# Why light does not travel in wires?

1. Jul 27, 2012

### Artlav

Asking in most general shape, why do signals from upper Ghz and above don't seem to travel in wires?

Radio frequency waves can propagate in wires - you can have a signal like that being induced from the wave interacting with the wire, being generated, measured directly, and so on.

But starting from low Thz you need special hardware - some sort of diode, bolometers, piroelectric, photoelectric, etc, that just measure energy level for specific bands instead of producing a thz level signal in a wire.

Is there a fundamental limit that prevents receiving light like radio, or is it a lack of suitably high-frequency active devices?

Is it even possible to transmit signals at high Thz, early Phz range in a metallic wire?

2. Jul 27, 2012

### phinds

One thing you might want to check out is the "skin effect"

3. Jul 27, 2012

### Artlav

The signal/current will travel in a progressively thinner layer as the frequency increase. I've also read that it will asymptotically approach some value as the frequency increase.

I still don't see what would stop the light frequency current from happening.

4. Jul 27, 2012

### Integral

Staff Emeritus
I am sort of guessing here.
Electrons have mass and must be accelerated by the energy in the EM field. It makes sense, that above some frequency they will not be able respond to the changes. Does it make sense that if the electrons cannot carry the energy it must be in the EM field.

5. Jul 27, 2012

### Drakkith

Staff Emeritus
At that frequency, could a wire/circuit even transmit the electric field any meaningful distance before the phase shifts 180 degrees?

6. Jul 28, 2012

### krd

No, it's the signal, not the radio wave itself.

When the signal is going through the wire it's not passing as light. It's a change in voltage on the wire that is analogous to the free air radio signal.

This is probably a horrible explanation. But when the radio signal is being passed through the wire, the electron clouds of the atoms, are compressing and decompressing like water with a ripple going through it. Maybe it's not a bad explanation - the charge passes along the surface of the wire - so if you imagine a ripple, that you can't see, passing in the electric field of the surface of the wire, then that's where your radio signal is. When you get it to hit and amplifier and a transmitting aerial, then you'll have your radio waves. And it works the other way around. When you have a radio aerial, the electrons shuffle in time to a frequency they're receiving and that gets turned into a changing current in the aerial, that you can amplify and then listen to if it's a radio station.

The thing about passing a signal, is if you can get it to come out the other end as you put it in, you don't have to worry much about how that happened. Like music is stored in transverse sine waves, but when you play it through your speakers, you get a longitudinal bloom (bloom is me. it's not a formal description, but I think it's good for visualising sound waves).

7. Aug 15, 2012

### curiousatlarg

The photons are absorbed and converted into electricity.

8. Aug 15, 2012

### Drakkith

Staff Emeritus
They are usually converted just to heat. An electrical current is only produced in solar cells.

9. Aug 15, 2012

### curiousatlarg

Oh, think of an incandescent light. The electricity passes throuhg the wire and photons are emitted.
If light were shined onto a filiment, it should produce heat and electricity I am guessing. It would be very small amount because of the dilution.

10. Aug 15, 2012

### Drakkith

Staff Emeritus
Nope. The current heats up the filament which emits black body radiation. This is the key point. The current isn't doing anything to release light itself, it merely heats up the filament.

When you shine light on the filament it simply heats up slightly and may kick some electrons out of the metal through the photoelectric effect. No electrical current will be produced.

11. Aug 16, 2012

### curiousatlarg

I was referring to the photoelectric effect. Some electricity should be produced shouldn't it? Some photons would be absorbed, causing electrons to be freed, while the electrons would then return to their origional state and then a photon would be released back. However, wouldn't some current be produced? Although the filament is producing light as blackbody radiation as current is applied, incidental light upon the filament should also produce either electricity or reflecting photons sholdn't it? What would be the breakdown for tungsten? I may have time to check. Wouldn't it depend on the energy of the photon? Back to lower frquency, the photons reflect, refract, or are converted to ac current right? Heat is produced also, no doubt, but how exactly? Could we assume the heat from metal is a result of photons being released at the IR energy level as the electrons return to their normal state?

