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Vibrational Frequency Co2 Global Warming


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#1 BanterinBoson

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Posted 12 September 2017 - 02:12 PM

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:

 

59b83310bd27a_atmospheric_transmission(1

 

Original graph found here https://commons.wiki...ransmission.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.   :confused:  Help!



#2 exchemist

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Posted 13 September 2017 - 01:08 AM

 

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:

 

59b83310bd27a_atmospheric_transmission(1

 

Original graph found here https://commons.wiki...ransmission.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.   :confused:  Help!

 

I'm not an expert on atmospheric chemistry but I may be able to help a little. 

 

Light of the wrong frequency to be absorbed by a substance will pass through without absorption. In other words, the substance is transparent to radiation of that frequency. CO2 absorbs in the IR, but not in the visible. 

 

When visible light hits the ground it is absorbed, not reflected (the ground, self evidently, being neither transparent nor shiny like a mirror.). This absorption leads to what is actually quite a complicated chain of events by which the initial excitation of electrons by the light, in the materials of which the ground is composed, is converted to vibrations and rotations of the molecules (and giant structures, in the case of minerals). Molecular vibrations and rotations (and translations) are of course what we know as heat.

 

These vibrationally and rotationally excited substances can radiate EM radiation in the IR and microwave regions of the spectrum and thereby drop down to lower energy states.

 

This is the spectroscopist's way of saying that the ground absorbs light radiation, is thereby warmed, and then gives off heat radiation. 

 

So what you are missing is that the incident higher frequency radiation passes through the CO2 in the air, whereas what is re-emitted by the warmed ground is lower frequency radiation, which is absorbed by CO2. 

 

Later footnote: I realise I did not address your attempted analogy with sound waves. You can only push such analogies up to a point, as the absorption and emission of EM radiation is governed by quantum theory, not classical wave theory. In the case of visible light, the "particle" absorbing is an electron in the molecule. In the case of IR, it is the changing electric dipole caused by a stretching, bending or twisting of the molecule, i.e. motion of whole atoms or sub-assemblies of atoms (which is why it occurs at longer wavelengths, due to the greater mass of the bodies in motion). We can get more deeply into all that if you like, but it gets quite involved if you do not already have some knowledge of quantum theory.


Edited by exchemist, 13 September 2017 - 01:47 AM.


#3 exchemist

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Posted 13 September 2017 - 02:28 AM

Thanks to whoever has removed the off-topic contribution :)

 

If BanterinBoson comes back, we might get good physical science thread here. 


Edited by exchemist, 13 September 2017 - 02:29 AM.


#4 exchemist

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Posted 13 September 2017 - 08:11 AM

In case anyone is interested, I've found a simplified explanation of how the vibration of the CO2 molecule absorbs and re-emits in the IR: http://butane.chem.u...ntal/L13/2.html

 

There has to be a change in dipole moment for absorption or emission to occur. The main constituents of the atmosphere (O2 and N2) are diatomic molecules with a symmetrical charge distribution. So when the bond between the atoms in these molecules is stretched or compressed, no change in dipole moment takes place. They are therefore inactive in the IR: i.e.  transparent to IR radiation.

 

With CO2 however, there is a bit more electron density on the 2 O atoms on the ends than on the central C atom, a bit like this: -  +  -. This leads to no net static dipole on the molecule, due to its O=C=O symmetry, but when it is stretched or bends, there is a change in dipole moment. So these vibrations can be excited by passing IR radiation of the right frequency. By the same token, if the molecule is already vibrating, it can emit an IR photon of the appropriate frequency and reduce its vibration. 

 

CO2 has no pure rotational spectrum, since the absence of a static dipole means there is no change in dipole moment as it spins. However this ceases to be the case if it is vibrating at the same time, so one does see rotational fine structure in the vibrational (IR) spectrum of CO2. 


Edited by exchemist, 13 September 2017 - 08:19 AM.


#5 BanterinBoson

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Posted 13 September 2017 - 07:06 PM

Thanks to whoever has removed the off-topic contribution :)

 

If BanterinBoson comes back, we might get good physical science thread here. 

 

I'm sorry, I wasn't aware I had replies.  Let me get caught up and we'll have a good conversation ;)



#6 BanterinBoson

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Posted 13 September 2017 - 08:00 PM

I'm not an expert on atmospheric chemistry but I may be able to help a little. 

 

You know more than I :)

 

 

Light of the wrong frequency to be absorbed by a substance will pass through without absorption. In other words, the substance is transparent to radiation of that frequency. CO2 absorbs in the IR, but not in the visible. 

 

I don't understand by what mechanism the co2 molecule can act as a medium for the transference of 10^14 frequency light while being a resonator at the frequency of 10^13.  Do you see my problem?  If a thing resonates at 10^13, it should resist vibrations at 10^14 by virtue of the mass/spring combination that causes the resonance at 10^13.

 

Btw, if it's ok with you, I strongly prefer talking in terms of frequency rather than wavelength if at all possible because that is the way I've come to understand sound and waves in general.

 

 

When visible light hits the ground it is absorbed, not reflected (the ground, self evidently, being neither transparent nor shiny like a mirror.). This absorption leads to what is actually quite a complicated chain of events by which the initial excitation of electrons by the light, in the materials of which the ground is composed, is converted to vibrations and rotations of the molecules (and giant structures, in the case of minerals). Molecular vibrations and rotations (and translations) are of course what we know as heat.

