Is it possible to create light?

[fredreload]

Is it possible to create light with electric field and magnetic field since light is a type of electromagnetic radiation?

P.S. With laser cooling you can slow down light to 38MPH.

 

[TeethWhitener]

Is it possible to create light with electric field and magnetic field since light is a type of electromagnetic radiation?

Yes it is. In fact, you might have posted the message above by creating light. All you need is an antenna and a radio transmitter (both commonly used in your cell phone's Wi-fi communication): https://en.wikipedia.org/wiki/Antenna_(radio)

If you mean visible light, we don't yet (as far as I'm aware) have a good way of directly generating oscillating electric fields at that frequency.

 

[DrStupid]

If you mean visible light, we don't yet (as far as I'm aware) have a good way of directly generating oscillating electric fields at that frequency.

https://www.uni-wuerzburg.de/en/sonstiges/meldungen/detail/artikel/die-erste-elektrisch-betriebene-lichtantenne-der-welt/

 

[TeethWhitener]

Well whaddaya know? Neat.

 

[berkeman]

https://www.uni-wuerzburg.de/en/sonstiges/meldungen/detail/artikel/die-erste-elektrisch-betriebene-lichtantenne-der-welt/

Very cool. Now if we can just turn that around into visible light rectennas, we might have the next step in photovoltaic efficiency improvement...

 

[DrStupid]

Now if we can just turn that around into visible light rectennas, we might have the next step in photovoltaic efficiency improvement...

http://www.nature.com/nnano/journal/vaop/ncurrent/full/nnano.2015.220.html

 

[berkeman]

http://www.nature.com/nnano/journal/vaop/ncurrent/full/nnano.2015.220.html

Wow, great recent progress that I didn't know about. Thanks!

Have you read the article? I wonder how efficient the conversion was. I'm tempted to buy the article...

 

[DrStupid]

I wonder how efficient the conversion was.

0,00005 %

 

[berkeman]

0,00005 %

Ouch! Okay, you just saved me $36...

 

[fredreload]

I was thinking by passing light through electric field or magnetic field the wavelength might change, since they are made from these two components, but I can't seem to find any source on it on Google. Now that we've slowed down light to 38mph, is there a change in relativity sense of light? I've heard that someone mentioned this concept can be used for data storage but I haven't really looked into it.

 

[phinds]

Now that we've slowed down light to 38mph, is there a change in relativity sense of light?

Nothing has changed about the properties of light. It still travels at c in a vacuum and at lower speeds in other materials.

 

[fredreload]

How about this formula? Just speculating, well I thought it is interesting to put 38^2 in the formula.

 

[jtbell]

You should not think of "c" in most relativity-related formulas fundamentally as the "speed of light (in a vacuum)", but rather as the "universal speed limit" or "universal invariant speed". Light happens to travel at that speed because it's associated with massless particles. Other massless particles also travel at that speed. If photons had mass, light (in a vacuum) wouldn't travel at speed c.

 

[Nugatory]

How about this formula? Just speculating, well I thought it is interesting to put 38^2 in the formula.

Interesting but meaningless, as the $c$ in that equation is the speed of light in vacuum, not a medium.

However, it's mostly for historical reasons that we say "$c$ is the speed of light". $c$ is a constant of nature in its own right, and if it ever turned out that light in a vacuum did not move at $c$ (which is not going to happen) we wouldn't say that we were wrong about the value of $c$, we'd say that we wrong about the behavior of light.

 

[fredreload]

How about this one? If I am a photon (it doesn't need to be in a vacuum right?) I would see that any distance I want to travel to is close to zero (if I remember what was taught in high school Physics). Then suddenly I am slowed down to 38mph and an entire universe appeared. Now this doesn't mean you need to be in an Einstein-Bose condensate to feel this? Then again why do I need to create more distance for myself by putting myself inside an Einstein-Bose condensate.

 

[phinds]

How about this one? If I am a photon (it doesn't need to be in a vacuum right?) I would see that any distance I want to travel to is close to zero (if I remember what was taught in high school Physics). Then suddenly I am slowed down to 38mph and an entire universe appeared. Now this doesn't mean you need to be in an Einstein-Bose condensate to feel this?

There is no such thing as the frame of reference of a photon so your question is not meaningful.

