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Linear antenna directed along z, photons (EM waves) propagate along x. Momentum of photons have only x component. Why electrons in antenna have z component of momentum?

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Aha, if electrons in an antenna have a momentum along that antnna, what do they do after reaching the end? –  Georg Jun 8 '11 at 15:16
    
...I think a vertical linear antenna will propagate waves in many directions, and the one thing I'm almost sure that it won't do is propagate preferentially in an x or y direction (not sure about z). It's still a valid question, but I maintain that we're searching for something different than an x-axis component, but instead an E&M pressure all around the wire that can then be turned into a net force with use of reflectors. –  AlanSE Jun 8 '11 at 16:28
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4 Answers

There are 2 different effects of the EM waves:

  1. Acceleration of charges due to EM field
  2. Radiation pressure of the EM field

If the EM wave propagates along x direction and your antenna is along the z direction, then electrons will be driven along the antenna by the electric field. On the other hand, the antenna will also receive an overall impact (momentum) along the x direction because of the radiation pressure (which does not affect the antenna practically since its effect is tiny and the antenna is fastened).

It is also worth pointing out that there are 2 descriptions of radiation:

  1. the classical description (EM waves) and
  2. the quantum description (photons)

The quantum description is more general. Both pictures are identical only in the limit of so many photons. Electromagnetic waves can be described in terms of photons, but individual photons cannot be described in terms of electromagnetic waves. Photons description is more fundamental.

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""The quantum description is more general. Both pictures are identical only in the limit of so many photons. Electromagnetic waves can be described in terms of photons, but individual photons cannot be described in terms of electromagnetic waves. Photons description is more fundamental."" This is fundamentally wrong. Ever heard of wave/particle dualism? –  Georg Aug 9 '11 at 15:06
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@Georg: The "wave" in the wave particle/duality has nothing to do with the electromagnetic waves. Wave/particle duality does not mean that Maxwell's equations are equivalent to quantum electrodynamics. QED is more general than the classical theory of radiation. –  Revo Aug 9 '11 at 15:27
    
Very interesting, but not physics. –  Georg Aug 9 '11 at 18:26
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@Georg He is correct. When we speak of wave particle duality the "wave" part is a probability wave, not a field wave. –  anna v Aug 9 '11 at 18:34
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@Anna, O sancta simplicitas! Is such rubbish thaught today? Read: en.wikipedia.org/wiki/Wave%E2%80%93particle_duality –  Georg Aug 9 '11 at 18:57
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Maybe interesting for some to take this a bit further:

This classical behavior goes all the way down to single cycle laser pulses which ionize atoms by accelerating them in the direction transverse to the motion of the laser pulse. Make sure to look at the very interesting (Realplayer) video...

http://www.cfa.harvard.edu/itamp/attosecondpdfs/paulus.pdf

http://cfa-www.harvard.edu/dvlwrap/itamp/0311/paulus.ram


Why is this behavior still classical?

(1) Because in this case even the single cycle of the ultrashort laser pulse has a wavelength much longer as the size of the field of the (bounded) electron.

(2) The single cycle laser pulse kicks the electron in one direction.


When, how and why do we get quantum mechanical Compton scattering effects?

When: Mainly if the wavelength of radiation becomes smaller (e.g. x-ray) as the volume in which the electron is contained.

How: The electron gets kicked in the direction parallel to the momentum of the photon.

Why: The electron can self-interfere when going from an initial state $\psi_i$ to a final state $\psi_f$. The interference current is a sinusoidal pattern of alternating charge and spin density with the same wavelength and direction as the incoming photon. The direction of the electron spin is essential in the interaction because it is the effective current from the alternating spin density which is the source of an alternating transverse electromagnetic field which compensates the electromagnetic field of the incoming photon: The photon is absorbed. Finally, the state of the electron after absorbing the photon isn't that of a free electron and another transition takes place under the emission of an outgoing photon which leaves the electron in a free (on-the-mass-shell) state. The math which describes these processes corresponds to the tree-level Feynman diagrams of the interaction

Regards, Hans

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Hans, your answer is useful, however it is still unclear why and how photons having only x component of momentum give to electron perpendicular to x momentum? Is it consistent with conservation of momentum? Best regards, Grigori –  grigori Jun 9 '11 at 17:57
    
@grigori. It's consistent with the conservation of momentum. The math to check this is given here: en.wikipedia.org/wiki/…, Hans –  Hans de Vries Jun 11 '11 at 23:34
    
Hans, the question was: why EM waves having only x momentum transfer to electron z momentum? Electron begins oscillating along z, and will radiate EM waves ~ sin(teta) from direction z, however it will not radiate along z direction, so its radiation field will not compensate electron z momentum. So it seems that in terms of EM waves, the conservation of momentum does not take place (but it takes place in terms of photons). Grigori. –  grigori Jun 19 '11 at 14:12
    
@Hans I'm sorry I didn't see this post when it first went up, or I would have asked you: why are the spin currents more important than the alternating charge layers? The charge layers are static but so are the current layers, so neither interacts with the incoming e-m wave until they are set in motion to some degree. If you are interested in explaining this, let me know and I will post it as a separate question. –  Marty Green Aug 9 '11 at 21:46
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The electrons in the antenna are moving perpendicular to the direction of motion of the emitted radiation. This is natural for a transverse wave. Since light is a transverse wave, as shown by Luboš Motl's beautiful graphic, there's nothing surprising here.

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Photons are quanta of electromagnetic waves which are transverse so if the momentum of the photon goes in the $x$-direction, then the magnetic and electric fields are in the transverse $yz$-plane, for example $B$ may be in the $y$ direction and $E$ may be in the $z$ direction. Because the electric field accelerates charged particles in the same (or opposite) direction, electrons will be accelerated in the $z$ direction as well.

enter image description here

If the wave is unpolarized, about 50 percent of its energy will be composed of the wave above and 50 percent will be composed of the other way where the directions of $B,E$ are interchanged. This other wave won't be able to shake the electrons.

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Could you elaborate more on the 50/50 breakdown of energy? When you say "the other way" I think direction and when you say "This other wave" I think the perpendicular wave. Because of that I can't pin down a clear meaning. –  AlanSE Jun 8 '11 at 16:33
    
@Zassounotsukushi: in the picture you have an EM wave with the electric field in the up-down direction and the magnetic field in the left-right direction. But there could also be a wave with the electric field in the left-right direction and the magnetic field in the up-down direction. An unpolarized wave would be a linear combination of both those possibilities, with the energy split evenly between them. But only the component of the electric field that is parallel to the antenna would actually be able to move the electrons in the antenna. –  David Z Jun 8 '11 at 18:00
    
In context of antennas and waves the near-field version of the wave had been better. –  Georg Jun 8 '11 at 20:03
    
Lubos, you explained what is plane wave, but did not answered the question. The question was: why (how) photons having only x momentum transform to electron perpendicular momentum? –  grigori Jun 9 '11 at 17:45
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