Simple physics disprove CO2 warming e...

Simple physics disprove CO2 warming effect

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Since: Sep 13

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#1 Sep 26, 2013
Imagine a blackbody at 300K surrounded by a perfect mirror which is also at 300K. The blackbody emits radiation according to its temperature. The radiation is then reflected by the mirror and returns to the blackbody. The blackbody absorbs this radiation. The mirror is also emitting its own radiation (due to being at 300K) which is also absorbed by the blackbody.

The blackbody is not warmed by the incident radiation despite receiving more than twice as much as it initially emitted. This is because the radiation is not of sufficiently short wavelength to cause it to increase its temperature.

Consider that if it were possible for this radiation to lead to a temperature increase in the blackbody then this would necessarily be a colder body heating a warmer body, which is forbidden by the second law of thermodynamics.

Applying this to AGW:
If we model the earth as a blackbody and the CO2 layer(s) as a half-silvered perfect mirror (CO2 absorbs and re-emits in both directions) we can see that the same scenario exists. For the absorption bands of CO2, the CO2 will 'reflect' 50% of the earth's emitted blackbody back to the ground, where it will have exactly no effect on the temperature of the ground, because it is of insufficiently high wavelength to cause a temperature increase. The earth is forced to re-emit this 'extra' radiation as soon as it is absorbed with no corresponding temperature increase. The re-emitted radiation travels back to the CO2 layer where it is split again, 50% goes into space and 50% returns to earth.

After several 'reflections' back and forth the 'trapped' radiation is undetectably small, and given the velocity of light the time taken for this 'trapped radiation' to escape is a tiny fraction of a second.

I believe this conclusively disproves AGW.

Comments?

Since: Sep 13

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#2 Sep 26, 2013
Correction:'insufficiently high wavelength to cause a temperature increase'

Should be:'insufficiently short wavelength to cause a temperature increase'

Since: Apr 08

"the green troll"

#3 Sep 26, 2013
Should be "simpleton physics..."

Since: Sep 13

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#4 Sep 26, 2013
Fair Game wrote:
Should be "simpleton physics..."
Do you disagree with any fact or reasoning presented?

Since: Apr 08

"the green troll"

#5 Sep 26, 2013
Aether wrote:
<quoted text>
Do you disagree with any fact or reasoning presented?
Fact? Reasoning?

Since: Apr 08

"the green troll"

#6 Sep 26, 2013
Now let me turn to the deniers. One of their favorite arguments is that the greenhouse effect does not exist at all because it violates the Second Law of Thermodynamics -- i.e., one cannot transfer energy from a cold atmosphere to a warmer surface. It is surprising that this simplistic argument is used by physicists, and even by professors who teach thermodynamics. One can show them data of downwelling infrared radiation from CO2, water vapor, and clouds, which clearly impinge on the surface. But their minds are closed to any such evidence.

Read more: http://www.americanthinker.com/2012/02/climat...
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Since: Apr 08

"the green troll"

#7 Sep 26, 2013
3) The greenhouse effect does not violate the 2nd Law of Thermodynamics. Just because the greenhouse effect (passively) makes the surface of the Earth warmer than if only (active) solar heating was operating does not violate the 2nd Law, any more than insulating your house more can raise its interior temperature in the winter, given the same energy input for heating. Very high temperatures in a system can be created with relatively small energy fluxes into that system *if* the rate of energy loss can be reduced (see #2, above). Again, energy input into a system does not alone determine what the temperature in the system will be.

http://www.drroyspencer.com/2013/01/misunders...

Since: Sep 13

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#8 Sep 26, 2013
Fair Game wrote:
Now let me turn to the deniers. One of their favorite arguments is that the greenhouse effect does not exist at all because it violates the Second Law of Thermodynamics -- i.e., one cannot transfer energy from a cold atmosphere to a warmer surface. It is surprising that this simplistic argument is used by physicists, and even by professors who teach thermodynamics. One can show them data of downwelling infrared radiation from CO2, water vapor, and clouds, which clearly impinge on the surface. But their minds are closed to any such evidence.
Read more: http://www.americanthinker.com/2012/02/climat...
Follow us:@AmericanThinker on Twitter | AmericanThinker on Facebook
That is not my argument. Go back and re-read it.

