This could have been included in the Earth’s Energy Budget series, but it deserved a post of its own.
First of all, what is “back-radiation” ? It’s the radiation emitted by the atmosphere which is incident on the earth’s surface. It is also more correctly known as downward longwave radiation – or DLR
What’s amazing about back-radiation is how many different ways people arrive at the conclusion it doesn’t exist or doesn’t have any effect on the temperature at the earth’s surface.
If you want to look at the top of the atmosphere (often abbreviated as “TOA”) the measurements are there in abundance. This is because (since the late 1970’s) satellites have been making continual daily measurements of incoming solar, reflected solar, and outgoing longwave.
However, if you want to look at the surface, the values are much “thinner on the ground” because satellites can’t measure these values (see note 1). There are lots of thermometers around the world taking hourly and daily measurements of temperature but instruments to measure radiation accurately are much more expensive. So this parameter has the least number of measurements.
This doesn’t mean that the fact of “back-radiation” is in any doubt, there are just less measurement locations.
For example, if you asked for data on the salinity of the ocean 20km north of Tangiers on 4th July 2004 you might not be able to get the data. But no one doubts that salt was present in the ocean on that day, and probably in the region of 25-35 parts per thousand. That’s because every time you measure the salinity of the ocean you get similar values. But it is always possible that 20km off the coast of Tangiers, every Wednesday after 4pm, that all the salt goes missing for half an hour.. it’s just very unlikely.
What DLR Measurements Exist?
Hundreds, or maybe even thousands, of researchers over the decades have taken measurements of DLR (along with other values) for various projects and written up the results in papers. You can see an example from a text book in Sensible Heat, Latent Heat and Radiation.
What about more consistent onging measurements?
The Global Energy Balance Archive contains quality-checked monthly means of surface energy fluxes. The data has been extracted from many sources including periodicals, data reports and unpublished manuscripts. The second table below shows the total amount of data stored for different types of measurements:
You can see that DLR measurements in the GEBA archive are vastly outnumbered by incoming solar radiation measurements. The BSRN (baseline surface radiation network) was established by the World Climate Research Programme (WCRP) as part of GEWEX (Global Energy and Water Cycle Experiment) in the early 1990’s:
The data are of primary importance in supporting the validation and confirmation of satellite and computer model estimates of these quantities. At a small number of stations (currently about 40) in contrasting climatic zones, covering a latitude range from 80°N to 90°S (see station maps ), solar and atmospheric radiation is measured with instruments of the highest available accuracy and with high time resolution (1 to 3 minutes).
Twenty of these stations (according to Vardavas & Taylor) include measurements of downwards longwave radiation (DLR) at the surface. BSRN stations have to follow specific observational and calibration procedures, resulting in standardized data of very high accuracy:
- Direct SW – accuracy 1% (2 W/m2)
- Diffuse radiation – 4% (5 W/m2)
- Downward longwave radiation, DLR – 5% (10 W/m2)
- Upward longwave radiation – 5% (10 W/m2)
Radiosonde data exists for 16 of the stations (radiosondes measure the temperature and humidity profile up through the atmosphere).
Click for a larger image
A slightly earlier list of stations from 2007:
Solar Radiation and Atmospheric Radiation
Regular readers of this blog will be clear about the difference between solar and “terrestrial” radiation. Solar radiation has its peak value around 0.5μm, while radiation from the surface of the earth or from the atmosphere has its peak value around 10μm and there is very little crossover. For more details on this basic topic, see The Sun and Max Planck Agree
.
What this means is that solar radiation and terrestrial/atmospheric radiation can be easily distinguished. Conventionally, climate science uses “shortwave” to refer to solar radiation – for radiation with a wavelength of less than 4μm – and “longwave” to refer to terrestrial or atmospheric radiation – for wavelengths of greater than 4μm.
This is very handy. We can measure radiation in the wavelengths > 4μm even during the day and know that the source of this radiation is the surface (if we are measuring upward radiation from the surface) or the atmosphere (if we are measuring downward radiation at the surface). Of course, if we measure radiation at night then there’s no possibility of confusion anyway.
Papers
Here are a few extracts from papers with some sample data.
Downward longwave radiation estimates for clear and all-sky conditions in the Sertãozinho region of São Paulo, Brazil by Kruk et al (2010):
Atmospheric longwave radiation is the surface radiation budget component most rarely available in climatological stations due to the cost of the longwave measuring instruments, the pyrgeometers, compared with the cost of pyranometers, which measure the shortwave radiation. Consequently, the estimate of longwave radiation for no-pyrgeometer places is often done through the most easily measured atmospheric variables, such as air temperature and air moisture. Several parameterization schemes have been developed to estimate downward longwave radiation for clear-sky and cloudy conditions, but none has been adopted for generalized use.
Their paper isn’t about establishing whether or not atmospheric radiation exists. No one in the field doubts it, any more than anyone doubts the existence of ocean salinity. This paper is about establishing a better model for calculating DLR – as expensive instruments are not going to cover the globe any time soon. However, their results are useful to see.
In another paper, Wild and co-workers (2001) calculated some long term measurements from GEBA:
This paper also wasn’t about verifying the existence of “back-radiation” – it was assessing the ability of GCMs to correctly calculate it. So you can note the long term average values of DLR for some European stations and one Japanese station. The authors also showed the average value across the stations under consideration:
And station by station month by month (the solid lines are the measurements):
Click on the image for a larger view
In another paper, Morcrette (2002) produced a comparison of observed and modeled values of DLR for April-May 1999 in 24 stations (the columns headed Obs are the measured values):
Click for a larger view
Once again, the paper wasn’t about the existence of DLR, but about the comparison between observed and modeled data. Here’s the station list with the key:
BSRN data
Here is a 2-week extract of DLR for Billings, Oklahoma from the BSRN archives. This is BSRN station no. 28, Latitude: 36.605000, Longitude: -97.516000, Elevation: 317.0 m, Surface type: grass; Topography type: flat, rural.
And 3 days shown in more detail:
Note that the time is UTC so “midday” in local time will be around 19:00 (someone good at converting time zones in October can tell me exactly).
Notice that DLR does not drop significantly overnight. This is because of the heat capacity of the atmosphere – it cools down, but not as quickly as the ground.
DLR is a function of the temperature of the atmosphere and of the concentration of gases which absorb and emit radiation – like water vapor, CO2, NO2 and so on.
We will look at this some more in a followup article, along with the many questions – and questionable ideas – that people have about “back-radiation”.
Update: The Amazing Case of “Back-Radiation” – Part Two
The Amazing Case of “Back Radiation” – Part Three
Darwinian Selection – “Back Radiation”
Notes
Note 1 – Satellites can measure some things about the surface. Upward radiation from the surface is mostly absorbed by the atmosphere, but the “atmospheric window” (8-12μm) is “quite transparent” and so satellite measurements can be used to calculate surface temperature – using standard radiation transfer equations for the atmosphere. However, satellites cannot measure the downward radiation at the surface.
References
Radiation and Climate, I.M. Vardavas & F.W. Taylor, International Series of Monographs on Physics – 138 by Oxford Science Publications (2007)
Downward longwave radiation estimates for clear and all-sky conditions in the Sertãozinho region of São Paulo, Brazil, Kruk et al, Theoretical Applied Climatology (2010)
Evaluation of Downward Longwave Radiation in General Circulation Models, Wild et al, Journal of Climate (2001)
The Surface Downward Longwave Radiation in the ECMWF Forecast System, Morcrette, Journal of Climate (2002)
That was very interesting to see the numbers and the charts showing daily and seasonal variations.
Probably most gardeners have an acquaintance with DLR without knowing the actual stats. They take into account visible ghg in their decisions about frost protection quite often.
Bryan is missing. There’s actual OBSERVATIONS here, so he should be lapping them up!
I would suggest that something from the observed spectral response of CO2 should be found too, something from the 1860’s should be accurate enough to show the narrower bands of CO2, even though their equipment wasn’t of high enough resolving power to get the fine structure ( maybe see if
Martin, P.E., and E.F. Baker (1932). “The Infrared Absorption Spectrum of Carbon Dioxide.” Physical Review 41: 291-303.
is appropriate)
“DLR is a function of the temperature of the atmosphere and of the concentration of gases which absorb and emit radiation – like water vapor, CO2, NO2 and so on.”
You might want to add cloud cover to that sentence. The difference in DLR between a cold clear night and a cold overcast night is large.
And be sure to point out in the next post that at some stations we measure longwave radiation emitted upward from the surface as well as DLR, and that the two quantities differ. I’ve seen the fallacious argument that the surface always emits exactly as much as it receives in DLR. Those making this argument either misapply Kirchhoff’s laws or mistakenly think that a microlayer at the surface heats up extremely rapidly to a temperature sufficient to produce an upwelling flux equal to the DLR.
replying to Rollins’ post of 2010 Jul 17, 11:44 am.
Of course you are right that land-sea surface upward (let us denote it Su) longwave radiation exceeds bottom of atmosphere downward emitted radiation. The difference is the amount of land-sea surface upward radiation that penetrates the atmosphere without interacting with it, and goes straight from the land-sea surface to space, let us denote it St. It is in practice difficult, or perhaps even in principle impossible, to directly measure St. Satellites cannot directly measure it because it is practically inseparable from the atmospheric upward emitted radiation to space.
As you note for the atmospheric downward emitted radiation, also the atmospheric radiation emitted upwards (Eu) direct to space can be estimated from knowledge of the atmospheric conditions. The measurable quantity, the total outgoing longwave radiation (OLR) is the sum of the two quantities St and Eu: we have OLR = Eu + St.
Su goes two ways: to be absorbed into the atmosphere (Aa) and to pass straight to space without interacting with the atmosphere (St). We thus have Su = Aa + St. Again, Aa cannot be directly measured. But like Ed and Eu, Aa can be estimated from atmospheric conditions.
It is an empirically observed fact that for calculations that use radiosonde data to estimate Aa and Ed, the two are empirically observed to be fairly nearly equal. There are diversities of atmospheric conditions in which one or the other is greater, but on average they are very nearly equal. See Ferenc Miskolczi’s paper indexed at http://multi-science.metapress.com/content/nm45w65nvnj3/?p=2037ca7987b54843bc4b526ff0b6ff7c&pi=1, The stable stationary value of the earth’s global average atmospheric Planck-weighted greenhouse-gas optical thickness,’Energy and Environment’ 21(4): 243- 262, doi 10.1260/0958-305X.21.4.243 .
As you rightly note, this is not to be explained simply by an isolated appeal to Kirchhoff’s law of radiation. The proper physical explanation is that besides radiation to carry heat from land-sea surface to atmosphere, and lower atmospheric absorption of solar radiation, there is also material transport, in the forms of conduction, evaporation, and convection, that carries energy from the land-sea surface to the lower atmosphere. These non-radiative transfers, when added to the two forms of radiative transfer, provide the lower atmosphere with just the amount of energy that is needed to maintain the relation Aa = Ed noted above. Most of this exchange takes place at low altitudes, within the turbulently mixed boundary layer of the atmosphere. Indeed most of it takes place at altitudes lower than 300 meters.
This empirically observed fact, that Aa = Ed, is, as you are well aware, not clearly recorded in textbooks. That seems to be not because they have examined the question and found to the contrary, but because they have not examined the question. You will note, however, in many texts (for example Goody and Yung 2nd edition 1989, page 250; Wallace and Hobbs 2nd edition 2006, page 138) there is emphasized what is called the the “cooling-to-space approximation”. This is very closely related to the fact that Aa = Ed, and is likely to have the same physical explanation.
You write of those who “mistakenly think that a microlayer at the surface heats up extremely rapidly to a temperature sufficient to produce an upwelling flux equal to the DLR”. This is not quite the thinking that provides the best physical explanation of how it comes about that Aa = Ed, and instead I commend the above explanation to you.
If you have, or know of someone else who has, some direct empirical observations of locally matched values of Aa and Ed, it will be of great value to science if you or they publish them to provide a check on Miskolczi’s observations and calculations, which seem so far not to have been checked empirically by other observers. There are of course plenty of speculative opinions and guesses and comments about what one would expect from one’s private reading and thinking, but it seems that so far no one has published an empirical check on Miskolczi’s observations and calculations.
You might also want to note that the emission of DLR cools the upper atmosphere as it warms the surface, and that the upper atmosphere and the surface are linked by convective coupling so as to maintain a fixed temperature difference.
When lifting oneself off the ground by your own bootstraps, the downward force on your arms is the same as the upward force on your boots.
During daylight surely the sun will be the greatest heater of the atmosphere and, where the atmosphere does not absorb, direct LW – even at greater than 10um wavelength
note that this is a log-log scale plot.
So downward long wave will be direct and re-radiated solar radiation (or is this absorbed in the upper atmosphere?) plus a very small potion of back radiation from the ground.
At night the mixed warmed atmosphere will back radiate. Presumably this will be from all the gasses (not just GHGs) since they must be all warmed by conduction???
At night heat radiated from the warm sea/earth will be again trapped by GHG molecules and reradiated in all directions equally. (will half or less or more be radiated to hit the earth?)
Radiation tangentially to the earth will hit more GHG molecules. Each will reradiate in all directions. The probability of a heat “particle” escaping to space rather than ending up as back radiation must be very low at low altitudes and very high at the last molecule in the atmosphere (but even here the “particle” has the finite possibility that it will radiate downwards and form anorher chain of passing the energy to the ground.
What is the overall downward to upward ratio for the atmosphere?
Mike
The TFK 2009 energy balance ( http://www.cgd.ucar.edu/cas/Trenberth/trenberth.papers/TFK_bams09.pdf ) has upward 396 and downward 333 W/m2. For low cloud cover, up and down are nearly equal. For clear sky and low humidity at night the upward radiation can exceed the downward radiation by more than 100 W/m2, leading to rapid ground cooling and a temperature inversion if there is no wind. You can freeze water in the desert if you expose it to the sky at night and insulate it well during the day, lots of straw and a shiny metal cover.
It just so happens, for global averages, that the difference between the TFK land-sea surface upward radiation and their atmospheric downward radiation, (396 – 333) W/m2, is about 63 W/m2, which happens to be rather near the amount of radiation from the land-sea surface transmitted direct to space, for the global average. The TFK diagram gives a value of 40 W/m2 for radiation direct to space from the land-sea surface, but this figure is not accurate, being stated by TFK to be merely found as a residual; they do not calculate it directly from radiosonde data, as is necessary to get a correct figure. That inaccurate residual figure of TFK is one source of some people’s mistaken denial of the empirical fact that Aa = Ed, noted above in my reply to Rollins.
Billings, OK is Central Time Zone which is UTC -6 hours. So noon would be 18:00 UTC with peak temperature approximately five hours later.
The model fit, while it’s in the ballpark, doesn’t look very good, lots of scatter and the slope is too low.
Back-radiation, or more precisely how it’s often described, is something I’m not entirely comfortable with. More specifically it’s effect on surface temperature. This is very much a chicken-egg problem (again), but I find it way less misleading to claim that Downward long wave radiation is dependent on temperature rather than the other way around, especially in the context of the effect of greenhouse gases. Essentially, if I’ve understood correctly, the DLR is the radiation emited by the gases in the atmosphere, with some reflected radiation added in (I’m guessing the cloudiness differences come in large part from reflected radiation due to the albedo of clouds rather than emitted radiation). As we know, atmosphere is quite opaque close to the surface where it is absorbing, which means that the DLR on the surface wouldn’t change that much if we increased the CO2 levels to 94,5% of the atmosphere and kept the same surface temperature.
“As we know, atmosphere is quite opaque close to the surface where it is absorbing, which means that the DLR on the surface wouldn’t change that much if we increased the CO2 levels to 94,5% of the atmosphere and kept the same surface temperature.”
Bear in mind that the temperature changes with altitude. So in an opaque atmosphere it is coming from a warm, low layer. In a more transparent atmosphere it is coming from a high, cold layer.
The troposphere (the turbulent lower layer of atmosphere where almost all the GHGs are) ranges from an average +14 C at the bottom to -54 C at the top, an amazing 68 C range over about 10-20 km.
Everybody knows that hot air rises, but they rarely go on to wonder about the tops of mountains being so cold and covered in snow.
The temperature drops in a straight line, although the density of atmosphere (and hence GHGs) drops almost exponentially. This 68 C difference has to be driven by something, to maintain it against a natural inclination to equalise. The fact that it is a single straight line suggests that the mechanism by which the top is cooled to -54 C is the same as that warming the surface to + 14 C.
So how is the top layer cooled? Is it radiating it away? Where to? Has anyone ever told you?
I think that thinking about and explaining the cooling of the top of the atmosphere is far more revealing than the warming of the surface. There are several possible explanations for the latter, and it’s hard to tell which one is right. There are fewer options for cooling.
The atmosphere cools to space by radiation, mainly from the troposphere, and a little from the stratosphere. The radiative activity of greenhouse gases is the mechanism of this. The main greenhouse gas is water vapour, with small contributions from CO2 and methane and ozone, and even smaller contributions from others. The global average Eu (emitted upwards atmospheric radiation to space, defined in more detail above in my reply to Rollins) contribution density profile with altitude is shown as figure 6 of Miskolczi (2010) cited above in my reply to Rollins. The Eu comes from all levels of the troposphere, more from lower tropospheric altitudes because they are warmer and have higher concentrations of water vapour, the main radiator to space, and higher concentrations of carbon dioxide. About half of the Eu comes from below 6 km. The reason for the linear drop in temperature with altitude is to do with rates of transfer of energy upwards by convection, and to do with the fact that radiation can escape directly to space more easily at higher altitudes because they have less optical thickness above them. At and above the tropopause, the absorption of solar radiation by ozone molecules reverses the temperature lapse rate, and the temperature increases with altitude up to the stratopause. At higher altitudes other factors are active. At altitudes above the tropopause, the air is so thin that the actual amounts of heat exchanged are relatively small, so that the main energy exchanges occur in the troposphere and at the land-sea surface.
Thanks.
The temperature difference between the surface and the top of the atmosphere, where it radiates to space, is 68 C. If there were no GHGs in the troposphere so the surface itself radiated directly to space, how much colder would it be?
The lapse rate would be the same, but the surface would be colder. Therefore the atmosphere would be thinner and colder withal and the TOA would be such that the emission would be balancing the reception.
Therefore, oddly enough, the TOA would actually be warmer.
I think you’ve got the figure wrong, mind
http://www.windows2universe.org/earth/Atmosphere/mesosphere.html
and it would be odd to have the TOA much hotter than the earth surface! 🙂
Christopher Game,
You can calculate Aa and Ed for various scenarios using MODTRAN. They’re close, but not exactly equal. In Sub-Arctic Winter conditions, for example, Ed is actually greater than Aa because there is a temperature inversion. But for most conditions, Ed is slightly less than Aa.
For total radiation directly to space, you have to include radiation from cloud tops. For the cloud covered surface, radiation to space from the surface is zero and tau (the optical depth of the atmosphere in the thermal IR as seen from the surface) is effectively infinite. Miskolzci gets around this problem by ignoring clouds completely in his calculation of the global average tau.
reply to DeWitt Payne’s post of 2010 Jul 18 5:01 pm
Thank you.
MODTRAN is a moderate-resolution calculator, that uses a band model, and therefore is likely not as accurate as HARTCODE, which is the high-resolution atmospheric radiative transfer code that Miskolczi (2010) wrote and uses. The high resolution is reached by use of the line-by-line method of calculation. The atmospheres offered by MODTRAN at the University of Chicago site (to which perhaps you refer?) are not very clearly defined as I see them on the website. Miskolczi’s radiosonde data seems more clearly defined in terms of a database of radiosonde records.
I see that you calculate Aa by use of MODTRAN. Please tell us precisely how you do this. The figures that I get from MODTRAN assume that the number labeled AVERAGE TRANSMITTANCE at the seventh non-blank line from the bottom of the text output is the same as
the quantity Ta = St / Su of Miskolczi, but I did not find a clear statement of exactly what that AVERAGE TRANSMITTANCE means. Will you very kindly show me such a clear statement? I downloaded and looked in the instruction manual for MODTRAN but did not find one.
I asked Miskolczi to do a few comparisons between MODTRAN calculations (as I perhaps mistakenly interpreted them) and HARTCODE calculations. There are differences which we did not explore in detail because I did not have a definite understanding of the meaning of the MODTRAN terms. But likely I would suspect that some of the differences could be due to the greater accuracy of HARTCODE.
In Miskolczi 2010, at Figure 5, Antarctic winter A, you will find a nice graph showing an example of the spectral density comparison between Aa and Ed,
with totals Aa – Ed = – 9.3 W/m2, that is to say, Ed > Aa, as you note in your post here for the MODTRAN Sub-Arctic Winter condition. Also, in Figure 5, Antarctic summer, you will see a case in which Aa > Ed, as you find for most conditions. These differences are slightly exceptional, and on average, near equality is found, as you say. For climatic studies, the global average is of particular importance, and shows very near equality. The small differences are due of course to the chaotic character of the atmospheric energy transport processes.
Miskolczi 2010 draws the reader’s attention to the fact that the presence or absence of clouds is ignored for the calculation in that paper. Here the focus of interest is the pure effect of greenhouse gases themselves without possible transient variations of feedbacks from clouds. For climatic considerations on a long time-scale, the assumption is that the clouds have settled their feedback effects to a steady value. The aim here is not to find the the total radiation direct to space, but to find a pure assessment of the virtual primary direct no-feedback effect of CO2.
As a detail, perhaps a little irrelevant here, the radiation from the top of the clouds direct to space is quite close to what would be the radiation direct to space from the land-sea surface underneath if the clouds were not there, given the same radiosonde record. This is because the atmosphere is drier and thinner above the clouds than it is near the land-sea surface, and this nearly compensates for the lower temperatures of the cloud tops.
I would be glad to hear of details and specific results of your MODTRAN investigations. Do you have your own copy, or do you use the University of Chicago facility on the internet, or what? What are the definitions of the terms? What dataset do you use? Would you consider publishing your methods and results?
Before we talk about “back-radiation”, we need to identify the average distance an IR photon can travel before it is re-absorbed/re-transmitted.
I’ve read that there is a 99.9% re-absorption of a 15 um IR photon (the main frequency for CO2) within 10 metres at the surface (50% within 1 metre) and there is 10^328 absorptions/retransmissions for a 15 um photon to go from the surface to 0.9 atm (the layer where clouds develop).
This would imply there is no “back-radiation”, there is just a delay for the photon to get from the surface to the 5 km to 10 km high layer where the photon starts to have a 50% chance of being emitted directly through the top of the atmosphere to space without being re-intercepted.
CO2, of course, plays a major role in this but there is just a gradual reduction in the amount of energy held in the atmospheric gases as one moves from the surface to layers higher in the atmosphere where the radiation can escape.
In other words, Stefan Boltzmann blackbody radiation and the lapse rate.
Photons operate in the world of quantum physics at the speed of light moderated by the few pico-seconds that a molecule holds onto them before transmitting them.
“Before we talk about “back-radiation”, we need to identify the average distance an IR photon can travel before it is re-absorbed/re-transmitted.”
Why?
Oh, because you can bluster a la Nassif about how something is impossible.
The more things change, the more they stay the same…
What you need to do, Bill, is consider how long it takes for a CO2 molecule to meet another molecule of the airmass and compare that with the emission time for a 15um emission from CO2.
You’ll find that CO2 doesn’t have time to re-emit before some other bugger comes along and inelastically collides with it, taking all that energy and making it thermal kinetic energy.
The average atmospheric molecule at sea level experiences 10 billion collisions per second.
When I wrote “re-emit”, I meant through emission or collisional exchange.
So, how does a 15 um IR photon emitted from the troposphere “back-radiate” to the surface? It doesn’t. It is intercepted by another molecule or transferred by collision in just a few metres.
The IR photons make a random walk through the atmosphere through millions of (CO2, N2, O2, H2O, O3 and Ar) molecules before they eventually escape.
“When I wrote “re-emit”, I meant through emission or collisional exchange.”
Then what did you mean by “Photons operate in the world of quantum physics at the speed of light ” and “we need to identify the average distance an IR photon can travel before it is re-absorbed/re-transmitted”
?
Because neither of those affect how kinetic thermalised energy is moved around.
“The IR photons make a random walk through the atmosphere through millions of (CO2, N2, O2, H2O, O3 and Ar) molecules before they eventually escape.”
1) And that random walk moves it around up or down. When it moves down, it’s back radiation.
2) IR is transparent in N2 and O2, the largest constituents of the atmosphere, so IR photons do not move the energy with these.
You’re parroting Nassif’s latest “work” because he’s written a puff piece on his vanity blog and he says it’s impossible to have AGW which suits you just fine.
What is moved with N2 and O2 is thermal kinetic energy. Not IR.
“So, how does a 15 um IR photon emitted from the troposphere “back-radiate” to the surface? ”
By the same mechanism it got to the troposphere.
[–moderator’s note – snipped for etiquette]
“Photons operate in the world of quantum physics at the speed of light moderated by the few pico-seconds that a molecule holds onto them before transmitting them.”
DK ALERT!!!
http://en.wikipedia.org/wiki/Slow_light
38mph light.
(they’ve managed to stop it too, briefly)
http://www.answers.com/topic/metastable-state-1
Picoseconds? How about 100minutes?
Dang. Both barrels.
Nice collection of studies on the subject. I did not know those indirect calculations through parameterizations. Thanks.
This is a fascinating post.
I checked to find that Billings, OK is about 36 1/2 degrees North. October 3 to 5 is close to the autumnal equinox,
so the inclination of the sun should be about
(90 -36.5) = 53.5 degrees. I wonder what the relationship between latitude and flux would be?
