Once Again, Skeptics do the Math that Warmists Won't Do....

[Billy_Bob]

You think no energy is lost with the emission of a photon?

That is correct. When an electron absorbs a photon and gains 10 eV, no energy is lost when it then emits a photon of 10 eV. Because energy is conserved.

Quantum Mechanics is primarily theroy as observations are impossible. Given your description above there should then be NO HEAT generated but this is not the case as empirical evidence proves otherwise. So where does your heat come from if there is no loss? The molecules can not warm up if there is no resistance and loss.

You think the energy is absorbed by the molecule, a photon is emitted, and no energy is lost in the process?

You think when a CO2 molecule gains 10 eV from a single photon, it can emit a single 9 eV photon and lose 1 eV, just because?

Actually, Yes! Or there is the loss of full photons (IE: Receives 10 and outputs 6) but there must always be a loss which generates heat (vibration). As this is quantum theroy we really dont know the answer to this question with certainty.

So tell me toddster, do you really think no energy whatsoever is lost?

So tell me SSDD, how does this energy magically get lost in your scenario?

Absorption causes the molecule to vibrate (become excited) and the interaction with other adjacent molecules is what creates the heat. Absorption means conversion from the original emitted vibration pattern -> into the electron causing it to vibrate -> When enough energy is stored the electron then sheds a photon vibrating at the new molecular temperature. Again this is QM theroy being postulated.

The photon could very well be moving at 10 eV or it could be moving at 9 eV, we dont know for sure. What we do know for sure is that any time energy is absorbed or transferred there is loss. We also know that the mass emitting the new photon is what determines its vibration or frequency.

We know for instance that cylinder full of CO2 molecules, at 1000ppm using argon as its inert counter, bombarded by photons at 0.4um will emit at 6.0um to 13.0um which is the bandwidth that is not able to reabsorb the lower vibrating photons in CO2. The narrow spectrum generator was tested in several bandwidths always ending with the same result up to 2.2um. Curiously as we hit areas of non-cunduction/absorbtion (or total band pass) we registered scatter and reflection. We didn't register how much heat was built up in the gas but that is for another day.

Again I dont really care about the terms, just be skeptical and ask questions. Remember, theory is just that, Theory!

There are a lot of questions and very few answers. I'll butt out now.

 

[SSDD]

You think no energy is lost with the emission of a photon?

That is correct. When an electron absorbs a photon and gains 10 eV, no energy is lost when it then emits a photon of 10 eV. Because energy is conserved.

Quantum Mechanics is primarily theroy as observations are impossible. Given your description above there should then be NO HEAT generated but this is not the case as empirical evidence proves otherwise. So where does your heat come from if there is no loss? The molecules can not warm up if there is no resistance and loss.

You think the energy is absorbed by the molecule, a photon is emitted, and no energy is lost in the process?

You think when a CO2 molecule gains 10 eV from a single photon, it can emit a single 9 eV photon and lose 1 eV, just because?

Actually, Yes! Or there is the loss of full photons (IE: Receives 10 and outputs 6) but there must always be a loss which generates heat (vibration). As this is quantum theroy we really dont know the answer to this question with certainty.

So tell me toddster, do you really think no energy whatsoever is lost?

So tell me SSDD, how does this energy magically get lost in your scenario?

Absorption causes the molecule to vibrate (become excited) and the interaction with other adjacent molecules is what creates the heat. Absorption means conversion from the original emitted vibration pattern -> into the electron causing it to vibrate -> When enough energy is stored the electron then sheds a photon vibrating at the new molecular temperature. Again this is QM theroy being postulated.

The photon could very well be moving at 10 eV or it could be moving at 9 eV, we dont know for sure. What we do know for sure is that any time energy is absorbed or transferred there is loss. We also know that the mass emitting the new photon is what determines its vibration or frequency.

We know for instance that cylinder full of CO2 molecules, at 1000ppm using argon as its inert counter, bombarded by photons at 0.4um will emit at 6.0um to 13.0um which is the bandwidth that is not able to reabsorb the lower vibrating photons in CO2. The narrow spectrum generator was tested in several bandwidths always ending with the same result up to 2.2um. Curiously as we hit areas of non-cunduction/absorbtion (or total band pass) we registered scatter and reflection. We didn't register how much heat was built up in the gas but that is for another day.

