static fields, but never the static fields themselves because these are artifacts of the classical description. At least this is what I read out of the cited answer. $\endgroup$
    – oliver
    Mar 25 at 17:55
  • 2
    $\begingroup$ I don't know what your point is. If a charge experiences no acceleration, the electromagnetic field must be zero where it's located. Nothing magical there that would invalidate the general field concept. The superposition of two static fields to obtain a zero field strength at some location is an artificial construction that serves as an excuse for not knowing the underlying physics (i.e. QFT). $\endgroup$
    – oliver
    Mar 25 at 20:00
  • 1
    $\begingroup$ I think the first answer of mine that you linked would be almost verbatim what I'd also write here as an answer - can you be a bit more specific what you think is lacking? Perhaps the problem is the word "mediating" - I think the answers you linked explain the specific sense in which the EM force is "mediated by" photons. Is your problem that this doesn't fit with a more colloquial or intuitive meaning of the word? $\endgroup$
    – ACuriousMind
    Mar 31 at 16:06
  • 1
    $\begingroup$ There is no notion of "exertion of force" in quantum mechanics. A force as something that is "exerted" is a rather classical concept. You're looking for something that doesn't exist. The meaning of "force" in the context of "four fundamental forces" is not the same as the meaning of "force" in Newton's laws. $\endgroup$
    – ACuriousMind
    Apr 2 at 10:41
2
$\begingroup$

I have never liked the "virtual particles" concept; they're the result of taking Feynman diagrams a tad bit too literally. Feynman diagrams are nice little pictorial representations of terms in a perturbation series designed to calculate scattering elements between two particles; there's one in anna v's answer. However, these should not be taken as literal pictures of what is going on.

Scattering in quantum mechanics is due to interactions between fields. Two particles (which are derivative from a field, see this answer) in approximate momentum states interact for some finite time and then asymptotically go to some other particle states in the far future (where interactions become negligible). This process is encoded in the LSZ reduction formula, and is our current understanding of how things go about; fields interact with each other. For example, the electron field might interact with the photon field, and this produces scattering.

In the non-relativistic limit, when scattering is not very large, we can neglect all but the leading order term in the perturbation series (encoded in the diagram in anna v's answer). Now, because we are in the non-relativistic limit, we can assume the Bohr approximation from regular QM holds. Thus, we can pretend the scattering is actually due to some potential, which comes out to be the Coulomb potential. However, in the end, it is (to the best of our knowledge) just fields interacting with other fields.

$\endgroup$
10
  • $\begingroup$ Fields interacting with other fields seems to be a very Classical model. Are you saying that QFT is, in this respect, a Classical field theory? Also, anna v talks of changing momentum vectors, yet a static field is just that, static. So this does not make clear to me how the quantum aspect explains the force exerted. $\endgroup$ Apr 2 at 10:33
  • $\begingroup$ @GuyInchbald this answer is quite correct. Fields interact with fields, whether you are talking about classical or quantum field theory. That's the entire point. The only sense in which particles are even defined is via Wigner's classification theorem for the Poincare group. Everything else if just fields doing things, and it happens that if in certain asymptotic limits the theory is free (which is what LSZ gets at) then the fields themselves can be given an interpretation in terms of "particles," so interaction of the fields is tantamount to interaction of the particles. $\endgroup$ Apr 4 at 1:26
  • $\begingroup$ @RichardMyers You appear to have jumped to the mistaken conclusion that I am confused over particles vs fields. I understand all that perfectly well, thank you. Please re-read and address what I ask more carefully. Some context for the term "force" would be rather more helpful; In particular, if a force necessarily involves an exchange of momentum vectors, as anna v stated, then how does that explain a static situation where a force is exerted but movement is restrained? $\endgroup$ Apr 4 at 7:14
  • 1
    $\begingroup$ @GuyInchbald Mr. Myers statement still holds; that's a classical situation, where it makes sense to discuss "force." There are, in reality, four fundamental interactions, not forces. These interaction fields allow for, well, interactions between quantum fields. In the proper classical limit, some of these interactions (like electromagnetism) can be interpreted as classical fields imposing classical forces on macroscopic objects, but at the fundamental level, it is, to the best of our knowledge, just fields interacting with one another through these interaction fields. $\endgroup$ Apr 5 at 16:18
  • 1
    $\begingroup$ There is no "$F=ma$" in quantum mechanics or field theory. This is a classical model, and a good approximation for some macroscopic situations at low velocities, but doesn't hold down here. For your levitating magnet bit, we have some field mediating interactions between matter fields, which classically looks like an electrostatic field exerting a force on the paper. This is not what is actually happening! That is a classical approximation, but is not reality; the reality is fields interacting with fields (AFAWK). Gravity is the result of another field, which classically pulls the scrap down. $\endgroup$ Apr 5 at 16:27
0
$\begingroup$

