Can a wire having a $610$-$670$ THz (frequency of blue light) AC frequency supply, generate blue light?
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We know that when we give alternating current across a wire then it will generate an electromagnetic wave which propagates outward.
But if we have a supply which can generate 610 to 670 terahertz of alternating current supply then does the wire generate blue light?
electromagnetism visible-light photons
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add a comment
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$begingroup$
We know that when we give alternating current across a wire then it will generate an electromagnetic wave which propagates outward.
But if we have a supply which can generate 610 to 670 terahertz of alternating current supply then does the wire generate blue light?
electromagnetism visible-light photons
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3
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You could shine light (which is supply) on the wire, and the wire reflects this light (technically emmiting). This is not meant in jest, but pointing out you have to couple the signal into your antenna, and we know you cannot use wires for that. So you would have to coupöe probably optically.
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– lalala
May 27 at 19:15
add a comment
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$begingroup$
We know that when we give alternating current across a wire then it will generate an electromagnetic wave which propagates outward.
But if we have a supply which can generate 610 to 670 terahertz of alternating current supply then does the wire generate blue light?
electromagnetism visible-light photons
$endgroup$
We know that when we give alternating current across a wire then it will generate an electromagnetic wave which propagates outward.
But if we have a supply which can generate 610 to 670 terahertz of alternating current supply then does the wire generate blue light?
electromagnetism visible-light photons
electromagnetism visible-light photons
edited Jun 21 at 10:08
knzhou
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asked May 26 at 13:56
user210956user210956
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You could shine light (which is supply) on the wire, and the wire reflects this light (technically emmiting). This is not meant in jest, but pointing out you have to couple the signal into your antenna, and we know you cannot use wires for that. So you would have to coupöe probably optically.
$endgroup$
– lalala
May 27 at 19:15
add a comment
|
3
$begingroup$
You could shine light (which is supply) on the wire, and the wire reflects this light (technically emmiting). This is not meant in jest, but pointing out you have to couple the signal into your antenna, and we know you cannot use wires for that. So you would have to coupöe probably optically.
$endgroup$
– lalala
May 27 at 19:15
3
3
$begingroup$
You could shine light (which is supply) on the wire, and the wire reflects this light (technically emmiting). This is not meant in jest, but pointing out you have to couple the signal into your antenna, and we know you cannot use wires for that. So you would have to coupöe probably optically.
$endgroup$
– lalala
May 27 at 19:15
$begingroup$
You could shine light (which is supply) on the wire, and the wire reflects this light (technically emmiting). This is not meant in jest, but pointing out you have to couple the signal into your antenna, and we know you cannot use wires for that. So you would have to coupöe probably optically.
$endgroup$
– lalala
May 27 at 19:15
add a comment
|
6 Answers
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It would be hard to generate such a current and harder still to get it to produce any blue light - though this is theoretically possible.
The main problem is that you are probably thinking of a metal wire. Metals absorb visible light, both reflecting it and turning it into lattice vibrations. This is because the wavelength of visible light is just a few thousand atoms long in size so it is in a "sweet spot" for exciting solid crystals. In fact, the tendency for solid objects to absorb, reflect and otherwise interact with visible light is why it is "visible".
In a normal radio wave, your metal wire will need to be on the order of a wavelength of the radio wave you want to produce. This is typically on the order of meters. Automobiles of the 20th century had metal wires sticking out of them, about 1 meter long, called "antennas", to catch such waves.
But for blue light the wavelength is only about 5 x $10^{-7}$ meters so any useful antenna would be very tiny because an "electron density wave" in your wire would be "turning around" before it got very far.
The electromagnetic spectrum is divided up not so much by "wavelength and frequency" as by the way that any given part of the spectrum interacts with matter. So, radio waves will interact via electron currents in long metal wires. But visible light interacts more with lattice vibrations and non-ionizing atomic transitions. So, "current in a wire" type emission works in frequency up to a thing called "the terahertz gap" https://en.wikipedia.org/wiki/Terahertz_gap . Above this frequency other emission techniques are usually required. Blue light is about three orders of magnitude higher in frequency than the terahertz gap.
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For normal radio it is efficient to have a wire that's Lambda/2 but not necessary; "any old wire" will emit some radiation at RF. So it's not prima facie impossible that any old wire will glow faintly blue at the right frequency.
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– Peter A. Schneider
May 27 at 5:40
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@PeterA.Schneider I think your comment exactly hits the main point of this question that the other answers aren't really addressing.
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– KF Gauss
May 27 at 7:03
4
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@PeterA.Schneider It's also not prima facie impossible that you might quantum tunnel to the opposite side of the Earth. In practice, it's effectively impossible. Electrons with sufficient energy to conduct a 600THz wave are immediately absorbed by conductive metals. We know of no means to generate a controlled electrical (ie: moving electrons) signal at that frequency either, so even if you had an idealized radiating nano-antenna we still would not be able to supply it a driving voltage.
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– J...
May 27 at 19:03
4
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@PeterA.Schneider Fundamentally, the wire itself cannot co-exist with an electromagnetic signal in the 600THz range. By removing the electrons from the interfering atoms, you can start to think about moving electrons that quickly in systems like Free Electron Lasers, but in a bulk conductor this is simply not possible.
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– J...
May 27 at 19:08
2
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@uhoh - definitely liquid mercury reflects blue light and diamond is transparent ... however I am focused on the triumvirate of waves (radio, micro) vs. lights (IR,visible, UV), vs. rays (x and gamma) in the EM spectrum - it is only the "lights" which have the ideal wavelengths to do things like couple with phonons ... the OP is thinking about applying "waves" techniques to the "lights" - and I just want to make clear why they are different beasts - rays "see" individual atoms, waves "see" large conductive seas in metals, lights "see" molecules to macroscopic things
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– Paul Young
May 28 at 13:35
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show 6 more comments
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An alternating voltage at that frequency is light. There's no 'generate' about it - the power supply is just a light source.
And if you have a wire, that is, a conductor made of metal, then the light won't propagate inside it at depths longer than the skin depth for that material at that particular frequency, which is generally tiny.
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I don't object to this, but a periodic change in the electromagnetic field and a periodic fluctuation in electron density and momentum are, to me, distinct concepts. Of course, you are right because the two hybridize as a collective excitation, but that kinda makes it hard to understand at the high school level.
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– Paul Young
May 26 at 14:38
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@PaulYoung It is hard to justify when beginners are taught (incorrectly) that "electric current" the same as "electrons moving". But in any case, there is no law of physics that says "the laws of physics must be easy to justify!"
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– alephzero
May 26 at 15:00
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This is technically accurate but potentially misleading. Would an observer looking at the wire see the color blue? That is what the OP wants to know.
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– Owen
May 27 at 0:08
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@alephzero : The idea that electric current is "electrons moving" isn't completely "wrong", it's incomplete : electric current is charge moving, which may or may not be electrons. In this case , however, it is electrons, and thus the question of the adequacy or not of that "high school" definition is unimportant here. The distinction between a moving electric field and moving charges is very important since it is at the heart of the oft-posed question when first encountering drift velocity - "why does the light turn on 'instantly' when I flip the switch if electrons crawl?"
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– The_Sympathizer
May 27 at 2:54
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Does this actually answer the question? It seems that it does not. Would the wire emit light? If the light would only propagate a tiny distance within the material, does that mean that light beginning at a place within the material that is closer to the edge than the limit of propagation would escape the material and be seen? I think this is where confusion is coming and will continue to come from.
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– Clonkex
May 27 at 3:20
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I second this Emilio Pisanty's point: the power supply you are
envisioning is a light source. Now the question that remains is: can
you propagate this light through a wire, just like you would do with a
regular low-frequency electric signal?
To get a hint of the answer, look at how people use wires to transport
high frequency signals, into the many MHz up to the multi-GHz range. A
single wire doesn't work, because it has a tendency to radiate all the
power you feed it into the air as free electromagnetic waves. The trick
is to use two wires carrying opposite currents. You can think of them
as one being the signal and the other being the return wire, but their
roles could be symmetric. If you keep them close enough, most of the
electromagnetic field will be confined between them, and you will be
able to transmit the power without too much loses. You can further
reduce the losses by twisting the wires together. At the highest
frequencies, you would get best results by putting one wire inside the
other which, shaped like a tube, functions like a shield. This is called
a coaxial cable, and some of them are good up to tens of GHz.
The think that is not so intuitive is that, while the metal wires carry
the current, the actual power is carried by the electromagnetic field
that propagates between the wires. So the main role of the metal wires
is thus to guide the electromagnetic waves and, for this reason, the
high-frequency cables are considered to be waveguides.
Could you adapt this waveguide technique to the propagation of light?
The answer is yes, some people have indeed built nanosized coaxial
cables for this very purpose.
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An optical fiber is exactly a waveguide for visible light.
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– Henning Makholm
May 27 at 11:11
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@HenningMakholm: Indeed, but it's usually made of non-conducting material. The nano-coax I linked to is then a better analogous to the typical metal wires the OP seems to have in mind.
