# Vibrational Frequency CO2 Global Warming

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I am unsure if I'm posting this in the right section, but nonetheless I require specialized assistance in verifying my understanding of vibrational frequencies of molecules and atoms with regard to the atmospheric insulative effects of various gases.

My current understanding is that "vibrational frequency" is what I've formerly called "resonant frequency" with regard to the effects of sound, but if I am in error, please correct me.

Now, relating light to sound, frequencies above resonance should reflect off of the particle since the particle is too massive to move at that frequency and frequencies below resonance should pass through the particle with minimal interference since the particle doesn't contain enough mass with which to interact with the wave.  I have edited a graph to aid in illustrating my problem:

Original graph found here https://commons.wikimedia.org/wiki/File:Atmospheric_Transmission.png

Am I correct?  Why or why not?

Summarily, the problem I am having is in visualizing how high-frequency light (ie visible) will pass through CO2, hit the earth, and then radiate back to the CO2 as lower-frequency IR-light where it then becomes trapped, seemingly forming a runaway heating effect.  So, what I need to know is how light interacts with particles differently than sound and why one energy-wave (light) behaves differently from another (sound) because my current understanding demands that if CO2 will insulate at IR frequencies (ie resonate), then it simply must reflect at visible (and higher) frequencies and therefore it would have a cooling effect, not warming, just like clouds of water vapor do.

To further clarify, high-energy light should be reflected away while only retaining low-energy IR light and therefore the overall effect is cooling.  What am I missing?  For if sound behaved in the way I am being asked to understand light, then a 20K tone would pass completely unimpeded through a wall, strike an object which would produce a vibration lower than 20K, say 100hz, then the lower frequency tone would somehow become trapped by the same wall that was transparent to the 20k tone.  It makes absolutely no sense.    Help!

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12 minutes ago, BanterinBoson said:

I am unsure if I'm posting this in the right section, but nonetheless I require specialized assistance in verifying my understanding of vibrational frequencies of molecules and atoms with regard to the atmospheric insulative effects of various gases.

My current understanding is that "vibrational frequency" is what I've formerly called "resonant frequency" with regard to the effects of sound, but if I am in error, please correct me.

Now, relating light to sound, frequencies above resonance should reflect off of the particle since the particle is too massive to move at that frequency and frequencies below resonance should pass through the particle with minimal interference since the particle doesn't contain enough mass with which to interact with the wave.  I have edited a graph to aid in illustrating my problem:

Am I correct?  Why or why not?

Reflection and transmission of light is not dictated by the frequency relative to the resonances. You have shorter wavelengths (below a micron) reflecting, but that includes visible light, which obviously is transmitted.

12 minutes ago, BanterinBoson said:

Summarily, the problem I am having is in visualizing how high-frequency light (ie visible) will pass through CO2, hit the earth, and then radiate back to the CO2 as lower-frequency IR-light where it then becomes trapped, seemingly forming a runaway heating effect.

A blackbody radiates a spectrum that depends on its temperature. The sun, whose surface is at around 6000K, radiates a lot of visible light (peak at ~550 nm). The earth, at around 300K, radiated strongly in the infrared (peak at ~15 microns). It's not a true runaway effect. Thermodynamics prevents the earth from getting hotter than the sun's surface, and geometry limits the temperature even further.

12 minutes ago, BanterinBoson said:

So, what I need to know is how light interacts with particles differently than sound and why one energy-wave (light) behaves differently from another (sound) because my current understanding demands that if CO2 will insulate at IR frequencies (ie resonate), then it simply must reflect at visible (and higher) frequencies and therefore it would have a cooling effect, not warming, just like clouds of water vapor do.

They are different kinds of waves. It mat be better to completely disregard sound waves completely.

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1 hour ago, swansont said:

Reflection and transmission of light is not dictated by the frequency relative to the resonances.

Why is that so?  What is it dictated by?

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You have shorter wavelengths (below a micron) reflecting, but that includes visible light, which obviously is transmitted.

Cloud particles do not transmit visible light, but visible light misses the particles and light may refract around them, but light does not go directly through the particles (at least, I don't understand how it could) unless the "light" is a radio wave or some long wave.

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A blackbody radiates a spectrum that depends on its temperature.

Is not temperature a measure of vibration?  So, the hotter and the more vibrations due to heat energy, the whiter the light and higher the frequencies emitted.

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The sun, whose surface is at around 6000K, radiates a lot of visible light (peak at ~550 nm). The earth, at around 300K, radiated strongly in the infrared (peak at ~15 microns). It's not a true runaway effect. Thermodynamics prevents the earth from getting hotter than the sun's surface, and geometry limits the temperature even further.

Yeah that makes sense.  I guess I meant "runaway" effect in a different way.  Perhaps I should have said "cumulative".

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They are different kinds of waves. It mat be better to completely disregard sound waves completely.

Why are they different waves?  That's the heart of my question.

It makes a lot of sense to relate radio waves passing through just about anything while gamma rays are reflected by just about anything to sound waves where high tones are reflected by anything while low bass travels through anything.  Therefore, how are they different?

I can understand an atom being different than a wall of a room and having different properties of reflection and resonance, but the wave itself seems analogous.  So I suppose my question boils down to the properties of carbon, oxygen, and co2 that affect the reflection or absorption of light.  Specifically, what is it about co2 that will allow transmission of high-frequency visible light while absorbing (ie blocking) low-frequency IR?

A capacitor is the only thing I am aware of that will pass high frequencies while restricting low frequencies, but a capacitor has a discontinuity.  A continuous resistor or inductor will restrict highs while passing lows.  Am I to understand co2 is acting like a capacitor?  If so, why is water not?  If water acted as a capacitor, then it would pass visible light while blocking IR and radio waves.  Clearly that is not the case, so it's hard for me to understand why co2 would not act the same as water vapor, which cools, not warms.

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2 hours ago, swansont said:

They are different kinds of waves. It mat be better to completely disregard sound waves completely.

Just to reinforce this.

