On 9/24/2012 22:10, glenn.tamblyn@bigpond.com wrote: Bob

Long time, no hear.

I have read through your paper and have some comments. As I commented in the past, the syntax of APL isn't the most readable of computer languages so my reply will be based on a more English like mathematics - more power to you for being able to get your head around APL

Comments
Strictly speaking, Kirchhoffs Equality only applies in conditions of thermal equilibrium. Outside those conditions, A & E are not equal.

Next your failure to distinguish between the differing Absorptivitys of the different components in the Earth system is a fundamental error.

If you wish to look at surface temperatures you need to be comparing the AE values of just the surface. But the 30% albedo figure for the Earth is a composite of surface reflection, cloud reflection and atmospheric reflection/scattering. To assess surface temperatures under differing conditions, you need to focus on surface AE values.

This statement "The ubiquitous null hypothesis seemingly accepted by all sides in the AGW debate is that the surface of the earth has an ae with respect to the sun of that observed 0.7 , but , that of an absolute black body , 1.0 , in the longer wavelengths" is simply incorrect.

No scientists thinks the Earth's surface Absorptivity is 0.7. They all understand quite well the existence of the different components of the Earth's aggregate Absorptivity. Similarly it is well recognized that the Earth's emissivity isn't 1, but is very close to that - usually estimated at 0.97. In some simpler models they make a simplifying assumption - it is given the name the 'Unity Assumption' - that Emissivity can be treated as 1. This avoids the need to attempt to calculate local emissivities for different points on the Earth.Although this will cause a miscalculation of absolute temperature due to the 4th power relationship in the SB Eqn, when one is interested in calculating deltaT, the impact will be much smaller. More sophisticated models all do include local variation in Emissivity, usually included in those components of the models that model soils, vegetation patterns, etc.

Absorptivity/emissivity aren't computed, they are measured for individual substances. It may be theoretically possible to derive the expected A/E vales for certain pure substances from first principles, but those first principles require an in-depth knowledge and calculations from Quantum Chemistry and a range of other sciences. And even then they are probably only meaningful for pure substances, near perfect crystals etc.

Here are some indicative numbers, taken from Wikipedia and other sources, for measured emissivity for various substances in the visible light bands.

Water is typically given as having an albedo of around 0.1. This is an average of the fact that it is even higher than this at angles of incidence up to around 50 deg - 0.03-0.07 or so. But then at high angles of incidence the albedo climbs markedly - see here http://en.wikipedia.org/wiki/File:Water_reflectivity.jpg. So averaged over the entire Earth, with most of the water at equatorial and mid latitudes but some at higher - and thus steeper angle of incidence latitudes, the figure I have seen most commonly for the average albedo of water, averaged over the entire planet is around 10%

Wikipedia has a general discussion of Albedo here: http://en.wikipedia.org/wiki/Albedo including a table of values for some indicative substances. To then derive a total surface Albedo for the entire planet - an averaged value of all surfaces based on their relative area - would need satellite data and area calculations. However with water making up 70% of the surface, forests having values of 8-18% albedo, bare Soil which is probably a reasonable proxy for rock at 17%, Sea Ice being 50-70% etc, this suggests to me that average albedo for the surface is above 10%, below 20% since there simply isn't enough surface area on land covered by very high albedo materials - fresh snow for example at 80 to 90% to move the average figure too far up from the baseline set by water.

Contrast that with the figures one can calculate from Kiehl & Trenberth 2009, Fig 1. Incoming solar reflected from the surface - 23 W/M^2, Total absorbed 161 W/M^2
So average albedo for the Earth = 23/(23+161) = 12.5%

I won't claim that the figures they cite are accurate to 3 decimal places but it certainly fits in reasonably well with the sort of back-of-an-envelope calculations one can make just with basic albedo data.

The most obvious thing arising from this number is that it isn't 30%. Yet that figure of 30% is quite well established from satellite observations. The reason for the disparity is that most of the Earth's albedo arises in the atmosphere. Reflection from clouds and to a lesser extent the atmosphere itself is substantially larger than surface reflection.

Again, the figures from K&T - reflected off the clouds & atmosphere - 79 W/M^2.

Their net figures for incoming solar are 341 W/M^2 at the edge of the atmosphere, 79 reflected by the atmosphere & clouds, 23 reflected by the surface, 161 absorbed by the surface. so (79+23)/341 = 29.9%  - there is the 30.