I am saying that a blackbody cannot 'heat itself' with its own reflected blackbody radiation. I am not saying anything about the atmosphere being a colder body.

According to the IPCC:
[quote]Greenhouse gases effectively absorb infrared radiation, emitted by the Earth's surface, by the atmosphere itself due to the same gases, and by clouds. Atmospheric radiation is emitted to all sides, including downward to the Earth's surface. Thus greenhouse gases trap heat within the surface-troposphere system. This is called the natural greenhouse effect.[/quote]

What I am pointing out is that the earth's surface is acting as a blackbody and the CO2 layer is acting as 50% reflector for wavelengths in its absorption bands. The proposed mechanism by the IPCC is that the earth is being 'warmed' by its own reflected radiation in these bands.

Re-read the post.

Since: Sep 13

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#9 Sep 26, 2013
Fair Game wrote:
3) The greenhouse effect does not violate the 2nd Law of Thermodynamics. Just because the greenhouse effect (passively) makes the surface of the Earth warmer than if only (active) solar heating was operating does not violate the 2nd Law, any more than insulating your house more can raise its interior temperature in the winter, given the same energy input for heating. Very high temperatures in a system can be created with relatively small energy fluxes into that system *if* the rate of energy loss can be reduced (see #2, above). Again, energy input into a system does not alone determine what the temperature in the system will be.
http://www.drroyspencer.com/2013/01/misunders...
The CO2 layer necessarily cannot retard the flow of heat from the planet because it doesn't trap heat between itself and the surface. It absorbs and isotropically re-emits thermal energy in certain bands. Of this radiation the portion which travels back down to the surface of the earth is PHYSICALLY UNABLE TO HEAT THE SURFACE and is absorbed and re-emitted instantly back to the sky, whereupon it is reabsorbed by the CO2, and again 50% of this radiation is lost to space.

If you actually took the time to follow the reasoning and the physics you would notice that all CO2 is doing is increasing the path length for escaping thermal radiation in it's absorption bands. This does *not* lead to a decrease in total thermal radiation escaping the earth.
LessHypeMoreFact

Toronto, Canada

#10 Sep 26, 2013
Aether wrote:
<quoted text>
That is not my argument. Go back and re-read it.
I am saying that a blackbody cannot 'heat itself' with its own reflected blackbody radiation. I am not saying anything about the atmosphere being a colder body.
You are making a STUPID argument. I agree that your example is in equilibrium. However it is wrong to say that 'The blackbody is not warmed by the incident radiation despite receiving more than twice as much as it initially emitted."

It receives exactly as much as it emits. You are trying to set up a paradox and failing. Both the mirror and the black body emit and receive exactly the same amount of radiation and are 'in equilibrium' with a temperature of 300K.

In the case of the black body, it's albedo is 1 and it absorbs and emits 100% of the radiation for a black body emitter/absorber. This radiation is then returned by the reflective surface in equal amounts by the sphere so it emits exactly as much as it recieves and is in balance.

In the case of the mirror, it's albedo is 0 so it absorbs and emits no radiation but does reflect the black body radiation impinging on it.
Aether wrote:
<quoted text>
I am saying that a blackbody cannot 'heat itself' with its own reflected blackbody radiation.
Yet it is doing so. The radiation it emits would NORMALLY lower it's total thermal energy and therefore cool it. But it receives the reflected radiation and this heats it up exactly by the same amount that it would 'cool'. Leaving it in a balance.
Aether wrote:
<quoted text>
According to the IPCC:
Actually according to the science of atmospheres and gas physics. The IPCC originates NO science
Aether wrote:
<quoted text>
Greenhouse gases effectively absorb infrared radiation, emitted by the Earth's surface, by the atmosphere itself due to the same gases, and by clouds. Atmospheric radiation is emitted to all sides, including downward to the Earth's surface.
GHGs do absorb LW IR radiation from all sources,generally in the form of a vibration within the molecule with a fixed quantum level.

