alpha2cen Posted January 27, 2011 Share Posted January 27, 2011 Theoritically we know well light is electromagnetic wave. We know light wave fellows Maxwell equations . How do we know light is electromagnetic wave by experiment? Light dose not interact with magnetic fields. Link to comment Share on other sites More sharing options...
ajb Posted January 27, 2011 Share Posted January 27, 2011 Historically the suggestion that light is an electromagnetic wave goes back to Maxwell and Hertz when it was discovered that electromagnetic waves travel at the speed of light, which was by then well measured. Philosophically, it is unlikely that this is just some coincidence. The original paper linking light and electromagnetism is Maxwell [1]. It is available here as a pdf. I honestly have no idea what experimental proof was offered. Modern proof can be found in aerial theory and practice. Accelerating electrons produces radio waves, all in accordance with Maxwell. This goes back to the pioneering work of Hertz. References [1] Maxwell, James Clerk (1865). "A dynamical theory of the electromagnetic field" (PDF). Philosophical Transactions of the Royal Society of London 155: 459–512 Link to comment Share on other sites More sharing options...
swansont Posted January 27, 2011 Share Posted January 27, 2011 Theoritically we know well light is electromagnetic wave. We know light wave fellows Maxwell equations . How do we know light is electromagnetic wave by experiment? Light dose not interact with magnetic fields. Light does interact with things that make the magnetic (and electric) fields. Link to comment Share on other sites More sharing options...
davey2222 Posted January 27, 2011 Share Posted January 27, 2011 An example of such interaction is the Faraday Effect. Link to comment Share on other sites More sharing options...
steevey Posted January 28, 2011 Share Posted January 28, 2011 Theoritically we know well light is electromagnetic wave. We know light wave fellows Maxwell equations . How do we know light is electromagnetic wave by experiment? Light dose not interact with magnetic fields. Perhaps because light only comes from particles which carry charges? Even a neutron is a composite of charges since its made of quarks which have color charge. Link to comment Share on other sites More sharing options...
alpha2cen Posted January 28, 2011 Author Share Posted January 28, 2011 We can make light by using many electric instruments. But it is not light. Same logic is like this, i.e., we can make flame from burning wood, but flame is not wood. We need a direct experimental evidence of electromagnetic wave. Link to comment Share on other sites More sharing options...
davey2222 Posted January 28, 2011 Share Posted January 28, 2011 Have a look at the Faraday effect. http://en.wikipedia.org/wiki/Faraday_effect Is this considered as a proof? Link to comment Share on other sites More sharing options...
alpha2cen Posted January 28, 2011 Author Share Posted January 28, 2011 Have a look at the Faraday effect. http://en.wikipedia..../Faraday_effect Is this considered as a proof? Was this experiment done at the vacuum condition? Link to comment Share on other sites More sharing options...
Klaynos Posted January 28, 2011 Share Posted January 28, 2011 Faraday effect. Magneto-optical Kerr effect. Kerr effect. Electro-optical Kerr effect. All of plasmonics. The simplest is the coupling of light to surface plasmons to form surface plasmon polaritons which you can propagate adn ten de-couple to have the light again... 1 Link to comment Share on other sites More sharing options...
davey2222 Posted February 3, 2011 Share Posted February 3, 2011 We know that alternating current in a conductor produces electromagnetic waves whose frequency is the frequency of the alternating current in the conductor. What if we can produce alternating current with a frequency in the optical region, will the conductor glow by giving off light? Can this be used to show that optical light is composed of magnetic field component? Has anyone done this before? Link to comment Share on other sites More sharing options...
swansont Posted February 3, 2011 Share Posted February 3, 2011 We know that alternating current in a conductor produces electromagnetic waves whose frequency is the frequency of the alternating current in the conductor. What if we can produce alternating current with a frequency in the optical region, will the conductor glow by giving off light? Can this be used to show that optical light is composed of magnetic field component? Has anyone done this before? They do it for x-rays, so you should be able to do it for visible light. It's not AC like your outlet supplies; it's an electron beam that is perturbed magnetically and (AFAICT) the period of the magnets and speed of the electrons dictates the frequency of the emitted light. http://news.sciencemag.org/sciencenow/2009/04/21-02.html?rss=1 Link to comment Share on other sites More sharing options...