Someone earlier said downwelling radiation should cool the atmosphere. I did a rough calculation of daily cooling of
the atmosphere.
mass atmosphere = 5* 10^18 kg=5*10^21gm
temp atmosphere 255K (effective radiating temp to space- underestimates heat content)
specific heat 1.01 joules/gm C
5* 10^21*1.01*255= 1.288 * 10^24 joules
radius earth = 6400km= 6.4*10^6 meters.
area earth = 4 pi r^2 =514,718,540,364,021.76
240 watts/sq meter = 240 joules/sec per square meter
60 sec/min*60 min/hr*24hr/day=86,400 secs per day
5.147* 10^14 sq meters*240 joules/sec/sq meter *8.64*10^4 secs/day= 1.067*10^22 joules per day radiated away
1.067*10^22/1.288*10^24 = 0.83%
So the atmosphere as a whole cools by less than 1% over the course of a day. That figure makes sense when you figure that the earth’s surface temperature my change by 10 C or more overnight far more than average changes over a week, but weather patterns persist for several days, and that’s why meteorologists can predict daily highs out a week or so. That cooling is obviously mostly from the
earth’s surface and air near the surface ,leaving most of the atmosphere unchanged.
The average Earth surface radiation level is 390 W/m2 but it peaks at 418 W/m2 during the day and falls to 364 W/m2 just before sunrise (or the change from Day peak to Night peak is just 54 W/m2.
The average incoming Solar (and other) radiation level, by contrast, increases from just above Zero at night to as high as 860 W/m2 at the peak of the day (with average Albedo).
So, the “back-radiation” would have to be a negative value (-400 W/m2 during the day) and a positive value (+364 W/m2 at night) to make those numbers work.
I have three questions to pose on this topic:
1) I have seen it repeatedly stated that when an atmospheric greenhouse gas molecule emits a photon there is an equal chance of it travelling up toward space as down toward the surface and I accept that to be true of spontaneous emission. This postulate figures centrally in the back of the envelope calculations of the “greenhouse effect” which can be found all over the blogosphere. However, there is another kind of emission – stimulated emission – (which is how lasers work) and I wonder whether in the atmosphere, which is subject to a continuous flux of upward long wave radiation from the surface, stimulated emission, which would take on the same upward direction as the stimulating surface radiation, could exist in such an amount as to significantly shift the balance of atmospheric long wave radiation away from 1/2 up and 1/2 down to some measurably greater amount up than down?
2) In order for downward long wave radiation to heat the surface its photons must be absorbed at the surface by molecules which are in an electronic configuration which allows them to absorb photons of that frequency. But if at the surface the temperature is higher than in the atmospheric source then might the molecules which might have absorbed such a photon be in fact unavailable because they have already moved to a higher energy configuration due to thermal collisions in the material which contains them?
3) If a downward travelling long wave photon is not absorbed at the surface will it not be reflected upward again without heating the surface and become indistinguishable from the general flux of surface emissions of long wave photons?
Stimulated emission requires that the molecule be in an excited state and remain in such an excited state without deexcitation by collision.
This doesn’t happen in our atmosphere.
Stimulated emission also requires an inverted population which requires “more than infinite” temperature which is not possible, so you have to work somewhere where temperature is badly defined, like an extremely rarefied gas.
This isn’t true in our troposphere.
And the emission if it happens (there are a lot of molecules, so even the impossible happens once in a blue moon), that IR photon is just going to get absorbed by another GHG or reflected off a surface (and where does reflection go?) well before it gets out.
The changes of it happening twice in a row is the multiplication of very improbable events so won’t happen in the universe in its entire life.
Bill Illis:
Are you saying the measurements are wrong because you can’t see how this radiation could be generated?
He’s saying they must be wrong because he KNOWS AGW is false. Therefore any measurement that is used to prove it must be wrong.
SoD – once again, nicely done, I’d only seen a few sample pyrgeometer numbers so your review of surveys is very useful.
On the obsession with Miskolczi some of your commenters seem to have – SoD, have you done a post reviewing any of his claims? His original paper that drew so much attention was badly flawed (the whole “virial theorem” argument particularly) and not really a theoretical paper at all. Apparently he has published some more but I’ve not paid any attention to it – as far as I’m aware his arguments are essentially yet another semi-zero-dimensional or one-dimensional model that has no mechanism for handling non-radiative effects or the determining relationships between fluxes and surface temperature that are the important question on climate sensitivity etc.
Bill Illis is making a similar error in believing radiation has to be instantaneously balanced. In fact, radiation doesn’t have to be balanced at all – except over long-term averages if you want your planet not to be greatly heating up and cooling. And even then the balance is only at altitudes (like the tropopause, or the “top of the atmosphere”) that effectively block all other heat transport mechanisms. Radiative energy flux balance is a very limited tool for telling you what’s going on in the atmosphere.
JT:
No. Thermalization of absorbed radiation dominates in the troposphere. This is where collisions transfer the energy to the thermal energy of the atmosphere.
Emission of radiation is isotropic = equal in all directions.
No. Emissivity and absorptivity are functions of wavelength. Not functions of the temperature of the source of the thermal radiation.
Take a look at Intelligent Materials and the Imaginary Second Law of Thermodynamics – and this is something I will cover more in the next post.
Yes. If it is not absorbed it will be reflected. After all, radiation incident on a surface can only be absorbed, reflected or transmitted. In the case of the earth “transmitted” isn’t an option so it is only absorption or reflection – and this is very simple to understand.
The properties of different surface types have been thoroughly studied. Take a look at The Dull Case of Emissivity and Average Temperatures.
What is wonderful is that is very easy to calculate what proportion of longwave radiation is absorbed by the surface of the earth – mostly it is very close to 100%.
More in the next installment.
JT’s question got me thinking. Let’s take a walk and tell me where I go astray.
In terms of emission, the photons will go in some random direction that is toward some uniformly distributed point on a sphere, where the center of that sphere is the molecule emitting the photon. Let’s suppose that the we define the radius of the sphere as the distance where there is a 50% chance that the photon will be absorbed at or before that distance. Now, it gets interesting. The rate of absorption depends on the density of the absorbing media. The density of an atmosphere continuously increases as the altitude decreases. So, a photon emitted downward encounters increasingly dense air, and a photon emitted upward encounters air that continuously decreases in density. If you define a surface at the 50% probability distance, that surface is not in a perfect sphere; it is distorted slightly. It will look a bit like an egg with the fat end down and the narrow end up. If you repeat this emission-absorption sequence a multitude of times with a multitude of molecules, the distortion of the absorption distribution function will, in effect, introduce a bias for upward movement of photons.
OK, tell me where I went astray.
Alan D McIntire:
I didn’t check your maths but it’s a very good point.
John Houghton shows in The Physics of Atmospheres (1986) that the radiative time constant of the atmosphere (in the troposphere) is approximately 6 days.
Science of Doom is correct for the Earth’s atmosphere, but amazingly, high up in the Martian atmosphere, stimulated emission is observed
http://laserstars.org/history/mars.html
Arthur Smith:
Thanks
Not so far. He gets a mention in New Theory Proves AGW Wrong!, but nothing of substance, however, it might be an interesting article for other readers.
Christopher Game,
While it doesn’t say, I’ve checked the calculation and the average transmittance is the integral of the surface transmitted column divided by the surface emission from 100-1500 cm-1, more or less. If you use the printed numbers, there’s rounding error that isn’t in the average transmittance number. Since the atmosphere is opaque below 100 cm-1 and above 1500 cm-1, one can calculate the total emission over all frequencies using Stefan-Boltzmann to get total Su. The surface emissivity in MODTRAN is 0.98.
Example:
1976 standard atmosphere, 100 km looking down, average transmittance 0.2527, TOA emission 196.627, S_U 360.427, S_T is then 91.09 and Aa is 269.38. At 0 km looking up, Ed is 258.673. So Aa/Ed is 1.04 and tau is ln(0.2527) = 1.376. But that’s not correct because it ignores the emission outside the 100-1500 cm range. S_U total at 288.2 K with an emissivity of 0.98 is 383.34 but St shouldn’t change (opaque for those wavelengths) so transmittance is 91.09/383.34 = 0.2376 and tau is then 1.437. Aa is then 292.25. Ed is a little trickier. I plotted the spectrum and eyeballed a Planck curve fit for the maximum emission parts of the curve and extended the range to 5-3000 cm-1. The best fit temperature was 286.4. Then I integrated 5-100 and 1500-3000 for the Planck curve giving a total of 22.3 W/m2 or 280.97 for Ed. The ratio is still 1.04.
SpectralCalc is line-by-line and they have atmospheric model capability (no clouds). I could run the numbers from SpectralCalc to compare, but it will take a while. I have to break the spectral range into segments and each segment takes several minutes to calculate. SpectralCalc is also limited to 6 ghg’s at a time and I don’t think they do stratospheric ozone, but I’m not sure.
reply to DeWitt Payne’s post of 2010 Jul 19 5:02 am.
Thank you for your very helpful and enlightening response.
It is good to learn what seems to be the definition of the AVERAGE TRANSMITTANCE of MODTRAN, from a serious expert in its use.
This definition assumes a simply spectrally defined atmospheric window. This definition is not the definition of Ta used by Miskolczi.
In contrast, the Miskolczi HARTCODE definition calculates the actual transmission assuming that the land-sea surface emits with a Planck black body spectrum. There are differences. For example, some “window wavelength” rays will travel obliquely and may need to traverse hundreds of km on their way direct to space, while others will pass vertically and have a shorter geometrical path through the denser parts of the atmosphere. Also there is slight “windowing” in the far infrared. And the emissivity of water vapour depends on the pressure, and so the apparent “window wavelengths” will also be pressure-, and thus altitude-dependent. The Miskolczi HARTCODE takes these factors into account explicitly, instead of guessing them as it seems MODTRAN does.
There are two slightly different questions. One is how closely do local single occasion measurements fit the law Aa = Ed ? The other is how closely does the global average fit the law? It seems that the local single occasion measurements mostly fit rather well. The global average is a much closer fit. It is not quite so easy to get a dataset for a proper weighting to get the global average.
To check the Aa = Ed law, it is important to study actual measured atmospheres. For example, the US 1976 standard atmosphere is postulatedly simplified and artificially constructed, and apparently does not actually obey the laws that really govern the dynamics of the atmosphere, so it cannot safely be used to check the law.
The emissivity of the land and sea, and the reflectivity of the land-sea/atmosphere interface, are not so easy to check precisely.
SoD states.. “What’s amazing about back-radiation is how many different ways people arrive at the conclusion it doesn’t exist or doesn’t have any effect on the temperature at the earth’s surface”…
Well, it exists and can be measured. So can the temperature of a bucket of warm water. And it contains a lot of heat (energy), but don’t try to heat up your hot coffee by adding warm water. It wont work, unless you believe in the amazing properties of back-radiation espoused on these pages.
JT states “In order for downward long wave radiation to heat the surface its photons must be absorbed at the surface by molecules which are in an electronic configuration which allows them to absorb photons of that frequency. But if at the surface the temperature is higher than in the atmospheric source then might the molecules which might have absorbed such a photon be in fact unavailable because they have already moved to a higher energy configuration due to thermal collisions in the material which contains them?”.. This is my understanding although I think vibrational state is the better model than electronic configuration when related to infra-red radiation.
By simple analogy imagine two adjacent tables of different heights with golf balls sitting on each. The golf balls from the higher table may roll off and bounce from the floor and end up on the lower table. But none of the golf balls on the lower table will have enough energy to bounce off the floor onto the higher table although the number and energy of these balls can be observed. (The heights of the table are indicative of the energy states of materials at different temperatures). .
I would suggest to those who consider the second law of thermodynamics “imaginary” to consider the energy available from back-radiation. At approximately 300 W/m2 a 10m x 10m “back-radiation” collector even at 33% efficiency would be good for a continuous 10KW power supply!…..
“And it contains a lot of heat (energy), but don’t try to heat up your hot coffee by adding warm water. ”
No, but you CAN heat up your coffee by putting it in the oven (where the air is heated). Or in a microwave (where the water is a good absorber of the photons).
“At approximately 300 W/m2 a 10m x 10m “back-radiation” collector even at 33% efficiency would be good for a continuous 10KW power supply!…..”
Indeed it is!
http://en.wikipedia.org/wiki/Solar_thermal_energy
For those who consider back radiation impossible, you’d have to ignore the power collectors that collect it.
Two words: Space Blanket.
If back radiation didn’t help keep you warm, then why do they make space blankets highly reflective? It’s basic heat transfer. The rate of heat transfer by radiation between two surfaces at temperature Ta and Tb is proportional to (Ta^4-Tb^4). Net heat transfer is always from higher temperature to lower temperature, but the rate of heat transfer is slower if the lower temperature is closer to the higher temperature.
See Radiation Basics and the Imaginary Second Law of Thermodynamics
Morris is talking about heating, not insulation, but as a demonstration of the power of “back radiation” it works very well.
Except the space blanket thing is that the heat from your body can’t conduct into space. Neither can convection take it away. So radiation is the only thing left. And a shiny surface TOWARD your body keeps that in well.
Where your point becomes analogous is the space blankets you buy for harsh conditions on earth.
Here on earth, unless you live in a vacuum flask, conduction and convection will always play *some* role. However, if you’re completely wrapped in a blanket, then the relevant temperature for energy losses is the temperature of the blanket.
The trick there is to make the blanket insulating. Reflective helps, but if it’s only that, then you’re still going to lose a lot of heat.
So the space blankets for earth have a good insulating layer between the shiny. The shiny cuts down on the radiation from your body on one side, the insulating layer in the middle increases the temperature drop that can be maintained (the thermal conductivity tells you how steep a drop will be maintainable across a temperature difference), then the other side will reflect the radiation from the (still warmer than the other side, but colder than your body) outside of the insulating layer, reducing the losses from THAT too.
In this case, there’s TWO layers that give “back radiation”.
Morris Minor:
Instead of simple analogy – which might be an incorrect analogy – why not explain what happens to the energy which is incident on the surface of the earth?
It will be covered in the next post, but you can help me write the article by explaining your point of view.
–The radiation is not absorbed so is reflected? Or vanishes?
–The radiation is absorbed but can’t have any effect on the energy content of that surface so vanishes in contravention of the first law of thermodynamics?
For reference see
Intelligent Materials and the Imaginary Second Law of Thermodynamics and The First Law of Thermodynamics Meets the Imaginary Second Law.
SoD says… “why not explain what happens to the energy which is incident on the surface of the earth?”
Well, a photon from a cool body will not be absorbed by a hot body.. it will be reflected (scattered) until it interacts with an IR active molecule at a lower vibrational excitation level (ie. a low table). This could be a GH molecule at a similar or higher altitude then that of the original molecule emitting the photon. Or it may continue into space. Of course, it is impossible to predict what will happen to a single photon, but statistically this would be the end outcome.
Please may I add here an answer to your question on my post in ‘First Law meets the imaginary 2nd Law’.
You state ‘When a body radiates at 10′C it emits radiation across a wide spectrum of wavelengths.
…
When a surface receives incident radiation of a particular wavelength it absorbs it according to its absorptivity at that wavelength.’
OK.. classical model meets quantum model.
‘When a body radiates .. it emits radiation across a wide spectrum of wavelengths.’
Well, yes and no. Consider a gas molecule say N2. At 10C, there will be no radiation emitted as N2 is not capable of a vibrational mode that has a changing dipole moment (it cannot produce an electro-magnetic wave by molecular vibration). A CO2 molecule has 3N-5 (4) possible vibration modes of which one is inactive (meaning no changing dipole) and two are identical. Therefore CO2 can emit radiation at two well defined wavelengths.
However, if we consider a complex molecular structure or a mixture of many different molecules, such as would be present on a typical black surface, then we would approximate a body that ‘emits radiation across a wide spectrum of wavelengths.’
I will change my golf balls on the table analogy to a stack of golf balls the height of which represents the temperature of our black body. We can envisage two adjacent stacks of golf balls of slightly different heights (eg. -10C and +10C). If a ball falls from the top of the ‘hot’ stack it will have enough energy to rebound and land on top of the cold stack. However the golf ball on top of the cold stack will not have the energy to increase the height of the hot stack. If our stack ‘emits radiation across a wide spectrum of wavelengths’ then, by analogy, a golf ball can fall from various positions in the stack. However this golf ball will be unlikely to have the energy necessary to increase the height of the other stack. Once removed from the stack the balls above will drop to fill this vacancy.
Is it possible that a ball of insufficient energy to reach the top of the adjacent stack may reach a height where it can dislodge and replace a ball from the other stack? Well maybe, but well known principles (Heisenberg comes to mind) suggest that you would not be able to observe this as you could not identify which ball came from where (ie the balls all look the same).
What if we considered our golf ball stack a convection model where each ball represents a molecule or small chunk in a large body where the height of the ball represents temperature, the temperature of the body being the average of all the balls. It is quite conceivable that a hot ball from the cold stack may fall off and replace a cold ball on the hot stack thus making the hot stack hotter. But does that happen? Does anyone think that a cold body can heat up a hot body by convection? If not, then why consider it possible by ‘back’ radiation?
Christopher Game,
Why do you think that? Do you have a reference other than Miskolczi? MODTRAN and line-by-line programs calculate atmospheric emission and transmittance exactly the same way as you describe the method for HARTCODE. The only difference is that MODTRAN uses a lower resolution band model. We know that MODTRAN does a pretty good job because MODTRAN calculated spectra look just like observed spectra. I started to compare SpectralCalc to MODTRAN, but it was going to take too long. I did calculate total emission from 100 to 530 cm-1 using similar conditions for both programs and the results were similar: SpectralCalc 26.1 and MODTRAN 26.7 W m-2 sr-1.
MODTRAN calculated spectra
Observed spectra 1
Observed spectra 2
I believe the pressure effect you refer to is known as the water vapor continuum. MODTRAN and many LBL programs include this, although not SpectralCalc yet for atmospheric paths. The water vapor continuum is most likely due to collision induced absorption by two water vapor molecules, but there is still some controversy about that and an empirical fit is used rather than an ab initio calculation.
If HARTCODE produces results that are greatly different from other LBL and band model programs using the same atmospheric profiles then Miskolczi is doing something wrong. But I don’t think that’s his problem and I would be very surprised if it were the case.
replying to DeWitt Payne’s post of 2010 Jul 19 3:11 pm.
I was referring to the definition of the term AVERAGE TRANSMITTANCE for MODTRAN that I accepted from you. I do not know how MODTRAN works. Your definition of AVERAGE TRANSMITTANCE that I accepted from you is not the definition of Ta that Miskolczi’s HARTCODE uses. I think my post gives an account of some, but not all, of the differences of the definitions.
Several published cooperative comparisons between HARTCODE and other LBL codes agree closely. There is no significant difference in relation to matters where the definitions are the same.
Perhaps you will read again and further consider my previous post which tries to clarify, at least partly, the difference in definitions between MODTRAN’s AVERAGE TRANSMITTANCE as you tell me of it, and HARTCODE’s Ta.
If this further consideration of what I wrote does not clarify things, perhaps you will tell me, and perhaps I will be able to write something more to try to clarify things.
Efforts to understand stratosphere cooling and downward radiation are shaking my confidence that I understand anything about radiative transfer. Could SOD or someone please let me know what part of this analysis, if any, is wrong?
The problem I began was to analyze the energy transfer between two altitudes with pressures NP and 1P and with temperatures T and T-t. I approached the problem from the perspective of molecules, not layers. Only the GHGs at each location absorb and emit. By microscopic reversibility, the absorption and emission spectra presumably are the same and SOD showed a partial spectrum of downward radiation in an earlier post. If the Stephan-Boltzmann law applied to individual molecules (see below), the rate of emission would vary with T^4 and (T-t)^4. To a first approximation, the difference would vary with t. Comparing the upward and downward flow of photons assuming t=0, there are N times as many GHG molecules to emit from NP, 1/N as many at 1P to absorb, and the same probability of making the journey without being absorbed. Therefore the energy flux will be proportional to t and (to a first approximation) independent of T, P and N.
The rate of energy flow after GHG doubling remains constant: emitters double, but the mean free path of each photon is halved. Wavelength doesn’t make much difference: If the emission rate of photons increases with wavelength, their mean free path decreases. However, when the mean free path is much longer than needed for a photon to escape into space (or reach the surface of the earth), cutting the mean free path in half by doubling GHGs won’t greatly reduce the probability that a photon leavies the atmosphere. This explains stratospheric cooling: Upward energy flux from the stratosphere to space increases with GHG doubling (doubled emitters), but a with smaller increase in the chance upwards photons won’t make it to space. (A photon with a 90% chance of escaping to space becomes an 81% chance of escaping through twice as much GHG not a 45% chance. For the mean free path, a 50% chance of escaping is reduced to 25%.)
An equivalent rational (which could be called “lower troposphere cooling”) may produce a greenhouse effect. Twice as many emitters send photons downward; but, for those photons emitted near the ground, the probability reaching the ground isn’t halved by doubled GHGs (and the resulting halved mean free path). Furthermore, the photons reaching the earth’s surface originate from closer to the ground, so their number probably will increase because the temperature is warmer. However, convection maintains a fixed lapse rate near the surface, so it isn’t clear why the extra energy the surface receives isn’t immediately recycled to the altitudes suffering from “lower tropospheric cooling”.
When I apply these methods to the concept of radiative forcing at the tropopause and the reduced probability of escape with increasing GHGs, everything gets FUBAR when I double the number of emitters. How is radiative forcing actually calculated?
Molecular and macroscopic physics appear to be in conflict. For an individual atom, the QM probability of emitting a photon depends on the probability of it being kicked into an excited vibrational state by random collisions with other molecules – which obviously depends on T, but not necessarily T^4. The total flux of energy from N individual molecules appears to vary directly with N. When we describe a layer of atmosphere using the Stephan-Boltzmann law, we say the emission varies with T^4 and don’t consider N at all. Statistical mechanics on a molecular level explains thermodynamics on a macroscopic level. An explanation should start with Planck’s Law, but most derivations don’t deal with isolated molecules.
Which leads to another dilemma: A blackbody radiates oT^4 in one direction (outward), while a layer of atmosphere radiates 0.5*oT^4 in two directions (up and down). The individual molecules on the surface of a black body obviously can’t distinguish in from out. Does half of the radiation from a black body really go inward and then get “reflected”. Real molecules do neither, they radiate in three dimensions.
“Which leads to another dilemma: A blackbody radiates oT^4 in one direction (outward), while a layer of atmosphere radiates 0.5*oT^4 in two directions (up and down).”
Nope, a blackbody radiates up and down together.
It’s just so optically thick that this only gets one way out left: up.
“The individual molecules on the surface of a black body obviously can’t distinguish in from out. ”
Indeed they don’t. They don’t have to. The photon travels its weary path, blocked.
Much the same thinking gets people into problems with thermos flasks: “A thermos keeps hot things hot and cold things cold. How does it know?”.
“Does half of the radiation from a black body really go inward and then get “reflected”.”
The scare quotes are accurate: they get “reflected” in so far as they never manage to get deep enough into the “blackbody” body to be trapped inside. Therefore all photons going “in” get stalled and any going “out” go free.
Those who tried to go “in” will try 50% to go “in again” and 50% “out this time”. The “out” will get out, the “in again” will get stalled.
Those who tried to go “in again”, 50% will try “in yet again” and 50% “out this time”. “out” will get out. “in yet again” will get stalled.
Repeat ad infinitum.
“Real molecules do neither, they radiate in three dimensions.”
Real molecules don’t radiate. They lose internal energy which is released as a photon which is radiation.
Come on, if you’re going to get pedantic, at least be pedantic about something that raises a conundrum with you.
“the QM probability of emitting a photon depends on the probability of it being kicked into an excited vibrational state by random collisions with other molecules”
Where did yo learn that, Frank?
It’s wrong, wherever you learned it.
The probability per unit time to emit a photon (or indeed anything bound in the system) depends on variously
1) the energy of the item to be radiated
2) the size of the potential wall to be bypassed
3) the occupancy rate of the state to be moved to
4) the number of such states that can be moved to
Collisions interrupt this transferal and can (because the collision may become inelastic) result in no release of the photon at all. Or possibly a release of a more powerful (or less powerful) one. Or moving that energy into the internal energy of the colliding (but unexcited) molecule.
“Regular readers of this blog will be clear about the difference between solar and “terrestrial” radiation. Solar radiation has its peak value around 0.5μm, while radiation from the surface of the earth or from the atmosphere has its peak value around 10μm and there is very little crossover.”
This is incorrect, there is, in fact, 100% crossover. This is hidden in your diagram by scaling the Sun’s curve by 10^6. As the temperature of the emitting body rises, each higher energy curve always encompasses the curve of lower energy – the peak moves to higher frequencies and higher energies.
What this means is that downward LWR during the day is swamped by the Sun, and only night time downward radiation is an accurate representation of back radiation
I ought to note that because the obvious potential field is gravity, that this wouldn’t work “in real life” because a gravitational potential field is always negative.
Therefore if you put that negative only field in place, you’ve reduced the energy available in the system, which means that there’s less available to keep the molecules moving.
However, if you had a positive field, you’d be adding more energy.
So you really need a “SpellJammer” gravity where the gravity plane as a positive (which keeps you walking on the floor by making you fall TOWARD the floor) and a negative (which makes you fall away from the floor when you fall through).
“This is incorrect, there is, in fact, 100% crossover.”
However, there’s less energy in the flux out here so far from the sun.
In fact, at the distance the TOA is from the earth compared to that from the Sun, the reducing ratio is the square of the distance between the Sun and the earth.
Since the proportion of photons at each wavelength do not change with the feeble gravitational field of our yellow dwarf, the reduction of the number of photons at ~4um is likewise reduced from the Sun.
But the earth is right here.
No (appreciable) loss in numbers.
What’s the ratio of the number of 10um photons (say) from the sun at earth distance compared to the earth?
How about 1.5 um photons?
