# A simple model of the CO2 greenhouse effect

For some time now I have wanted to better understand exactly  how the CO2 greenhouse effect works. The only real way to do this  is to calculate how  heat is radiated out from the Earth’s surface through an IR absorbing atmosphere to space.  I set myself the task of writing from scratch  a simple programme to do this, focussing only on CO2 and  ignoring other greenhouse gases. – You can download the code here..

Fig 1. IR spectra of Earth showing dominance of a single CO2 band.

I soon realised that some basic assumptions need to be made to make any progress. At the heart of the problem is yet again the lapse rate. We have to assume a pre-existing lapse rate, since without one there can be no greenhouse effect. At the same time it is claimed that a lapse rate cannot be sustained without a greenhouse effect, since the top of the atmosphere must radiate energy to space in order to to enable heat flow through it. The lapse rate itself is caused  by convection and depends only on Cp of mainly N2 and O2. To make progress on a model of the real atmosphere some assumptions must be made:

1. A flat stationary Earth surface with constant incident solar radiation and  Ts=288K
2. An environmental lapse rate DT/DZ = 6.5 C/km up to a height of 11km with constant temperature from 11 to 15km.
3. Hydrostatic equilibrium: $P_0 e^{-\frac{z}{Scale}}$ where Scale is the scale height of the atmosphere
4. The Beer-Lambert law for absorption of radiation:  $\tau = e^{-kPL}$ ,   where $\tau$ is the transmission coefficient (0-1)  and A=(1-$\tau$) is the absorption by CO2, k is the absorption coefficient, P=partial pressure of CO2, and L is the distance travelled by radiation through air.
5. Kirchoff’s Law:  This simply states that a black body in thermal equilibrium will emit the same amount of energy as it absorbs. We will assume this to be the case for a layer of air containing CO2 at a given height z.

To simplify things further we only  take the main absorption band for CO2 between 13-17 microns. Then we simply guestimate that 20% of the total black body radiation emitted by the surface lies within this CO2 band – see figure 1.

The absorption coefficient k for the 13-17 micron  band has been measured to be 1.48 m-1 atm-1 (Essenhigh 2001). For the partial pressure of CO2 we take the concentration in ppm scaled to the atmospheric pressure at a given height. Atmospheric pressure then varies according the hydrostatic balance, and the partial CO2 pressure is in atmospheres i.e.

$P = CO_2P_0e^{\frac{-z}{scale}}$  where scale is the atmospheric scale height.

The Model

The atmosphere is divided into 100m thick layers above the ground with a surface temperature of 288K. As the surface radiates IR upwards, it is partly absorbed in each layer by CO2 molecules. It is assumed that Kirchoff’s law applies and that each layer re-emits an equal amount of IR in CO2 bands as that absorbed for a given temperature T. The atmosphere is assumed to be dry with no other greenhouse gas present. How does the radiation balance vary with CO2 concentration ?  Figure 1 shows a schematic.

Figure 2: Model schematic of radiation transfer from surface at 288K through 100m wide slices of atmosphere. Each slice is a grey body absorbing and emitting IR. A lapse rate of 6.5K/km and a hydrostatic scale height of 8.6km is assumed.

The surface radiates heat as a black body of temperature 288K upwards. The atmosphere is divided into 150×100 m layers. Each level absorbs photons from any direction  according to its local temperature, and numbers of CO2 molecules( partial CO2 pressure). The absorption(transmission) rates  are governed by the Beer-Lambert law depending on T(lapse rate) and P(hydrostatic balance). The lapse rate is  taken as 6.5 deg.C /km and the Pressure variation as exp(-z/scale), where scale is the “scale height” in the atmosphere where the pressure drops by the factor e (2.71828..). The scale height varies with temperature but on Earth the average scale height is about 8.6km.  The calculation starts at level 0 and works upwards. At each level the transmitted and absorbed IR from below are calculated. The actual net IR flux upwards is the difference between the up-going radiation and the integral of all the higher levels of down-going radiation.  Thermal equilibrium is assumed for each level, so at level j we must sum up all the contributions from all higher levels applying Beer-Lambert’s law.  The IR flux up and down are then stored for each level.

Results

The model was run several times with varying concentrations of CO2. Figure 3 shows the net IR energy flux for varying for the different CO2 concentrations.

Fig3. Radiation fluxes versus height in the atmosphere for the 13-17 micron CO2 band and varying CO2 concentrations.

Notice how rapidly the reduction in ground radiation occurs for low densities of CO2. At around 100 ppm the atmosphere becomes opaque to IR, as >90% of outgoing radiation becomes absorbed. The decrease in radiation is now logarithmic and the shape of the distribution changes. The shape depends on the density profile and where the atmosphere absorbs up-going radiation. In this simple model the ground radiation actually increases slightly when concentrations reach 1000 ppm as more radiation gets absorbed in the atmosphere.

