0
$\begingroup$

In antenna technology we distinguish between nearfield and widefield. In the nearfield the electric and the magnetic fields are shifted by 90°. If you look closer you can see that there are two possibilities of this shift, 90° and -90°.

To explain it you have to remember what this 90° means. Let (in vacuum) the coordinate systems X axis be parallel to the E field, let the Y axis be parallel to the B field and the Z axis is parallel to c * t . In Z equal zero let the E field be maximum and directed in the X direction. The B field is zero.

90° later (in terms of E = E(max) * cos α and B = B(max) * sin α) and this is a quarter of the wavelength the B field can be directed to the left or to the right. And this is natural because B = B(max) * - cos α is the second possible state of the nearfield radio waves.

Regardless of the approach to see radio waves as one electromagnetic wave (statistical method) it is obvious that all radio waves are made from photons which are emitted during the electrons acceleration in the antenna rod.

My question is, do these photons all have the same sequence of the E and B fields? The same question appears for quantum dots which produce single photons.

Edit: There has to be a right or left hand rule because if half the photons have B-field with 90° to E and half have -90° then there wouldn't be magnetic field at all.

Update: I get it. It's the right hand grip rule (conventional direction of current) because there is no principal difference to a straight wire.

$\endgroup$
  • $\begingroup$ The right hand grip rule doesn't apply for waves. $\endgroup$ – Per Arve Aug 25 '14 at 18:18
2
$\begingroup$

You're asking a classical electromagnetism question. You don't benefit from thinking about photons.

If the antenna's electromagnetic field (calculated from Maxwell's equations) is right-circularly-polarized at a certain point, then you can say that every photon is right-circularly-polarized. If the field is linearly polarized, then you can say that every photon is linearly polarized. It's only in mixed states (like unpolarized light) where you have to say that different photons have different properties, or better yet describe the photons using density matrices. This is usually not relevant for a classical antenna.

Another thing:

If you have a normal antenna and you lower the power lower, lower, lower, until it's so weak that on average only one photon per second is emitted...Nothing really changes!

You can still use classical electromagnetism to describe the (expectation value of) electric and magnetic fields. They won't be any different, just weaker. There is no line you cross where fields become too weak to use classical electromagnetism.

Remember, photons don't interact with each other! So it doesn't matter whether an antenna is emitting a gazillion photons per second versus one photon per second.

$\endgroup$
0
$\begingroup$

Short answer: yes.

Regarding quantum fields, one can think of the "particles" associated with them as sort of a minimum currency for exchange, i.e. to carry an interaction, it can only be done in discrete units. But like currency, it is still a currency of some type, so a photon represents the unit of interaction of the field it comes from. Different arrangements of EM fields "exchange" photons with different characteristics --- when you pay in US dollars or Japanese yen, you are exchanging different types of currency.

I know that it is a bit difficult to understand as the issue is rather complex part of Quantum Field Theory (QFT). A couple of very good textbooks that I can recommend and cite as a reference are Peskin & Schroeder, "An Introduction to Quantum Field Theory", West View Press, 1995, ISBN-13: 978-0201503975, and Greiner & Reinhardt, "Field Quantization", Springer-Verlag, 1996, ISBN-13: 978-3540780489. For the educated layman I would also recommend Richard Feynman, "QED: The strange theory of light and matter", Princeton University Press, ISBN 0-691-08388-6.

I hope this is at least partly helpful as an answer, and I can elaborate more if you wish --- though the full mathematical machinery of QFT can be quite elaborate, and not knowing your background exactly I don't know where to start at this moment

$\endgroup$
  • $\begingroup$ Ok, let us start with the first two sentences. Are they right? $\endgroup$ – HolgerFiedler Aug 21 '14 at 7:58
  • $\begingroup$ The "frivolous" information which was edited out pertained to the "personal experience" which was requested in the question originally, so it seemed relevant at the time. Sorry... Regarding some other answers: About not "benefiting from thinking about photons", I agree that for a classical E&M question it may be overkill, but an explanation with photons should still fit in the framework of classical E&M. Of course there are many choices of basis for the quantum states of photons. Coherent or wavenumber states are two; but the actual fields should be a superposition of those "basis photons" $\endgroup$ – gildardo Aug 25 '14 at 21:46
0
$\begingroup$

Photons as described by QED (quantum electro dynamics) are based on wave solutions of maxwells equations in free space (no charges, no currents). In antenna theory these may be called far field solution.

