EM wave interaction with an antenna has been covered, e.g.: How do we visualise antenna reception of individual radiowave photons building up to a resonant AC current on the antenna? Most answers focus on antenna atoms individually absorbing a photon's energy. But photons have oscillating EM fields. When light passes through water it loses velocity (but not energy we are told) because of the interaction of photon EM fields with electron electric fields. Might this play any role in the interaction of radio waves with a metal antenna? Can the photon EM oscillation induce its frequency on the antenna's delocalized valence electrons? Does the antenna absorb EM energy this way and then re-radiate so there is no net loss to the original EM field (but if the antenna heats up there is loss...). ************************* My antenna example/question introduces too many (conduction) interactions. My real interest centered on the reduced velocity of the photon in glass or water? Is that absorption and reemission as HF suggests? Or does the photon's electric field interact with electron charge in the medium?

anna v writes: " there is nothing oscillating in the photon except the quantum mechanical probability of detecting it. A photon just has energy, spin and momentum. The energy of the photon is E=hν where ν is the frequency...." To characterize photon energy as an inert, quiescent, non-oscillatory quantity is to embrace the original concept of kinetic energy (Coriolis, Watt, Lord Kelvin) from the mid-19th century! Have we come no further? Physicists will make the photon into whatever is convenient for the question at hand, particle, field or mathematical operator. In fact, the photon – no mass, no trajectory - is a 'particle' only by analogy: it terminates at a point with momentum.

  • $\begingroup$ Antennas need to be terminated (i.e. connected to something) for the energy to dissipate ..... otherwise it reflects or becomes transparent (depends on position). A lot of engineering/physics goes into terminating antennas with the proper "load" (called matching) so the most energy can be gained from the antenna. $\endgroup$ Apr 15, 2023 at 23:50
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    $\begingroup$ Presumably, it is a quote starting with "To characterize photon energy as an inert, quiescent, non-oscillatory quantity is to embrace the original concept of kinetic energy". Where does it come from? What is the source? $\endgroup$ Apr 16, 2023 at 9:40

4 Answers 4


Yes, indeed it can. If the antenna is not tuned to the photon's wavelength then the photon may bounce off it. A simple example of this is provided by a waveguide with a poorly-connected end plate. Some of the forward power (EM photons) will excite the end plate and radiate out beyond it, while others will bounce off it and travel back the way they came, as what is known as reflected power. For a waveguide in the form of a coaxial cable, the ratio between the forward and reflected waves is referred to as the voltage standing wave ratio or VSWR (pronounced "vizwah").

This isn't a cheesy interpretation thing which vanishes when the oscillating wave collapses on measurement. It is eminently measurable and survives the measurement process. In other words, it is a real, concrete phenomenon.

On the other hand, one may argue that actually the forward photon was absorbed and the reverse photon reemitted. This is actually the standard theoretical model. But it is just one of those cheesy and unfalsifiable interpretation things. It adds nothing to the classical picture, no new predictions, and nothing like that. So how much credence you give it is a philosophical issue, not a physical lab test one.

  • $\begingroup$ " forward photon was absorbed and the reverse photon re-emitted. " real electron photon interactions in order to keep the energy ,which means the frequency of the incoming wave, must not be absorbed and re-emitted.If this happened the coherence of the incoming beam would be lost, because re-emittance is random in 360 degrees. What happens is that in the center of mass "lattice electron- photnn" the scattering is elastic , and in the lab a small part (not measurable) of the lab energy and momentum is transferred to the electron . These many photon scatterings add coherently . $\endgroup$
    – anna v
    Jun 8, 2021 at 4:48

I will focus on what you wrote in the title:

Can EM (radio) wave oscillation interact with an antenna without photon absorption?

Production of a radio wave

To understand what a radio wave is made of, you have to remember how it was generated:

  • on the surface of a metal rod, many electrons are accelerated back and forth
  • during these accelerations, they emit photons
  • for each period of the oscillation, the number of emitted photons increases to a maximum and then goes back to zero.

The technically useful thing about this is that the electrons, which are all accelerated in the same direction, all emit their photons with the same directions of their electric and magnetic field components:

  • the electric field is directed parallel to the electric field potential of the rod
  • the magnetic field component is directed perpendicular to both the electric field and the direction of propagation

We get a stream of a periodically swelling number of photons from zero to a maximum and back to zero. During each half period, the directions of the electric and the magnetic field component are changing.

How does that help to transfer information from the emitting to the receiving antenna?

Receiving a radio wave

With the above said the, answer to your question is the following. But, as Guy Inchbald said, „it is just one of those cheesy and unfalsifiable interpretation things, it adds nothing to the classical picture, no new predictions, nothing like that. So how much credence you give it is a philosophical issue not a physical lab-test one.“

The receiving antenna is hit by numerous photons, since all bodies that are not at absolute zero Kelvin permanently emit (and receive) photons. These photons that hit the rod interact with a surface electron and are absorbed and reemitted with different directions and frequencies. This is noise for the antenna.