Figure 4 shows the back radiation component at each level in the atmosphere showing how it increases and changes shape with increasing CO2 concentrations.

Fig4: Back radiation at each level from the atmosphere above that level. For comparison the net upward IR is shown for 300ppm and 600ppm. Note how the shape changes at higher concentrations.

The total greenhouse effect can be seen easily by plotting how the radiation flux varies with CO2 concentration in the atmosphere. Figure 5 shows how both the ground radiation and  the TOA (Top of Atmosphere)  vary. The difference between the two is the energy absorbed by the atmosphere itself. There is a more or less logarithmic dependency of outgoing radiation on CO2 levels above 100ppm giving approximately TOA=13lnCO2   and  Ground = 17lnCO2.

Fig 5. Dependence of the radiation from for the ground and TOA on CO2 levels in the atmosphere. Also shown is the absorbed energy flux by the atmosphere. Above 100ppm the dependence becomes logarithmic with CO2 levels.

Now we can estimate the surface temperature change induced by the greenhouse effect to maintain energy balance.  Differentiating the Stefan-Boltzmann equation we get

$\frac{DS}{DT} = 4\sigma T^3$

$DT = \frac{DS}{4\sigma T^3}$    equation 1.

The  reduction in outgoing radiation from the surface to space caused by a doubling of CO2 levels from  300ppm to 600 ppm is ~5 6 watts/m2 – see fig 4. Alternatively the radiative forcing at the surface by the increased “back radiation” is also ~5 6 watts/m2. The surface temperature will increase to counteract this change by warming just enough to balance the outgoing radiation.

The predicted warming from a doubling of CO2 is  1.2 1.5 deg.C using equation 1.

1. This “direct” warming caused by a doubling of CO2 to 600ppm is very similar to other estimates and is encouraging. Note also that this includes a “guestimate” of 20% of SB radiation falling within the main CO2 absorption band. This should be calculated more precisely.
2. The flattening of the radiation profiles with height for higher CO2 levels implies a warming of the upper troposphere in line with more sophisticated models. This is of course yet to be confirmed experimentally.

Conclusion

A simple model of the physics of the greenhouse effect due to CO2 gives results similar to more sophisticated radiative transfer codes. There is no need to calculate an effective radiation height as the shape of the radiation profile itself changes with CO2 levels.

The real atmosphere is far more complex than this simple model and H2O dominates weather and climate. The chief uncertainty in predicting future warming is the net feedback of water. Extra CO2 causes a declining radiative forcing with increasing concentration. It is essentially logarithmic dependence with  CO2 levels. Does extra evaporation of H2O enhance a small CO2 induced  warming or do increased  clouds counteract it ? What role if any do solar cycles play? Are these effects more dominant than the direct warming of ~1 degree due to a doubling of CO2? I hope we don’t have to wait 20 years for nature to answer these basic questions !

I welcome all criticism – including those who think  I have got this completely wrong !

Reference: Check out also the previous inspired work of   !

PhD High Energy Physics Worked at CERN, Rutherford Lab, JET, JRC, OSVision
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### 51 Responses to A simple model of the CO2 greenhouse effect

1. Richard111 says:

As a layman I am baffled by the assumption that CO2 in the atmosphere only seems to radiate if it has already absorbed energy from the surface. Since CO2 is in a temperature range of 288K close to the surface to say 193K at whatever altitude the local lapse rate reaches, I would expect the CO2 to be able to radiate MORE than it absorbs due to kinetic collisions with other molecules in the atmosphere. Am I totally screwed up here?

2. Clive Best says:

No – You are absolutely right! CO2 in the atmosphere radiates proportional to the local temperature**4 at each level. So it does indeed radiate more than it directly “absorbs” from the surface or from other CO2 molecules. That is why the lapse rate is crucial to the greenhouse effect, without it the atmosphere would be isothermal and there could be no greenhouse effect. So in reality it is convection (and evaporation on Earth ) that powers the heat engine that causes the lapse rate in the atmosphere. Greenhouse gases really just fine tune it !

3. Richard111 says:

Thank you. That gives me the confidence to look deeper into your model. I would like to understand it in detail. Years ago when I first heard about ‘global warming’ due to CO2 I was instantly a sceptic because I could not find any reports of change in the lapse rate under clear sky conditions. I have lived and worked in the deserts of Africa and the Middle East and almost on the equator in Singapore. I was very aware of the rate of temperature change in the different environments. I had expected the desert regions to respond to ‘back radiation’ at night by minimum temperatures slowly increasing. Haven’t found anything positive yet. In Singapore daytime temperatures never aproached desert temperatures because of water vapour actually shielding the surface from some of the insolation. This effect is occasionally visible in the solar record at this site: http://www.milfordweather.org.uk/solar.php Clear skies are shown as a smooth trace. Passing clouds show spikes. At times, when there are periods of no clouds you can see the different levels of solar energy when the relative humidity changes. This alone proves water vapour controls local climate.