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– Edgar Bonet
May 27 at 12:14
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This is an interesting paper, I would be curious to know how they couple into and out of that nano-coax. While it may use conductive components in its construction, I can't help but feel that it effectively acts in much the same way as a normal optical fiber. I don't believe we have any means of exciting an electrical signal at ~600THz into such a cable. Other than shining a blue laser at it and coupling it in with optical techniques, how could you put energy into that nano-coax cable?
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– J...
May 28 at 15:05
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@J.. As a thought experiment, imagine taking any standard RF oscillator, and scaling it way down so it's a few tens of nanomatres across and oscillates in the frequency range of blue light. Now you connect the output to your scaled down coax. At the other end of the tiny coax you make a shielded dipole antenna and look at it. Is it glowing blue?
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– immibis
May 28 at 22:15
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@immibis It's not my imagination that's lacking - my intuition tells me that physics will protest before your scaling experiment gets to the required size. If you've read a paper where this has actually been done, or at least has been even theorized to be possible, I'd be interested to read it.
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– J...
May 28 at 22:18
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We do have a power supply which can generate current oscillations at optical frequencies: light. It won't transmit any distance along a wire, but if your "current source" and antenna are the same object, you have what is called an optical antenna, and the study of these is an active field of research. I don't know if any of them have any meaningful efficiency down at blue light frequencies, but they do work in the green which is not too far off. See, for example, this review article.
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Even at tens of gigaherts, one does not carry current "in" a conductor -- it is carried along the outside (Google "skin effect").
There are transmission lines for high-frequency RF that basically launch an RF wave along a single naked wire, and catch it at the other end -- think of a coax without the outer shield. If you take this analogy and pursue it into absurdity and beyond, then if you take a really well polished wire, and really carefully launch blue light along its length, then as long as the wire doesn't bend too suddenly, the light -- or some portion of it -- will be refracted and "stick*" to the wire.
I think you could achieve a setup in a lab that involved people looking at a blue-glowing end of a carefully-maintained copper wire or gold wire and going "oooh!". I doubt there is much potential for practical use here.
* Imprecise language used on purpose -- I'd have to do a lot of work to do the math on this one!
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"Imprecise language used on purpose" - indeed ... I am waiting for someone to say "frequency dependent permittivity", "maxwell's equations", "boundary conditions at the interface" and "quantum effects" ... the OP's question seems to be at the gnarly coal-face where electrical engineers meet physicists and can't talk to one another
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– Paul Young
May 30 at 15:39
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Yeah definitely . You can create light corresponding on any frequency by this method. It’s just that creating this circuit will be very challenging. To give you a perspective, the highest frequency that we have obtained with modern electronic circuits is around 10^11 Hz.
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a) Source? b) If we can create 10^11 Hz, how does that relate to tetrahertz? Is it more or less?
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– Clonkex
May 27 at 3:13
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@Clonkex kilo, mega, giga, tera? Steps of 3.
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– Peter A. Schneider
May 27 at 5:37
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@PeterA.Schneider I mean, I know how to read numbers in scientific notation, and I know what tera means (ugh, I wrote tetra by accident in my last comment), but it's not from immediately obvious whether 10^11 is smaller or larger than the required 610-670 THz, given I commonly work directly with such large numbers. I'm sure others will know without any effort whatsoever that 1 THz is 10^12 Hz, but I had to sit and think about it.
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– Clonkex
May 27 at 7:00
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@Ishan I guess the easiest way to create such an oscillation in the metal electrons is ... to shine a blue light at them. And lo and behold, the metal will shine blue right back ;-). As Emilio so insightfully observed, the oscillations won't penerate the metal very deeply, it's kind of a surface effect.
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– Peter A. Schneider
May 27 at 10:55
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At a few nanometres, you'd be hand soldering molecules.
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– rackandboneman
May 28 at 23:19
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6 Answers
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6 Answers
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It would be hard to generate such a current and harder still to get it to produce any blue light - though this is theoretically possible.
The main problem is that you are probably thinking of a metal wire. Metals absorb visible light, both reflecting it and turning it into lattice vibrations. This is because the wavelength of visible light is just a few thousand atoms long in size so it is in a "sweet spot" for exciting solid crystals. In fact, the tendency for solid objects to absorb, reflect and otherwise interact with visible light is why it is "visible".
In a normal radio wave, your metal wire will need to be on the order of a wavelength of the radio wave you want to produce. This is typically on the order of meters. Automobiles of the 20th century had metal wires sticking out of them, about 1 meter long, called "antennas", to catch such waves.
But for blue light the wavelength is only about 5 x $10^{-7}$ meters so any useful antenna would be very tiny because an "electron density wave" in your wire would be "turning around" before it got very far.
The electromagnetic spectrum is divided up not so much by "wavelength and frequency" as by the way that any given part of the spectrum interacts with matter. So, radio waves will interact via electron currents in long metal wires. But visible light interacts more with lattice vibrations and non-ionizing atomic transitions. So, "current in a wire" type emission works in frequency up to a thing called "the terahertz gap" https://en.wikipedia.org/wiki/Terahertz_gap . Above this frequency other emission techniques are usually required. Blue light is about three orders of magnitude higher in frequency than the terahertz gap.
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16
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For normal radio it is efficient to have a wire that's Lambda/2 but not necessary; "any old wire" will emit some radiation at RF. So it's not prima facie impossible that any old wire will glow faintly blue at the right frequency.
$endgroup$
– Peter A. Schneider
May 27 at 5:40
3
$begingroup$
@PeterA.Schneider I think your comment exactly hits the main point of this question that the other answers aren't really addressing.
$endgroup$
– KF Gauss
May 27 at 7:03
4
$begingroup$
@PeterA.Schneider It's also not prima facie impossible that you might quantum tunnel to the opposite side of the Earth. In practice, it's effectively impossible. Electrons with sufficient energy to conduct a 600THz wave are immediately absorbed by conductive metals. We know of no means to generate a controlled electrical (ie: moving electrons) signal at that frequency either, so even if you had an idealized radiating nano-antenna we still would not be able to supply it a driving voltage.
$endgroup$
– J...
May 27 at 19:03
4
$begingroup$
@PeterA.Schneider Fundamentally, the wire itself cannot co-exist with an electromagnetic signal in the 600THz range. By removing the electrons from the interfering atoms, you can start to think about moving electrons that quickly in systems like Free Electron Lasers, but in a bulk conductor this is simply not possible.
$endgroup$
– J...
May 27 at 19:08
2
$begingroup$
@uhoh - definitely liquid mercury reflects blue light and diamond is transparent ... however I am focused on the triumvirate of waves (radio, micro) vs. lights (IR,visible, UV), vs. rays (x and gamma) in the EM spectrum - it is only the "lights" which have the ideal wavelengths to do things like couple with phonons ... the OP is thinking about applying "waves" techniques to the "lights" - and I just want to make clear why they are different beasts - rays "see" individual atoms, waves "see" large conductive seas in metals, lights "see" molecules to macroscopic things
$endgroup$
– Paul Young
May 28 at 13:35
|
show 6 more comments
$begingroup$
It would be hard to generate such a current and harder still to get it to produce any blue light - though this is theoretically possible.
The main problem is that you are probably thinking of a metal wire. Metals absorb visible light, both reflecting it and turning it into lattice vibrations. This is because the wavelength of visible light is just a few thousand atoms long in size so it is in a "sweet spot" for exciting solid crystals. In fact, the tendency for solid objects to absorb, reflect and otherwise interact with visible light is why it is "visible".
In a normal radio wave, your metal wire will need to be on the order of a wavelength of the radio wave you want to produce. This is typically on the order of meters. Automobiles of the 20th century had metal wires sticking out of them, about 1 meter long, called "antennas", to catch such waves.
But for blue light the wavelength is only about 5 x $10^{-7}$ meters so any useful antenna would be very tiny because an "electron density wave" in your wire would be "turning around" before it got very far.
The electromagnetic spectrum is divided up not so much by "wavelength and frequency" as by the way that any given part of the spectrum interacts with matter. So, radio waves will interact via electron currents in long metal wires. But visible light interacts more with lattice vibrations and non-ionizing atomic transitions. So, "current in a wire" type emission works in frequency up to a thing called "the terahertz gap" https://en.wikipedia.org/wiki/Terahertz_gap . Above this frequency other emission techniques are usually required. Blue light is about three orders of magnitude higher in frequency than the terahertz gap.
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16
$begingroup$
For normal radio it is efficient to have a wire that's Lambda/2 but not necessary; "any old wire" will emit some radiation at RF. So it's not prima facie impossible that any old wire will glow faintly blue at the right frequency.
$endgroup$
– Peter A. Schneider
May 27 at 5:40
3
$begingroup$
@PeterA.Schneider I think your comment exactly hits the main point of this question that the other answers aren't really addressing.