Sound waves affect molecules considered as point particles acting on the entire mass of the molecules as one to creating pressure waves (the sound).
The upper limit of such activity is a couple of hundred kilohertz in normal circumstances.

EM radiation affects the relationship between the individual parts of the molecule (ie its constituent atoms).
The best way to imagine this (It's too late tonight for me to draw the vibrational modes of molecules, you will have to look them up) is to imagine the  (much lighter) atoms as balls with mass connected by springs.
As such they perform various dances known as waggling, twisting, stretching and so on.

Edit there are also modes of absorbtion and emission due to interaction with the subatomic parts of molecules, particularly the electrons for the present purposes.

The frequencies of these motions are 10 to 100 gigahertz for IR and 100 to a few thousand gighertz for visible light.
Many order of magnitude different from the sound vibrations.

Edited by studiot

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24 minutes ago, studiot said:

Just to reinforce this.

Sound waves affect molecules considered as point particles acting on the entire mass of the molecules as one to creating pressure waves (the sound).
The upper limit of such activity is a couple of hundred kilohertz in normal circumstances.

EM radiation affects the relationship between the individual parts of the molecule (ie its constituent atoms).
The best way to imagine this (It's too late tonight for me to draw the vibrational modes of molecules, you will have to look them up) is to imagine the  (much lighter) atoms as balls with mass connected by springs.
As such they perform various dances known as waggling, twisting, stretching and so on.

The frequencies of these motions are 10 to 100 gigahertz for IR and 100 to a few thousand gighertz for visible light.
Many order of magnitude different from the sound vibrations.

Yes, I understand that the orders of magnitude are different, but the properties of waves are the same.  If we have a collection of atoms known as a wall, then the wall will have its specific resonance as well as the individual atoms within the wall (depending how tightly bound by the springs the atoms are and how much mass each atom contains).

Also, just to add that radio waves can go as low as 3 Hz or lower https://en.wikipedia.org/wiki/Extremely_low_frequency

Anyway, if we have an atom that resonates at 10^14 hz, then raising the frequency of the light to 10^15 should find the atom too heavy to vibrate that fast and therefore the atom will reflect the light.  If we lower the frequency to 10^13 hz, the light passes through the atom and does not reflect.

That is analogous to a tuning fork vibrating at 1000 hz and being too heavy to vibrate faster.  Or a wall constructed of drywall that is too heavy to vibrate at 1000hz, so it reflects as well.  If the frequency is reduced to 100 hz, the fork and wall pass the sound without much resistance.

The difference is the order of magnitude, but the principle is the same.

Microwave ovens, for instance, emit frequencies well below resonance of the individual atoms, but instead the waves act on the whole water molecule.  So water has several resonances: that of hydrogen, that of oxygen, that of the union of both H and O as a whole molecule, and that of a collection of molecules.

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Can you try to narrow the focus of your question?

Edited by studiot

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13 minutes ago, studiot said:

Can you try to narrow the focus of your question?

If IR light (10^13 hz) causes co2 to resonate, then why doesn't co2 reflect visible light (10^14 hz)?

The narrative I'm being asked to buy is that visible light passes unimpeded by co2, but then the same co2 absorbs the IR from the earth.  That is the mechanism by which the earth supposedly warms by co2 (light can get in, but it can't get out due to a lowering in frequency).

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Here is the absorption spectrum of Carbon Dioxide.

As you can see it does not absorb frequencies higher than the mid infra red.

Why should it reflect anything it is not dense enough, even if the individual molecules could do so. (It is only .04% of the atmosphere)

Note that water and glass are much denser and also transparent to ordinary light, but also absorb in the infra red.

That is why they make good greenhouses too.

Note most transparent plastics do not show the same level of IR absorbtion and make correspondingly poorer greenhouses.

The CO2 molecules absorb the IR radiation from Earth, but excited molecules soon drop back to lower states, re-emitting that radiation.
They do not reflect it back to Earth.

Edited by studiot

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4 minutes ago, studiot said:

As you can see it does not absorb frequencies higher than the mid infra red.

Right, it does not absorb, but it reflects visible, UV, and most certainly gamma rays.

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Why should it reflect anything it is not dense enough, even if the individual molecules could do so. (It is only .04% of the atmosphere)

Well, as small as they are, they do resonate and therefore they will reflect frequencies higher than their resonance.  Wiki says the density of an atomic nucleus is about 2.3×1017 kg/m3.  That seems quite dense.  https://en.wikipedia.org/wiki/Nuclear_density

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Note that water and glass are much denser and also transparent to ordinary light, but also absorb in the infra red.

Glass is a good example.  Glass will reflect UVB, which has a higher frequency than the UVA that glass passes.  So if glass passes frequencies lower than UVB, then why does it reflect IR?  It can only be due to the rigid atomic structure of the glass because one molecule of glass wouldn't have the support of its neighbors and therefore wouldn't have the rigidity to resist the IR frequencies.  IR must be affecting the molecular structure of the glass much like microwaves affect the whole molecule of water in food.  Once the frequency decreases beyond that point, they should once again find that glass is transparent.

Rigidity of a solid panel of co2 isn't a concern in the atmosphere.  As far as I know, co2 just floats around.

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That is why they make good greenhouses too.

I think they make good greenhouses because they pass visible light and stop convection in order to retain heat.  Glass is also somewhat reflective of IR and a poor thermal conductor.

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The CO2 molecules absorb the IR radiation from Earth, but excited molecules soon drop back to lower states, re-emitting that radiation.
They do not reflect it back to Earth.

Right, they radiate the IR in all directions, some of which come back to earth, but most head off into space.  That's another argument against the warming of a planet by co2.

Maybe pics will help.  Below are the two cases: a direct hit and a miss by solar radiation.

Direct-impact solar ray:

Solar ray misses CO2:

Btw, I love this forum software!

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43 minutes ago, BanterinBoson said:

but it reflects visible, UV

Then why can I see through it?