Their figure of 341 is essentially what your trigonometric calculations would produce based on surface emissions at the Sun's surface, Earth's distance from the Sun etc. And dividing by 4 to compare the frontal area the Earth presents to the Sun to its total surface area.

If you add up the numbers above, 23+79+161 doesn't equal 341. What I haven't included in this is 78 W/M^2 absorbed by the atmosphere before it reaches the surface.

So, Albedo for the surface is around 12.5%. So Absorptivty for the surface is 1-0.125 = 0.875.

Now to Absorptivity/Emissivity in the Infra Red.

Now to Absorptivity/Emissivity in the Infra Red.
This website lists the emissivities for a range of materials. http://www.engineeringtoolbox.com/emissivity-coefficients-d_447.html

They are quoted as emissivity at 300K so essentially in the temperature, and thus IR frequency range we are concerned with when looking at radiation from the Earth's surface.

Water - 0.95 - 0.963
Wood - 0.885 - 0.95
Sand - 0.76
Marble - 0.95
Ice - 0.966 - 0.985
Gypsum - 0.85
Granite - 0.45
Basalt - 0.72

Here is another site quoting emissivities for various materials http://www.omega.com/literature/transactions/volume1/emissivityb.html#s1
Also the abstract from this paper discusses emissivity of land surfaces: http://journals.ametsoc.org/doi/abs/10.1175/JCLI3720.1. Important note about the abstract - they discuss broadband emissivity. Then there is this plot from the UCSB: http://www.icess.ucsb.edu/modis/EMIS/images/seawat10.gif. This is emissivity of seawater over a range from 3 to 14 microns. That doesn't cover all the IR spectrum we are interested in but it certainly covers a large proportion of the emitted energy. They have a range of other plots: http://www.icess.ucsb.edu/modis/EMIS/html/water.html

Importantly in the IR range that matters here, both water and ice have a very high emissivity, as does wood. It is mainly rocks and soils that would tend to drag any average emissivity value for the planet down.

The figure I have heard quoted for average surface emissivity is 0.97. I can't absolutely confirm that but with most of the Earth being water, ice and forests, it is hard to imagine exposed rocks etc being able to pull emissivity down significantly below this.

So, for an idealized grey sphere at the Earth's distance from the Sun and very importantly with no atmosphere, the A/E ratio will be 1 as you mention . The calculation is:

    341 = 5.67 10^-8 * Tsurf^4. Solve for Tsurf and we get 278.5, much in line with your calc.

But an idealized grey body is still just that, an idealization. Lets make it a bit more realistic. Lets assume that the 2 emissivities values can be used - a 2 tone grey sphere. One for incoming sunlight, one for outgoing IR. Now we have

    341 * 0.875 = 0.97 * 5.67 10^-8 * Tsurf^4. Solve for Tsurf and we get 271.4.

So the emissivity difference cools the temp somewhat, now down just below freezing. This is what a non-ideal grey body with approximately the Earth's emissivity profile would be at. Importantly, it has no atmosphere.

Now lets slowly add more components and see what happens. This is a totally un-physical situation. I am doing this as a thought experiment to unpack the various aspects of what happens in the system.

First lets add an atmosphere but without any absorption of EM radiation. Neither of incoming Solar or outgoing IR. But we will allow this atmosphere to reflect sunlight, just as the current atmosphere does.

Since the atmosphere & clouds reflect 79 W/M^2 of the incoming 341 (from K&T 09), the Albedo of the Atmosphere is 79/341 = 23.2%

So now our energy balance eqn for the surface is

    341 * (1-0.232) * 0.875 = 0.97 * 5.67 10^-8 * Tsurf^4. Solve for Tsurf and we get 254.0 K

This result is un-physical in several ways. Firstly by artificially 'turning off' the absorption properties of the gases in the atmosphere. Next, in such a colder world there would be much less evaporation so much less cloud cover so actually not as much reflection off the atmosphere. However such a world is none the less colder than now, colder than during the depths of an Ice Age. So the figure for surface Albedo in the Solar spectrum (12.5%) will be too low because there would be much more ice reflecting sunlight.

The important point is that adding an atmosphere with some level of additional reflection will have a net cooling effect. Firstly due to whatever level of cloud cover does apply (and lets not forget that the atmosphere itself reflects some sunlight due to reflection off aerosols and scattering). Secondly due to increased ice cover in a colder world.