$endgroup$
– KF Gauss
May 27 at 7:03
4
$begingroup$
@PeterA.Schneider It's also not prima facie impossible that you might quantum tunnel to the opposite side of the Earth. In practice, it's effectively impossible. Electrons with sufficient energy to conduct a 600THz wave are immediately absorbed by conductive metals. We know of no means to generate a controlled electrical (ie: moving electrons) signal at that frequency either, so even if you had an idealized radiating nano-antenna we still would not be able to supply it a driving voltage.
$endgroup$
– J...
May 27 at 19:03
4
$begingroup$
@PeterA.Schneider Fundamentally, the wire itself cannot co-exist with an electromagnetic signal in the 600THz range. By removing the electrons from the interfering atoms, you can start to think about moving electrons that quickly in systems like Free Electron Lasers, but in a bulk conductor this is simply not possible.
$endgroup$
– J...
May 27 at 19:08
2
$begingroup$
@uhoh - definitely liquid mercury reflects blue light and diamond is transparent ... however I am focused on the triumvirate of waves (radio, micro) vs. lights (IR,visible, UV), vs. rays (x and gamma) in the EM spectrum - it is only the "lights" which have the ideal wavelengths to do things like couple with phonons ... the OP is thinking about applying "waves" techniques to the "lights" - and I just want to make clear why they are different beasts - rays "see" individual atoms, waves "see" large conductive seas in metals, lights "see" molecules to macroscopic things
$endgroup$
– Paul Young
May 28 at 13:35
|
show 6 more comments
$begingroup$
It would be hard to generate such a current and harder still to get it to produce any blue light - though this is theoretically possible.
The main problem is that you are probably thinking of a metal wire. Metals absorb visible light, both reflecting it and turning it into lattice vibrations. This is because the wavelength of visible light is just a few thousand atoms long in size so it is in a "sweet spot" for exciting solid crystals. In fact, the tendency for solid objects to absorb, reflect and otherwise interact with visible light is why it is "visible".
In a normal radio wave, your metal wire will need to be on the order of a wavelength of the radio wave you want to produce. This is typically on the order of meters. Automobiles of the 20th century had metal wires sticking out of them, about 1 meter long, called "antennas", to catch such waves.
But for blue light the wavelength is only about 5 x $10^{-7}$ meters so any useful antenna would be very tiny because an "electron density wave" in your wire would be "turning around" before it got very far.
The electromagnetic spectrum is divided up not so much by "wavelength and frequency" as by the way that any given part of the spectrum interacts with matter. So, radio waves will interact via electron currents in long metal wires. But visible light interacts more with lattice vibrations and non-ionizing atomic transitions. So, "current in a wire" type emission works in frequency up to a thing called "the terahertz gap" https://en.wikipedia.org/wiki/Terahertz_gap . Above this frequency other emission techniques are usually required. Blue light is about three orders of magnitude higher in frequency than the terahertz gap.
$endgroup$
It would be hard to generate such a current and harder still to get it to produce any blue light - though this is theoretically possible.
The main problem is that you are probably thinking of a metal wire. Metals absorb visible light, both reflecting it and turning it into lattice vibrations. This is because the wavelength of visible light is just a few thousand atoms long in size so it is in a "sweet spot" for exciting solid crystals. In fact, the tendency for solid objects to absorb, reflect and otherwise interact with visible light is why it is "visible".
In a normal radio wave, your metal wire will need to be on the order of a wavelength of the radio wave you want to produce. This is typically on the order of meters. Automobiles of the 20th century had metal wires sticking out of them, about 1 meter long, called "antennas", to catch such waves.
But for blue light the wavelength is only about 5 x $10^{-7}$ meters so any useful antenna would be very tiny because an "electron density wave" in your wire would be "turning around" before it got very far.
The electromagnetic spectrum is divided up not so much by "wavelength and frequency" as by the way that any given part of the spectrum interacts with matter. So, radio waves will interact via electron currents in long metal wires. But visible light interacts more with lattice vibrations and non-ionizing atomic transitions. So, "current in a wire" type emission works in frequency up to a thing called "the terahertz gap" https://en.wikipedia.org/wiki/Terahertz_gap . Above this frequency other emission techniques are usually required. Blue light is about three orders of magnitude higher in frequency than the terahertz gap.
edited May 28 at 19:58
answered May 26 at 14:30
Paul YoungPaul Young
2,4408 silver badges27 bronze badges
2,4408 silver badges27 bronze badges
16
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For normal radio it is efficient to have a wire that's Lambda/2 but not necessary; "any old wire" will emit some radiation at RF. So it's not prima facie impossible that any old wire will glow faintly blue at the right frequency.
$endgroup$
– Peter A. Schneider
May 27 at 5:40
3
$begingroup$
@PeterA.Schneider I think your comment exactly hits the main point of this question that the other answers aren't really addressing.
$endgroup$
– KF Gauss
May 27 at 7:03
4
$begingroup$
@PeterA.Schneider It's also not prima facie impossible that you might quantum tunnel to the opposite side of the Earth. In practice, it's effectively impossible. Electrons with sufficient energy to conduct a 600THz wave are immediately absorbed by conductive metals. We know of no means to generate a controlled electrical (ie: moving electrons) signal at that frequency either, so even if you had an idealized radiating nano-antenna we still would not be able to supply it a driving voltage.
$endgroup$
– J...
May 27 at 19:03
4
$begingroup$
@PeterA.Schneider Fundamentally, the wire itself cannot co-exist with an electromagnetic signal in the 600THz range. By removing the electrons from the interfering atoms, you can start to think about moving electrons that quickly in systems like Free Electron Lasers, but in a bulk conductor this is simply not possible.
$endgroup$
– J...
May 27 at 19:08
2
$begingroup$
@uhoh - definitely liquid mercury reflects blue light and diamond is transparent ... however I am focused on the triumvirate of waves (radio, micro) vs. lights (IR,visible, UV), vs. rays (x and gamma) in the EM spectrum - it is only the "lights" which have the ideal wavelengths to do things like couple with phonons ... the OP is thinking about applying "waves" techniques to the "lights" - and I just want to make clear why they are different beasts - rays "see" individual atoms, waves "see" large conductive seas in metals, lights "see" molecules to macroscopic things
$endgroup$
– Paul Young
May 28 at 13:35
|
show 6 more comments
16
$begingroup$
For normal radio it is efficient to have a wire that's Lambda/2 but not necessary; "any old wire" will emit some radiation at RF. So it's not prima facie impossible that any old wire will glow faintly blue at the right frequency.
$endgroup$
– Peter A. Schneider
May 27 at 5:40
3
$begingroup$
@PeterA.Schneider I think your comment exactly hits the main point of this question that the other answers aren't really addressing.
$endgroup$
– KF Gauss
May 27 at 7:03
4
$begingroup$
@PeterA.Schneider It's also not prima facie impossible that you might quantum tunnel to the opposite side of the Earth. In practice, it's effectively impossible. Electrons with sufficient energy to conduct a 600THz wave are immediately absorbed by conductive metals. We know of no means to generate a controlled electrical (ie: moving electrons) signal at that frequency either, so even if you had an idealized radiating nano-antenna we still would not be able to supply it a driving voltage.
$endgroup$
– J...
May 27 at 19:03
4
$begingroup$
@PeterA.Schneider Fundamentally, the wire itself cannot co-exist with an electromagnetic signal in the 600THz range. By removing the electrons from the interfering atoms, you can start to think about moving electrons that quickly in systems like Free Electron Lasers, but in a bulk conductor this is simply not possible.
$endgroup$
– J...
May 27 at 19:08
2
$begingroup$
@uhoh - definitely liquid mercury reflects blue light and diamond is transparent ... however I am focused on the triumvirate of waves (radio, micro) vs. lights (IR,visible, UV), vs. rays (x and gamma) in the EM spectrum - it is only the "lights" which have the ideal wavelengths to do things like couple with phonons ... the OP is thinking about applying "waves" techniques to the "lights" - and I just want to make clear why they are different beasts - rays "see" individual atoms, waves "see" large conductive seas in metals, lights "see" molecules to macroscopic things
$endgroup$
– Paul Young
May 28 at 13:35
16
16
$begingroup$
For normal radio it is efficient to have a wire that's Lambda/2 but not necessary; "any old wire" will emit some radiation at RF. So it's not prima facie impossible that any old wire will glow faintly blue at the right frequency.
$endgroup$
– Peter A. Schneider
May 27 at 5:40
$begingroup$
For normal radio it is efficient to have a wire that's Lambda/2 but not necessary; "any old wire" will emit some radiation at RF. So it's not prima facie impossible that any old wire will glow faintly blue at the right frequency.
$endgroup$
– Peter A. Schneider
May 27 at 5:40
3
3
$begingroup$
@PeterA.Schneider I think your comment exactly hits the main point of this question that the other answers aren't really addressing.