4. Clive Best says:

This is because water is more important than CO2 in regulating Earth temperatures. 70% of Earth’s surface is ocean and small increases in radiative forcing – whether that be solar, diurnal, seasonal or even rising CO2 levels will induce evaporation and latent heat cooling. The Earth’s surface sweats in order to keep cool !

Is the net effect a positive feedback as we have been led to believe ? No – I think the evidence over 4 billion years is the opposite – negative feedback – see here https://clivebest.com/blog/?p=3659

5. I think you disabled comments with the Jetpack Comments and some anti-spam feature of WordPress. See: http://www.technize.net/solution-invalid-security-token-error-with-jetpack-comments/

• So “disabled” is too strong a word. More like “hobbled”.

6. Nice work, and thank you for the flattering credit to my own efforts. I will try to keep up with you here. I have a few questions. (I’m going to post them one at a time because the site has been rejecting them for some reason, but I’m not sure what parts are offensive.)

(1) I’d like to clarify your use of Kirchoff’s law. There are several versions, so I’ll drop the word “Kirchoff”. There is the principle of radiative symmetry, where the absorption and emission spectra of a body are intimately coupled. In your code, I believe this is what you are doing. There is the assumption that atmospheric layers are thermal equilibrium as gas passes through them. I agree with that as a basis for a simulation. But a cell of gas rising through the atmosphere is not in thermal equilibrium. It is losing heat by radiating into space. Do you agree?

• Clive Best says:

Yes – by Kirchoff’s law I really mean radiative symmetry. Each 100m layer is assumed to be in thermal equilibrium at a temperature defined by the lapse rate. CO2 molecules contained within each layer absorb and emit the same amount of IR. The simple model ignores bulk movement of gas (convection) between adjacent layers. The lapse rate and barometric pressure are baseline assumptions. The result is an idealized CO2 only radiative transfer estimate. This energy flow is insufficient to maintain the lapse rate

I agree that a proper model which includes convection and latent heat increases energy flow through the atmosphere, and will enhance radiation loss to space – mainly through H2O bands.

• I’m getting hung up on this statement. “CO2 molecules contained within each layer absorb and emit the same amount of IR.” I don’t see that in your code, and I believe that’s not correct. There is a net heat gain by transport of warm air into the layer and there is a net heat loss by radiation into space. So it’s not true to say that the same amount of heat is absorbed as emitted. But I don’t think that’s what you actually mean to say.

• Clive Best says:

I think it should be something like: “CO2 molecules contained within each layer are in thermal equilibrium at the local temperature. They absorb and emit IR in radiative balance. The net upward flux is the integral of the transmitted flux from all warmer lower levels minus downward flux from all higher levels”

7. (2) Your plots show radative flux. It looks like this is the flux radiated by the atmosphere, not the sum of atmospheric plus planetary surface. Is that right? Even when you are plotting the flux from the ground, you are plotting only the heat in the CO2-absorbing band that is emitted by the ground. Is that right?

• Clive Best says:

Yes that’s right. It is just the net flux at each level for the radiated heat emitted from the ground. We use Beers-Lambert law to calculate the % absorbed and re-emitted at a given level.

8. (3) In your calculation of the down-welling radiation, you don’t appear to respect the end to the lapse rate that you assume at 11 km for the up-welling radiation.

$TU=$TS – $zz*$lapse;

Should this be conditional upon the altitude being less than 11 km?

9. (4) What is the following line for? I’m just curious.

$SB =$frac*$sigma*(4*$T**2*$lapse*$a-2*$T**2*($lapse*$a)**2-($lapse*$a)**4); It appears to be a de-bugging calculation, some derivative of the radiative flux with altitude, similar to the one we looked at in the comments here. 10. (The website rejected this passage until I removed the special character like mu and degree.) (5) You use the same$frac for all temperatures. I agree that this is a worthwhile assumption. Indeed, we make the same assumption in the “heat” routine of our ” rel=”nofollow”>peak emission wavelength as temperature decreases. As you can see in our Total Escaping Power series of posts, the peak emission wavelength of CO2 at 290K is 10 um, well below the CO2 absorption band at 15 um. At 210K, however, the peak emission is at around 14 um. Thus you can see that $frac will increase from 0.2 to some around 0.3 as the temperature drops from 290K to 210K. Less power will escape the upper atmosphere than your simulation would suggest, so the CO2 doubling effect will be greater. When we calculate the effect using the continuous spectrum for CO2, and in increments of wavelength, but for layers 3-km thick (not 100 m), we get 1.6C for CO2 doubling instead of your 1.2C. Thus we see that the error introduced by your assumption is not so big, and thus I claim that your assumption of constant$frac is well worth it for the simplification it brings to the simulation.

