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How radio waves create the current in antenna in terms of photons? If it is Compton scattering then why is not changed the freuency of photons?

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1 you have any preparation in quantum field theory? Does the phrase "coherent scattering" mean anything to you? It would help anyone trying to prepare an answer to know from whence you are coming. – dmckee May 20 '11 at 14:08

An elementary explanation, at high school level:

The beam of radio wave photons are coherent, as Vladimir said. Coherent means that the electric and magnetic field of each individual photon has a fixed phase with all the others.

When the wave reaches an antenna, some of the photons are absorbed, pushing the electrons to a slightly higher energy level (energy h*nu) in the conduction band. Thus it is not scattering but absorption that generates the current with the frequency of the incoming beam.

It is coherence that , as the photon is absorbed, pushes or repulses the electrons in step, so that a current that has the frequency of the impinging beam is built up.

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Suppose linear antenna directed along z, photons (EM waves) propagate along x. Momentum of photons have only x component. Why electrons have z component momentum? – Grigori May 22 '11 at 12:06
I thought photons were electrically neutral. How can they have an electric field? Also, when the photons are absorbed, what decides in which direction the current will flow? – Steve Apr 12 '14 at 3:42
Photons are neutral but in bulk, if they are coherent ( in step) they built up electromagnetic waves which do have electric and magnetic fields. The way the build up happens is illustrated here… . In a hand waving sense the potentials entering the maxwell equations enter the equations for photons, so they have the rudiments for the build up. – anna v Apr 12 '14 at 3:46

No, it is not a Compton scattering - the electrons in antenna are not really free.

A radio wave is a flux of coherent photons, they act together, not one by one.

Any EMW makes charges move but bound charges may make a work, so the incident wave may be partially absorbed.

Besides, a low-frequency Compton scattering is quite the same as the classical EMW scattering. So the "scattered" part of the resulting wave is nearly of the same frequency.

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We can consider free electron (probe charge in rest in free space). Photons propagating along x axis will not push charge perpendicular to x axis. However EM waves propagating along x axis will cause oscillations of charge mainly perpendicular to x. What description is correct? – Grigori May 22 '11 at 11:58
Yes, photon will push a probe charge perpendicular to x axis. – Vladimir Kalitvianski May 22 '11 at 12:11
Thanks for answers. Why photon does not give momentum to charge along x axis? – Grigori May 24 '11 at 14:58
To give the momentum along the x axis is quite difficult. It is possible only in frontal collisions. Consider a collision of two billiard balls. They scatter aside in most of the cases due to collision being non frontal. Why collision with a photon should be always frontal? It's not frontal normally so there is a perpendicular to x motion of the scattered photon and the target electron. – Vladimir Kalitvianski May 24 '11 at 15:19
close to perpendicular to x motion of the target electron means that momentum of scattered photon remains close to x. In such cases very low portion of photon energy is transformed to electrom. It seems not true situation for antennas. – Grigori May 26 '11 at 13:11

The beam of radio waves with coherent photons may be produced only by MASER. Without that photons do not have correlated phases, e.g. famous 21 cm hydrogen line in radio-astronomy is not consist from coherent photons. For radio waves, produced by usual transmitters situation even “worst” - here there is hardly even to suggest a strict justification to find photons with frequency related with frequency of transmitter. I do not have exact calculations, but to understand the reason of the problem it is enough to recall, that radio-transmitter may emit waves of different shapes and simply described by classical physics, but photon is emitted due to quantum transition between some energy levels and it is very different process. After all, photon by definition is described via time dependent function like $exp(-i\omega t)$ and we may not emit “triangle” photons instead of “sinusoidal” one using transmitter with electric current with tricky time dependance.

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Current is moving electrons. A photon with a frequenzy will give its energy to the electrons in the antenna. The energized electrons then travel thru the wire to the radio at the same frequenzy as the photons will give of its energy. When the photon field, or rather the electromagnetic field oscillates from positive+ to negative-, as in the AC used to produce the field, the current in the recieving antenna will also produce current, or moving electrons in a oscillating way, from + to - acording to frequenzy.

The photon to electrons energy "delivery" is due to Einsteins photoelectric effect.

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"When the photon field, or rather the electromagnetic field..." OP is asking how to explain the phenomenon using photons instead of the electromagnetic field. Therefore, this doesn't really address the question. – AccidentalFourierTransform Jan 28 at 21:38

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