$endgroup$
– KF Gauss
May 27 at 7:03
$begingroup$
@PeterA.Schneider I think your comment exactly hits the main point of this question that the other answers aren't really addressing.
$endgroup$
– KF Gauss
May 27 at 7:03
4
4
$begingroup$
@PeterA.Schneider It's also not prima facie impossible that you might quantum tunnel to the opposite side of the Earth. In practice, it's effectively impossible. Electrons with sufficient energy to conduct a 600THz wave are immediately absorbed by conductive metals. We know of no means to generate a controlled electrical (ie: moving electrons) signal at that frequency either, so even if you had an idealized radiating nano-antenna we still would not be able to supply it a driving voltage.
$endgroup$
– J...
May 27 at 19:03
$begingroup$
@PeterA.Schneider It's also not prima facie impossible that you might quantum tunnel to the opposite side of the Earth. In practice, it's effectively impossible. Electrons with sufficient energy to conduct a 600THz wave are immediately absorbed by conductive metals. We know of no means to generate a controlled electrical (ie: moving electrons) signal at that frequency either, so even if you had an idealized radiating nano-antenna we still would not be able to supply it a driving voltage.
$endgroup$
– J...
May 27 at 19:03
4
4
$begingroup$
@PeterA.Schneider Fundamentally, the wire itself cannot co-exist with an electromagnetic signal in the 600THz range. By removing the electrons from the interfering atoms, you can start to think about moving electrons that quickly in systems like Free Electron Lasers, but in a bulk conductor this is simply not possible.
$endgroup$
– J...
May 27 at 19:08
$begingroup$
@PeterA.Schneider Fundamentally, the wire itself cannot co-exist with an electromagnetic signal in the 600THz range. By removing the electrons from the interfering atoms, you can start to think about moving electrons that quickly in systems like Free Electron Lasers, but in a bulk conductor this is simply not possible.
$endgroup$
– J...
May 27 at 19:08
2
2
$begingroup$
@uhoh - definitely liquid mercury reflects blue light and diamond is transparent ... however I am focused on the triumvirate of waves (radio, micro) vs. lights (IR,visible, UV), vs. rays (x and gamma) in the EM spectrum - it is only the "lights" which have the ideal wavelengths to do things like couple with phonons ... the OP is thinking about applying "waves" techniques to the "lights" - and I just want to make clear why they are different beasts - rays "see" individual atoms, waves "see" large conductive seas in metals, lights "see" molecules to macroscopic things
$endgroup$
– Paul Young
May 28 at 13:35
$begingroup$
@uhoh - definitely liquid mercury reflects blue light and diamond is transparent ... however I am focused on the triumvirate of waves (radio, micro) vs. lights (IR,visible, UV), vs. rays (x and gamma) in the EM spectrum - it is only the "lights" which have the ideal wavelengths to do things like couple with phonons ... the OP is thinking about applying "waves" techniques to the "lights" - and I just want to make clear why they are different beasts - rays "see" individual atoms, waves "see" large conductive seas in metals, lights "see" molecules to macroscopic things
$endgroup$
– Paul Young
May 28 at 13:35
|
show 6 more comments
$begingroup$
An alternating voltage at that frequency is light. There's no 'generate' about it - the power supply is just a light source.
And if you have a wire, that is, a conductor made of metal, then the light won't propagate inside it at depths longer than the skin depth for that material at that particular frequency, which is generally tiny.
$endgroup$
12
$begingroup$
I don't object to this, but a periodic change in the electromagnetic field and a periodic fluctuation in electron density and momentum are, to me, distinct concepts. Of course, you are right because the two hybridize as a collective excitation, but that kinda makes it hard to understand at the high school level.
$endgroup$
– Paul Young
May 26 at 14:38
9
$begingroup$
@PaulYoung It is hard to justify when beginners are taught (incorrectly) that "electric current" the same as "electrons moving". But in any case, there is no law of physics that says "the laws of physics must be easy to justify!"
$endgroup$
– alephzero
May 26 at 15:00
15
$begingroup$
This is technically accurate but potentially misleading. Would an observer looking at the wire see the color blue? That is what the OP wants to know.
$endgroup$
– Owen
May 27 at 0:08
5
$begingroup$
@alephzero : The idea that electric current is "electrons moving" isn't completely "wrong", it's incomplete : electric current is charge moving, which may or may not be electrons. In this case , however, it is electrons, and thus the question of the adequacy or not of that "high school" definition is unimportant here. The distinction between a moving electric field and moving charges is very important since it is at the heart of the oft-posed question when first encountering drift velocity - "why does the light turn on 'instantly' when I flip the switch if electrons crawl?"
$endgroup$
– The_Sympathizer
May 27 at 2:54
9
$begingroup$
Does this actually answer the question? It seems that it does not. Would the wire emit light? If the light would only propagate a tiny distance within the material, does that mean that light beginning at a place within the material that is closer to the edge than the limit of propagation would escape the material and be seen? I think this is where confusion is coming and will continue to come from.
$endgroup$
– Clonkex
May 27 at 3:20
|
show 12 more comments
$begingroup$
An alternating voltage at that frequency is light. There's no 'generate' about it - the power supply is just a light source.
And if you have a wire, that is, a conductor made of metal, then the light won't propagate inside it at depths longer than the skin depth for that material at that particular frequency, which is generally tiny.
$endgroup$
12
$begingroup$
I don't object to this, but a periodic change in the electromagnetic field and a periodic fluctuation in electron density and momentum are, to me, distinct concepts. Of course, you are right because the two hybridize as a collective excitation, but that kinda makes it hard to understand at the high school level.
$endgroup$
– Paul Young
May 26 at 14:38
9
$begingroup$
@PaulYoung It is hard to justify when beginners are taught (incorrectly) that "electric current" the same as "electrons moving". But in any case, there is no law of physics that says "the laws of physics must be easy to justify!"
$endgroup$
– alephzero
May 26 at 15:00
15
$begingroup$
This is technically accurate but potentially misleading. Would an observer looking at the wire see the color blue? That is what the OP wants to know.
$endgroup$
– Owen
May 27 at 0:08
5
$begingroup$
@alephzero : The idea that electric current is "electrons moving" isn't completely "wrong", it's incomplete : electric current is charge moving, which may or may not be electrons. In this case , however, it is electrons, and thus the question of the adequacy or not of that "high school" definition is unimportant here. The distinction between a moving electric field and moving charges is very important since it is at the heart of the oft-posed question when first encountering drift velocity - "why does the light turn on 'instantly' when I flip the switch if electrons crawl?"
$endgroup$
– The_Sympathizer
May 27 at 2:54
9
$begingroup$
Does this actually answer the question? It seems that it does not. Would the wire emit light? If the light would only propagate a tiny distance within the material, does that mean that light beginning at a place within the material that is closer to the edge than the limit of propagation would escape the material and be seen? I think this is where confusion is coming and will continue to come from.
$endgroup$
– Clonkex
May 27 at 3:20
|
show 12 more comments
$begingroup$
An alternating voltage at that frequency is light. There's no 'generate' about it - the power supply is just a light source.
And if you have a wire, that is, a conductor made of metal, then the light won't propagate inside it at depths longer than the skin depth for that material at that particular frequency, which is generally tiny.
$endgroup$
An alternating voltage at that frequency is light. There's no 'generate' about it - the power supply is just a light source.
And if you have a wire, that is, a conductor made of metal, then the light won't propagate inside it at depths longer than the skin depth for that material at that particular frequency, which is generally tiny.
answered May 26 at 14:28
Emilio PisantyEmilio Pisanty
91.2k23 gold badges230 silver badges475 bronze badges
91.2k23 gold badges230 silver badges475 bronze badges
12
$begingroup$
I don't object to this, but a periodic change in the electromagnetic field and a periodic fluctuation in electron density and momentum are, to me, distinct concepts. Of course, you are right because the two hybridize as a collective excitation, but that kinda makes it hard to understand at the high school level.
$endgroup$
– Paul Young
May 26 at 14:38
9
$begingroup$
@PaulYoung It is hard to justify when beginners are taught (incorrectly) that "electric current" the same as "electrons moving". But in any case, there is no law of physics that says "the laws of physics must be easy to justify!"
$endgroup$
– alephzero
May 26 at 15:00
15
$begingroup$
This is technically accurate but potentially misleading. Would an observer looking at the wire see the color blue? That is what the OP wants to know.
$endgroup$
– Owen
May 27 at 0:08
5
$begingroup$
@alephzero : The idea that electric current is "electrons moving" isn't completely "wrong", it's incomplete : electric current is charge moving, which may or may not be electrons. In this case , however, it is electrons, and thus the question of the adequacy or not of that "high school" definition is unimportant here. The distinction between a moving electric field and moving charges is very important since it is at the heart of the oft-posed question when first encountering drift velocity - "why does the light turn on 'instantly' when I flip the switch if electrons crawl?"