12. Aug 16, 2012

### curiousatlarg

OK, for tungsten, UV light will liberate electrons (292 nm). Other wavelengths will not, however, there will be re-emitted photons. I am trying to figure out how many IR photons would be relased for tungsten exposed to a nominal full visible spectrum.

13. Aug 16, 2012

### nsaspook

14. Aug 16, 2012

### Drakkith

Staff Emeritus
In a normal metal the light gives energy to electrons but the electrons are kicked in random directions. Some may go one way in the wire, others may go the other way, etc. This means that no current is produced, merely a heating effect. (Net current I mean. The electrons have random movements)

Heat in a material is simply the random motion of its particles, which includes the electrons. So when light hits the metal and excites electrons the random nature of the effect means that the wire would just heat up.

By the way, the term "electricity" doesn't actually mean anything. Instead use things like electric current, voltage, etc.

15. Aug 16, 2012

### curiousatlarg

OK, yes, the electrons would be moving at random. And the random motion of particles is heat. In the wire, how much of that is IR photons? On the one hand you have motion of the atoms, and on the other, you hve IR photons, which could be detected by using a vacuum and detecting the temperature change of an exposd object. How is the heat added to the atoms in the wire?

At longer wavelengths, the electrons are aligned in the wire and generate AC. At the visible spectum, how much of the light is converted into random electron movement, and how much is converted into lower energy photons?

A filament with light shining on it could be used to generate current though, with a voltage applied to a cathode which is not exposed to the light, and connected electrically to the filament, the electrons would jump to the cathode.

What about photons? If electrons are kicked off of the tungsten when they are struck with 293 nm, what of other wavelengths? The electrons would still absorb the energy wouldn't they? They would not be able to escape, however, they would still absorb the photon, and re-emit right?

Do metals emit photons when they are struck by photons, even though the threshhold energy has not been met?

16. Aug 16, 2012

### Drakkith

Staff Emeritus
What do you think IR photons have to do with this?

Why would it be converted to lower energy photons? What effect are you referring to?

You would spend energy to create the voltage for practically zero return. I would question if you even needed the voltage.

Other wavelengths above the red end of the spectrum would eject some electrons if they aren't buried too far into the wire, as they need energy to get to the surface and then to be ejected. Higher energy photons deliver more energy and make it easier. If they aren't ejected the electrons are simply excited and the material heats up. But I have to ask what you think they are emitting afterwards, as the energy is now thermal energy and will be availble to the bulk material, which contributes to its temperature.

Did you mean electrons? Objects do not generally emit photons just because they are struck by other photons. At least not in a direct step. It usually requires that the absorbed photons energy be converted to thermal energy first before being released as a photon within the black body spectrum of the object.

17. Aug 17, 2012

### curiousatlarg

Photons striking elecrons in metal will add energy and excite the electron. If the energy of the photon is greater than the threshhold energy for that metal, the electron will be ejected. However, if the photon is not, the electron will return to it's origional orbit, but in the process, shouldn't it release a photon? The reason I mentioned IR is because it is lower energy, and it is comon as thermal radiation under these conditions. Also, since IR is readily emitted by the tungsten, shoudn't it be readiy absorbed? The main thing I am not clear on is the converson of energy from photons to metal. If we assume all of the energy striking the filament is from photons, then, what are the principle ways that the energy is conducted in the filament? As for the generation of current, I was referring to Herts's experiment. As for the conversion to thermal energy, wouldn't IR be one of the principle forms? Other forms relate to the motion of the atoms. Sorry that I am not more concise.

18. Aug 17, 2012

### Drakkith

Staff Emeritus
Metals contain valence electrons that don't have a simple atomic orbital, but are delocalized, occupying the whole material at once. Exciting one of these simply promotes it to one of a near infinite number of energy levels that metals have. It doesn't return to it's energy level by emitting a photon, or at least it doesn't have to, it can give energy to the rest of the metal. Because of the huge amount of energy levels available to a metals electrons, they generally absorb practically all EM radiation wavelengths.

