Guest Post by Tom Vonk
In a recent post I considered the question in the title. You may see it here : http://wattsupwiththat.com/2010/08/05/co2-heats-the-atmosphere-a-counter-view/
The post generated great deal of interest and many comments.
Even if most of the posters understood the argument and I answered the comments of those who did not, I have been asked to sum up the discussion.
Before starting, I will repeat the statement that I wished to examine.
“Given a gas mixture of CO₂ and N₂ in Local Thermodynamic Equilibrium (LTE) and submitted to infrared radiation, does the CO₂ heat the N₂?”
To begin, we must be really sure that we understood not only what is contained in the question but especially what is NOT contained in it.
- The question contains no assumption about the radiation. Most importantly there is no assumption whether a radiative equilibrium does or does not exist. Therefore the answer will be independent from assumptions concerning radiative equilibrium. Similarly all questions and developments concerning radiative transfer are irrelevant to the question.
- The question contains no assumption about the size or the geometry of the mixture. It may be a cube with a volume of 1 mm³ or a column of 10 km height. As long as the mixture is in LTE, any size and any geometry works.
- The question contains no assumption about boundary conditions. Such assumptions would indeed be necessary if we asked much more ambitious questions like what happens at boundaries where no LTE exists and which may be constituted of solids or liquids. However we do not ask such ambitious questions.
Also it is necessary to be perfectly clear about what “X heats Y” means.
It means that there exists a mechanism transferring net (e.g non zero) energy unidirectionaly
from X to Y .
Perhaps as importantly, and some posters did not understand this point, the statement
“X heats Y” is equivalent to the statement “Y cannot cool X”.
The critical posts – and here we exclude posts developing questions of radiative transfer which are irrelevant as explained in 1) above – were of 2 types.
Type 1
The argument says “LTE never exists or alternatively LTE does not apply to a mixture of CO₂ and N₂.”
The answer to the first variant is that LTE exists and I repeat the definition from the original post : “A volume of gas is in LTE if for every point of this volume there exists a neighborhood in which the gas is in thermodynamic equilibrium (TE)”
2 remarks to this definition:
- It is not said and it is not important how large this neighborhood of every point is. It may be a cube of 1 mm³ or a cube of 10 m³ . The important part is that this neighborhood exists (almost) everywhere.
- LTE is necessary to define local temperature. Saying that LTE never exists is equivalent to saying that local temperatures never exist.
The second variant admits that LTE exists but suggests that a mixture of CO₂ and N₂ cannot be in LTE.
The LTE conditions are given when energy at every point is efficiently spread out among all available degrees of freedom (translation, rotation, vibration).
The most efficient tool for energy spreading are molecular collisions.
Without going in a mathematical development (see statistical thermodynamics for those interested), it is obvious that LTE will exist when there are many molecular collisions per volume unit.
This depends mostly on density – high density gases will be often in LTE while very low density gases will not.
For those not yet convinced, hold out a thermometer in your bedroom and it is probable that it will show a well defined temperature everywhere – your bedroom is in LTE .
We deal here with a mixture of CO₂ and N₂ in conditions of the troposphere which are precisely conditions where LTE exists too.
Type 2
The argument says “The mean time between collisions is much shorter than the mean decay time (e.g time necessary to emit a photon) and therefore all infrared energy absorbed by the CO₂ molecules is immediately and unidirectionaly transferred to the N₂ molecules.”
In simple words – the CO₂ never has time to emit any IR photons because it loses vibrational energy by collisions instead.
This statement is indeed equivalent to the statement “CO₂ heats N₂”.
Now let us examine the above figure.
The good understanding of this figure will do much better than only answering the original question. It will also make clear to everybody what is really happening in our gas mixture in LTE.
The figure shows the distribution of the kinetic energy (Ox axis) among the N₂ molecules (Oy axis).
This typical curve is called the Maxwell Boltzmann distribution, has been known for more than 100 years and experimentally confirmed with high accuracy.
We know that the temperature is defined by <E>, the energy average.
Hence it is the curve shown in the figure that defines the temperature of a gas.
Another way to say the same thing is to say that the curve depends only on temperature. If we wanted to have the distribution for another gas than N₂ , f.ex CO₂ or O₂, it would be given by an identical curve.
The blue curve gives the distribution of kinetic energy at 25°C while the red curve gives the distribution at 35°C.
The minimal energy is small but non-zero and there is no maximal energy.
A very important point on the Ox axis is the energy of the first vibrationally excited state of a CO₂ molecule.
