As most readers know, clouds are still poorly understood and under-represented in climate models. This new research may help.
From: DOE/Pacific Northwest National Laboratory
Effect of cloud-scattered sunlight on earth’s energy balance depends on wavelength of light
Accounting for wavelength effects will likely improve climate models
RICHLAND, Wash. — Atmospheric scientists trying to pin down how clouds curb the amount of sunlight available to warm the earth have found that it depends on the wavelength of sunlight being measured. This unexpected result will help researchers improve how they portray clouds in climate models.
Additionally, the researchers found that sunlight scattered by clouds — the reason why beachgoers can get sunburned on overcast days — is an important component of cloud contributions to the earth’s energy balance. Capturing such contributions will increase the accuracy of climate models, the team from the Department of Energy’s Pacific Northwest National Laboratory reported in Geophysical Research Letters earlier this month.
“The amount of the sun’s energy that reaches the earth’s surface is the main driver of the earth’s temperature. Clouds are one of the least understood aspects of climate change. They can block the sun, but light can also bounce off one cloud into another cloud’s shadow and increase the solar energy hitting earth,” said PNNL atmospheric scientist Evgueni Kassianov.
White clouds
Clouds both cool down and warm up the earth’s surface. They cool the earth by reflecting some sunlight up into outer space, and they warm it by bouncing some sunlight down to the surface. Overall, most clouds have a net cooling effect, but atmospheric scientists need to accurately measure when they cool and warm to produce better climate models that incorporate clouds faithfully.
But it’s a hard number to get. Fair-weather clouds are big puffy white objects that bounce a lot of light around. They can make the sky around them look brighter when they’re there, but they float about and reform constantly. Cloud droplets and aerosol particles in the sky — tiny bits of dirt and water in the air that cause haziness — scatter light in three dimensions, even into cloud shadows.
To determine the net cloud effect, researchers need two numbers. First they need to measure the total amount of sunlight in a cloudy sky. Then they need to determine how bright that sky would be without the clouds, imagining that same sky to be blue and cloudless, when aerosols are in charge of a sky’s brightness. The difference between those numbers is the net cloud effect.
Rainbow energy
Researchers have traditionally estimated the net cloud effect by measuring a broad spectrum of sunlight that makes it to the earth’s surface, from ultraviolet to infrared. But clouds are white — that’s because the large water droplets within them scatter light of all colors almost equally in the visible spectrum, the part of the electromagnetic spectrum that includes the colors of the rainbow.
On the other hand, aerosols — both within clouds and in the open sky — bounce different-colored light unequally. Broadband measurements that fail to distinguish color differences might be covering up important details, the researchers thought.
Instead of taking one broadband measurement that covers everything from ultraviolet to infrared, Kassianov and crew wanted to determine how individual wavelengths contribute to the net cloud effect. To do so, the team used an instrument that can measure brightness at four different wavelengths of color — violet, green, orange, red — and two of infrared.
In addition, this instrument, a spectral radiometer at DOE’s Atmospheric Radiation Measurement Climate Research Facility located on the southern Great Plains in Oklahoma, allowed the team to calculate what the brightness would be if the day sported a cloudless, blue sky. The spectral measurements taken by the radiometer can be converted into the amount and properties of aerosols. Then aerosol properties can be used to calculate clear blue sky brightness.
Clouds Gone Wild
Comparing measured values for cloudy sky to the calculated values for clear sky, the researchers found that, on average, puffy fair-weather clouds cool down the earth’s surface by several percent on a summer day. Although clouds cool overall, two components that the researchers looked at — from direct and scattered sunlight — had opposite effects.
The direct component accounts for the shade provided by clouds and cools the earth. The second component accounts for the sunlight scattered between and under clouds, which makes the sky brighter, warming the earth.
“The sunlight scattered by clouds can heat the surface,” said Kassianov. “We all know that we can still get sunburned on cloudy days. This explains why.”
In the Oklahoma summer, the scattered-light effect measured by the researchers could be quite large. For example, if a cloud passed over the instrument, the measured cloudy sky brightness exceeded calculated clear sky value by up to 30 percent. Kassianov attributes that large difference to scattered sunlight being “caught on tape” by the radiometer.
“Sunlight scattered by three-dimensional, irregular clouds is responsible for the observed large difference. The one-dimensional cloud simulations currently used in large-scale climate models don’t capture this diffuse light,” said Kassianov.
Aerosols’ Day in the Sky
The team also found that the effect changed depending on the measured visible-spectrum wavelength, and whether the light was direct or scattered.
With direct light, the cooling caused by clouds was weakest on the violet end of the spectrum and strongest at infrared. With scattered light, warming caused by clouds was also weakest at violet and the strongest at infrared. Overall, the least cooling and warming occurred at violet, and the most cooling and warming occurred at infrared.
Because large droplets in clouds scatter sunlight almost uniformly across the spectrum, the clouds themselves can’t be the reason why different wavelengths contribute differently to the net cloud effect. Compared to cloud droplets, aerosols are more than 100 times smaller and scatter wavelengths differently. These results suggest that aerosols — which not only cause haziness but contribute to cloud formation as well — are responsible for the wavelength differences, something researchers need to be aware of as they study clouds in the sky.
“If you want to study how aerosols and clouds interact,” said Kassianov, “you need to look in the region of the spectrum where aerosol effects are significant. If you want to fish, you go where the fish are biting.”
This work was supported by the U.S. Department of Energy Office of Science.
Pacific Northwest National Laboratory is a Department of Energy Office of Science national laboratory where interdisciplinary teams advance science and technology and deliver solutions to America’s most intractable problems in energy, national security and the environment. PNNL employs 4,900 staff, has an annual budget of nearly $1.1 billion, and has been managed by Ohio-based Battelle since the lab’s inception in 1965. Follow PNNL on Facebook, LinkedIn and Twitter.