Even a high pressure tank of CO2 is transparent (apart from the tank itself)

43 minutes ago, BanterinBoson said:

Right, they radiate the IR in all directions, some of which come back to earth, but most head off into space.

For most CO2 molecules the preponderance of all directions is either back to Earth or towards another CO2 molecule above it or to the side of it.
I don't think most is lost to space, since most of the CO2 (a heavier gas than air) is beneath the upper atmosphere.

43 minutes ago, BanterinBoson said:

Glass is a good example.  Glass will reflect UVB, which has a higher frequency than the UVA that glass passes.  So if glass passes frequencies lower than UVB, then why does it reflect IR?  It can only be due to the rigid atomic structure of the glass because one molecule of glass wouldn't have the support of its neighbors and therefore wouldn't have the rigidity to resist the IR frequencies.  IR must be affecting the molecular structure of the glass much like microwaves affect the whole molecule of water in food.  Once the frequency decreases beyond that point, they should once again find that glass is transparent.

Does glass reflect IR?

Are you not getting mixed up between reflection and absorbtion?

Google has lots of IR absorbtion curves for glass.

Of course there is always a question of angle of incidence when it comes to reflection.

Edited by studiot

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10 minutes ago, studiot said:

Then why can I see through it?

Even a high pressure tank of CO2 is transparent (apart from the tank itself)

Water is transparent, yet we look up in the sky and, behold, we can't see through the clouds.  And from space, the clouds seem quite reflective.  It's hard to drive in fog and having bright lights only exacerbates the problem due to reflectivity.

The pressure in the tank gives the molecules different resonant properties from those of co2 floating loosely in the atmosphere.  Also, dry ice is less transparent than water ice indicating that co2 is less likely to pass visible light than water, probably because it's heavier and more resistant to those frequencies.

Heavier things will vibrate faster if fixed to stiffer springs.  If not, then they will resist faster vibrations.  CO2 in the atmosphere has no springs attached, therefore it's too heavy to pass visible light.  CO2 in a tank will be compressed and be bound by rigid springs giving it a faster resonance.

The variables in resonance are the spring constant and the mass.  Change either and the resonance changes.  So, comparing compressed gas and glass to co2 at .04% concentration in the atmosphere isn't a fair comparison.

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For most CO2 molecules the preponderance of all directions is either back to Earth or towards another CO2 molecule above it or to the side of it.
I don't think most is lost to space, since most of the CO2 (a heavier gas than air) is beneath the upper atmosphere.

If most are not lost to space, then most come back to earth, therefore co2 can effectively reflect IR back to earth.  But I can't see it that way.  There are many more paths to space than to earth by virtue of the earth's curvature; therefore, it's a net-loss of radiation.  That would be true for any point above earth, except in a valley between hills which are places where there are more paths to earth than sky.

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12 hours ago, BanterinBoson said:

Yes, I understand that the orders of magnitude are different, but the properties of waves are the same.

Sound is a longitudinal wave which requires a medium in order to be transmitted. Light is a transverse wave which does not.

10 hours ago, BanterinBoson said:

Right, it does not absorb, but it reflects visible, UV, and most certainly gamma rays.

UV and gammas will ionize molecules, or pass right through, a material. Very little reflection. (You can't make traditional mirrors for high-frequency light). But when you are talking about individual molecules, reflection isn't the right terminology (that's for bulk material). It's absorption and re-emission.

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Glass is a good example.

Glass is kind of an anomaly. Not many substances are transparent in the visible spectrum. Many substances, especially comprised of light atoms, are transparent in the x-ray/gamma range.

9 hours ago, BanterinBoson said:

The pressure in the tank gives the molecules different resonant properties from those of co2 floating loosely in the atmosphere.

No, not anything that matters here.

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Heavier things will vibrate faster if fixed to stiffer springs.  If not, then they will resist faster vibrations.  CO2 in the atmosphere has no springs attached, therefore it's too heavy to pass visible light.  CO2 in a tank will be compressed and be bound by rigid springs giving it a faster resonance.

Your mental model of how this works is flawed.

13 hours ago, BanterinBoson said:

Why is that so?  What is it dictated by?

For bulk materials, below any resonances, it's dictated by the index of refraction. It's purely classical.

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Cloud particles do not transmit visible light, but visible light misses the particles and light may refract around them, but light does not go directly through the particles (at least, I don't understand how it could) unless the "light" is a radio wave or some long wave.

Water tends to absorb light.

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Is not temperature a measure of vibration?  So, the hotter and the more vibrations due to heat energy, the whiter the light and higher the frequencies emitted.

Yes. The sun is hotter and gives visible light (along with a little UV and IR). The earth is cooler and emits IR.

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It makes a lot of sense to relate radio waves passing through just about anything while gamma rays are reflected by just about anything to sound waves where high tones are reflected by anything while low bass travels through anything.  Therefore, how are they different?

Gammas aren't reflected. They are absorbed or transmitted. (Radiation shielding would be so much easier if gammas were easily reflected)

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I can understand an atom being different than a wall of a room and having different properties of reflection and resonance, but the wave itself seems analogous.  So I suppose my question boils down to the properties of carbon, oxygen, and co2 that affect the reflection or absorption of light.  Specifically, what is it about co2 that will allow transmission of high-frequency v

The CO2 absorbs in the IR. It then re-rediates the light, some of which goes back to the earth. Thinking in terms of reflection is a mistake here. It's transmission vs absorption/re-emission (or scattering)

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7 hours ago, BanterinBoson said:

Water is transparent, yet we look up in the sky and, behold, we can't see through the clouds.  And from space, the clouds seem quite reflective.  It's hard to drive in fog and having bright lights only exacerbates the problem due to reflectivity.

The pressure in the tank gives the molecules different resonant properties from those of co2 floating loosely in the atmosphere.  Also, dry ice is less transparent than water ice indicating that co2 is less likely to pass visible light than water, probably because it's heavier and more resistant to those frequencies.