davey2222 Posted February 3, 2011 Share Posted February 3, 2011 Interesting. But how does it tie magnetic field component and EM wave together as a proof that EM waves are made up of a magnetic field component? Link to comment Share on other sites More sharing options...
swansont Posted February 3, 2011 Share Posted February 3, 2011 Interesting. But how does it tie magnetic field component and EM wave together as a proof that EM waves are made up of a magnetic field component? It doesn't. That's not the question I was answering. Light can give magnetic dipole transitions. They are weaker than electric dipole transitions, but can occur under conditions where an ED transition is forbidden. http://en.wikipedia.org/wiki/Electric_dipole_transition 1 Link to comment Share on other sites More sharing options...
lemur Posted February 3, 2011 Share Posted February 3, 2011 (edited) We know that alternating current in a conductor produces electromagnetic waves whose frequency is the frequency of the alternating current in the conductor. What if we can produce alternating current with a frequency in the optical region, will the conductor glow by giving off light? Can this be used to show that optical light is composed of magnetic field component? Has anyone done this before? Let me go turn on my electric range elements to high and see if they glow . . . actually, wouldn't an incandescent lightbulb work just as well? Edited February 3, 2011 by lemur Link to comment Share on other sites More sharing options...
swansont Posted February 3, 2011 Share Posted February 3, 2011 Let me go turn on my electric range elements to high and see if they glow . . . actually, wouldn't an incandescent lightbulb work just as well? How is that an example of giving off light at the frequency of the alternating current? ———————— Visible-spectrum synchrotron radiation has also been observed. Modern accelerators operate at too high of an energy to see this, I would think, but the size and energy of earlier versions were. http://solidstate.physics.sunysb.edu/teach/phy518/pollock_ajp_51_278_83.pdf Link to comment Share on other sites More sharing options...
lemur Posted February 3, 2011 Share Posted February 3, 2011 (edited) How is that an example of giving off light at the frequency of the alternating current? Yes, I actually thought about that after I posted it but I wasn't sure so I thought I'd wait and see what others said. How does Ohm's law go again? Current/amps = voltage/resistance? So isn't the frequency of the AC current related to the current/amps, if it's not the same thing? I suppose a lower voltage current would have to alternate at a higher frequency to deliver the same amount of energy, right? So the temperature of the conductor, and therefore the frequency of photons emitted, would be a product of the AC frequency as well as the voltage/resistance relationship, no? Or was the other post saying that the frequency of the AC current produces a certain amount of radiation at that frequency regardless of the voltage/resistance of the system? How could the frequency of photon emissions from a conductor be determined by something other than the temperature? Is this an example of electric current getting translated directly into EM emissions without heating the material? Edited February 3, 2011 by lemur Link to comment Share on other sites More sharing options...
davey2222 Posted February 4, 2011 Share Posted February 4, 2011 Yes, I actually thought about that after I posted it but I wasn't sure so I thought I'd wait and see what others said. How does Ohm's law go again? Current/amps = voltage/resistance? So isn't the frequency of the AC current related to the current/amps, if it's not the same thing? I suppose a lower voltage current would have to alternate at a higher frequency to deliver the same amount of energy, right? So the temperature of the conductor, and therefore the frequency of photons emitted, would be a product of the AC frequency as well as the voltage/resistance relationship, no? Or was the other post saying that the frequency of the AC current produces a certain amount of radiation at that frequency regardless of the voltage/resistance of the system? How could the frequency of photon emissions from a conductor be determined by something other than the temperature? Is this an example of electric current getting translated directly into EM emissions without heating the material? What you are assuming is the Ohmic heating which gives of heat radiation and visible light. That can occur with direct current as well. Frequency of AC current is unrelated to its voltage and current magnitudes. What I was proposing was to jack up the frequency of an AC supply into the optical range. According to Maxwell's equations the EM waves produced should be in the visible spectrum also which we can then see with our naked eyes. This was supposed to prove that EM waves were made up of a magnetic field component. However there is a problem in regard to the skin effect at such optical frequency. This can cause heating and make the conductor glow. Separating the light produced from heating and Maxwell's is another problem. I suppose covering the conductor with an opaque layer can block out the glow due to heating. Link to comment Share on other sites More sharing options...
