Talk:Electromagnetic radiation

Magic quantum effects?

I do not understand the last edits from today. [1] When I said that photons can be created at will, that doesn't mean by power of thought alone-- the phrase in English certainly covers more than that. What do you mean that "photons are not really particles"? When a positron meets an electron two of something are produced (not three or four or one). What do YOU call them? I also said that there is only one way EMR can be created, and that is by acceleration of charged particles (true). This was changed to be "in classical physics." No, that is wrong. In any type of physics, you must accelerate a charge in some way to create EMR, or a photon. There are no exceptions (if you disagree, provide one). Quantum physics may be needed to explain the line spectra of (say) nuclear gamma emission (just as it is needed to explain the line spectra of atomic emission!) but the fact of the emission itself is not a quantum process per se, only the quantization of it is. IOW, quantum physics is necessary to explain the quantization, not the process itself. The emission and absorption of gamma rays is heavily dependent on the electric dipole and quadrupole moment of nuclei as well as the charge contributions of electrons that penetate them [2] and this means that gamma rays are always (no exceptions) produced and absorbed by rearrangment of nuclear charges (which are certainly accelerated in the process). When positrons are annihilated and when neutral pions decay, again charged particles and antiparticles are pulled into each other (that means accelerated), before they disappear into photons. Same process. In QED, real photons (which are necessary for EMR) couple to charges. They do not couple to non-accelerated charges, which instead proceed in straight world lines, with no transfer of momentum. SBHarris 06:51, 19 January 2012 (UTC)

Regarding "at will": sorry if my attempt at humour made my meaning less clear. My objection to "at will" is that as far as I can tell, it doesn't mean anything here. EMR can be produced or destroyed by many different processes, and energy is always conserved. What does it mean to say that it can be produced or destroyed at will?? This seems to imply an arbitrariness to whether EMR is created or destroyed. If you had some specific meaning in mind that the current wording doesn't capture, let me know and perhaps we can find a clearer way to express it.

You seem to be taking a classical point of view, thinking of quantum mechanics as a perturbation on top of classical physics. While that's often a fine analytical method for solving problems, it's not really accurate. The world is quantum mechanical. Classical mechanics is an approximation, valid in certain limits. Nuclear and atomic emission are both fundamentally quantum processes. An electron in an atom does not have well-defined velocity. When the atom absorbs a photon, the electron changes to a state with higher energy, and still does not have well-defined velocity. The change is instantaneous; there is no well-defined change in velocity with respect to time. Describing this as "acceleration" is not, from a quantum point of view, a very useful description. I found your statement "The creation of EMR happens by only one known mechanism..." much, much too strong, and misleading.

I didn't like the wording "Photons are particles, but their properties become more 'particle-like'..." Light is not really a particle or a wave. It is what it is; some aspects of light's behaviour are conveniently described using a particle model, while others are conveniently described using a wave model. Quantum mechanics requires neither model; one can predict the results of measurements without presuming that light is either a particle or a wave. It's better just to say "Photons' properties become more 'particle-like'...". I don't object in general to calling photons particles, anymore than I would object to calling light a wave. It's just that in this specific context saying "photons are particles" added nothing useful to the sentence.--Srleffler (talk) 05:15, 20 January 2012 (UTC)

Wave model chart is wrong

The Electric and the Magnetic components of the wave should have phase difference of pi/2. Then the wave can carry energy in the direction of propagation, represented by the Poisson's vector, rotating in the plane, perpendicular to the direction of propagation. bspasov@yahoo.com — Preceding unsigned comment added by 98.154.17.2 (talk) 04:32, 13 March 2012 (UTC)

You are mistaken. The electric and magnetic components are in phase. If they were not, the time average of the Poynting vector would be zero, and no net energy could be carried by the wave. This has been discussed before.--Srleffler (talk) 04:40, 13 March 2012 (UTC)

It is YOU who are mistaken! Consult the Maxwell equations! rotE=−B The rotation of the E field is maximal at the zero (or inflexion) point. (And vice versa.) It is horrible to distribute such a poisoning stupidity in the Wiki. (after a due consideration which make it even worse)

On the other hand: I found some 30 pictures related to the topics in the google and EVERY ONE OF THEM WAS WRONG. So you are not alone with your misconception.