You notice that at 25°C the majority of N₂ molecules has insufficient kinetic energy to excite this vibrational state.
Only the proportion of them given by the dark blue surface has enough energy to excite the vibrational state by collision.
When the temperature increases to 35°C, you notice that the proportion of N₂ molecules able to excite the vibrational CO₂ state by collision has significantly increased .
This proportion is given by the sum of the dark blue and light blue surface.
You also notice that as there exists no maximal energy, there will be a proportion of N₂ molecules able to excite the vibrational CO₂ state at any temperature.
Trivial so far? Well it will not get much more complicated.
First 2 technical points which play no role in the argument but which I would like to mention for the sake of completness.
- The figure shows the translational kinetic energy. Even if in some (popular) literature the temperature is defined as being an average of the translational kinetic energy, this is not strictly true.
The temperature is really defined as an average of all energy modes. So what about the vibrational and rotational energy?
At the low tropospheric temperatures we are considering, the distribution of the vibrational energy is extremely simple : about 5% or less of the molecules are in the first excited state and 95% or more are in the ground state.
As for the rotational energy, it can be computed classically without quantum corrections and the result is that it also follows a Maxwell Boltzmann distribution.
Therefore if we wished to plot the total energy (Etranslational + Evibrational + Erotational) we would rescale the Ox axis and obtain exactly the same curve as the one that is shown.
However as we are interested in studying the T/V interactions, it is the curve of the translational kinetic energy that interests us.
- We find the omnipresence of LTE again. This curve has been derived and experimentally confirmed only, and only if, the gas is in TE. Therefore the following 2 statements are equivalent :
“The gas is in LTE” , “The energy distribution at every point is given by the Maxwell Boltzmann distribution” .
If you feel that these statements are not equivalent, reread carefully what is above.
Now we can demonstrate why the Type2 argument is wrong.
Imagine that you mix cold N₂ represented by the blue curve in the Figure with highly vibrationally excited CO₂. The mixture would then not be in LTE and a transient would take place.
In the molecular process (1) CO₂* + N₂ → CO₂ + N₂⁺ which says that a vibrationally excited CO₂ molecule (CO₂*) collides with an N₂ molecule , decays to the ground state (CO₂) and increases the translational kinetic energy of N₂ (N₂⁺) , there would be a net energy transfer from CO₂* to N₂ .
As a result of this transfer the temperature of N₂ would increase and the blue curve would move to the red one.
However doing that, the number of molecules able to excite CO₂ vibrationally would increase (see the blue surfaces in the figure).
That means that during the increase of the temperature of N₂ , the rate of the opposite molecular process (2) CO₂ + N₂⁺ → CO₂* + N₂ where N₂ molecules (those from the blue surface in the figure) vibrationally excite CO2 molecules, will increase too.
Of course the transient net energy transfer from CO₂ to N₂ will not continue forever because else the mixture would transform into superheated plasma.
A local equilibrium will be established at each point and in this equilibrium the rate of the process (1) will be exactly equal to the rate of the process (2).
The curve of energy distribution will stop moving and the Maxwell Boltzmann distribution will describe this distribution at every point.
This is exactly the definition of LTE.
The transient will stop when the mixture reaches LTE and its characteristic feature is that there is no local net energy transfer from CO₂ to N₂.
This result demonstrates both that the Type2 argument is wrong and that the answer on the question we asked at the beginning is “No”.
In very simple words, if you take a small volume (for example 1 m³) of the CO₂ and N₂ mixture in LTE around any point , then there cannot be any net energy transfer from CO₂ to N₂ within this volume.
To establish the last step we will take the following statements.
- The result obtained for the CO₂ and N₂ mixture in LTE is equally true for a mixture containing 78% of N₂ , 21% of O₂ , x% of CO₂ and 1-x % H₂O in LTE.
- The mixture defined above approximates well the troposphere and the troposphere is indeed in LTE
- From the 2 statements above and the demonstrated result follows :
“The CO₂ does not heat the troposphere” what is the answer on the question asked in the title.
Caveat1
I have said it both in the initial post and in this one.
Unfortunately, I know that it can’t be avoided and that some readers will still be confused about the result established here and start considering radiative transfers or radiative equilibriums.
That’s why I stress again that LTE and the result established here is totally independent of radiative equilibriums and radiative transfer properties.
However it does falsify one misconception concerning radiative properties of CO₂ that has also figured in the comments and that is that “CO₂ does not radiate at 15µ because it “heats” N₂ instead”.