Ross:
Have a read of The Sun and Max Planck Agree.
The solar radiation which is scaled down by a factor of 10^6 is the solar radiation measured when you are parked just off the surface of the sun.
The solar radiation received at the surface of the earth is a tiny fraction of this because we are 150M km away.
The easy way to convince yourself that the solar radiation and the terrestrial radiation are in approximate balance is to consider what would happen if you were right.
In that case, we receive over 1,000,000 times more energy than we radiate out so the earth will heat up very rapidly.
In fact, the numbers balance at the top of atmosphere where the outgoing radiation from the climate system is about 239W/m^2 and the incoming solar radiation which is absorbed, when averaged over the surface area of the earth is also about 239W/m^2. (Also there are some instrumentation accuracy issues with some uncertainty in the order of 5W/m^2).
See Earth’s Energy Budget for an introduction to these basics.
Frank:
It’s a big subject and I’m not really sure I understand what your question is.
For a calculation of radiative transfer through the atmosphere:
-absorption is calculated by the number of absorbers of each type.
-their measured absorption at each wavelength (if the most accurate method of line by line calculation is used)
— the absorption of each type of molecule for a given line varies with pressure and temperature, proportional to [P/T]^0.5.
-this absorbed energy is thermalized via collisions so that the “local” environment is at the same temperature (LTE) and temperature change can therefore be calculated from the heat capacity of the atmosphere at that level and the absorbed energy
-emission is calculated by the planck function via the temperature of that layer and the concentration of each emitter, combined with the lines at which emission can take place
So there is an integral in two “dimensions” – vertically through the atmosphere, and across all wavelengths.
The atmosphere needs to be divided into layers because the temperature and pressure varies which changes the planck function, the quantity of absorbers, and the line width of each absorber.
Once the atmosphere is divided into 20 or so layers the accuracy of radiative transfer doesn’t improve much by dividing it into more.
Explaining the complete process in detail including the quantum level takes many chapters of a textbook.
Trying to do a back of envelope calculation of radiative transfer will require many large envelopes, a calculator and many weeks of work.
So when you say things like:
-I’m not really sure what you are referring to, as most derivations start with the planck function for emission, formulae for absorption at a monochromatic wavelength, then look at invidividual absorption lines of each type of molecule along with how these lines vary with temperature and pressure.
Christopher Game,
I went back to Miskolczi’s original paper and according to that, Ta is the vertical flux fraction of the emitted vertical surface flux transmitted from the surface to space (see Fig. 1). Any LBL program can calculate the flux at other angles, but Miskolczi’s theory deals only with the vertical fluxes. So MODTRAN average transmittance and Miskolzci’s Ta are, in fact, identical.
Re: Mark 7/18
“… it would be odd to have the TOA much hotter than the earth surface!”
The top of the atmosphere is hotter than the surface.
http://apollo.lsc.vsc.edu/classes/met130/notes/chapter1/vert_temp_all.html
But then, what is our definition of the top of the atmosphere?
And, is that definition based more on radiative properties or density?
Sorry, but I’m also wondering why DLR is so interesting. Photons rattle around (are absorbed and emitted), molecules collide, equilibrium, more or less, is established, and in the end, energy is neither created nor destroyed. This has always happened. What matters is the altitude where the sum of the outbound energy not redirected downward is balanced with the sum of the inbound energy not redirected outward. I suppose you could look at it as an integration of probabilities. Who cares what route it took to get there or how many transformations it went through?
Below that altitude, convective forces keep things stirred up, and you can pretty much follow the lapse rate and latent heat effects to the surface to find the temperature there. Please tell me where I went majorly wrong in this.
Chris G
Because so many people think it doesn’t exist, isn’t caused by “greenhouse” gases, or doesn’t change the temperature of the surface.
Many things written about on this blog are not of interest to people in atmospheric physics because they are so basic and not in dispute..
Chris G, as I said, you can get all sorts of weird effects when you have a situation where temperature is badly defined.
E.g. where there’s so little gas that you can’t average them out.
See also: Solar Corona (1 million degrees C, cf 6000C for the photosphere)!
And I quote:
“so you have to work somewhere where temperature is badly defined, like an extremely rarefied gas.
This isn’t true in our troposphere.”
The thermosphere is not the troposphere and it’s 1/10,000th an atmosphere. Pretty thin.
After all, what’s the temperature of the vaccum of deep space?
replying to DeWitt Payne’s post of 2010 Jul 20 at 4:57 pm.
Thank you for your reply.
It is good that you have looked again at Miskolczi’s original paper. I see that you have here quoted from Miskolczi (2007) Időjárás 111(1): 1-40, section 3 at page 5. It seems that the precise meaning of what Miskolczi wrote has not become clear at this point. In this article, Appendix A: Flux optical depth, at page 36, is explicit. Since I don’t know how to write the mathematical expressions for this medium, I will not quote here the full appendix, but it may be enough to clarify things if I quote the following from page 37: “The first step is to compute the directional mean transmittances … … … is the local zenith angle … …The Planck-weighted hemispheric mean transmittances were computed ….” Miskolczi’s appendix here uses the usual Greek letter μ to denote the cosine of the zenith angle. HARTCODE uses 9 zenith angles to get the full hemispheric transmission. In HARTCODE, the vertical transmission means the transmission that would be seen by an altitude-specified vertically-downwards-looking measuring instrument that collects rays from the whole downward hemisphere with a vertical axis of symmetry. The word vertical refers to the symmetry axis of the hemisphere of integration, and does not mean that only vertical rays are studied. I think this appendix clarifies things. If you are interested, the detailed description of HARTCODE can be downloaded from http://miskolczi.webs.com/hartcode_v01.pdf
For detectors at high specified altitudes, the curvature of the surface of the earth comes into it, and HARTCODE takes that into account.
As I understand what you tell us about the definition of the AVERAGE TRANSMITTANCE of MODTRAN, it is calculated by considering only vertical rays?
Your post writes of “vertical fluxes”. As I read things, there are two conventions for defining the word “flux” in the present context. Let us consider the monochromatic flux. One convention is that used, for example, by Chandrasekhar 1950, by Mihalas 1978, and by Goody and Yung 1989; this is a vector found from full spherical integration of the specific radiative intensity; the specific intensity, ranging over all angles of a sphere, determines the direction and sense of the vector. The other convention is that used, for example, by Paltridge and Platt 1976 and by Liou 2002; it requires the user to specify the direction and sense of an axis of symmetry of the desired hemisphere of integration of the specific radiative intensity. The latter convention is of particular interest for work with a two-stream model of a uniform flat stratified medium, with the axis of symmetry normal to the strata, as for example in the present work.
I’m still not convinced there’s a difference. Modtran actually calculates W cm-2 sr-1. To get the radiance (their terminology) over the whole hemisphere one multiplies by pi steradians. But when calculating ratios, the answer is the same. I’ll need to see actual numbers before I’ll believe that Miskolczi’s calculations are significantly different from Modtran.
The pyrgeometers referenced above (example, Eppley Labs PIR ) used to measure DLR have a hemispherical field of view, so they’re seeing the whole sky. Looking at Kwajalein, for example, that’s tropical atmosphere. The observed DLR is 416 W/m2. The average temperature there is 83 F or 301.5 K. Clear sky DLR calculated using Modtran tropical atmosphere from 100-1500 cm-1 at 301.5 K (offset 1.8 degrees) and constant RH gives 362.4 W/m2 adding in the tails from 5-100 and 1500-300 cm-1 is another 34 W/m2 for a total of 396.4 W/m2. That’s pretty good agreement considering that the observed data includes cloudy and partly cloudy days which increases DLR significantly.
SoD, Mark,
Thanks, well and good, no real arguments here.
But, I’ve run into trouble before in discussions about density, saturation, and rates of absorption before, and I think it had a lot to do with differing definitions of the TOA. My questions there were, and still are, in earnest.
I can think of a several:
I think 50 miles used be the definition for qualifying as an astronaut. (Pretty arbitrary.)
Where the total pressure is less than X. (Doesn’t change much if all you are doing is swapping O2 for CO2.)
Where the partial pressure of LW absorbing gasses falls below X. (Changes significantly as PPMs of GHGs change.)
Where the total absorption between there and space falls below X. (Doesn’t really distinguish between SW and LW effects. )
Where the chance that a LW photon traveling outward will be absorbed and remitted in a direction where it will intercept the earth falls below X.
Stratopause
Mesopause
Or is there no clear definition, just a general acknowledgment that there exists some altitude above which the thermodynamic effects of the atmosphere are negligible?
Though, if there exist measurable differences in temperature across a region, and those differences are caused by radiation being absorbed and/or emitted, I’m not sure when/where/at what point the effects that occur within that region can be safely ignored.
“I think 50 miles used be the definition for qualifying as an astronaut. (Pretty arbitrary.)”
That is the top of the atmosphere from a human point of view.
Not TOA in the case of radiative output to space.
That is where you get your problems from.
See also “Optical Depth” on any course book or Google for it. This explains how the TOA changes depending on the absorptivity of the gas being considered.
Note that as height increases in our troposphere, temperature decreases, therefore a TOA that is colder at one wavelength than the TOA at a different wavelength radiates less power at that wavelength.
50 miles is the top of the atmosphere as far as humans are concerned.
TOA radiatively is a different thing altogether.
For the purpose of calculating LW emission and absorption, the TOA is where additional emission/absorption becomes insignificant. That’s somewhere in the range of 70-100 km. MODTRAN stops at 100 km. At 70 km looking up, the total LW emission is 0.049 W/m2. That’s pretty small compared to 258.67 W/m2 at 0 km looking up (1976 standard atmosphere, other settings default).
I do have to ask, i have read that the average altitude of the long wave that leaves the troposphere is 6km??? Some is absorbed higher, and some back radiation, down from the stratosphere… but not really significant?
The altitude varies a lot with frequency. For the window region, ~800-1200 cm-1, except for the ozone band near 1000 cm-1, the altitude is close to zero. For the center of the CO2 band, the altitude is a lot higher. In the tropics, the CO2 emission altitude is greater than 12 km. Water vapor is mostly gone by 6 km, so that tends to fix the average altitude.
The absorption tables for the CO2 band suggest that the majority of absorption of surface photons by CO2 is occurring very close to the ground. At wavenumber 650, for example, over 50% are absorbed in the first 25m, and at the peak of the band (wavenumber 670) Nicol has calculated over 90% extinction by 5m. (see http://www.ruralsoft.com.au/ClimateChange.doc ).
The back-radiation is similarly affected – most wavenumber 670 photons originating from higher than 5m above the ground won’t make it to the surface.
So the temperature of the back-radiating CO2 is approximately the same as the surface, and in the CO2 bands Back Radiation approximately equals Surface emission. Additional CO2 does not alter this – if anything the two become closer. There is therefore very little heating of the atmosphere due to absorption in the CO2 band.
In the water bands, absorption occurs at a higher level (I think – I have no tables – perhaps DeWitt Payne can comment?) so the Back Radiation is coming from a cooler place, and there is a net difference between the two fluxes. This net difference (according to Kiehl & Trenberth, around 25W/m^2) warms the atmosphere.
As the surface temperature increases, the evaporation rate increases, and the net heat transfer into the atmosphere from surface radiation correspondingly decreases – the increased water vapour decreases the effective height of the back-radiation in the water bands.
Does anyone have a calculation of the height of the net heating by surface radiation in the atmosphere? Is the net heating by condensation of water vapour (a flux which is 3 times as large) also occur at roughly the same altitudes?
Morris Minor
Referencing your comment of July 21, 2010 at 2:35 am. The original article you mention wasn’t about a gas radiating. However, the same problem for your explanation arises there – which you should think about.
If we move to talking about a gas radiating – let’s take CO2 – one of the principle bands of emission is around 15um.
The same question applies which I don’t think you understand.
CO2 at -10’C radiates some energy in the 15um band.
CO2 at +10’C radiates some energy in the 15um band.
Do you believe a surface of 0’C will absorb the photon emitted from CO2 at +10’C?
Do you believe a surface of 0’C will absorb the photon emitted from CO2 at -10’C?
The energy of the individual photon in question is hv = hc/l, where h = planck constant, v = frequency, c= speed of light and l =wavelength.
E = 1.33×10^-20 J for a 15um photon.
Notice no dependence on the temperature of the original source body.
The only question that needs to be asked is what is the absorptivity of the surface at 15um.
And whatever the answer is, it is the same for a 15um photon emitted from a -10’C as for a +10’C body.
SoD says..
The only question that needs to be asked is what is the absorptivity of the surface at 15um. And whatever the answer is, it is the same for a 15um photon emitted from a -10′C as for a +10′C body.
Yes but No. I can see the problem!
The emissive power of Body 1 relative to Body 2 is :- E = s (T1 – T2)^4.
The Stefan-Boltzmann model is usually given relative to 0K which of course doesn’t really exist. But the model is a relative construct. The Emissivity of the hot body relative to the cold body is a positive value and that of the cold body relative to a hot body is a negative value, meaning that there is no net emission from the cold body. It absorbs radiation from the hotter body.
Take a simpler example where T1 = T2. Between the surfaces there is no emission or absorption. I anticipate that some will say that the cold body will emit sT^4 and absorb sT^4 from the other surface but what does that mean? Will a body absorb a photon and emit one of the same energy simultaneously? Well, maybe, but you could never measure this. A photon may be reflected but how would you know the difference? Both the Stefan-Boltzmann Law and the Second Law of Thermodynamics are statistical models. They do not predict the path of a single photon. It appears that a small misconception in the application of the Stefan-Boltzmann law has led some to disregard the Second Law of Thermodynamics, always a dangerous thing to do.
Chris G from July 21, 2010 at 1:00 am:
There’s nothing wrong with that.
In fact, the standard approach to calculation of radiative transfer through the atmosphere starts with the integral over all solid angles. However, with some clever maths, “it can be shown” that the atmosphere can be treated as a plane parallel problem (ie lots of horizontal layers stacked one on top of the other) if we use the new optical thickness, X* = 1.66 X (where X is the optical thickness as usually defined)
So the oversimplification of half up-half down isn’t used to calculate anything of importance. If that was what you were asking..
SoD,
Bingo, that’s what I was asking. Thanks.
I suffered a moment of hubris, but having slept on it, I realize that my mental model was just my own internalization of the integration formulas I’ve been seeing off and on for some time. It’s a similar situation to how, even in a vacuum, projectiles do not really follow a parabolic curve, and in groups above a certain level of understanding, this is common knowledge, but parabolic is close enough for laymen.
DeWitt’s definition of TOA makes the most sense to me. My troubles in this area generally happen when others consider the TOA to be a constant, and I do not.
replying to DeWitt Payne’s post of 2010 Jul 21 at 1:10 am.
Thank you for your reply. I am learning about MODTRAN from you.
Accepting your statement that MODTRAN calculates a quantity with unit W cm-2 sr-1: I would call that a spectrally integrated specific radiative intensity. As you say, to get a flux, with unit being power per unit area, one will integrate over a suitable solid angle. By my reckoning, the total solid angle of a hemisphere, the relevant range of integration here, is 2 pi steradians. But we are looking at a radiative field that is far from isotropic, and therefore an actual explicit integration is needed. HARTCODE does that actual explicit integration with nine zenith angle moieties, taking into account the curvature of the earth’s surface.
Christopher Game:
Is this the situation in question – total hemispherical power from a planar surface?
If so, then the W cm^-2 sr^-1 would be multiplied by pi to convert to W/cm^2.
replying to Colin Davidson’s post of 2010 Jul 21 at 1:49 am.
He asks “Does anyone have a calculation of the height of the net heating by surface radiation in the atmosphere?”
In noting for the CO2 band the very low and nearly equal altitudes of absorption of radiation from the land-sea surface and emission from the atmosphere that reaches the land-sea surface, you have given an important part of the explanation of why, in the terminology of Miskolczi 2010, Energy and Environment 21(4): 243-262, one has overall Aa = Ed. The other main part of the explanation is also given in your post, in the form of a question. The water vapour land-sea/atmosphere radiative exchange is also at low altitudes, mostly less than 300 m. There are further dynamic factors that make for the equality Aa = Ed: the lower atmosphere absorbs some solar radiation, and is turbulently mixed and so gains energy from material transport from the land-sea surface, by conduction and evaporation. These are enough to make the balance of the land-sea/atmosphere radiative exchange nearly zero overall on average and in many individual location occasions. Nearly all the radiative cooling of the land-sea surface is by way of radiation direct to space, through the atmospheric window, which was discovered by George Simpson in 1928.
The near zero balance of the land-sea/atmosphere radiative exchange means that the transfer of energy from land-sea surface to atmosphere is nearly all by way of conduction and evaporation. This conclusion does not agree well with two of the figures that one might read on the diagram of Trenberth, Fasullo, and Kiehl 2009. They say that their figure of 40 W/m2 is calculated as a residual. This means that it is subject to error. Nevertheless the difference (396 – 333) W/m2 = 63 W/m2 is rather close to the right value. It seems likely that the Miskolczi finding that Aa = Ed is reliable, because it is explicitly investigated in its own right, and calculated directly, both for the global average and for particular occasions of radiosonde measurement. A graph of the altitude profile of the global average spectrally integrated contribution densities to Aa and Ed is shown as figure 3 in that Miskolczi 2010 paper. Also some examples of particular occasional radiosonde data and calculations of spectral densities of Aa and Ed are shown in figure 5 of that paper. You can see the matches, which agree with your comments.
The heating of the atmosphere by condensation of water vapour occurs mostly at much higher altitudes that the radiative exchange. These water vapour condensation heating altitudes are shown in William Gilbert’s 2010 paper ‘The thermodynamic relationship between surface temperature and water vapor concentration in the troposphere’, Energy and Environment 21(4): 263-275.
replying to scienceofdoom’s post of 2010 Jul 21 at 4:03 am.
For the present purpose an actual explicit integration is needed because the radiation is far from isotropic; there is no reliability in ‘converting’ by multiplying by a constant. The solid angle over a hemisphere is 2 pi, over a full sphere it is 4 pi.
Regarding Miskolczi vs MODTRAN:
Well it is not terribly clear to me, particularly as I do not know how the MODRAN transmission is calculated, how to compare the two.
But the use of multipliction of their INTEGRATED RADIANCE by pi (actually they use 3.14) to get their value of Iout, suggests an assumption that the radiation is Lambertian (varies with the cosine of the zenith angle). Now I do not think that is the case with regard to the surface direct radiation due to the change in path length with zenith angle. Would not the direct radiation fall off more quickly with high zenith angles?
I have checked the Miskolczi paper and in the appendix it outlines a method for calculating the optical depth that appears to take account of variance with zenith angle. The method appears to be the one for the flat plane (not for a spherical earth) but I do not think that this matters much.
So how much difference would it make?
Here I find the concept of optical depth not very helpful as it could lead one to thinking that doubling the path length would double the depth in a straightforward way. For discrete frequencies it does but once they are integrated a doubling of path length may only make a marginal difference to the total.
Enough of that! What I would like to know is the consequence of Ed = Aa?
That the atmosphere radiates to the surface precisely the amount that the surface radiates to the atmosphere would seem queer as the bulk of the atmosphere is colder than the surface. It would imply that radiative heating is a never to happen event. It would also seem to imply that cooling the atmosphere (and hence making it emit less) would lower its absorptivity. Which would imply that molecules can read a thermometer and judge how often to dodge. I think that absorptivity can only decrease with increasing temperatures and then normally only marginally.
Perhaps the atmosphere adopts a strategy whereby the temperature profile is just right so the Ed = Aa.
Doesn’t Ed = Aa imply that the whole atmospheric column and the surface are at the same temperature?
I rather think it does.
Now how is this Ed sourced?
Does each layer radiate dirctly back precisely the amount of radiation it absorbed directly from the surface?
Well I suppose it must as I don’t think that a cooler layer could make a net contribution to the warmer surface. If none of them can make a net contribution then each must send back precisely that which it received.
And what is so special about the surface anyway?
What about the layers themselves, do they maintain the same relationships with each other?
Somehow this does seem to indicate that radiative heating is a never happening event.
But I am forgetting something, there is a specifc value of optical depth involved 1.841 (84% absorption) that makes all this possible. Or perhaps this all happens in the long run and is an average sort of thing, but I do not think that on average a cooler body can transmit back to a warmer body to the same extent that a warmer can transmit to a cooler.
If I can manage not being whimsical for a bit I can say that:
It is true that to a first approximation adjacent layers of the atmosphere are in radiative equilibrium, as is the lowest with the surface, but you can neither use this, nor LTE, to induce that the surface is in, or even close to, radiative equilibrium with non-adjacent layers, nor between any non-adjecent layers in the atmosphere. Radiative heating is a happening event.
As any sustained period when Ed > Aa would imply radiative heating of a warmer body by a cooler body and pose a bit of a problem for the law, we must have Ed = Aa for all sustained periods so just one example of Ed < Aa would be enough to show that the equility cannot hold. Perhaps the answer lies in the clouds. They are almost perfect LW blackbodies (they will absorb all the upwelling radiation) and in order for them to be in radiative equilibrium with the surface they must be at the same temperature as the surface. Whereas this is true for fog, I just do not think it is true in general. I think clouds tend to be cooler than surface and it only needs one cool cloud to pass overhead to show that the equality can not hold in general.
Alex
SOD: You say: “emission is calculated by the planck function via the temperature of that layer and the concentration of each emitter, combined with the lines at which emission can take place”. However, the planck function tells us about energy output per unit area, not about the energy output per emitting molecule. The integrated form of Planck’s function over wavelength is W = eoT^4. Does emissivity depend on concentration? How does dimensional analysis work here? At a molecular level, a) total emission should depend on the total number of emitters and b) emission per unit volume should depend on the number of molecules per unit volume (pressure times the mixing ratio). Energy flux is measured per unit area, not per unit volume. Somehow energy per unit volume is turning into energy per unit area.
A second related area of confusion arises from the relationship between absorption and emission. At a molecular level, these processes are the reverse of each other – the probability of a GHG molecule absorbing a photon of the right energy and entering an excited vibrational state is directly linked to the probability of a GHG molecule in that excited vibrational state emitting photon. The probability of the GHG molecule being in an excited state depends on exp(-hv/kT), not T^4. The close relationship between emission and absorption suggests to me that the length of the mean free path of a photon of a particular wavelength is inversely related to the number of photons carrying energy at that wavelength, a relationship that would have important consequences for energy flux through a thick atmosphere.
further reply to scienceofdoom’s post of 2010 Jul 21 at 4:03 am.
The further factor of one half that is being used for the ‘conversion’ procedure for the MODTRAN quantities comes about as follows.
For the case of a hemispherically integrating detector at a high altitude, looking down, one is in effect using a model in which the land-sea surface is considered to be an isotropic emitter, with no intervening atmosphere. The vertical value of the spectrally integrated specific radiative intensity is the quantity supplied by MODTRAN, if I rightly understand what DeWitt Payne has written. It is assumed that this quantity arises from a perfectly diffusely emitting (Lambertian) horizontal land-sea surface of infinite extent. The Lambert cosine law tells how this will be seen by the detector as it takes in the full range of upcoming rays, vertical and non-vertical. There is no source of downward IR radiation that comes into this model of the high altitude case. The factor 1/2 comes from the contribution to the integrand of the Lambert cosine law. I think this is in agreement with the foregoing comment of Alexander Harvey, as well as in agreement with the conclusions of DeWitt Payne and of scienceofdoom.
This model is perhaps a reasonable rough estimate, but it does not take into account the real physical situation in the atmosphere, for which the non-isotropy requires an explicit integration over zenith angles. As I understand DeWitt Payne’s post, the information for the latter integration is not supplied by MODTRAN. That is one reason why use of MODTRAN may lead to answers different from those of HARTCODE.
HARTCODE does take into account that the surface of the earth is not flat and of infinite extent.
reply to Alexander Harvey’s post of 2010 Jul 21 at 7:10 am.
Alexander Harvey is talking about how it comes about that Aa = Ed.
The equality is not exact at every place and occasion, but it is very close on average and close on most occasions and places, with relatively small exceptions.
The near equality comes about because most of the radiative exchange between the atmosphere and the land-sea surface takes place in the turbulently mixed boundary layer of the troposphere.
The energy supplied to each place in the turbulently mixed boundary layer of the troposphere by radiation from the land-sea surface is equal to the energy radiated from that place direct to the land-sea surface. This is called radiative exchange equilibrium.
The concept of radiative exchange equilibrium should be distinguished from, and is definitely different from, the concept of pointwise radiative equilibrium in the atmosphere, which, at each point of the atmosphere, takes into account also the radiation from the atmsophere direct to space as well as radiative exchange between the place and other parts of the atmosphere. Pointwise radiative equilibrium does not occur in the turbulently mixed boundary layer of the troposphere.
It is helpful to keep in mind this important distinction between the concept of radiative exchange equilibrium, which refers to a relation between two spatially distinct bodies, and the concept of pointwise radiative equilibrium, which refers to just one point or infinitesimal volume. Radiative exchange equilibrium describes the radiative exchange between the land-sea surface and the lower troposphere. Pointwise radiative equilibrium does not occur in the lower troposphere.
How can there be such radiative exchange equilibrium when there is a temperature gradient in the lower troposphere? It happens because radiation from the land-sea surface is only one of the sources of energy for each place in the lower troposphere. The other sources are lower tropospheric absorbed solar radiation and material transport (conduction and convection) of energy from the land-sea surface into the lower troposphere. These other sources provide energy that allows radiative exchange equilibrium even though the land-sea surface temperature is in general equal to the lowest atmospheric temperature only at the very lowest altitude, and there is a temperature lapse rate above that. These other sources do not actually supply energy for radiation from atmosphere to land-sea surface; they supply energy for radiation from the atmosphere direct to space, but this latter balance provides conditions that allow the radiative exchange equilibrium to hold.