They then re-radiate that heat as LW IR photons of exactly the same frequency in ALL directions.
Aether wrote:
<quoted text>
Thus greenhouse gases trap heat within the surface-troposphere system. This is called the natural greenhouse effect.
Since the majority of IR is coming from the hotter surface, but the emitted radiation is omnidirectional including downward back to the surface, this effectively 'reflects' a portion of the upward traveling radiation and thus forms an 'insulating effect' that retards heat flow.
Aether wrote:
<quoted text>
What I am pointing out is that the earth's surface is acting as a blackbody and the CO2 layer is acting as 50% reflector for wavelengths in its absorption bands. The proposed mechanism by the IPCC is that the earth is being 'warmed' by its own reflected radiation in these bands.
Re-read the post.
Aether wrote:
<quoted text>
What I am pointing out is that the earth's surface is acting as a blackbody and the CO2 layer is acting as 50% reflector for wavelengths in its absorption bands.
Ok. So say you are right or at least use those figures. This shows that ONLY 50% of the heat is 'getting out' which means that the surface must radiate TWICE as much to balance the incoming flow. In other words, the surface must be hotter to 'balance' incoming radiation from the sun.

This temperature difference is about 33C and called the 'natural greenhouse effect'.
LessHypeMoreFact

Toronto, Canada

#11 Sep 26, 2013
Aether wrote:
<quoted text>
The proposed mechanism by the IPCC is that the earth is being 'warmed' by its own reflected radiation in these bands.


If you look at the equilibrium *of the surface* you will find that it is hotter than can be explained by the incoming solar radiation alone. But it is still in balance at a 33C warmer temperature because it receives the 'back radiation' from the reflection of the GHGs.

And again is in 'equilibrium'.

While the planet is in equilibrium because the amount of radiation that gets BY the GHGs is enough to balance the incoming solar insolation.

Hope that helps. But you really should try to ask a local teacher. The physics isn't that hard to understand, but you seem to think that YOU know more than anyone else. This arrogance is impeding your learning.
LessHypeMoreFact

Toronto, Canada

#12 Sep 26, 2013
In the case of the black body, it's albedo is 1 ..

In the case of the mirror, it's albedo is 0

Oops. Always get those confused. The numbers are reversed as the albedo is the amount of light that is REFLECTED instead of absorbed. A black body has an albedo of 0..

And note that the same number defines the ability to absorb OR emit.

My explanation above could be modified slightly to deal with the Earths albedo of about .3 but that would complicate it somewhat without it adding anything..

Since: Sep 13

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#13 Sep 26, 2013
LessHypeMoreFact wrote:
<quoted text>
You are making a STUPID argument. I agree that your example is in equilibrium. However it is wrong to say that 'The blackbody is not warmed by the incident radiation despite receiving more than twice as much as it initially emitted."
It receives exactly as much as it emits. You are trying to set up a paradox and failing. Both the mirror and the black body emit and receive exactly the same amount of radiation and are 'in equilibrium' with a temperature of 300K.[/quote]

No. If hot mirrors are too hard for you to comprehend then consider the following equivalent scenario: a small object inside a balloon, both are at 300K. The small object has less mass and less surface area and less heat capacity than the balloon. The small object will receive a higher intensity of radiation from the balloon than it gives up to the balloon due to their surface areas. The pea will not be heated by the balloon despite receiving a much higher intensity of radiation than it emits.

[QUOTE who="LessHypeMoreFact "]<quoted text>
Ok. So say you are right or at least use those figures. This shows that ONLY 50% of the heat is 'getting out' which means that the surface must radiate TWICE as much to balance the incoming flow. In other words, the surface must be hotter to 'balance' incoming radiation from the sun.
This temperature difference is about 33C and called the 'natural greenhouse effect'.
No. The surface cannot have its temperature increased by receiving a portion of its own reflected backbody radiation. This is a pure fiction.