11. Clive Best says:

I think you are right here. Yes I should update the model to include variation of the peak emission with temperature. At the same time we could also include the effect of pressure broadening. The width of the CO2 absorption band shrinks with pressure the higher you go in the atmosphere.

I think it should also be possible to plot “effective” temperature change with height in the atmosphere due to increasing CO2. But then I also need a model for convection and evaporation !

12. As far as I know, the CO2 absorption is a function only of the density of a layer in kg/m2. You don’t need to look up the spectrum at different pressures and temperatures. The equation you use with partial pressure in your code is accurate.

It is the water-vapor spectrum that is a strong function of pressure, due to interactions between water-vapor molecules. I had to get the water vapor spectra from the Spectral Calculator (requires payment). If you are interested, we present the combined CO2 and water vapor spectra for 3-km atmospheric layers here.

If you change $frac from 0.2 to 0.3 or something as temperature decreases from 290K to 210K, this adjustment will go a long way towards accounting for the change in the peak emission wavelength. If you re-calculate with that adjustment and find you get 1.4C or 1.5C from doubling, surely that’s good enough? 13. Richard111 says: Okay, one lost and baffled layman here. I thought each and every CO2 molecule is an independant entity in the atmosphere. Even during collisions there is no interaction between the electron paths of the different molecules. Absorption or emission of radiation directly effects or is effected by the change in electron paths of the individual molecule. Hence the reason scientists use gases of any element to acquire a well defined radiative spectrum. A statement above – “peak emission wavelength of CO2 at 290K is 10 um” is where I fell apart. From my reading of the physics of radiation, CO2 can radiate at 15 um from as low as 170K until temperature aproaches 600K where weak emission at 4.3 um will be noted. Peak emission for the 4.3 um band for CO2 is 673K (400C). As the temperature approaches about 1,000K emission at 2.7 um will be noted. All the emission spectrums I have seen do not show CO2 emitting at other temperatures. I have a photocopy from Perry’s if interested. The source temperature for the complete trace was 1,500K. • My statement “peak emission wavelength of CO2 at 290K is 10 um” is incorrect. You are quite right that CO2 will emit more power at 15 um than at any other wavelength regardless of where it is in the atmosphere. I should have said, “The peak emission wavelength of a black body at 290K is 10 um, and at 210K is 14 um.” In case you are still confused, I offer this further explanation. To obtain the emission by a non-black body we multiply the black-body radiation spectrum by the non-black body’s absorption spectrum. So suppose an atmospheric layer absorbs 90% of 15 um. It will emit 90% as much 15-um radiation as would a black body at the same temperature. If it absorbs only 1% of 10 um radiation, then it emits only 1% as much 10 um radiation as a black body at the same temperature. This is what we call radiative symmetry, and can be deduced by thought experiment from the Second Law of Thermodynamics. As CO2 gets colder, two things happen. First, the black-body total radiated power drops as T^4. Second, the peak emission wavelength of a black body increases as 1/T. But, as you so rightly point out, the absorption spectrum of the CO2 remains constant: it absorbs 14-16 um. If a layer of CO2 at 290K radiates 20% as much power as a black body at 290K, then it will radiate a larger fraction, say 30%, as much power as a black body at 210K. Given the complexity of implementing this correction, however, I don’t think it’s worth doing in Clive’s existing program, which demonstrates the Greenhouse effect in an astonishingly compact piece of code. • Richard111 says: Thank you for that confirmation. Back to my layman quibles. 🙂 Before I look at what the CO2 in the atmosphere can absorb I want to know how much energy is in the 15 um band in the radiation from the surface. I am assuming that the surface is at 15C (288K) and also assume that at night, and a wet surface, emissivity is so close to 1 that blackbody thinking is acceptable. I look here for my answer as my math is, to say the least, lacking. http://profhorn.aos.wisc.edu/wxwise/AckermanKnox/chap2/planck_curve.html Now on to the 14-16 um band. Again I have problems but so as not to clutter up this thread I have posted here: http://www.globalwarmingskeptics.info/thread-249-post-12061.html#pid12061 I don’t expect an involved reply as there is no way my contribution will help the work done here, but I hope other layman might feel they can follow the discussion. 