Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

As I have understood it, the Standard Model includes particles that carry the different forces, e.g. the electromagnetic (EM) force, the gravitational (G) force. When talking about EM fields such as visible light or microwaves, the associated particle is said to be the photon. But what about a static EM field without any electricity, like a common household magnet? How does that magnet communicate its force? Via photons?

share|cite|improve this question

Yes -- the carrier for any electromagnetic interaction is a photon. Including ordinary magnets.

You might ask: "Why then can't I see those photons?" Well, even from classical point view, the photons can have different wavelenghts. And with the magnet you deal with really large wavelenghts, while one cannot "see" radiowaves. While also I think that one cannot even use the concept of "interaction carrier" without working in a context of quantum field theory anyway.

share|cite|improve this answer
1  
so photons of really large wavelengths means photons of really low energies? Is the magnet constantly loosing energy by emitting these photons or does it also absorb photons of the same wavelength? Are the photons emitted isotropically or do they follow the field lines in some way? Should I include these questions in the question rather than in a comment? – Dirty Harry Jan 19 '11 at 11:27
    
edit your question at your leasure, be sure to follow the faq – Argus Jul 10 '12 at 0:12
    
@DirtyHarry The energy is conserved by quantum uncertainty and i read that the field lines are caused by the visualization particles sticking together, kinda like how planets clear their orbits while the Sun's gravity is a continuum. – Cees Timmerman Apr 7 at 14:15

Amplifying the fine answer of Kostya, to address Dirty Harry's question. Static forces, such as slow moving magnets, like the one you mention involve extraordinarily low energy photons, so, since these are massless, extraordinarily long wavelength photons: their technical name in relativistic Quantum Field Theory, where the carrier notion is most useful, is "infrared" photons.

In macroscopic, classical settings, they don't add much intuition to the picture, and, as the questions suggest, they confuse the landscape more than illuminate it. Nevertheless, they connect seamlessly with our low energy, small $\hbar$, normally non relativistic world.

Indeed, technically, the infrared limit of the tree-level ("Born") amplitude of particle interactions yields the Coulomb potential, as detailed in, e.g. Peskin and Schroeder's book (M. Peskin and D. Schroeder, An Introduction to Quantum Field Theory, Ch 6.6.1, eqn 6.97) or this answer; (for a n.r. treatment, see this answer here ), and the associated Maxwell's equations, and so all the phenomena the question would contemplate. These amplitudes conserve energy and momentum, except for ultra-short times not of relevance here, so, no, the magnet is not constantly losing or gaining energy by emitting and absorbing these zero energy limit photons, unless the test object moves... infinite wavelength photons... they are a limit. A roiling sea of nothings.

Nevertheless, this sea of nothings informs the very active and responsive vacuum of QFT. If, however, there is movement of the test body subject to the magnet, yes, the force at a distance is carried by gadzillions of such photons that take energy and momentum from the magnet and transfer it to the object, anisotropically, and collectively shaping what we perceive as the classical force.

Mind you, even in non-relativistic quantum mechanics, most of the time one need not consider them, in calculating energy spectra, and one works with a semiclassical Coulomb field of, e.g., an atomic nucleus, the 1/r potential of the Schroedinger eqn. Only in subtle cases of relativistic corrections, like the celebrated Lamb shift (1947, that started utilization of QFT) would one see subtle correctional effects to energy levels due to this picture.

Nevertheless, smoothing out the correspondence limit of QFT has been, historically, sufficiently intriguing to have Feynman suggest to a student to calculate planetary motions through (hypothetical) massless graviton exchanges (it actually works!), in Newton's gravity, not EM. (see WP article. )

As indicated above, however, intuition is tricky with delocalized (λ⟶∞) entities such as these: so, when you think you can describe repulsion of two particles with a metaphor of two boats where a boat throws a basketball at the other, you run into deep perplexity when you consider that the two boats could also attract each other, by the very same exchange of a basketball!! I suspect the wisest attitude is to leave particle mediation to QFT, where it was born, and where it does most good.

share|cite|improve this answer
    
But field lines return to the source, so a boomerang would fit better than a basketball. – Cees Timmerman Apr 7 at 14:00
    
Well, the energy-momentum transfer goes one way, and the boomerang interacts with the air/medium---it would not work in vacuum... In any case, I brought these imperfect analogies up to discourage their use, as they tend to confuse the actual crystal-clear mathematical situation, not as a challenge for seeking better ones! – Cosmas Zachos Apr 7 at 14:23
    
Didn't even Einstein use visual analogies? Can toroidal vortices deflect each other? – Cees Timmerman Apr 7 at 14:36
    
Indeed, E did, and so did Maxwell, but only kept those that made sense... So did Feynman, who introduced much of this picture of particle mediation... So, had a metaphor worked half-decently, he would be sure to have used it... He did use the bombardier analogy for positrons, but stayed away from boats on a lake... He appreciated the delocalization involved is at the heart of the uncertainty principle, and few good metaphors for that have shed more light than confusion... – Cosmas Zachos Apr 7 at 14:54
    
As a consolation prize, for EM radiation, you may think of real photons, instead of virtual ones. In that case, you can estimate that a typical microwave oven producing about 700 W of microwaves of λ~3cm is spewing out about $10^{26}$ real photons of that wavelength per second. – Cosmas Zachos Apr 8 at 22:44

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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