Instead of insulting me and my argument, please go back and re-read the original argument, go do some research on blackbody physics then come back.

Since: Sep 13

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#14 Sep 26, 2013
Sorry the quoting system is messed up:

If hot mirrors are too hard for you to comprehend then consider the following equivalent scenario: a small object inside a balloon, both are at 300K. The small object has less mass and less surface area and less heat capacity than the balloon. The small object will receive a higher intensity of radiation from the balloon than it gives up to the balloon due to their surface areas. The pea will not be heated by the balloon despite receiving a much higher intensity of radiation than it emits.

This is how blackbodies work. You cannot increase the temperature of a blackbody by exposing it to its own reflected radiation. Nor can you increase its temperature by exposing it to another blackbody of the same temperature -- no matter how large in size of surface area or intensity of radiation.
LessHypeMoreFact

Toronto, Canada

#15 Sep 26, 2013
Aether wrote:
<quoted text>
No. The surface cannot have its temperature increased by receiving a portion of its own reflected backbody radiation. This is a pure fiction.
Again, illogical nonsense. My point was that the temperature, in your example, stayed the same because the emitted radiation was exactly the same as the absorbed (and reflected) return. No change.

Please try to think and understand before responding.
Aether wrote:
<quoted text>
Instead of insulting me and my argument, please go back and re-read the original argument, go do some research on blackbody physics then come back.
I have read and understood your claims. They do not hold water. When one person thinks that they know more than the majority of those with expertise in the area, he is either VERY VERY smart or missing the boat. YOU are no Eistein. Even I can point out your errors. But it does no good if you don't think about what I said.
LessHypeMoreFact

Toronto, Canada

#16 Sep 26, 2013
Aether wrote:
If hot mirrors are too hard for you to comprehend
There is no confusion about 'hot mirrors'. The issue is your failure to understand 'black body' and albedo. A perfect mirror will return 100% of the radiation it receives. Thus a perfect mirror will not change temperature (neither absorbing nor emitting radiation) and the temperature OF the mirror is irrelevant.
Aether wrote:
consider the following equivalent scenario: a small object inside a balloon, both are at 300K. The small object has less mass and less surface area and less heat capacity than the balloon. The small object will receive a higher intensity of radiation from the balloon than it gives up to the balloon due to their surface areas. The pea will not be heated by the balloon despite receiving a much higher intensity of radiation than it emits.
This scenario is NOT equivalent.

Assuming that the two are black bodies, the smaller mass WILL be(come) hotter. The total radiation it EMITS, HAS TO equal the radiation it receives to achieve an 'equilibrium'. i.e the intensity of radiation times the area must be equal for both.

You have set up a closed system so this must apply. Initially, the two objects may be at the same temperature but the two will exchange heat energy by IR radiation until the temperature of the smaller ball is high enough to balance the radiation it receives from the larger balloon (which will have cooled).

Think of it this way. Say we suddenly had a 'second sun' at 1AU. Would the temperature of the Earth stay the same? Why or why not. Note that the second sun would be a DUPLICATE of the existing sun and the temperature would be the same.

Since: Sep 13

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#17 Sep 26, 2013
LessHypeMoreFact wrote:
<quoted text>
There is no confusion about 'hot mirrors'. The issue is your failure to understand 'black body' and albedo. A perfect mirror will return 100% of the radiation it receives. Thus a perfect mirror will not change temperature (neither absorbing nor emitting radiation) and the temperature OF the mirror is irrelevant.
<quoted text>
This scenario is NOT equivalent.
Assuming that the two are black bodies, the smaller mass WILL be(come) hotter. The total radiation it EMITS, HAS TO equal the radiation it receives to achieve an 'equilibrium'. i.e the intensity of radiation times the area must be equal for both.
You have set up a closed system so this must apply. Initially, the two objects may be at the same temperature but the two will exchange heat energy by IR radiation until the temperature of the smaller ball is high enough to balance the radiation it receives from the larger balloon (which will have cooled).
Think of it this way. Say we suddenly had a 'second sun' at 1AU. Would the temperature of the Earth stay the same? Why or why not. Note that the second sun would be a DUPLICATE of the existing sun and the temperature would be the same.
No sir, you are wrong. If the smaller body were to become hotter then this would be a violation of the second law of thermodynamics. The smaller body emits more radiation but does not increase in temperature. In other words the smaller body becomes a perfect emitter and absorber for those frequencies, retaining none of the heat energy from incident radiation below its cut off.