Reference: Kassianov E., Barnard J., Berg L.K., Long C.N., and C. Flynn, Shortwave Spectral Radiative Forcing of Cumulus Clouds from Surface Observations, Geophys Res Lett, April 2, 2011, DOI 10.1029/2010GL046282 (http://www.agu.org/pubs/crossref/2011/2010GL046282.shtml).
Abstract:
The spectral changes of the shortwave total, direct and diffuse cloud radiative forcing (CRF) at surface are examined for the first time using spectrally resolved all-sky flux observations and clear-sky fluxes. The latter are computed applying a physically based approach, which accounts for the spectral changes of aerosol optical properties and surface albedo. Application of this approach to 13 summertime days with single-layer continental cumuli demonstrates: (i) the substantial contribution of the diffuse component to the total CRF, (ii) the well-defined spectral variations of total CRF in the visible spectral region, and (iii) the strong statistical relationship between spectral (500 nm) and shortwave broadband values of total CRF. Our results suggest that the framework based on the visible narrowband fluxes can provide important radiative quantities for rigorous evaluation of radiative transfer parameterizations and also can be applied for estimation of the shortwave broadband CRF.
Discover more from Watts Up With That?
Subscribe to get the latest posts sent to your email.
it depends on the wavelength of sunlight being measured. This unexpected result will help researchers improve how they portray clouds in climate models.
Unexpected? There’s the problem… climate researchers know very little about the underlying science.
Aside from the reflectance/trapping effects discussion with regard to clouds, I have seen something that appears to be a lens-focusing effect. I have a 1KW PV system that has a typical maximum output of 8-900 watts on a clear day. We experience lenticular clouds here in Colorado, and I have seen on a few occasions where the PV output goes to 1.2-1.3KW when the sun is behind a solitary lenticular cloud. Of course this would only be a local concentrating of light, not affecting the average, but I find it interesting that this appears to happen.
Ed Barbar says:
April 24, 2011 at 11:53 am
….
Thinking simply, it would seem if the earth were to heat up, there would be more water evaporation, as the atmosphere heats up the clouds will be higher, etc.
I have no idea whether this is an “in the noise” effect or actually makes a difference. Anyone know?
Yes, please read “The Thermostat Hypothesis”, an essay by Willis Eschenbach, June 14 2009. At http://www.oarval.org/Thermostat.htm
“”””” cba says:
April 24, 2011 at 8:57 am
Nonegatives says:
April 23, 2011 at 7:53 pm
The fact that CO2 and H2O absorb different wavelengths of light has been known for a while now. Did they just now figure out that clouds are actually water vapor?
Clouds are droplets. vapor is individual molecules. where ever you’ve got droplets, you’ve got additional degrees of freedom capable of absorbing and emitting continuum like a solid or liquid black body radiator versus a discrete spectrum like an atom or molecule. “””””
Well some clouds are water droplets, and soem are probably ice crystals; presumably the high ones ar emostly the latter.
But the liquid droplet ones are NOT interracting with the sunlight, in any simple “reflectance” mode. They are lenses, with very definable optical properties, and it is trivial to show that a spherical droplet of water, illuminated by a collimated parallel beam of light, such as sunlight (1/2 degree divergence), simply refratcs the beam into a broadly diverging cone of rays. Well the droplet of course is a “Converging” lens, but the focus is less than the dop diameter, behind the drop, and from there on the light diverges into a cone that has most of its energy inside about a 45 degree cone half angle, but some light is scattered to very wide angles beyond 90 degrees from the sun. So the fluffy white light you see, is scattered light, not reflected light. Two or three droplets in succession, and you have a totally diffused isotropic radiation pattern. We can’t post pictures here, or else I could show exactly what the refracted output from a spherical droplet looks like.
My gut feel for the albedo component of clouds versus polar ice, is more in tune with your view cba, that it would be with Bill Illis picture. Fresh snow, less than a few hours old can have high reflectance, but after just a few hours, it becomes somewhat transparent, due to surface melting, and once the light gets into the snow/ice, it is largely trapped by TIR, so the reflectance drops to where it is no more reflective than a lot of rocks or greenery.
But the key indicator of the net effect of clouds, is simply the total reflectance of a cloud covered earth compared to a cloudless earth. If it wasn’t for the blue sky scattering; the blue planet, would be the black planet, as the oceans have about 3% max sunlight reflectance; and they are over 70% of the surface, and an even larger fraction of the earth area that receives the bulk of the solar energy. The area inside the Antarctic circle contains only 8.3% of the Southern Hemisphere surface area (1-cos 23.5 deg).
My quarrle with the authors of this study/paper/whatever, would be that the sides of those fluffy clouds, that they say are scattering light down to the surface (that would otherwise miss the surface), are also going to scatter light that was headed for the surface anyway, and redirect it out to the rest of the universe. I doubt that they can prove any significant bias towards, an increase, downwards.
I don’t have a quarrel with the statement that the underside of thin white clouds is radiometrically brighter than the clear blue sky. After all, the normal BB spectrum contains only 25% of its total energy at wavelengths below the spectral peak wavelength, and the dominantly blue color of the sky is well below the solar spectral peak wavelength. So the blue sky adreesses less than 25% of solar energy, and it is isotropic, so no more than half of it can be directed downwards.
Liquid water droplets on the other hand are highly transparent to the bulk of the solar energy spectrum; including that same blue light, and therefore those water droplets, can optically scatter a much broader spectrum, than the Blue Sky Raleigh scattering.
But don’t try to josh us, that because the cloud is brighter than the sky, then it must warm the earth. Try pointing your radiaometer to the approximately half of one degree direction that contains that local area star known as “the sun”, and then tell us the clouds are brighter than that. That half of one degree of the sky is what is warming the earth; not the blue skylight, nor the clouds.