 

[fredreload]

Well, there is a frame of reference for an object close to the speed of light right? Is having a mass a requirement?

Credit:

 

[ZapperZ]

There is a series of severe misunderstanding here that should have been corrected.

First of all, one needs to clearly understand what is meant by "the speed of light" that the OP is referring to, and this is especially the case when we are talking about the speed of light in a medium. This is the group velocity of light! That is what being slowed down in a medium.

Secondly, light has been slowed down, but has been slowed down to ZERO m/s. This means that it has been stopped, held for some time, and then "replayed" back (this is different than light being absorbed completely by an opaque object).

Thirdly, an antenna generating EM radiation is the same as having a bunch of charge, such as a bunch of electrons, being jiggle up and down (or left and right, etc.). This is similar to what is being done in the numerous synchrotron light sources all over the world, and in the many FEL facilities around the world. Most of the light being generated (including, in principle, visible, UV, IR, etc.) are bunches of electrons passing through an insertion device such as a wiggler or undulator that causes the passing bunches to jiggle back and forth, just like what you do in an antenna due to the moving current.

Zz.

 

[phinds]

Well, there is a frame of reference for an object close to the speed of light right?

Yes there is. What does that have to do with the current discussion?

Is having a mass a requirement?

A requirement for what? Having a frame of reference? Then, yes, in that massless objects travel at c and have no frame of reference.

 

[fredreload]

Well, here's a discussion about Photon's frame of reference. So I suppose I'll just leave it at laser cooling, sorry about that. How about anything at high temperature? Anyone got a video on that?

 

[ZapperZ]

Well, here's a discussion about Photon's frame of reference.

That discussion simply re-enforced the idea that such a reference frame doesn't exist in Relativity (look at the postulates of Special Relativity). You need to understand that first and foremost before attempting to use anything from Relativity. Otherwise, you'll be using it in places where it wasn't meant to be used.

Zz.

 

[fredreload]

Well yes, that's why I left my idea at laser cooling and laser cooling is done before. And now I want some information about atom at super high temperature.

 

[ZapperZ]

Well yes, that's why I left my idea at laser cooling and laser cooling is done before. And now I want some information about atom at super high temperature.

That's super vague. Atoms at "super high temperatures" are no longer atoms. They could be a plasma!

Zz.

 

[fredreload]

Cool, I'll look into it!

 

[phinds]

Well, here's a discussion about Photon's frame of reference.

No, this quickly became (after the misleading title) a thread about the fact that a photon does not HAVE a frame of reference.

 

[fredreload]

Hmm, it seems light can be condensed in the same way.

 

[DaveC49]

The most obvious source for light is an atom in an excited state and the emission of a photon as the excited state. The emission of energy as a photon by the excited electron as it changes state is the result of the interaction of the excited atom with the electromagnetic vacuum field. See Weisskopf-Wigner QED theory of spontaneous emission or Quantum Optics, Scully and Zubiary,pp23- (https://books.google.com.au/books?i...en#v=onepage&q=atom transition photon&f=false) should give you a link to the eBook.

 

[ZapperZ]

The most obvious source for light is an atom in an excited state and the emission of a photon as the excited state. The emission of energy as a photon by the excited electron as it changes state is the result of the interaction of the excited atom with the electromagnetic vacuum field. See Weisskopf-Wigner QED theory of spontaneous emission or Quantum Optics, Scully and Zubiary,pp23- (https://books.google.com.au/books?id=20ISsQCKKmQC&pg=PA430&dq=atom+transition+photon&hl=en#v=onepage&q=atom transition photon&f=false) should give you a link to the eBook.

I'm not sure by what you mean as "the most obvious source for light". For most people, the most obvious source is the old incandescent light bulb. This is definitely not a source that is due to atomic transition (look at the spectrum from such a source). One may make an argument that the fluorescent light bulbs may be such a source, but now with LEDs starting to be more prevalent, we are again moving away from sources involving atomic transitions.

Zz.

 

[DaveC49]

In the case of an incandescent filament, the emission is black-body radiation which in Einsteins explanation of blackbody emission is as the result of excitation and relaxation of a set of discrete energy states in the surface of the metal. These are transitions of electrons associated with the conduction band of the metal and the electrons in the conduction band have a distribution of energies given by the Bose-Einstein distribution. It is still an atomic transition not from a single isolated atom but from an ensemble of atoms in such close proximity that their outer electron shells overlap to form a band of energy states.