Again I dont really care about the terms, just be skeptical and ask questions. Remember, theory is just that, Theory!

There are a lot of questions and very few answers. I'll butt out now.

Not much more point in talking to luke warmers than there is in talking to warckaloon warmers....they all believe in magic....just different levels of power in the magic. Ian, for example accepts that gravity and convection play a part (unsure how large a part he thinks) but also believes in warming via CO2...never quite realizing that if you accept one, you negate the other. The acceptance of both changes all of the numbers....the greenhouse hypothesis has no place within for anything but itself....it is balanced on the edge of a razor...change one thing and suddenly it is failing. The fact that sea water is a poor absorber of LW radiation which happens to be what CO2 emits could well be a death blow for the greenhouse hypothesis...gravity and convection having an effect on temperature would be more than the greenhouse hypothesis as it stands now could withstand.

 

[SSDD]

You think no energy is lost with the emission of a photon?

That is correct. When an electron absorbs a photon and gains 10 eV, no energy is lost when it then emits a photon of 10 eV. Because energy is conserved.

Quantum Mechanics is primarily theroy as observations are impossible. Given your description above there should then be NO HEAT generated but this is not the case as empirical evidence proves otherwise. So where does your heat come from if there is no loss? The molecules can not warm up if there is no resistance and loss.

You think the energy is absorbed by the molecule, a photon is emitted, and no energy is lost in the process?

You think when a CO2 molecule gains 10 eV from a single photon, it can emit a single 9 eV photon and lose 1 eV, just because?

Actually, Yes! Or there is the loss of full photons (IE: Receives 10 and outputs 6) but there must always be a loss which generates heat (vibration). As this is quantum theroy we really dont know the answer to this question with certainty.

So tell me toddster, do you really think no energy whatsoever is lost?

So tell me SSDD, how does this energy magically get lost in your scenario?

Absorption causes the molecule to vibrate (become excited) and the interaction with other adjacent molecules is what creates the heat. Absorption means conversion from the original emitted vibration pattern -> into the electron causing it to vibrate -> When enough energy is stored the electron then sheds a photon vibrating at the new molecular temperature. Again this is QM theroy being postulated.

The photon could very well be moving at 10 eV or it could be moving at 9 eV, we dont know for sure. What we do know for sure is that any time energy is absorbed or transferred there is loss. We also know that the mass emitting the new photon is what determines its vibration or frequency.

We know for instance that cylinder full of CO2 molecules, at 1000ppm using argon as its inert counter, bombarded by photons at 0.4um will emit at 6.0um to 13.0um which is the bandwidth that is not able to reabsorb the lower vibrating photons in CO2. The narrow spectrum generator was tested in several bandwidths always ending with the same result up to 2.2um. Curiously as we hit areas of non-cunduction/absorbtion (or total band pass) we registered scatter and reflection. We didn't register how much heat was built up in the gas but that is for another day.

Again I dont really care about the terms, just be skeptical and ask questions. Remember, theory is just that, Theory!

There are a lot of questions and very few answers. I'll butt out now.

In the magical theory, it doesn't require any energy to cause a vibration. The energy passes through unchanged but expends a bit of magic to cause the vibration. One would think that even magic requires energy though. Wonder where that magic comes from? Gaia perhaps?

 

[Old Rocks]

One way magical photons. Just ask SSo DDumb.

 

[CrusaderFrank]

One way magical photons. Just ask SSo DDumb.

CO2 absorbs the energy like a sponge and multiplies it, then shoots its Heat Death ray toward Earth

CO2 -- it melts lead on Venus so you DENIERS!!!! better shape the fuck up!

 

[Toddsterpatriot]

You think no energy is lost with the emission of a photon?

That is correct. When an electron absorbs a photon and gains 10 eV, no energy is lost when it then emits a photon of 10 eV. Because energy is conserved.

Quantum Mechanics is primarily theroy as observations are impossible. Given your description above there should then be NO HEAT generated but this is not the case as empirical evidence proves otherwise. So where does your heat come from if there is no loss? The molecules can not warm up if there is no resistance and loss.