Heavier things will vibrate faster if fixed to stiffer springs.  If not, then they will resist faster vibrations.  CO2 in the atmosphere has no springs attached, therefore it's too heavy to pass visible light.  CO2 in a tank will be compressed and be bound by rigid springs giving it a faster resonance.

The variables in resonance are the spring constant and the mass.  Change either and the resonance changes.  So, comparing compressed gas and glass to co2 at .04% concentration in the atmosphere isn't a fair comparison.

Come on, are we talking sense or not?

I asked why we could see through a carbon dioxide atmosphere, either as dilute as the Earth's atmosphere or pure, if carbon dioxide reflects visible light.

Your answer apppears to be that we can't see a suspension of water droplets (clouds or fog) which is true but irrelevant.

One of the important characteristics of wave motion is that reflection occurs at a boundary.
The boundary is not at the atomic scale but water doplets provide an ideal boundary to disperse the light rays by reflection.

I didn't notice much solid carbon dioxide around when I got up this morning either, so what is the relevance of the transparancy or opacity of the stuff?

Edited by studiot

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7 hours ago, studiot said:

Come on, are we talking sense or not?

I asked why we could see through a carbon dioxide atmosphere, either as dilute as the Earth's atmosphere or pure, if carbon dioxide reflects visible light.

Your answer apppears to be that we can't see a suspension of water droplets (clouds or fog) which is true but irrelevant.

One of the important characteristics of wave motion is that reflection occurs at a boundary.
The boundary is not at the atomic scale but water doplets provide an ideal boundary to disperse the light rays by reflection.

I didn't notice much solid carbon dioxide around when I got up this morning either, so what is the relevance of the transparancy or opacity of the stuff?

I don't want to seem unappreciative of your help, but I don't see the relevancy to any of this talk of getting up this morning and not seeing co2 or any talk of glass or any talk of compressed gases.  I want to focus on one molecule of co2.  Will that one molecule increase or decrease the temp of the earth?  And why?  It's a theoretical, hypothetical thought experiment to help me learn.

Once we discover what the one molecule will do, then we can string them together and speculate from there.

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27 minutes ago, BanterinBoson said:

I don't want to seem unappreciative of your help, but I don't see the relevancy to any of this talk of getting up this morning and not seeing co2 or any talk of glass or any talk of compressed gases.  I want to focus on one molecule of co2.  Will that one molecule increase or decrease the temp of the earth?  And why?  It's a theoretical, hypothetical thought experiment to help me learn.

Once we discover what the one molecule will do, then we can string them together and speculate from there.

No need to reinvent the wheel, i.e. speculate. This is pretty well understood.

Each CO2 molecule can absorb infrared radiation in the absorption bands. It can then re-emit that photon back to the earth. It does not absorb any of the incoming visible light. When something absorbs more radiation than it emits, it heats up. It then radiates more energy, so you reach steady-state at a higher temperature.

Each new CO2 molecule will add to this effect, but it is not a linear result.

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7 hours ago, swansont said:

Sound is a longitudinal wave which requires a medium in order to be transmitted. Light is a transverse wave which does not.

I don't think that is true.  Light travels through an energy field... I believe it is called the higgs field, but I could be wrong on the nomenclature.  Anyway, light can't travel through nothingness and it's a wave traveling through a medium very much like sound.

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UV and gammas will ionize molecules, or pass right through, a material. Very little reflection. (You can't make traditional mirrors for high-frequency light). But when you are talking about individual molecules, reflection isn't the right terminology (that's for bulk material). It's absorption and re-emission.

That's interesting.  I haven't thought about that.  Gamma rays are on the order of 10^19 hz and so could act on the electrons themselves rather than the nucleus, which is far too massive to be affected by such frequencies.  That would explain the ionization by knocking electrons off atoms.  The penetration of gamma rays would be an artifact of the atom being mostly empty space.

I understand absorption and re-emission, but not what happens when the frequency falls below or rises above that of absorption.  That is what I'm trying to grasp.

If the frequency falls from 10^19 to 10^16 (UV region), then it no longer affects electrons, but the nucleus.  It's by virtue of mass and frequency.  Is that right?  So when the frequency falls even further to 10^13 (the IR region), then it no longer affects the nucleus, but bonded molecules by virtue of the mass and bond strength.  Is that right?

So the different frequencies affect differing collections of masses and springs.

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Glass is kind of an anomaly. Not many substances are transparent in the visible spectrum. Many substances, especially comprised of light atoms, are transparent in the x-ray/gamma range.

Yes, that makes sense now that you mention it.  I believe it is because of the above-stated... that atoms are mostly empty and the x, gamma rays travel through like a fishing net.  Once the frequencies fall from 10^19 to 10^16, then atoms become more opaque to the light.

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The CO2 absorbs in the IR. It then re-rediates the light, some of which goes back to the earth. Thinking in terms of reflection is a mistake here. It's transmission vs absorption/re-emission (or scattering)

So what does co2 do with visible light?

8 minutes ago, swansont said:

No need to reinvent the wheel, i.e. speculate. This is pretty well understood.

Each CO2 molecule can absorb infrared radiation in the absorption bands. It can then re-emit that photon back to the earth. It does not absorb any of the incoming visible light.

So co2 is completely transparent to visible light?  As if it were not there?

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1 minute ago, BanterinBoson said:

I don't think that is true.  Light travels through an energy field... I believe it is called the higgs field, but I could be wrong on the nomenclature.  Anyway, light can't travel through nothingness and it's a wave traveling through a medium very much like sound.

No, the vacuum is not a medium as the word is used in physics.

1 minute ago, BanterinBoson said:

That's interesting.  I haven't thought about that.  Gamma rays are on the order of 10^19 hz and so could act on the electrons themselves rather than the nucleus, which is far too massive to be affected by such frequencies.  That would explain the ionization by knocking electrons off atoms.  The penetration of gamma rays would be an artifact of the atom being mostly empty space.

Nuclei can and do absorb gamma rays.

1 minute ago, BanterinBoson said:

I understand absorption and re-emission, but not what happens when the frequency falls below or rises above that of absorption.  That is what I'm trying to grasp.