It is also to be noted that we consider only the T/V process because it is only the vibrational modes that interact with IR radiation.
There are also rotational/translational and rotational/vibrational transfers.
The same argument used for T/V applies also for the R/T and R/V processes in LTE – e.g there is no net energy transfer between these modes in LTE even if for example the R/T process has a much higher probability than a T/V process.
For the sake of clarity we don’t mention specifically the R/T and R/V processes.
Caveat2
The result established here is a statistical thermodynamics bulk property.
This property is of course not sufficient to establish the whole dynamics of a system at all time and space scales.
If that was our ambition – and it is not – then we would have to consider boundary conditions and macroscopic mass, momentum and energy transfers, e.g convection, conduction, phase changes, lapse rates etc.
More specifically this result doesn’t contradict the trivial observation that if one changes the parameters of the system, for example composition, pressure, radiation intensity and spectrum, etc, then the dynamics of the system change too.
Yet it contradicts the notion that once these parameter are fixed there is a net transfer of energy from CO₂ to the troposphere. There is not.
Caveat3
It will probably appear obvious to most of you but it has also to be repeated.
This result says little about comparisons between the dynamics of 2 very different systems such as, for example, an Earth without oceans and atmosphere, and an Earth with oceans and atmosphere. Clearly the dynamics will be very different but it stays that in the case of the real Earth with an atmosphere in LTE, there will be no net energy transfer from the CO₂ to the atmosphere.
It seems that LTE has been defined such that these events are excluded and the process of heating causes continual equilibration to a new LTE states. Some of this seems like asking if a window heats your house when it lets the sunlight come through.
Phil. says:
September 3, 2010 at 9:27 pm
The existence of a black body type radiation from gases too shows there is a continuum statistically of photon energies. It is not necessary to think of “appropriate energy levels” because the continuous motion and change of direction of molecules integrates over all phase space and creates a continuum. Individual molecules need appropriate energy to get excited and deexcited (except when at the “continuum” of their potential well spectrum, hi l and n,..) . Collectively, because of the motion of each molecule, which is of commensurate energy, the effect is a continuum and it is confusing people to keep talking as if there is no continuum of photons, i.e. electromagnetic waves, appropriate for the classical thermodynamic formulation.
I remind you that a monatomic gas also has black body type radiation which comes from the continuous collisions which are exchanged soft photons.
Oliver Ramsay says:
September 3, 2010 at 9:11 pm
Surely, that photon would have been long gone into the void if CO2 hadn’t absorbed it and therefore it’s reasonable to say that CO2 introduced that energy to the atmosphere.
Put a brick next to a fireplace. When hot enough wrap it and place it in the bed. Is the heat coming from the brick, or is it just that it has a large heat capacity and releases the heat slower than if it were made of iron?
CO2 changes the heat capacity of the atmosphere.
I have to echo Dave Springer’s aversion to “might”. That does sound implausibly haphazard.
I hope my second post clarified that it was a rhetorical “might”.
anna,
as I’ve come to understand it, there are problems with that continuum concept. While a gas has all sorts of kinetic energies involved, they are not going to smear out sets of lines into a continuum until extremely high temperatures are reached. What causes the continuum from the photosphere of the sun, which corresponds to having a pressure similar to 50km above Earth’s surface, is the presence of free electrons caused by the metals present in the gas composition and H- ions. Note metals are defined differently in astronomy.
The same sort of thing is observed in vapor lamps where one gets a spectrum of the gas until the current density rises to a somewhat higher point.
Matter in solid or liquid form continuums readily and clumps of atoms called dimers in the atmosphere do form at least broadened wings. Gases behave with spectrums. That would suggest the continuum is associated with the interactions of structures of atoms and molecules.
Also, when it comes to the notion of deactivating internal energy states by collisions, there is reciprocity with creating higher energy states by collisions.
BTW, what continuum can you have from monatomic hydrogen at temperatures like room temperature that do not raise the electron above the ground state?
cba says:
September 4, 2010 at 5:18 am
anna,
as I’ve come to understand it, there are problems with that continuum concept. While a gas has all sorts of kinetic energies involved, they are not going to smear out sets of lines into a continuum until extremely high temperatures are reached. What causes the continuum from the photosphere of the sun, which corresponds to having a pressure similar to 50km above Earth’s surface, is the presence of free electrons caused by the metals present in the gas composition and H- ions. Note metals are defined differently in astronomy.