(István Dániel, physicist, Hungary) — Preceding unsigned comment added by Dániel I fiz (talk • contribs) 18:26, 15 April 2012 (UTC)

Sigh. The curl (also called the rotation or vorticity) of a moving wave is 90 degrees out of phase with its simple amplitude. See those max E and max B points, at the crests of their waves? There is zero rot or curl there at those points, which is to say that vorticity is zero on the x axis under the crest of an E or B wave. The max rot or curl (vorticity) actually occurs where the amplitude is zero (the node), where it's changing the fastest in space (along the axis if propagation). Vorticity there is maximal. You must be able to visualize this if you can visualize curl or vorticity. Remember that the capping surface for Stoke's law for an EM plane wave can't be orthogonal to the direction of the wave, since all curls in that direction are always zero (which is why a loop antenna can't "see" an EM wave directed exactly "through" it.). Instead, your capping surface, or loop antenna, must be directed so it lines up with all wire in the same direction as the E field, so that the magnetic flux lines of the polarized wave go exactly through the loop area orthogonally, as the wave passes. Then, it's easy to see that when a crest of magnetic field amplitude is directly in the center of such a loop, there's no change in flux momentarily according to Maxwell, and thus no EMF around the loop. This, at a time when the E field at the loop is largest! But E and induced EMF are 90 degrees out of phase in time, in the far field. They only work like Faraday induction (in phase) in the near-field, where current and B are strongest at times when E (due to minimal charge separation in the antenna) is smallest. (And vice versa, as when charge separation in the antenna is largest and E is largest in the near field, current is smallest and B is smallest in th near-field)

If you don't believe it, write out the 3 two-term vector cross products for curl E and B for the components of the E and B fields in 3D and free-space (no currents or charges) and set the components of dEx/dx and dBx/dx in the direction of x propagation, to zero (since of course they are, or can be in the absense of static sources) and then pick a linear polarization so that you're only looking at By and Ez for propagation direction x. Thus, for pure linear polarization set Bz = 0 and Ey = 0. Then the remaining x spacial derivatives of By and Ez per Maxwell are equal to the time derivatives of each other, on the other side of the equation, per Maxwell. That's Maxwell's equations for a plane wave in free space. Set c = 1. Answer is below.

Of course ALSO (easier to visualize), the rate of change of traveling wave with time at any point, is ALSO 90 degrees out of phase with the amplitude. This is easier to see right off, as clearly dB/dt and dE/dt must be zero at the crest of the wave, since for an instant there, the wave is not changing in either time or space. And also dE/dt and dB/dt are maximal at nodes where E and B are zero.

Since both rot and time derivative of fields are out of phase with the amplitude by the same amount, it turns out the simple E and B amplitudes are exactly in phase with each other! If rotE = -dB/dt and rotB = -dE/dt (in units where c = 1), then it turns out that dEz/dx = dBy/dt, and dBy/dx = dEz/dt. You can see that derivatives with regard to x (x = axis of propagation) are zero at the top of sine waves and cosine waves on axis x. If the wave is moving, you can see that the time-derivative of the same waves must momentarily be zero at the crests also. So, basically, Ez = By at any time t, and any place x. And there you are. SBHarris 23:29, 15 April 2012 (UTC)

• Later: Okay, here's the simple version, noting first that Maxwell, like all good relativistically correct equations, treats space and time on equal footing. The formula for an E field of EMR is something like E = A sin(argument), like E = A sin (kx-wt). Okay. Maxwell's exquations for free space say :$\nabla \times \mathbf{E} = -dB/dt$. The left side reduces to simple first-order derivatives of E with respect to space, and the right side is a first order derivative with respect to time. Thus, both derivatives of such a wave function result in the same 90 degree sin -> cosine phase change, or the reverse, and the phase stays the same for E and B on the other side. The same happens for the other curl B equation. Voila. A real difficulty here is that the "natural language" statement of the two relevant free-space Maxwell equations often goes something like this: a change in the E field results in a B field. Not quite-- in fact that's sort of wrong and misleading and results in problems like this. More correctly, a spacial change in the E field results in a time change in the B field and vice versa. But both are first derivatives, so whatever B is produced from E, has the same phase as E, and vice versa. SBHarris 01:49, 16 April 2012 (UTC)
  • Yes. The rate of change of E in space equals -1 times the rate of change of B in time. This clearly forbids a 90° phase difference, since the rate of change of E at a maximum is zero, and the rate of change of B where it crosses zero is not zero, but rather maximum. --Srleffler (talk) 02:19, 16 April 2012 (UTC)

OK. You are perfectly right. I made 2 errors at the same time. First I forgot about the time derivative. And I also thought something wrong about the shape of the standing waves. Sorry about this. (DI) — Preceding unsigned comment added by 193.224.139.4 (talk) 08:37, 16 April 2012 (UTC)