Lets now add back the next feature of the atmosphere. Absorption of incoming solar by the atmosphere. This is a little more complicated since the atmosphere can't just absorb this energy. It has to go somewhere. As a simplifying assumption, I will assume that once the atmosphere has absorbed energy, 50% of this eventually gets radiated to space without reaching the surface, the other 50% does reach the surface. But this radiation reaching the surface from the atmosphere is IR, so we need to apply the IR band absorptivity

So our energy balance at the surface for this example is:


    (((341 * (1-0.232)) -78) * 0.875) + (0.97 * 78/2) = 0.97 * 5.67 10^-8 * Tsurf^4. Solve for Tsurf and we get 245.2 K


So the fact that some incoming Solar gets absorbed by the atmosphere and that a certain percentage never reaches the surface (here I have assumed the split is 50/50) causes more cooling. And again we would need to reduce reflection from the atmosphere due to fewer clouds, but increase reflection from the surface due to more ice.

Now lets add back another feature. Evapo-transpiration (ET) & Convection (C) from the surface. Radiation isn't the only method available for removing heat from the surface although these processes can't ship heat all the way out to space; they only add it to the atmosphere. Lets make the same simplifying assumption we used with absorption of incoming Solar. 50% of the heat added to the atmosphere by these processes goes to space. 50% is radiated back to the surface. K&T 09 put the value for ET and C combined at 97 W/M^2. So lets run the calc again, taking that into account.

    (((341 * (1-0.232)) -78) * 0.875) + (0.97 * (78+97)/2) = 97 + (0.97 * 5.67 10^-8 * Tsurf^4). Solve for Tsurf and we get 236.1 K

Again this is assuming that the ET & C values that apply in current climate would have the same magnitude in this hypothetical world. Obviously they would be much smaller. But still a net cooling effect on surface temperatures of some amount. And this would increase ice sheet expansion and drive temps lower still due to even more reflection.

Importantly, so far I haven't introduced any effect from GH gases into this thought experiment. And the Earth is frigid.

Now lets start adding the effect of the GH gases. K&T 09 put the absorption of outgoing IR radiation by GH gases at 356/396 = 89.9%. So lets factor that in but retain our 50/50 split of where radiation goes when it is emitted by the atmosphere, up or down.

    (((341 * (1-0.232)) -78) * 0.875) + (0.97 * (78+97)/2) + (0.899 * 0.5 * (0.97 * 5.67 10^-8 * Tsurf^4)) = 97 + (0.97 * 5.67 10^-8 * Tsurf^4).

Solve for Tsurf and we get 264.8 K a huge increase.

Now lets add back the final component. The distribution of radiation up and down isn't 50/50. According to K&T09, the atmosphere radiates 199 W/M^2 up vs 333 W/M^2 down. 333/(199+333) = 62.6% down. Plug this into my eqn and we get:

    (((341 * (1-0.232)) -78) * 0.875) + (0.97 * (78+97)/2) + (0.899 * 0.626 * (0.97 * 5.67 10^-8 * Tsurf^4)) = 97 + (0.97 * 5.67 10^-8 * Tsurf^4).

Solve for Tsurf and we get 285.0 K - another increase. And within the ball park of actual accepted surface temps.


My point Bob is that,starting with an idealized grey body with no atmosphere, then progressively adding more and more real world features results in continually lower and lower temperatures, until we start adding in GH effect features. The exact values of some of these stages may be questionable, but they all have a cooling impact. So too does the impact on Albedo through expanded ice extent. Starting with an idealized grey body, every factor in the system makes the Earth colder. Only the effect of the GH gases has a warming impact.

The -33 C figure often quoted is referenced as an indicative number, an 'all else being equal' number. It is not intended as anything precise. More along the lines of the thought experiment I performed here. Some effects would certainly not have as much cooling effect - cloud cover, ET & C particularly. But they all still have some net cooling effect compared to your airless grey ball. And equally, ice cover certainly wouldn't be 'equal' either. Consider that the first temp, for the airless grey ball, is already down to the sorts of temps we see at the bottom of an Ice Age, when Ice sheets reached down to Kansas City. Cool the Earth 10-20 C colder than that due to the absence of the GH Effect and how much further would the ice stretch - Galveston, Mexico City? Cover the Earth with enough ice and it most certainly could get that cold.

And we know that in the very distant past, 100's of millions to billions of years ago, there were instances of such Snowball Earth's. Turn off the heating system that keeps the ice at bay and the Earth freezes over.

Best Regards
Glenn