NOBODY, ever observed the Temperature to increase, (signifying the effect of warming), when a cloud (ANY cloud) passes between the observer and the sun. Iy ALWAYS cools down to a LOWER Temperature; no matter what.
“”””” Craig Goodrich says:
April 24, 2011 at 11:09 pm
… if a cloud passed over the instrument, the measured cloudy sky brightness exceeded calculated clear sky value by up to 30 percent. Kassianov attributes that large difference to scattered sunlight being “caught on tape” by the radiometer.
This is crazy. Photographers have known this since before the First World War (that’s why they used to carry so many slightly different yellow and orangeish filters), and the lowered contrast on cloudy days was often utilized for studio portraits. This is bleeding-edge science? A 1930s-era Weston photographic meter would have told the same story — though not, of course, peer-reviewed or tax-funded… “””””
The clear blue sky, is NOT the source of earth’s energy budget. On a perfectly cloudless day, when there are no clouds around to increase the sky brightness, by 30%, a careful inspection of the entire sky, will generally find one particular direction encompassing about one half of one degree of solid angle (cone full diameter), where the brightness suddenly rises to a value many thousands of times brighter than either the clear blue sky; or any cloud that might happen on the scene.
Experimental evidence suggests that this very high brightness radiant energy is coming from a nearby star in our galaxy.
George E. Smith says:
April 25, 2011 at 11:11 am
George, it looks like we’re in pretty close agreement. BTW, there is a little program I found that will do Mie scattering calculations and show various things like a fogbow. Some of the reflections do split up wavelengths and some do not. The Mie scattering is far stronger than wavelength dependent clear sky scattering and it produces nonuniform results by direction for a single scattering. Suffice to say one could use a supercomputer just to tabulate this stuff. It appears that a fogbow intensifies the apparent optical thickness and increases the reflectance at around 180 deg from the direction or within 60 deg. of that point. Note that I use the term reflectivity to refer to the scattering of incoming light by all those millions of water droplets as it really is not going to be calculable or distinguishable in any meaningful way.
Of course at high enough altitudes, there will be ice rather than liquid. That leads to the possibility of non spherical particles also. Combine that with the tiny size of many of the ice or liquid particles and the potential for seeding particulates being close in size to that of the very small droplets and you’ve got a devils brew that cannot be calculated realistically.
It’s interesting that J London, 1957 as reference in the physical meteorology book I referenced above covered these various aspects that you don’t find in KT97. Scattering (rayleigh) was determined to be about 17% of incoming with only 7% escaping the atmosphere. Clouds tended to transmit about 10% or 20% or more and absorb only 1 or 2% (if I recall the numbers properly).
Note that the modelers of warming think they can imitate the stuff assuming 1 scattering event.
Somewhere I came across the notion that total overcast conditions tended to allow around only 10% of the incoming power to reach the surface.
As for your 3% water surface reflection, I’d put that closer to 4 or 3.5 %. If the sun is at a high zenith angle, it tends to be quite low. As that angle increases towards 90, the surface becomes quite reflective – but then again, this is not where most of the TSI is going and in fact, there is much smaller amounts per unit area there and thicker atmospheric pathlengths. While this occurs at higher latitudes always, but not that much towards the equator – except that early morning and late afternoon gives one a daily equivalent to that higher latitude zenith condition. Realistic average land conditions including ice and snow provide albedo reflectivity in the order of 17 to 19 % which leaves surface albedo to be less than 10% and 62% almost totally opaque cloud cover drops that down towards 3.5% or about 11 w/m^2 surface reflection that makes it back to the TOA. It seems to me that the KT97 errors were quite convenient to trenberth’s position on the subject. However, it does put the atmospheric scattering/cloud reflectivity (scattering too) up in the 27-28% albedo realm rather than merely the 24% range and their proclaimed factors of land use change and snow/ice cover loss into the realm of the ludicrous.
what I find interesting is that of the LWIR cloud situation. Regardless of ice or liquid, one now has solids or liquids (or sort of supermolecules) capable of continuum emissions rather than spectral emissions.
I just did a crude model test of this sort of thing assuming that a cloud top at 3km would have the same T as ambient and that it was opaque in the LWIR – leading to a full BB emission at temperature T. The result was a somewhat familiar spectral pattern with a BB temperature at 3km ambient (using the 1976 std atm value). I got around 235 w/m^2 power at the TOA (instead of almost 270w/m^2 for the clear sky 288.2k case). In the cloudy case, the co2 doubling had an effect of less than 3 w/m^2 versus the 3.7w/m^2 clear sky co2 doubling case. In fact, assuming the case of a 30% h2o vapor increase along with the co2 doubling, the result was hardly over 4w/m^2 total, again significantly less than the clear sky version. Of course, the 30% h2o increase would require a 5 deg C rise in the whole atmospheric column if one assumes constant relative humidity as a constraint. One is only missing the vast majority of w/m^2 necessary to cause such a 5 deg C increase.
“”””” As for your 3% water surface reflection, I’d put that closer to 4 or 3.5 %. If the sun is at a high zenith angle, it tends to be quite low. As that angle increases towards 90, the surface becomes quite reflective – “””””
Water has a refractive index over most of the solar energy spectrum of 1.333. The normal incidence Fresnel reflection coefficient is ((n-1/(n+1))^2 =2.037% The total reflectance (both polarisations) remains almost constant up to the Brewster angle which is arctan (n) = 53.1 deg for water; that is the angle from the zenith; not the elevation angle. Cosine of 53 deg is 0.6 ( large angle of a 3-4-5 triangle is 53 deg 8 minutes). So at 53 degrees, your have 1/0.6 = 1.666 air mass, so quite a lot more atmospheric absorption than the zenith air mass one value. And don’t forget that as that sun elevation angle drops, to get those higher reflection coefficients, the obliquity factor is spreading the light out over a much larger earth surface area; so the resultant reflected energy per unit area isn’t as large as you would think. I believe a full integration using the full Fresnel Polarisation equations, would show that the full hemispherical surface reflectance does not exceed 3% (off the oceans). That’s for a uniform diffuse (whiteout) irradiance.