 

When light hits the ground, some of it is reflected otherwise we wouldn't see the ground ;)  But yes, I understand that the ground absorbs the light and grass absorbs all but the green part of light; hence why it is green.  Temperature is a measure of vibrations (kinetic energy), is that right?  And so the hotter a body becomes, the higher the frequency of the light emitted.

 

 

These vibrationally and rotationally excited substances can radiate EM radiation in the IR and microwave regions of the spectrum and thereby drop down to lower energy states.

 

Isn't it true that they can emit frequencies of EMR correlated to their temperatures?  It doesn't always have to be IR and MW radiation, does it?

 

 

So what you are missing is that the incident higher frequency radiation passes through the CO2 in the air, whereas what is re-emitted by the warmed ground is lower frequency radiation, which is absorbed by CO2. 

 

Yes, millions of webpages says that.  It's the "why" that I am after ;)  My problem is that if co2 resonates at IR frequency, then it should not transfer visible light.  Therefore the addition of co2 would be a net-reduction of heat on the ground by virtue of resisting the passage of visible light... and by a second mechanism which is that the re-emittance of IR will have more paths leading to space than to earth due to the curvature of the earth.

 

 

 

Later footnote: I realise I did not address your attempted analogy with sound waves. You can only push such analogies up to a point, as the absorption and emission of EM radiation is governed by quantum theory, not classical wave theory. 

 

I'm far from an expert on quantum theory, but isn't that just the probability of finding a particle in any one location?  So it's a probability distribution wave and not an actual wave through a real medium like light and sound.  I understand that light travels through an energy field medium (perhaps the higgs field) because a wave without a medium doesn't make sense.

 

So all waves are propagations through mediums and will have relatable and analogous properties.

 

 

 

In the case of visible light, the "particle" absorbing is an electron in the molecule.  

 

I thought it was resonance causing the absorption, which is a function of mass/spring combinations of the whole atom or molecule(s).  For instance, gamma rays (10^19) will ionize atoms by acting on the electrons due to their extremely high frequencies, but UV (10^16) is too slow (vibrations or too slow) to act on electrons and will begin to transition to acting on the whole atom.

 

"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...cular_vibration

 

 

 

In the case of IR, it is the changing electric dipole caused by a stretching, bending or twisting of the molecule, i.e. motion of whole atoms or sub-assemblies of atoms (which is why it occurs at longer wavelengths, due to the greater mass of the bodies in motion).

 

I can see that.  IR and MW are probably the transitions from atomic resonances to molecular and bonded-molecular resonances.  By "bonded" I mean like a pane of glass... rigidly bonded and held tightly in position.

 

"The frequency of the periodic motion is known as a vibration frequency, and the typical frequencies of molecular vibrations range from less than 1013 to approximately 1014 Hz"  https://en.wikipedia...cular_vibration

 

10^14 is visible and 10^13 is IR.

 

A heavy atom bound tightly should resonate faster than in free space and therefore would absorb different frequencies of light, but gases in the atmosphere I'm assuming are free space except for the bonds between the C and O.

 

So if we have O=C=O, then the C in the middle will have different resonant properties than a C by itself.  It must be that the collection of C and O just happen to resonate in the 10^13 range by virtue of the mass/spring combination and if true, then it should also resist vibrations in the 10^14 range which will prevent passage of visible light.

 

The whole O=C=O molecule should resonate in the MW range (maybe 10^9) like water.

 

 

 

We can get more deeply into all that if you like, but it gets quite involved if you do not already have some knowledge of quantum theory.

 

I'd like to learn as much as I can.



#7 BanterinBoson

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Posted 13 September 2017 - 08:38 PM

In case anyone is interested, I've found a simplified explanation of how the vibration of the CO2 molecule absorbs and re-emits in the IR: http://butane.chem.u...ntal/L13/2.html

 

There has to be a change in dipole moment for absorption or emission to occur. The main constituents of the atmosphere (O2 and N2) are diatomic molecules with a symmetrical charge distribution. So when the bond between the atoms in these molecules is stretched or compressed, no change in dipole moment takes place. They are therefore inactive in the IR: i.e.  transparent to IR radiation.

 

With CO2 however, there is a bit more electron density on the 2 O atoms on the ends than on the central C atom, a bit like this: -  +  -. This leads to no net static dipole on the molecule, due to its O=C=O symmetry, but when it is stretched or bends, there is a change in dipole moment. So these vibrations can be excited by passing IR radiation of the right frequency. By the same token, if the molecule is already vibrating, it can emit an IR photon of the appropriate frequency and reduce its vibration. 

 

CO2 has no pure rotational spectrum, since the absence of a static dipole means there is no change in dipole moment as it spins. However this ceases to be the case if it is vibrating at the same time, so one does see rotational fine structure in the vibrational (IR) spectrum of CO2. 

 

OK, that is confusing.  So can we infer that it's impossible to heat O2 and N2 because there is no dipole moment?  I'm lost because on one hand the article is saying that the differential of charges is what causes interaction with the EMR, but if there are no differential in charges then the molecule can never get hot by EMR.  Therefore O2 and N2 can never absorb energy and will forever be cold.  That doesn't seem right.



#8 exchemist

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Posted 14 September 2017 - 01:26 AM

OK, that is confusing.  So can we infer that it's impossible to heat O2 and N2 because there is no dipole moment?  I'm lost because on one hand the article is saying that the differential of charges is what causes interaction with the EMR, but if there are no differential in charges then the molecule can never get hot by EMR.  Therefore O2 and N2 can never absorb energy and will forever be cold.  That doesn't seem right.