You think the energy is absorbed by the molecule, a photon is emitted, and no energy is lost in the process?

You think when a CO2 molecule gains 10 eV from a single photon, it can emit a single 9 eV photon and lose 1 eV, just because?

Actually, Yes! Or there is the loss of full photons (IE: Receives 10 and outputs 6) but there must always be a loss which generates heat (vibration). As this is quantum theroy we really dont know the answer to this question with certainty.

So tell me toddster, do you really think no energy whatsoever is lost?

So tell me SSDD, how does this energy magically get lost in your scenario?

Absorption causes the molecule to vibrate (become excited) and the interaction with other adjacent molecules is what creates the heat. Absorption means conversion from the original emitted vibration pattern -> into the electron causing it to vibrate -> When enough energy is stored the electron then sheds a photon vibrating at the new molecular temperature. Again this is QM theroy being postulated.

The photon could very well be moving at 10 eV or it could be moving at 9 eV, we dont know for sure. What we do know for sure is that any time energy is absorbed or transferred there is loss. We also know that the mass emitting the new photon is what determines its vibration or frequency.

We know for instance that cylinder full of CO2 molecules, at 1000ppm using argon as its inert counter, bombarded by photons at 0.4um will emit at 6.0um to 13.0um which is the bandwidth that is not able to reabsorb the lower vibrating photons in CO2. The narrow spectrum generator was tested in several bandwidths always ending with the same result up to 2.2um. Curiously as we hit areas of non-cunduction/absorbtion (or total band pass) we registered scatter and reflection. We didn't register how much heat was built up in the gas but that is for another day.

Again I dont really care about the terms, just be skeptical and ask questions. Remember, theory is just that, Theory!

There are a lot of questions and very few answers. I'll butt out now.

Not much more point in talking to luke warmers than there is in talking to warckaloon warmers....they all believe in magic....just different levels of power in the magic. Ian, for example accepts that gravity and convection play a part (unsure how large a part he thinks) but also believes in warming via CO2...never quite realizing that if you accept one, you negate the other. The acceptance of both changes all of the numbers....the greenhouse hypothesis has no place within for anything but itself....it is balanced on the edge of a razor...change one thing and suddenly it is failing. The fact that sea water is a poor absorber of LW radiation which happens to be what CO2 emits could well be a death blow for the greenhouse hypothesis...gravity and convection having an effect on temperature would be more than the greenhouse hypothesis as it stands now could withstand.

You think no energy is lost with the emission of a photon?

That is correct. When an electron absorbs a photon and gains 10 eV, no energy is lost when it then emits a photon of 10 eV. Because energy is conserved.

You think the energy is absorbed by the molecule, a photon is emitted, and no energy is lost in the process?

You think when a CO2 molecule gains 10 eV from a single photon, it can emit a single 9 eV photon and lose 1 eV, just because?

So tell me toddster, do you really think no energy whatsoever is lost?

So tell me SSDD, how does this energy magically get lost in your scenario?

 

[Toddsterpatriot]

You think no energy is lost with the emission of a photon?

That is correct. When an electron absorbs a photon and gains 10 eV, no energy is lost when it then emits a photon of 10 eV. Because energy is conserved.

Quantum Mechanics is primarily theroy as observations are impossible. Given your description above there should then be NO HEAT generated but this is not the case as empirical evidence proves otherwise. So where does your heat come from if there is no loss? The molecules can not warm up if there is no resistance and loss.

You think the energy is absorbed by the molecule, a photon is emitted, and no energy is lost in the process?

You think when a CO2 molecule gains 10 eV from a single photon, it can emit a single 9 eV photon and lose 1 eV, just because?

Actually, Yes! Or there is the loss of full photons (IE: Receives 10 and outputs 6) but there must always be a loss which generates heat (vibration). As this is quantum theroy we really dont know the answer to this question with certainty.

So tell me toddster, do you really think no energy whatsoever is lost?

So tell me SSDD, how does this energy magically get lost in your scenario?