People try to argue that the 3% reflectanvce from the oceans, is only valid for a perfectly flat (optical) ocena, and that waves will tip the surface away from the sun creating a higherr reflectance. Unfortunately, they forget, that evbery wave has a back side too, and for every surface element tilted to higher incidence angles, there is another tilted to lower incidence angles. The net result is that the reflectance off at least a mildly rough sea, is no different from a perfectly flat sea.
Glass with a Refractive index of 1.5 has a normal reflectance coefficient of 4%; and you only get that low a refractive index for glasses that are almost pure silicon dioxide.
Schott BK-7 which is a Boro-slicate crown glass has a nominal refractive index of 1.517. 5% reflectance (normal) is more typical of things like ordinary plate glass, which are higher index, than Borosilicate crowns.
I don’t have a lot of problem with the clouds being quite high absorbers of the LWIR surface emissions. But whereas the incoming blockage is of the direct, almost colimated sun beam, the outgoign LWIR is at least Lambertian, and most likely isotropic, since the surfaces aren’t optically flat. Even if it was Lambertian, then the irradiance of the cloud bottome, would go as cosine^4 of the obliquity angle (from surface to cloud), and also drop as inverse square of the cloud height.
So those nice puffy cotton ball clouds, only get to intercept a miniscule fraction of the LWIR emissions from their shadow zone; yet they block ALL (times their absorptance) of the incoming solar beam.
If the extra scattered light off the bottom of the cloud exceeded, the amount of solar energy incoming that they block; then the Temperature would go up in the shadow zone. It doesn’t; it always goes down; no matter what.
George E. Smith says:
April 25, 2011 at 1:56 pm
George,
I think we’ve got a disagreement there. According to the old 1957 London paper, actual absorption of visible is only a couple of percent of total incoming as I recall. A big portion of the incoming TSI is ‘reflected’ away as albedo and the remaining is going to be scattered downward, granted significantly less than clear sky. I think a typical number is around 10% for total overcast conditions. I don’t see LWIR penetrating the cloud to any depth. What comes out the top is going to be BB radiation from the top small fraction of thickness that constitutes one optical path length. If it were dust, I’d believe there would be substantial penetration. This though has got to be close to the equivalent of a surface of water or clear ice that yields the low albedo, even at visible wavelengths. I think that low reflectivity gets worse at longer wavelengths too.
Generally, the reality follows the theory pretty well. There’s nothing significant starting to happen with visible light reflectivity of water by the time you hit 60 deg from the zenith. It’s only the last little bit where the reflectivity shoots up and by that time, the atm. thickness is getting thicker than 1.6 x zenith angle = 0 and the surface area of a square meter of insolation is spread out over more than that.
It all still leaves KT97 numbers at around 0.033 for average surface albedo and all the rest of the roughly 0.3 albedo from atmospheric scattering (small potatoes) and clouds (the 800 pound gorilla in the room). This also tends to leave hansen, lacis, dessler, trenberth and the like grasping at exagerations and grasping at straws, apparently hoping no one has noticed their little errors.
cba says:
April 25, 2011 at 6:01 pm
What comes out the top is going to be BB radiation from the top small fraction of thickness that constitutes one optical path length.
———————————————————–;
Assuming BB denotes Black Body radiation, the Stefan-Boltzmann law (or BB) results from the integral of the Planck radiation law.
Planck law requires a continuous frequency distribution, e.g., from a solid or a liquid.
Would you happen to have the frequency or wavelength spectrum of a cloud?
And by one optical length, are we talking about 10 microns – or maybe 20 microns?
Agile,
you are right about BB, wrong about optical path length. It’s not a wavelength but rather a thickness associated with absorption.
Since the droplets are either ice or liquid, there should be a continuum absorption and emission and not a molecular spectral response like a single molecule or a gas exhibits. It’s energy quantum states existing due to the fact that many molecules are interacting as a solid or as a liquid. Hence, apply Planck’s law for the temperature present. Then, apply the absorption spectrum to the curve to deal with the gas between the cloud and the ‘measurement’ location because it’s a log function and there’s going to be h2o molecules present that will exhibit the molecular spectral absorption/emission response.
How close this is to reality may depend somewhat on droplet size. I doubt it can really be actually measured very well either, except from the surface or TOA. I’ve made a synthetic spectrum sample using a line by line database but I haven’t been able to compare this to any TOA measurements yet.
“”””” cba says:
April 25, 2011 at 6:01 pm
George E. Smith says:
April 25, 2011 at 1:56 pm
George,
I think we’ve got a disagreement there. According to the old 1957 London paper, actual absorption of visible is only a couple of percent of total incoming as I recall. A big portion of the incoming TSI is ‘reflected’ away as albedo and the remaining is going to be scattered downward, granted significantly less than clear sky. I think a typical number is around 10% for total overcast conditions. I don’t see LWIR penetrating the cloud to any depth. What comes out the top is going to be BB radiation from the top small fraction of thickness that constitutes one optical path length. “””””
Well I can’t disagree with you; or agree, for that matter, since I can’t determine just what it is you are saying:- “”””” According to the old 1957 London paper, actual absorption of visible is only a couple of percent of total incoming as I recall. “”””” So what are they (the old 1957 london paper) saying is absorbing what part of the “visible” portion of TSI.
