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How do photons interact in magnetism and how is the force felt

  1. Sep 1, 2012 #1
    I understand that as one person here puts it "The electromagnetic interaction is mediated by the constant exchange of photons from one charged object to another.
    Some electromagnetic interactions involve "real" photons or "virtual" photons instead."

    But suppose I have 2 magnets and push their positive poles toward each other. I feel a force repelling them. Is that force photons then? The magnets can feel that force as the magnets repel each other from several inches away, so there are photons traveling 2 inches away? And how they repel the 2 magnets? Like what is happening between the 2 magnets? How do they travel in order to repel each other? Do they bounce and physically push the other magnet because I thought they had no mass or very small mass so they most be many of them to produce such a force? Also, if I put the other pole and now the magnets attract each other how do the photons now move? Has anyone been able to detect photons from a magnetic field? What is the photon frequency for the magnet?
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  3. Sep 3, 2012 #2

    Jano L.

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    Welcome at PF, jmmy.

    It is often said that the EM field is made of photons. But it is important that this idea was invented primarily for description of radiation.

    In case of two magnets, there is no detectable radiation. I would be surprised if you find a comprehensive explanation of magnetic force between two magnets in terms of photons.

    If you want to understand how magnets work and why they attract, you will do better with classical electromagnetic theory. There we imagine magnets as composed of little current loops made of circulating electrons. The circulating electrons are subject to forces due to other circulating electrons and also the forces of constraint. The result is, that magnets act on each other.
  4. Sep 3, 2012 #3


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    Quantum field theory describes the repulsion (or other forces) as exchange of virtual photons between the magnets.

    The classical description is easier though.
  5. Sep 3, 2012 #4


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    This is a Very Frequently Asked Question, and I think the reason is partly because in the popular vein, everything about electromagnetism needs to be re-expressed in terms of photons!

    Whereas in fact, photons are useful in a limited domain where quantum effects are present. (high frequency, low intensity) Otherwise, there still remain many valid and useful concepts: antennas, electromagnetic waves, index of refraction, capacitance, etc., along with all the other things you learned in E&M and optics.

    One of the jobs that quantum field theory does have is to show that the presence of virtual longitudinal photons leads to the familiar properties of electromagnetic fields. This can be done quite generally, and in several ways. Many books show that the photons can be "integrated out", leaving a direct interaction between current/charge densities that reproduces Coulomb's Law and so on. Another approach is to show that they lead to a term in the Hamiltonian, the familiar ½(E2 + B2), and the electric and magnetic forces result from the gradient of this potential.
  6. Sep 3, 2012 #5


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    People often ask similar questions here concerning the electrostatic force between two charges. As has been posted here repeatedly by tom.stoer, virtual photons are not necessary for this interaction. Their presence or absence depends on the "gauge" in which you set up the calculation. In the Coulomb gauge there are no photons involved.


    Don't think of photons as being sort of like little tiny balls flying around. They have energy and momentum, but their size or position is not well-defined at all, at least not while they're propagating.
  7. Sep 3, 2012 #6


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    This is what I meant by "integrating out".
  8. Sep 3, 2012 #7
    Besides a mathematical definition what are the virtual photons actually doing?
  9. Sep 3, 2012 #8

    Vanadium 50

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    But that's the whole point. Virtual photons are a "mathematical definition" and they aren't "actually doing" anything.
  10. Sep 3, 2012 #9


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    Take the Coulomb field as an example. It's a static field, right? Part of the reason that virtual photons are a poor description for a Coulomb field is that they arise from time-dependent perturbation theory, and the Coulomb field is time-independent.

    So when you read things like those virtual photons are "constantly being emitted and absorbed", or "flying rapidly back and forth", it's based on a misconception. If this sort of description were valid, the logical followup questions would be, "How often are they emitted?" "How rapidly do they fly back and forth?" And these questions have no answers, because the model is wrong.
  11. Sep 3, 2012 #10
    So then what is the correct model that involves photons in electromagnetism?
  12. Sep 9, 2012 #11
    It seems to me that nobody can answer the question. there is no shame in a simple "I don't know".

    First, qed and modern physics states that the force carriers for em force are photons.
    So photons should be observable. If there are not there that means that they do not carry that force. So no matter what their frequency (ir or whatever) there should be a way to measure the photons. If nobody has maybe someone should.
  13. Sep 10, 2012 #12


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    Sorry jmmy, Try reading the responses again. We've tried to explain why the question is a meaningless one.
  14. Sep 10, 2012 #13

    If you wanted to hear that there is no classical model of the solidity of matter, i'd say you are right. There is no such model. The 'model' that gets presented is a very weak approximation and under closer examination is found to be wrong, as Bill_K pointed out.
    Last edited: Sep 10, 2012
  15. Sep 10, 2012 #14
    thanks guys so what would be some simple recommended readings to understand this better? or even online classes? for example it is easy in my mind to understand gravitation as curved space time. However, I have a hard time picturing this. Also, everything I read says that photons are the force carriers so I do not get why they can't be observed.
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