Absorption causes the molecule to vibrate (become excited) and the interaction with other adjacent molecules is what creates the heat. Absorption means conversion from the original emitted vibration pattern -> into the electron causing it to vibrate -> When enough energy is stored the electron then sheds a photon vibrating at the new molecular temperature. Again this is QM theroy being postulated.

The photon could very well be moving at 10 eV or it could be moving at 9 eV, we dont know for sure. What we do know for sure is that any time energy is absorbed or transferred there is loss. We also know that the mass emitting the new photon is what determines its vibration or frequency.

We know for instance that cylinder full of CO2 molecules, at 1000ppm using argon as its inert counter, bombarded by photons at 0.4um will emit at 6.0um to 13.0um which is the bandwidth that is not able to reabsorb the lower vibrating photons in CO2. The narrow spectrum generator was tested in several bandwidths always ending with the same result up to 2.2um. Curiously as we hit areas of non-cunduction/absorbtion (or total band pass) we registered scatter and reflection. We didn't register how much heat was built up in the gas but that is for another day.

Again I dont really care about the terms, just be skeptical and ask questions. Remember, theory is just that, Theory!

There are a lot of questions and very few answers. I'll butt out now.

You think the energy is absorbed by the molecule, a photon is emitted, and no energy is lost in the process?

You think when a CO2 molecule gains 10 eV from a single photon, it can emit a single 9 eV photon and lose 1 eV, just because?

Actually, Yes! Or there is the loss of full photons (IE: Receives 10 and outputs 6) but there must always be a loss which generates heat (vibration). As this is quantum theroy we really dont know the answer to this question with certainty.
If heat is generated, the energy is not lost.

 

[Toddsterpatriot]

So lets review. Any molecule which absorbs energy from an incoming photon or other exciting item, which creates heat within the molecule looses the rate at which it was vibrating on entry and vibrates at the rate of the molecules heat which then emits it. The reduction in amplitude and speed (heat output) is the deciding factor in the new wavelength.

In any energy transfer there is loss.

Source

As energy passes through CO2 the wavelength changes to a length that CO2 can not reabsorb.

As energy passes through CO2 the wavelength changes to a length that CO2 can not reabsorb.

Where do you get this? I didn't see it at your source. I did see the following.

Kirchhoff's Law says that good absorbers of a particular wavelength are also good emitters of that wavelength, and poor absorbers of a wavelength are also poor emitters at the same wavelength.

 

[SSDD]

Where do you get this? I didn't see it at your source. I did see the following.

It is called actually grasping a process. I doubt that you will find much documentation that tells you explicitly that when you turn the handle on a faucet water is supposed to come out....

 

[SSDD]

If heat is generated, the energy is not lost.

Not lost, but the remainder must be at a longer wavelength...

 

[Toddsterpatriot]

Where do you get this? I didn't see it at your source. I did see the following.

It is called actually grasping a process. I doubt that you will find much documentation that tells you explicitly that when you turn the handle on a faucet water is supposed to come out....

We've seen your failure to grasp already.

 

[Billy_Bob]

So lets review. Any molecule which absorbs energy from an incoming photon or other exciting item, which creates heat within the molecule looses the rate at which it was vibrating on entry and vibrates at the rate of the molecules heat which then emits it. The reduction in amplitude and speed (heat output) is the deciding factor in the new wavelength.

In any energy transfer there is loss.

Source

As energy passes through CO2 the wavelength changes to a length that CO2 can not reabsorb.

As energy passes through CO2 the wavelength changes to a length that CO2 can not reabsorb.

Where do you get this? I didn't see it at your source. I did see the following.

Kirchhoff's Law says that good absorbers of a particular wavelength are also good emitters of that wavelength, and poor absorbers of a wavelength are also poor emitters at the same wavelength.

A piece of steel laying on the ground is a good absorber of 0.4-2.9um but radiates at 3.6um during the day and 12.2um at night. That portion of Kirchhoff''s Law is subjective.



CO2's primary absorption bands are 0.4um, 1.2um, 2.6-2.9um, 6.1-7.5um and 9.5-15.6um. During sunlight it is not uncommon for CO2 to radiate its incoming power to its primary band of 9.5-15.6um. Power loss is inevitable. CO2 during the day will reflect or scatter outside its active bands. Once emitted CO2 will be unable to reabsorb because of entropy (cascading loss of frequency). Everyone forgets that there are other gases in our atmosphere which are in play and they do not radiate their energy at band that CO2 can reabsorb.