I’m still using, and still seeing published, the exact same solar spectral graphs for air mass zero (TSI) and ground level air mass one (Zenith sun) that used in school 55 years ago; well even before that old 1957 London paper came out. The AM-1 curve always assumes some standard average water vapor content for the atmosphere, since it includes the water absorption bands that start at about 700 nm. But before you get to 700 nm, you already have a significant loss even at the spectral peak. Well the curve I can most easily get my hands on today, is from the Aero-Jet General Infra-Red Engineering Handbook, 25 August 1961, and it is republished in The Infrared Handbook from ERIM., in their chapter 3 on the sun. For those graphs they plot a best fit black body radiation peak spectral irradiance of 2.0 kW/m^2/micron, and the actual anomalous AM-0 sun is about 2.25 around 480 nm. The AM-1 sea level peak, is at the4 same 480 bn but is only1.5 kW/m^2/micron, so that is a 25% loss at the center of the spectrum. Now that isn’t atmospheric absorption or water vapor absorption; it is mostly Raleigh Scattering, that produces the blue sky. I haven’t done the actual calculation of the blue scattered energy, but one thing I do know from actual observation, is that the blue sky looks exactly the same looking down as it does looking up, so the Raleigh scattered blue source is quite isotropic, which means that about half of the short wavelengths that are scattered, ends up returning to space (scattered, not reflected), and the other half reaches the ground. If you look down from 36,000 feet in a cloudless sky over the ocean, you see the same blue sky because the ocean looks quite black (in the visible). Of course if you happen to look in the direction of the specular reflection of the sun, you will get 2% of the sunlight, but away from the sun, you get black from the ocean. At LWIR wavelengths the (calm) ocean is going to be a Lambertian thermal emitter, when seen from altitude.
Ozone has a weak absorption in the visible from about 500 to 650 nm and O2 even has a band around 780nm but I can’t separate that from the coincident water band. Then a number of H2O bands occur, and CO2 doesn’t kick in until about1.4 microns.
But actual water droplets in clouds are quite different, because they are large enough to behave optically. Ordinary geometrical optics predicts that a spherical droplet, will convert colimated sunlight into a strongly divergent beam (after focussing) that puts most of the energy into about a 90 degree full cone angle; but is still reflracting loight out to 90 degrees from the sun. So in a cloud, just a few consecutive droplet refractions, and you have a perfectly isotropic distribution of scattered light.
I’ve never said that the LWIR can penetrate the cloud to any depth; of course it is absorbed. But that simply results in subsequent emission of a thermal spectrum from the cloud, which also gets absorbed, and then re-emitted. So the original ground sourced LWIR may not come out the top of the cloud, but the cloud itself will become a source of isotropic LWIR emissions.
My point was that because the surface emission is totally diffuse, the cloud can only intercept that fraction of the diffuse energy that actually hits the lcoud, and the higher those puff balls are the smaller that fraction is. Yes of course with a total overcast, you have a different result, including that a lot less solar energy reaches the surface to heat it.
But as to the polar ice albedo, you can’t have it both ways. You can’t combine a non-existent zenith sun irradiance, with a near grazing incidence high surface reflectance, and claim a large albedo due to the ice, or the open sea water either.
George,
The 1-2 % absorption was referring to cloud absorption and I recall that the number is for the total TSI absorbed by clouds over the cloud fraction present. The London paper is quite different in cloud reflectivity and cloud fraction from that of the KT97 crowd. The amount turns out similar but the two values are virtually swapped. KT97 assumes a 0.62 cloud fraction and a reflectivity back to space of around 35%. London has a cloud fraction assumed to be around 35% and an average cloud reflectivity of about 60% +. These all refer to the the visible spectrum or at least to the TSI.
I think the London paper assumes a rayleigh scattering in clear sky of around 17% with about 7% escaping the atmosphere as albedo. The rest gets absorbed by the atmosphere and the surface.
What I use is the 1976 std atmosphere values and I assume a true BB spectrum from the Sun. Note that the BB spectrum is close but not actually correct for what comes from the sun. Dealing with LWIR means though that I am not using a solar spectrum. As for the atmosphere, it has plenty of visible light absorption lines present. These are telluric lines found in spectral imaging of stars. Of course plenty of lines doesn’t mean plenty of power absorption in this case.
I’m not sure of the reference to snow and water albedo at angles far from the zenith. Ice and snow tend to be at high latitude. These do not require large zenith angles to be large. Ice cannot be clear ice though. Otherwise, it’s going to be similar to water in reflectivity. It’s mostly fresh snow that offers high albedo. Of course, at the greater angles from the zenith plain old water is going to increase in reflectivity so there will be little advantage in having the fresh snow and ice instead of open water. Hence, those claiming the loss of snow and ice having a significant effect on albedo is totally bogus for the world today.
your comment on the ground lwir not making through but rather the top emitting at its characteristic temperature was the point I was trying to make with the other gentleman earlier. That should be true as long as the optical thickness in the LWIR qualifies as thick or that it isn’t transparent to LWIR.
The net result for my little model I’m playing with is that the spectrum seen for cloud cover should be the top of the cloud temperature BB attenuated by the presence of the atmospheric contents above that surface.
I seem to recall seeing spectrums for TOA looking down (and surface looking up) that came from ARM. First attempts to find my links or to find them on a site though has not been going well. Until then, I cannot compare my model assumptions to what is being measured.
It would seem we have less in disagreement than I originally thought
“”””” I seem to recall seeing spectrums for TOA looking down (and surface looking up) that came from ARM. First attempts to find my links or to find them on a site though has not been going well. Until then, I cannot compare my model assumptions to what is being measured.
It would seem we have less in disagreement than I originally thought “””””
I’m sure we are not in great disagreement. For example, your “London Paper talks of a 17% Raleigh Scattering with about 7% lost as albedo. And I mentioned that the Raleigh scattered blue light (amount unknown) was essenitally isotropic, so about half of it escapes to space, with half coming down as blue sky. I do believe 7% is half of 17%; give or take some error bands. And yes I did simply ignore the atomic spectral lines of the atmosphere, on the premise that though they may be many, they are also narrow compared to molecular bands. I’m not sure of the origin of the 500-650 band of Ozone. Assuming that the ozone is in a high cold thin layer, then the broadening of molecular lines shouldn’t be too large; but I don’t quite see how such a shrt wavelength band occurs. I’m not a chemist; so pretty shaky on molecular spectroscopy. I’d tend to believe Phil’s interpretation of what causes that absorption, although it is not a prominent dip.