Some detailed graphs of the radiative exchange equilibrium between the land-sea surface and the lower troposphere are shown in Miskolczi (2010), Energy and Environment, 21(4): 243-262.
The law Aa = Ed is not yet recognized in textbooks. They do not provide a proof that it must occur, but it is nevertheless an empirically observed fact, first made clear in Miskolczi (2007), Időjárás 111(1): 1-40, shown there in its Figure 2.
There are, however, some fragmentary empirical anticipations of this law in textbooks, especially in what is called the “cooling to space approximation”, mentioned for example in Paltridge and Platt (1976) ‘Radiative Processes in Meteorology and Climatology’, Elsevier, Amsterdam, at page 172, and in Goody and Yung (1989) ‘Atmospheric Radiation. Theoretical Basis’, Oxford University Press, Oxford, at page 250, and in Wallace and Hobbs (2006) ‘Atmospheric Science. An Introductory Survey’, 2nd edition, Elsevier, Amsterdam, at page 138.
It is not the case that the Planck-weighted or true greenhouse-gas optical thickness of the atmosphere has a universal constant value at every place on every occasion. Far from it. But the constant value occurs for the global average at each time. At the poles the Planck-weighted or true greenhouse-gas optical thickness is often about 1.0 . At the equator it can easily be as high as about 3.5 . The global average is close to 1.87 at all times. Some details of this are also shown in Miskolczi (2010).
I would like to thank Christopher Game (0436, 21July) for his comprehensive reply to my questions.
I was most interested in his statement that the Back Radiation from water vapour is coming from around 300m and below, and wondered if Science_Of_Doom or DeWitt Payne have any comments.
Is the Kiehl & Trenberth figure for NET surface radiation absorbed by the atmosphere (ie Surface Radiation minus Back Radiation minus Surface Radiation which makes it out through the window) derived from the wings of the absorption bands, where absorption (and radiation) is weak and (on average) from high in the atmosphere? Is it believable that this would amount to around 8% of the absorption?
“Does emissivity depend on concentration? ”
No.
“How does dimensional analysis work here? ”
Same as anywhere else.
“Somehow energy per unit volume is turning into energy per unit area. ”
Temperature is a macro level measure and doesn’t apply at the individual molecule level. At that level, there’s momentum and kinetic energy.
“suggests to me that the length of the mean free path of a photon of a particular wavelength is inversely related to the number of photons carrying energy at that wavelength,”
But if emission is interrupted by collision then it doesn’t happen.
You need to work out the time it takes for an excited state to decay back into unexcited state. And how long it takes between collisions. That time depends on all molecules in a volume, not those involved in emission alone.
In most of the troposphere, CO2 doesn’t get a chance to re-emitt except in a small fraction of opportunities. Most of the energy is in the gas, not in excited states of CO2.
“Nearly all the radiative cooling of the land-sea surface is by way of radiation direct to space, through the atmospheric window, which was discovered by George Simpson in 1928.”
Simpson discovered the atmospheric window. He didn’t discover most of the radiative cooling is by radiating through that window.
And ~30W from the earth goes out through that window. About 10% of the radiative loss from the earth into the atmosphere. A long way from “most”.
This failure to understand and apply may be the reason why Miskolczi’s 2010 paper is incorrect and could only find outlet through the self-admittedly partisan E&E journal, where peer review has many times before been eschewed in favour of notoriety and declaiming the climate science currently known.
“there is no reliability in ‘converting’ by multiplying by a constant. ”
When calculating the standard deviation from the mean of a value whose linear change is the square of the read value (e.g. kinetic energy, referenced by velocity which is measurable) the standard deviation of the read value is multiplied by 1.1 to attain a very good estimate of the deviation of the square from the mean.
This has been a staple of engineering and operational statistical practice for nearly as long as the “Normal Distribution” has existed.
The random walk distance is calculated not by deconvolving into the independent axes (x and y) but by multiplying the mean distance traveled per two steps by the square root of 2, because geometry attains that in two independent movements, the average movement in the second stop is at right angles to the original vector and the actual distance moved in those two steps will “on average” be the length of the hypotenuse.
Such constructs are used to determine the contribution of multi-dimensional (where the dimensions are independent) problems of the same nature to be third root for three dimensions, fourth root for four, and so on.
Such multiplications for wave theory of light emission uses the spherical plane where instead of 1/sqrt(2) being used, the geometrical construct gives a value based (IIRC) on the average of the diameter of a planar section on that sphere parallel to the originating plane surface, and therefore (approximately) 1.33 is the correct value.
Please also explain how you arrive at the statement that emission is not isotropic.
Chris G from July 21, 2010 at 7:43 pm:
This is known as the “diffusivity approximation”.
From “Radiation and Climate”, Vardavas and Taylor (2007)
replying to Mark’s post of 2010 Jul 21 at 3:41 pm
Thank you for your comment, Mark.
Yes, indeed, that’s right: Simpson discovered the atmospheric window in 1928. It was not till the work of Miskolczi in 2004 and 2007 that it was discovered that practically all the radiative cooling of the land-sea surface is by radiation direct to space.
You give a figure of ~30W from the earth going out through that window. You mean, I suppose, a global average of ~30 W/m2. I wonder where you get that figure from?
You write that your figure is about 10% of the radiative loss from the earth into the atmosphere. This implies that you have a figure for the radiative loss from the earth into the atmosphere. Again I wonder what is that figure and how did you get it?
Perhaps if you will kindly tell us where you got those figures from, or how you found them, we can make some progress together.
“Is the Kiehl & Trenberth figure for NET surface radiation absorbed by the atmosphere (ie Surface Radiation minus Back Radiation minus Surface Radiation which makes it out through the window) derived from the wings of the absorption bands”
No, because the NET surface radiation includes Surface Radiation which has nothing to do with the CO2 absorption wings.
E.g. Although your car travels slower with more people in it, your car doesn’t go slower just because several people left your house.
The ground emission is not from CO2.
If you want to see if your question needs answering without that element, please restate and someone who has more knowledge of radiative transfer will probably be able to answer that where I would not.
Christopher Game:
My sides are still hurting.. You are having us on, right?
What was discovered in 2004 that no one knew before???
replying to Mark’s post of 2010 Jul 21 at 3:52 pm.
Thank you for this comment, Mark.
My statement that “there is no reliability in ‘converting’ by multiplying by a constant” seems to be a problem. The way you write about this makes me think that my statement, that “there is no reliability in ‘converting’ by multiplying by a constant”, was not clearly expressed.
Let me try again. What I mean is that the ‘conversion’ ought to be done by an explicit integration, not just by multiplying by a constant. The method of multiplying by a constant is suitable only for isotropic radiation, but the present radiation is far from isotropic.
You ask also “Please explain how you arrive at the statement that emission is not isotropic.” You are referring, I think, to my statement in my post of 2010 Jul 21 at 4:44 am that “the radiation is far from isotropic”.
What I mean here is that an observer in a craft at an altitude of say 70 km will see infrared radiation that is not isotropic. This means that when he points his direction-sensitive narrow-aperture telescopic infrared radiation detector in different directions, the readings will vary with variation in direction, in a systematic way. When he points the telescope to outer space he will detect very very little radiation. When he points it directly vertically down at the earth he will see the intensity that I think DeWitt Payne is telling us will be reported by MODTRAN, which is the most intense for any earthwards direction. When he points it downwards but more or less sideways, he will see radiation that is weaker than the vertical value that MODTRAN reports, but stronger than that from outer space. This is what I mean by radiation that is not isotropic.
There is more here, to complete my meaning: Because the atmosphere is not homogeneous and the atmosphere is not “grey”, our observer at 70 km cannot reliably use a simple model that supposes that the radiation he sees from the earth comes from a uniformly and perfectly diffusely (Lambertian) radiating land-sea surface, with no intervening atmosphere. (Only by use of such a model could he arrive at the idea of ‘converting’ by multiplying by a constant.) The key physics here is that the optical properties of the earth’s atmosphere vary with altitude and with wavelength, that is to say, the atmosphere is not homogeneous and “grey”.
When the radiation is not isotropic, the hemispheric integral that is needed for the hemispheric flux value (that we are seeking) must be found by actual explicit integration over the hemisphere, which requires values of specific radiative intensity at all zenith angles. The desired hemispheric integral cannot reliably be found by assuming that all the relevant information is provided by just one value of specific radiative intensity, the one for the symmetry axis of the hemisphere.
reply to Mark’s post of 2010 Jul 21 at 9:20 pm.
Thank you for sharing your mirth, Mark.
But no, I am not having you on. I really mean it that “It was not till the work of Miskolczi in 2004 and 2007 that it was discovered that practically all the radiative cooling of the land-sea surface is by radiation direct to space.”
As I understand your post of 2010 Jul 21 at 3:41 am, you did not, at that time of writing accept that “It was not till the work of Miskolczi in 2004 and 2007 that it was discovered that practically all the radiative cooling of the land-sea surface is by radiation direct to space.” I think there are many who do not yet understand that practically all the radiative cooling of the land-sea surface is by radiation from that surface direct to space. This means that it is new to them. It was first made clear in the work of Miskolczi in 2004 and 2007, and now again in 2010. This is the meaning of the law Aa = Ed.
Colin Davidson:
I’m not certain I understand this question but I will have a go at what I think it means..
The values are worked out like this:
1. Surface radiation is calculated from σT^4 – using the known values of T around the world, and the σT^4 is averaged to get 396W/m^2
2. Outgoing longwave radiation (OLR) is measured by satellite, which provide very comprehensive coverage.
3. Backradiation is calculated via models, obviously tested against the relatively sparse measurements (as noted in this article). The calculations are done using a band model (narrowband Malkmus model) rather than line by line calculations.
4. The radiation through the atmospheric window is ad-hoc “after the fact” addition to the diagram – ie not an integral part of the calculation – which is because the atmospheric window does actually absorb some radiation.
So I’m not sure what you mean by “ derived from the wings of the absorption bands..“.
And what the atmosphere absorbs it also radiates – over the long term global average.
A note from the Trenberth and Kiehl 1997 paper, for the many readers, seeing as this paper has so much misunderstanding surrounding it:
At some stage fairly soon I will do an article about the paper.
Christopher Game:
I’m not Mark.
I haven’t read about this amazing discovery. So I have a few questions.
Does “practically all” the radiation from a 20’C land surface go direct to space?
What proportion would you estimate is not absorbed by the atmosphere?
Does “practically all” the radiation from a 20’C ocean surface go direct to space? (It appears you think yes)
What proportion would you estimate is not absorbed by the atmosphere?
Now I’ve read it again my question isn’t quite right, I thought you were saying the ocean surface..
I’m trying to regain my bearings..
– The surface radiates on average 396 W/m^2?
– We measure 239 W/m^2 by satellite from the TOA?
I have to check these absolute basics because I can’t get hold of how the statement could be so wrong..
Maybe I misunderstood what you were saying. But it seemed plain – so what proportion is “practically all”?
At the very least, you’ve made my day.
replying to Colin Davidson’s post of 2010 Jul 21 at 12:14 pm.
Thank you for your reply, Colin Davidson.
Just a little point. I did not mean that the radiation absorbed by atmospheric water vapour from the land-sea surface is absorbed “around 300m and below”. I meant that it is nearly all absorbed below 300m, and hardly any of it reaches 300m, except in the most exceptionally dry circumstances.
I would also like to comment further on the relations we are here considering. I will use the notation of Miskolczi. Let us by Su denote the land-sea surface upward radiation that enters the lowest atmosphere. Let us by St denote the radiation from the land-sea surface transmitted direct to space. Let us by Aa denote the radiation from the land-sea surface that is absorbed by the atmosphere. Let us by Ed denote the atmospheric radiation that is emitted downwards and reaches the land-sea surface. Let us by Eu denote the atmospheric radiation that is emitted upwards and reaches space directly without further interaction with the atmosphere. Let us by OLR denote the total outgoing longwave radiation.
Then we have by definition the identities;
Su = Aa + St
OLR = Eu + St.
It is widely known that Aa is of the same order of magnitude as Ed. But what is new in Miskolczi’s work is that the relation is not just one of being of the same order of magnitude: it is one of equality or near equality.
The Trenberth Fasullo Kiehl 2009 figures may be considered here. They are an update on the Kiehl Trenberth 1997 figures. I will state all figures here without their proper unit of W/m2, for the sake of visual clarity. As I read TFK2009’s figure 1, they say that Su = 396, Ed = 333, OLR = 239, Eu = 169 + 30 = 199, Aa = 356, St = 40.
It is notable that the TFK2009 figures lead to Su – Ed = 396 – 333 = 63. The TFK2009 figures state that St = 396 – 356 = 40. However, private emails to Kiehl have led to them saying that this figure St = 40 was open to adjustment, considering that it was found as a residual. After consultation with NASA, they proposed the adjusted value of St = 66. With this newly adjusted figure we have from their figures that Su – Ed = St – 3 for the global average. But by the above identity Su = Aa + St we have Su – Aa = St. It follows that, by these global average figures, Aa = Ed – 3. Thus these global average figures actually do not very much disagree with the Miskolczi law Aa = Ed. This approximate agreement refers only to the case of the global average, while the Miskolczi Aa = Ed law applies also rather well for particular places and occasions as assessed by particular radiosonde records, with only small errors; TFK2009 do not comment on this case of particular place and occasion radiosonde records.
With the adjusted St, we expect a TFK2009 adjusted Aa = 396 – 66 = 330. Also we expect that the adjusted TFK2009 Eu will be given by Eu = OLR – St = 239 – 66 = 173.
The adjusted TFK2009 figure St = 66 may perhaps be not perfectly exact, but it is very much nearer the Miskolczi value than is stated in the original TFK2009 figure 1. It is thus a mistake to rely on the old estimate that St = 40, which was found only as a residual, and not directly estimated in its own right.
replying to scinceofdoom’s posts of 2010 Jul 21 at 10:01am and at 10:06 am.
Thank you for your comments.
Both Mark and scienceofdoom please excuse me for mixing names or identities. My mistake.
When I wrote “practically all the radiative cooling of the land-sea surface is by radiation direct to space”, I meant by radiative cooling the net IR radiative balance of the land-sea surface. Most of the radiation from the land-sea surface, Su say about 396 W/m2, is absorbed by the very low altitude troposphere (Aa is say about 333 W/m2). The difference Su – Aa = St is about 63 W/m2. This 63 W/m2 is the only actually effective radiative cooling of the land sea-surface, because Aa = Ed means that the land-sea surface does not actually cool radiatively by radiative exchange with the atmosphere. The remaining cooling of the land-sea surface is into the atmosphere, but it is non-radiative, by conduction and evaporation and convection.
I think this may clarify things. If not, I will try again.
From Science_Of_Doom’s and Mark’s responses, it is clear that I have confused what I was trying to clarify. My apologies – I will try to make my question a little clearer.
1. It is plain from the K&T numbers that the NET transfer of energy from the surface to the atmosphere by radiation is about 25W/m^2.
[Net Surface Radiation Absorbed By Atmosphere = Surface Radiation – Surface Radiation Which Gets Out To Space Through The Window – Back Radiation
Some have objected that it is not the NET that is important, but the gross. To which I would respond “what energy transfer would you calculate from a dry surface to an atmosphere separated by a 1mm vaccuum from the surface?”]
2. I want to know at what height this NET energy enters the atmosphere. It is only at that height that the atmosphere will heat due to radiation.
3. The Earth radiates in the IR Band as an almost Black Body at every frequency. Let’s assume that CO2 and Water Vapour also emit as perfect black bodies in their emission bands. In those bands the net radiation absorbed by the atmosphere will be purely dependant on the temperature difference between the surface and the absorbing gas.
4. The CO2 absorption numbers suggest that surface radiation is almost extinguished very close to the surface – within the first couple of hundred metres at the outside, but mostly within the first 50 metres. Because this is close to the ground there is virtually no temperature difference and therefore virtually no NET surface radiated energy is being absorbed by CO2.
5. I have no figures for H2O absorption. Christopher Game stated that the height of absorption (and therefore back-radiation) is 300m and below. In that case also, there is very little temperature difference between the absorbing gas and the surface, and once again virtually no NET surface radiated energy is being absorbed by H2O.
6. There is very little absorbing material left! I therefore postulate that the wings of the absorption bands (ie where emission by the gases is not blackbody) is where the NET Surface Radiation is heating the atmosphere, as the mean height of emission to earth means that the emitting gas is significantly cooler than the surface.
7. I wondered if Science_Of_Doom or DeWitt Payne had any figures on this – is it reasonable that the wings of the gas emission BANDS are NET absorbing a minimum of 8% of the surface radiation into the atmosphere?
[That looks wrong to me, but I don’t have any numbers or tools…]
1. It is plain from the K&T numbers that the NET transfer of energy from the surface to the atmosphere by radiation is about 25W/m^2.
Well 26.
“2. I want to know at what height this NET energy enters the atmosphere.”
Optical depth changes with wavelength. So the answer is “all through the atmosphere, with different rates at different heights”.
“Because this is close to the ground there is virtually no temperature difference and therefore virtually no NET surface radiated energy is being absorbed by CO2.”
300m still has a greater than 3C difference (more because there is a greater adiabatic lapse rate right at the surface because energy is going in more at the lower level than the upper level).
The power difference between those two is given by
sigma(T1^4-T2^4)
T2=surface temp, say 300K
T1=air temp at 300m, say 303K.
sigma=5.7×10-8
or 18W
But optical depth changes, remember. And the difference between surface temp and even 10m air temp is often quite a bit larger than 3C. Both of which increase that figure markedly.
26W isn’t all that far out of the running is it.
PS that 18W is for the entire absorption. Not just the wings.
I apologise to Christopher Game, as I wrote my above question before I had seen his interesting post at 1118, 21July.
Christopher is claiming, on the basis of the TFK numbers and subsequent adjustment of them by Kiehl in response to new NASA data, that there is very little NET transfer of energy from the Surface to the Atmosphere by the mechanism of Radiation.
I consider, based on the absorption patterns described in general terms in my post at 1133, that this is a quite likely situation. But I can’t prove it – one would need to do the absorption calculations across the frequency range.
In other words the planet cools by Evaporation of Water and Conduction and Direct Radiation to Space only, the NET Surface Radiation to Atmosphere being an insignificant term.
If this is correct, then other adjustments would be required to the TFK numbers. The general equation is:
Sunlight_Absorbed_at_the_Surface = Surface_Radiation_through_the_Window + NET_Surface Radiation_Absorbed_by_the_Atmosphere + Energy_Used_to_Evaporate_Water + Energy_Conducted_from_the_Surface _to_the_Atmosphere
The other implication is that if Ed=Aa, then the “Greenhouse Effect” is only the slight closing of the Window brought about by the higher gas concentrations. My understanding is that this would only have a minor effect on the temperature of the planet.
Christopher,
If I have you right, Ed = Aa, is not a law but a tendency.
That is: (As-Ed)/Ed tends to be small.
If so, I would concur, but might like to now how small is small.
In situations of this type where you have to fluxes Fu and Fd, it is true that (Fu-Fd)/Fu may tends to be small.
It certainly the case with diffusion and when Fu=Fd, the net flow ceases and the system is considered to have equalised.
If it is just a tendency then it does not seem to me to have a great deal new to offer as we know that to be the case already.
In the atmospheric case, an analogous diffusion process would be a many species one. The highly absorbed species would have very small diffusion coefficients and would have a marginal effect on the net flow of all species and (Fu-Fd)/Fu would be very very small in their case. If you were interested the primary flow you would only consider the weakly absorbed species whose diffusion coefficients are large and for which (Fu-Fd)/Fu is still small but palpable. It has its problems as an analogy but that is life, it does I hope highlight that the difference between the highly absorbed wavelengths that make up the bulk of the interaction in the near surface layer and for whom (Fu-Fd)/Fu is going to be very very small, and the weakly reacting species for which (Fu-Fd)/Fu may be significant and measurable. For these weakly absorbed species the Temperatures of the regions producing the two fluxes may be signifcantly different, and this is important.
The case involving clouds is significant as they are regarded as being close to black bodies in the LW region, they tend to be cooler than the surface (when averaged diurnally) and I believe give rise to a measurable difference between Fu and Fd.
The differences are going to be small, I do not know what they are but I hear that they are of the order of 10% of the flux upward from the surface, but it really does not matter. If it could be shown that this ration has an invariant value (albiet when averaged over reasonably short interval say days or even years) then that would be novel. Apparently the proposed value is 0% and I do not think that this can be the case.
I believe I understand that for the vast majority of radiative interactions between adjacent layers the fluxes are very very close to being equal between up and down direction, but that is not were the bodies are buried, they are to be found in the direct interactions between layers hundreds or thousands of metres appart and for these weakly absorbed species I do not think the equality holds or can hold. I doubt that a general 0% value could hold unless the atmosphere was isothermal or that it was almost totally opaque to LW. I do not think I have done more than restate some part of the basis of the assumptions behind the cooling to space approximation.
Alex
Apologies for the number of posts getting stuck in the moderation queue in the last couple of days.
Some angry exchanges elsewhere on the blog had me ratchet up the spam/moderation filter which consequently has started trapping normal comments.
If your posts are held up, please bear with me. Normal service will soon be resumed.
Regarding the diffusivity factor (commonly 1.66) I believe that this is a bit of a kludge in that it is a convenience that is used where it is appropriate.
I believe that when it is not appropriate other values are used, or perhaps the function is calculated in some other way.
I believe it holds fairly true for value of tau from zero to one(ish) and is exactly true for some value around 0.5 but it is known to be an approximation and is treated as such. It is not a magic number it is just the value that gives what is considered to be the best general agreement between the required function and one more easy to calculate.
Alex
It seems to be a general derivation, at least from looking at the original equation and the maths, however, I’m happy for someone explain the limitations if they exist.
Let us make sure that we are talking about the same thing.
I estimated an integral of what I believe is the appropriate radiation function for planar geometry using discrete summation, and I did not get an invariant value for the diffusivity factor (or parameter).
What I got looks like it agrees with the following overview:
http://langley.atmos.colostate.edu/courses/at622/lectures/lecture_04.ppt#285,35,Slide 35
If this is the factor under discussion then it would be an approximation and I would leave it up to you to determine the existence of the limitation.
Hopefully the link will work, I will check after posting.
Alex
The link was close, but it was the next page that I wished to highlight:
Second try:
http://langley.atmos.colostate.edu/courses/at622/lectures/lecture_04.ppt#285,35,Slide%2035
If It works that is the result and the preceeding slides show how it is derived.
The slide show goes on to discuss sounders and the GIFTS (Geosynchronous Imaging Fourier Transform Spectrometer) Remote Sensor, which is interesting.
Alex
No luck it has got a screwy URL, it needs a space but it is after the # so it really needs a space and not a %xx type substitute.
Just skip to the next slide.
Sorry
Alex
Re: Colin Davidson on July 21, 2010 at 11:33 pm
In answer to your Q2:
“I want to know at what height this NET energy enters the atmosphere. It is only at that height that the atmosphere will heat due to radiation.”
I believe that it is primarily due to radiative interactions directly between the surface and layers much cooler in temperature. The effect should increase with height in the atmosphere due to the lapse rate but decrease if the falloff in absorbing molecules. I also suspect that a good deal may be due to the interaction between the surface and the bottom of clouds. Were I to hazard a guess I would favour the cloud base to be the clearest place where one could say “that is where one could observe it happening”, but that is not to say that I know it to be the majority effect.
Perhaps someone knows better. I may be fixated on clouds because I do beleive that they will clearly give rise to a measurable difference between Ed and Aa for in the cloudy sky case the absorption of LW is more or less total and the down flux will be less than Ed due to the lower temperature of the cloud base compared to the surface. The photon carriers which conspire to produce this flux difference will be weakly aborbed ones and most importantly the “window” ones as a cloud is effective in absorbing all of these and emitting these same species directly back to the surface but from a radiator at a lower temperature than the surface.
Alex
Oops typo:
… decrease “with” the falloff in absorbing molecules.
Alex
reply to Colin Davidson’s post of 2010 jul 22 at 12:02 am.
Colin writes “I consider, based on the absorption patterns described in general terms in my post at 1133, that this is a quite likely situation. But I can’t prove it – one would need to do the absorption calculations across the frequency range.”
The calculation you want has been done by Ferenc Miskolczi. His paper is indexed at http://multi-science.metapress.com/content/nm45w65nvnj3/?p=2037ca7987b54843bc4b526ff0b6ff7c&pi=1, The stable stationary value of the earth’s global average atmospheric Planck-weighted greenhouse-gas optical thickness,’Energy and Environment’ 21(4): 243- 262, doi 10.1260/0958-305X.21.4.243 .
It is a pity that others have not bothered to do the calculation that interests you. Then there would be some basis for agreement or disagreement with Miskolczi’s calculation on the two radiosonde datasets that he has examined. As things stand today we are looking at a situation in which Miskolczi alone has done the calculation and the results have surprised people. Some have been interested and glad to learn the results, others have preferred to ridicule, ignore, or deny them because they don’t like their implications. What is needed is independent actual empirical checking. Then we could have a real basis for agreement or disagreement.
Colin Davidson also writes “In other words the planet cools by Evaporation of Water and Conduction and Direct Radiation to Space only, the NET Surface Radiation to Atmosphere being an insignificant term.” This is the right description of the cooling of the land-sea surface of the planet.
Colin also writes “The other implication is that if Ed=Aa, then the “Greenhouse Effect” is only the slight closing of the Window brought about by the higher gas concentrations. My understanding is that this would only have a minor effect on the temperature of the planet.”
This statement, I think, needs some tidying up.
When Colin here writes of “the Greenhouse effect”, others wanting to express the same meaning would write of “the contribution of CO2 to the greenhouse effect” because they would say that the greenhouse effect is a fundamental part of the atmospheric energy transport process, both natural and perhaps hypothetically human-induced.