swansont Posted February 4, 2011 Share Posted February 4, 2011 The biggest problem would be getting an oscillator at such a frequency. AFAIK the cutting edge ones are still a few orders of magnitude short. Link to comment Share on other sites More sharing options...
Klaynos Posted February 4, 2011 Share Posted February 4, 2011 I know we don't use oscillators for THz spectrometers because you can't get them fast enough currently. Visible light is about 2 orders of magnitude faster... Link to comment Share on other sites More sharing options...
lemur Posted February 4, 2011 Share Posted February 4, 2011 (edited) What you are assuming is the Ohmic heating which gives of heat radiation and visible light. That can occur with direct current as well. Frequency of AC current is unrelated to its voltage and current magnitudes. What I was proposing was to jack up the frequency of an AC supply into the optical range. According to Maxwell's equations the EM waves produced should be in the visible spectrum also which we can then see with our naked eyes. This was supposed to prove that EM waves were made up of a magnetic field component. However there is a problem in regard to the skin effect at such optical frequency. This can cause heating and make the conductor glow. Separating the light produced from heating and Maxwell's is another problem. I suppose covering the conductor with an opaque layer can block out the glow due to heating. First off, are you proposing that somehow a conductor, such as the filament of an incandescent light bulb or the element of an electric range, could be induced to generate visible light without producing the same frequency-spectrum that it gives off when powered by a lower-frequency AC current (including all the infrared/heat)? Second, if an AC current's frequency would be raised to the speed of visible frequencies, wouldn't its voltage have to be very low (i.e. low enough not to produce an amount of current that would melt the conductor)? In that case, wouldn't Ohm's law still apply, and the amount of light produced would be equivalent to the amount of current flowing through a certain amount of resistance? In other words, regardless of the frequency of the current-oscillation, wouldn't the determinant factors be the amount of power/current and the amount of resistance? Related question: if the AC current oscillated at the frequency of the EM waves it was causing, would the voltage of the oscillations be equivalent to the amount of force in each wave? Edited February 4, 2011 by lemur Link to comment Share on other sites More sharing options...
swansont Posted February 4, 2011 Share Posted February 4, 2011 The frequency does affect the impedance for a capacitive or inductive circuit. If it's purely resistive it won't. Ohm's law isn't the issue here. Link to comment Share on other sites More sharing options...
davey2222 Posted February 4, 2011 Share Posted February 4, 2011 (edited) Lemur, We are discussing how to prove that light is an EM wave. We can generate x-rays from deceleration of fast electrons but this doesn't tie in with electromagnetism in that sense. We can get optical light from LED but this is just excited electrons losing their excitation and emitting photons in the process, hence still doesn't tie in with EM. Since we know a current produces a circulating magnetic field around it, according to Maxwell's, an AC current produces EM waves. If we can have current frequency in the optical range then the produced EM waves should be in the optical range and could be seen with the naked eye. As was pointed out by others, producing THz is a challenge. We don't even have such an oscillator to begin with. At such a high frequency the skin effect comes into play, causing surface heating, but this is not the issue. And then the inductive reactance becomes significant at this frequency and this can stop AC from flowing. Perhaps there is an indirect way of doing this? Edited February 4, 2011 by davey2222 Link to comment Share on other sites More sharing options...