14. Richard111 says: Clive, I am struggling to resolve your CO2 bandwidth of 13-17 um. My layman’s mind is stuck on the fact that there are a specific number of CO2 molecules in any parcel of air. Also the average kinetic speed is set by the local temperature. The kinetic distribution curve indicates very few molecules will be moving at speeds sufficient to absorb a 13 um or 17 um photon bearing in mind of all the air molecules with this kinetic speed only 0.04% will be CO2. So my gut feeling is that treating the 13-17 um band, from the point of view of the CO2 in any air parcel, as an actual blackbody block of radiation is over stating the performance of CO2 in the atmosphere. Problem is I don’t have the math to express this. I am currently looking for a simple explanation which would find the actual kinetic speed (within any local temperature air parcel) in metres per second, that actual fraction of light speed, which equates with the CO2 molecule meeting a 17 um photon head on or being overtaken by a 13 um photon and absorbing said photon, assuming it is in a receptive condition. My next mental block is that the CO2 molecule can either pass that energy to another air molecule by collision, which seems more likely given its speed relative to bulk of air molecules, or emit a 15 um photon in a completely unknown direction. Take heart your post is giving me lots to think about. I have the time. 🙂 15. Clive Best says: Sorry for not replying earlier – I have been traveling for a few days. Quantum Mechanics dictates at which frequencies CO2 (or H2O) molecules can absorb or emit IR photons. The main CO2 absorption band corresponds to vibrations of the outer oxygen atoms “>. In that sense the local temperature of the layer at which IR photons are absorbed is not fundamentally important.The higher the temperature and pressure the slightly broader the absorption spectrum becomes, but this is a minor effect. So the total energy absorbed or emitted at any level in the atmosphere depends more on the number of CO2 molecules than anything else. Note however that this is just for the narrow “quantum” bands for CO2. H2O has a far wider spectrum of absorption bands. So for example Microwave ovens are tuned to the frequency that causes molecules of water to rotate, and the heating is through friction to the food. Likewise some warming from CO2 absorption from the surface occurs. On Earth however greater heat transfer is caused by convection and evaporation which drive the lapse rate. There is still effective warming of the atmosphere by radiative transfer. The simple model above basically ignores all these effects , assuming a pre-existing lapse to estimate the additional effect of increased CO2 levels in the atmosphere. 16. Ken Gregory says: You have two Figure 1s, and no Figure 2. Under “Results” you write, “Figure 2 shows the net IR energy flux for varying for the different CO2 concentrations.”, but this is Figure 3. Under the second Figure 1 you write, “The lapse rate is taken as 6.8 deg.C /km”. This should be 6.5 deg.C/km. as written in point 2. You write, “the sale height on Earth of 8.6 km.”. This should be “the scale height …” You might consider adding a definition of “scale height” = the height in the atmosphere where the pressure drops by the factor e (2.71828…). Scale height depends on temperature, so it changes with height. Under figure 3 you write, “At around 200 ppm the atmosphere becomes opaque to IR, as >90% of outgoing radiation becomes absorbed.” You assume no downward radiation at 15 km altitude, so the outgoing radiation would equal the net radiation at 15 km altitude. The outgoing radiation at 0 ppm is 78 W/m2, and at 200 ppm is 45 W/m2 as per Figure 3. In this 13 to 17 micron band, 42% of the outgoing radiation becomes absorbed. I don’t see why you say >90%. The RT-model.txt file says, “The atmosphere is divided into 15 x 100m layers”. This should be 150 x 100 m layers. Please show the derivation of the equation for net radiation up:$SB = $frac*$sigma*(4*$T**2*$lapse*$a-2*$T**2*($lapse*$a)**2- ($lapse*$a)**4)