For photons below the ionization energy (visible and IR are below) if the atom or molecule does not have an absorption resonance at that energy, the photon is not absorbed. It passes by the atom.

1 minute ago, BanterinBoson said:

If the frequency falls from 10^19 to 10^16 (UV region), then it no longer affects electrons, but the nucleus.  It's by virtue of mass and frequency.  Is that right?  So when the frequency falls even further to 10^13 (the IR region), then it no longer affects the nucleus, but bonded molecules by virtue of the mass and bond strength.  Is that right?

Below ionization, the absorption causes the electron to go to an excited state. The nuclear structure is not involved.

1 minute ago, BanterinBoson said:

So what does co2 do with visible light?

Nothing. It's transparent — light passes through it. (Though the light might change direction via Rayleigh scattering, but this is AFAIK unimportant for this discussion)

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5 minutes ago, swansont said:

No, the vacuum is not a medium as the word is used in physics.

A wave makes no sense without a medium.

"The Higgs field is an energy field that is thought to exist everywhere in the universe

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Nuclei can and do absorb gamma rays.

Is there evidence of that?  I know they can emit gamma rays, but I couldn't find anything on google about absorption of them.

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For photons below the ionization energy (visible and IR are below) if the atom or molecule does not have an absorption resonance at that energy, the photon is not absorbed. It passes by the atom.

Passes by?  Or through?

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Below ionization, the absorption causes the electron to go to an excited state. The nuclear structure is not involved.

"Vibrational excitation can occur in conjunction with electronic excitation in the ultraviolet-visible region. The combined excitation is known as a vibronic transition, giving vibrational fine structure to electronic transitions, particularly for molecules in the gas state." https://en.wikipedia.org/wiki/Molecular_vibration

As I was saying, as the frequency falls past 10^16 (UV), other mass combinations come into play.  If 10^19 affects electrons, then 10^16 affects things bigger than electrons.

"the typical frequencies of molecular vibrations range from less than 10^13 to approximately 10^14 Hz," so that's visible and IR with the transition occurring in the UV band.

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Nothing. It's transparent — light passes through it. (Though the light might change direction via Rayleigh scattering, but this is AFAIK unimportant for this discussion)

I don't understand how visible light (10^14) can pass straight through it.  Can you explain further?

CO2 is composed of C and O which are composed of E, P, N, and Quarks.

CO2 can resonate as a molecule (probably in the microwave band, similar to water ie 10^9).  C can resonate as an atom bonded to 2 other atoms as springs.  O can resonate bonded to the C.  The Electrons can resonate as well as the nuclei and Protons, Neutrons, and I suppose the Quarks as well (all irrelevant to visible light).

So it seems the focus is narrowed to what the bonded C and O will do with 10^14 light.  If 10^14 is too fast to vibrate (absorb into) the C or O, but not fast enough to vibrate subatomics, then how can it pass through?  Because "passing through" is defined as "using it as a medium for which to pass through".

On the other hand, IR light (10^13) will absorb into (resonate) either C or O or both, therefore the mass/spring combination is such that 10^14 visible light will be impeded by the same mass/spring combination that absorbs 10^13.  What is the mechanism for the transference of the 10^14 wave through the medium which is the co2 molecule?  That little bit of understanding is the holy grail to all of this discussion.

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The reason is best found in the wave theory of light and is the same reason we cannot see atoms or electrons with a light microscope.
The wavelength  of visible light is just too long by several orders of magnitude to resolve distance this small.

The radius of an electron is of the order of 10-16 metres
The radius of a carbon or oxygen atom is of the order of 10-10 metres

The wavelength of visible light is of the order of 10-7 metres

Since you like sound analogies, it is the same reason low frequency sound can diffract round objects, whilst higher frequencies sounds are blocked.

In your sketch you have shown the wave being much smaller than the CO2 molecule.
In fact it is the other way round. A visible light wave is ten thousand times larger.

Edited by studiot

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13 minutes ago, BanterinBoson said:

A wave makes no sense without a medium.

"The Higgs field is an energy field that is thought to exist everywhere in the universe

Is there evidence of that?  I know they can emit gamma rays, but I couldn't find anything on google about absorption of them.

Passes by?  Or through?

"Vibrational excitation can occur in conjunction with electronic excitation in the ultraviolet-visible region. The combined excitation is known as a vibronic transition, giving vibrational fine structure to electronic transitions, particularly for molecules in the gas state." https://en.wikipedia.org/wiki/Molecular_vibration

As I was saying, as the frequency falls past 10^16 (UV), other mass combinations come into play.  If 10^19 affects electrons, then 10^16 affects things bigger than electrons.

"the typical frequencies of molecular vibrations range from less than 10^13 to approximately 10^14 Hz," so that's visible and IR with the transition occurring in the UV band.

I don't understand how visible light (10^14) can pass straight through it.  Can you explain further?

CO2 is composed of C and O which are composed of E, P, N, and Quarks.

CO2 can resonate as a molecule (probably in the microwave band, similar to water ie 10^9).  C can resonate as an atom bonded to 2 other atoms as springs.  O can resonate bonded to the C.  The Electrons can resonate as well as the nuclei and Protons, Neutrons, and I suppose the Quarks as well (all irrelevant to visible light).

So it seems the focus is narrowed to what the bonded C and O will do with 10^14 light.  If 10^14 is too fast to vibrate (absorb into) the C or O, but not fast enough to vibrate subatomics, then how can it pass through?  Because "passing through" is defined as "using it as a medium for which to pass through".

On the other hand, IR light (10^13) will absorb into (resonate) either C or O or both, therefore the mass/spring combination is such that 10^14 visible light will be impeded by the same mass/spring combination that absorbs 10^13.  What is the mechanism for the transference of the 10^14 wave through the medium which is the co2 molecule?  That little bit of understanding is the holy grail to all of this discussion.

"A wave makes no sense without a medium.  "
The universe isn't obliged to follow your ideas of what "makes sense".

