Is it possible to create light?

[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.