The same sort of thing is observed in vapor lamps where one gets a spectrum of the gas until the current density rises to a somewhat higher point.
Matter in solid or liquid form continuums readily and clumps of atoms called dimers in the atmosphere do form at least broadened wings. Gases behave with spectrums. That would suggest the continuum is associated with the interactions of structures of atoms and molecules.
Also, when it comes to the notion of deactivating internal energy states by collisions, there is reciprocity with creating higher energy states by collisions.
BTW, what continuum can you have from monatomic hydrogen at temperatures like room temperature that do not raise the electron above the ground state?
The problems you quote are of the type ” for gases at atmospheric temperatures and pressures there are holes/windows in the black body type distribution”. Still, that looks continuous enough for calculations, as far as I am concerned. Vapor lamps and 50km high are not normal atmospheric conditions , where one is describing the greenhouse effect.
This is not my field of physics, for sure. I am taking a step at a time. Lets take helium, a true monatomic gas.
1) Will a helium balloon in space loose temperature according to some black body type radiation?
The answer must be yes, because otherwise we would have had the perfect insulator.
2) How is it loosing energy, considering that there are no molecular vibrational and rotational levels?
The answer must be by collisions.
3) What do collisions mean at the quantum mechanical level?
They mean scattering
4)How does scattering happen between two atoms with tightly bound electrons?
By something like Wan Der Waals forces: Virtual photons are exchanged with the electron fields during the collision, when the two atoms kinetically are close enough for the fields to be seen, and part of the kinetic energy is lost by a real soft photon to space. A diagram similar to Compton scattering. The electrons stay at the energy levels where they are. If they were excited, there would be no energy loss for the gas.The loss has to be with radiation to space, and is mainly a continuum because the kinetic energies are a continuum.
I believe this is the only QM picture of black body radiation that makes sense.
The sad thing is that one has to go through all this yoga positions of QM unnecessarily, and confusingly, because classical thermodynamics and statistical mechanics are adequate to describe what is happening with gases at normal atmospheric pressures and temperatures.
anna v says;
“Put a brick next to a fireplace. When hot enough wrap it and place it in the bed. Is the heat coming from the brick, or is it just that it has a large heat capacity and releases the heat slower than if it were made of iron?
CO2 changes the heat capacity of the atmosphere.”
———–
I was aware that people around the world had pretty much given up filling their beds with hot rocks, but I thought that was due to the advent of central heating and electric blankets.
You’re suggesting that it was actually the realization that it didn’t work that led them to abandon the centuries-old practice.
I’m now wondering if I should throw out the batteries from my flashlight, since they’re just storing energy and are unable to produce light.
Are you saying, too, that the milkman doesn’t deliver milk, a cow does?
Oliver Ramsay says:
September 4, 2010 at 7:19 am
I guess I am saying a milkman does not make milk.
Batteries I would consider a different story, in the saga of energy transformations.
CO2 is not a battery. It stores energy for a delta( t )and releases it, it increases the heat capacity of the gas. H2O of course does that much more efficiently.
@ur momisugly anna v
So, if the atmosphere were 99% CO2 and 1% N2 and subjected to EMR of 85nm, would the N2 not heat the CO2? That is to say, would it not be the medium by which energy from the source was introduced into the CO2?
anna v says:
September 4, 2010 at 1:24 am
Phil. says:
September 3, 2010 at 9:27 pm
The existence of a black body type radiation from gases too shows there is a continuum statistically of photon energies.
Thanks for confirming that it wasn’t a misunderstanding due to language, you really don’t understand the physics of the absorption/emission of electromagnetic radiation by gases.
About LTE. I recently went through a 36 lecture long thermodynamics coarse at MIT Open Course Ware just to make sure I stay on the same wavelength and I have found that terminology may be causing most of the discord between the many comments above on others it is when speaking of different scales. To some it seems think LTE, let’s say a one cubic meter, cannot be in LTE if any energy is flowing through it, especially radiation. Since all matter has energy flowing in and out of it all of the time, that is a non-physics state that would only be true “no where in our universe”. What I have always thought and reinforced by that coarse is that cubic meter is in LTE as long as every sub-volume is at the same temperature. That means no changes of any kind are happening that would cause one of those sub-areas to have a different temperature. I imagine those sub-areas to be something like a cubic millimeter, not a square yoctometer so it gets so incredibly small that quantum single-molecular actions do affect the “local” sub-volume state. I think that is what Tom Vonk was saying. That is also not saying energy is not flowing through this cubic meter, it is just that that flow never affects the temperature at any one place in that cubic meter. Now if some of you learned a different meaning to LTE, put a name, let’s say ‘sam’ on this definition and let’s talk about sam.