$endgroup$
– The_Sympathizer
May 27 at 2:54
9
$begingroup$
Does this actually answer the question? It seems that it does not. Would the wire emit light? If the light would only propagate a tiny distance within the material, does that mean that light beginning at a place within the material that is closer to the edge than the limit of propagation would escape the material and be seen? I think this is where confusion is coming and will continue to come from.
$endgroup$
– Clonkex
May 27 at 3:20
|
show 12 more comments
12
$begingroup$
I don't object to this, but a periodic change in the electromagnetic field and a periodic fluctuation in electron density and momentum are, to me, distinct concepts. Of course, you are right because the two hybridize as a collective excitation, but that kinda makes it hard to understand at the high school level.
$endgroup$
– Paul Young
May 26 at 14:38
9
$begingroup$
@PaulYoung It is hard to justify when beginners are taught (incorrectly) that "electric current" the same as "electrons moving". But in any case, there is no law of physics that says "the laws of physics must be easy to justify!"
$endgroup$
– alephzero
May 26 at 15:00
15
$begingroup$
This is technically accurate but potentially misleading. Would an observer looking at the wire see the color blue? That is what the OP wants to know.
$endgroup$
– Owen
May 27 at 0:08
5
$begingroup$
@alephzero : The idea that electric current is "electrons moving" isn't completely "wrong", it's incomplete : electric current is charge moving, which may or may not be electrons. In this case , however, it is electrons, and thus the question of the adequacy or not of that "high school" definition is unimportant here. The distinction between a moving electric field and moving charges is very important since it is at the heart of the oft-posed question when first encountering drift velocity - "why does the light turn on 'instantly' when I flip the switch if electrons crawl?"
$endgroup$
– The_Sympathizer
May 27 at 2:54
9
$begingroup$
Does this actually answer the question? It seems that it does not. Would the wire emit light? If the light would only propagate a tiny distance within the material, does that mean that light beginning at a place within the material that is closer to the edge than the limit of propagation would escape the material and be seen? I think this is where confusion is coming and will continue to come from.
$endgroup$
– Clonkex
May 27 at 3:20
12
12
$begingroup$
I don't object to this, but a periodic change in the electromagnetic field and a periodic fluctuation in electron density and momentum are, to me, distinct concepts. Of course, you are right because the two hybridize as a collective excitation, but that kinda makes it hard to understand at the high school level.
$endgroup$
– Paul Young
May 26 at 14:38
$begingroup$
I don't object to this, but a periodic change in the electromagnetic field and a periodic fluctuation in electron density and momentum are, to me, distinct concepts. Of course, you are right because the two hybridize as a collective excitation, but that kinda makes it hard to understand at the high school level.
$endgroup$
– Paul Young
May 26 at 14:38
9
9
$begingroup$
@PaulYoung It is hard to justify when beginners are taught (incorrectly) that "electric current" the same as "electrons moving". But in any case, there is no law of physics that says "the laws of physics must be easy to justify!"
$endgroup$
– alephzero
May 26 at 15:00
$begingroup$
@PaulYoung It is hard to justify when beginners are taught (incorrectly) that "electric current" the same as "electrons moving". But in any case, there is no law of physics that says "the laws of physics must be easy to justify!"
$endgroup$
– alephzero
May 26 at 15:00
15
15
$begingroup$
This is technically accurate but potentially misleading. Would an observer looking at the wire see the color blue? That is what the OP wants to know.
$endgroup$
– Owen
May 27 at 0:08
$begingroup$
This is technically accurate but potentially misleading. Would an observer looking at the wire see the color blue? That is what the OP wants to know.
$endgroup$
– Owen
May 27 at 0:08
5
5
$begingroup$
@alephzero : The idea that electric current is "electrons moving" isn't completely "wrong", it's incomplete : electric current is charge moving, which may or may not be electrons. In this case , however, it is electrons, and thus the question of the adequacy or not of that "high school" definition is unimportant here. The distinction between a moving electric field and moving charges is very important since it is at the heart of the oft-posed question when first encountering drift velocity - "why does the light turn on 'instantly' when I flip the switch if electrons crawl?"
$endgroup$
– The_Sympathizer
May 27 at 2:54
$begingroup$
@alephzero : The idea that electric current is "electrons moving" isn't completely "wrong", it's incomplete : electric current is charge moving, which may or may not be electrons. In this case , however, it is electrons, and thus the question of the adequacy or not of that "high school" definition is unimportant here. The distinction between a moving electric field and moving charges is very important since it is at the heart of the oft-posed question when first encountering drift velocity - "why does the light turn on 'instantly' when I flip the switch if electrons crawl?"
$endgroup$
– The_Sympathizer
May 27 at 2:54
9
9
$begingroup$
Does this actually answer the question? It seems that it does not. Would the wire emit light? If the light would only propagate a tiny distance within the material, does that mean that light beginning at a place within the material that is closer to the edge than the limit of propagation would escape the material and be seen? I think this is where confusion is coming and will continue to come from.
$endgroup$
– Clonkex
May 27 at 3:20
$begingroup$
Does this actually answer the question? It seems that it does not. Would the wire emit light? If the light would only propagate a tiny distance within the material, does that mean that light beginning at a place within the material that is closer to the edge than the limit of propagation would escape the material and be seen? I think this is where confusion is coming and will continue to come from.
$endgroup$
– Clonkex
May 27 at 3:20
|
show 12 more comments
$begingroup$
I second this Emilio Pisanty's point: the power supply you are
envisioning is a light source. Now the question that remains is: can
you propagate this light through a wire, just like you would do with a
regular low-frequency electric signal?
To get a hint of the answer, look at how people use wires to transport
high frequency signals, into the many MHz up to the multi-GHz range. A
single wire doesn't work, because it has a tendency to radiate all the
power you feed it into the air as free electromagnetic waves. The trick
is to use two wires carrying opposite currents. You can think of them
as one being the signal and the other being the return wire, but their
roles could be symmetric. If you keep them close enough, most of the
electromagnetic field will be confined between them, and you will be
able to transmit the power without too much loses. You can further
reduce the losses by twisting the wires together. At the highest
frequencies, you would get best results by putting one wire inside the
other which, shaped like a tube, functions like a shield. This is called
a coaxial cable, and some of them are good up to tens of GHz.
The think that is not so intuitive is that, while the metal wires carry
the current, the actual power is carried by the electromagnetic field
that propagates between the wires. So the main role of the metal wires
is thus to guide the electromagnetic waves and, for this reason, the
high-frequency cables are considered to be waveguides.
Could you adapt this waveguide technique to the propagation of light?
The answer is yes, some people have indeed built nanosized coaxial
cables for this very purpose.
$endgroup$
4
$begingroup$
An optical fiber is exactly a waveguide for visible light.
$endgroup$
– Henning Makholm
May 27 at 11:11
3
$begingroup$
@HenningMakholm: Indeed, but it's usually made of non-conducting material. The nano-coax I linked to is then a better analogous to the typical metal wires the OP seems to have in mind.
$endgroup$
– Edgar Bonet
May 27 at 12:14
$begingroup$
This is an interesting paper, I would be curious to know how they couple into and out of that nano-coax. While it may use conductive components in its construction, I can't help but feel that it effectively acts in much the same way as a normal optical fiber. I don't believe we have any means of exciting an electrical signal at ~600THz into such a cable. Other than shining a blue laser at it and coupling it in with optical techniques, how could you put energy into that nano-coax cable?
$endgroup$
– J...
May 28 at 15:05
1
$begingroup$
@J.. As a thought experiment, imagine taking any standard RF oscillator, and scaling it way down so it's a few tens of nanomatres across and oscillates in the frequency range of blue light. Now you connect the output to your scaled down coax. At the other end of the tiny coax you make a shielded dipole antenna and look at it. Is it glowing blue?
$endgroup$
– immibis
May 28 at 22:15
2
$begingroup$
@immibis It's not my imagination that's lacking - my intuition tells me that physics will protest before your scaling experiment gets to the required size. If you've read a paper where this has actually been done, or at least has been even theorized to be possible, I'd be interested to read it.
$endgroup$
– J...
May 28 at 22:18
|
show 2 more comments
$begingroup$
I second this Emilio Pisanty's point: the power supply you are
envisioning is a light source. Now the question that remains is: can
you propagate this light through a wire, just like you would do with a
regular low-frequency electric signal?
To get a hint of the answer, look at how people use wires to transport
high frequency signals, into the many MHz up to the multi-GHz range. A
single wire doesn't work, because it has a tendency to radiate all the
power you feed it into the air as free electromagnetic waves. The trick
is to use two wires carrying opposite currents. You can think of them
as one being the signal and the other being the return wire, but their
roles could be symmetric. If you keep them close enough, most of the
electromagnetic field will be confined between them, and you will be
able to transmit the power without too much loses. You can further
reduce the losses by twisting the wires together. At the highest
frequencies, you would get best results by putting one wire inside the
other which, shaped like a tube, functions like a shield. This is called
a coaxial cable, and some of them are good up to tens of GHz.