You can’t estimate the no-feedback warming effect of CO2 in the real atmosphere, even with assuming water vapour remains constant, by this model because the water vapour absorption overlaps the CO2 absorption band. The actual effect of doubling CO2 is MUCH less than shown here because much of the radiation that could have been absorbed by CO2 is actually absorbed by water vapour. Figure 3 shows the change in outgoing radiation from 300 ppm to 600 ppm is 9 W/m2, but we know from full radiation models the actual no-feedback number is 3.7 W/m2.

You write, “Alternatively the radiative forcing at the surface by the increased “back radiation” is ~5 watts/m2.” No, the forcing is at the top of the atmosphere. The change in outgoing radiation due to an increase in CO2 will cause the climate system to increase in temperature until the outgoing radiation is restored to the original value, so it equals the incoming absorbed radiation.

• Clive Best says:

Sorry for the miss-types, I have corrected them now.
I have also included above the derivation of the net up radiation component between a layer at temperature T and the next level $a meters higher at temperature T-$lapse*\$a.

$\sigma T^4 - \sigma (T - la)^4$

$= \sigma T^4 - \sigma (T^4 - 4T^2la + 2T^2la^2 + la^4)$

$= \sigma (4T^2la - 2T^2la^2 - la^4)$

Regarding the no feedback warming of a doubling of CO2. Yes I am essentially ignoring water and am using only a very rough estimate of the fraction of IR within the CO2 band. For sure the estimate is not accurate. However my aim was to understand how the radiation transfers upwards through the atmosphere, and how it changes with increasing CO2. Looking at the exact values we get DS(ground) = 6 watts/m2 and DS(TOA) = 8 watts/m2 between 300ppm and 600ppm. So you’re right that using frac=0.2 should really give DT = 1.5K (ground) or 1.9K(for TOA). I would argue that the ground figure is the correct one to use because the atmosphere itself absorbs radiation and warms up ~ 0.4K in this model. I think that top of the atmosphere forcing for surface temperatures ignores the atmosphere itself. A real calculation would also have to iterate the lapse rate changes especially from H2O. The lower the lapse rate the lower the greenhouse effect.

I will now go and do the calculation again properly using Wien’s displacement law to better estimate of the fraction of radiation within the CO2 band at a given temperature T. Ideally then we can then include water vapor.

• You have a sharp eye. I missed all of those edits.

You say, “The actual effect of doubling CO2 is MUCH less than shown here because much of the radiation that could have been absorbed by CO2 is actually absorbed by water vapour. ”

Can you back that up with some absorption spectra? Here is the combined CO2 and H20 absorption spectra for 3-km layers of the Earth’s atmosphere on a clear, spring day in a temperate part of the world.

http://homeclimateanalysis.blogspot.com/2010/07/earths-atmosphere.html

We obtained the spectra from the on-line Spectral Calculator used by astronomers. As you can see, the water vapor absorption does not dominate the CO2 absorption. Only in the lowest 3-km layer is the water vapor absorption competitive with CO2 in the 14-16 um band.

17. Ken Gregory says:

The expansion of (T – la)^4 is incorrect. The expansion should be in the form of:
(x + y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4 ;where x = T, y = -la.
See http://www.ltcconline.net/greenl/courses/154/seqser/binomial.htm

The expansion should be:
(T – la)^4 = T^4 – 4T^3 la +6T^2 (la)^2 -4T(la)^3 + (la)^4

(Don’t forget the brackets around la, so (la)^4, not la^4.)
I see no reason to do this expansion.
Just use SB = frac * sigma * (T^4 – (T- la)^4)

With T = 288, SB = 0.702 W/m2
Your incorrect formula gives SB = 0.00165 W/m2

• Clive Best says:

In fact this is a red herring of my own making – I also got confused. SB is never actually used in the calculation. Originally the hope was to use an analytical formula so I derived (wrongly) the SB term and this was a hangover . However I then went to integrating over to integrating the individual layers using basic sigma*T^4 for each layer. I forgot this because I was away for a week on another project ! I will try in the next few days to apply the correct fraction to the CO2 line. I will also clean up the code.

Sorry for confusion !

18. Clive Best says:

You’re absolutely right ! There is a stupid error caused by on wrong sign in my expansion. I should have just looked it up ! I will now correct it ASAP. My basic aim was to get understandable results which can be compared to a black box R-T model like MODTRAN.

19. Richard111 says:

Well done, Clive. I wish I could learn as much from my mistakes as you are doing. I shall be disapearing to study as much as I can of Kevan Hashemi’s writings. Hopefully I will become a wiser layman.

20. Ken Gregory says:

I am not familiar with the PERL code, so to better understand what you have done, I created an Excel file to match your procedure.
The Excel file is at :

I created graphs from the Excel file to correspond to your Figures 3 and 4:
Fig. 3
Fig. 4

The figures only shows the profiles for the 100, 300, 450 and 600 ppm CO2 concentrations.
The curves look close, but not exactly the same as your figures.
Could you please review the Excel file and determine why my file doesn’t quite exactly replicate your results?

• Clive Best says:

That’s really great and Excel is a lot simpler for people to understand than PERL.
The results are very nearly the same, but I think they differ slightly because of the way the net flux up and down are calculated. My net flux down is that coming from above a given layer whereas yours is the same but includes the layer itself. So I think it is just a definition of whether level i is the bottom of the level or the top. Probably it should be in the middle. Anyway I like your spreadsheet as it gives a simple overview. It would also be easy to increase number of levels and extend up to say 18km.

This weekend I hope to :
1) calculate “frac” properly for each temperature using Wein’s displacement – see http://en.wikipedia.org/wiki/Wien's_displacement_law.
2) calculate the temperature rise in each layer due to absorbed IR. I think it should be DQ/Cp.
3) Get a more realistic estimate for DT for a doubling of CO2. Where DT1= temp change at surface, DT2(i) the change in temperature at each level

21. Gerry Parker says:

Hello Clive!