Now to what is in that cubic meter in LTE, air. I imagine one stationary cubic meter of it above my patio where I’m taking a break and about 100 meters off of the ground. It is dead quiet, no convection or wind to mess up our static example, i.e., drop the weather effects. Are the CO2 molecules in that cubic meter absorbing infrared from the surface. Of course, so are H2O molecules. Are they transferring that energy immediately to the translational speed of N2 and any other molecules? Yes. That has been thoroughly discussed and most agree. Can a portion of the N2 molecules have enough velocity to vibrationally re-excite CO2 molecules. Yes. Does this happen in equilibrium, yes for this cube is in equilibrium, it never changes on a second scale. But these are so small that there are billions or many trillions in any one cubic millimeter so the temperature of all of the cubic millimeters in that cubic meter are always at the same temperature, at least to many digits of precision, say 0.0000001ºC, that is ignorable by convention.
Now the big question, in that cubic meter does the CO2 heat the N2? I follow Tom Vonk on his logic. I would say no. At a single-molecule or quantum level, yes, but at a thermodynamically level, no. At the single molecule there is heating and cooling between these molecules in an equal amount and gazillion times per second locally. If a photon leaves this cubic meter one instantly is replaced, that is, out = in and smoothly distributed.
Here’s another thing I feel is left out many times, anytime you speak about radiation heating or thermalization, there is always cooling also, when you speak of cooling, there is heating. A photon absorbed by CO2 cooled the surface where it came from and warms the CO2 molecule. If it then thermalizes a N2 molecule, the CO2 gets cooler and the N2 gets warmer, no net warming. If a high-velocity N2 molecule excites a CO2 molecule, the N2 cools and the CO2 warms. So where exactly is the net warming? I think that was Tom’s simple point. To actually make it warmer you would have to make my patio warmer below it, then we would no longer be in equilibrium for a while until it is re-established but he wasn’t speaking of that.
In that manner I am still not sold that an increase of CO2 or H2O molecules make that cubic meter warmer if the influx of radiation from below doesn’t increase. Not in this narrow example. However, does it increase the radiation going out of this cube to cause a warmer surface below? Once again I say no. That would cause more output without more input and we just determined that the vast amount of transfer out of CO2 molecules is via thermalization, not radiation. Make no sense right there to me. However, one more jab, does more CO2 widen the wings on the absorption bands therefore closing a no-interaction window? Now that would cause some net warming but only if H2O doesn’t intersect at the wings and the change it seems would be very tiny.
Oliver Ramsay says:
September 4, 2010 at 11:44 am
@ur momisugly anna v
So, if the atmosphere were 99% CO2 and 1% N2 and subjected to EMR of 85nm, would the N2 not heat the CO2? That is to say, would it not be the medium by which energy from the source was introduced into the CO2?
Physics is an exact science, i.e. statements made have to lead to calculations and calculable quantities. I would accept the statement that it would be the medium. Being the medium is not the same as heating.
“”wayne says:
September 4, 2010 at 3:24 pm “”
I agree with wayne’s post. Consider a cubic meter of air directly under the cubic meter of air wayne talks about: it will have a tiny bit more thermal energy (lapse rate), a tiny bit more density (pressure gradient due gravity), and a tiny bit more total mass (total number of molecules will be that much greator). Under the same conditions of the air above the lower cubic meter will radiate that tiny bit MORE ENERGY than the cubic meter above. Convention has it that in a column of air a parcel of that air (our cubic meter) will radiate half the available energy up and half down.
Therefore the lower cubic meter will radiate slightly more energy upwards than is coming down from the cubic meter of air above.
CO2 in the atmosphere provides a mechanism to transport energy UPWARDS from the surface to space. Increasing CO2 improves the mechanism.
When CO2 absorbs a 15um photon the energy is instantaneously thermalised into v2 bending vibration.
The translational KE of the CO2 molecule is unaffected so its temperature is unchanged.
However the vibrational KE can be transferred (most likely to N2) in a V to T exchange increasing the local temperature.
Emission is a lot less likely than absorption (5 emissions for 100 absorptions at 260K).
Hence the radiative effect of CO2 in the troposphere seems to be a tiny heating effect from this trace gas.