No no no, hold your horses! What you can infer is that you cannot heat oxygen and nitrogen by means of IR or microwave radiation. That's all. 

 

You most certainly can heat them by higher frequency (UV) radiation, as this will stimulate electronic transitions in the molecule - which can then cascade down (via various interesting but complicated processes) into vibrational and rotational excitation.

 

And of course there are other means of heating a gas than radiation. The 3 means of heat transfer that we all learn at school are: conduction, convection and radiation, yes?     


Edited by exchemist, 14 September 2017 - 01:26 AM.


#9 OceanBreeze

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Posted 14 September 2017 - 01:57 AM

 

 

.  It must be that the collection of C and O just happen to resonate in the 10^13 range by virtue of the mass/spring combination and if true, then it should also resist vibrations in the 10^14 range which will prevent passage of visible light.

 

 

 

 

Why do you think that? (the bolded part)

 

As an analog, we can use the simple form of the differential equation for an RLC circuit:

 

[math] L\frac { di }{ dt } +Ri+\frac { q }{ C } =E[/math]

 

Where E is some constant voltage

L is the inductance and C the capacitance

 

This exact same relationship also applies to mechanical systems, and extends also to molecules such as CO2 where mass stands in for inductance and elasticity stands in for capacitance.

 

(I think it was Boyle who first demonstrated that gases are elastic, euphemistically calling it the “spring of air”)

 

Anyway, resonance occurs when the two reactive components have equal and opposite response and the resonant frequency becomes trapped, as in the classic tank circuit. Both higher and lower frequencies can pass through unimpeded.



#10 exchemist

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Posted 14 September 2017 - 02:56 AM

You know more than I :)

 

 

I don't understand by what mechanism the co2 molecule can act as a medium for the transference of 10^14 frequency light while being a resonator at the frequency of 10^13.  Do you see my problem?  If a thing resonates at 10^13, it should resist vibrations at 10^14 by virtue of the mass/spring combination that causes the resonance at 10^13.

 

Btw, if it's ok with you, I strongly prefer talking in terms of frequency rather than wavelength if at all possible because that is the way I've come to understand sound and waves in general.

 

 

When light hits the ground, some of it is reflected otherwise we wouldn't see the ground ;)  But yes, I understand that the ground absorbs the light and grass absorbs all but the green part of light; hence why it is green.  Temperature is a measure of vibrations (kinetic energy), is that right?  And so the hotter a body becomes, the higher the frequency of the light emitted.

 

 

Isn't it true that they can emit frequencies of EMR correlated to their temperatures?  It doesn't always have to be IR and MW radiation, does it?

 

 

Yes, millions of webpages says that.  It's the "why" that I am after ;)  My problem is that if co2 resonates at IR frequency, then it should not transfer visible light.  Therefore the addition of co2 would be a net-reduction of heat on the ground by virtue of resisting the passage of visible light... and by a second mechanism which is that the re-emittance of IR will have more paths leading to space than to earth due to the curvature of the earth.

 

 

I'm far from an expert on quantum theory, but isn't that just the probability of finding a particle in any one location?  So it's a probability distribution wave and not an actual wave through a real medium like light and sound.  I understand that light travels through an energy field medium (perhaps the higgs field) because a wave without a medium doesn't make sense.

 

So all waves are propagations through mediums and will have relatable and analogous properties.

 

 

I thought it was resonance causing the absorption, which is a function of mass/spring combinations of the whole atom or molecule(s).  For instance, gamma rays (10^19) will ionize atoms by acting on the electrons due to their extremely high frequencies, but UV (10^16) is too slow (vibrations or too slow) to act on electrons and will begin to transition to acting on the whole atom.

 

"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...cular_vibration

 

 

I can see that.  IR and MW are probably the transitions from atomic resonances to molecular and bonded-molecular resonances.  By "bonded" I mean like a pane of glass... rigidly bonded and held tightly in position.

 

"The frequency of the periodic motion is known as a vibration frequency, and the typical frequencies of molecular vibrations range from less than 1013 to approximately 1014 Hz"  https://en.wikipedia...cular_vibration

 

10^14 is visible and 10^13 is IR.

 

A heavy atom bound tightly should resonate faster than in free space and therefore would absorb different frequencies of light, but gases in the atmosphere I'm assuming are free space except for the bonds between the C and O.

 

So if we have O=C=O, then the C in the middle will have different resonant properties than a C by itself.  It must be that the collection of C and O just happen to resonate in the 10^13 range by virtue of the mass/spring combination and if true, then it should also resist vibrations in the 10^14 range which will prevent passage of visible light.

 

The whole O=C=O molecule should resonate in the MW range (maybe 10^9) like water.

 

 

I'd like to learn as much as I can.

Thanks for the comprehensive reply. However I find that on discussion forums it is hard to deal effectively with a lot of points in parallel, so forgive me if I go slowly and answer only one thing at a time.

 

Reading your response, it seems to me the basic difficulty you have is your attempt to stretch your analogy between EM radiation and the transmission, absorption and reflection of sound waves into areas where it does not hold up. It seems to me we need to look at some of the differences between the interaction between matter and sound waves on the one hand, and between matter and EM radiation on the other. 

 

Here are 3 differences between sound waves and light waves to start with:

1) sound is a longitudinal wave whereas EM radiation is a transverse wave;

 

2) sound propagates in a compressible medium. Light requires no medium, being an oscillation of electric and magnetic fields, at right angles to each other and to the direction of propagation;

 

3)  light, unlike sound wave, can only be absorbed, emitted or reflected in quanta, which we call photons. 