Yes, CO2 does tend a little to close the window. But it depends exactly how you define “window” to say exactly what this means. Roughly speaking one can think of the spectral window as a static characteristic of the spectrum of various gases. Or one can think of it rather differently, as a dynamic characteristic of the whole atmosphere. It is best to use two different terms for these two different conceptions. One can speak of the ‘spectral window’ and of the ‘atmospheric window’ to distinguish the two concepts.
The way Miskolczi approaches the problem is to define the Planck-weighted or true greenhouse-gas optical thickness of the atmosphere, and to calculate its global average.
This tells how much of the radiation from the land-sea surface is absorbed by the atmosphere and how much of it passes direct to space without interacting with the atmosphere. It is not necessary to go into detail about the exact use of the word “window” when one has this information, because this information tells what one wants to know. (Some finer points can be made about clouds here, but for the present it is convenient simply to assume that the clouds maintain their steady state, and to consider their effects as a steady background.)
The Anthopogenic Global Warming (AGW) doctrine can then be stated that human-induced CO2 increases in atmospheric CO2 will increase the true greenhouse-gas optical thickness of the atmosphere. The doctrine then states that there will be consequent delay of passage of heat from the land-sea surface to space, and that there will be consequences of this. One of the most important consequences, essential to the AGW story, according to the doctrine, is what they call “positive feedback by water vapour effect” (on the true greenhouse-gas optical thickness). This latter part of the doctrine means that increase in global average climate temperature has as a main part of its mechanism an increase in the water vapour content of the atmosphere such as to cause a further increase in the true greenhouse-gas optical thickness, by about twice the virtual no-feedback effect of CO2. It is the combined increases that the doctrine asserts are the basis of the AGW thing. The sum of the no-feedback effect and the “water vapour feedback” increment will now have a total magnitude of three times the calculated virtual no-feedback effect of CO2 alone.
Miskolczi said “Let’s look at the past 61 years of NOAA world average radiosonde data and see how much the true greenhouse-gas optical thickness has increased.”
He first calculated, using that dataset, how much increase would be expected on the assumption that the increase was instantaneous and had not had time to affect the atmosphere by any other process than merely what we may call the virtual no-feedback increase of true greenhouse-gas optical thickness due to CO2 increase.
Then he calculated how much actual increase there was in the true greenhouse-gas optical thickness over the 61 years from 1948 to 2008.
Then he compared the virtual no-feedback effect with the actual effect over the past 61 years. The actual effect was not three times the virtual no-feedback effect as postulated by the doctrine. It was about one quarter of it. Indeed the actual increase in true greenhouse-gas optical thickness was so small as to be not statistically significantly different from zero.
The increase in true greenhouse-gas optical thickness due to CO2, hypothesized by the AGW doctrine, just didn’t happen in the real physical world. Not a significant trace of it.
What went wrong for the AGW doctrine?
The true greenhouse-gas optical thickness of the earth’s atmosphere seems to have a naturally governed stable steady value. It seems to have a naturally maintained constant value, in other words. The NOAA dataset shows this being maintained over 61 years. This is a simple fact of observation. No theory of climate comes into establishing this observed fact over the 61 year period from 1948 to 2008.
The theory of the exact mechanisms that keep the value stable and steady is perhaps not so easy to pin down. Water vapour and temperature and pressure vary and they probably determine the outcome. However it may be mediated, the empirically observed result is that the AGW hypothesized increase did not break through the natural stable steady state.
As it happens the NOAA 61-year dataset shows more. It shows that while the temperature was recorded as increasing on a linear estimate over the 61 years, the water vapour content on a linear estimate decreased. This means that there could not have been a positive water vapour feedback on the climate time scale such as is postulated by the AGW doctrine.
This leaves the AGW doctrine without a leg to stand on. Thus you can see why AGW people feel a need to ignore, ridicule, or deny the findings of Miskolczi.
replying to Alex Harvey’s post of 2010 Jul 22 at 6:10 am.
Thank you, Alex, for your reply.
Alex writes:
“If I have you right, Ed = Aa, is not a law but a tendency.
That is: (As-Ed)/Ed tends to be small.
If so, I would concur, but might like to now how small is small.”
Christopher replies:
If you like to call it a tendency and not a law I don’t see a problem there. What matters is that Aa – Ed is small in comparison with Aa or Ed, as you say. But I think its smallness is more remarkable than you imply here.
Just how small is something that can be investigated further. You have access to Ferenc’s calculations of it in his papers. Sad to say, no one else has bothered to provide the answer to your question to check Ferenc’s values. You might be able to contribute such a check. We would all welcome it.
Alex writes:
“In situations of this type where you have to fluxes Fu and Fd, it is true that (Fu-Fd)/Fu may tends to be small.
It certainly the case with diffusion and when Fu=Fd, the net flow ceases and the system is considered to have equalised.
If it is just a tendency then it does not seem to me to have a great deal new to offer as we know that to be the case already.”
Christopher replies:
You are at this point trivializing the result. Others are more offended by it and prefer to ignore, ridicule or deny it, as you do at another point in your post.
But you are not really saying anything in your trivializing comment.
According to the TFK2010 unadjusted figures, that many rely on, Aa – Ed = (356 – 333) W/m2 = 23 W/m2. According to the adjusted figures, Aa – Ed = – 3 W/m2, in fairly near agreement with the Aa = Ed rule. If you think an adjustment from 23 to 3 is not worth talking about, then Ferenc’s work doesn’t have much to offer you.
Alex writes:
“I do not think I have done more than restate some part of the basis of the assumptions behind the cooling to space approximation.”
Christopher replies:
The cooling to space approximation I think started as an empirical finding, not derived from a priori assumptions as you seem to suggest on the basis of what looks like guessing. It started with the work of Rodgers and Walshaw 1966. You might like to look at the original papers or the account of them in Paltridge and Platt 1976, pages 165-172, or in Goody and Yung 1989 page 250, who state that “its importance cannot be overstated”, or in Wallace and Hobbs page 138, who state that it works “surprisingly well”. What is worth seven pages to Paltridge and Platt, and has an importance that cannot be overstated for Goody and Yung, and works surprisingly well for Wallace and Hobbs, is perhaps trivial to you, but not to me.
What is remarkable is that instead of people saying “Oh, we all knew that Aa is very nearly equal to Ed, what’s new about that?” as you seem at times to be saying, many of them actively or even busily deny it or ridicule it. I will not name names.
It is said of new ideas that the establishment reacts to them in three steps: first they ignore, then they ridicule, then they say “Oh, what’s new about that, we knew it all along.”
Alex writes:
“for the vast majority of radiative interactions between adjacent layers the fluxes are very very close to being equal between up and down direction, but that is not were the bodies are buried, they are to be found in the direct interactions between layers hundreds or thousands of metres appart and for these weakly absorbed species I do not think the equality holds or can hold.”
Christopher replies:
This, combined with several other of your comments, puts you in several camps at once: in the camp that trivializes or ignores; in the camp that says “Oh, we knew that all along”; and in the camp that says “I do not think it is true”, the active deniers.
The Aa = Ed equality is not derived, as you mischievously suggest, merely from an assumption of small differences between nearby places, though of course we all admit the truth of that with suitable provisos (which you carelessly neglect to state).
The Aa = Ed equality is derived from the calculations of direct interactions between layers hundreds or thousands of metres apart (indeed actually tens of kilometres apart in fact), as you indicate would impress you.
You seem to put your intuition ahead of the empirical facts. I can hardly argue against that kind of thinking. Which side of the bed did you get out of this morning?
reply to Alexander Harvey’s post of 2010 Jul 22 at 12:06 am.
Alex, I wrote a reply to your post, but something went wrong with the posting mechanism and that reply of mine seems to have vanished. Perhaps it will reappear.
“The increase in true greenhouse-gas optical thickness due to CO2, hypothesized by the AGW doctrine, just didn’t happen in the real physical world. Not a significant trace of it.”
The radiative balance is determined by the TOA.
Therefore the important thing is where the 1 optical depth occurs from space down through the earth’s atmosphere.
NOTE: at heights, there’s much less water vapour because it quickly passes 0C where ice will start to form, solid H20, not merely liquid. CO2 doesn’t freeze out for a hundred or so more degrees. Therefore increase in CO2 has a greater effect since CO2 at height will intercept H2O bands it shares from lower down.
The IPCC report into the science (WG1) goes into great depth in this, referring to papers that have more information. See
http:/www/ipcc.ch
Colin Davidson from July 21, 2010 at 11:33 pm:
The question seems odd, I think we have had a little of this discussion before, both not understanding each other’s perspective..
Radiation from the surface heats the atmosphere. Different wavelengths are absorbed to X% at different heights. (There isn’t one height that absorbs radiation of any given wavelength).
Regardless, the absorbed radiation heats the whole atmosphere in that region (thermalization), and that region of the atmosphere radiates back to the ground and up to other levels in the atmosphere.
Which is why I haven’t really tried to understand how much radiation originating from the surface in each wavelength makes it to what height before being attenuated by X% (e.g. to 36% for an e^-τ relationship).
Picture a different atmosphere where the concentration of radiative absorbers (across a wide range of wavelengths) is very high and convection doesn’t take place. In that case you might have “almost zero” difference between the radiation up from the ground and down from the first 100m of the atmosphere. Likewise for each arbitrary layer.
But in that strange case, the surface temperature would be extremely high.
So I can’t picture why the information helps in solving anything – but that doesn’t mean it doesn’t help, it might be my lack of imagination. And for its own sake it might be very interesting.
What am I not getting?
further reply to Alexander Harvey’s post of 2010 Jul 22 at 12:06 pm.
For IR at TOA, Fu = 238 W/m2, Fd = 0, and you want me to accept that (Fu – Fd) / Fu is small? Try the other one, it’s got bells on it.
That it not a situation of the same type as in:
“In situations of this type where you have to fluxes Fu and Fd, it is true that (Fu-Fd)/Fu may tend to be small.”
I cannot see how the situation at TOA is similar to that at the surface, unless you take Fu to be the radiation generated at TOA where the relationship becomes indeterminate (Fu=Fd=0). But for any layers below TOA where there is absorption it would be defined and may tend to be small if the temperature differences are small between interacting layers.
This is in contrast to the situation between the surface a cloud base where the radiative interaction should be strong (the absorption of the “window” radiation) and the temperature difference should be significant and where I think Ed=Aa cannot hold, and not hold by a significant margin.
How do you view the situation with respect to interaction between the surface and the cloud base?
Alex
reply to Mark’s post of 2010 Jul 22 at 8:02 am.
Mark, I think it may help if you read posts carefully before replying to them.
comment on scienceofdoom’s post of 2010 Jul 22 at 8:14 am.
All the following refers only to infrared radiation.
When one grasps that Aa = Ed, one sees that the land-sea surface heats the atmosphere only by conduction, evaporation, and subsequent convection, not by radiation. The land-sea surface does not cool by radiation to the atmosphere. The land-sea surface and the atmosphere are in radiative exchange equilibrium. Radiation from the land-sea surface that actually removes heat from it, unbalanced, goes straight to space; there is no back radiation from space.
Christopher, I think you should read the WG1 paper:
http://www.ipcc.ch
before continuing in your defence of the un-reviewed paper in E&E.
Christopher, I think you should read the WG1 paper:
http://www.ipcc.ch
the science that that E&E paper proposes to be incorrect is in there and I would suggest you find out what the IPCC considers the science to be before you start reading single papers in E&E.
Regarding Mark July 19, 2010 at 9:07 pm, Frank responds:
I was trying to describe the effect temperature has on the probability that a single GHG molecule will emit a photon. According to Boltzmann’s distribution law, “if the energy associated with some state or condition of a system is ε then the frequency with which that state or condition occurs, or the probability of its occurrence, is proportional to exp(-E/kT)”. (http://assets.cambridge.org/97805218/11194/excerpt/9780521811194_excerpt.pdf; page 2). So the probability of a GHG molecule having enough energy to be in a vibrationally excited state from which it can emit a photon appears to depend on exp(-hv/kT). If none of the others factors that affect the probability of a photon being emitted have any temperature dependence, then the overall process could have this temperature dependence. After we integrate over all wavelengths, we arrive at energy flux that varies with T^4. However, a single GHG only emits
This is regarding the unkowns (there are more values on the energy balance illustration than in the data tables) in the TKF2009 paper.
Can anyone tell me if the following relations are all claimed by Miskolczi to hold for the Earth:
“The mechanism of the greenhouse effect may
better be explained as the ability of a gravitationally bounded atmosphere to
convert F0 + P0 to OLR in such a way that the equilibrium source function
profile will assure the radiative balance ( F0 + P0 = OLR), the validity of the
Kirchhoff law ( ED = SU A), and the hydrostatic equilibrium ( SU = 2EU ).”
…
“Long term balance between F0 + P0 and OLR can only exist at the Su =(F0+P0) /(1- 2A/5) ~ 3(F0+P0)/2 planetary equilibrium surface upward radiation.”
(I could not paste the equations from the paper so I retyped them and may have made an error)
That is:
Ed=SuA = Aa
Su = 2Eu
Su = (F0+P0)/(1-2A/5)
Are these all claimed to hold simultaneously? If so I think St can be deduced from the available data as a direct consequence.
Also am I right in that Eu = OLR – St so that the TFK2009 data gives us:
Su = 2(OLR – St) and St = OLR -Su/2 and hence Aa = Ed = 3Su/2 – OLR?
I think that these would give St = 238.5 – 396/2 = 40.5W (as a direct consequence of the Miskolczi relations)
giving tau = -ln(40.5/396) = 2.28, Not 1.86
1.86 would imply St = 61.6W and either a much greater value for OLR or a smaller value for Su.
The problem is I do not know which of the Miskolczi relations are held to be claims (i.e invariant or laws) as he does not seem to include them in either the abstract or the conclusions.
The two vital ones SU = 2EU & ED = AA are emphasised in the summary of this slide presentation:
Click to access ZM_v10_eng.pdf
so I presume they are meant to be laws.
The first gives us Eu = 198 and hence St = 238.5 – 198 = 40.5W as before, which is all we need.
Can anyone comment on both what Miskolczi claims to hold (and whether I can do my sums)?
Alex
“I was trying to describe the effect temperature has on the probability that a single GHG molecule will emit a photon.”
I was trying to point out that you have to be much more careful in how you deal with this because it is mixing models: macro and micro scales.
“After we integrate over all wavelengths, we arrive at energy flux that varies with T^4.”
This merely means that the occupancy effect is higher than the black body curve suggests. This is not new, lasers do this all the time.
So does the sun:
Note how the visible spectrum is more intense than the blackbody curve projects.
This is because there is a lot of interactions.
So if, for example, a CO2 molecule with slightly higher velocity (but not enough to excite IR emission) is hit by another molecule OF ANY SORT with enough or more energy to reach excitation, the collision will have a chance of causing IR emission.
Therefore as well as those molecules excited enough to generate IR excitations themselves emitting, there are ones that are collisionally excited adding to the flux.
And just such energy abundance is found because these two or more) molecules have an excess of energy because they cannot punt some of that to the other degrees of freedom.
This is how equipartitioning of energy in all states occurs.
The actual calculations to *prove* this are postgraduate studies, if you wish to prove to your own satisfaction, I suggest you find a University that does radiative Physics and contact them for either papers or materials in the subject.
It is not easy reading..!
reply to Mark’s post of 2010 Jul 22 9:50 am.
Thank you Mark for your valuable advice.
“This is regarding the unkowns (there are more values on the energy balance illustration than in the data tables) in the TKF2009 paper.”
I would suggest you contact the journal. They often get more information (especially data) that cannot be printed (like data) or would be far too expensive to print as a matter of course (almost all supplemental materials) but retain them for any scientist with sufficient interest to ask.
There will likely be a cover charge, at least P&P, unless you’re in a suitably accredited university (NY Academy for the Performing Arts would likely not have a quid-pro-quo with the NAS, for example…).
PS the problem with your sums is that you need to explain the derivation of the terms and the provenance of the equations used.
Unless you happen to come across Trenberth here, or someone similarly au fait with the subject, that is needed to determine what you’re doing and THEN whether it’s right (or how it’s wrong, which may be as fixable as “you’re using 181 incorrectly, it should be used in this … equation”)
reply to Alexander Harvey’s post of 2010 Jul 22 at 7:42 pm.
We have a school of red herrings here.
No worries, Christopher.
Remember, though, even when it sounds like I’m certain, that certitude is because this is far too limited a communication medium. If you look back in places where I’ve said “you’re wrong, this is what…” you can see that in the responses to that that there are plenty of cases where, maybe I’m not *wrong*, but there’s definitely nuances.
In short, the information I give should be used to help find your answer, not the answer itself (rather like Budda, which I resemble in some regards… 🙂 )
But given no better clue, look to consensus: that’s where lots of heavy thinkers asked “did I get that wrong?” and found out that they did and the model improved.
Until it’s stood the test of contact with other scientists, single papers that *say* they’ve “overturned” that consensus will not have proven their utility and most probably haven’t.
reply to Alexander Harvey’s post of 2010 Jul 22 at 8:42 pm
Alexander writes:
“Can anyone comment on both what Miskolczi claims to hold (and whether I can do my sums)?”
I will comment on what Miskolczi claims, but not on whether you can do your sums.
Miskolczi offers a first-glance model and a more refined model. By mixing the two models, anything can be proved, such as 0 = 1. One must stick to one model at a time.
We have some definitional identities:
Su = Aa + St
OLR = Eu + St
tau = – ln (St / Su)
The first-glance model is
(1) Aa = Ed
(2) Su = 2 Eu
(3) 3 Ed = 5 Eu.
The more refined model is
(1) Aa = Ed
(2) Su = OLR (1 + tau +exp(-tau)) / 2
(3) 3 Ed = 5 Eu
The first-glance model, by a little elementary algebra, leads to Su = 6 St and thus to tau = – ln (1/6) = 1.79176…
The more refined model, by a little elementary algebra, leads to a transcendental equation for tau, which has the solution tau = 1.86756… . In this more refined model we find Su = 1.97 Eu. For an empirical observational demonstration of this more refined model, see Ferenc Miskolczi’s paper indexed at http://multi-science.metapress.com/content/nm45w65nvnj3/?p=2037ca7987b54843bc4b526ff0b6ff7c&pi=1, The stable stationary value of the earth’s global average atmospheric Planck-weighted greenhouse-gas optical thickness, ’Energy and Environment’ 21(4): 243- 262, doi 10.1260/0958-305X.21.4.243 . The earlier papers contain a lot of empirical observations and a lot of theory and are not so easy to read.
further note on the two Miskolczi models described in my post of 2010 Jul 22 at 9:46pm.:
For each of the two models, the first-glance model and the more refined model, the empirical observations show that
(1) holds rather well for every radiosonde ascent,
(2) holds approximately for every radiosonde ascent, and
(3) holds only for the global average, not by any means for every radiosonde ascent.
The results deduced for tau therefore hold only for the global average.
The case for ‘back radiation’ is weak. Nothing here gives evidence that ‘back-radiation’ has been detected or measured. The papers quoted do not show measurement of back-radiation. Instead they measure upward radiation and subtract this value from a hypothetical value relative to abs. zero. The difference is the so-called back-radiation. Surely a value of 300W/m2 would seem implausible to most, it is comparable in magnitude to radiance from the sun!
Here is how it is explained by the makers of a typical Pyrgeometer..
“In order to calculate the incoming LW irradiance at the detector, the temperature of the pyrgeometer body must be known. … The downward longwave radiation is then calculated using the following formula :-
LW = Uemf/S + ( 5.67*10-8 * Tb4 )
where Uemf is the output voltage from the thermopile, S is the calibration constant of the instrument, and Tb is the pyrgeometer body temperature, measured by the thermistor, in degrees Kelvin. Note that for an upward facing pyrgeometer, the thermopile output voltage will in most instances be negative. This is because the upwelling irradiance from the pyrgeometer is likely to be greater then the incoming irradiance from the sky. “
I therefore challenge anyone on this site to show me direct evidence of back-radiation. Better, show me a method to power a 40W light globe using back-radiation. It should be easy as there is 300 W/m2 available. If back-radiation is able to raise the temperature of the Earths surface then it will be possible to harness this energy to produce a small electrical generator. The first person with a solution wins a multi-million dollar share of the IP rights.
Sorry for any confusion, I will try again being a little more brief:
If the Miskolczi equalities
Ed = Aa
Su = 2Eu
are claimed to be true then from the second equality and the Trenberth data:
OLR = 238.5W
Su = 396W
Ed = 333W
we get
Eu = Su/2 = 396W/2 = 198W
St = OLR – Eu = 238.5W – 198W = 40.5W
(I believe that the Trenberth values for Su and OLR are not seriously disputed, e.g. not by 10W or more)
but from the second equality Ed = Aa we get
Aa = Ed = 333W
so
St = Su – Aa = 396W-333W = 63W
So two of the Miskolczi claims lead to two quite different (23W) values of St.
Which indicates that thes two claims are incompatible with the Trenberth data and/or each other.
Hence I wonder if we are intended to consider these two equalities to be laws, and if not what use are they.
Alex
reply to Morris Minor’s post of 2010 Jul 22 at 9:55 pm.
Morris Minor writes:
“Take a simpler example where T1 = T2. Between the surfaces there is no emission or absorption.”
Dear Morris Minor, I suggest you read something about Pierre Prevost’s 1791 work that demonstrates the Prevost exchange principle, the foundation of all modern work on radiation.
reply to Morris Minor’s post of 2010 Jul 22 at 6:10 pm.
Morris Minor writes:
“If back-radiation is able to raise the temperature of the Earths surface then it will be possible to harness this energy to produce a small electrical generator. The first person with a solution wins a multi-million dollar share of the IP rights.”
You can do this for yourself beside a nice cool stream of water from the melting snow. Use the stream water to cool your pyrgeometer, face your pyrgeometer upwards, lead off the output of the detecting circuitry, and you will have a small electrical generator.
When you have done this, think about the implications of Aa = Ed, in the light of your reading about the Prevost Exchange Principle.
OK, I think I can see.
Either there’s something missing in the Miskolczi equalities listed or it’s just wrong in that they don’t apply.
It’s entirely possible to get very different values for the same thing, but these usually have to be abnormal (lasing, ultra low pressure [allows the OIII emission lines attributed to ‘nebulium’ until we got extreme vacuum], and other oddities not often found in nature).
The figures from Trenberth are, if not definitive, at least appropriate, so the only option for the equality holding true is if balance is highly unstable.
I’m afraid I’m not certain whether I’ve got the physicalities of the properties Ed and Su as he defines them to get this, but if Ed = Aa, then there’s no net transfer of heat into the atmosphere, which is going to be a problem.
Also if Su = 2Eu then you’re assuming either Sg=Aa which is not going to be true either (unless you throw out the IR window).
There’s a few other problems in the diagram right at the end:
It mixes upward and downward together into the atmosphere, but not into the ground. Apples and kumquat comparison.
It ignores reflection (but then assumes emissivity not 1)
P and P0 aren’t useful physicalities (on average, deep temperatures don’t change: thermal inertia).
Doesn’t seem to require that outgoing equals incoming TOA.
Ignores TOA requirements and has the surface radiation balance as the sine qua non of thermal determination of the temperature.
As we’ve had before: changes are driven by adiabatic lapse, but this doesn’t define the constant (surface temp).
And I’m made less happy with Christopher’s talk of “a more refined model”. If you have to use a more refined model, then your model was wrong.
In all, that powerpoint seems more like an 18thCentury paper on thermodynamics, where they seemed to collect any old set of figures and give them a name: multiplying entities is NEVER a good idea. Each one you add gets to multiply the chances of getting something wrong (not just add).
In short, the presentation makes some assumptions not based in physics (remember, correlation is not causation!), multiplies entities uselessly and is unclear as to its terms of reference.
Add that to
a) the apparent (it’s not easy to read) failure to add in known effects
b) the incorrect energy balance required
c) mixing up and down fluxes but only in places
d) Needing a “more refined model”
I would say, ignore this and see if it stands the test of time.
If someone can make useful predictions out of it and they can be shown to be the case, then it may pick up some interest.
And the bottom line, Alex, seems like both myself and Christopher don’t think your calculations are wrong nor your conclusion.
“I therefore challenge anyone on this site to show me direct evidence of back-radiation.”
We’d need to give you a pyrometer and a LW/SW IR filter to do that.
But if there’s no such thing as back radiation, then that means the atmosphere is at absolute zero.
I challenge you to show me direct evidence of this…
Morris says: “I therefore challenge anyone on this site to show me direct evidence of back-radiation.”
Try this one. It has not only measured the LDR, but also has found it to be changing in agreement with what’s known about atmospheric physics and increasing GHE.
I’m sure some of the papers pointed by SOD in the main text has direct measurements too, but I don’t have access to the full paper.
Mark (0757,22Jul) responded to one of my posts. In part he said:
“300m still has a greater than 3C difference (more because there is a greater adiabatic lapse rate right at the surface because energy is going in more at the lower level than the upper level).”
Is this right?
I thought that the lapse rate decreased when energy enters the atmosphere, ie the added energy increases the temperature, rather than decreasing it. As it does with Conduction (around 20W/m^2) right at the surface.
I had a look at some radiosonde data. In many cases the 300m temperature was higher than the surface, while in others it was up to 5DegC cooler. I think the average would be perhaps around 1DegC cooler.
SOD: If one considers a sphere of atmosphere of radius R containing some emitting GHG’s, the number of emitters inside varies with R^3, but the flux according to Planck’s function with R^2. If I were to reduce the radius by a factor of 10, the number of emitters would drop by factor of 1000, while the power/surface area emitted by Planck’s function drops by 100. It doesn’t make much sense that the emission per molecule goes up as R decreases. (If I make R small enough, the probability of a photon being absorbed before it reaches the surface of the sphere is negligible.) I had thought that the Planck function had units of power/surface area, but I now see that the proper units are power/surface area/solid angle, but I’m not sure this adds anything more than a factor of 4*Pi or a factor of 0.5 to split upward emission from downward.