For CO2 above the troposphere where temperatures increase then emission becomes increasingly more possible and a cooling effect can occur especially at night.
However the combined effects are very much on the margins of heat transfer in the atmosphere.
anna v said;
“Physics is an exact science, i.e. statements made have to lead to calculations and calculable quantities. I would accept the statement that it would be the medium. Being the medium is not the same as heating.”
————-
Well, I can’t say that I wasn’t told!
I just couldn’t believe the point to be made was so trivial, especially amid abstruse discussion of pressure broadening etc.
“Physics is an exact science”, you say, but language is not and although definitions abound for the noun ‘heat’, that’s not the case for the verb, either transitive or intransitive.
Tom Vonk took great pains to describe LTE, when he could have just declared that he had a problem with the word “heat”. In a subsequent comment, he said this piece was for the benefit of non-physicists; I didn’t realize he meant ‘idiots’.
ana, this cavilling is particularly galling since you unapologetically employed the language of rhetoric in your own assertions.
Fortunately, there are plenty of interesting discussions that have arisen on the side and provided food for thought, and the insight into the perversity of academic punctiliousness at the expense of perspicuity is painful but potent.
At 288 K the black-body radiation or total emissive power from Earth is 391W/m^2 (124 Btu/h-ft^2). CO2 in the atmosphere absorbs 79.8W/m^2 after 3600m (CO2 absorptivity 0.184 at 350ppm or PcL of 1.24 m.atm). H2O in the atmosphere absorbs 248W/m^2 after 120m (H2O absorptivity 0.573 for a PwL of 2.77 m.atm).
Doubling CO2 to 700ppm would absorb the same 79.8W/m^2 after only 2000m ( CO2 absorptivity 0.195 at 700ppm 0r PcL of 1.4 m.atm).
The total radiant heat absorption in the atmosphere is unaffected by CO2 increases. Any increases seem only to have slight effects on heat distribution.
wayne says:
September 4, 2010 at 3:24 pm
“…To some it seems think LTE, let’s say a one cubic meter, cannot be in LTE if any energy is flowing through it, especially radiation. Since all matter has energy flowing in and out of it all of the time, that is a non-physics state that would only be true “no where in our universe”. What I have always thought and reinforced by that coarse is that cubic meter is in LTE as long as every sub-volume is at the same temperature.”
If the gas is subjected to non-equilibrium radiation, then it is is not in LTE, because there is no volume, however small, in which all the energy modes of the gas are in equilibrium or “at the same temperature”. Equipartition is violated. The degree to which this is true may be very small (at tropospheric densities thermalisation is very fast), but you still cannot use the assumptions of LTE and equipartition to analyse the radiative properties of the system, as Vonk erroneously attempts, because it is precisely those small departures from LTE that are important.
“R Stevenson says:
September 5, 2010 at 5:07 am
At 288 K the black-body radiation or total emissive power from Earth is 391W/m^2 (124 Btu/h-ft^2). CO2 in the atmosphere absorbs 79.8W/m^2 after 3600m (CO2 absorptivity 0.184 at 350ppm or PcL of 1.24 m.atm). H2O in the
”
Those look a bit strong but it is not relevent as the atmosphere is radiating absorbed energy as well as absorbing it. That means power gets transferred over greater distances. Or put another way, there’s not nearly that much absorption going on because some of it continues on being radiated again.
Richard111 says:
September 4, 2010 at 10:02 pm
…
CO2 in the atmosphere provides a mechanism to transport energy UPWARDS from the surface to space. Increasing CO2 improves the mechanism.
Richard, thanks for the reply. Everything you said seems true but possibly the last sentence… maybe. I’m trying hard to stay perfectly correct in physics and I’ll have to think hard on that last one. If you are speaking of N2 or O2 that has touched the warmer surface and by conduction become warmer then, more CO2 would allow that to release it’s increased energy faster by reverse thermalization and then radiated, I see that. Without the CO2 or water molecules that heat would have no way to cool.
Paul Birch says:
September 5, 2010 at 9:17 am
“If the gas is subjected to non-equilibrium radiation, then it is is not in LTE, because there is no volume, however small, in which all the energy modes of the gas are in equilibrium or “at the same temperature”. Equipartition is violated. The degree to which this is true may be very small (at tropospheric densities thermalisation is very fast), but you still cannot use the assumptions of LTE and equipartition to analyse the radiative properties of the system, as Vonk erroneously attempts, because it is precisely those small departures from LTE that are important.”