 

The crux of your difficulty seems to be why EM waves of higher frequency than the natural frequency of the oscillator are not reflected from it. To answer that, one has to consider the mechanism of reflection. With sound, when the waves encounter a solid surface, the compression of the medium at a wave peak raises the air pressure against the surface and that higher pressure in the medium is then relieved by a flow of air away from the surface, causing the wave to change direction and be reflected.

 

But with light there is no medium. So this mechanism of reflection cannot - and does not - apply.

 

All you have is individual photons encountering the electron cloud of an individual molecule. If the photon has an energy (related to frequency by E=hν)  identical to the energy gap between two of the allowed states in the molecule, it will be absorbed. If not, it will just pass by, perhaps polarising the electron cloud a bit, temporarily, as it does so.Those are its only options. (This polarisation can however affect the phase velocity of light, thereby influencing the refractive index of a transparent substance.) 

 

As I say, you cannot push the analogy of sound and light too far, or it breaks down. 

 

Note added later: Having considered OceanBreeze's post, I now wonder abut your analysis of what happens in the case of sound. I have read your initial post more carefully now and I think you may be relying too much on mass of the oscillator. You seem to have been saying that the mass of the oscillator is too great for it to move when the frequency is high and that it therefore reflects the sound. You can get an oscillator to resonate at higher frequency either by decreasing the mass or by increasing the restoring force acting on the mass when it is displaced, e.g. a stronger spring. Whereas what determines whether it reflects the sound or the sound just goes round it is a function of its physical size, surely, rather than its mass? Bear in mind that a gas is mostly empty space, so it does not present a solid wall to the incoming photons.   


Edited by exchemist, 14 September 2017 - 07:27 AM.


#11 BanterinBoson

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Posted 14 September 2017 - 02:19 PM

No no no, hold your horses! What you can infer is that you cannot heat oxygen and nitrogen by means of IR or microwave radiation. That's all. 

 

I still don't understand the distinction between IR and any other band of frequencies as it relates to dipole moments, but I'll take your word for it in this case.

 

 

And of course there are other means of heating a gas than radiation. The 3 means of heat transfer that we all learn at school are: conduction, convection and radiation, yes?     

 

Conduction is merely the transference of vibrations by means of "touch" (like one guitar string transferring its vibrations to another string by touching).  "Touch" in this case is defined as the van der waals contact distance https://en.wikipedia...der_Waals_force

 

Convection is the "stirring up" of vibrational bodies where the resulting heat transfers are still radiative and conductive.

 

So there are only 2 means of transferring vibrations (ie heat): touch and radiative.

 

One means of transferring vibrations from one atom to another in the atmosphere is by radiative waves propagating through a ubiquitous energy field which is directly analogous to the transference of sound from a speaker to an eardrum through the ubiquitous atmosphere on earth.  The other means is the collision of atoms where vibrations are directly transferred which is analogous to the guitar string example above.

 

(I have to admit that I didn't consider "touch" in questioning the heating of O and N, but in my defense, I was just considering the heating by radiation and, in hindsight, should have phrased my question differently.)



#12 exchemist

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Posted 14 September 2017 - 02:29 PM

I still don't understand the distinction between IR and any other band of frequencies as it relates to dipole moments, but I'll take your word for it in this case.

 

 

Conduction is merely the transference of vibrations by means of "touch" (like one guitar string transferring its vibrations to another string by touching).  "Touch" in this case is defined as the van der waals contact distance https://en.wikipedia...der_Waals_force

 

Convection is the "stirring up" of vibrational bodies where the resulting heat transfers are still radiative and conductive.

 

So there are only 2 means of transferring vibrations (ie heat): touch and radiative.

 

One means of transferring vibrations from one atom to another in the atmosphere is by radiative waves propagating through a ubiquitous energy field which is directly analogous to the transference of sound from a speaker to an eardrum through the ubiquitous atmosphere on earth.  The other means is the collision of atoms where vibrations are directly transferred which is analogous to the guitar string example above.

 

(I have to admit that I didn't consider "touch" in questioning the heating of O and N, but in my defense, I was just considering the heating by radiation and, in hindsight, should have phrased my question differently.)

Yes, so you can heat the air by contact with the warm ground (conduction/convection) or with UV absorption high up, where it is done mainly by ozone (an allotrope of oxygen). This explains why the air is warmer near the ground than higher up - until you get to the very top of the atmosphere, that is. 

 

The thing about dipole moments is to do with the excitation of vibrational and rotational modes. EM radiation has an oscillating electric field. This  can tend to align, or stretch, a molecule in which there is a slightly +ve end and a -ve end. There is much more to this and it is governed by quantum effects, but you may be able to see that radiation of the right frequency can set a dipole spinning, just as a compass needle can be made to spin by an alternating magnetic field.  



#13 BanterinBoson

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Posted 14 September 2017 - 03:18 PM

Why do you think that? (the bolded part)

 

I'm glad you asked.  It is because frequencies higher than resonance are attenuated.  If a mass vibrates at resonant frequency, then trying to vibrate it faster is going to be progressively harder.

 

 

As an analog, we can use the simple form of the differential equation for an RLC circuit:

 

[math] L\frac { di }{ dt } +Ri+\frac { q }{ C } =E[/math]

 

Where E is some constant voltage

L is the inductance and C the capacitance

 

This exact same relationship also applies to mechanical systems, and extends also to molecules such as CO2 where mass stands in for inductance and elasticity stands in for capacitance.

