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Goldilocks Frequencies And Photon Packet Frequencies

Radio waves

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#18 Mattzy

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Posted 09 June 2019 - 02:27 AM

Ah, OK. I would imagine this relates to some happy medium, balancing the snags of the more line-of-sight behaviour of high frequencies (need more towers to cope with blocking by buildings etc) against whatever the advantage is of higher frequency (rhertz can probably fill you n on that side of things - I am not a radio engineer).

Yes, reducing cost by minimising hardware. I think its about penetration of walls and windows etc. I think in future there will have to be more frequencies in use within that range.



#19 Mattzy

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Posted 09 June 2019 - 02:29 AM

Yes it IS all photons. When a dipole antenna radiates, there are moving charges. When an electron drops to a lower level in the atom, it is a moving charge. In fact if you do quantum theory, you learn about something called the "transition dipole", which describes the process by which the electron emits a photon. It is just a question of scale, really.  

 

I really recommend this Wiki link: https://en.wikipedia...gnetic_spectrum. I suggest reading this and then maybe asking questions, if you have any. 

Thanks, I'll check it out.



#20 Mattzy

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Posted 09 June 2019 - 03:57 AM

Don't lose your peace of mind because of this dual behavior (explanation I should say).

 

The only thing that a purist physicist can criticize you is your first comment: they don't talk about orbits (Bohr) or shells (Schrodinger) anymore. It's a

thing of the past. The new explanation used at the core of Quantum Physics today is: a bounded electron (the one stuck at an atom) changes its

energy levels up (absorbing a photon) or down (emitting a photon).

 

Any other concept that I remarked in red is CORRECT, basicly.

 

If you want to be cool and always right about this topic, follow the master/creator of the Quantum Physics: Max Planck.

On his words "electromagnetic energy travels along the space as waves and are absorpted or emitted by matter in discrete amounts or

quantum of actions with E = hf".

 

You'll keep hearing about photons travelling the space, frequency or size of photons, etc. This is due to decades of sloppy education, nothing more.

 

Photons are used as an easy mean to explain discretness of energy of light. It's easy for people to visualize massless particles moving around with

hf energy per photon. Is not that it's wrong, only that it's incorrect to think that way. Using photons will prevent you to understand phenomena in the

macroworld like: reflection, refraction, difraction, radiation of energy, etc.

 

Planck created by his own the basis of the quantum physics when, at his Jan. 1901 paper, he explained the radiation inside a black body cavity. It

was known for decades that such radiation was electromagnetic, but any attempt to find a formula to describe it failed, because classical (Maxwell)

electromagnetism was used. Planck found the right equation when he asserted that any electromagnetic wave (a single monochromatic one)

emitted by his resonators (he didn't know that atoms existed by then) had a minimum value of energy E = h.f (h is the later called Planck's constant).

and also asserted that not any other EM wave could have an energy value lower than hf. Whith this discretization of energy in quantums, he

created the basis for the Quantum Physics (and photons).

 

This is the curve of the radiation of a black body cavity, from the times of Planck and valid today: You'll observe that cover any electromagnetic

frequency from AM radio (and lower yet) up to the lethal gamma rays. Visible light is in the region between 450 and 750 nanometers. Larger

wavelengths are called infrared radiation (in three groups), below which are the microwaves, etc. Above visible light (shorter wavelengths)

there is the region of UV radiation, X-ray radiation and, finally, gamma radiation (usually present at nuclear process and cosmic radiation).

 

1.png

 

These curves graph the radiation measured through a small hole in a black body cavity, and is a function of the temperature

within the cavity. The temperature is one variable that is changed easily.

 

Today, black body radiation generators of different ranges are manufactured and sold for industrial, medical, scientific and military

applications.

 

This is an example: https://www.newport.com/p/67032

 

LS-pg5-35a_800w.jpg?9

 

This is how the cavities are done (there are many materials and techniques) but this is the most frequent (using grafite):

 

 

2.png

Great stuff rhertz! And it cheers me up that I at least have the excuse to blame the education system of the 1970's (we had a lot of black-outs!). I will now have to re-confirm a very simple baseline to start from. Could you give me just true or false and I will go away and ask why or why not. I don't want to take too much of your time. 1. The radio wave is not a microwave.  2. When we charge an antenna, we increase energy in its atoms (electrons). 3. When we discharge an antenna we decrease energy in its atoms and quanta are emitted from some of its electrons. 4. The quanta unify into a single monochromatic EM wave? (this is where I'm really lost - I keep seeing an increasing gradient of quanta that make up a wave - which I was calling a pulse)  5. This wave is just pure EM energy and that's all there is to it, and it can't be thought of as a microwave or visible photon - this is the macroworld?

If I conceive a microwave as a dot of energy traveling on a path describing a concertinaed rattle snake - as I have seen depicted - I can see how it can't pass through a 1mm hole, so here we are thinking photons. But it would have to be a single quanta or photon? The visible light photon is depicted as a tiny rattle snake that will pass through the hole. The radio wave is much bigger, so here we are in the macro world where we think about reflection and diffraction etc. ??? Thanks again rhertz, please let me know as briefly as you like if I'm starting to get it.