"Is there evidence of that?  I know they can emit gamma rays, but I couldn't find anything on google about absorption of them."

""the typical frequencies of molecular vibrations range from less than 10^13 to approximately 10^14 Hz," so that's visible and IR with the transition occurring in the UV band."

10^14 Hz is rather high f or  a vibrational transition.

Visible light has frequencies that are higher than the vibrations of CO2.

That's why the light goes through it.

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1 hour ago, studiot said:

The reason is best found in the wave theory of light and is the same reason we cannot see atoms or electrons with a light microscope.
The wavelength  of visible light is just too long by several orders of magnitude to resolve distance this small.

The radius of an electron is of the order of 10-16 metres
The radius of a carbon or oxygen atom is of the order of 10-10 metres

The wavelength of visible light is of the order of 10-7 metres

Since you like sound analogies, it is the same reason low frequency sound can diffract round objects, whilst higher frequencies sounds are blocked.

In your sketch you have shown the wave being much smaller than the CO2 molecule.
In fact it is the other way round. A visible light wave is ten thousand times larger.

I don't know what prompted the idea that I'm being unappreciative since I specially said I do not want to seem that way.  I definitely appreciate the opportunity to learn, but you have to understand that I can't just accept things by faith or on someone's authority... I have to understand the rationale.

If the argument is that the light is too big to "see" (reflect off of) the co2 molecule, then it is bigger in the IR band (wavelength longer) and so how can it be absorbed?  I'm still not understanding the rationale.

What is the cross-sectional width of a photon?  It can't be the wavelength because that is a measure along the x-axis (axis of travel), which is perpendicular to the cross-section.  The measure of the y-axis is the amplitude of the wave, but is that a physical distance?  This page says photons do not have amplitudes https://www.quora.com/If-you-have-two-photons-with-equal-wavelengths-and-one-has-a-bigger-amplitude-what-would-that-mean-would-the-photon-have-more-energy  So then what is the width of a photon?  And why would it be bigger than an electron or quark?

If a photon is smaller than an electron, then my drawing has the ray of light many times too big.  I'm sorry if I seem argumentative because it's not my intention to drive you guys crazy, but simply that I haven't understood why the width of a photon would be = to its wavelength.  And I haven't understood why co2 would absorb (resonate) in the 10^13 band while being a zero-resistance medium for the transference of 10^14 visible light.

I learned that co2 can be a near-zero-resistance medium for the transference of gamma rays (10^19), which I previously thought wasn't possible.  So that's evidence that I am not seeking to be obtuse.

Surely talking to me isn't that bad.  What else would you be utilizing this forum for if it weren't for me?   Echo-chambers are boring and you need someone to banter with to make it fun.

2 hours ago, John Cuthber said:

"A wave makes no sense without a medium.  "
The universe isn't obliged to follow your ideas of what "makes sense".

I'm not arguing with the universe; I'm arguing with someone's claim about the universe.

1) There is sufficient evidence for the claim of a ubiquitous energy field medium.

2) The idea of a wave makes no sense outside of a medium through which the wave can propagate.

Claiming a wave can propagate through absolute nothingness is a self-defeating statement.

Quote

"Is there evidence of that?  I know they can emit gamma rays, but I couldn't find anything on google about absorption of them."

"This means that nuclear resonance (emission and absorption of the same gamma ray by identical nuclei) is unobservable with free nuclei, because the shift in energy is too great and the emission and absorption spectra have no significant overlap."

So far there is evidence that if the nuclei is held sufficiently strong, then it can be forced to accept gamma radiation.

"Just as a gun recoils when a bullet is fired, conservation of momentum requires a nucleus (such as in a gas) to recoil during emission or absorption of a gamma ray."

That seems a lot like a "bounce" to me.  It also seems like the thing that would be doing the absorption would be subatomic particles and not the whole nucleus because the whole nucleus has too much mass to be affected by such energetic vibrations while quarks are sufficiently small to accommodate them.

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""the typical frequencies of molecular vibrations range from less than 10^13 to approximately 10^14 Hz," so that's visible and IR with the transition occurring in the UV band."

10^14 Hz is rather high f or  a vibrational transition.

You trimmed the part of my quote that was relevant:

"Vibrational excitation can occur in conjunction with electronic excitation in the ultraviolet-visible region. The combined excitation is known as a vibronic transition, giving vibrational fine structure to electronic transitions, particularly for molecules in the gas state." https://en.wikipedia.org/wiki/Molecular_vibration

10^14 is not the transition; it's even higher in the 10^16 range (UV).

Quote

Visible light has frequencies that are higher than the vibrations of CO2.

That's why the light goes through it.

That doesn't explain anything, but is merely a statement.

If visible light has frequencies higher than the vibrations of co2, then they should bounce.  If that is not true, then why is it not true?

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13 hours ago, BanterinBoson said:

I don't know what prompted the idea that I'm being unappreciative since I specially said I do not want to seem that way.  I definitely appreciate the opportunity to learn, but you have to understand that I can't just accept things by faith or on someone's authority... I have to understand the rationale.

If the argument is that the light is too big to "see" (reflect off of) the co2 molecule, then it is bigger in the IR band (wavelength longer) and so how can it be absorbed?  I'm still not understanding the rationale.

What is the cross-sectional width of a photon?  It can't be the wavelength because that is a measure along the x-axis (axis of travel), which is perpendicular to the cross-section.  The measure of the y-axis is the amplitude of the wave, but is that a physical distance?  This page says photons do not have amplitudes https://www.quora.com/If-you-have-two-photons-with-equal-wavelengths-and-one-has-a-bigger-amplitude-what-would-that-mean-would-the-photon-have-more-energy  So then what is the width of a photon?  And why would it be bigger than an electron or quark?

If a photon is smaller than an electron, then my drawing has the ray of light many times too big.  I'm sorry if I seem argumentative because it's not my intention to drive you guys crazy, but simply that I haven't understood why the width of a photon would be = to its wavelength.  And I haven't understood why co2 would absorb (resonate) in the 10^13 band while being a zero-resistance medium for the transference of 10^14 visible light.