 

(I think it was Boyle who first demonstrated that gases are elastic, euphemistically calling it the “spring of air”)

 

Anyway, resonance occurs when the two reactive components have equal and opposite response and the resonant frequency becomes trapped, as in the classic tank circuit. Both higher and lower frequencies can pass through unimpeded.

 

That is interesting.  The mass itself is a low-pass filter and the spring is a high pass.  Hmm... I guess I never thought of it that way (thanks for sharing).  But saying the whole atmosphere acts as a capacitor isn't correct, I think.

 

What is at issue is a mass/spring combination that resonates at 10^13 should resist vibrations progressively higher (10^14), but not so high as to vibrate the subatomics within the co2 (10^19 gamma rays) which is an entirely new system of mass/springs.

 

This can be illustrated easily if you have a ported subwoofer box.  As the frequency starts low (10-20hz), you should feel air moving in the port.  As the frequency rises to the port frequency, you will feel lots of air moving.  As the frequency rises further, you will feel less air until no air movement is felt at all.  Frequencies higher than resonance will not move the air because the column of air has too much mass to be moved so quickly.  (The spring is the volume of air inside the box)

 

One could fashion a bandpass box, but the same principles would still apply.  It would merely be a dual port-frequency emitter while all other frequencies would be attenuated by the mass of air in the ports.  I can't remember how to make an acoustic notch filter, but the same principles would certainly apply.  If the spring is a volume of air and the mass is a column of air, then frequencies higher than the port frequency will be attenuated.



#14 BanterinBoson

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Posted 14 September 2017 - 04:15 PM

Yes, so you can heat the air by contact with the warm ground (conduction/convection) or with UV absorption high up, where it is done mainly by ozone (an allotrope of oxygen). This explains why the air is warmer near the ground than higher up - until you get to the very top of the atmosphere, that is. 

 

The thing about dipole moments is to do with the excitation of vibrational and rotational modes. EM radiation has an oscillating electric field. This  can tend to align, or stretch, a molecule in which there is a slightly +ve end and a -ve end. There is much more to this and it is governed by quantum effects, but you may be able to see that radiation of the right frequency can set a dipole spinning, just as a compass needle can be made to spin by an alternating magnetic field.  

 

I still having difficulty.  

 

Maybe we should start at one end of the spectrum and move progressively to the other and describe what is happening.  Here is what I currently think:

 

Gamma rays (10^19) will affect the subatomic particles within the atoms because the frequency is too high to move masses larger than those.

Xrays (19^18)  will affect subatomics less than gamma but are still too fast to vibrate larger masses.

UV rays (10^16) is the transition point where whole nuclei are being affected more-so than subatomics.

Visible (10^14) mainly affects the atomic masses.

IR (10^13) begins to transition to whole molecules 

Microwave (10^10) will affect larger molecules (the microwave oven is about 10^9)

Radio waves (1hz up to 10^7 I guess).  Not sure what they affect.

 

My numbers may be off by a point or two, but that's the gist of how I understand reality.

 

I previously thought IR was tuned perfectly to affect something within the co2 molecule and it was by virtue of the mass/spring combination inside that made it so.  Now you're saying that it's the dipole moment and the separation of charges that causes IR to act on the whole co2 molecule and it's not by virtue of simply existing as a mass/spring setup.

 

The source of my confusion could be from here: https://en.wikipedia...oven#Principles

 

 

 

A microwave oven heats food by passing microwave radiation through it. Microwaves are a form of non-ionizing electromagnetic radiation with a frequency higher than ordinary radio waves but lower than infrared light. Microwave ovens use frequencies in one of the ISM (industrial, scientific, medical) bands, which are reserved for this use, so they do not interfere with other vital radio services. Consumer ovens usually use 2.45 gigahertz (GHz)—a wavelength of 12.2 centimetres (4.80 in)—while large industrial/commercial ovens often use 915 megahertz (MHz)—32.8 centimetres (12.9 in).[20] Water, fat, and other substances in the food absorb energy from the microwaves in a process called dielectric heating. Many molecules (such as those of water) are electric dipoles, meaning that they have a partial positive charge at one end and a partial negative charge at the other, and therefore rotate as they try to align themselves with the alternating electric field of the microwaves. Rotating molecules hit other molecules and put them into motion, thus dispersing energy. This energy, when dispersed as molecular vibration in solids and liquids (i.e. as both potential energy and kinetic energy of atoms), is heat. Sometimes, microwave heating is explained as a resonance of water molecules, but this is incorrect;[21] such resonances occur only at above 1 terahertz (THz).[22] Rather it is the lag in response of the polar water molecule to the impinging electromagnetic wave. This type of dielectric loss mechanism is referred to as dipole interaction.

 

So if 2.45 x 10^9 hz acts on water molecules because of the dipole moment we were discussing and if atomic resonances occur only above 10^12, then how does IR (which are mostly above 10^12) act the same on co2 as a microwave does on water?

 

(Specifically, IR is 3x10^11 to 4.3x10^14 which is just to the side of the visible. https://en.wikipedia.org/wiki/Infrared )

 

Just for reference, dielectric heating is:

 

 

 

Dielectric heating, also known as electronic heatingRF (radio frequency) heating, and high-frequency heating, is the process in which a high-frequency alternating electric field, or radio wave or microwave electromagnetic radiation heats a dielectric material. At higher frequencies, this heating is caused by molecular dipole rotation within the dielectric.

 

No mention of IR.  Am I to understand that IR heats by dielectric means?  Or is IR acting upon the atoms inside molecules?  If so, then that article was wrong.  It seems I have conflicting information.  If so, then maybe O2 and N2 can be heated by IR (but not MW).  Anyway, I'm thoroughly confused by this.