lemur Posted February 4, 2011 Share Posted February 4, 2011 The frequency does affect the impedance for a capacitive or inductive circuit. If it's purely resistive it won't. Ohm's law isn't the issue here. I may be confusing Ohm's law with something else similar. The issue I'm talking about is that energy going through a conductor has to go somewhere. If there is little resistance, the energy will keep moving until in encounters resistance, no? When there is resistance, the energy dissipates in the form of heat (kinetic and/or radiation), right? The frequency of radiation always corresponds with temperature, I thought, because of black-body behavior, no? So doesn't any photon emission from a conductor have to correlate with the amount of energy it is failing to conduct (i.e. the amount it resists)? Then wouldn't the frequency of the photons emitted have to do with the ability of the material to dissipate the energy via conduction and/or convection? Whatever doesn't get lost through conduction (electric current or heat) or convection would have to be emitted as EM waves, correct? So what basis is there for the EM emission frequency except the amount of energy getting emitted? And what basis is there for the amount of energy being emitted except the amount of resistance in the conductor and the amount of energy lost as heat? We can get optical light from LED but this is just excited electrons losing their excitation and emitting photons in the process, hence still doesn't tie in with EM. If an electron is generating the light in an LED, how is that not tied in with EM? What is electromagnetism except electrons? Since we know a current produces a circulating magnetic field around it, according to Maxwell's, an AC current produces EM waves. If we can have current frequency in the optical range then the produced EM waves should be in the optical range and could be seen with the naked eye. How does a radio-transmitter work? It sounds like you are essentially talking about making a radio transmitter that can transmit visible frequencies. What about microwave ovens and doppler radar? What about an incandescent filament? Does that also produce a circulating magnetic field around it other than the photons themselves? Link to comment Share on other sites More sharing options...
swansont Posted February 4, 2011 Share Posted February 4, 2011 I may be confusing Ohm's law with something else similar. The issue I'm talking about is that energy going through a conductor has to go somewhere. If there is little resistance, the energy will keep moving until in encounters resistance, no? When there is resistance, the energy dissipates in the form of heat (kinetic and/or radiation), right? The frequency of radiation always corresponds with temperature, I thought, because of black-body behavior, no? So doesn't any photon emission from a conductor have to correlate with the amount of energy it is failing to conduct (i.e. the amount it resists)? Then wouldn't the frequency of the photons emitted have to do with the ability of the material to dissipate the energy via conduction and/or convection? Whatever doesn't get lost through conduction (electric current or heat) or convection would have to be emitted as EM waves, correct? So what basis is there for the EM emission frequency except the amount of energy getting emitted? And what basis is there for the amount of energy being emitted except the amount of resistance in the conductor and the amount of energy lost as heat? For blackbody radiation, the spectrum is dependent on temperature. We aren't talking about blackbody radiation, or Ohm's law. We're talking about the same effect that gives rise to radio waves — electrons oscillating at a specific frequency will radiate at that frequency. If an electron is generating the light in an LED, how is that not tied in with EM? What is electromagnetism except electrons? It's not the process under discussion How does a radio-transmitter work? It sounds like you are essentially talking about making a radio transmitter that can transmit visible frequencies. What about microwave ovens and doppler radar? What about an incandescent filament? Does that also produce a circulating magnetic field around it other than the photons themselves? Yes to radio and microwave — that's the same process we're talking about. No to incandescent filaments. That's blackbody. Link to comment Share on other sites More sharing options...
lemur Posted February 4, 2011 Share Posted February 4, 2011 For blackbody radiation, the spectrum is dependent on temperature. We aren't talking about blackbody radiation, or Ohm's law. We're talking about the same effect that gives rise to radio waves — electrons oscillating at a specific frequency will radiate at that frequency. It's not the process under discussion Yes to radio and microwave — that's the same process we're talking about. No to incandescent filaments. That's blackbody. Ok, thanks for clarifying. Now can you please say whether it is possible for a conductor to emit photons at a visible frequency without generating a corresponding amount of heat/infrared? What is the difference between using heat to oscillate electrons and using AC current? Is the issue getting the electrons to move without the nuclei vibrating? Isn't that what electricity is? Aren't photons the way that electrons release energy when they can't transfer it as kinetic momentum to adjacent particles? Link to comment Share on other sites More sharing options...
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