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What Medium is Propagating Electromagnetic Waves?

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posted on Aug, 17 2019 @ 01:14 AM
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a reply to: kwakakev




This is why gamma rays can use it too, they have a charge too.

No.
Gamma rays are very high frequency (and thus, very high energy) electromagnetic radiation.
No charge.




posted on Aug, 17 2019 @ 01:38 AM
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a reply to: Phage

Then how do you define the differences in the medium of travel between a visible light photon and a gamma ray photon?



posted on Aug, 17 2019 @ 01:43 AM
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a reply to: kwakakev

There is no difference. Both are self-propagating.



posted on Aug, 17 2019 @ 02:04 AM
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a reply to: Phage

I agree that a photon does not have a charge, but it does have momentum that moves through the field of charge,



posted on Aug, 17 2019 @ 02:16 AM
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a reply to: kwakakev

I agree that a photon does not have a charge,
Then why did you say this?

This is why gamma rays can use it too, they have a charge too.




but it does have momentum that moves through the field of charge,

Charged field or magnetic field or no field, the photon does propagate. Because electromagnetic radiation consists of an electric field and a magnetic field (electromagnetic radiation, get it?). They each cause the other to propagate. Self-propagation.



edit on 8/17/2019 by Phage because: (no reason given)



posted on Aug, 17 2019 @ 02:47 AM
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a reply to: Phage

That was a good catch. Having some troubles expressing and forming just how all this works. Electron and Protons have a charge. We can define and measure it, thats clear.

How this charge is generally defined is as the electromagnetic force. When we try and look very closely at it we find points in space that bounce around in signal strength. Things are moving very fast with lost of interactions going on. I do see the actual area of affect for the charge of the proton and electrons extending along the inverse square rule at they interact with each other. It is where these interactions between charged elements take place that supports the movement of the photon through space.



posted on Aug, 17 2019 @ 02:51 AM
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a reply to: kwakakev

A photon, electromagnetic radiation, has no charge. It is not affected by electrical or magnetic fields unless those fields have pretty much unbelievable energy density, such that they induce gravitational effects. But in that case the effects are only observable from a frame of reference outside of them. As far as the photon is concerned, it's self-propagating itself in a very straight line.



edit on 8/17/2019 by Phage because: (no reason given)



posted on Aug, 17 2019 @ 03:51 AM
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a reply to: Phage

A photon does not have a charge, but it does have a frequency or color. How this can self propagate for billions of years does hint at unbelievable energy density. I am a bit lost with how electromagnetic radiation is not affected by electromagnetic fields.



posted on Aug, 17 2019 @ 04:11 AM
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a reply to: kwakakev




How this can self propagate for billions of years does hint at unbelievable energy density.
A supernova produces very high energy densities. And gamma rays.


I am a bit lost with how electromagnetic radiation is not affected by electromagnetic fields.
I know.

edit on 8/17/2019 by Phage because: (no reason given)



posted on Aug, 17 2019 @ 05:48 AM
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originally posted by: kwakakev
a reply to: Phage

A photon does not have a charge, but it does have a frequency or color. How this can self propagate for billions of years does hint at unbelievable energy density.
If you're talking about "A photon", it has an energy according to one of the simplest equations in physics, Planck's constant times the frequency.

That's the energy, and it doesn't appear to change unless the photon passes through stretching space which can lower the frequency.

I don't know of any well-defined concept for the energy density of "A photon" since I don't know of any well defined concept for the size of a photon to calculate a meaningful energy density number for a single photon. So when someone says energy density, I automatically think of multiple photons, where it is possible to calculate meaningful numbers for energy density, based on the number of photons, their individual energies, and the density of the photons (how many there are per cubic centimeter for example).