I learned that co2 can be a near-zero-resistance medium for the transference of gamma rays (10^19), which I previously thought wasn't possible.  So that's evidence that I am not seeking to be obtuse.

Surely talking to me isn't that bad.  What else would you be utilizing this forum for if it weren't for me?   Echo-chambers are boring and you need someone to banter with to make it fun.

No talking to you isn't all that bad, in fact you seem to be progressing towards proper discussion.

I try very hard to read and address the contents of  the posts of others inlcuding yourself.

Which I hope you will agree is frustratingly irrelevant.

But this last time you seem to be actually responding to what I wrote, so let us start again, without preconceptions.

I think you seek a mechanism to explain how EM waves can pass by, or be absorbed by, or reflected by objects in their path.
Ideally this mechanism should offer the conditions when each possibility will happen and concur with the results of real world observations.

Please confirm that this is the case because I am trying to explain this to you.

I also said that the easist explanation for this is wave theory. By this I meant classical wave theory, which in my opinion is sufficient to discuss the matter.

My starting premises, based on real world observations, are:

The structures of aerials for TV and radio broadcasts allow the broadcast waves to pass by and are sometimes reflected by these structures, but only extract the energy (ie absob at least some waves) for certain specific wavelengths which they are said to be tuned to.

For simplicity I will take the simplest possible aerial, a length of conductor such as a straight aluminium or copper tube or wire.

It is an observational fact that the wavelengths this aerial is tuned to depend upon its length in much the same way as and organ pipe can be tuned to a particular sound wavelength.

Please confirm that you consider there observational facts.

I am saying it like this because you chose to simply quibble by ignoring various helpful comments I made to enliven and enrich the discussion.

I propose the development and extension of this theory to atmospheric carbon dioxide molecules.

Working together this can be achieved.

Continuing to work in seoparate compartments, it is unlikely the discussion can progress.

The choice is yours.

Edited by studiot

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12 hours ago, studiot said:

No talking to you isn't all that bad, in fact you seem to be progressing towards proper discussion.

I try very hard to read and address the contents of  the posts of others inlcuding yourself.

Which I hope you will agree is frustratingly irrelevant.

The difficulties could be an artifact of text-based communication.  It's not my intention to cause frustrations.

Quote

I think you seek a mechanism to explain how EM waves can pass by, or be absorbed by, or reflected by objects in their path.

I seek a mechanism to describe the propagation of visible light (10^14 hz) through the co2 molecule.

I have learned that gamma rays (10^19 hz) can propagate through the co2 molecule using the subatomic particles as a medium, but I do not believe 10^14 hz is sufficient to utilize subatomics in order to propagate through the molecule (since 10^16 was identified as the transition).

I have noted that IR (10^13 hz) can vibrate atoms within the co2 molecule and therefore lower-than-resonance frequencies have a mechanism for which to pass.

Microwaves (10^10) can vibrate the entire co2 molecule utilizing the dipole moment and therefore lower-than-resonance frequencies have a mechanism for which to pass.

But visible light seems to be of the perfect frequency to neither vibrate the subatomics nor the atomics nor whole molecules and therefore there is no medium for propagation.

Quote

The structures of aerials for TV and radio broadcasts allow the broadcast waves to pass by and are sometimes reflected by these structures, but only extract the energy (ie absob at least some waves) for certain specific wavelengths which they are said to be tuned to.

For simplicity I will take the simplest possible aerial, a length of conductor such as a straight aluminium or copper tube or wire.

It is an observational fact that the wavelengths this aerial is tuned to depend upon its length in much the same way as and organ pipe can be tuned to a particular sound wavelength.

Please confirm that you consider there observational facts.

Yes, those appear to be observational facts.

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On 9/13/2017 at 7:08 PM, BanterinBoson said:

What is the cross-sectional width of a photon?  It can't be the wavelength because that is a measure along the x-axis (axis of travel), which is perpendicular to the cross-section.  The measure of the y-axis is the amplitude of the wave, but is that a physical distance?  This page says photons do not have amplitudes https://www.quora.com/If-you-have-two-photons-with-equal-wavelengths-and-one-has-a-bigger-amplitude-what-would-that-mean-would-the-photon-have-more-energy  So then what is the width of a photon?  And why would it be bigger than an electron or quark?

Cross-sectional width of a photon is not well-defined. It's not a rigid sphere. Modeling it as one won't work.

There are instances where the wavelength does indicate a "size" of a photon. e.g. trying to send a photon through a waveguide of some sort. The cutoff is related to the wavelength. Photons with long wavelengths can't be transmitted. Photons around the size of the waveguide are attenuated. Photons with shorter wavelengths are transmitted.

For atomic interactions, if the photon is at a resonant frequency, the atom "looks big" and there is a significant interaction probability.  If not, then it "looks" small. The probabilities do not, in general have a nice relationship with wavelength.

Quote

If a photon is smaller than an electron, then my drawing has the ray of light many times too big.  I'm sorry if I seem argumentative because it's not my intention to drive you guys crazy, but simply that I haven't understood why the width of a photon would be = to its wavelength.  And I haven't understood why co2 would absorb (resonate) in the 10^13 band while being a zero-resistance medium for the transference of 10^14 visible light.

Electrons are point particles. But again, this is QM and physical size is not necessarily a property that has any bearing on interactions.

CO2 does not have an energy structure that allows it to absorb a photon other than at certain allowed values.

Quote

I'm not arguing with the universe; I'm arguing with someone's claim about the universe.

1) There is sufficient evidence for the claim of a ubiquitous energy field medium.

2) The idea of a wave makes no sense outside of a medium through which the wave can propagate.

Claiming a wave can propagate through absolute nothingness is a self-defeating statement.

This was settled over 100 years ago. A medium was looked for but not found, and was shown to not exist. Special relativity confirmed that no medium was needed, nor was a physical medium even possible.

Nobody is claiming "absolute nothingness" but "medium" has a specific meaning with regard to optics. The Higgs field is not a medium in that sense.