Here’s something for you to consider. On a cold, clear night, one can go outside with an umbrella and stay warmer by opening the umbrella over your head. Perhaps it is possible to measure the actual effect of water vapor on a cold winter evening in the same way. Imagine you measure the down-ward radiation on the evening before a cold front comes through, while the humidity is still high, and then on a subsequent evening after the dry air from the cold front has established itself. It seems to me this difference should give a good measurement and allow a calculation of the residual downward heat radiation, i.e. the amount from other atmospheric components like CO2.

22. Clive Best says:

On a cold clear night with no solar radiation the ground is radiating upwards and cooling rapidly. So yes a black umbrella would trap some of that outgoing IR radiation keeping you warmer. There would have to be no wind to mix the atmosphere.
Total net cooling of the surface will depend on the water vapour present in the atmosphere above the surface. Your suggestion to measure the downward radiation at the surface with varying humidity is good – surely there must be data on this somewhere?
I used a similar type of argument to “measure” the feedback from water vapour by comparing deserts with the tropics at similar latitudes. Deserts have very low water content in the atmosphere whereas the tropics are very humid. I calculated the area averaged temperature rises since 1900 using Crutem4 for the two regional areas and compared them. Then assuming that CO2 greenhouse effects are the same for both, any differences can only be due to the water content in the atmosphere.
Sure enough the dry deserts have warmed more than the wet tropics, which is exactly the opposite of IPCC quoted positive feedbacks. In fact the data supports a negative feedback from water of F = – 1.5 +/- 0.8 W/m2K-1

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25. mark says:

dear al have been doing a little thinking about the models used to calculate “co2 warming” and have to query why all the models use a constant incident solar flux, after all the only places on a real earth where one would expect to see this over a 24 hour period this is near the poles. if you use a model of 12 hours darkness followed by 12 hours sinusoidal (0 to pi) normalized so the total flux over 24 hours matches then you can achieve much higher peak (noon) tempatures certainly above 15 degrees centigrade, and maybe you dont need co2 back scattering to force the tempature up.

26. glenncz says:

Is the greenhouse effect working during the winter when there is no sunshine in the Arctic?

• Clive Best says:

That is a good question !
I think the answer is no – there can be no greenhouse effect above the Arctic winter. There is no temperature gradient (lapse rate) and essentially the stratosphere touches the ground. Therefore heat cannot flow from the surface to space.

27. Clive,

Reposting my comment here from your question at Science of Doom:

I applaud your efforts to get a better understanding by writing your own code. I found for myself that the bits I didn’t understand only became clearer as I turned them around, looked at them in different directions, tried to model them – and read different textbooks.

Playing around with your own model will give you much better understanding of how various gases affect heat transfer in the atmosphere.

One item that seems wrong in your description is:

Kirchoff’s Law: This simply states that a black body in thermal equilibrium will emit the same amount of energy as it absorbs. We will assume this to be the case for a layer of air containing CO2 at a given height z.”

Later in the comments in response to some questioning over this point you said:

..CO2 molecules contained within each layer absorb and emit the same amount of IR..

I’ve explained why emission does not equal absorption in Part Two. In brief, it’s because absorption amount depends on the absorptivity and the intensity of the incident radiation, but emission depends on the emissivity (= absorptivity) and the temperature of the local gas.

And it’s even more true for a specific gas, just as it is for a particular wavelength – or band of wavelengths.

From other comments you made it is possible you are confusing LTE (local thermodynamic equilibrium) with TE (thermodynamics equilibrium). They are easy to confuse because the names are almost the same. However, they are quite different animals.

TE is where Kirchhoff’s law applies – it’s not a very useful condition for the atmosphere.

LTE essentially means that there are lots of collisions so that normal energy states are distributed how you would expect and consequently the Planck law is correct.

In a gas, the redistribution of absorbed energy occurs by various types of collisions between the atoms, molecules, electrons and ions that comprise the gas. Under most engineering conditions, this redistribution occurs quite rapidly, and the energy states of the gas will be populated in equilibrium distributions at any given locality. When this is true, the Planck spectral distribution correctly describes the emission from a blackbody..

Anyway, I hope the more detailed explanation along with the accompanying diagrams of flux and spectra at different heights is clear – if not please comment here or in Part Two.

• clivebest says:

Yes, probably one of my faults is that I use imprecise language, but I hope my physics is more precise.

“I’ve explained why emission does not equal absorption in Part Two. In brief, it’s because absorption amount depends on the absorptivity and the intensity of the incident radiation, but emission depends on the emissivity (= absorptivity) and the temperature of the local gas.”

I agree with that – emission is thermal but incident absorption/transmission is not !

So LTE applies to (thermal) emission of radiation by CO2 but not to absorption or transmission of incident radiation. However, I think it can be assumed that thermodynamics ( convection/latent heat + absorbed radiation) results in the measured lapse rate.

