- #1
mee
- 213
- 1
why is light refereed to as the electromagnetic spectrum when light carries no charge and is unaffected by magnets?
mee said:why is light refereed to as the electromagnetic spectrum when light carries no charge and is unaffected by magnets?
humanino said:light IS affected by magnets !
ZapperZ said:It is historically called "electromagnetic" because from Maxwell equation, the description of light contains two oscillating components: an electric field vector and a magnetic field vector, thus the "electro" and "magnetic" parts.
Zz.
what_are_electrons said:Mee's photon question is the same question I posed to another forum very recently.
Perhaps the question should be rephrased to:
"What is the experimental evidence / proof that light has either electric or magnetic properties?"
Maxwell's equations do not prove that light (photons or quanta) are made of electromagnetic radiation/disturbance/waves/... Maxwell's equations were derived from experimental evidence (from Faraday, Lorentz and Gauss) involving solid state materials with electric currents.
Maxwell only suggested that light was made of EM.
Hertz took his suggestion and made an antenna.
Microwaves (one frequency of photons) are routinely used to transfer momentum to electrons in synchrotrons, but this is not proof that photons have electric or magnetic properties.
Antenna emit radio waves (another frequency), but this does not prove that radio waves have electric or magnetic properties.
Photon entanglement is not a proof of E or M properties.
Electrons absorb and emit photons just like an antenna, but that is not proof.
Electrons are influenced by E and M fields, which shows that electrons have both E and M fields, but electrons are not photons.
Protons can also absorb and emit photons. This is shown by NMR.
Magnetic fields and electric fields are known to produce a change in the energy (frequency) of the photons that excited electrons emit as they decay to ground state, but that is not proof that photons are affected by E or M fields. The change caused by the magnet or electric field is due to work (energy added or lost) done to the electron before or after it emits or absorbs the photons. There is no proof that the magnetic or electric fields have influenced the photons once they are freed from the electrons.
So, the question remains:
"What is the experimental evidence / proof that light has either electric or magnetic properties?"
It may be an extreme case, but magnetic fields do affect photons, http://www.journals.uchicago.edu/cgi-bin/embpcgi.pl/cgi-bin/res-page.epl?objid=374334 , which should be observable in magnetar atmospheres.vanesch said:No.humanino said:light IS affected by magnets !
Well.
If you want to nitpick, I guess it is, as photon-photon interactions.
But for all practical purposes, it isn't.
cheers,
Patrick.
Nereid said:It may be an extreme case, but magnetic fields do affect photons, http://www.journals.uchicago.edu/cgi-bin/embpcgi.pl/cgi-bin/res-page.epl?objid=374334 , which should be observable in magnetar atmospheres.
ZapperZ said:Visit a particle accelerator. Look at the RF accelerating "cells" that are used in both the photoinjector and the linac. The particles (typically electrons) are accelerated purely via either standing wave or traveling wave RF fields (which is of course, an EM wave). Without the presence of the E-field in the correct geometry, this would not have been possible. In fact, practically all calculations and design of photoinjectors and accelerating structures for accelerators use classical E&M theory.
Secondly, the fact that Maxwell Equations were derived from experimental observations is in itself, an experiment proof of the validity of the idea that EM radiation contains both E and B field. I mean, what stronger empiricial evidence can we have than having theory that is consistent with practically ALL of classical E&M observation?
Thirdly, visit any synchrotron center and look at experiments that make use of the intense light coming off one of the insertion device. The polarization of that light (the evolution of the E-field as defined in Maxwell Eqn) plays a significant role in experiment ranging from optical conductivity to photoemission, to x-ray diffraction! If light has no E and B field, it will then be doing something completely mysterious in experiments such as angle-resolved photoemission, and the semiconductors in your modern electronics should not work (the band structures of most widely-used semiconductors were experimentally confirmed using photoemission measurements).
