I have read in wikipedia this statement

"The energy carried from one winding of a transformer to another, in quantum terms is carried by virtual photons, not real photons" (wikipedia src: virtual particle)

Of course anything in wikipedia could be solved "changing it" =), but I wonder, someone put it for a reason, then in the case it have sense, I don't know what is the limit for that, what are the frequencies for virtual or real photon interactions. (if is the frequency what makes the difference, or what ?)

Perhaps it's related with the kind of interaction, I mean photoelectric effect and the induction are different mechanism , but I don't see clearly how "real photons" disappear from the picture, if they are the lesser energy that can be transfered anyway.

I have put that into a comment for this answer Low frequency electromagnetic waves but I think is better to open an specific question.


1 Answer 1


In your case it is a near field that stands for "virtual" photons. The near field does not propagate like EMW in the sense it is "attached to the charges and currents. In fact, it is the Coulomb and quasi-static magnetic interactions of charges and currents. Despite being time-dependent, the near field decays with distance differently (faster) and does not carry away any energy-momentum. Real photons do.

There is no quantum terms for the near field. It is not quantized. It exists, as I said, as a potential interaction $\propto \frac{1}{|\vec{r}_1 - \vec{r}_2|}$.

  • $\begingroup$ If it's not quantized, its fascinating, but could we then exchange energy "in near field" without plank energy limitations? $\endgroup$
    – HDE
    Sep 9, 2011 at 16:01
  • $\begingroup$ Yes, there is radiationless mechanism of energy exchange. If you scatter one cold atom from another, they may push each other away without any photons, so no $\hbar\omega$ is involved. $\endgroup$ Sep 9, 2011 at 16:06
  • 1
    $\begingroup$ This answer is correct about near field, but it is using an electrostatic near-field interaction to illustrate. For the case of transformers, the near field should be magnetic B-flux inducing an E-field at near zero frequency, which is not described by electrostatics or magnetostatics, but requires induction. $\endgroup$
    – Ron Maimon
    Sep 9, 2011 at 17:33
  • $\begingroup$ @Ron Maimon: You are right, as usual. I am just not sure if the magnetic interaction cannot be completely "explained" via the Coulomb one. $\endgroup$ Sep 9, 2011 at 19:48

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.