#21 exchemist

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Posted 09 June 2019 - 05:35 AM

Great stuff rhertz! And it cheers me up that I at least have the excuse to blame the education system of the 1970's (we had a lot of black-outs!). I will now have to re-confirm a very simple baseline to start from. Could you give me just true or false and I will go away and ask why or why not. I don't want to take too much of your time. 1. The radio wave is not a microwave.  2. When we charge an antenna, we increase energy in its atoms (electrons). 3. When we discharge an antenna we decrease energy in its atoms and quanta are emitted from some of its electrons. 4. The quanta unify into a single monochromatic EM wave? (this is where I'm really lost - I keep seeing an increasing gradient of quanta that make up a wave - which I was calling a pulse)  5. This wave is just pure EM energy and that's all there is to it, and it can't be thought of as a microwave or visible photon - this is the macroworld?

If I conceive a microwave as a dot of energy traveling on a path describing a concertinaed rattle snake - as I have seen depicted - I can see how it can't pass through a 1mm hole, so here we are thinking photons. But it would have to be a single quanta or photon? The visible light photon is depicted as a tiny rattle snake that will pass through the hole. The radio wave is much bigger, so here we are in the macro world where we think about reflection and diffraction etc. ??? Thanks again rhertz, please let me know as briefly as you like if I'm starting to get it.

Mattzy, if I can just interject on one point, the emission of radio-frequency photons from an antenna is not the result of electrons changing levels within the atoms. That is how visible light is produced, but the energy gaps between atomic levels are far too big (i.e too much energy is emitted each time) to make radio waves. (As you may know by now the higher the frequency, the more energy the photon has, by Planck's relation E=hν.  So a radio-frequency photon has a lot less energy than a visible light photon.)

 

In a solid metal, as a consequence of the chemical bonding between atoms (what we call "metallic bonding"), the outermost orbitals (energy levels) of the atoms overlap and merge together, into a band of energies called the "conduction band". Electrons in the conduction band are no longer bound to any particular atom and can travel fairly freely through the solid (this is what makes metals conduct electricity and heat so well). In the antenna they are accelerated by the alternating voltage, first in one direction and then the other and it is this that leads to them emitting radio-frequency photons. You can view it as very tiny changes in energy level, but all taking place within the conduction band. 

 

The wave issues I'll leave to rhertz. :)


Edited by exchemist, 09 June 2019 - 05:38 AM.

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#22 Mattzy

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Posted 10 June 2019 - 03:02 AM

Thanks rhertz and exchemist. I'll do some reading on the links you recommend. It's all very interesting - and a bit mysterious at this stage - but that's good. I'm glad I can go back to saying photon! I'll be back after a while on this one.



#23 Mattzy

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Posted 10 June 2019 - 03:08 AM

Mattzy, if I can just interject on one point, the emission of radio-frequency photons from an antenna is not the result of electrons changing levels within the atoms. That is how visible light is produced, but the energy gaps between atomic levels are far too big (i.e too much energy is emitted each time) to make radio waves. (As you may know by now the higher the frequency, the more energy the photon has, by Planck's relation E=hν.  So a radio-frequency photon has a lot less energy than a visible light photon.)

 

In a solid metal, as a consequence of the chemical bonding between atoms (what we call "metallic bonding"), the outermost orbitals (energy levels) of the atoms overlap and merge together, into a band of energies called the "conduction band". Electrons in the conduction band are no longer bound to any particular atom and can travel fairly freely through the solid (this is what makes metals conduct electricity and heat so well). In the antenna they are accelerated by the alternating voltage, first in one direction and then the other and it is this that leads to them emitting radio-frequency photons. You can view it as very tiny changes in energy level, but all taking place within the conduction band. 

 

The wave issues I'll leave to rhertz. :)

Very good interjection exchemist! All new to me too. I think it's going to help me on the issue of photons that do and don't go through a microwave screen (maybe not). I'll go to those links first.



#24 exchemist

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Posted 10 June 2019 - 03:29 AM

Very good interjection exchemist! All new to me too. I think it's going to help me on the issue of photons that do and don't go through a microwave screen (maybe not). I'll go to those links first.

I'm not sure it will help much with that. You could read this, though: http://hyperphysics....ves/mwoven.html

 

But it is useful for you to know that EM radiation can be generated by a large variety of different things, depending on the wavelength. For X-rays and visible light, it is electrons changing levels in atoms and molecules. For infra red it is vibrations of molecules. For microwaves it is rotations of molecules (and motion of electrons in cavity magnetrons etc). For radio waves it is motion of electrons in an antenna.  The common feature is that In all cases you have motion of an electric charge.


Edited by exchemist, 10 June 2019 - 03:30 AM.


#25 exchemist

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Posted 10 June 2019 - 05:05 AM

I'm not sure it will help much with that. You could read this, though: http://hyperphysics....ves/mwoven.html

 

But it is useful for you to know that EM radiation can be generated by a large variety of different things, depending on the wavelength. For X-rays and visible light, it is electrons changing levels in atoms and molecules. For infra red it is vibrations of molecules. For microwaves it is rotations of molecules (and motion of electrons in cavity magnetrons etc). For radio waves it is motion of electrons in an antenna.  The common feature is that In all cases you have motion of an electric charge.

Later note: I have found in several internet sources (e.g. this one: https://physics.stac...-by-small-holes) the information that the transmitted power through a hole is proportional to (d/λ)⁴, where d is the diameter of the hole and λ is the wavelength.

 

So you can see from this that when λ>>d, d/λ will be much less than 1 and, raised to the 4th power, it is absolutely minuscule. So if your microwave has a wavelength of 12cm and the holes are 0.5cm diameter, d/λ will be about 0.04, and raised to the 4th power it will be 0.00000004.  


Edited by exchemist, 10 June 2019 - 05:10 AM.

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