The transmitted radiation is the most interesting ( for me at least) because it increases with altitude. The net radiation to space for 13-17 microns is about half that radiated from the surface ( 37 watts/m2 /compared to 72 watts/m2) which is more or less in agreement with satellite data.

Once the atmosphere dominates radiation losses over direct radiation to space from the surface, then greenhouse gas dynamics must also tend to maximise cooling of the planet (maximum entropy).

For the Earth’s atmosphere today in the 13-17 micron band this coincides with current CO2 concentrations of 300-400 ppm. I find this more than a coincidence. As far as I know no-one can explain why natural CO2 levels are so low in on Earth.

28. Clive, finally I found your site, by chance directed from ResearchGate.net discussing under topic “how can sun sun influences in earths climate. Ever since the first IPCC assessment report was published in 1990 I have been very critical towards the CO2 myth and you have strengthened me to continue fighting the global warming myth despite strong opposition here in Finland.

I do have a peculiar question to which I am not able to find an answer. The question arose when discussing the hypothetical Earths temperature without any greenhouse gases.

Let’s assume a planet with inert gases like N2 and O2. The Sun will warm the surface, but the heat will dissipate to space without any blockage. However, the inert gas molecules in proximity to the warm surface will naturally warm up due to conduction. Because solar warming will not be uniform (Earth is round) using inevitably there will be differential warming of the near surface “air” and should lead to wind and also convection.

This way the entire atmosphere would eventually warm up, but could not cool down because neither H2 or O2 can emit IR. What temperature would this atmosphere acquire?

• Clive Best says:

You have asked one of the most difficult questions I know of. I don’t think there is a standard answer as to what would happen to the climate on Earth if you removed all greenhouse gases from the atmosphere. Whatever happens overall energy would be balanced and nearly all radiation to space would be from the surface

Some people argue that the Earth’s surface would be equal to Teff (255K) and that the atmosphere would be isothermal also at 255K

Other people argue that the lapse rate would still be generated since as you say differential heating and night/day gradients would generate convective cells of rising and falling air transporting heat around the planet. These people argue that the surface would be 255K and the lapse rate would fall at ~10deg per km below that. The tropopause would be much lower.

I favor a position somewhere between the two. There would still be some radiation loss from the atmosphere because even N2 and O2 radiate IR during collisions. Their emissivity is just very small. The moon is roughly the same distance from the sun and has no atmosphere. Its temperature changes rapidly from night to day and can reach 20 deg C or so at the equator. For Earth without greenhouse gases, the N2 atmosphere would even out temperature variations and raise the temperature slightly. There would be a lapse rate but the tropopause would be much lower. With O2 present there would still be ozone acting as a greenhouse gas. So average temperatures would be perhaps ~260K with ~280K at the equator and fall with height as before.

An earth without water would be a dead planet. Water is the key to the Earth’s thermostat control. CO2 is just a side issue in comparison.

• R Graf says:

Clive, I am wondering about your aside above that ozone is a greenhouse gas. Since ozone has a concentration in the top of the atmosphere, (forgetting pollution for now), and it famously absorbs shortwave UV, isn’t it a negative greenhouse gas? It’s shading the rest of the Earth from this energy and must get very hot in the ozone layer. That elevated temperature would greatly increase the rate of black body emission back to space, correct?

If so, can CO2 have a little of this effect also when intense sun rays are at the zenith and even CO2’s 2.8 micron band is getting a stream of energy from the solar spectral tail?

29. Thanks Clive for your attempt to answer. I should have elaborated a little more on this issue, because as you may recall it has often been noted that our Earth would be 33 degrees colder without greenhouse gases.

I do agree that the question itself is stupid and in my view it is just as stupid to insist on a 33 C colder Earth devoid of greenhouse gases, except for water vapour which according to the early IPCC doctrine was only a feedback and did not participate in the dangerous global warming of the planet as do the man made gases.

I guess we both are fixed on the importance of the “water thermostat” and that is what should be hammered into our politicians.

30. sayema says:

I want to know absorptivity and emissivity of carbod dioxide from solar radiation to earth surface ?

32. Alan B says:

If I understand Fig 5 correctly it indicates that at a CO2 concentration above 400ppm, there is essentially no change in absorption of energy by atmospheric CO2, so why does the temperature increase if CO2 concentration is doubled?

• Clive Best says:

The argument goes that it is the TOA ( top of atmosphere) IR flux that matters. If slightly less IR energy escapes to space then that heat must remain in the atmosphere. This extra ‘heat’ warms the surface just enough so that the IR energy to space again equals the incoming solar radiation. That is because the source of all IR is the surface and the flux depends on the 4th power of Temperature (Stefan-Boltzmann).

33. Alan B says:

But Fig 5 shows the energy absorbed by the atmosphere decreases at CO2 concentration above 400ppm, so I do not understand why the temperature would increase.