I can quite easily prove this mathematically if you would care to look: https://docs.google.com/file/d/0BzIpKaysqkJfY...

Notice that if you pick R = 10 m, r = 1 m, for example (so a shell of radius 10m for the outter blackbody and a sphere of radius 1 meter for the inner body) then the ratio of power received by the inner body to power emitted (in the first few instants of time) is 10000 to 1. Yet the bodies are forbidden to heat eachother due to the second law of thermodynamics. What actually happens thereafter is the inner body will receive 10000 times more energy than it is putting out, then absorb and immediately re-radiate this energy back out. Despite having greater intensity of outgoing radiation it will not become any hotter.

Your analogy with a duplicate sun is completely wrong. The earth has a much lower blackbody temperature than the sun, so the sun is able to heat the earth. If two suns of exactly the same temperature were side by side they would not be able to heat eachother, even if one was 10,000,000 times larger. This is a direct consequence of the second law of thermodynamics. I feel you need to seriously think this through before dismissing it out of hand. Please read the mathematics. Any questions, feel free to ask.
LessHypeMoreFact

Toronto, Canada

#18 Sep 27, 2013
Aether wrote:
<quoted text>
No sir, you are wrong. If the smaller body were to become hotter then this would be a violation of the second law of thermodynamics.
I agree with you on the second law. I fooled myself on that one. The system has to be in a balance between energy emitted and absorbed. Hmm.. I went wrong somehow. Where? To stay at a fixed temperature the flux received must balance the flux emitted. The flux per unit area is the same but the area differs. What am I missing??
Aether wrote:
<quoted text>
The smaller body emits more radiation but does not increase in temperature.
Now you are talking nonsense. The level of flux radiation emitted by a black body is a clear function of temperature. The hotter the object, the higher the radiation flux from it.

i.e. "Black-body radiation has the unique absolutely stable distribution of radiative intensity that can persist in thermodynamic equilibrium in a cavity.In equilibrium, for each frequency the total intensity of radiation that is emitted and reflected from a body (that is, the net amount of radiation leaving its surface, called the spectral radiance) is determined solely by the equilibrium temperature, and does not depend upon the shape, material or structure of the body."

Hmm. This implies that the two black bodies, while differing in shape and area, emit the same amount of heat energy?
http://en.wikipedia.org/wiki/Black-body_radia...
Aether wrote:
<quoted text>
In other words the smaller body becomes a perfect emitter and absorber for those frequencies, retaining none of the heat energy from incident radiation below its cut off.
It does not BECOME a perfect emitter and absorber. It is a black body which is DEFINED as a perfect absorber and emitter.
Aether wrote:
<quoted text>
Notice that if you pick R = 10 m, r = 1 m, for example (so a shell of radius 10m for the outer black body and a sphere of radius 1 meter for the inner body) then the ratio of power received by the inner body to power emitted (in the first few instants of time) is 10000 to 1.
The surface area of a sphere is 4\pi r^2 so with the same temperature, the flux would be 10^10 or 100, not 10,000 times.
Aether wrote:
<quoted text>
Yet the bodies are forbidden to heat each other due to the second law of thermodynamics. What actually happens thereafter is the inner body will receive 10000 times more energy than it is putting out, then absorb and immediately re-radiate this energy back out. Despite having greater intensity of outgoing radiation it will not become any hotter.
This conflicts with the fact that "In equilibrium, for each frequency the total intensity of radiation that is emitted and reflected from a body (that is, the net amount of radiation leaving its surface, called the spectral radiance) is determined solely by the equilibrium temperature, and does not depend upon the shape, material or structure of the body"