The photons of the transmitter of a radio wave act together (as a group) and the electrons on the surface of the receiving antenna are periodically influenced by the common field components of the incoming photon stream.
The electrons do not dissipate the incoming energy (the noise) in all directions; instead the electrons start to move along the rod!

Regarding your question: Photons are indivisible particles from their emission to their absorption. They transfer energy—and you get a signal at the receiver. Or they do not interact with anything and continue their propagation.


A much simpler way of visualising how radio waves enter the antenna is to just use what’s observed and forget theoretical assumptions like the Lorentz force, photons, electrons, etc. These are not actually observed.

But what is observed is as follows: A conducting wire is also the antenna. Rotate a magnetic field around the wire and we rotate the magnetic fields of the atoms in the wire. (Faraday and others called it induction and pretended without evidence that it was tiny electrons moving about.) Those rotating magnetic fields are essentially an AC current. Farther down the conducting wire, these rotating magnetic fields induce alternating magnetic fields in the air at the same frequency as the AC current. Those are called EMR. And when they hit the antenna, they induce the antenna metal conductor to oscillate the metal atoms magnetic fields in sympathy. Induction really.

That is amplified and sent to the rest of the system to be played by the speakers, etc. Now most establishment theorists will not like this simple explanation. But the fact is it’s the closest explanation to the actual mechanism you will ever get. Because this explanation is what’s observed, not assumed.

For instance, a low-frequency AC conductor will induce very small pollen particles to rotate their magnetic fields when suspended in a solution around the conducting wire. The best evidence for the assumption an electric current is not a flow of electrons, but rather that it is the conductor atoms magnetic fields which are rotating. Which themselves were rotated by the observed rotating magnetic field of the induction motor.


But photons have oscillating EM fields

Wrong. There is nothing oscillating in the photon, except the quantum mechanical probability of detecting it. A photon just has energy, spin and momentum. The energy of the photon is $E=hν$ where $ν$ is the frequency of the classical EM wave a quantum superposition of a large number of photons will make.

This is just a special case to see how different photons and classical light are:

Enter image description here

On the left is the buildup of a polarized classical electromagnetic wave, the red arrow depicting the electric field maximum. The middle describes the photon, which individually has only forward and backward spin. Nevertheless, in confluence of many photons, it builds up the classical wave.

The classical Maxwell mathematics for electromagnetic waves is accurate when used with classical electricity mathematics.

When one discusses atoms and valence electrons, one is in the frame of quantum mechanics. Radio wave photons are very low energy photons and can only scatter with conduction band electrons in a conductor. The photons scatter electromagnetically with the conduction band electrons, giving a small part of their energy and creating a current. The loss of energy for the photon is minimal. The mathematics is such that the probability of this scattering has the frequency of the radio wave in its formulation (even though the classical wavelength of radio waves is from centimeters to kilometers). It needs a study of quantum electrodynamics to see how this happens (example).

  • $\begingroup$ "The classical Maxwell mathematics for electromagnetic waves is accurate when used with classical electricity mathematics" is only half the story. In quantum field theory, the field variation from the zero-point level also follows much the same equations. Photons do oscillate as waves, and in big enough numbers coherent (laser) photons can be used to move atoms around in the troughs of the waves. $\endgroup$ Jun 7, 2021 at 19:22
  • $\begingroup$ @GuyInchbald if you are talking of quantum optics photons, they follow a different field theory than the standard model QED, the standard model photons are point elementary particles axiomatically, any wave properties they have belong to the probability calcualations of QED. $\endgroup$
    – anna v
    Jun 7, 2021 at 19:34
  • $\begingroup$ That "standard model photons are point elementary particles" is news to me. I have long been told that they are disturbances of the zero-point photon field, and that wave-particle duality rules out such simplistic notions as point particles. But perhaps your standard model is different? $\endgroup$ Jun 7, 2021 at 19:39
  • $\begingroup$ @GuyInchbald The particle physics standard model is based on the same postulates of quntum mechanics as all quantum mechanical models. Quantum field theories are a higher level structure on the basic quantum mechanics solutions to allow modeling of many particle interactions. The "they are disturbances of the zero-point photon field, " is , like the electron field, the neutrino field etc a plane wave solution of the corresponding QM potential equation with 0 potential. Creation and annihilation operators on these "fields" create and annihilate particles, photon included. The wave nature is $\endgroup$
    – anna v
    Jun 8, 2021 at 4:04
  • $\begingroup$ a probability wave for detection in space time, that is why a lot of particles, including photons, are needed to display the wave nature , as with your laser example. See this simple experiment for photons sps.ch/artikel/progresses/… and this answer of mine for electrons physics.stackexchange.com/questions/238855/… $\endgroup$
    – anna v
    Jun 8, 2021 at 4:14

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