#15 BanterinBoson

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Posted 14 September 2017 - 06:28 PM

Thanks for the comprehensive reply. However I find that on discussion forums it is hard to deal effectively with a lot of points in parallel, so forgive me if I go slowly and answer only one thing at a time.

 

Reading your response, it seems to me the basic difficulty you have is your attempt to stretch your analogy between EM radiation and the transmission, absorption and reflection of sound waves into areas where it does not hold up. It seems to me we need to look at some of the differences between the interaction between matter and sound waves on the one hand, and between matter and EM radiation on the other. 

 

Here are 3 differences between sound waves and light waves to start with:

1) sound is a longitudinal wave whereas EM radiation is a transverse wave;

 

2) sound propagates in a compressible medium. Light requires no medium, being an oscillation of electric and magnetic fields, at right angles to each other and to the direction of propagation;

 

3)  light, unlike sound wave, can only be absorbed, emitted or reflected in quanta, which we call photons. 

 

The crux of your difficulty seems to be why EM waves of higher frequency than the natural frequency of the oscillator are not reflected from it. To answer that, one has to consider the mechanism of reflection. With sound, when the waves encounter a solid surface, the compression of the medium at a wave peak raises the air pressure against the surface and that higher pressure in the medium is then relieved by a flow of air away from the surface, causing the wave to change direction and be reflected.

 

But with light there is no medium. So this mechanism of reflection cannot - and does not - apply.

 

All you have is individual photons encountering the electron cloud of an individual molecule. If the photon has an energy (related to frequency by E=hν)  identical to the energy gap between two of the allowed states in the molecule, it will be absorbed. If not, it will just pass by, perhaps polarising the electron cloud a bit, temporarily, as it does so.Those are its only options. (This polarisation can however affect the phase velocity of light, thereby influencing the refractive index of a transparent substance.) 

 

As I say, you cannot push the analogy of sound and light too far, or it breaks down. 

 

Note added later: Having considered OceanBreeze's post, I now wonder abut your analysis of what happens in the case of sound. I have read your initial post more carefully now and I think you may be relying too much on mass of the oscillator. You seem to have been saying that the mass of the oscillator is too great for it to move when the frequency is high and that it therefore reflects the sound. You can get an oscillator to resonate at higher frequency either by decreasing the mass or by increasing the restoring force acting on the mass when it is displaced, e.g. a stronger spring. Whereas what determines whether it reflects the sound or the sound just goes round it is a function of its physical size, surely, rather than its mass? Bear in mind that a gas is mostly empty space, so it does not present a solid wall to the incoming photons.   

 

Would you prefer that I not breakup the post into a point-by-point format and instead write a large body of text?  I can go either way.

 

Yes, sound is longitudinal whereas light is transverse, but that's where I believe the differences end.  I should say that I don't believe matter exists as a tangible thing, but instead is energy of sufficient content and frequency.  E=Mc^2 says that all energy has mass, so even a 1hz radio wave will have mass (about 7.4 x 10^-51 kg according to my calculations).  So what differentiates what we commonly call energy and what we commonly call matter is simply the content of energy and the frequency; that's it.  Essentially, there is no such thing as matter.

 

This is a good video describing what matter actually is 

 

My favorite part is the bit about 3:00 that says if you pull a quark apart from another quark, the added energy to do that actually creates another quark pair out of the energy field that exists everywhere.  It demonstrates how matter is created by the addition of energy.  Matter is basically a bottleneck of energy within the energy field and the speed of light is simply the "tipping point" where added energy no longer translates so much to frequency as it does to becoming a "particle" and that transition slows the speed.  So if we begin with a radio wave at 1 hz and start adding energy, the frequency will increase until we pass the gamma rays and begin making "matter" as neutrinos, electrons, quarks which can't quite travel the speed c anymore even though they contain much more energy than a photon (and should have a higher frequency as well ie > 10^20).

 

Here is a video showing how a photon can be slowed to walking speed: 

 

The takeaway from that video is noticing how light compresses into a particle-looking thing.  So if matter is just a bottleneck of energy, then a photon can resemble a particle if it has difficulty traveling less than infinite speed because the leading edge will be compressed due to the resistance of travel through the energy field just like it does when traveling through the BEC.  That resistance to travel is just the speed of causality.

 

The speed of light is the speed of causality: 

 

So all of that just to state that I believe light is a wave traveling through a universal medium analogous to sound.  And to state that there is no fundamental differentiation between matter and energy.  So then light traveling through the energy field and encountering a particle is really just encountering a dense part of the energy field.  Analogous to sound, we could say that sound encountering an object is really just encountering a dense section of air (the medium thru which sound travels).  Other than the transverse/longitudinal objection, the analogy is perfect.  The photon itself is also just a dense section of the energy field and the sound wave is just a dense section of air.

 

___________________________________________

 

On the issue of reflection, light has to reflect... because it's a wave.  If light didn't reflect, we couldn't see anything but light-emitters because it would mean all light had been absorbed.  As I pointed out before, grass is green because green is the only frequency that is reflected.  All others are absorbed and converted into sugar and IR radiation (which we can't see).  

 

Light has to reflect.  The question is does it reflect off of co2 in the 10^14 band?  When light hits a co2 molecule, there are only 3 options:  It reflects, absorbs, or passes straight through.  We already know it does not absorb and I can't imagine how it could pass straight through because the frequency is not low enough to move the whole molecule like microwaves would, nor is it high enough to travel through the subatomic particles like gamma rays would... and that leaves no mechanism for propagation through the molecule.  The only option left is to say it reflects.