The think that is not so intuitive is that, while the metal wires carry
the current, the actual power is carried by the electromagnetic field
that propagates between the wires. So the main role of the metal wires
is thus to guide the electromagnetic waves and, for this reason, the
high-frequency cables are considered to be waveguides.
Could you adapt this waveguide technique to the propagation of light?
The answer is yes, some people have indeed built nanosized coaxial
cables for this very purpose.
$endgroup$
4
$begingroup$
An optical fiber is exactly a waveguide for visible light.
$endgroup$
– Henning Makholm
May 27 at 11:11
3
$begingroup$
@HenningMakholm: Indeed, but it's usually made of non-conducting material. The nano-coax I linked to is then a better analogous to the typical metal wires the OP seems to have in mind.
$endgroup$
– Edgar Bonet
May 27 at 12:14
$begingroup$
This is an interesting paper, I would be curious to know how they couple into and out of that nano-coax. While it may use conductive components in its construction, I can't help but feel that it effectively acts in much the same way as a normal optical fiber. I don't believe we have any means of exciting an electrical signal at ~600THz into such a cable. Other than shining a blue laser at it and coupling it in with optical techniques, how could you put energy into that nano-coax cable?
$endgroup$
– J...
May 28 at 15:05
1
$begingroup$
@J.. As a thought experiment, imagine taking any standard RF oscillator, and scaling it way down so it's a few tens of nanomatres across and oscillates in the frequency range of blue light. Now you connect the output to your scaled down coax. At the other end of the tiny coax you make a shielded dipole antenna and look at it. Is it glowing blue?
$endgroup$
– immibis
May 28 at 22:15
2
$begingroup$
@immibis It's not my imagination that's lacking - my intuition tells me that physics will protest before your scaling experiment gets to the required size. If you've read a paper where this has actually been done, or at least has been even theorized to be possible, I'd be interested to read it.
$endgroup$
– J...
May 28 at 22:18
|
show 2 more comments
$begingroup$
I second this Emilio Pisanty's point: the power supply you are
envisioning is a light source. Now the question that remains is: can
you propagate this light through a wire, just like you would do with a
regular low-frequency electric signal?
To get a hint of the answer, look at how people use wires to transport
high frequency signals, into the many MHz up to the multi-GHz range. A
single wire doesn't work, because it has a tendency to radiate all the
power you feed it into the air as free electromagnetic waves. The trick
is to use two wires carrying opposite currents. You can think of them
as one being the signal and the other being the return wire, but their
roles could be symmetric. If you keep them close enough, most of the
electromagnetic field will be confined between them, and you will be
able to transmit the power without too much loses. You can further
reduce the losses by twisting the wires together. At the highest
frequencies, you would get best results by putting one wire inside the
other which, shaped like a tube, functions like a shield. This is called
a coaxial cable, and some of them are good up to tens of GHz.
The think that is not so intuitive is that, while the metal wires carry
the current, the actual power is carried by the electromagnetic field
that propagates between the wires. So the main role of the metal wires
is thus to guide the electromagnetic waves and, for this reason, the
high-frequency cables are considered to be waveguides.
Could you adapt this waveguide technique to the propagation of light?
The answer is yes, some people have indeed built nanosized coaxial
cables for this very purpose.
$endgroup$
I second this Emilio Pisanty's point: the power supply you are
envisioning is a light source. Now the question that remains is: can
you propagate this light through a wire, just like you would do with a
regular low-frequency electric signal?
To get a hint of the answer, look at how people use wires to transport
high frequency signals, into the many MHz up to the multi-GHz range. A
single wire doesn't work, because it has a tendency to radiate all the
power you feed it into the air as free electromagnetic waves. The trick
is to use two wires carrying opposite currents. You can think of them
as one being the signal and the other being the return wire, but their
roles could be symmetric. If you keep them close enough, most of the
electromagnetic field will be confined between them, and you will be
able to transmit the power without too much loses. You can further
reduce the losses by twisting the wires together. At the highest
frequencies, you would get best results by putting one wire inside the
other which, shaped like a tube, functions like a shield. This is called
a coaxial cable, and some of them are good up to tens of GHz.
The think that is not so intuitive is that, while the metal wires carry
the current, the actual power is carried by the electromagnetic field
that propagates between the wires. So the main role of the metal wires
is thus to guide the electromagnetic waves and, for this reason, the
high-frequency cables are considered to be waveguides.
Could you adapt this waveguide technique to the propagation of light?
The answer is yes, some people have indeed built nanosized coaxial
cables for this very purpose.
answered May 27 at 8:28
Edgar BonetEdgar Bonet
2,0511 gold badge11 silver badges12 bronze badges
2,0511 gold badge11 silver badges12 bronze badges
4
$begingroup$
An optical fiber is exactly a waveguide for visible light.
$endgroup$
– Henning Makholm
May 27 at 11:11
3
$begingroup$
@HenningMakholm: Indeed, but it's usually made of non-conducting material. The nano-coax I linked to is then a better analogous to the typical metal wires the OP seems to have in mind.
$endgroup$
– Edgar Bonet
May 27 at 12:14
$begingroup$
This is an interesting paper, I would be curious to know how they couple into and out of that nano-coax. While it may use conductive components in its construction, I can't help but feel that it effectively acts in much the same way as a normal optical fiber. I don't believe we have any means of exciting an electrical signal at ~600THz into such a cable. Other than shining a blue laser at it and coupling it in with optical techniques, how could you put energy into that nano-coax cable?
$endgroup$
– J...
May 28 at 15:05
1
$begingroup$
@J.. As a thought experiment, imagine taking any standard RF oscillator, and scaling it way down so it's a few tens of nanomatres across and oscillates in the frequency range of blue light. Now you connect the output to your scaled down coax. At the other end of the tiny coax you make a shielded dipole antenna and look at it. Is it glowing blue?
$endgroup$
– immibis
May 28 at 22:15
2
$begingroup$
@immibis It's not my imagination that's lacking - my intuition tells me that physics will protest before your scaling experiment gets to the required size. If you've read a paper where this has actually been done, or at least has been even theorized to be possible, I'd be interested to read it.
$endgroup$
– J...
May 28 at 22:18
|
show 2 more comments
4
$begingroup$
An optical fiber is exactly a waveguide for visible light.
$endgroup$
– Henning Makholm
May 27 at 11:11
3
$begingroup$
@HenningMakholm: Indeed, but it's usually made of non-conducting material. The nano-coax I linked to is then a better analogous to the typical metal wires the OP seems to have in mind.
$endgroup$
– Edgar Bonet
May 27 at 12:14
$begingroup$
This is an interesting paper, I would be curious to know how they couple into and out of that nano-coax. While it may use conductive components in its construction, I can't help but feel that it effectively acts in much the same way as a normal optical fiber. I don't believe we have any means of exciting an electrical signal at ~600THz into such a cable. Other than shining a blue laser at it and coupling it in with optical techniques, how could you put energy into that nano-coax cable?
$endgroup$
– J...
May 28 at 15:05
1
$begingroup$
@J.. As a thought experiment, imagine taking any standard RF oscillator, and scaling it way down so it's a few tens of nanomatres across and oscillates in the frequency range of blue light. Now you connect the output to your scaled down coax. At the other end of the tiny coax you make a shielded dipole antenna and look at it. Is it glowing blue?
$endgroup$
– immibis
May 28 at 22:15
2
$begingroup$
@immibis It's not my imagination that's lacking - my intuition tells me that physics will protest before your scaling experiment gets to the required size. If you've read a paper where this has actually been done, or at least has been even theorized to be possible, I'd be interested to read it.
$endgroup$
– J...
May 28 at 22:18
4
4
$begingroup$
An optical fiber is exactly a waveguide for visible light.
$endgroup$
– Henning Makholm
May 27 at 11:11
$begingroup$
An optical fiber is exactly a waveguide for visible light.
$endgroup$
– Henning Makholm
May 27 at 11:11
3
3
$begingroup$
@HenningMakholm: Indeed, but it's usually made of non-conducting material. The nano-coax I linked to is then a better analogous to the typical metal wires the OP seems to have in mind.
$endgroup$
– Edgar Bonet
May 27 at 12:14
$begingroup$
@HenningMakholm: Indeed, but it's usually made of non-conducting material. The nano-coax I linked to is then a better analogous to the typical metal wires the OP seems to have in mind.
$endgroup$
– Edgar Bonet
May 27 at 12:14
$begingroup$
This is an interesting paper, I would be curious to know how they couple into and out of that nano-coax. While it may use conductive components in its construction, I can't help but feel that it effectively acts in much the same way as a normal optical fiber. I don't believe we have any means of exciting an electrical signal at ~600THz into such a cable. Other than shining a blue laser at it and coupling it in with optical techniques, how could you put energy into that nano-coax cable?
$endgroup$
– J...