However, mathematically they constitute a complete set of solutions to maxwells equations. This implies that by making linear combinations of the wave solutions we can produce any free field solution, thus also near field solutions. For example the magnetic field from a magnetic dipole.

Points of complication:

  • In QED the photon is formulated with the vector potential A, and scalar potential V (or rather the 4-vector potential).
  • In QED, the complex solutions are needed to describe a photon with certain momentum $p = \hbar k$, where $k$ is the wavenumber.
  • A single photon has a fixed shape in space of its 4-vector potential, but the amplitude has a distribution of values like that of the quantum mechanical oscillator. A state with fixed amplitude consist of a linear combination of all photon numbers from zero to infinity.

EDIT: I think I understand your question now! You are interested in the far-zone field. You have to try to understand the EM-waves from Maxwells Equations. For a wave traveling in the positive $z$-direztion, the fields are related as $$ k\mathbf{\hat{z}}\times \mathbf{E}=\omega \mathbf{B},\,\,\,\mathbf{\hat{z}\cdot E} = 0 $$ where $k$ is the wavenumber and $\omega$ is the angular frequency. The $\mathbf{B}$-field points in the opposite direction if the wave travels in the opposite direction.

$\endgroup$
  • $\begingroup$ Let us discuss the first two sentences. Are they right? $\endgroup$ – HolgerFiedler Aug 21 '14 at 9:11
  • $\begingroup$ Maxwells equations in free space is just wave equations. A wave equation may be viewed as an equation for a set of coupled harmonic oscillators. Both in the classical and quantal description the same transformation will bring the system inte uncoupled harmonic oscillations. For waves in a finite volume the new coordinates will be the amplitudes of the standing waves. $\endgroup$ – Per Arve Aug 21 '14 at 9:59
  • $\begingroup$ The standing wave mode tells how the individual oscillators swing together. What I tried to describe is normal mode theory. $\endgroup$ – Per Arve Aug 21 '14 at 10:08
  • $\begingroup$ @HolgerFiedler Yes, a single photon is the first excited state of a normal mode oscillator of the electromagnetic fields.(F.Mandl, G.Shaw, Quantum Field Theory, John Wiley and Sons) Far-zone field from an antenna can be of various kinds depending on the shape of the antenna, for example electric dipole. Actually there is an infinite series of different electric and magnetic multipole patterns that are possible far-zone fields. The same multipoles are possible in photon emission, se e.g. A.Bohr, B.Mottelson, Nuclear Structure vol I. $\endgroup$ – Per Arve Aug 21 '14 at 17:29
  • $\begingroup$ We agree that the two first sentences are right. What is about the third sentence about the two possible shifts of +90° and -90° of one of the fields to the other? See this and that picture. $\endgroup$ – HolgerFiedler Aug 21 '14 at 17:48
0
$\begingroup$

Yes, all radio waves have inthe nearfield the same sequence of the E and the B field. It's the right hand grip rule (conventional direction of current) because there is no principal difference to a straight wire.

$\endgroup$
-3
$\begingroup$

For a macroscopic antenna, there are no benefits in thinking by quanta. Anyway, these emitted photons are bosons, and their bosonic behaviour dominates.

By contrast individual antennas like a rotating gas molecule, or the NH3 molecule or atomic hydrogen at 21 cm, are tight bond by quantic rules, so they emit or receive an exact quantum h of action-by-cycle ; however what is emitted or absorbed remains strictly electromagnetic wave, which never transmute into corpuscles.

There is no validated physics for such a postulated transmutation, anyway.

$\endgroup$
  • $\begingroup$ Welcome to Physics SE. Look around and take the tour. As for your answer, quantum mechanics is marvelously validated. $\endgroup$ – Jon Custer Oct 6 '15 at 22:15

protected by Qmechanic Oct 6 '15 at 23:13

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.