 

And if that is true, then it necessarily means that co2 cools the earth by reflecting higher-energy photons away from earth while only passing lower-energy photons.  That's what clouds do and it's why cloudy days are cooler than sunny days.

 

I'm going to address this more specifically:

 

 

 

All you have is individual photons encountering the electron cloud of an individual molecule. If the photon has an energy (related to frequency by E=hν)  identical to the energy gap between two of the allowed states in the molecule, it will be absorbed. If not, it will just pass by, perhaps polarising the electron cloud a bit, temporarily, as it does so.Those are its only options. (This polarisation can however affect the phase velocity of light, thereby influencing the refractive index of a transparent substance.) 

 

If the idea is that the atom is mostly empty and a photon could sneak by in what can only be described as a "vastness of space", then why does IR not simply pass by?  If the photon never touches the atom, then how does the atom know what frequency the photon is and whether to absorb it or not?  Conversely, if the atom knows what frequency the light is, then it must somehow "touch".  And if it touches, then it can reflect.

 

Gamma rays can propagate through an atom because they act on the subatomic particles, so atoms are transparent to gamma rays.  But visible light cannot act on subatomics and have no mechanism for propagation through the atom unless the visible light can move the whole atom.  Since IR moves the whole atom, then visible should not be able to and therefore should reflect.  Visible light just happens to be of such frequency that it is too fast to move the whole molecule or atoms within the molecule and it's too slow to move the subatomics.  So it's perfectly tuned to bounce.

 

 

 

Whereas what determines whether it reflects the sound or the sound just goes round it is a function of its physical size, surely, rather than its mass? Bear in mind that a gas is mostly empty space, so it does not present a solid wall to the incoming photons.  

 

Sound travels in all directions until the frequency increases and it begins to act more directional, so I suspect the same is true with radio waves being omnidirectional as opposed to gamma rays and especially electrons being so super-directional that they are referred to as particles.  Anyway, I don't think volume factors in and it's just a function of mass and the spring.  If a wave "wraps" around an object, the object has still "stolen" some of the wave's energy either through absorption or reflection.

 

So let's say we have a tuning fork that resonates at 200 hz and we play a tone through a speaker that is also 200hz.  The fork should resonate, which is absorbing and re-emitting the sound.  Now we play 2000hz and find the fork is bouncing the sound.  If we play a 20hz tone, we may find that the whole fork rattles in our hand as well as the rest of the house.  So in this case the fork would be like a gas molecule in a cloud that re-emits resonances, reflects above-resonances, and transmits below-resonances.

 

There is no way to increase the surface area of the fork without increasing the mass of the fork or decreasing the density.  If the mass is increased, then it won't resonate at 200.  If the density is decreased in order to preserve mass, then it won't reflect as well at 2000.  So it seems reflectivity is also a function of mass and not area.

 

I haven't given much thought to directionality and it raises the question of how big is a photon?  I don't think anyone really knows.  I suspect it would have to be smaller than an electron, but no one really knows how big that is either.  We know the mass, but the physical size of a particle that isn't really a particle is a bit of a mystery.



#16 exchemist

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Posted 15 September 2017 - 02:20 AM

I still having difficulty.  

 

Maybe we should start at one end of the spectrum and move progressively to the other and describe what is happening.  Here is what I currently think:

 

Gamma rays (10^19) will affect the subatomic particles within the atoms because the frequency is too high to move masses larger than those.

Xrays (19^18)  will affect subatomics less than gamma but are still too fast to vibrate larger masses.

UV rays (10^16) is the transition point where whole nuclei are being affected more-so than subatomics.

Visible (10^14) mainly affects the atomic masses.

IR (10^13) begins to transition to whole molecules 

Microwave (10^10) will affect larger molecules (the microwave oven is about 10^9)

Radio waves (1hz up to 10^7 I guess).  Not sure what they affect.

 

My numbers may be off by a point or two, but that's the gist of how I understand reality.

 

I previously thought IR was tuned perfectly to affect something within the co2 molecule and it was by virtue of the mass/spring combination inside that made it so.  Now you're saying that it's the dipole moment and the separation of charges that causes IR to act on the whole co2 molecule and it's not by virtue of simply existing as a mass/spring setup.

 

The source of my confusion could be from here: https://en.wikipedia...oven#Principles

 

 

So if 2.45 x 10^9 hz acts on water molecules because of the dipole moment we were discussing and if atomic resonances occur only above 10^12, then how does IR (which are mostly above 10^12) act the same on co2 as a microwave does on water?

 

(Specifically, IR is 3x10^11 to 4.3x10^14 which is just to the side of the visible. https://en.wikipedia.org/wiki/Infrared )

 

Just for reference, dielectric heating is:

 

 

No mention of IR.  Am I to understand that IR heats by dielectric means?  Or is IR acting upon the atoms inside molecules?  If so, then that article was wrong.  It seems I have conflicting information.  If so, then maybe O2 and N2 can be heated by IR (but not MW).  Anyway, I'm thoroughly confused by this.

On the contrary, the references you have turned up make my point very well and should be starting to dispel your confusion.

 

Perhaps it may become clearer if you keep in mind 3 points of quantum theory that apply to these processes:-

 

1) EM radiation is quantised, and consists of undivisible "packets" of radiation called photons. Planck's relation between the energy of a photon and frequency, E=hν, says that the higher the frequency, the more energy there is in each photon.