I run into the same dilemma when I think about a layer of atmosphere. Absorption in the layer depends on thickness of the layer and the total number of GHGs inside. Emission from the layer (eoT^4) depends on surface area. For example, see Figure 6.8 in Fred Taylor’s book (the source for some of your posts) assumes that e is 1.
A more complete approach may be found in Wallace&Hobbs, Atmospheric Science 2nd Edition, p134-5, which discusses Schwarzschild’s equation for how the intensity of light of ONE specific wavelength is changed (dI) by an absorption term (-Ikpr*ds) and emission term (+B(T)*kpr*ds) when passing through a path length ds of atmosphere, where I is the incident intensity, k appears to be an absorption coefficient (apparently identical to the emission coefficient by Kirchhoff’s Law), p (rho) is density, r is presumably mixing ratio, and B(T) is the Planck function for that wavelength. The term kpr*ds should be dimensionless and pr is proportional to the number of emitters in a unit volume. Now we have emission which varies directly with the number of emitters (wild applause) and emission per GHG molecule directly linked to absorption per molecule (standing ovation).
The following sections of the book (not yet mastered) integrate Schwarzcshild’s equation to get an optical thickness (which I intuitively called “mean free path” and didn’t apply to emission), discuss a “plane parallel approximation” to get vertical component of flux, and only then integrate over wavelength. It seems like other approaches skip ahead to integrate Planck’s function over wavelength to get oT^4, then go back to correct for varying emission with wavelength by adding an emissivity term (eoT^4). Since the pr varies 1,000,000X between Earth and Venus and e must be between 0 and 1, there is no obvious way (in my current state of ignorance) to have emission depend on the density of emitters.
I’m sorry if any of this has been properly discussed elsewhere at your site. I ran a search for Schwarzschild and didn’t find it mentioned at your site.
Some questions, if you can stand more from someone who is trying to learn:
1) I may be confusing absorption and emission WITHIN a layer of atmosphere with emission FROM a layer. Are drawings showing atmospheric layers with a flux of eoT^4 fundamentally correct? Does a thin layer of Venusian atmosphere at 275 degK emit like a thin layer of the earth’s atmosphere at the same temperature except for a change in e? Does e come from the Schwarzschild eqn by some mathematics I haven’t encountered?
2) Absorption (and emission?) lines are broadened by pressure and temperature, possibly accounting for the [P/T]^0.5 factor you mention at 7/20 12:25 am. Do P and T change the total absorption (and emission) of a single line or do they simply redistribute it within the width of that single line? If not, is the above “absorption coefficient” k in Schwarzschild’s equation a function of P or T for a single line?
3) Is the breakdown in the assumptions underlining LTE high in the atmosphere important to predicting stratospheric cooling with increasing GHG’s? (A simple yes or no will be adequate.)
reply to Alexander Harvey’s post of 2010 Jul 22 at 9:59 pm.
Alexander writes:
“Sorry for any confusion, I will try again being a little more brief:
If the Miskolczi equalities
Ed = Aa
Su = 2Eu
are claimed to be true then from the second equality and the Trenberth data:
OLR = 238.5W
Su = 396W
Ed = 333W
we get
Eu = Su/2 = 396W/2 = 198W
St = OLR – Eu = 238.5W – 198W = 40.5W
(I believe that the Trenberth values for Su and OLR are not seriously disputed, e.g. not by 10W or more)
but from the second equality Ed = Aa we get
Aa = Ed = 333W
so
St = Su – Aa = 396W-333W = 63W
So two of the Miskolczi claims lead to two quite different (23W) values of St.
Which indicates that thes two claims are incompatible with the Trenberth data and/or each other.
Hence I wonder if we are intended to consider these two equalities to be laws, and if not what use are they.”
Christopher replies:
You are right to point to these problems.
The Miskolczi quasi-all-sky models (that I have sketched above in my post of 2010 Jul 22 at 9:46 pm) that you are talking about here, refer to a quasi-all-sky protocol, that ignores variability due to cloud variability, assuming that the clouds set a stable steady background condition.
For the sake of definiteness and greater refinement let us consider the more refined model (defined in my post of 2010 Jul 22 at 9:46 pm, above) in which tau = 1.87. Let us work from the measured value OLR = 238 W/m2, and accept TFK2009’s value of 78 W/m2 for the solar radiation absorbed by the atmosphere, and 160.5 W/m2 for the solar radiation absorbed by the land-sea surface. (Gobal averages all over.) Then, assuming an overall steady state, with the earth in all-wavelength radiative exchange equilibrium with the rest of the universe, the more refined Miskolczi model leads to Su = 360.4 W/m2 which does not agree with the TFK2009 value of 396 W/m2.
According to the TFK2009 figures, the right value for St is 63 W/m2 as you found above, or the adjusted value, 66 W/m2 as proposed by the emails referred to above. The value for the more refined Miskolczi model referred to above, using the figures cited above, is 55.7 W/m2. Other values found more directly by Miskolczi are about 61 W/m2. If you add the TFK2009 window wavelength radiation from the cloud tops, 30 W/m2, to the no-clouds TFK2009 value of 40 W/m2, you get a total window wavelength emission to space of 70 W/m2. TFK2009 give Eu from non-cloudy sky as 169 W/m2, compared with the more refined model’s value of 183 W/m2 on these assumptions.
This is because the quasi-all-sky model ignores the presence or absence of clouds.
As it happens, the presence or absence of clouds is not too important in the present considerations. While cloud bottoms absorb window wavelength radiation from the land-sea surface, the cloud tops emit window wavelength radiation. The temperature of emission from cloud tops is lower than from the land-sea surface, but also the atmospheric window is much much wider open above clouds than it is above the land-sea surface, because most of the water vapour is below and within the clouds. The result is that clouds do not have too much effect in “closing the window”. You can read details of this at http://miskolczi.webs.com/kt97_comments.pdf, as you likely already have done long ago.
Perhaps you will say “What’s the use of a model that ignores the presence or absence of clouds!”
The present use is that the more refined model correctly gives the empirically observed stable stationary value of the atmospheric global average Planck-weighted or true greenhouse-gas optical thickness, close to 1.87, and that it provides a simple practical empirical monitor of the putative hypothetical virtual direct primary no-feedback effect of CO2.
The global average true greenhouse-gas optical thickness tells immediately the possible primary total direct effect of CO2 in delaying the passage of infrared radiation from the land-sea surface to space, because it tells directly how much energy is potentially able to be absorbed by the atmosphere, without complications of feedback due to such things as possible effects on clouds, due to the changes of CO2 that are of interest.
When we find by empirical examination of the NOAA re-analyzed 61-year radiosonde dataset that the global average true greenhouse-gas optical thickness is not changed by the increase of CO2 in the atmosphere, we can know that the climate variations over that time cannot be due to CO2 changes. With no primary direct no-feedback effect actually realised in the radiosonde records, there can be no secondary feedback effects. This is useful information.
You are concerned about the use of the word “law”, which you think might or might not be used in this context. In his latest paper (indexed at http://multi-science.metapress.com/content/nm45w65nvnj3/?p=2037ca7987b54843bc4b526ff0b6ff7c&pi=1, The stable stationary value of the earth’s global average atmospheric Planck-weighted greenhouse-gas optical thickness, ’Energy and Environment’ 21(4): 243- 262, doi 10.1260/0958-305X.21.4.243), Miskolczi uses the word “relationship” to refer to the three propositions, (1), (2), (3), of the more refined model that is demonstrated there, as it is reported above.
Science_Of_Doom (22Jul, 0814) kindly responded to one of my posts, in part:
“[Colin Davidson from July 21, 2010 at 11:33 pm:
2. I want to know at what height this NET energy enters the atmosphere. It is only at that height that the atmosphere will heat due to radiation.]
The question seems odd, I think we have had a little of this discussion before, both not understanding each other’s perspective..”
The height at which surface energy enters the atmosphere has consequences for the detail of the lapse rate. If it is low down, then the lower atmosphere will be hotter than it otherwise would be.
We know where Conduction (On the KT numbers about one fifth of the transfer) enters the atmosphere (the surface).
We can guess where heat from condensation of Water Vapour(about three fifths of the transfer) enters the atmosphere (below the cloud tops).
What about the remaining fifth – the 25 or so W/m^2 of NET Radiation?
Christopher Game (0323, 22Jul) kindly responded to one of my posts, entering the discussion “What is the Greenhouse Effect?”
My thoughts on this are below, and assume in all cases a transition from one equilibrium state to another (ie transients are completed, and an “average” KT planet is assumed):
1. The Sun drives the Surface and also provides some energy to the atmosphere.
2. The Surface drives the atmosphere.
3. The Greenhouse Effect is the change in Surface temperature which occurs because of the IR active gases.
4. The Surface air temperature is tightly bound to the temperature of the Surface. There can be no change in Surface Air temperature without a corresponding change in the Surface temperature.
5. The Suface energy flux balance must be maintained:
Forcings (Absorbed_Sunlight + Back_Radiation) = Consequences(Surface_Radiation + Evaporation + Conduction.)
Assuming Conduction is a constant then the Surface sensitivity to a change in forcing is between 0.1 (high rate of change of evaporation) and 0.2 (zero change in evaporation) DegC/W/m^2.
6. Note that the ONLY way to change the Surface Temperature is to alter the Surface Forcings – either the sunlight or the back radiation. An energy imbalance at the tropopause, and Radiative Forcing (also at the Tropopause) is of no interest or consequence to the Surface unless such changes are manifested in a change in Surface Forcing – either sunlight or back_radiation.
7. The surface is very insensitive to a change in forcing – or to put it another way there has to be a large increase in forcing to change the surface temperature. To MAINTAIN a Surface temperature increase of 3DegC (the IPCC “projection”, guess, or non-prediction) requires an increase in forcing (presumably back-radiation) of 16 to 32 W/m^2.
8. To get this the models assume a very small rate of change of evaporation of around 1% per DegC, but an increase of water vapour content of about 7% per DegC. But recent real world measurements suggest the evaporation rate change assumed by the models is understated, and more in line with the WV content increase. In other words, the evaporation assumption could be causing a large overestimate in temperature response.
9. Christopher has stated that NET surface radiation absorbed into the atmosphere is very small, and that this was confirmed by detailed calculations by Miskolczi. (I agree with Christopher that someone should repeat these calculations before discounting them.) If that is the case then the only ways Back-Radiation can increase are if the Atmosphere heats up (but it can’t unless the surface does) or if the Window closes.
RE: Christophe & Mark:
First of all thanks for replying:
So I start with:
The more refined model is
(1) Aa = Ed
(2) Su = OLR (1 + tau +exp(-tau)) / 2
(3) 3 Ed = 5 Eu
and
tau = 1.86756
so
(1+tau+Exp(-tau))/2 = (1 + 1.86756 + 0.1545)/2 = 1.51103
and given OLR is the tranberth figure 238.5
Su = 238.5 * 1.51103 = 360.38
St = Su * 0.1545 = 55.7
Aa = Su – St = 304.7
Eu = OLR – St = 238.5 – 55.7 = 182.82
Ed = 5Eu/3 = 304.7 = Aa (as required)
But 360.38W/m^2 corresponds to an effective radiating temperature of 283.8K which seems a little cold.
Alternatively
given Trenberths Su = 396
OLR = 396/1.51103 = 262.08
St = Su * 0.1545 = 61.18
Aa = Su – St = 334.82
Eu = OLR – St = 200.89
Ed = 5Eu/3 = 334.82 = Aa (as required)
But OLR = 262.08 is a long way from Trenberth’s (238.5) value
So either Trenberth has the Su wrong by:
396 – 360.38 = 35.62
or the OLR out by
262.08 – 238.5 = 23.58
or both out by some smaller factors.
The smallest pair of deviations, that I can find, that meet the Miskolczi criteria correspond to an error of ~14.2 in Trenberth’s values for both Su and OLR
and occurs when the
Su = 381.8 and
OLR = 252.7
It would seem to me that 14W would be a big error to occur in either Trenberth’s value for OLR or Su.
So either I am making a mistake, or there are big errors in the OLR and or the Su (and by implication surface temps/emissivities),
or the Miskolczi relations from the refined model do not hold.
What is of interest is that only two of the Trenberth data values are of importance here:
OLR and Su.
Given OLR Miskolczi gives us Su,
given Su Miskolczi gives us OLR.
The value of St is derived directly so it does not matter what value Trenberth puts on it.
As I understand it OLR is measured (and reasonably accurately) and Su is calculated by integrating over all local T^4 * local emissivity (but I could well be wrong about Su). So if there is a substantial error it would be in Su.
If that be the case then either the world is a lot cooler (about 7C) or the emissivities are a lot lower (~.89 as opposed to ~0.98) than we believe to be the case.
If Su is measured either it is done very inaccurately or Miskolczi’s equalities fail at that hurdle.
I hope the above be an empirical argument.
Alex
My last post crossed with a reply from Christopher that I had not seen:
So it restates a lot of his values, this is regretable but I am a very slow writer, and hence somewhat inevitable.
I am pleased to have someone explain what Miskolczi meant, not what I might assume.
Sorry if it confuses.
Alex
reply to Mark’s post of 2010 Jul 22 at 10:34 pm.
Mark gives a long list of objections to Miskolczi’s work, some of which I will address here.
Mark writes:
“And I’m made less happy with Christopher’s talk of “a more refined model”. If you have to use a more refined model, then your model was wrong.”
Christopher’s reply:
The AGW AOGCMs have been subject to repeated refinements for about a quarter of a century now.
Mark writes:
“18thCentury paper on thermodynamics”
Christopher replies:
Thermodynamics began with Carnot in the 19th century, and the name ‘thermodynamics’ was coined in 1849 by Joule. Thermodynamics is about the distinction between two kinds of energy transfer, as heat and as work.
Mark writes;
“And the bottom line, Alex, seems like both myself and Christopher don’t think your calculations are wrong nor your conclusion.”
Christopher’s reply:
I pointed out that I would not comment on Alex’s “sums” as he called them. But Mark reads into that that I don’t think they are wrong.
Mark writes:
“Ignores TOA requirements”
Christopher replies:
It is not clear to me where this remark “Ignores TOA requirements” comes from.
Mark writes:
“Also if Su = 2Eu”
Christopher replies:
According to TFK2009 figure 1, Eu = (169 + 30) W/m2 = 199 W/m2, and Su = 398 W/m2. The equality Su = 2 Eu belongs to the first-glance model. The more refined model yields Su = 1.97 Eu. Both models use the quasi-all-sky protocol. They are not exhaustive of all detail, as Mark notes. The merit of the Miskolczi work is that it posits general relationships, perhaps not the final and ultimate ones, but it is an advance over unprincipled lists of numerical estimates.
Mark writes:
“but if Ed = Aa, then there’s no net transfer of heat into the atmosphere, which is going to be a problem.”
Christopher’s reply:
I am glad to see Mark take issue here. Now we are getting somewhere.
Mark is right to say that Aa = Ed implies that there is no net transfer of heat into the atmosphere, if he means no net infrared radiative transfer of heat into the atmosphere from the land-sea surface. Aa = Ed describes radiative exchange equilibrium between the land-sea body and the atmophere. But of course there are conductive and evaporative transfers, followed by convection, of heat into the atmosphere from the land-sea surface; no problem.
But let me ask Mark, does he know of direct empirical tests of Aa = Ed, besides Miskolczi’s, or is he happy to take the TFK2009 approach that requires them in effect to be calculated as “residuals”? Direct empirical checks are the appropriate way to deal with empirical claims. Mark can really help if he can produce such a check.
“But let me ask Mark, does he know of direct empirical tests of Aa = Ed,”
I don’t know that Miskolczi has done the test either.
I *do* know that there has to be a net transfer into the atmosphere else we have a perfectly lagged earth and we’re all dead already.
“If that be the case then either the world is a lot cooler (about 7C) or the emissivities are a lot lower (~.89 as opposed to ~0.98) than we believe to be the case.”
Alex, this is why “proving” is the term used in physics.
It doesn’t mean “show it’s right” but “to test”. If the physics doesn’t agree with the reality measured, then the physics has been tested and failed.
As I said before, there may be something to the model in that powerpoint presentation, but it doesn’t seem to be that section of it.
And physics, based as it is on *utility of application* is the only place where this will be sorted out. If many citations of this work occur in papers that advance the science, then it has shown utility. If it never gets cited again, it has shown none.
So the best idea is to leave it to stand the test of time.
“Christopher’s reply:
I pointed out that I would not comment on Alex’s “sums” as he called them. But Mark reads into that that I don’t think they are wrong.”
I came to that conclusion that you stated that it would be better to use a more refined model.
If the unrefined model could be used to show Alex’s maths wrong, then that would be a distraction.
I apologise if I was wrong in assuming you were not trying to distract.
C.G. said:
“As you rightly note, this is not to be explained simply by an isolated appeal to Kirchhoff’s law of radiation.”
WHAT?
You can dis Kirchoff, or Bunsen and Stokes and Maxwell, but they were right.
Emission must equal absorption, and vice versa. What part of that is difficult to understand? The surface absorbs 161W, therefore it emits 161W. The atmosphere absorbs 78W, therefore it emits 78W. 161+78=239W, exactly the measured emission at the TOA per T&K 2008. It could not be otherwise.
You people are chasing your tails with all of this thermodynamics nonsense, the answer is found in EM theory, it is simple, it explains all the observed phenomena, and it is what makes the universe work. It even explains the surface temperature of Venus to within a couple of hundred (not several thousand) watts. But I am not going to explain it to you. [deleted comment -moderator’s note: please read the Etiquette ].
BTW, SOD, would you please stop repeating that tired old line about SW coming solely from SI and DLW coming solely from surface back”radiation” (NOT an EM term). If the 78W of SI absorbed by the atmosphere does not show up in the measured SI at the surface, and is not part of the DLW, does that mean it just DISAPPEARED? It is part of the DLW, in spite of having originated in the incoming SI. This is basic resonance line A/E behavior (spectroscopy 101). If it is not subtracted from the DLW, the values become meaningless and confusing, even if not explaining this is an oversight on your part.
reply to Mark’s posts of 2010 Jul 23 at 8:24 am.
Mark writes:
““But let me ask Mark, does he know of direct empirical tests of Aa = Ed,”
I don’t know that Miskolczi has done the test either.”
Christopher’s reply:
Mark is selectively quoting me, truncating my sentence, by leaving out my modifier phrase “besides Miskolczi’s”.
Let me thank Mark, for his candour. He says doesn’t know of a direct empirical test of Aa = Ed; I infer that he means besides Miskolczi’s, which he does not comment on at that point. We can agree that we don’t know of a direct empirical test besides Miskolczi’s.
And then Mark says he doesn’t know that Miskolczi has done the test either. Does he mean by this that he hasn’t made a serious effort to read any of Miskolczi’s papers, or what?
Mark writes:
“I *do* know that there has to be a net transfer into the atmosphere else we have a perfectly lagged earth and we’re all dead already.”
Christopher’s reply:
This makes it seem that Mark hasn’t made a serious effort to read any of Miskolczi’s papers of 2004, 2007, or 2010. It is quite clear even at an initial reading of any of those papers, that Miskolczi holds that there is a net transfer of energy from land-sea surface to atmosphere, but what Mark writes above clearly claims that Miskolczi denies this. Miskolczi shows in those papers that such net transfer is conductive, evaporative, and then convective, not radiative.
In my post above, of 2010 Jul 23 at 1:03 am, addressed to Mark, I wrote: “Mark is right to say that Aa = Ed implies that there is no net transfer of heat into the atmosphere, if he means no net infrared radiative transfer of heat into the atmosphere from the land-sea surface. Aa = Ed describes radiative exchange equilibrium between the land-sea body and the atmophere. But of course there are conductive and evaporative transfers, followed by convection, of heat into the atmosphere from the land-sea surface; no problem.” Did Mark not read that, or is it somehow unclear?
reply to Mark’s post of 2010 Jul 23 at 11:30 am.
Apology accepted.
“BTW, SOD, would you please stop repeating that tired old line about SW coming solely from SI and DLW coming solely from surface back”radiation” (NOT an EM term).”
Why do that? It’s a tired old line because it’s true.
Repeating it doesn’t make it any less true.
And just because something is not an EM term, doesn’t mean it doesn’t exist.
The problem is that the observations cannot be understood unless you accept backradiation.
The physical process that causes that phenomenon IS and EM term: radiation going the other way.
If the air has a temperature, it radiates. And since near the surface, space is hard to get to, most of it goes down to the earth, where it warms the earth just the same as sunlight does.
“I infer that he means besides Miskolczi’s,”
You infer incorrectly.
That powerpoint presentation did not measure up in any concrete form of clarity the premise it is attempting to explain.
Since it’s so poorly written, it is best to leave it to see if there is any utility in it. Which can be measured by how many papers use its contents to extend the scientific base.
The presentation is attempting to overturn nearly two centuries of work by other scientists. This is an extraordinary claim.
It requires extraordinary proof.
However, Alex’s work shows that the system presented is internally inconsistent and your attempts to explain how to make it consistent are unphysical.
This is fairly safe proof that the paper is wrong.
reply to Colin Davidson’s post of 2010 Jul 23 at 12:37 am.
Colin writes:
“My thoughts on this are below, and assume in all cases a transition from one equilibrium state to another (ie transients are completed, and an “average” KT planet is assumed):
1. The Sun drives the Surface and also provides some energy to the atmosphere.
2. The Surface drives the atmosphere.
3. The Greenhouse Effect is the change in Surface temperature which occurs because of the IR active gases.
4. The Surface air temperature is tightly bound to the temperature of the Surface. There can be no change in Surface Air temperature without a corresponding change in the Surface temperature.”
Christopher replies: Agreed.
Colin writes:
“5. The Suface energy flux balance must be maintained:
Forcings (Absorbed_Sunlight + Back_Radiation) = Consequences(Surface_Radiation + Evaporation + Conduction.)”
Christopher replies:
This is not quite right. There is no straightout law of nature that says that the Surface energy flux balance must be maintained.
It is an assumption of a steady-state model that the Surface energy balance is maintained, with the additional assumption implied that heat is not passing net energy from depths of land-and-sea to atmosphere nor the other way. Your equation expresses these assumptions.
I am very wary of the term “forcings”. It is a trademark of Hansenism. It is an expression of model assumptions, sometimes not made clear, and can be confusing. For example, changes of radiation, conduction, and evaporation can, under some circumstances, be considered as virtual drivers. Your equation would be much better without the terms ‘Forcings’ and ‘Consequences’.
No comment on the rest of Colin’s post till he writes:
“9. Christopher has stated that NET surface radiation absorbed into the atmosphere is very small, and that this was confirmed by detailed calculations by Miskolczi. (I agree with Christopher that someone should repeat these calculations before discounting them.)”
Christopher replies: Agreed.
“There is no straightout law of nature that says that the Surface energy flux balance must be maintained.”
However, to hold in abeyance the changes that keep it out of balance requires extra forcings.
Do you have any?
“I am very wary of the term “forcings”.”
How about “force”? Same thing. But what is a “thing that forces actions that would naturally occur to fail to happen” but “forcings”?
reply to Alexander Harvey’s post of 2010 Jul 23 at 12:45 am.
Thank you for your careful work here.
I have checked and agree with your results starting with OLR = 238.5 for the more refined model. I have not checked your results for the case starting with Su = 396, but I trust you have them right.
This model is a quasi-all-sky model. It is not a full model of the atmospheric energy transport process, because it does not consider variations due to cloud changes. Miskolczi 2007 on page 5 writes that, from the ERBE 2004 data product, he gets that the global average clear-sky OLR to be 266.4 W/m2, not too far from your value here of 262.08. I think this supports the idea that the discrepancies that you note are likely due to the exclusion of cloud changes from consideration.
UncertaintantyRunAmok
Perhaps I haven’t explained the subject very well.
This article is about demonstrating that “back radiation” exists. I didn’t invent the term “back radiation”, which is why I use the term DLR as much as possible in this article.
The next article on this subject will be explaining the source of the radiation.
Your comment is slightly confused but hints at an important point which probably everyone who knows that DLR exists accepts.
Solar radiation heats the atmosphere – approximately 78 W/m^2. And so does radiation from the surface.
The atmosphere itself emits longwave radiation. This term – “longwave” – is just a conventional description, not some special type of radiation. It simply means radiation greater than 4um.
The reason that atmosphere emits 99.9% of its radiation at greater than 4um is because of its temperature. It doesn’t matter whether the atmosphere in turn was heated by a body at 300K or 3000K or 3000000000K – once the atmosphere is at a temperature around a few hundred K it radiates mostly above 4um.
DLR is emitted by the atmosphere – but the source of the energy was both shortwave from solar radiation and longwave from terrestrial radiation. In turn, the source of the energy that generated terrestrial radiation was also solar. That doesn’t stop the DLR being longwave.
It is important to be clear about these basic points.
So I haven’t claimed that the solar radiation heating the atmosphere has disappeared. If there’s something I have written that has led you to think that, please let me know and I will see if I can make it clearer.
reply to UncertaintyRunAmok’s post of 2010 Jul 23 at 8:50 pm.
UncertaintyRunAmok writes:
“C.G. said:
“As you rightly note, this is not to be explained simply by an isolated appeal to Kirchhoff’s law of radiation.”
WHAT?
You can dis Kirchoff, or Bunsen and Stokes and Maxwell, but they were right.”