Zz.
what_are_electrons said:There is no EM radiation (photons of light) within a solid.
what_are_electrons said:We both know that light (microwaves in this case) can transfer momentum to electrons, but that is not proof that light has EM properties. Tell me - how is it that classical EM theory can be applied to a quantum particle when classical defines a band or a sea of electrons with all states filled?
Sorry, but the experimental observations used by Maxwell were achieved using solid state materials. There is no EM radiation (photons of light) within a solid. It seems that we need to discern EM fields generated by currents within solid state and EM radiation which is outside the solid state. Being consistent with a theory is not proof. I need experimental evidence.
I agree that the magnets cause the electrons to emit light by forcing the electrons to change energy state, but that does not prove that light has EM properties. It is not necessary to use polarized light to produce those effects so I don't see your point. I agree that light is indeed absorbed by electrons and/or parts of the core, but that does not prove that light has EM properties.
My objectives in this topic are to:
(1) begin a serious effort to better define the inherent properties of light within a classical view and to combine that with a quantum field theory if possible and needed
(2) in conjunction with the objective of (1) is a need to define the true nature of the electron within a classical framework, not the statistical probabilities now using in QM
It's time to bring physics back to reality. That is one reason for the quote I use from Einstein.
One person's 'exotica' is another's everyday fare?ZapperZ said:Well, do you think it is actually fair to bring in exotica such as vacuum polarization/fluctuations, etc., simply to show this?
D'accord. However, the original question was done and dusted by post#4 (yours, I think); as often happens in threads here, it started to move onto something else.If we have to invoke such a picture, then we shouldn't be talking about an EM wave (and classical field) in the first place, since in QFT, there are no such thing as a classical field. So the original question then (how "electromagnetic" radiation got its name) becomes moot and irrelevant.
Yes.Can we at least agree upon the idea that CLASSICAL E&M indicates no apparent interaction of EM-radiation with magnetic fields?
ZapperZ said:What does a "filled state" have anything to do with momentum transfer from light to electrons? Secondly, how do you think this momentum transfer occur? Since light has no mass, the ONLY means of transfer is via an interaction of the electron with an electric field. Since there IS an energy transfer, it cannot be just from a magnetic field, so there has to be an electric field present. It is this axial electric field within a photoinjector cell and the linac that is the mechanism of acceleration for these electrons. And the fact that the GEOMETRY of the system and the direction of the E-field in the RF system MATTERS on whether we have an acceleration or not is clearly a proof. And keep in mind that these are FREE electrons. They are not confined to any solid and therefore have no "band or sea of filled states"!
Er.. hello? Look up an area of condensed matter physics (of which solid state physics is a part of) called optical conductivity. In it, you will see propagation of light/EM radiation/photons through solids. Experimental techniques such as Raman spectroscopy and FTIR all use this phenomena! It is one of the techniques we use in condensed matter to study the phonon structure of solids!
I was hoping you would at least try to figure out what "angle-resolved photoemission" is before you replied to me. It would at least give you an indicaton why the polarization of the photon matters in such experiments. I guess I was wrong. My avatar shows the photoelectron intensity of electrons moving in a particular direction in a crystal. It was done with a plane-polarized light having the SAME direction as the momentum of the electrons. If I use the identical light, but with the polarization rotated by 90 degrees (but keeping the same Poynting vector direction), I get almost nothing! Both classical and quantum theories clearly show that the most effective transfer of energy and momentum between an electric field and electrons in solids is when they the E-field and the electronic momentum are parallel to each other. This is the fundamental ingredient of the description of a photoemission phenomena - the coupling between the E-field vector of the light incident light with the crystal lattice. This is such a well-known phenomena that we USE it to study materials with - thus my statetement that your semiconductors in your modern electronics would not work if what we learned from photoemission results are wrong!