I am a bit lost with how electromagnetic radiation is not affected by electromagnetic fields.
It's because electromagnetic fields affect charges, and photons don't have any charge. You can take a flashlight in each hand and cross the beams, and they pass right through each other. Where the beams cross, you could put a sheet of paper and see the light is twice as bright there but it's simply because you have two beams instead of one, so at a given location they can add up but this isn't the beams affecting each other, it's simply the sum of beam1 plus beam2.


originally posted by: Phage
a reply to: kwakakev

A photon, electromagnetic radiation, has no charge. It is not affected by electrical or magnetic fields unless those fields have pretty much unbelievable energy density, such that they induce gravitational effects.
Yes, I don't think visible light or lower energy photons can interact, but if the photons have very high energy, apparently one of the the photons can fluctuate into a pair of virtual charged particles which can also allow some type of interaction with another photon, other than gravitational. This paper can explain what was observed at the CMS experiment better than I can:

Evidence for light-by-light scattering...

I think it's safe to say no such photon-photon effects occur when you cross two flashlight beams. For example, red light photons have about 2eV of energy, while the two photons described in that paper each had about 2 GeV energy, as much energy as a billion visible light photons.



posted on Aug, 17 2019 @ 09:38 AM
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a reply to: Arbitrageur

The Planck length helps make sense of quantum operation. It was defined as the smallest measure that made sense for Mr Planck and is used in measuring the progression of EM waves. It works for measuring many other aspects in the quantum realm. Does this mean that this is where the fabric of the universe stops or is there another lower layer below? Just how much information can be involved in a photon?

The Planck length represents each cycle in a moving frequency. Is this right?

E=hv means The higher the frequency the higher the energy of the photon. Interesting to hear the energy reduces as space stretches. Also means that energy increases as space contracts. So this results in a color change. How the background radiation of the big bang is measured is an example of this.

It sounds like a photon does have a charge from your description, with its energy density about one billion times less that of the electron. Is it possible that each color represents a very small change in this very small value of an electrons charge?



posted on Aug, 17 2019 @ 12:14 PM
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originally posted by: kwakakev
a reply to: Arbitrageur

The Planck length helps make sense of quantum operation. It was defined as the smallest measure that made sense for Mr Planck and is used in measuring the progression of EM waves. It works for measuring many other aspects in the quantum realm. Does this mean that this is where the fabric of the universe stops or is there another lower layer below? Just how much information can be involved in a photon?

The Planck length represents each cycle in a moving frequency. Is this right?
No, it's not right. As the name "length" suggests, it's a length. Like the meter is a length. If you understand any fundamentals about light/EM radiation you know that higher frequencies mean shorter wavelengths so more cycles in a given length, whether using a meter or planck length.

Maybe we should stick to fundamentals since you're struggling so much with those, I think it's too much for you at this point to get into how much information a photon has because it gets complicated.


E=hv means The higher the frequency the higher the energy of the photon. Interesting to hear the energy reduces as space stretches. Also means that energy increases as space contracts.
I have yet to run across any example of space contracting, but we do have observations of space expanding.


How the background radiation of the big bang is measured is an example of this.
Correct, those wavelengths are stretched a lot from expanding space.


It sounds like a photon does have a charge from your description

Here's what I said:


originally posted by: Arbitrageur
photons don't have any charge.

So if you conclude from that a photon has a charge you misread something. I think your misunderstanding was also illustrated in a previous post asking something about what was inside the proton when it was smashed in a particle smasher.


originally posted by: kwakakev
If you can smash open a proton and it does not produce any photons, it does suggest that some other kind of energy is going on there. And there is, gamma rays.

That's ok, it's a common misconception, but that's not how high-energy physics works. When we smash protons at the LHC, it's not to crack them open and see what's inside. On the contrary, when we smash them we observe things that aren't any part of a proton at rest.