Quote

"This means that nuclear resonance (emission and absorption of the same gamma ray by identical nuclei) is unobservable with free nuclei, because the shift in energy is too great and the emission and absorption spectra have no significant overlap."

So far there is evidence that if the nuclei is held sufficiently strong, then it can be forced to accept gamma radiation.

Mössbauer spectroscopy is a specific instance, where the recoil is minimized so there is no Doppler shift, which would move you off-resonance. That's what the passage about unobservable with free nuclei means.

But gamma absorption was famously used in the Pound-Rebka experiment (and followups) which confirmed part of general relativity (the first stages of which, aka special relativity, confirmed that light requires no medium)

Quote

You trimmed the part of my quote that was relevant:

"Vibrational excitation can occur in conjunction with electronic excitation in the ultraviolet-visible region. The combined excitation is known as a vibronic transition, giving vibrational fine structure to electronic transitions, particularly for molecules in the gas state." https://en.wikipedia.org/wiki/Molecular_vibration

10^14 is not the transition; it's even higher in the 10^16 range (UV).

That doesn't explain anything, but is merely a statement.

If visible light has frequencies higher than the vibrations of co2, then they should bounce.  If that is not true, then why is it not true?

Because light doesn't behave like sound.

8 hours ago, BanterinBoson said:

I seek a mechanism to describe the propagation of visible light (10^14 hz) through the co2 molecule.

The CO2 molecule is unable to absorb that light. There are no allowable transitions at those frequencies.

Quote

But visible light seems to be of the perfect frequency to neither vibrate the subatomics nor the atomics nor whole molecules and therefore there is no medium for propagation.

If there are no molecules at all, the light propagates just fine. Light does not need the molecules in order to propagate. Free space has a value for magnetic permeability and electrical permittivity. The presence of molecules modifies that value and can change the overall propagation speed of the light (proportional to the index of refraction)

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8 hours ago, swansont said:

Cross-sectional width of a photon is not well-defined. It's not a rigid sphere. Modeling it as one won't work.

I agree and I think it is because the photon is a wave, not a particle.

What do you think about the analogy of the directionality of sound increasing as frequency increases to that of EMWs?  Are radio waves more omnidirectional than higher frequencies of light?  And the electron is so super-directional that we describe it as a particle?  I'm trying to equate "particle-ness" to directionality.

Quote

There are instances where the wavelength does indicate a "size" of a photon. e.g. trying to send a photon through a waveguide of some sort. The cutoff is related to the wavelength. Photons with long wavelengths can't be transmitted. Photons around the size of the waveguide are attenuated. Photons with shorter wavelengths are transmitted.

Isn't that also true of sound through a tube?

Quote

For atomic interactions, if the photon is at a resonant frequency, the atom "looks big" and there is a significant interaction probability.  If not, then it "looks" small. The probabilities do not, in general have a nice relationship with wavelength.

I think I can see what you're saying.  The atom "looks big" if the photon is at resonant frequency because the photon interacts significantly and it "looks small" if the photon is at a frequency higher than resonance because there is little interaction, but then I'm having trouble seeing it "looking small" if the photon is below resonance because lower frequencies will interact.

For instance, holding a weight suspended by a rubber band in your hand and then slowly move your hand up and down, the weight should follow the movement of your hand.  As you move faster, you'll find the resonant frequency where the weight and your hand move in opposite directions.  As you move your hand faster than resonance, the weight will not move.

If the weight is a co2 molecule and the motion of your hand is the light, then the light is able to pass up to the point of resonance.  Beyond that, the weight does not move and the light is stopped from propagating further (until the frequency is so high that it can propagate through the constituents that form the weight).

Quote

Electrons are point particles. But again, this is QM and physical size is not necessarily a property that has any bearing on interactions.

Don Lincoln says here (at 1:30) that electrons and top quarks have no size:

Quote

CO2 does not have an energy structure that allows it to absorb a photon other than at certain allowed values.

I can see that, but what happens outside those values is what I'm having trouble seeing.

Quote

This was settled over 100 years ago. A medium was looked for but not found, and was shown to not exist. Special relativity confirmed that no medium was needed, nor was a physical medium even possible.

I'm not sure light having a constant speed regardless of the speed of the observer necessarily disproves a medium.  I can see how it would seem so, but space is actually spacetime in 4 dimensions and having zero velocity with respect to the medium necessitates maximum velocity through time and maximum velocity through space necessitates zero velocity through time, so it seems not so easy to disprove an aether with relativity due to the extra time variable and the fact that, from the point of view of light, there is no time or space.  How do we use something that sees no space or time to disprove something that exists within spacetime?

Quote

Nobody is claiming "absolute nothingness" but "medium" has a specific meaning with regard to optics. The Higgs field is not a medium in that sense.

Maybe it's not the Higgs field, but it has to be something:

Quote

Mössbauer spectroscopy is a specific instance, where the recoil is minimized so there is no Doppler shift, which would move you off-resonance. That's what the passage about unobservable with free nuclei means.

But gamma absorption was famously used in the Pound-Rebka experiment (and followups) which confirmed part of general relativity (the first stages of which, aka special relativity, confirmed that light requires no medium)

Yes I concede that gamma rays travel through nuclei and it was an error on my part not seeing the mechanism of utilizing the subatomics as a medium for propagation.  That realization was indeed insightful.

Quote

If there are no molecules at all, the light propagates just fine. Light does not need the molecules in order to propagate. Free space has a value for magnetic permeability and electrical permittivity. The presence of molecules modifies that value and can change the overall propagation speed of the light (proportional to the index of refraction)

If you're saying that light will pass straight through a molecule if there are no resonances, then how does light reflect off of chlorophyll?  We know that grass is green because it absorbs all frequencies of visible light except green, therefore green is reflected and that is what we see.  All other frequencies are absorbed and converted to sugar and IR radiation which is re-emitted.  So if green light reflects off of chlorophyll, then why can't light reflect off of co2?