As for the thermal continuum emission from water/ice (in clouds), I’m quite interested in what real observed spectra are since I am quite uncomfortable with the notion that only ghg molecular species, are emitting LWIR from the atmosphere. I don’t see why gas molecules in collision (acceleration events) can’t emit BB like radiation; but such emission from a single molecule presumably changes the Temperature; since the heat capacity of a single molecule is small. So I’m not sold on the notion that gases don’t emit thermal continuum spectra; but of course I wouldn’t expect the emissivity to be very high because of the lower molecular density. Thermal emission from solids (at earth environmental Temperatures) also lowers the Temperature; but in that case, the heat capacity is very much higher so the Temperature inertia is more robust.
I haven’t looked into Mie scattering much, because I get the impression that it is a consequence of crud in the atmosphere; whereas Raleigh scattering occurs even in pollution free atmosphere.
George,
gas molecules are colliding at tremendous rates. Sometimes they raise a molecule to a higher state and sometimes they lower it from a higher state. Also photons being absorbed and emitted will change the state. You have to have a transition of the right energy to absorb or emit a given photon. Also, times for a state to decay and emit a photon may be longer than typical collision interaction times and all these go into the nature of a strong or weak line.
A state change will change the energy of a molecule, but then it won’t take long for it to come back to ‘normal’ Of course normal is quite a range of possible energies. If enough energy is coming in to balance what is going out, the temperature wont change overall. If not, then the T will change until the energy transfer rates balance again.
Ionized gases can emit continuum. The sun’s photosphere is emitting a continuum, mostly due to the ‘metals’ and not the H or He. A typical response would be that sure a gas will emit a continuum if you’re talking about an optically thick amount at the wavelengths of interest. And, it’s true. It’s a bit of the old ‘catch 22’. Thick means it can absorb as well as emit. That won’t happen where the gas is transparent. Ultimately that might mean that the wings of the lines get smeared out due to all the molecules bouncing around at different velocities in difference directions and red shifts and blue shifts and just plain old statistical liklihoods of being on the fringes.
mie scattering is you cloud droplets and also potentially your dust particles. rayleigh is individual molecules or atoms.
other than to see it, you might search for Iris, a Mie scattering display program. I doubt it can produce much information of interest beyond the graphs.
“”””” cba says:
April 26, 2011 at 4:37 pm
George,
gas molecules are colliding at tremendous rates. Sometimes they raise a molecule to a higher state and sometimes they lower it from a higher state. Also photons being absorbed and emitted will change the state. You have to have a transition of the right energy to absorb or emit a given photon. “””””
As I understand the continuum emission of ionised gases; it is a simple consequence of the fact that an arriving electron that “drops in” to one of the atom’s available (and permitted) energy levels, can have any value of energy, so the energy change in the reaction is a continuous function; in a sense the vast plasma doesn’t have discrete energy levels for its free electrons.
When it comes to the Black Body spectrum, and the Planck Radiation formula; which is the best known continuum spectrum, I’m under the impression that the derivation is purely one of classical Physics involving “particles” having a certain number of degrees of freedom to which an amount of energy can be assigned. There aren’t any atomic energy levels involved; and Planck’s contribution to remove the Ultra-Violet catastrophe of the Raleigh Jeans formula, was simply to require that the allowable energy assigned to each degree of freedom, under the equi-partition Principle had to be an integral multiple of a discrete or quantized amount (kT0 rather than any possible continuous value, that Raleigh Jeans assumed, in their derivation.
The purported origin of the radiation was simply the Classical Maxwellian Electromagnetism requirement, that an an accelerated electric charge; being thus a varying electric current, must radiate electro-magnetic waves. The whole purpose of the Stanford two mile linear accelerator, was to eliminate the charge acceleration due to electrons travelling in a circle in a magnetic field. Even in this post Quantum Mechanical world, Maxwell’s equations still insist that accelerated charges radiate; it’s just an electric current in an antenna, situation.
So how does this apply to BB radiation
“”””” cba says:
April 26, 2011 at 4:37 pm
George,
gas molecules are colliding at tremendous rates. Sometimes they raise a molecule to a higher state and sometimes they lower it from a higher state. Also photons being absorbed and emitted will change the state. You have to have a transition of the right energy to absorb or emit a given photon. “””””
As I understand the continuum emission of ionised gases; it is a simple consequence of the fact that an arriving electron that “drops in” to one of the atom’s available (and permitted) energy levels, can have any value of energy, so the energy change in the reaction is a continuous function; in a sense the vast plasma doesn’t have discrete energy levels for its free electrons.
When it comes to the Black Body spectrum, and the Planck Radiation formula; which is the best known continuum spectrum, I’m under the impression that the derivation is purely one of classical Physics involving “particles” having a certain number of degrees of freedom to which an amount of energy can be assigned. There aren’t any atomic energy levels involved; and Planck’s contribution to remove the Ultra-Violet catastrophe of the Raleigh Jeans formula, was simply to require that the allowable energy assigned to each degree of freedom, under the equi-partition Principle had to be an integral multiple of a discrete or quantized amount (kT0 rather than any possible continuous value, that Raleigh Jeans assumed, in their derivation.
The purported origin of the radiation was simply the Classical Maxwellian Electromagnetism requirement, that an an accelerated electric charge; being thus a varying electric current, must radiate electro-magnetic waves. The whole purpose of the Stanford two mile linear accelerator, was to eliminate the charge acceleration due to electrons travelling in a circle in a magnetic field. Even in this post Quantum Mechanical world, Maxwell’s equations still insist that accelerated charges radiate; it’s just an electric current in an antenna, situation.