Christopher replies:
Dear UncertaintyRunAmok, you have misread or misunderstood what I wrote. I did not write that Kirchhoff, Bunsen, Stokes, or Maxwell were wrong, nor did I dis them. Of course the factors described by them come into this in important ways. I just said that there were other factors besides theirs that come into this. Evaporation and convection for example. I think you did not read carefully what I wrote.
No comment on the rest of your post.
reply to Mark’s post of 2010 Jul 23 at 9:46 pm.
No comment.
reply to Mark’s post of 2010 Jul 23 at 9:51 pm.
No comment.
Morris Minor from July 22, 2010 at 9:56 pm:
So I think what you are claiming is that it would be impossible to measure if it did exist.
What spectral shape would you expect of this radiation that isn’t really DLR?
If it is actually radiation emitted from the surface of the earth then you will have something that closely follows the Planck function, like these examples:
I look forward to finding out your views before I post the next article with the answer.
SoD asks, ‘What spectral shape would you expect of this radiation that isn’t really DLR?’
The spectral shape I would expect would be what you get if you subtract the pink line from the blue line. Due to the similarity in temperatures there will a low level radiation emitted from the source of the blue line (+10C emitter).
What happens to the pink radiation? (I guess that question is coming). Does it just disappear? I don’t know. But it can’t be observed or measured. We can only observe the net result (which is the spectral shape in my answer above).
See also my water tank model below. Here it is clear that the flow from the low level tank can no longer be seen or measured.
Well, watch out for the next article on this subject where the spectrum of the “it’s not really DLR” will be revealed.
Some further thoughts on back-radiation, consider two tanks of water, one with a low water level, H1, the other tank with a higher water level H2.
The simplified flow rate from a pipe at the base of the tank is given by:-
F = k.H
Where:
k is a constant which allows for radius and length of pipe, viscosity etc.
H is the Head or water level above our reference point (the ground).
The flow from Tank1 to ground would be F1 = k*H1 and the flow from Tank2 to ground:- F2 = k*H2.
What would happen if we joined these tanks together by attaching both pipes to the other tank? According to the thinking on this site, F1 would flow from Tank1 into Tank2 and F2 would flow from Tank2 into Tank1 simultaneously.
If we put a flow meter in the pipe we could measure the flow between the tanks which we can call Fr. This is our water flow “pyrgeometer”. If we know the level of Tank2 (H2) and ‘k’ then we can use the equation: F2 = Fr – k.H1 to find the flow from the lower tank to the higher!
This is how the pyrgeometer obtains a value for back-radiation. F2 is a measure of ‘backflow’, analogous to back radiation. Does backflow really exist?
And to follow the reasoning on this site in regard to the imaginary second law, flow from the lower tank would be able to increase the level of the higher tank!
Christopher Game wrote (0946, 23Jul):
“Colin writes:
“5. The Suface energy flux balance must be maintained:
Forcings (Absorbed_Sunlight + Back_Radiation) = Consequences(Surface_Radiation + Evaporation + Conduction.)”
Christopher replies:
This is not quite right. There is no straightout law of nature that says that the Surface energy flux balance must be maintained.
It is an assumption of a steady-state model that the Surface energy balance is maintained, with the additional assumption implied that heat is not passing net energy from depths of land-and-sea to atmosphere nor the other way. Your equation expresses these assumptions.
I am very wary of the term “forcings”. It is a trademark of Hansenism. It is an expression of model assumptions, sometimes not made clear, and can be confusing. For example, changes of radiation, conduction, and evaporation can, under some circumstances, be considered as virtual drivers. Your equation would be much better without the terms ‘Forcings’ and ‘Consequences’. ”
I would like to thank Chris for his comments.
At the beginning of my post I wrote:
[I]”assume in all cases a transition from one equilibrium state to another (ie transients are completed, and an “average” KT planet is assumed)…”
in other words CHristopher is correct to say that the equation assumes there are no land/sea/ice/oceandepth exchanges.
If the system is in equilibrium (my assumption) then the equation has to balance (a consequence of the assumption).
I agree with Christopher that the terms “forcings” and “feedbacks” are rather strange. Apparently the Surface Radiation is a “feedback” but the Back Radiation is not! But what is true is the statement of the Surface Flux Balance in the form:
Absorbed_Sunlight + Back_Radiation =
Surface_Radiation + Evaporation + Conduction.
And if we assume Conduction to be a constant term (this seems reasonable – what we are assuming here is that if the planet changed temperature the same temperature difference between surface and atmosphere would be maintained) then the Surface temperature is constrained by the surface response to “forcing” changes. This response is quite weak, and highly sensitive to the evaporative term.
If you choose a weak evaporative change, then you get a surface which is twice as responsive as one with a strong change. The “IPCC” modellers I think choose a weak change, but a strong (Clausius-Clapeyron) Water Vapour change. In this way they can have a huge “Feedback” from added water vapour, without “paying” for it with increased clouds and decreased Surface sensitivity.
I’m not saying that is wrong, but instead of a lot of obsfuscation (try getting this fact from the IPCC WG1 report!) it would be nice to see it highlighted as an issue, and some justification for the very low level of change of evaporation used in the models.
Morris Minor wrote at 0247, 24Jul:
“Some further thoughts on back-radiation, consider two tanks of water, one with a low water level, H1, the other tank with a higher water level H2.
The simplified flow rate from a pipe at the base of the tank is given by:-
F = k.H
Where:
k is a constant which allows for radius and length of pipe, viscosity etc.
H is the Head or water level above our reference point (the ground).
The flow from Tank1 to ground would be F1 = k*H1 and the flow from Tank2 to ground:- F2 = k*H2.
What would happen if we joined these tanks together by attaching both pipes to the other tank? According to the thinking on this site, F1 would flow from Tank1 into Tank2 and F2 would flow from Tank2 into Tank1 simultaneously. ”
I would like to thank Morris Minor (great moniker!) for his thoughts, but I don’t agree that in this thread an energy flow is assumed from the weaker to the stronger source. As far as I can see in this thread it is accepted that the NET flow from the stronger to the weaker is the result.
So I think we all agree with MM’s point. I do.
reply to Colin Davidson’s post of 2010 Jul 24 at 3:10 am.
Colin writes:
“I’m not saying that is wrong, but instead of a lot of obsfuscation (try getting this fact from the IPCC WG1 report!) it would be nice to see it highlighted as an issue, and some justification for the very low level of change of evaporation used in the models.”
You are letting yourself be lured into thinking in Hansenese. This is not a good language for scientific analysis of the climate process. It has built-in distortions of understanding. It is built around the Bode 1945 theory of amplifier design (see for example Hansen, J., Lacis A., Rind, D., Russell, G., Stone, P., Fung, I., Ruedy, R., Lerner, J. (1984) Climate Sensitivity: Analysis of Feedback Mechanisms, pages 130-163 in ‘Climate Processes and Climate Sensitivity’, of the Maurice Ewing Series, Geophysical Monograph, vol. 5 no. 29, ISBN 0875904041, or Schlesinger, M.E. (1986) Equilibrium and transient climatic warming induced by increased atmospheric CO2, ‘Climate Dynamics’ 1: 35-51). The Bode theory is very suitable for its purpose, the design of negative feedback amplifiers, but it is not a suitable theory for the physics of climate. Its main attribute is its ability to confuse and baffle and mislead, which of course is a virtue in some eyes. Better to try the methods of physics for a physical problem. For a discussion of some aspects of the Hansenist formalism, see Kimoto, K., (2009) On the confusion of Planck feedback parameters, ‘Energy and Environment’ 20(7): 1057-1066.
Further to Morris Minor’s analogy, the true situation is analagous to two water tanks passing water to the ground, and their respective flows then being magically returned into the top of the other tank.
In that case there would be flow from the lower to the higfher tank, and flow from the higher to the lower tank, and the result would be the same as if the tanks were joined together. In that latter case however, there is only flow in one direction.
In the case we are considering, there is flow in both directions. But the result is the same as if the flow was unidirectional.
reply to Colin Davidson’s post of 2010 Jul 24 at 3:28 am.
Colin is talking about a hydrodynamic model for the study of radiative processes. This is very often not a safe plan, sad to say, because while in hydrodynamics, local mechanisms are governed by ‘local’ variables, in radiation dynamics, local mechanisms depend not only on ‘local’ but also on ‘remote’ variables; that is the essence of the notion of radiation, as distinct from ponderable material processes.
Christopher Game kindly responded to my post at 0328,24Jul stating in part:
“You are letting yourself be lured into thinking in Hansenese. This is not a good language for scientific analysis of the climate process. It has built-in distortions of understanding. It is built around the Bode 1945 theory of amplifier design (see for example Hansen, J., Lacis A., Rind, D., Russell, G., Stone, P., Fung, I., Ruedy, R., Lerner, J. (1984) Climate Sensitivity: Analysis of Feedback Mechanisms, pages 130-163 in ‘Climate Processes and Climate Sensitivity’, of the Maurice Ewing Series, Geophysical Monograph, vol. 5 no. 29, ISBN 0875904041, or Schlesinger, M.E. (1986) Equilibrium and transient climatic warming induced by increased atmospheric CO2, ‘Climate Dynamics’ 1: 35-51). The Bode theory is very suitable for its purpose, the design of negative feedback amplifiers, but it is not a suitable theory for the physics of climate. Its main attribute is its ability to confuse and baffle and mislead, which of course is a virtue in some eyes.”
My response is straying off-topic (Back Radiation), so I hope SOD will forgive me. I’m not sure what Christopher is objecting to in my post, which was essentially about the role of evaporation in the Greenhouse.
With regards to whether classical Control Theory is a useful tool in analysing the climate, I am not expert. But I recognise that there may well be multi-order differential equations in a model which seeks to emulate the climate-weather system. I agree that a simple Bode model will not be adequate to explain the full dynamic behaviour of a complex system, a system that many have described as being non-deterministic, due to the impossibility of describing the starting conditions accurately and to sufficient detail.
Part of this problem is “What is Climate?” Is it Average Weather – and if so over what timescale? Years? Decades? Centuries? The last Hurricane? If the timescale is long enough do the chaotic system behaviours average out into a KT style model? Or is the steady state instead some form of multi-sinusoid?
reply to Colin Davidson’s post of 2010 Jul 24 at 4:06 am.
Thank you Colin for your moderate reply. Yes, I was a bit unnecessarily zealously flag-waving in my comment of 2010 Jul 24 at 3:10 pm. There wasn’t anything to really object to in your post of 2010 Jul 24 at 3:10 am to which I was replying.
I think the important thing is to try to clearly specify the physical model, preferably including a flow diagram of some kind, in terms of which one is thinking, on each occasion of discussion. I was just waving a flag that (as you perhaps might partly seem to agree?) the Bode theory, beloved of the Hansen-IPCC gang, is often harmful when misused as they misuse it. I am not too sure exactly what constitutes “Classical control theory”; I think ‘control theory’ is mostly an engineering term. I am happy to speak of dynamical systems theory, and for simple models, as you note, a system of ordinary differential equations can be helpful. I agree with your comments after that.
further reply to Colin Davidson’s post of 2010 Jul 24 at 4:06 am.
Colin, I stepped back a too far. I do have something to say about your post of 2010 Jul 23 at 12:37 am. The latter was the target of my shot about Hansenese.
In my terms, your post of 2010 Jul 23 at 12:37 am was spoken in Hansenese, and the accompanying thinking was consequently in terms of Hansenism. The key here is that your model, though perhaps not explicitly stated, was implicitly pivoted on the land-sea surface temperature. You wrote:
“6. Note that the ONLY way to change the Surface Temperature is …” and
“7. The surface is very insensitive to a change in forcing …” and
“8. To get this the models assume ….”
This is leading with your chin. It is accepting the Hansenist paradigm, that surface temperture is the pivotal variable around which to construct one’s model; the surface temperature is the only one that is sensed for all feedbacks in the Hansenist paradigm; that is how they trick people into misusing the inappropriate Bode theory.
Use of the Hansenist paradigm, pivoted on feedback sensed only from the land-sea surface temeperature, is an example of what the mediaeval scholars called ‘petitio principii’: asking the principal question too early in the discussion, before the proper analytic and investigative groundwork has been done (I learnt of this historical usage of the term by reading the scholarly work of Jaakko Hintikka). This error of reasoning skews the thinking entirely, and practically give the game away to the Hansen-IPCC gang before the discussion even starts.
True, the focus of interest is in the outcome for the land-sea surface temperature. True, if the earth’s energy transport process were an engineering problem, one would give a pivotal place to the focus of interest. But this is physics, not engineering, no matter how realistically oriented and practically beneficial is the valuable art of engineering. Science is not quite the same as engineering, though the cunning misuse of the Bode theory by Hansesism might lead one to think otherwise.
Physically thinking, the earth’e energy transport process is about flows of energy. Perhaps, for example, and this is only an example for the sake of my present comments, perhaps, for example, the best model of the process is in terms of the circulations that carry energy from the equator to the poles, or from the land-sea surface to the upper troposphere. It is likely then that the best model will take as pivotal, not as the Hansenist paradigm does, the land-sea surface temperature, but, rather, the best model will take as pivotal one of the variables that more directly describe that circulatory transport, though I think that the best simple model will likely need at least two pivotal variables.
The desired information about the land-sea surface temperature can be derived as an output of the best model, but history seems to show that it is not the variable about which to pivot the very dynamics of one’s model. The gang has been using the Hansensist paradigm since perhaps 1984 or so, making no useful progress in understanding since then, but working a treat for them in the sense that it has misled the world and blocked the progress of understanding all that time.
So my comment was intended to draw your attention to the dangers of the mistake of accepting the Hansenist paradigm as a result of speaking Hansenese. I am saying that we need to use a physically conceived model, not an ad hockery such as Hansenism.
I would like to thank Christopher Game for his most interesting posts, and the courteous manner in which he has written them.
While reading his last post, I realised that I had slipped into error in the wording of my post at 1237, 23JUL.
My post is only correct if it is viewed as an attempt to describe a new KT diagram at a new higher temperature.
If you try to redraw the diagram for 3DegC higher, then the only way to achieve a surface energy flux balance (a balance which is forced upon us if we are describing a ficticious average equilibrium planet) is to raise the amount of sunlight absorbed by the surface and/or the back radiation. And that amount is dependant to a great extent on the amount of increased evaporation you pick. If you pick low, you get half the increase compared to when you pick high.
Is this a valid approach? I think it is because it teaches us several things about the real world:
1. The rate of evaporation is very important in the determination of the surface temperature.
2. The rate of change of evaporation with surface temperature is very important in the determination of the temperature response to changes in albedo, cloud cover, the sun and back radiation.
3. The surface is nowhere near as responsive to such changes as the atmosphere because:
a. It is hotter, so more energy is required to heat it up, and
b. energy is expended in evaporating water.
4. If there is an increase in back radiation, the NET radiation to the atmosphere DROPS as the evaporation increases, the outward energy from the surface always being equal to the absorbed sunlight (as required by this trivially simple model).
Based on this simple model, which can be described as a sanity check on the far more complex models in present use, I again ask the question:
The simple model shows that a sustained large temperature change due to changes in energy reaching the surface is unlikely unless a very low sensitivity is assumed for evaporation while a high sensitivity is assumed for water vapour content. What is the justification for this?
Responding to Christopher Game’s post at 2135, 24 July, on my assumption of a Hansenite paradigm.
This was a most interesting post which gave me great food for thought (discomforting but necessary!)
I agree that the terms of the debate and my part in it are being driven by one paradigm: CO2 causes “forcing” at the Tropopause which is manifested as a temperature increase at the surface, with the alarming subtext that this is all bad for us.
I also agree that there are other ways to look at the system, for instance the work of Spencer is of very great interest as he has made an attempt to measure the system using parameters which are not surface temperature.
And there is also the work of Miskolczi, central to which (as Christopher rightly points out) is the hypothesis that Ed =Aa (Net Surface radiation into the atmposphere is zero).
reply to Colin Davidson’s post of 2010 Jul 24 at 10:41 pm.
Thank you Colin.
Colin writes:
“If you try to redraw the diagram for 3DegC higher, then the only way to achieve a surface energy flux balance (a balance which is forced upon us if we are describing a ficticious average equilibrium planet) is to raise the amount of sunlight absorbed by the surface and/or the back radiation. And that amount is dependant to a great extent on the amount of increased evaporation you pick. If you pick low, you get half the increase compared to when you pick high.”
Christopher replies:
There is a third factor in the surface energy balance, non-material transport in the form of conduction, evaporation, and subsequent convection, which appears in your second sentence but unaccountably not in your first sentence above. I think I understand you aright, that the evaporation occurs in the diagram that you refer to.
Colin writes:
“Is this a valid approach? I think it is because it teaches us several things about the real world:
1. The rate of evaporation is very important in the determination of the surface temperature.”
Christopher agrees.
Colin writes:
“2. The rate of change of evaporation with surface temperature is very important in the determination of the temperature response to changes in albedo, cloud cover, the sun and back radiation.”
Christopher replies:
This is still Hansenist thinking. It assumes that surface temperature is the pivotal determinant of rate of evaporation, and fails to consider wind.
Colin writes:
“3. The surface is nowhere near as responsive to such changes as the atmosphere because:
a. It is hotter, so more energy is required to heat it up”
Christpher replies:
A loose statement, no comment.
Colin writes:
“, and
b. energy is expended in evaporating water.”
Christopher replies:
In the evaporation of water, energy is carried from land-sea surface into the atmosphere.
Colin writes:
“4. If there is an increase in back radiation, the NET radiation to the atmosphere DROPS as the evaporation increases, the outward energy from the surface always being equal to the absorbed sunlight (as required by this trivially simple model).”
Christopher replies:
I think this is misworded. Do you mean that ‘the NET energy transport to the atmosphere DROPS’? No further comment at this point.
Colin writes:
“Based on this simple model, which can be described as a sanity check on the far more complex models in present use, I again ask the question:
The simple model shows that a sustained large temperature change due to changes in energy reaching the surface is unlikely unless a very low sensitivity is assumed for evaporation while a high sensitivity is assumed for water vapour content. What is the justification for this?”
Christopher replies:
This is still Hansenist thinking. It is written around the pivotal concept of temperature sensitivity. Who cares about trying to justify the Hansenist picture? It has failed for about a quarter of a century. Better to make a fresh physical model with emphasis perhaps on circulations.
replying to Colin Davidson’s post of 2010 Jul 24 at 11:00 pm.
Thank you Colin.
By the Hansen-IPCC paradigm I mean a model in which the land-sea surface temperature is the one and only sensed variable that governs feedback. The metaphor of “forcing” at the tropopause is perhaps part of that paradigm too, but note that it is only a metaphor for a supposed indirect effect, not an actual fact.
Actually, physically, CO2 potentially directly contributes to the true greenhouse-gas optical thickness. That is how it is supposed to work, immediately and directly. The next step is to find out how much it does so. Miskolczi 2010 shows that the historical answer, over the past 61 years linear trend, is nil. Who is interested in something that hasn’t actually happeed?
Christopher Game wrote (0001, 24Jul):
“There is a third factor in the surface energy balance, non-material transport in the form of conduction, evaporation, and subsequent convection, which appears in your second sentence but unaccountably not in your first sentence above. I think I understand you aright, that the evaporation occurs in the diagram that you refer to.”
The omission of convection was deliberate. Convection is an energy transport mechanism which is totally within the atmosphere. The surface only tranfers energy to the atmosphere radiantly, convectively or materially (as evaporated water molecules).
Christopher Game wrote (0001,24Jul):
“Colin writes:
“2. The rate of change of evaporation with surface temperature is very important in the determination of the temperature response to changes in albedo, cloud cover, the sun and back radiation.”
Christopher replies:
This is still Hansenist thinking. It assumes that surface temperature is the pivotal determinant of rate of evaporation, and fails to consider wind.”
It may be Hansonist thinking (I wouldn’t know!). But the calculation inherently captures wind as the KT diagram is for an “average” planet with average surface at an average temperature with average humidity and average wind (by definition).
So whether or not it is Hansonist, I think the statement is actually correct.
I must thank Christopher for the time he has taken to respond to my comments. He wrote at 0001, 25Jul:
“Colin writes:
“3. The surface is nowhere near as responsive to such changes as the atmosphere because:
a. It is hotter, so more energy is required to heat it up”
Christpher replies:
A loose statement, no comment.”
I’m sorry not to have been more explicit, and agree it was an extremely loose statement.
It takes relatively more energy to MAINTAIN the temperature of a hot body than a cold one, as the radiation from a body is sT^4 W/m^2.
For the top of the Troposphere, at about -55DegC, the radiant energy loss is 128W/m^2 *emissivity (which is much less than 1, probably around 0.3 to 0.4 as there isn’t any water vapour). If we increased the atmospheric temperature here by 1 DegC, it would only require additional energy input of around 1W/m^2/DegC to MAINTAIN that temperature.
For the planetary surface things are quite different. Emissivity is about 1. If we increase the temperature here, and if radiation were the only heat loss mechanism, it would require in excess of 5W/m^2 additional energy input to MAINTAIN the new temperature.
Christopher Game wrote:
“Colin writes:
“4. If there is an increase in back radiation, the NET radiation to the atmosphere DROPS as the evaporation increases, the outward energy from the surface always being equal to the absorbed sunlight (as required by this trivially simple model).”
Christopher replies:
I think this is misworded. Do you mean that ‘the NET energy transport to the atmosphere DROPS’? No further comment at this point.”
No.
The Surface Balance can be rewritten as:
Absorbed Sunlight = (Surface Radiation – Back radiation) + Evaporation + Conduction , or
Absorbed Sunlight = NET Radiation + Evaporation + Conduction
If the system heats up, and the Evaporative term increases, with Conduction and Absorbed Sunlight assumed constant, then NET Radiation drops.
This is a different way of saying that an increase in either Absorbed Sunlight or Back Radiation is required to MAINTAIN the surface at an elevated temperature – not just an increase to cover the increased Surface Radiation, but also the increased evaporation.
Christopher Game wrote (0001,25JUL):
“Colin writes:
“Based on this simple model, which can be described as a sanity check on the far more complex models in present use, I again ask the question:
The simple model shows that a sustained large temperature change due to changes in energy reaching the surface is unlikely unless a very low sensitivity is assumed for evaporation while a high sensitivity is assumed for water vapour content. What is the justification for this?”
Christopher replies:
This is still Hansenist thinking. It is written around the pivotal concept of temperature sensitivity. Who cares about trying to justify the Hansenist picture? It has failed for about a quarter of a century. Better to make a fresh physical model with emphasis perhaps on circulations.”
My question is a very reasonable one – I’m trying to understand the “Hansonite” physics, and I hope I have shown that this is not a trivial point. I would think that the difference in “Hansonite” physics between a large evaporative change and a small one is the difference between 0.5 DegC and 3DegC. (Not a minor chinklet – and I don’t see the armour plate. However, no doubt SOD will have some Kevlar secreted under his surcoat…)
However I agree that fresh viewpoints and models are necessary and applaud your championship of that cause.
This post comes as something of a let-down following the hugely informative and civil conversation between Colin and Christopher. Thank you, gentlemen!
Sod has produced a variety of DLR data as impregnable proof of back-radiation spontaneously warming the surface, relegating the 2nd law of thermodynamics to the imaginations of a few deluded souls. It ain’t quite so simple. The measurements tell us nothing about the source of the DLR. The 2nd law tells us that it must be warmer than the receptor. This, in fact, is what wintering researches at the South Pole found where temperature inversion prevailed. No conflict with the 2nd law there. Sod notes only a small drop in DLR between day and night in one data-set, explaining that at night the atmosphere cools more slowly than the surface. Is temperature inversion the nocturnal norm? In general, the domed receptors could be expected to capture lateral radiation from a near-by natural or man-made feature (taking N,S,E,W,uP and Dn as the cardinal directions in space, four fifths of surface radiation and two-thirds of atmospheric radiation begin laterally). Again, no conflict. Nor would there be conflict were the receptor capturing surface radiation reflected by the atmosphere.
John Millett:
This article is just the first step in the process. First of all many people have questioned whether the DLR exists and suggested that it is zero or mW/m^2 for example.
This certainly hasn’t yet demonstrated what the radiation does or doesn’t do, just that it exists.
Part Two covers more about the source of the radiation.
Perhaps you can comment on the source of the radiation. It might be very interesting.
And perhaps you can explain how the atmosphere stops radiating when the surface is warmer and starts radiating when the surface is cooler.
Other believers in the Imaginary Second Law of Thermodynamics think that the atmospheric radiation “reaches” the surface, but then are vague about what happens next. One of our most illustrious commenters for a while claimed that it was “mostly reflected”.
The idea that gases have some kind of mechanism for only radiating towards colder surfaces is quite fascinating.
I really hope you flesh this one out some more, if only to help others understand the strength, or otherwise, of your argument.
And perhaps you can comment on this diagram from Fundamentals of Heat and Mass Transfer, 6th edition, Incropera and DeWitt:
Do Incropera and DeWitt not understand the 2nd law of thermodynamics?
“And perhaps you can explain how the atmosphere stops radiating when the surface is warmer and starts radiating when the surface is cooler.”
Why would the intensity of atmospheric radiation depend on the temperature of the surface? The issue that needs to be resolved is why absorption and thermalisation at the surface occurs at temperatures lower than the atmosphere’s but not at higher ones. I note that the textbook excerpt says that the warmer body would cool. It doesn’t say that it would initially get warmer, as a result of absorbing radiation from its colder surroundings, and then cool down.
reply to Colin Davidsons’s post of 2010 Jul 25 at 3:51am
typo for conductively.
reply to Colin’s post at 3:56 am.
Your reply now exemplifies the evil character of Hansenism. You write:
“Christopher Game wrote (0001,24Jul):
“Colin writes:
“2. The rate of change of evaporation with surface temperature is very important in the determination of the temperature response to changes in albedo, cloud cover, the sun and back radiation.”