19. Aug 17, 2012

### curiousatlarg

OK, that makes sense. Next, what happens to the EM radiation? In some cases, electrons are ejected (for wavelengths below 293 nm), other cases, photons are emitted. In others a current is produced (if the metal object is the right length. For example, would a nano antenna generate electricity from light? Heat is also produced. In other cases, the EM passes around the metal or is reflected.

20. Aug 17, 2012

### curiousatlarg

I think I should make a table of wavelengths and predicted responses.

21. Aug 17, 2012

### Drakkith

Staff Emeritus
In the photoelectric effect electrons are only ejected if they have enough energy to get out of the metal. This depends on the metal in question and where in the metal the electron is. If it's far underneath the surface then it pretty much wont be able to make it out and be ejected as the required energy is too high. I want to say the maximum wavelength that can eject electrons is around 500-550 nm. Wavelengths shorter than this can eject electrons while wavelengths longer than this cannot.

As for photons, they aren't emitted in this case. As for antennas, they dont work with optical frequencies, only microwave and radio frequencies. IR and visible light is simply too high of a frequency.

22. Aug 17, 2012

### f95toli

Firstly, dielectic waveguides are more or less "wires" and can be used even for visible light; the most obvious example is an optial fibre; but you can also create waveguides on chips etc. The main problem tends to be that they are quite lossy.

Secondly. you CAN use wires even a very high frequencies as long as you circuit is small enough; I've seen example of people uing twised wires at 1 THz which is essentially far-infrared radiation. Small antennas can also be used up to a couple of THz.

The point is that normal metal wires simply become impractial because of the increasing losses as you go to higher frequencies which is why they are rarely used above say 100 GHz or so. But there is a transitional region fram say tens of GHz up to about 1-2 Thz where you can use EITHER wires or optical elements (lenses etc) depending on the application.

23. Aug 17, 2012

### ZapperZ

Staff Emeritus
Strangely enough, your question here isn't really about "light travelling in wires", but rather the details of a photoemission process! You are asking what happened to the electrons that never made it out, and to be able to answer that, you need to understand what exactly is going on in the photoemission process at every step of the way.

We can do this by following Spicer's 3-Step model for the photoemission process.

Step 1: A photon is absorbed by the metal, and a conduction electron is excited to the vacuum state (i.e. above the work function of the metal).

Step 2: via random diffusion/scattering process, the "lucky" electron migrates to the surface.

Step 3: the electrons that make it to the surface then escapes the metal.

Now, what happens to the ones that did not escape. There are several scenarios:

1. In Step 2, since this is a purely random process, it means that statistically, HALF of the electrons will end up moving in the wrong direction, i.e. they will have components of their velocity moving away from the surface. Those will not make it out. This is why in a simple, standard metal, you'll never get a quantum efficiency (QE - number of electrons out per photon) greater than 50%. In fact, for metals, the QE is often 0.01% - 0.0001%! These "lost" electrons will eventually lose their energy and decay back into the conduction band.

2. Also in Step 2, these excited electrons can bump into other electrons (both excited and not excited). Remember, these are metals, and there are a lot of conduction electrons. Electron-electron scattering tends to remove energy from the energetic ones. So in just one single collision with another electron, that excited electron can lose its energy and drops back into the conduction band, below the work function. Here, you can already see that the primary mechanism for energy loss isn't the emission light, but a transfer of energy to the rest of the conduction electron.

Zz.

24. Aug 17, 2012

### curiousatlarg

Thanks!

25. Aug 17, 2012

### curiousatlarg

Ah, yes, the origional question was why light does not transfer signals to wires. I was supposing that a nano antenna made of wire could work somehow, and also wondering what happens when light strikes any metal wire.