At the quantum framework, everything in the world is made up by the particles in the standard model of particle physics and their interactions. All other theories can be shown mathematically to be emergent from this level.

The macroscopic electric and magnetic fields are the result of the addition of a high number of charged quantum particles, what is a "force" at the quantum level is built up to "force" of classical level electric and magnetic fields.

What is the electric "force" at the quantum level? It is the exchange of dp/dt, (p a momentum vector) between two interacting charged particles, for example two electrons, a first order term in the expansion in series for the solution :

e-e-

The diagram is a recipe for writing the integrals necessary for a calculation of the probability of interaction between two electrons.

The dp/dt is assigned to the virtual photon connecting the two vertices. Virtual particles are a mnemonic to help in conserving the quantum numbers characterising the interaction.

In principle, no matter how far away are two electrons , they will be interacting with such a virtual photon. In order to detect an electric field there must be an interaction at the quantum level. When distances are large, the classical "force" can be defined within classical electrodynamics.

So in a nonmathematical description one can say that a static field is built up by virtual photons. A test charge measuring classically the static electric field reacts with F=ma , and F is the sum of the dp/dt for each individual quantum interaction . Considering that the number of fundamental charges are of the order of ~ $10^{23}$ (avogadro number) it is better to use classical electrodynamics.

$\endgroup$
9
  • $\begingroup$ If charged particles in a static field are exchanging momentum vectors, why are they not changing their velocities (e.g. protons in a solid lattice)? Are you saying that the answer I link to, denying virtual photons, is in fact wrong? Can you be explicit about that? How can F=ma occur with, say, a pair of magnets nailed to the wall? $\endgroup$ Mar 25 at 19:45
  • 1
    $\begingroup$ If a force can be measured, there exists a dp/dt. which is acceleration and is part of the definition of a force. Momentum is exchanged and contrary forces ( of the latice creating the fields) hold everything in place because the individual electron atom interactions at macroscopic distances are very very very very weak. It is the summation that builds up the classical force. Virtual particles are just a mathematical tool, to remember the quantum numbers that must be exchanged between real particles. Real partilcles have their four momentum on their mass shell, i.e. the "length of the four $\endgroup$
    – anna v
    Mar 26 at 5:41
  • $\begingroup$ momentum vector is the mass of the real particle. hyperphysics.phy-astr.gsu.edu/hbase/Relativ/vec4.html ( special relativity). Real particles are the incoming and outgoing in the diagram. $\endgroup$
    – anna v
    Mar 26 at 5:42
  • $\begingroup$ Sorry, you seem to be repeating your position and not answering the concerns I raised about it. $\endgroup$ Mar 26 at 6:01
  • 1
    $\begingroup$ "denying virtual photons, is in fact wrong?" virtual photons are a mathamatical tool, so a tool cannot be denied . That it represents a measurable four vector is what is denied,as the mass of the transfer vector is variable and not on mass shell, in the photon case it is different than zero.. $\endgroup$
    – anna v
    Mar 26 at 7:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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