It is the interaction of gases which makes it nearly impossible for CO2 to reabsorb its emitted energy.

 

[Toddsterpatriot]

So lets review. Any molecule which absorbs energy from an incoming photon or other exciting item, which creates heat within the molecule looses the rate at which it was vibrating on entry and vibrates at the rate of the molecules heat which then emits it. The reduction in amplitude and speed (heat output) is the deciding factor in the new wavelength.

In any energy transfer there is loss.

Source

As energy passes through CO2 the wavelength changes to a length that CO2 can not reabsorb.

As energy passes through CO2 the wavelength changes to a length that CO2 can not reabsorb.

Where do you get this? I didn't see it at your source. I did see the following.

Kirchhoff's Law says that good absorbers of a particular wavelength are also good emitters of that wavelength, and poor absorbers of a wavelength are also poor emitters at the same wavelength.

A piece of steel laying on the ground is a good absorber of 0.4-2.9um but radiates at 3.6um during the day and 12.2um at night. That portion of Kirchhoff''s Law is subjective.

View attachment 34743

CO2's primary absorption bands are 0.4um, 1.2um, 2.6-2.9um, 6.1-7.5um and 9.5-15.6um. During sunlight it is not uncommon for CO2 to radiate its incoming power to its primary band of 9.5-15.6um. Power loss is inevitable. CO2 during the day will reflect or scatter outside its active bands. Once emitted CO2 will be unable to reabsorb because of entropy (cascading loss of frequency). Everyone forgets that there are other gases in our atmosphere which are in play and they do not radiate their energy at band that CO2 can reabsorb.

It is the interaction of gases which makes it nearly impossible for CO2 to reabsorb its emitted energy.

CO2's primary absorption bands are 0.4um, 1.2um, 2.6-2.9um, 6.1-7.5um and 9.5-15.6um. During sunlight it is not uncommon for CO2 to radiate its incoming power to its primary band of 9.5-15.6um.

Which CO2 emission bands are not absorbed by CO2?

 

[Old Rocks]

Ah, Billy Boob, thank you for the comedy routine.

 

[IanC]

Actually, Yes! Or there is the loss of full photons (IE: Receives 10 and outputs 6) but there must always be a loss which generates heat (vibration). As this is quantum theroy we really dont know the answer to this question with certainty.
......
Absorption causes the molecule to vibrate (become excited) and the interaction with other adjacent molecules is what creates the heat. Absorption means conversion from the original emitted vibration pattern -> into the electron causing it to vibrate -> When enough energy is stored the electron then sheds a photon vibrating at the new molecular temperature. Again this is QM theroy being postulated.

you are confusing kinetic energy of molecules with the excited state of CO2 which is called vibration. a CO2 molecule can absorb a specific photon and vibrate, and then emit the same photon and stop vibrating. there is no energy loss when jumping back and forth from these two states. this type of energy exchange is not dependent on temperature as it is just a single molecule.

average molecular speed, eg kinetic energy, is one of the definitions of temperature. when two molecules collide, their electron clouds are deformed and the combined speed of both is slightly less after the collision. when the electron clouds pop back into place they give off at least a photon each. this is the blackbody radiation. the hotter the object (faster moving molecules), the harder the possible collision, the more energetic the ejected photon. these collisions cause the typical Planck curve of radiation that has a large range of possible outcomes. this type of energy exchange is dependent on temperature because it involves more than one molecule and the speed (kinetic energy) of the molecules.

there are also hybrid events when an excited molecule exits the collision in an unexcited state. that energy has been added to the kinetic energy or the radiation energy emitted, and is how CO2 warms the atmosphere. this type of event is dependent on both kinetic energy and the availability of excited GHGs.

this is a massively simplified explanation.

 

[CrusaderFrank]

Ah, Billy Boob, thank you for the comedy routine.