 

2) Rotation and vibration in molecules is quantised. They cannot vibrate or rotate as much or as little as they want, but can only do so in set, fixed increments or steps. Diagrammatically this is often shown by a series of permitted "energy levels", rising up in energy from a bottom level which is called the "ground state". 

 

3) Absorption of a photon by a molecular rotation or vibration can only take place when the energy of the photon exactly corresponds to the energy difference between rotational or vibrational energy levels. E=hν will then give you the "resonance frequency", if you like.  

 

IR and microwave are different regions of the EM spectrum. The references you have turned up are all about microwaves, not IR. And so they are referring to the EM oscillating electric field interacting with the dipole in a molecule (if there is one) to make it rotate. This is the basis of microwave heating. So the rotation bit at least should be becoming clear to you, I hope.

 

Whether you can call IR heating "dielectric heating" is a moot point. IR is not generally thought of as a radio wave, so the term "dielectric" is not really used. At this region of the spectrum, the frequency is far too high to stimulate rotation of molecules, because the energy of each photon is far greater than the energy gaps between rotational levels in molecules, so rotational changes cannot absorb them.  But the energy is about right for the gaps between vibrational energy levels. For IR photons to be absorbed, the radiation has to couple somehow with the vibrational distortions of the molecule. For this, there needs to be a change in dipole moment when the molecule goes from one vibrational state to another, i.e. as it stretches,  bends or twists. Hence, molecules with no dipole won't absorb IR. But, in terms of your ball and spring model you are right: the mass of the atoms and and stiffness of the chemical bonds between them is what determines the spacing of the energy levels and consequently the "resonance frequencies" that will be absorbed.  The dipole is the thing that the radiation is able to influence, in order to transfer energy to the molecule.  Without a dipole change in the course of a vibration, this is not possible, as there is nothing for the electric field of the radiation to grab hold of.   

 

It is worth saying at this point that if you excite rotations only , or vibrations only, the process of constant collisions between molecules will soon redistribute the energy from the states originally excited to a whole mixture of both rotations and vibrations. (There is something called the Equipartition principle in statistical thermodynamics, which governs how the energy ends up shared among all the possible modes of motion in an assembly of molecules in thermal equilibrium.) 


Edited by exchemist, 15 September 2017 - 08:33 AM.


#17 exchemist

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Posted 15 September 2017 - 03:12 AM

Would you prefer that I not breakup the post into a point-by-point format and instead write a large body of text?  I can go either way.

 

Yes, sound is longitudinal whereas light is transverse, but that's where I believe the differences end.  I should say that I don't believe matter exists as a tangible thing, but instead is energy of sufficient content and frequency.  E=Mc^2 says that all energy has mass, so even a 1hz radio wave will have mass (about 7.4 x 10^-51 kg according to my calculations).  So what differentiates what we commonly call energy and what we commonly call matter is simply the content of energy and the frequency; that's it.  Essentially, there is no such thing as matter.

 [videos snipped]

 

So all of that just to state that I believe light is a wave traveling through a universal medium analogous to sound.  And to state that there is no fundamental differentiation between matter and energy.  So then light traveling through the energy field and encountering a particle is really just encountering a dense part of the energy field.  Analogous to sound, we could say that sound encountering an object is really just encountering a dense section of air (the medium thru which sound travels).  Other than the transverse/longitudinal objection, the analogy is perfect.  The photon itself is also just a dense section of the energy field and the sound wave is just a dense section of air.

 

___________________________________________

 

On the issue of reflection, light has to reflect... because it's a wave.  If light didn't reflect, we couldn't see anything but light-emitters because it would mean all light had been absorbed.  As I pointed out before, grass is green because green is the only frequency that is reflected.  All others are absorbed and converted into sugar and IR radiation (which we can't see).  

 

Light has to reflect.  The question is does it reflect off of co2 in the 10^14 band?  When light hits a co2 molecule, there are only 3 options:  It reflects, absorbs, or passes straight through.  We already know it does not absorb and I can't imagine how it could pass straight through because the frequency is not low enough to move the whole molecule like microwaves would, nor is it high enough to travel through the subatomic particles like gamma rays would... and that leaves no mechanism for propagation through the molecule.  The only option left is to say it reflects.

 

And if that is true, then it necessarily means that co2 cools the earth by reflecting higher-energy photons away from earth while only passing lower-energy photons.  That's what clouds do and it's why cloudy days are cooler than sunny days.

 

[snip]

Oh dear. Now you seem to be coming out of the closet as a crank. All this "I believe" stuff. What a pity. Perhaps I have been wasting my time.

 

The thing is, in science you can't just make sh1t up.  The models we have are based on agreement with observation. 

 

If you say "I believe" that the only difference between sound and light waves is that one is longitudinal and the other transverse, and then you go on to claim that CO2 must reflect visible light, can't you see how silly that is? It is the work of a moment to get a glass tube, fill it with CO2 gas and observe that you can see straight through it. So your model is obviously wrong.  

 

All I can do is explain the model we have for all this in chemistry. Our model, unlike yours, works - it agrees with observation and experiment. If you want to explore other, non-quantum, explanations, they may be a fun game but they won't work. That is why quantum theory was invented, you see. Because scientists found that without it, they could not account for what they found in nature.

 

If you are interested in learning more about chemistry and chemical physics I am here to help. If you want to indulge a naive alternative that is demonstrably wrong, you will have to find someone else to discuss it with.

 

Over to you.  


Edited by exchemist, 15 September 2017 - 03:22 AM.