So how does this apply to BB radiation “DUE SOLELY TO TEMPERATURE” ? Well Temperature is characterized by a collection of molecules/atoms having a Maxwell Boltzmann distribution of kinetic energies ,with the mean energy being Temperature dependent; square root of T, as I recall. As a result of this, the molecules/atoms are in constant collisions covering the whole range of collision energies due to the MB distribution. In those collisions, you have electric charges in free flight (constant velocity( neglecting gravity)) entering into collisions that result in a follwing free flight (constant velocity) at some totally different and unpredictable velocities; and for the duration of the interraction time of the two colliding atoms/molecules, those electric charges are undergoing acceleration, and therefore must radiate EM waves during the collision time. Now of course for non-polar atoms/molecules, such as Ar, or O2, or N2, etc the charge centre of the nuclear protons, and the orbital electrons, is essentially a single point; so what is this NET charge acceleration ?
Well the -ve charge of course is due to the orbital electrons, and the +ve charge is due to the nuclear protons; which are nearly a thousand times more massive. so when the particles collide, the acceleration of the electron cloud, and the nucleus are not the same, so the atom/molecule gets deformed during the collision time; and that is why there is a net charge acceleration because the more massive nucleus, is alot more sluggish tna the elctrons; so for all practical purposes, it is only the electron acceleration that matters.
So BB radiation from gases is explainable without quantum mechanics, using just Maxwell’s equations of Electro-Magnetism. Planck only needed to have the equipartitioned energies assigned to the degrees of freedom to be multiples of kT, rather than a continuous energy spectrum, to eliminate the UV catastrophe of Raleigh-Jeans. Other than that the derivation was just statistical mechanics.
It is pretty hard to argue that the Classical Physics of Maxwell’s Equations of Electro-magnetism was destroyed by the Quantum Theory of atomic structure; when those equations are enshrined via (c), (mu-naught) (epsilon-naught) as the only fundamental physical constants to have absolute values; with zero uncertainty.
But your point about the optically thick gaseous source is just a different way of expressing my comment that the Temperature must drop with the emission due to the low thermal mass associated with the gas sample, and that due to the low total number of radiating particles (density), the emissivity is low, which is why the gaseous continuum emission is of low intensity; but I don’t believe it is absent; it just takes a hell of a lot of gas to get any intensity.
Someone else referred to the thermal emission from gases, as being collision induced radiation. I like that description; because I believe it is simply the transient acceleration of the electron cloud charge during however many femtosecond (or maybe attoseconds) is involved in the atomic/molecular collision event, that is the physical origin of that emission.
Just remember that theoretically, the spectrum of (BB) thermal emissions goes from zero frequency to infinite frequency. You can’t explain that with any sort of energy level structure even in the solid state.
Niels Bohr, sort of screwed things up, by declaring that Maxwell’s insistence that acelerated charges radiate, was null and void, and that they didn’t have to radiate, if they were part of an energy level orbital structure; only when they jumped from one state to another. It was quite arbitrary, and quite without foundation, and if it were accepted today, then Maxwell’s electromagnetism would be dead. Of course it isn’t because the quantum mechanical picture was changed so that the atomic energy level structure was not one of planetary orbits of moving electrons; but simply a probability distribution of electron location (orbitals). Get rid of the planetary orbits with their constant acceleration, and Maxwells is resurrected to become King of the heap with his derivation of the Velocity of light.
Anyway; that is the way I view it; but your exposition does not create any trauma for me; I can adapt to your imagery.
George,
well you’re pretty much describing stuff. Maybe the question should be why do electrons in these various states – like the ground state – not radiate even though they seem to be in motion with continual acceleration. Note, I seem to recall one hypothesis somewhere about maybe it was actually the change in acceleration rather than the acceleration that caused the radiation.
Of course, for IR we’re not talking electron states but rotation and vibrational molecular activities and for the continuum of solids and liquids, one must even add in the interactions between molecules.
The boltzman (or planck curve) distribution of energy states defines the classical BB. It’s not the only continuum distribution possible although it’s the most likely one. Synchrotron radiation is another and it can be found apparently as the dominant emission from galaxies with active galactic nuclei. But that has nothing to do with the topic at hand.
cba says:
April 26, 2011 at 5:29 am
————————————————————;
How large is a water molecule?
How large is a CO2 molecule?
How large is a raindrop?
What is the density of water vapor?
What is the density of water?
Since the Earth’s surface temperature is roughly 1/3 radiative and 2/3 thermodynamic – and the radiative part has a small variation, then IMHO, any explanation which isn’t based fluid properites (interacting with external forces) is doomed to failure.
What needs to be explained is why the Earth is acting like a refrigerator.
“”””” cba says:
April 26, 2011 at 7:06 pm
George,
well you’re pretty much describing stuff. Maybe the question should be why do electrons in these various states – like the ground state – not radiate even though they seem to be in motion with continual acceleration. Note, I seem to recall one hypothesis somewhere about maybe it was actually the change in acceleration rather than the acceleration that caused the radiation. “””””
Well cba, that would seem to be the whole point. Bohr in his “planetary” description of the atom simply assumed that the electrons are in motion; and therefore must be accelerating if in circular orbits; later generalized to elliptical orbits by Arnold Sommerfeld, and that picture enabled Bohr to calculate the (presumed) kinetic energy of the orbiting electron, and in the end explain the Balmer and other series of atomic spectral lines (for Hydrogen atom at least). That discovery was one of the truly great achievements of the golden age of Physics. I used to think that Atomic spectra were the greatest thing since long before sliced bread.