Translation: The science fails my AGWCult Theory so I have to resort to insults

 

[Billy_Bob]

Actually, Yes! Or there is the loss of full photons (IE: Receives 10 and outputs 6) but there must always be a loss which generates heat (vibration). As this is quantum theroy we really dont know the answer to this question with certainty.
......
Absorption causes the molecule to vibrate (become excited) and the interaction with other adjacent molecules is what creates the heat. Absorption means conversion from the original emitted vibration pattern -> into the electron causing it to vibrate -> When enough energy is stored the electron then sheds a photon vibrating at the new molecular temperature. Again this is QM theroy being postulated.

you are confusing kinetic energy of molecules with the excited state of CO2 which is called vibration. a CO2 molecule can absorb a specific photon and vibrate, and then emit the same photon and stop vibrating. there is no energy loss when jumping back and forth from these two states. this type of energy exchange is not dependent on temperature as it is just a single molecule.

average molecular speed, eg kinetic energy, is one of the definitions of temperature. when two molecules collide, their electron clouds are deformed and the combined speed of both is slightly less after the collision. when the electron clouds pop back into place they give off at least a photon each. this is the blackbody radiation. the hotter the object (faster moving molecules), the harder the possible collision, the more energetic the ejected photon. these collisions cause the typical Planck curve of radiation that has a large range of possible outcomes. this type of energy exchange is dependent on temperature because it involves more than one molecule and the speed (kinetic energy) of the molecules.

there are also hybrid events when an excited molecule exits the collision in an unexcited state. that energy has been added to the kinetic energy or the radiation energy emitted, and is how CO2 warms the atmosphere. this type of event is dependent on both kinetic energy and the availability of excited GHGs.

this is a massively simplified explanation.

We are NEVER dealing in single molecules.... Maybe that is the problem, the hypothetical is overruling commonsense. QM theroy is still theroy.

 

[Toddsterpatriot]

So lets review. Any molecule which absorbs energy from an incoming photon or other exciting item, which creates heat within the molecule looses the rate at which it was vibrating on entry and vibrates at the rate of the molecules heat which then emits it. The reduction in amplitude and speed (heat output) is the deciding factor in the new wavelength.

In any energy transfer there is loss.

Source

As energy passes through CO2 the wavelength changes to a length that CO2 can not reabsorb.

As energy passes through CO2 the wavelength changes to a length that CO2 can not reabsorb.

Where do you get this? I didn't see it at your source. I did see the following.

Kirchhoff's Law says that good absorbers of a particular wavelength are also good emitters of that wavelength, and poor absorbers of a wavelength are also poor emitters at the same wavelength.

A piece of steel laying on the ground is a good absorber of 0.4-2.9um but radiates at 3.6um during the day and 12.2um at night. That portion of Kirchhoff''s Law is subjective.

View attachment 34743

CO2's primary absorption bands are 0.4um, 1.2um, 2.6-2.9um, 6.1-7.5um and 9.5-15.6um. During sunlight it is not uncommon for CO2 to radiate its incoming power to its primary band of 9.5-15.6um. Power loss is inevitable. CO2 during the day will reflect or scatter outside its active bands. Once emitted CO2 will be unable to reabsorb because of entropy (cascading loss of frequency). Everyone forgets that there are other gases in our atmosphere which are in play and they do not radiate their energy at band that CO2 can reabsorb.

It is the interaction of gases which makes it nearly impossible for CO2 to reabsorb its emitted energy.

CO2's primary absorption bands are 0.4um, 1.2um, 2.6-2.9um, 6.1-7.5um and 9.5-15.6um. During sunlight it is not uncommon for CO2 to radiate its incoming power to its primary band of 9.5-15.6um.

Which CO2 emission bands are not absorbed by CO2?

 

[IanC]

absorption equals emission.

 

[Billy_Bob]

absorption equals emission.

If we were a single gas planet this would have meaning. We are not. Now lets add those other gases and see what happens.. This is precisely where AGW goes wrong. People get all giddy that CO2 passes and emits in specific wavelengths, and it may well be able to absorb that same wavelength (again QM theroy) again but they always leave out the other gases and water vapor. How they affect CO2 is not what they expected.

Water vapor is expected to increase temperature but what we have found in empirical evidence is that it has no bearing whatsoever in accelerating warming but it does accelerate cooling.