What the LHC does is accelerate protons to give them momentum, and in the equation Eros gave you before where total energy is the sum of the mass related energy plus the momentum related energy, at 7 TeV the momentum related energy of the photons is over 7000 times as great as the mass related energy. When the protons collide that energy is released, and that is how they were able to make Higgs bosons that are 133 times more massive than a proton. There was never any Higgs boson inside a proton before it was smashed, it ended up forming as a consequence of all the momentum in the proton released as energy in the collision.

So please abandon any misconception that if you see a Higgs boson come out of a proton-proton collision, that means there was a Higgs boson in the protons. There was not. Likewise you should also abandon the notion that just because a very high energy photon at the LHC can fluctuate into a pair of charged virtual particles, that means it has charge. That's just as wrong as far as I can tell.

There are numerous experiments placing limits on the charge of a photon, so look those up. The limits are very close to zero, only because it's probably impossible to devise an experiment to show photon charge is exactly zero.

In quantum mechanics, nature appears to follow certain rules or laws. We don't make the laws, we watch what nature does and try to see what patterns nature follows consistently, thus we observe conservation laws like conservation of charge, and conservation of energy and so on. When you put all those rules or laws together, they define what is allowed and what is not allowed to happen. So a photon with over 1.022 MeV of energy can produce a positron and electron each with half that much energy; it's allowed because charge is conserved and provided the photon interacted with something else then momentum can also be conserved.

But I think you will have a very muddled view of physics or nature if you then conclude that because an electron and positron came out of the photon in some sense, that the photon contained those particles. Why? Many reasons, the properties are dramatically different. Electrons and positrons can exist at rest and have mass and charge, while photons can't exist at rest and don't have mass or charge, and so on.

So I think the only way to look at it that makes sense to me is to say the electron and positron "pair" were not "inside" the photon before the "pair production", rather, before the pair production, the photon was a massless, chargeless particle. Then in "pair production", the energy in that particle was converted into something else by something allowed in nature's rules which had an equivalent amount of mass-energy content. It's a quantum mechanical phenomenon that doesn't seem to be intuitive like much of classical physics, so it's based on observation.

edit on 2019817 by Arbitrageur because: clarification



posted on Aug, 18 2019 @ 12:15 AM
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a reply to: Arbitrageur

My understanding of the source of the Planck value is as a length. This value provided the grid structure to measure and understand quantum mechanics. The variations in frequency of the EM field does not directly fit into a consistent Planck length. How all these frequencies do move at the speed of light does fit with a consistent Planck length.



the two photons described in that paper each had about 2 GeV energy, as much energy as a billion visible light photons.


This was the piece that gave me a clue towards a photon having a charge. It is very small, but with enough energy in condensing photons in travel, it can produce a new photon.



The limits are very close to zero, only because it's probably impossible to devise an experiment to show photon charge is exactly zero.


Yeah this is the area I am looking at, on the way to 0 or infinity. It is impossible to devise and experiment with that kind of thinking. We are talking about it. Sure it is hard, but maybe one day we can if we dare to dream.

I see the Higgs bosons as a theory to help explain gravity. The working going on puts it in the lead.



posted on Aug, 18 2019 @ 07:38 AM
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The plank constant is not the planck length. The planck constant is the quanta of energy per wavelength

It's value is 6.626 × 10^-34 m2 kg / s the units of which can be converted to J s

J = kg⋅m2/s2

Do not get confused because the unit of Joule has kg in it, which i sort of predict will be your next point.

The planck length is defined using the planck constant the gravitational constant G and the speed of light. It is a very small number and isn't really to do with how many oscillations you can get in a space. it is more over that our understanding of normal geometry and behaviour becomes ill defined, or completely non-defined in a classical manner. It is for example tied in with the uncertainty principle.



On condensing photons, the photon still doesn't appear to have any charge, we shouldn't need to reiterate this

Here is the particle data groups 4 page summary of the Photon, study it

Mass - There have been many many limits placed on the photon mass... the current accepted value is 'Less than 10^-18 eV' Note the electron mass is 511,000 eV

Charge - Dependant upon method less than 10^-35 e or 10^-46 e where e is the electron charge.