Biological effects

The content of the first paragraph of the "Biological effects" section has been disputed. There is a discussion at Wikipedia:Dispute resolution noticeboard#Electromagnetic radiation. Alternate proposed text for the section can be seen in this diff, and this one. Help would be appreciated in locating reliable sources that indicate what the current consensus is in the scientific community.--Srleffler (talk) 04:20, 7 April 2012 (UTC)

Typo in Wave model

About half way through the Wave Model paragraph there is the word electromagnctic

Fixed: electromagnctic --> electromagnetic

Reference with dead link

I have reverted variants of this edit twice, because of the edit summary "drop unconventional statement backed by dead link". The validity of a reference is not affected by a link becoming dead. The statement is supported by a citation. If you want to remove it, let's discuss it here first.--Srleffler (talk) 06:00, 21 June 2012 (UTC)

The statement that observing light being absorbed in discrete quantities is not by itself evidence of the quantization of light is not really controversial. Demonstrating that the light itself was quantized, not merely just the interaction with matter, won Einstein a Nobel prize. The leap from the observation to the conclusion is not a trivial one.

Perhaps we can rephrase the statement to make it clearer?--Srleffler (talk) 06:12, 21 June 2012 (UTC)

My apologies -- wasn't familiar with the guidelines in this case. My thinking was that the reference needed to be a legitimate one, and in this case a dead link didn't seemed to disqualify it. I realize now that the reference can be valid even if the link is dead. It does seem like the link should be dropped, however. modify 06:30, 21 June 2012 (UTC)

I looked up the guidelines on this to make sure I'm right. From Wikipedia:Link rot: "Do not delete factual information solely because the URL to the source does not work any longer. WP:Verifiability does not require that all information be supported by a working link, nor does it require the source to be published online." Also from that page, the dead link should remain, because having the URL may aid future editors in locating the original source. It's also possible that the link may become active again later—not unlikely in this case since it looks like a site configuration error has taken out most of this research group's online presentations.--Srleffler (talk) 06:56, 21 June 2012 (UTC)

Sounds good. modify 07:18, 21 June 2012 (UTC)

The phenomenon of the quantization of light is not in question for me. It's the thesis that it's matter, per se, that is causing the quantum behavior that seems novel to me, and for that reason, possibly suspect. My knowledge of the subject is not adequate to confirm or challenge this point, however, so I have no issue with the reversion. modify 06:37, 21 June 2012 (UTC)

I see. The statement is not very clear. The fact that one observes that absorption of light is quantized is not by itself evidence that light is quantized; it is only evidence that the absorption process operates in discrete steps. If all one knew was that light is absorbed in discrete quanta, the simplest theory would be that the quantum nature of matter causes this phenomenon. In fact, it is also true that light itself is quantized; the issue is just that observing that absorption is quantized is not by itself sufficient to prove this.--Srleffler (talk) 06:56, 21 June 2012 (UTC)

I like that clarification. From your explanation I gather that the statement is not about the nature of light or matter, it's about what can be concluded from the particular experiment in question. That makes sense. modify 07:18, 21 June 2012 (UTC)

And as a historical note, Einstein's explanation of the photoelectric effect didn't really put the nail in the coffin, either, since it could as well be explained as just another example of quantum absoption by matter (in this case an ionizing atom) while the light itself continued as a field which was available as a bank from which you could take any withdrawal so long as you had the right sized "bag". Even the dim light of the photoelectric experiment contained far too large an intensity to see single "photon" events in real time, so you're free to blame which ever side (or both sides) of the transaction for the discrete nature. Einstein didn't convince Planck. What did convince Planck (and the world) was the Compton effect, which is pretty hard to explain without single photons interacting with single electrons. In the Compton effect, an electron can absorb any energy it likes, and does, but is limited not by its own quantum nature, but by conservation of momentum for ingoing and outgoing photon, as well as itself. So, two particles. And end of controversy. SBHarris 18:13, 21 June 2012 (UTC)

E and B fields in/out phase

In traveling waves in-phase ("real"), but in standing waves out-of-phase ("quadrature").[3] Also in near field (or in the antenna) the electric and magnetic field components are 90 degrees out of phase for traveling waves[4]. (see also Near and far field or Near-field electromagnetic ranging or experimental [5] or for sound waves [6]). — Preceding unsigned comment added by 195.113.87.138 (talk) 09:51, 7 January 2013 (UTC)

Sure, but this is an article on EM radiation, not the EM field, which has its own article. The facts you mention are in the article on EMF and perhaps need more emphasis. But here, EMR already *means* far field and traveling wave. So in-phase it must be. In order to radiate it must be in phase for B and E. SBHarris 18:36, 7 January 2013 (UTC)