This is where I went wrong. The area is not an issue for thermal equilibrium (i.e shape) otherwise I could create a hollow ball to increase emission.
Aether wrote:
<quoted text>
If two suns of exactly the same temperature were side by side they would not be able to heat each other, even if one was 10,000,000 times larger.
Agreed. But the case of the earth being heated is not analogous to your example since the sun IS hotter. And the flux that is received by the earth must be balanced by the flux emitted by the earth. This is NOT comparable to the 'thermal equilibrium' of your example.
Aether wrote:
<quoted text>
Any questions, feel free to ask.
I agree that I made a mistake on your example. I do learn things here. But you should apply the same rule yourself.

Since: Sep 13

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#19 Sep 27, 2013
LessHypeMoreFact wrote:
<quoted text>
I agree with you on the second law. I fooled myself on that one. The system has to be in a balance between energy emitted and absorbed. Hmm.. I went wrong somehow. Where? To stay at a fixed temperature the flux received must balance the flux emitted. The flux per unit area is the same but the area differs. What am I missing??
<quoted text>
Now you are talking nonsense. The level of flux radiation emitted by a black body is a clear function of temperature. The hotter the object, the higher the radiation flux from it.
i.e. "Black-body radiation has the unique absolutely stable distribution of radiative intensity that can persist in thermodynamic equilibrium in a cavity.In equilibrium, for each frequency the total intensity of radiation that is emitted and reflected from a body (that is, the net amount of radiation leaving its surface, called the spectral radiance) is determined solely by the equilibrium temperature, and does not depend upon the shape, material or structure of the body."
Hmm. This implies that the two black bodies, while differing in shape and area, emit the same amount of heat energy?
http://en.wikipedia.org/wiki/Black-body_radia...
<quoted text>
It does not BECOME a perfect emitter and absorber. It is a black body which is DEFINED as a perfect absorber and emitter.
<quoted text>
The surface area of a sphere is 4\pi r^2 so with the same temperature, the flux would be 10^10 or 100, not 10,000 times.
<quoted text>
This conflicts with the fact that "In equilibrium, for each frequency the total intensity of radiation that is emitted and reflected from a body (that is, the net amount of radiation leaving its surface, called the spectral radiance) is determined solely by the equilibrium temperature, and does not depend upon the shape, material or structure of the body"
This is where I went wrong. The area is not an issue for thermal equilibrium (i.e shape) otherwise I could create a hollow ball to increase emission.
<quoted text>
Agreed. But the case of the earth being heated is not analogous to your example since the sun IS hotter. And the flux that is received by the earth must be balanced by the flux emitted by the earth. This is NOT comparable to the 'thermal equilibrium' of your example.
<quoted text>
I agree that I made a mistake on your example. I do learn things here. But you should apply the same rule yourself.
I'm sorry can you please identify the error in the math in the paper I wrote rather than simply saying it's wrong.

Secondly, you have no addressed the issue at all. The emission is a function only of temperature and wavelength, agreed. However if the blackbody is a perfect emitter and absorber then incident radiation on it below the cut-off frequency must be absorbed. And if it is absorbed by cannot add heat energy then it must be immediately re-emitted. You haven't addressed this point.

If you are correct in your analysis then with a star 10,000 times the size of the smaller star, the smaller star should heat up due to the presence of the larger star because the incoming flux is greater than the outgoing flux -- according to you.

This is a clear violation of the second law, so explain.

Since: Sep 13

Location hidden

#20 Sep 27, 2013
Sorry, correction:

If you are correct in your analysis then with a star 10,000 times the size of the smaller star -- which are both at the same temperature -- the smaller star should heat up due to the presence of the larger star because the incoming flux is greater than the outgoing flux -- according to you.

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