May 28 at 15:05
$begingroup$
This is an interesting paper, I would be curious to know how they couple into and out of that nano-coax. While it may use conductive components in its construction, I can't help but feel that it effectively acts in much the same way as a normal optical fiber. I don't believe we have any means of exciting an electrical signal at ~600THz into such a cable. Other than shining a blue laser at it and coupling it in with optical techniques, how could you put energy into that nano-coax cable?
$endgroup$
– J...
May 28 at 15:05
1
1
$begingroup$
@J.. As a thought experiment, imagine taking any standard RF oscillator, and scaling it way down so it's a few tens of nanomatres across and oscillates in the frequency range of blue light. Now you connect the output to your scaled down coax. At the other end of the tiny coax you make a shielded dipole antenna and look at it. Is it glowing blue?
$endgroup$
– immibis
May 28 at 22:15
$begingroup$
@J.. As a thought experiment, imagine taking any standard RF oscillator, and scaling it way down so it's a few tens of nanomatres across and oscillates in the frequency range of blue light. Now you connect the output to your scaled down coax. At the other end of the tiny coax you make a shielded dipole antenna and look at it. Is it glowing blue?
$endgroup$
– immibis
May 28 at 22:15
2
2
$begingroup$
@immibis It's not my imagination that's lacking - my intuition tells me that physics will protest before your scaling experiment gets to the required size. If you've read a paper where this has actually been done, or at least has been even theorized to be possible, I'd be interested to read it.
$endgroup$
– J...
May 28 at 22:18
$begingroup$
@immibis It's not my imagination that's lacking - my intuition tells me that physics will protest before your scaling experiment gets to the required size. If you've read a paper where this has actually been done, or at least has been even theorized to be possible, I'd be interested to read it.
$endgroup$
– J...
May 28 at 22:18
|
show 2 more comments
$begingroup$
We do have a power supply which can generate current oscillations at optical frequencies: light. It won't transmit any distance along a wire, but if your "current source" and antenna are the same object, you have what is called an optical antenna, and the study of these is an active field of research. I don't know if any of them have any meaningful efficiency down at blue light frequencies, but they do work in the green which is not too far off. See, for example, this review article.
$endgroup$
add a comment
|
$begingroup$
We do have a power supply which can generate current oscillations at optical frequencies: light. It won't transmit any distance along a wire, but if your "current source" and antenna are the same object, you have what is called an optical antenna, and the study of these is an active field of research. I don't know if any of them have any meaningful efficiency down at blue light frequencies, but they do work in the green which is not too far off. See, for example, this review article.
$endgroup$
add a comment
|
$begingroup$
We do have a power supply which can generate current oscillations at optical frequencies: light. It won't transmit any distance along a wire, but if your "current source" and antenna are the same object, you have what is called an optical antenna, and the study of these is an active field of research. I don't know if any of them have any meaningful efficiency down at blue light frequencies, but they do work in the green which is not too far off. See, for example, this review article.
$endgroup$
We do have a power supply which can generate current oscillations at optical frequencies: light. It won't transmit any distance along a wire, but if your "current source" and antenna are the same object, you have what is called an optical antenna, and the study of these is an active field of research. I don't know if any of them have any meaningful efficiency down at blue light frequencies, but they do work in the green which is not too far off. See, for example, this review article.
answered May 27 at 19:04
llamallama
4112 silver badges6 bronze badges
4112 silver badges6 bronze badges
add a comment
|
add a comment
|
$begingroup$
Even at tens of gigaherts, one does not carry current "in" a conductor -- it is carried along the outside (Google "skin effect").
There are transmission lines for high-frequency RF that basically launch an RF wave along a single naked wire, and catch it at the other end -- think of a coax without the outer shield. If you take this analogy and pursue it into absurdity and beyond, then if you take a really well polished wire, and really carefully launch blue light along its length, then as long as the wire doesn't bend too suddenly, the light -- or some portion of it -- will be refracted and "stick*" to the wire.
I think you could achieve a setup in a lab that involved people looking at a blue-glowing end of a carefully-maintained copper wire or gold wire and going "oooh!". I doubt there is much potential for practical use here.
* Imprecise language used on purpose -- I'd have to do a lot of work to do the math on this one!
$endgroup$
$begingroup$
"Imprecise language used on purpose" - indeed ... I am waiting for someone to say "frequency dependent permittivity", "maxwell's equations", "boundary conditions at the interface" and "quantum effects" ... the OP's question seems to be at the gnarly coal-face where electrical engineers meet physicists and can't talk to one another
$endgroup$
– Paul Young
May 30 at 15:39
add a comment
|
$begingroup$
Even at tens of gigaherts, one does not carry current "in" a conductor -- it is carried along the outside (Google "skin effect").
There are transmission lines for high-frequency RF that basically launch an RF wave along a single naked wire, and catch it at the other end -- think of a coax without the outer shield. If you take this analogy and pursue it into absurdity and beyond, then if you take a really well polished wire, and really carefully launch blue light along its length, then as long as the wire doesn't bend too suddenly, the light -- or some portion of it -- will be refracted and "stick*" to the wire.
I think you could achieve a setup in a lab that involved people looking at a blue-glowing end of a carefully-maintained copper wire or gold wire and going "oooh!". I doubt there is much potential for practical use here.
* Imprecise language used on purpose -- I'd have to do a lot of work to do the math on this one!
$endgroup$
$begingroup$
"Imprecise language used on purpose" - indeed ... I am waiting for someone to say "frequency dependent permittivity", "maxwell's equations", "boundary conditions at the interface" and "quantum effects" ... the OP's question seems to be at the gnarly coal-face where electrical engineers meet physicists and can't talk to one another
$endgroup$
– Paul Young
May 30 at 15:39
add a comment
|
$begingroup$
Even at tens of gigaherts, one does not carry current "in" a conductor -- it is carried along the outside (Google "skin effect").
There are transmission lines for high-frequency RF that basically launch an RF wave along a single naked wire, and catch it at the other end -- think of a coax without the outer shield. If you take this analogy and pursue it into absurdity and beyond, then if you take a really well polished wire, and really carefully launch blue light along its length, then as long as the wire doesn't bend too suddenly, the light -- or some portion of it -- will be refracted and "stick*" to the wire.
I think you could achieve a setup in a lab that involved people looking at a blue-glowing end of a carefully-maintained copper wire or gold wire and going "oooh!". I doubt there is much potential for practical use here.
* Imprecise language used on purpose -- I'd have to do a lot of work to do the math on this one!
$endgroup$
Even at tens of gigaherts, one does not carry current "in" a conductor -- it is carried along the outside (Google "skin effect").
There are transmission lines for high-frequency RF that basically launch an RF wave along a single naked wire, and catch it at the other end -- think of a coax without the outer shield. If you take this analogy and pursue it into absurdity and beyond, then if you take a really well polished wire, and really carefully launch blue light along its length, then as long as the wire doesn't bend too suddenly, the light -- or some portion of it -- will be refracted and "stick*" to the wire.
I think you could achieve a setup in a lab that involved people looking at a blue-glowing end of a carefully-maintained copper wire or gold wire and going "oooh!". I doubt there is much potential for practical use here.
* Imprecise language used on purpose -- I'd have to do a lot of work to do the math on this one!
answered May 29 at 20:32
TimWescottTimWescott
2264 bronze badges
2264 bronze badges
$begingroup$
"Imprecise language used on purpose" - indeed ... I am waiting for someone to say "frequency dependent permittivity", "maxwell's equations", "boundary conditions at the interface" and "quantum effects" ... the OP's question seems to be at the gnarly coal-face where electrical engineers meet physicists and can't talk to one another
$endgroup$
– Paul Young
May 30 at 15:39
add a comment
|
$begingroup$
"Imprecise language used on purpose" - indeed ... I am waiting for someone to say "frequency dependent permittivity", "maxwell's equations", "boundary conditions at the interface" and "quantum effects" ... the OP's question seems to be at the gnarly coal-face where electrical engineers meet physicists and can't talk to one another
$endgroup$
– Paul Young
May 30 at 15:39
$begingroup$
"Imprecise language used on purpose" - indeed ... I am waiting for someone to say "frequency dependent permittivity", "maxwell's equations", "boundary conditions at the interface" and "quantum effects" ... the OP's question seems to be at the gnarly coal-face where electrical engineers meet physicists and can't talk to one another
$endgroup$
– Paul Young
May 30 at 15:39
$begingroup$
"Imprecise language used on purpose" - indeed ... I am waiting for someone to say "frequency dependent permittivity", "maxwell's equations", "boundary conditions at the interface" and "quantum effects" ... the OP's question seems to be at the gnarly coal-face where electrical engineers meet physicists and can't talk to one another
$endgroup$
– Paul Young
May 30 at 15:39
add a comment
|
$begingroup$
Yeah definitely . You can create light corresponding on any frequency by this method. It’s just that creating this circuit will be very challenging. To give you a perspective, the highest frequency that we have obtained with modern electronic circuits is around 10^11 Hz.