 

[DrClaude]

Ithe electrons in the conduction band have a distribution of energies given by the Bose-Einstein distribution.

You mean Fermi-Dirac, right?

 

[ZapperZ]

In the case of an incandescent filament, the emission is black-body radiation which in Einsteins explanation of blackbody emission is as the result of excitation and relaxation of a set of discrete energy states in the surface of the metal. These are transitions of electrons associated with the conduction band of the metal and the electrons in the conduction band have a distribution of energies given by the Bose-Einstein distribution. It is still an atomic transition not from a single isolated atom but from an ensemble of atoms in such close proximity that their outer electron shells overlap to form a band of energy states.

This is not right. To be able to make a transition to produce light emission, it cannot be in intraband transition. Otherwise, you violate conservation laws.

Incandescent light bulb has vibrational states transitions. You cannot call it "atomic states" anymore than you can call metallic bands as atomic states. It is why atomic/molecular physics is different than solid state/condensed matter physics.

Zz.

 

[DaveC49]

@Dr Claude Agree. The original explanation by Einstein was in terms of Bose Einstein statistics, but when Dirac developed the quantum mechanical treatment of electrons as fermions where the electrons cannot occupy exactly the same energy state. This of course is what creates the band of states known as the conduction band in metals.
@ZapperZ I have to admit to not knowing the exact mechanism involved. As the filament is at a high temperature, I would guess that phonon (lattice vibration) interactions are producing vacancies in the atomic states which are localised on each atom and that these vacancies are in turn filled by transitions from the conduction band states which are not localised on an ndividualatom which result in the emission of light. As the conduction band states have a Fermi-Dirac distribution of states as a function of energy, the resulting emission spectrum has a correponding energy-frequency-wavelength distribution. I am trying to check this out further. You're right in that any transitions must obey the selection rules which are essentially conservation rules ( energy, angular momentum(spin and orbital)).

 

[ZapperZ]

Again this is incorrect.

If you have solved the simplest 1D chain that is a standard exercise in any undergraduate solid state class, you would have seen two types of phonon modes: acoustic and optical. The optical mode has a vibration similar to the electric dipole oscillation. In other words, this mode is optically active. It can absorb and it can emit photons.

This is not an atomic transition.

Zz.

 

[TeethWhitener]

If you have solved the simplest 1D chain that is a standard exercise in any undergraduate solid state class, you would have seen two types of phonon modes: acoustic and optical.

This is, strictly speaking, incorrect. Acoustic phonon modes are inter-unit cell motions, while optical modes are intra-unit cell motions. A system with only one atom per unit cell necessarily has only inter-unit cell modes, so it's all acoustic phonons by definition.

The optical mode has a vibration similar to the electric dipole oscillation. In other words, this mode is optically active. It can absorb and it can emit photons.

If optical modes were the important ones in blackbody emission, then the blackbody spectrum would be 1) discrete, and 2) material-dependent. Instead, what we observe is that blackbody radiation is (to a good approximation) material-independent and continuous across the entire frequency range, meaning that once something is heated to roughly 800K, it'll glow dull red no matter what it's made of. This tells us that the emission is incoherent (there's a roughly continuous distribution of emitting modes), which means that the important modes for blackbody radiation are typically acoustic phonons (because their spectrum is continuous to a good approximation, especially at lower energies). Basically, in order for Rayleigh-Jeans to hold at low temps, you need a spectrum that has a roughly continuous density of states all the way down to zero energy. Acoustic modes fulfill this role rather elegantly.

EDIT: The above is true at low temperatures. Obviously, once temps start to get much higher, you do see optical modes, and eventually atomic transitions, being excited. But for blackbody radiation at more mundane temperatures, the continuous spectrum of the acoustic modes puts out the most power.

 

[ZapperZ]

I didn't specify the configuration of the 1D chain. I stated it in general to illustrate that there are 2 distinct modes that are possible.

Why would the optical mode be discrete at low temperatures? The phonon dispersion is continuous. You can see this when you do a UV-VIS measurement. You do not get a discrete spectrum.