Then it appears that you are trying to bring up unverified ideas of your own. This then belongs in the Theory Development section. BTW, have you ever considered that the reason why you do not "see" this is because you haven't studied it, nor done it? I'd say my advice to you to visit an accelerator and synchrotron facilities was timely. I have or am working at both, and I have no problem "seeing" the principle of physics at work everyday.
Zz.
meteor said:the Faraday effect rotates the plane of polarization of plane polarized light. This rotation is achieved applying a magnetic field to light
what_are_electrons said:First things first.
Please clarify your statement about transfer. Does light transfer momentum or energy?
I am trying to resolve how light transfers momentum or energy. That is part of the point of these discussions. You mentioned energy transfer. Do you really mean energy transfer or momentum transfer? Once the photon transfers that momentum or energy, where does the photon go? Back in time?
To observe, measure, record or prove the existence of the E or M properties of light, they must be tested in a vacuum, not within condensed matter which is a completely different medium that is filled with particles that are well proven to have electric and magnetic properties. So, let's drop solid state phenomena from these discussions.
(5)(8) Not sure what you mean by Optical Conductivity? Is this photoconductivity?
When we talk about linear polarization, we are talking about light that has passed through the electron and atomic core structure of a solid state (condensed matter) crystal that produces the observed change (linear polarization) in the light.
A crystal is a highly complex set of electric and magnetic fields which are not well defined. (I say this because QM can only do a "good" ab initio projection of hydrogen which has only a single electron and single proton.)
The questions that come to mind in the production of linear polarized light are: What happened in the atom to produce that change? Did one or more of the electrons produce that change? Was it just the electric field of a single electron or was it a set of electrons that produced the change? What is the shape of the electric and magnetic fields of the electron that presumeably absorbed part of the original beam and passed the emitted beam of light that is now "linear" polarized? Does the EM field of a single electron have a peculiar shape that causes light to be emitted in mainly one direction when the excited state is decaying and emitting the photon?
I'm saying that we make good use of polarization in many different spectroscopies, but unfortunately, we still don't know what is going on because we have used solid state matter to produce light polarization.
Again, we must not use solid state physics to study light. I still need an example of experimental proof that shows that light has E and/or M properties.
Hmmm, for some reason I thought we were talking about the transfer of momentum from microwaves to free electrons in the synchrotron beam, not the electrons that are in the process of emerging from an atom due to photons from the beam.ZapperZ said:Both. In a photoemission, It's energy is transferred directly to the KE of the photoelectrons, while it's momentum is the reason why photoemission does not occur to uncouple, free elecrons due to conservation of momentum.
Zz.
I still hope to abandon the solid state and discuss just vacuum based EM.ZapperZ said:Both. Photon number is not conserved. The photon need not go anywhere. The phonon modes of a solid are able to take up the slack very well, thank you.
Zz.
Does this mean gaseous stars are not a valid source of light?ZapperZ said:But you will understand why I find this insistance rather silly. All you need to do is look up, for example, the standard accepted values of "e" and "h" from CODATA to realize that those two fundamental values were obtained from condensed matter experiments! Secondly, the whole definition of what light/photons are, are based on its interaction with matter! And in fact, photons came into being BECAUSE of its solid state phenomena. This is similar to insisting that we discuss charged particles but without invoking that they have a charge!
Zz.
Tell me. Why is it that accelerated electrons emit light? And is there some inherent property within the electron that is the source of this polarization or is the arrangement of the magnets?ZapperZ said:Common, but not necessary. An insertion device such as an undulator or wiggler in the electron synchrotron storage ring can also produce highly polarized light. And it need not be just linearly polarized light either! Thus, I can produce polarization in light without having to use any "solid state" device at all!
Zz.
At this time, I do not know what polarization is. I know what I've been taught and have read, but I'm still searching for proof. Just like the E and B fields of light. I'm looking for proof they exist.ZapperZ said:You talk freely about "polarization" without bothering to define what it is, since you do not believe there are E and B fields in light.
Zz.