I do want your input but I can’t follow some terms above so I can’t follow you:
“subjected to non-equilibrium radiation” – in my example there is never “non-equilibrium” anywhere at any time, if a photon comes in one is going out, everywhere, you can speak of the whole cubic meter or a square millimeter portion of the cubic meter.
“there is no volume” – what do you mean here about volume when tied to my example, so why ‘equipartition is violated’. You then speak of troposphere (huge scale) and I was just speaking of one simple cubic meter above my patio. But yes, thermalization is nearly instantaneous.
“because it is precisely those small departures from LTE” – what departures in my example?
Help me understand what you mean exactly.
anna v says:
September 4, 2010 at 6:46 am
This is not my field of physics, for sure. I am taking a step at a time. Lets take helium, a true monatomic gas.
1) Will a helium balloon in space loose temperature according to some black body type radiation?
The answer must be yes, because otherwise we would have had the perfect insulator.
2) How is it loosing energy, considering that there are no molecular vibrational and rotational levels?
The answer must be by collisions.
3) What do collisions mean at the quantum mechanical level?
They mean scattering
4)How does scattering happen between two atoms with tightly bound electrons?
By something like Wan Der Waals forces: Virtual photons are exchanged with the electron fields during the collision, when the two atoms kinetically are close enough for the fields to be seen, and part of the kinetic energy is lost by a real soft photon to space. A diagram similar to Compton scattering. The electrons stay at the energy levels where they are. If they were excited, there would be no energy loss for the gas.The loss has to be with radiation to space, and is mainly a continuum because the kinetic energies are a continuum.
I believe this is the only QM picture of black body radiation that makes sense.
I so glad you got into this area of continuum and scattering. I found last night in a book on atmospheric physics that this small underlying radiative trait of the atmosphere as a whole does have a component of a continuous black body radiation. It is tied to electron interactions as collisions occur. As you said, if not N2 would be a perfect insulator, never able to shed it’s heat, and that is not real. This book did admit that this is poorly understood area in radiative physics. Darn, no numbers.
Regarding radiative cooling, you might want to look at the article that Dr. Spencer wrote on that subject a few months ago:
Spencer: Earth sans greenhouse effect – what would it be like?
http://wattsupwiththat.com/2010/01/02/spencer-earths-sans-greenhouse-effect-what-would-it-be-like/
or
http://www.drroyspencer.com/2009/12/what-if-there-was-no-greenhouse-effect/
I believe most of what we call ‘weather’ is due to the fact that the Earth can expel heat more effectively by radiation in the thin air of the upper atmosphere than it can from the surface. If this were not the case, the upper atmosphere would be progressively warmed by up-welling convection, much like that indicated on the left-hand chart at the top of this article.
dammit: I may be wrong again, but I think all this discussion of “blackbody radiation” totally ignores the simple fact that the “blackbody radiator” is a purely theoretical construct. There are no black bodies in reality. The REAL WORLD substances “soak up” heat, as well as radiate energy. BBs don’t do that–they irradiate EXACTLY what they receive. If I’m correct, this makes one very BIG difference in how the atmosphere and planet interact with solar radiation. No?
jae
The infamous K&T diagram shows the Earth Surface radiation as 396W/m2.
To get this figure they used the emissivity of the Earth surface at its maximum “perfect” value of 1.
Even Wikipedia do not go as far as that:
Emissivity is…..
0.96 to 0.99[5][6] (except for some small desert areas which may be as low as 0.7). Clouds, however, which cover about half of the earth’s surface, have an average emissivity of about 0.5[7]“…….
Now you might think that K&T did a rough and ready calculation and just picked the nearest whole number.
However they give the surface radiation as 396W/m2 that is to three figures of accuracy.
Now I suspect that in every well run university physics department the students are told that if they give a result to three significant then all data used to calculate that result should be accurate to three figures or better.
What conclusion can we come to?
K&T don’t know any better and just muddle along?
Or is it a deliberate exaggeration so as to inflate figures to make the so called “Greenhouse effect” appear to have some reality!
For instance you take water which covers 70% of the planets surface it is not some perfectly flat laboratory surface.
If you sail a boat you will be well aware that the surface of the sea has a turbulent surface with breaking waves and spray.