$endgroup$
$begingroup$
a) Source? b) If we can create 10^11 Hz, how does that relate to tetrahertz? Is it more or less?
$endgroup$
– Clonkex
May 27 at 3:13
$begingroup$
@Clonkex kilo, mega, giga, tera? Steps of 3.
$endgroup$
– Peter A. Schneider
May 27 at 5:37
5
$begingroup$
@PeterA.Schneider I mean, I know how to read numbers in scientific notation, and I know what tera means (ugh, I wrote tetra by accident in my last comment), but it's not from immediately obvious whether 10^11 is smaller or larger than the required 610-670 THz, given I commonly work directly with such large numbers. I'm sure others will know without any effort whatsoever that 1 THz is 10^12 Hz, but I had to sit and think about it.
$endgroup$
– Clonkex
May 27 at 7:00
5
$begingroup$
@Ishan I guess the easiest way to create such an oscillation in the metal electrons is ... to shine a blue light at them. And lo and behold, the metal will shine blue right back ;-). As Emilio so insightfully observed, the oscillations won't penerate the metal very deeply, it's kind of a surface effect.
$endgroup$
– Peter A. Schneider
May 27 at 10:55
1
$begingroup$
At a few nanometres, you'd be hand soldering molecules.
$endgroup$
– rackandboneman
May 28 at 23:19
|
show 1 more comment
$begingroup$
Yeah definitely . You can create light corresponding on any frequency by this method. It’s just that creating this circuit will be very challenging. To give you a perspective, the highest frequency that we have obtained with modern electronic circuits is around 10^11 Hz.
$endgroup$
$begingroup$
a) Source? b) If we can create 10^11 Hz, how does that relate to tetrahertz? Is it more or less?
$endgroup$
– Clonkex
May 27 at 3:13
$begingroup$
@Clonkex kilo, mega, giga, tera? Steps of 3.
$endgroup$
– Peter A. Schneider
May 27 at 5:37
5
$begingroup$
@PeterA.Schneider I mean, I know how to read numbers in scientific notation, and I know what tera means (ugh, I wrote tetra by accident in my last comment), but it's not from immediately obvious whether 10^11 is smaller or larger than the required 610-670 THz, given I commonly work directly with such large numbers. I'm sure others will know without any effort whatsoever that 1 THz is 10^12 Hz, but I had to sit and think about it.
$endgroup$
– Clonkex
May 27 at 7:00
5
$begingroup$
@Ishan I guess the easiest way to create such an oscillation in the metal electrons is ... to shine a blue light at them. And lo and behold, the metal will shine blue right back ;-). As Emilio so insightfully observed, the oscillations won't penerate the metal very deeply, it's kind of a surface effect.
$endgroup$
– Peter A. Schneider
May 27 at 10:55
1
$begingroup$
At a few nanometres, you'd be hand soldering molecules.
$endgroup$
– rackandboneman
May 28 at 23:19
|
show 1 more comment
$begingroup$
Yeah definitely . You can create light corresponding on any frequency by this method. It’s just that creating this circuit will be very challenging. To give you a perspective, the highest frequency that we have obtained with modern electronic circuits is around 10^11 Hz.
$endgroup$
Yeah definitely . You can create light corresponding on any frequency by this method. It’s just that creating this circuit will be very challenging. To give you a perspective, the highest frequency that we have obtained with modern electronic circuits is around 10^11 Hz.
answered May 26 at 14:06
Ishan JawaleIshan Jawale
598 bronze badges
598 bronze badges
$begingroup$
a) Source? b) If we can create 10^11 Hz, how does that relate to tetrahertz? Is it more or less?
$endgroup$
– Clonkex
May 27 at 3:13
$begingroup$
@Clonkex kilo, mega, giga, tera? Steps of 3.
$endgroup$
– Peter A. Schneider
May 27 at 5:37
5
$begingroup$
@PeterA.Schneider I mean, I know how to read numbers in scientific notation, and I know what tera means (ugh, I wrote tetra by accident in my last comment), but it's not from immediately obvious whether 10^11 is smaller or larger than the required 610-670 THz, given I commonly work directly with such large numbers. I'm sure others will know without any effort whatsoever that 1 THz is 10^12 Hz, but I had to sit and think about it.
$endgroup$
– Clonkex
May 27 at 7:00
5
$begingroup$
@Ishan I guess the easiest way to create such an oscillation in the metal electrons is ... to shine a blue light at them. And lo and behold, the metal will shine blue right back ;-). As Emilio so insightfully observed, the oscillations won't penerate the metal very deeply, it's kind of a surface effect.
$endgroup$
– Peter A. Schneider
May 27 at 10:55
1
$begingroup$
At a few nanometres, you'd be hand soldering molecules.
$endgroup$
– rackandboneman
May 28 at 23:19
|
show 1 more comment
$begingroup$
a) Source? b) If we can create 10^11 Hz, how does that relate to tetrahertz? Is it more or less?
$endgroup$
– Clonkex
May 27 at 3:13
$begingroup$
@Clonkex kilo, mega, giga, tera? Steps of 3.
$endgroup$
– Peter A. Schneider
May 27 at 5:37
5
$begingroup$
@PeterA.Schneider I mean, I know how to read numbers in scientific notation, and I know what tera means (ugh, I wrote tetra by accident in my last comment), but it's not from immediately obvious whether 10^11 is smaller or larger than the required 610-670 THz, given I commonly work directly with such large numbers. I'm sure others will know without any effort whatsoever that 1 THz is 10^12 Hz, but I had to sit and think about it.
$endgroup$
– Clonkex
May 27 at 7:00
5
$begingroup$
@Ishan I guess the easiest way to create such an oscillation in the metal electrons is ... to shine a blue light at them. And lo and behold, the metal will shine blue right back ;-). As Emilio so insightfully observed, the oscillations won't penerate the metal very deeply, it's kind of a surface effect.
$endgroup$
– Peter A. Schneider
May 27 at 10:55
1
$begingroup$
At a few nanometres, you'd be hand soldering molecules.
$endgroup$
– rackandboneman
May 28 at 23:19
$begingroup$
a) Source? b) If we can create 10^11 Hz, how does that relate to tetrahertz? Is it more or less?
$endgroup$
– Clonkex
May 27 at 3:13
$begingroup$
a) Source? b) If we can create 10^11 Hz, how does that relate to tetrahertz? Is it more or less?
$endgroup$
– Clonkex
May 27 at 3:13
$begingroup$
@Clonkex kilo, mega, giga, tera? Steps of 3.
$endgroup$
– Peter A. Schneider
May 27 at 5:37
$begingroup$
@Clonkex kilo, mega, giga, tera? Steps of 3.
$endgroup$
– Peter A. Schneider
May 27 at 5:37
5
5
$begingroup$
@PeterA.Schneider I mean, I know how to read numbers in scientific notation, and I know what tera means (ugh, I wrote tetra by accident in my last comment), but it's not from immediately obvious whether 10^11 is smaller or larger than the required 610-670 THz, given I commonly work directly with such large numbers. I'm sure others will know without any effort whatsoever that 1 THz is 10^12 Hz, but I had to sit and think about it.
$endgroup$
– Clonkex
May 27 at 7:00
$begingroup$
@PeterA.Schneider I mean, I know how to read numbers in scientific notation, and I know what tera means (ugh, I wrote tetra by accident in my last comment), but it's not from immediately obvious whether 10^11 is smaller or larger than the required 610-670 THz, given I commonly work directly with such large numbers. I'm sure others will know without any effort whatsoever that 1 THz is 10^12 Hz, but I had to sit and think about it.
$endgroup$
– Clonkex
May 27 at 7:00
5
5
$begingroup$
@Ishan I guess the easiest way to create such an oscillation in the metal electrons is ... to shine a blue light at them. And lo and behold, the metal will shine blue right back ;-). As Emilio so insightfully observed, the oscillations won't penerate the metal very deeply, it's kind of a surface effect.
$endgroup$
– Peter A. Schneider
May 27 at 10:55
$begingroup$
@Ishan I guess the easiest way to create such an oscillation in the metal electrons is ... to shine a blue light at them. And lo and behold, the metal will shine blue right back ;-). As Emilio so insightfully observed, the oscillations won't penerate the metal very deeply, it's kind of a surface effect.
$endgroup$
– Peter A. Schneider
May 27 at 10:55
1
1
$begingroup$
At a few nanometres, you'd be hand soldering molecules.
$endgroup$
– rackandboneman
May 28 at 23:19
$begingroup$
At a few nanometres, you'd be hand soldering molecules.
$endgroup$
– rackandboneman
May 28 at 23:19
|
show 1 more comment
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$begingroup$
You could shine light (which is supply) on the wire, and the wire reflects this light (technically emmiting). This is not meant in jest, but pointing out you have to couple the signal into your antenna, and we know you cannot use wires for that. So you would have to coupöe probably optically.
$endgroup$
– lalala
May 27 at 19:15