But in the modern quantum view of the atom; there isn’t any need to assign movement and hence acceleration to the elctrons; the “Orbital” shapes are simply probability maps of where the electron is likely to be found. Bohr’s original atom doesn’t allow the Principal quantum number (n) to be zero, since that implied in his model, an orbit that went right through the center of the nucleus. Today the probability of that being the location of the electron, is not considered a problem.
Maxwells EM radiation of course arises naturally from applying his four equations of the EM field to varying (sinusoidal) currents in a “dipole” antenna; which implies separated electric charges (dipole) and movement of those charges (current) which alters the dipole moment, and hence the EM field. The force of electromagnetism is one of the two forces of nature that has infinite range; the other being gravity, so that implies that the EM field, like the gravitation field, can spread infinitely far from the source (antenna) . But once the field is way out there, and the source current is varying rapidly enough, the finite value of the velocity of propagation of the field, prevents the entire field from collapsing back on the antenna, and it starts to disconnect from the antenna, and create closed fields that now propagate away at the speed of the EM radiation which is (c). So it is not unlike blowing soap bubbles. The bubble stays attached to the wire ring, so long as you don’t blow too hard. But when you blow hard enough, the soap film tears off the ring, and if the surface tension can close the gaping hole in the film (natural surface area minimisation) faster than the air can rush out of the bubble through the hole (velicity of sound limitation), then a disconnected closed bubble is generated, and can propagate away.
Somewhat analagously the :bubbles of electromagnetic field disconnect from the antenna and close up both the electric field lines, and the magnetic field lines so both are continuous, and the bubble escapes at 3E8 m/s. To achieve that disconnect and propagation, the current in the antenna has to increase, to establish the field, and then rapidly decrease, to reverse the field before the established field can collapse back onto the antenna. That’s why the requirement for charge acceleration (or current variation) which is the same thing.
My Radio-Physics Major comes in handy, when trying to understand how EM radiation works.
The key distinction between the continuum thermal radiation, and the absorption band spectra of molecules, is that the former is a consequence entirely of the Temperature of the assemblage of particles, and somewhat independent of what those particles are; Planck’s formula is a consequence of statistical mechanics, and involves no assumptions about the energy level structure of atoms or molecules. It does require the assumption of only quantised energy values being allowed for the kinetic energies of the degrees of freedom of the energetic particles; but needs no knowledge of atomic numbers or electron masses or anything like that.
The LWIR emissions from molecules (or absorptions) are “resonance” phenomena, that are entirely dictated by the structure of those molecules, and things like the strength of chemical bonds, and the possible modes of oscillation. All the common GHGs are known to have molecular structures that either have a non zero electric charge dipole moment, as H2O does (CO2 doesn’t); or are capable of exhibiting a non zero electric dipole moment, when oscillating in some of the possible modes of oscillation permitted by the physical structure of the molecule.
In the case of CO2, we have the “symmetrical stretch” mode, where the two Oxygens maintain equal and opposite vibrations along the linear axis of the molecule while the C remains stationary (well it is the origin of the local co-ordinate system). In that case, the dipole moment remains identically zero, so that modfe is not IR active. The assymmetrical stretch mode, where all three atoms move along the molecule axis of symmetry, and only their center of mass remains stationary, the electric dipole moment now is not always zero so this mode is IR active at 4 microns. The important mode for the GHG phenomenon, is the degenerate bending mode, where two identical modes at right angles are possible (the degeneracy), and the molecule can do an elbow bend about the carbon atom. Since that sort of oscillation involves a higher moment of inertia for the components, the resulting resonance frequency is lower, so we get the common 15 micron band of CO2 (or is it 14.74 or some such number).
But you see those are discrete resonance oscilations that are a consequence of atomic or molecular structure; and are pretty much independent of Temperature. The thermal continuum radiation is entirely a consequence of Temperature because of the inter molecular collisions; which distort the atomic or molecular electric charge dipole moment during the collision and permit the radiation of EM radiant energy.
Well that’s a somewhat stick in the sand description of how I see it; which takes a lot longer to describe than to understand.
The molecular band spectra cannot absorb or emit EM radiation from down to but non including DC, and up to beyond the upper reaches of the gamma ray frequencies; but a heated body can.
George,
”
Just remember that theoretically, the spectrum of (BB) thermal emissions goes from zero frequency to infinite frequency. You can’t explain that with any sort of energy level structure even in the solid state.
”
there’s lots of other stuff though other than molecular or atomic. Once you go beyond a single molecule, you’ve got molecular interactions plus broadeing due to the doppler effects.
“The LWIR emissions from molecules (or absorptions) are “resonance” phenomena, that are entirely dictated by the structure of those molecules, and things like the strength of chemical bonds, and the possible modes of oscillation. All the common GHGs are known to have molecular structures that either have a non zero electric charge dipole moment, as H2O does (CO2 doesn’t); or are capable of exhibiting a non zero electric dipole moment, when oscillating in some of the possible modes of oscillation permitted by the physical structure of the molecule.
”
Here is where we are not in agreement exactly. I think you’re missing a little tidbit.
The BB spectrum curve at a temperature is also the boltzman distributions of energy states. What one has with something like the Hitran database information is the absorption coefficient for a given line under the defined circumstances used. This is not directly dependent on temperature for the absorption. However, this is the liklihood of a molecular interaction at the particular wavelength (or wavelengths given the line width). Multiplying the liklihood of the number of molecules in a given state (the BB curve) by the liklihood of a molecule undergoing an absorption transition gives us our emission spectrum, which depends both on the nature of the materials and upon the temperature of those molecules. After all, if there are no molecules in the higher energy excited states there can’t be any emissions from them. This combination gives us the emission spectrum for those molecules and it is temperature dependent even though the absorption portion of this concept is almost independent of temperature. Net result is that a gas blocking a bb emitter at the same temperature will result line emission rates = line absorption rates and the gas will be essentially invisible. It also can’t happen for llong because of energy balance considerations.