How could anyone consider this as a perfect black body?
wayne says:
September 5, 2010 at 3:31 pm
Paul Birch says:“If the gas is subjected to non-equilibrium radiation, then it is is not in LTE, because there is no volume, however small, in which all the energy modes of the gas are in equilibrium or “at the same temperature”. Equipartition is violated. The degree to which this is true may be very small (at tropospheric densities thermalisation is very fast), but you still cannot use the assumptions of LTE and equipartition to analyse the radiative properties of the system, as Vonk erroneously attempts, because it is precisely those small departures from LTE that are important.”
“I do want your input but I can’t follow some terms above so I can’t follow you: “subjected to non-equilibrium radiation” – in my example there is never “non-equilibrium” anywhere at any time, if a photon comes in one is going out, everywhere, you can speak of the whole cubic meter or a square millimeter portion of the cubic meter.”
We have to distinguish between equilibrium and mere steady state. In thermal equilibrium, we would have black body radiation at the same temperature as the bulk gas. If the radiation is anything other than this (as it is everywhere in the atmosphere, because it is coming from places with different temperatures) we do not have thermal equilibrium. In this case the number of photons coming out will in general be more than the number coming in, and the photons coming out will in general have lower average energies (will be at a longer wavelength) than the photons coming in. There is an entropy increase continually taking place. This is true so long as there is any absorption whatsoever.
““there is no volume” – what do you mean here about volume when tied to my example, so why ‘equipartition is violated’. You then speak of troposphere (huge scale) and I was just speaking of one simple cubic meter above my patio. But yes, thermalization is nearly instantaneous.”
I mean that in this non-equilibrium state, caused and continually replenished by the absorption of non-equilibrium radiation, the various molecular species and their vibrational and rotational modes are in general at different temperatures (have different distributions of energy) even in the same place. The assumption of LTE requires equipartition – that all these energy modes have the same distribution of energy, in equilibrium with each other. There is no single thermodynamic temperature that can reproduce the patterns of excitation obtained under this non-equilibrium radiation. Thermalisation may be very fast, but it misleading to think of it as (nearly) instantaneous, because it is the fact that it is not, and never can be, instantaneous, that permits non-equilibrium populations to persist. The presence of those populations changes the way the gas emits and absorbs radiation; in particular, it means that, contrary to Vonk’s assertion, emission and absorption are not symmetric.
Note that even in a single-species gas, absorption of the non-equilibrium radiation will still cause a departure from LTE; the absorbing molecules (that is, those molecules that happen to have absorbed the last microsecond’s radiation) will be continually heating up the gas, while the emitting molecules (those molecules that happen to have radiated the last microsecond’s thermal radiation at the temperature of the gas) will be continually cooling the gas. This can be a steady state, but not a thermal equilibrium. If thermalisation is slow, the gas will have two or more populations, radiating a mixed spectrum – partly at the temperature of the incoming radiation, and partly at the temperature of the bulk gas. If thermalisation is fast (approaching instantaneity in the limit) the gas will only radiate at its own bulk temperature, which in general will be different from that of the incoming radiation.
Paul Birch says:
September 6, 2010 at 4:11 am
“If thermalisation is fast (approaching instantaneity in the limit) the gas will only radiate at its own bulk temperature, which in general will be different from that of the incoming radiation.”
I mean, of course, that in (never attained) limit of instant thermalisation, the gas would then radiate only at its own temperature. In practice, there is always some thermalising delay, thus some of the non-equilibrium populations, and thus some re-radiation at the temperature of the incoming radiation; the faster the thermalisation process, the less of this there will be.
R Stevenson says:
September 5, 2010 at 5:07 am
At 288 K the black-body radiation or total emissive power from Earth is 391W/m^2 (124 Btu/h-ft^2). CO2 in the atmosphere absorbs 79.8W/m^2 after 3600m (CO2 absorptivity 0.184 at 350ppm or PcL of 1.24 m.atm). H2O in the atmosphere absorbs 248W/m^2 after 120m (H2O absorptivity 0.573 for a PwL of 2.77 m.atm).
Doubling CO2 to 700ppm would absorb the same 79.8W/m^2 after only 2000m ( CO2 absorptivity 0.195 at 700ppm 0r PcL of 1.4 m.atm).
If we accept these figures as accurate one thing puzzles me.
The main wavelength said to be absorbed by CO2 is around 15um.
However H2O also absorbs this wavelength admittedly not as strongly as CO2.
However because on average there are about 30 H2O molecules for every CO2 then the absorption must be mainly by H2O in the 15um region
CO2 takes 3600m to completely absorb 15um
H2O completely absorbs 15um in 240m.
Question is;
How does any 15um radiation make it past say 250m?