Note also that the photon being charged is highly inconsistent to pull out of the measurements an and if it is added that some might be positive, negative or neutral the issue is even worse.

All this points at the Mass being zero and the charge also being zero.


Also the Higgs boson is a manifestation of a mechanism that explains how particles get mass



posted on Aug, 18 2019 @ 08:54 AM
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originally posted by: kwakakev
The variations in frequency of the EM field does not directly fit into a consistent Planck length. How all these frequencies do move at the speed of light does fit with a consistent Planck length.
Those statements seem to contradict each other. using the speed of light, you can convert EM frequency to wavelength. Once you have the wavelengths, you can compare those to any measurement unit for length. If the wavelengths are different, they are different, whether you compare them to a meter or a millimeter or a Planck length.


This was the piece that gave me a clue towards a photon having a charge. It is very small...

Yeah this is the area I am looking at, on the way to 0 or infinity. It is impossible to devise and experiment with that kind of thinking. We are talking about it. Sure it is hard, but maybe one day we can if we dare to dream.
I take it you don't know much about metrology, or calculating measurement errors, etc.

Let's take the oceans of Earth which contain approximately 20 trillion trillion drops of water. Now let's say you want to add one drop of water, and measure the change in the ocean level, do you think it could be done? I'd say that's beyond hard.

Yet if the charge of an electron was like the volume of water in all the earth's oceans, the charge of a photon if there is any, would be like less than a billionth of a drop of water, that's the kind of mind blowing accuracy the photon charge measurements have been able to achieve. The 2018 particle Data Group published value for photon charge is "less than 1E-35e", which is less than 100 billionth of a trillionth of a trillionth of the charge of an electron.

So even if the photon actually did have such a charge below that experimentally measured limit, that small amount of charge wouldn't have any measurable effect on anything we can measure currently, so it doesn't help to explain any observations even if it was there, so you're completely ignoring the measurements that determined photon charge is "less than 1E-35e" if you try to explain anything with photon charge.


I see the Higgs bosons as a theory to help explain gravity. The working going on puts it in the lead.
Matt Strassler calls that a reasonable guess but says it's completely wrong.

Why the Higgs and Gravity are Unrelated

One of the questions I get most often from my readers is this:

Since gravity pulls on things proportional to their mass, and since the Higgs field is responsible for giving everything its mass, there obviously must be a deep connection between the Higgs and gravity… right?

It’s a very reasonable guess, but — it turns out to be completely wrong. The problem is that this statement combines a 17th century notion of gravity, long ago revised, with an overly simplified version of a late-20th century notion of where masses of various particles comes from.



originally posted by: ErosA433
Charge - Dependant upon method less than 10^-35 e or 10^-46 e where e is the electron charge.
Yes I thought there were studies suggesting a tighter limit on charge than what's on the summary sheet, but this is what I got from the summary sheet the PDG published last year:



Even the less than 10^-35e seems like an extremely tight constraint.

edit on 2019818 by Arbitrageur because: clarification



posted on Aug, 18 2019 @ 09:53 AM
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a reply to: Arbitrageur

A EM wave length does have an indirect relationship to the Planck length. It is the Energy of the photon that contains the frequency as it travels. The difference between a red and blue light is that the blue light is described as a higher frequency or has more energy as defined by E=hv.

With the ongoing exponential growth of computing power, the processing capabilities to define one drop in an ocean is coming online. If you have your procedural programming together the run time code is not much, maybe? Having the storage and processing capabilities for very large numbers is what is needed if we are to achieve full spectrum domination. When it comes to exploring space, having that 'beam me up scotty' technology, we need to know what light is before we can break it.

There are number class that can contain numbers as big as the memory available, millions of digits in size. If we are going to explore and colonize space we need to make sure the logic is solid. Minor errors can have you drifting in space for a very long time.



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