Sorry. Did not mean to imply that.ZapperZ said:Read above why I find this last insistance rather strange, and the example of a wiggler/undulator. Furthermore, for some odd reason, you appear to not consider the well-ordered modes that we can get in waveguides as a valid way to generate not only polarized light, but light with exact geometry. If you think such RF signal in an accelerator was generated by some "solid state" process, give me a call since I'd like to buy one to replace all our bulky klystrons.
Zz.
Have not read about this yet. But I ask: What is a plasma if not a collection of electrons, ions and neutral atoms.ZapperZ said:As an added note: recent development in new scheme of particle acceleration is a clear example of the presence of E and B field in light. Both the plasma wakefield generation using intense laser, and the dielectric-loaded wakefield scheme (of which I am working on) would be zilch without E and B field in EM radiation - which IS the initial question in this string anway, isn't it?
Zz.
what_are_electrons said:Does this mean gaseous stars are not a valid source of light?
Have not read about this yet. But I ask: What is a plasma if not a collection of electrons, ions and neutral atoms.
Yep, this string is about E & B in EM radiation. My objective is to find proof that light by itself, free of any solid state phenomena, does have E & B properties.
Am looking for a definitive way to prove that light, in and of itself, has E & B properties as "demanded" by the original question (restated below). To achieve this, it seems reasonable to eliminate systems (ie all states of matter) that are expected to complicate the experiment.ZapperZ said:Where did I say that? You, on the other hand, consider any light interaction with solids as not valid per light description. So the exclusion philosophy was on YOUR part, not mine.
Zz.
If I'm moaning, then yes, I'm moaning about the use of any state of matter to resolve whether or not light has E & B fields or as some prefer to think an electromagnetic field.ZapperZ said:It certainly is NOT a "solid state" device, which is what you were moaning about. So now, not only do you want to eliminate light interaction with solids, you want to eliminate its interactions with all matter! How convenient..
Zz.
ZapperZ said:No, it ISN'T about E and B in EM radiation. It was about why EM radiation was called that, way back before QFT, before photons, etc.. If you believe that light has no E and B components, then I suggest you come up with a formulation of your version of the theory of light to explain ALL of the observed phenomena and publish it. Then we'll talk. Till then, any further discussion on this belongs in the Theory Development section.
Zz.
Light is referred to as electromagnetic spectrum because it is a form of energy that can travel through space in the form of waves. These waves have both electric and magnetic components, hence the term "electromagnetic". The spectrum refers to the range of wavelengths of electromagnetic radiation, including visible light, radio waves, microwaves, infrared radiation, ultraviolet radiation, X-rays, and gamma rays.
The electromagnetic spectrum is a continuum of all the possible wavelengths of electromagnetic radiation, with light being a small portion of this spectrum. Light is a type of electromagnetic radiation that falls within the visible range of the spectrum, with wavelengths ranging from 400 to 700 nanometers. This means that light is a form of energy that is emitted and absorbed by charged particles and can travel through a vacuum at the speed of light.
The electromagnetic spectrum is important because it encompasses all forms of electromagnetic radiation, which are essential to our daily lives. Different forms of electromagnetic radiation have various uses, such as visible light for vision, radio waves for communication, microwaves for cooking, and X-rays for medical imaging. Understanding the spectrum allows us to harness these forms of energy for various applications.
The electromagnetic spectrum is measured in units of length called nanometers (nm) or meters (m), depending on the scale. The smaller the wavelength, the higher the frequency and energy of the electromagnetic radiation. The spectrum is divided into different regions based on these measurements, with visible light falling between the range of 400-700 nm.
No, we cannot see the entire electromagnetic spectrum with our naked eyes. Our eyes are only sensitive to a small portion of the spectrum, which is visible light. However, with the help of specialized equipment, we can detect and measure other forms of electromagnetic radiation, such as infrared and ultraviolet light, that are not visible to us. This allows us to study and utilize the full range of the electromagnetic spectrum.