Zz.

 

[TeethWhitener]

The phonon dispersion is continuous.

Yes. I should not have said discrete. I don't know if you typed this before, during, or after my edit (re: the low temperature part). But the point that I was trying to make is that blackbody radiation is largely material-independent, whereas optical phonon modes are not. Of course, acoustic modes technically aren't either, but for a bulk material, their spectrum extends all the way to zero energy, so that those modes are always populated (i.e., the density of states at low energy is essentially material independent).

 

[DaveC49]

Again this is incorrect.

If you have solved the simplest 1D chain that is a standard exercise in any undergraduate solid state class, you would have seen two types of phonon modes: acoustic and optical. The optical mode has a vibration similar to the electric dipole oscillation. In other words, this mode is optically active. It can absorb and it can emit photons.

This is not an atomic transition.

Zz.

Thanks for that. The optical phonon modes look like the go. Undergrad solid state was 40 years ago and I haven't touched SS much since then. I'll dig up a copy of Kittel and refresh my failing memory.

 

[sophiecentaur]

Thanks for that. The optical phonon modes look like the go. Undergrad solid state was 40 years ago and I haven't touched SS much since then. I'll dig up a copy of Kittel and refresh my failing memory.

HAHA. I have a copy on my shelf. I found the SS hard enough in those days so, today, it's really going to make my brain ache,

 

[nasu]

@Dr Claude Agree. The original explanation by Einstein was in terms of Bose Einstein statistics, but when Dirac developed the quantum mechanical treatment of electrons as fermions where the electrons cannot occupy exactly the same energy state. This of course is what creates the band of states known as the conduction band in metals.
@ZapperZ .

I am afraid you confuse the statistics of the phonons in a cavity with that of the electrons.
The photons, beeing bosons, can be described by a Bose-Einstein type of distribution. Sometimes called Plank distribution as it was first introduced by Plank.

But electrons has to be described by Fermi-Dirac distribution, even for Einstein. I don't think he used a BE distribution for electrons. Do you have any reference to this?

 

[Mark Harder]

https://www.uni-wuerzburg.de/en/sonstiges/meldungen/detail/artikel/die-erste-elektrisch-betriebene-lichtantenne-der-welt/

Is this device an electrical diode? In other words, does the asymmetry in its structure produce an asymmetry in conductivity? Tunnel diodes are semiconductor devices in which electrons penetrate a classically insulating junction by means of the QM tunnelling effect. I've never heard of it, but do/can tunnel diodes emit light?

 

[rootone]

There are many ways in which light can be created by using electrical energy.
Light generated from magnetic fields is much less well known, but you might want to check this out.
http://www.nature.com/articles/srep00492

 

[Mark Harder]

Electron beam synchrotrons also emit light. Magnets are used to steer the electrons in curved paths. The electrons are therefore accelerating toward the center of curvature. Maxwell's laws guarantee that accelerating charges emit electro-magnetic waves. Special relativity demands that at their near-luminal velocities the electrons emit photons in a focused forward directed beam. Synchrotron light sources radiate in the extreme UV to X-ray region of the EM spectrum. The tightly focused, intense radiation finds many uses in chemical and physical studies, x-ray crystallography for instance. If the electrons are circulating in bunches, then the radiation is emitted in very short but high-frequency pulses; hence, such devices can be used to time extremely fast molecular processes. I don't know if synchrotrons are used to generate IR to visible wavelengths. I would think the existence of lasers would make this application less important. Related electron beam devices called 'wigglers' or 'undulators' work according to the same principles; but in these cases, the beam passes through a gauntlet of permanent magnets that alternate in polarity along the beam path. This causes the particles to follow a zig-zag path. At each turn, they emit EMR in the forward direction. Unlike the synchrotron, this arrangement causes the forward-directed photons to interact with the particles further on down the device. I seem to recall that this has some sort of laser-like stimulated emission effect. I'm afraid the explanation gets a little fuzzy at that point, so I'll stop there. Except to say that the EMR produced has applications similar to synchrotron radiation. However, in the case of wigglers/undulators, the only application is EMR generation. These devices can also be made on a much smaller scale than synchrotrons, the size of small labs in some cases.