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I Antennas and electromagnetic radiation

  1. Jul 25, 2016 #1
    Antennas work with variable current that leads to accelerations and deceleration ofor the electrons, the frequency of the photon or the electromagnetic radiation I want to generate depends on the change in kinetic energy of the electron E= hf= change of kinetic energy of the electron, according to bremsstrahlung and electromagnetic radiation can be created without an harmonic motion of the charge we are considering

    So if we create an antenna that makes electrons undergoing harmonic motion we will get that the change in kinetic energy will be 1/2 m w^2 A^2

    So hf= 1/2 m w^2 A^2

    Andi know that is wrong cos I know that the frequency of the emited wave is equal to the frequency of the harmonic motion of the electrons, but Why?

    Also seems logical that is equal to the frequency of the electrons because the electric field oscillates with that frequency but where is the mistake I made?
    I mean the frequency of the photons emited by the electrons doesn't depend by the frequency the electrons were oscillating but the change in kinetic energy of those no?
    How can I connect the quantum mechanics relation E= hf and the bremsstrahlung correctly to the antenna fenomenon? Please help, this is driving me crazy
     
  2. jcsd
  3. Jul 25, 2016 #2

    tech99

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    The energy of each photon depends only on the frequency emitted. So for an antenna driven by a sinusoidal voltage, that is the emitted frequency. And for Bremsstrahlung the emitted frequency depends on the duration of the collision. As it is a "step function", the emitted frequencies extend in a broadband manner from zero to this frequency.
    If we radiate more power, the number of photons changes, not their individual energy.
     
  4. Jul 25, 2016 #3

    Dale

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    An antenna is classical, there is no need or benefit to treating it quantum mechanically. There isn't any way to identify a particular RF photon as having originated from a particular electron, so there is not a physical meaning behind calculating the change in KE of a single electron and trying to relate that to the frequency of the RF. It has no bearing on anything either classically or quantum mechanically.
     
  5. Jul 25, 2016 #4
    Why is the emitted frequency of the photon the same as the frequency of oscillation of the electron?
     
  6. Jul 25, 2016 #5

    sophiecentaur

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    When EM waves interact with condensed matter (solids and liquids) the interaction is not between an incident photon and an oscillation of a single electron. It is better to talk in terms of an oscillating current (the totality of charges in a region of the metal) in the case of the metal in an antenna (not all antennae are actually metal and dielectric antennae are not totally uncommon). The mean displacement of electrons which have a drift velocity of, say 1mm per second (typical) is really going to be pretty small if the incident radiation has a frequency of just 10MHz. How far would an electron with a speed of 1mm/s get in half a cycle of RF (period 0.1μs)? Yet you could have an oscillating current of many Amps.
    Free electrons in the Ionosphere are a different matter and they can be thought of as oscillating back and forth in response to the incident field. In the presence of a magnetic field, they actually perform elliptical orbits!
    If you want to pursue a model of individual charges moving in step with the incident radiation then what other frequency could they have than that of the radiation? What would make them move in any other way?
     
  7. Jul 25, 2016 #6
    So... analising the fenomenon with quantum mechanics isn't right because in the fenomenon are implied a large number of electrons... and in macroscopic situations the EM waves is just the oscillation of electromagnetic field created directly by the oscillation of an electron electric field E= k e/ R^2 where R is a function R (t)= Ro sin (ωt + φ) , that leads to a sinusoidal cjanging in the magnetic field either, can someone check if what i said is right Please?
     
  8. Jul 26, 2016 #7

    tech99

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    Looking at the macro scale, when a radio frequency voltage is applied to a wire, so that a wave travels along it, I think we can regard it as a moving charge. It is also worth mentioning that the free charge on a conductor is very great, something like 10 A-h in 1 cc of copper, so the slightest movement has a large effect.
     
  9. Jul 26, 2016 #8

    sophiecentaur

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    It's perfectly 'right' as long as you use the correct "Quantum Mechanics". In a solid, the electron doesn't behave like it does in the isolated H atom. QM still applies but you don't have discrete energy levels. you have Energy Bands (continuous values) because of the way the whole structure works. Read around about Solid State Physics and if you can find a site that suits your level (start with Wiki) you may find an answer to your question that makes sense - it is fairly steamy stuff and needs a bit of background knowledge to get this subject straight. Worth a try though.
     
  10. Jul 26, 2016 #9

    Dale

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    As @sophiecentaur mentioned, it is not that it isn't right, it is that you have to do it right. And when you do it right you just get more complicated math to describe the standard classical treatment.
     
  11. Jul 26, 2016 #10

    sophiecentaur

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    I just re-read this and part of your problem is that you feel the need to connect an individual photon with an individual electron - as if the photon is a little bullet that strikes the antenna in a particular place and causes an electron to vibrate. If you take the basic Hydrogen model, with which you are familiar, when a wave interacts with the atom, it will raise its energy level (if it's of the right frequency). There is then a pause, of an indeterminate length of time, before the atom decays and emits a photon of the same frequency. No two adjacent atoms will not take the same to decay so the result will be that a beam of light will simply be scattered randomly if that's what happened to the light. In fact, when EM strikes a conductor (or a dielectric like glass), the incident wave is coherent and so is the reflected (or refracted) wave (or the signal passed to a receiver down the wire). So, what happens must be a mechanism that maintains the coherence of the signal with our adding a massive random element into the process. So the whole of the region of the substance comprising the antenna must have a coherent response to the incoming signal. A photon has no 'extent' and cannot be thought of as starting at its source and making a B line for a particular destination. It is 'anywhere' in space until it actually interacts with an object. Only then does it really have a definite place and time. The diagrams you can find throughout the world of Physics (particularly the less informed attempts to 'educate' people) do not help because the little bullet model does not die easily! You will find many many discussions about how big a photon is and they are all on very dodgy territory; best to avoid.
    OH yes - and Bremstrahlum is strictly an Emission phenomenon and doesn't relate to a receiving antenna that's made of a solid. Don't worry about any apparent contradiction between the two.
     
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