Christopher replies:
This is still Hansenist thinking. It assumes that surface temperature is the pivotal determinant of rate of evaporation, and fails to consider wind.”
It may be Hansonist thinking (I wouldn’t know!). But the calculation inherently captures wind as the KT diagram is for an “average” planet with average surface at an average temperature with average humidity and average wind (by definition).
So whether or not it is Hansonist, I think the statement is actually correct.” ”
Christopher here replies:
Hansen is so spelt.
I just ask our very civil commenters to please stay within the etiquette.
We may mock others for poor science but attributing motives and evil intent is not in the spirit of this blog. Adding -ist doesn’t really avoid the issue.
Condemning poor science is fine, you just have to stay with that approach.
continuing my reply to Colins post oat 3:51 am (I pressed the enter key at he wrong moment)
Christopher here replies:
Hansen is so spelt.
Wind is very important in determining the rate of evaporation and you did not mention it, mentioning only the beloved pivot of Hansenism. I am not saying that temeperature is not important. I am saying that Hansenism leads to distortions of emphasis. Literally, what you wrote is not false, but it focuses on a model which I think is confusing and dangerous. Your post was saying that your calculation, on a model that you do not explicitly specify, implicitly captures the wind; you say “with … average wind” when talking about change of temperature; do you mean that the wind does or doesn’t change when the temperature changes? This is a good example of how Hansenist assumptions lead to muddle, whether literally “correct” or not. I was concerned with the use of the paradigm as a source of muddle.
reply to Colin’s post at 4:16 am.
Back of the envelope calculations. No comment.
reply to Colin’s post at 4:24 am.
Was absorbed sunlight assumed constant and was conduction assumed constant? I forget the thread.
reply to Colin’s post at 4:36 am.
Colin writes”My question is a very reasonable one – I’m trying to understand the “Hansonite” physics, and I hope I have shown that this is not a trivial point.”
Christopher replies:
It is a waste of time trying to understand the Hansen-IPCC paradigm because it is not physics. You can waste your entire life struggling with nonsense trying to make sense of it; that’s how they keep the upper hand. Better to spend your valuable time and valuable intellect by starting afresh with a really physical model.
reply to scienceofdoom’s post of 2010 Jul 25 at 8:10 am.
Thank you for your kind warning. Message received.
reply to Jon Millett’s post of 2010 Jul 25 at 5:52 am.
John Millett writes:
“The 2nd law tells us that it must be warmer than the receptor.”
Christopher’s reply.
Radiative exchange was proven by Prevost in 1791. The difference between radiation and conduction is that radiation travels to remote places with some of it passing right through and not interacting thermally with the media in its path. The conduction model doesn’t work for radiation, and implicitly you are thinking in terms of a conduction model which has a successful local variable theory with no need to consider remote variables. More explicitly, you are confusing net transfer which is governed by the second law in appropriate conditions, on the one hand, with a picture of exchange with two-way transfer on the other. DLR is one member of a two-way exchange and is not directly and simply accounted for by the second law. Recall the precise statement of the second law. I suggest you read Prevost 1791 and Balfour Stewart 1858.
I have never seen a definition or description of thermodynamic laws which specifically exempts radiative heat transport. Does Prevost or Stewart provide one? Nor have I seen one that mentions “net heat”, an accounting term, only the physical property “heat’.
further comments on the second law of thermodynamics in his context
Classically the second law is about entropy. Entropy is classically defined only for systems in thermodynamic equilbrium.
The earth’s energy transport process is not an equilibrium system. To use the second law for systems not in equilibrium is a far far far more difficult thing than to use it for systems in equilibrium. The very definition of entropy for non-equilibrium systems is very very difficult. If you find someone who thinks it is easy, you have found someone who doesn’t understand the difficulties.
“Entropy is classically defined only for systems in thermodynamic equilbrium.”
Sorry, entropy exists out of equilibrium.
s=k Log(omega).
Entropy is the number of ways a system outwardly identical to the one under investigation can be arranged.
It has naff all to do with equilibrium, thermal or otherwise.
“The very definition of entropy for non-equilibrium systems is very very difficult.”
No, I just did it up there
s=k Log(omega)
reply to Mark’s post of 2010 Jul 25 at 6:10 am.
Dear Mark, I suggest you read a good textbook about entropy for non-equilibrium systems. A good start would be with Walter T. Grandy Jr’s 2008 ‘Entropy and the Time Evolution of Macroscopic Systems’, Oxford University Press, Oxford, ISBN 9780199546176.
oops, I should have written Mark’s post of 2010 Jul 25 at 9:58 am, not at 6:10 am as I did.
“Dear Mark, I suggest you read a good textbook about entropy ”
Dear Chrisopher, thank you kindly for that advice which has been appropriated actioned.
Please also consider for yourself the equation of entropy available from any good 19th Century textbook or later. Within you will find the definition of entropy wherein there lies not one jot of thermal equilibrium, just the ways of ordering things so they have the same outcomes.
Such investigation would prove most helpful for your future postings on the subject.
Thank you all for giving me a valuable blog encounter. I have some very pressing business that I must attend to this week and it will deprive me of the leisure to blog over the week.
Dear Mark, in future I will not be posting much about entropy because it is too difficult for me in this context.
Christopher Game wrote at 0825, 25Jul:
“Christopher here replies:
Hansen is so spelt.
Wind is very important in determining the rate of evaporation and you did not mention it, mentioning only the beloved pivot of Hansenism. I am not saying that temeperature is not important. I am saying that Hansenism leads to distortions of emphasis. Literally, what you wrote is not false, but it focuses on a model which I think is confusing and dangerous. Your post was saying that your calculation, on a model that you do not explicitly specify, implicitly captures the wind; you say “with … average wind” when talking about change of temperature; do you mean that the wind does or doesn’t change when the temperature changes? This is a good example of how Hansenist assumptions lead to muddle, whether literally “correct” or not. I was concerned with the use of the paradigm as a source of muddle.
reply to Colin’s post at 4:16 am.
Back of the envelope calculations. No comment.
reply to Colin’s post at 4:24 am.
Was absorbed sunlight assumed constant and was conduction assumed constant? I forget the thread.”
I would like to thank Chris for these comments.
At the start of my original post, and subsequently, I indicated I was working with a Kiehl-Trenberth world. This is a static, “average” world, in equilibrium. It is indeed a strange world. It has average day/night. It has average Latitude and Season. It has average land/sea surface. It has average cloud and average wind. It is quite valid to attack this extremely simple model as being too silly, nevertheless I think it can be instructive, particularly to apply sensibility tests to climate predictions.
I then moved the equilibrium, by increasing surface temperature. Implicitly I kept the absorbed sunlight the same. I stated my assumption of an unchanged conductive term, on what seem to be conservative grounds.
What I found was that to maintain a 3 DegC increase in the KT world, you need a very high level of back-radiation. An equilibrium at this new temperature can only be maintained if a very low change in the evaporation rate (1% per DegC) but a high change in the amount of water vapour content (6.5% per DegC, which you get by assuming a constant Relative Humidity) in the atmosphere are assumed.
If instead you use a high change of evaporation rate (6.5% per DegC) then the temperature increase which can be maintained is only 0.5DegC. This highlights an area which needs some careful and detailed explanation from those claiming a large temperature increase for a doubling of CO2.
Also implicit in the model is that if the surface temperature is increased, the NET radiation into the atmosphere decreases by the same amount that the evaporation rate increases. Put another way, the increase in back radiation must not only balance the increased surface radiation but also the increased evaporation.
Christopher doesn’t like the model. Fair enough.
Christopher objected to my omission of wind as a major factor in the change of evaporation rate. He is of course correct that wind is a major influence and that if wind changes with temperature the evaporation rate will be severly affected. In the KT world there is no place for wind, which is also assumed constant.
I don’t have figures for the change of global wind with temperature. But there are recent (as always, disputed!)numbers for evaporation rate change, suggesting that it is closer to 6.5% per DegC than 1% per degC.
Complex isn’t it? The following are interesting:
1. Relative Humidity. Does it change with temperature? Why?/What data exists?
2. Wind. Does it change with temperature? and where? Why?/What data exists?
Finally I apologise for getting Dr Hansen’s name wrong. (I agree with SOD about labelling third parties).
John Millett:
I said: “And perhaps you can explain how the atmosphere stops radiating when the surface is warmer and starts radiating when the surface is cooler.”
John Millett responded:
That was my point, it was what you appeared to be saying earlier.
So we can conclude that, in your opinion, the atmosphere does radiate towards the surface regardless of the surface temperature?
And also in your opinion, this radiation reaches the surface?
So you also agree that the radiation from the colder body is absorbed by the hotter body?
This is what the textbook shows. Do you agree with it?
Yes, radiation leaves the atmosphere directed towards the surface.
It may reach the surface if it is not absorbed or reflected in the intervening atmosphere in its path.
It may be absorbed at the surface if it is not reflected there.
It may be thermalised at the surface if it has the energy to push an electron into a higher energy state.
The excerpt from the textbook doesn’t say that the body absorbs all of the radiation from the surroundings. It leaves open the thermalising question by not qualifying the rate of surface cooling as being slower than it otherwise would have been in the absence of atmospheric radiation.
Apart from that I can’t be as categorical as you would like.
“What I found was that to maintain a 3 DegC increase in the KT world, you need a very high level of back-radiation.”
All you need is time and an imbalance.
A perfectly insulated leaden block will melt with even 3W of heating power supplied.
In the case of the earth, all you’d need is a small increase in the opacity of the atmosphere.
If the atmosphere is 50% opaque, then you have a sum limit of 2x incident solar as feedback.
If the atmosphere is 66% opaque, then you have a sum limit of 3x incident solar as feedback. For +16%.
At 75% 4x. For +9%
At 80% 5x. For +5%
At 83.3% 6x. For +3.3%
Mark posted at 1012, 26 Jul:
““What I found was that to maintain a 3 DegC increase in the KT world, you need a very high level of back-radiation.”
All you need is time and an imbalance.
A perfectly insulated leaden block will melt with even 3W of heating power supplied.
In the case of the earth, all you’d need is a small increase in the opacity of the atmosphere.
If the atmosphere is 50% opaque, then you have a sum limit of 2x incident solar as feedback.
If the atmosphere is 66% opaque, then you have a sum limit of 3x incident solar as feedback. For +16%.
At 75% 4x. For +9%
At 80% 5x. For +5%
At 83.3% 6x. For +3.3%”
Perhaps my original phrasing was (again) inexact.
We are not dealing with the transition from one state to another, but with two stable equilibrium states.
If the surface temperature in a KT world is 3 Degrees higher than now, ie 18DegC, then the back radiation needs to be significantly higher to MAINTAIN the surface at this new temperature (otherwise the system will not be in equilibrium but in transition.)
What Mark has described is TRANSITION between states, not the EQUILIBRIUM state finally reached. He is of course quite right that a small imbalance applied for a long enough period can have a large effect.
“What Mark has described is TRANSITION between states, not the EQUILIBRIUM ”
No, what Mark has shows is the effect of a positive feedback.
Planet warms.
Planet warms atmosphere.
Atmosphere warms planet.
Planet warms more.
Planet warms atmosphere more.
Atmosphere warms planet more.
And round and round we go.
As long as the atmosphere is less than 100% opaque (perfect insulation), there is a feedback.
The process toward that endpoint takes time because of the heat capacity of the system. But the endpoint listed above IS the equilibrium point, not the transition.
The point I’m trying to get across and I’m worried your phraseology is going to confuse is that it isn’t that there’s generation of energy in the atmosphere, but that it is merely holding on to it longer.
A large bucket will hold more water than a small one, even if you have the same rate of water pouring in.
This doesn’t mean the bucket is creating water.
I apologise but I don’t understand Mark’s posts at 1331 and 1333, 26Jul.
Is Mark saying that positive feedback goes on and on forever? And that that is the final system state? (see 1331 post, which sems to me to be describing a transition process – but maybe I’m misunderstanding?)
That is, the 1997 KT diagram (see IPCC AR4 WG1, Chapter 1), which is in Equilibrium is the last such diagram which can be drawn?
Perhaps Mark could explain it in more detail. At the moment I don’t understand his objections, so I can’t make any meaningful response.
On second thoughts Mark’s 1333 post may explain his difficulty.
Mark considers that if the atmosphere becomes more optically opaque, energy transfer from the Surface via the Atmosphere to Space will be slowed down, and the Atmosphere will contain more energy, and will heat up.
Then that warmed atmosphere will heat the surface because the back-radiation will have increased (because the atmosphere is hotter).
For the purposes of this discussion only, I agree. But the Back Radiation has to warm more than the surface. It also has to balance the increased evaporation. This means that, providing the warmed system is in equilibrium, and assuming that Conduction is unaltered (ie the temperature difference between the Surface and the Surface Air is the same) the system heats up less than you might think.
Unless you assume a large Water Vapour increase but a miniscule Evaporation increase. Then you can get very large temperature changes. A 20% Water Vapour change for an expenditure of only 3% additional evaporation could get you 3DegC temperature rise. But is that really realistic? Recent measurements suggest not.
“Mark considers that if the atmosphere becomes more optically opaque, energy transfer from the Surface via the Atmosphere to Space will be slowed down”
Think the escape is retarded.
Rather like the mirror on the ends of a laser.
If the mirrors do not reflect often enough, you get no laser.
“I agree. But the Back Radiation has to warm more than the surface. It also has to balance the increased evaporation.”
Yes. But it is doing that too.
Thing is, more evaporation, more GHG and the thicker the atmosphere is. Therefore the greater the effect of backradiation.
“This means that, providing the warmed system is in equilibrium, and assuming that Conduction is unaltered (ie the temperature difference between the Surface and the Surface Air is the same) the system heats up less than you might think.”
See above.
“Unless you assume a large Water Vapour increase but a miniscule Evaporation increase.”
Only at low levels of opacity.
If you’re already at 8%, you can get another multiplier if you increase the opacity 3.3%. In the case of the earth, each percent increase in opacity would give you ~50W extra backradiation from an 80% opaque atmosphere.
That extra 1% at 50% opaque atmosphere would give you 168/16=10W extra.
The effect of a small increase in water vapour gets higher the higher the opacity is before the increase. And alarmingly so. five times the effect at 80% opacity than at 50% opacity.
I am still having great difficulty understanding Mark’s view.
It appears from his post that evaporation doesn’t matter. If there is more evaporation (ie more clouds and more rain) it doesn’t matter. It’s the same as less clouds and less rain. It doesn’t matter. Positive feedback will accommodate whatever surface condition we choose.
Here is my take:
1. It is unlikely that you will get a 20% increase in atmospheric water vapour (for a 3DegC temperature rise) content without a similarly large increase in cloudiness and rainfall. If you don’t get the 20% you don’t get much “positive feedback”.
2. If you get increased rainfall, you have to have increased evaporation. How much is uncertain, but the 3% used in the models seems to be very low.
3. The scientific literature seems to agree that if Evaporation is high, then the surface doesn’t balance, because the back radiation is insufficient to maintain the 3DegC elevated temperature.
4. I would like to see a detailed explanation of why the assumed evaporation rate in the models is so low, and measurement data to back that case.
“It appears from his post that evaporation doesn’t matter. If there is more evaporation (ie more clouds and more rain) it doesn’t matter.”
From models and evidential records, yes.
If anything, there’s a *slight* positive feedback from extra clouds.
“Positive feedback will accommodate whatever surface condition we choose.”
Positive feedbacks are the likely result of the overall effect of clouds, which are the only possible negative feedback mechanism. But the effect is small and mostly neutral.
We can’t choose the result.
“content without a similarly large increase in cloudiness and rainfall.”
1) clouds are a positive (if small) feedback
2) rain allows dumped energy in the atmosphere directly. Therefore heating the atmosphere and, since the lower atmosphere sees the earth more easily than space, will radiate more fiercely toward the surface than before. I.e. increased back radiation
“If you get increased rainfall, you have to have increased evaporation.”
Which reduces the radiative loss from the surface, but increases the back radiation from the atmosphere.
“The scientific literature seems to agree”
What scientific literature?
“I would like to see a detailed explanation ”
So would I. Why haven’t you supplied one?
“why the assumed evaporation rate in the models is so low, ”
Because it doesn’t accumulate.
It falls out.
“and measurement data to back that case.”
In your case, you are talking about a hypothetical situation where there’s a 3degree warming. It’s rather hard to get measurement data to back up a hypothetical case.
I would like to thank Mark for his responses.
My thesis is this:
1. The Surface air temperature is tightly coupled to the Surface temperature, due essentially to the large conduction flux at the interface (amounting to nearly one fifth of the total surface emergy fluxes into the atmosphere).
2. The only ways the atmosphere can influence the Surface temperature are by:
A. Varying the absorbed solar radiation by:
i. Changing the cloud cover, which changes sunlight
ii. Changing the atmospheric absorbtion of sunlight
B. Varying the Back Radiation by:
i. Changing the intensity of radiation due to a change in the density of a greenhouse gas (if a GHG density increases, the mean height from which an emitted photon strikes the surface drops, this is at a higher temperature so the intensity will be higher).
ii. Changing the temperature of the atmosphere.
The surface response to change in sunlight or back-radiation is to change its temperature until a flux balance is restored. If you look at the very simple equations involved, the Change in “Forcing” from one stable state to a second stable state is balanced by the sum of the changes in Surface Radiation and Evaporation.
I would think that all the above is agreed.
If the Back Radiation increases and there is no other forcing change:
1. The Surface temperature increases (by between 0.1 and 0.2 W/m^2)
2. The temperature increase is limited by the Surface Radiation (nearly Black Body) and the evaporation increase.
3. Because there is an increase in evaporation, the NET surface radiation into the atmosphere falls – the atmosphere is getting the same total flux from the surface, but the relative importance of radiation has fallen.
Mark has been concentrating on the IR opacity of the atmosphere. I have been concentrating on the necessary conditions for the surface. These are not necessarily incompatible – they seem to me to be completely separate topics. I’m not too interested in feedbacks in these posts – it’s a thorny thicket anyway, in great dispute. I’m interested in the final steady state at the surface, and whether it’s reasonable.
1) Incorrect
Thermal (==Convection, sorry used the wrong figure in the other post) 24W/m^2
Total flux out 390W/m^2
390>>24
2) Missed out: Increasing the temperature by cloud creation at low levels, reducing losses yet again
The net difference between this and cloud reduction of solar is that it will be slightly more low cloud effect than high cloud effect, a net positive increase, which is counter to your hypothesis.
” I would think that all the above is agreed.”
Would you agree with the amendments above?
“If the Back Radiation increases and there is no other forcing change:
1. The Surface temperature increases (by between 0.1 and 0.2 W/m^2)”
Why those figures? If backradiation increases 1W/m^2 and nothing else, then the surface temperature will increase to up the output by 1W/m^2.
You’re missing something out.
“3. Because there is an increase in evaporation, the NET surface radiation into the atmosphere falls – the atmosphere is getting the same total flux from the surface, but the relative importance of radiation has fallen.”
Only from the earth. The atmosphere still gets hotter and the opacity increases, ensuring that the TOA reduces and total retained energy is imbalanced and there will have to be heating.
Therefore backradiation will become of higher relative importance.
Despite it having been the very highest importance before (324 >> 24, and about the same as the 390 from the earth).
PS if the earth radiates less than before, the temperature would be less than before and therefore the evaporation will be less than before. The hypothesis you have is that the earth warms by 3 Degrees.
But how will a 3 degree rise cause a drop?
3 is greater, not less than, zero.
Mark responded to my post at 1138, 28Jul which stated:
“1. The Surface air temperature is tightly coupled to the Surface temperature, due essentially to the large conduction flux at the interface (amounting to nearly one fifth of the total surface emergy fluxes into the atmosphere).”
Mark said:
“1) Incorrect
Thermal (==Convection, sorry used the wrong figure in the other post) 24W/m^2
Total flux out 390W/m^2
390>>24”
Mark appears to be confusing all sorts of numbers here. The three fluxes from the Surface which are absorbed into the atmosphere [numbers are from Kiehl & Trenberth 1997, IPCC AR4 WG1 Chap1] are:
Conduction: 24W/m^2
Evaporated water: 78W/m^2
NET Surface Radiation Absorbed In Atmosphere = 26W/m^2
(Surface Radiation – Back Radiation – Surface Radiation through the window to Space = 390-324-40 = 26 W/m^2)
The total is 128W/m^2. Conduction is 18.75% of this, which as I meant to say is “nearly one fifth of the total NET surface energy fluxes absorbed into the atmosphere”.
Mark kindly responded to my post at 2338, 28Jul, with the question:
“2) Missed out: Increasing the temperature by cloud creation at low levels, reducing losses yet again
The net difference between this and cloud reduction of solar is that it will be slightly more low cloud effect than high cloud effect, a net positive increase, which is counter to your hypothesis.
” I would think that all the above is agreed.”
Would you agree with the amendments above?”
No.
The original post listed the causes of changes to forcing. I do agree that Mark has a point – that Back Radiation may also increase due to reflection of Surface Radiation from cloud bases, ie a change in cloud cover might affect back radiation.
However whether that surface radiation reaches the clouds or the reflection reaches the ground is moot. We have seen from earlier posts that the fierce absorption by CO2 and Water Vapour means most surface radiation is highly attenuated above 300m. And the same is true in reverse – any reflection of IR from the cloud base has to make it through the CO2 and H2O molecules lying in wait, and this is highly unlikely unless the cloud bases are below 300m.
Mark responded to my post at 2338, 28Jul in part:
““If the Back Radiation increases and there is no other forcing change:
1. The Surface temperature increases (by between 0.1 and 0.2 W/m^2)”
Why those figures? If backradiation increases 1W/m^2 and nothing else, then the surface temperature will increase to up the output by 1W/m^2.
You’re missing something out.”
Mark is quite correct. The original should read:
““If the Back Radiation increases and there is no other forcing change:
1. The Surface temperature increases (by between 0.1 and 0.2 DegC/W/m^2)”
My apologies for the omission.
The numbers are derived from the Surface Enegy Flux Balance equation, and assume that Conduction is a constant quantity in all equilibrium states. The strange world of Kiehl & Trenberth 1997 is assumed.
The calculation:
Change In Forcing = (4sT^3 + 78y) Change in Temperature’
where s = 5.67×10^-8,
T = temperature in DegK
y=Percentage change in evaporation rate per DegC
Mark wrote:
““3. Because there is an increase in evaporation, the NET surface radiation into the atmosphere falls – the atmosphere is getting the same total flux from the surface, but the relative importance of radiation has fallen.”
Only from the earth. The atmosphere still gets hotter and the opacity increases, ensuring that the TOA reduces and total retained energy is imbalanced and there will have to be heating.
Therefore backradiation will become of higher relative importance.
Despite it having been the very highest importance before (324 >> 24, and about the same as the 390 from the earth).”
Perhaps some numbers may help.
At a surface temperature of 15DegC, the numbers from Kiehl & Trenberth 1997 (all figures are W/m^2)are:
Surface Radiation : 390
Back Radiation: 324
Surface Radiation through the window: 40
Thus NET SR absorbed by atmosphere: 26
Conduction: 24
Evaporation: 78
At 18DegC, and assuming a modest 6W/m^2 increase in evaporation, the numbers are:
Surface Radiation : 406
Back Radiation: (calculated so that energy balance is maintained): 366
Surface Radiation through the window: 40
Thus NET SR absorbed by atmosphere: 20
Conduction: 24
Evaporation: 84
Mark and I have been discussing the Surface Flux balance, and the consequences implied for a final steady state at a higher temperature.
It is indisputable that if GHG concentrations increase, the IR opacity of the atmosphere increases. I am not qualified to quantify by how much.
However it is as well to remember what the sources of energy into the atmosphere are (Numbers are from Kiehl & Trenberth , 1997, see IPCC WG1, AR4, Chapter 1):
1. 67W/m^2 Absorbed sunlight, mostly in the upper strata of the atmosphere above the clouds. All of the wavenumber 670 photons and all of the UV are absorbed in the Stratosphere.
2. 78W/m^2 (Three fifths of the Surface energy absorbed by the atmosphere) from condensing water vapour, mostly in the clouds (but some as dew, frost, fog, mists etc).
3. 26W/m^2 (One fifth of the surface energy absorbed by the atmosphere) from NET radiation from the surface, which enters the atmosphere with an unknown height profile, probably weighted towards the surface.
4. 24W/m^2 (One fifth of the Surface energy absorbed by the atmosphere) by direct Conduction at the Surface.
If the IR opacity increases:
1. The Absorbed sunlight will be absorbed at a higher altitude.
2. The NET Surface Radiation will be absorbed at a lower altitude.
3. The mean altitude from which radiation occurs to Space will be higher for the gas whose density has increased.
However it is generally agreed that all the energy going into the atmosphere, from whatever source, ends up as Atmospheric heat (KE of the air molecules) and that energy transport within the atmosphere is by CONVECTION. This is unaffected by IR opacity and continues unimpeded.
“and that energy transport within the atmosphere is by CONVECTION. This is unaffected by IR opacity and continues unimpeded.”
No, it isn’t generally agreed.
The temperature of the atmosphere indicates its radiative losses. And the opacity inequality between the two bounding conditions: Free Space / Solid Earth indicate whether it will be mostly going down or going up, or equal.
Convective processes manage to transport 78W/m^2.
Thermal radiation back down (so not the entire radiative loss) is 324W/m^2.
324>>78
Even considering net flow, the difference isn’t much, though it reverses (66 vs 78).
And increasing evaporation increases the opacity, increasing the downward flux by increasing the feedback proportion and increasing the temperature (ratev aries as T^4). There is a reduction in the earth outflow by radiation, but if the earth cools then evaporation doesn’t happen so quick, so it still has to be a net positive, just not as positive as it would have been without evaporation).
Mark wrote:
“The temperature of the atmosphere indicates its radiative losses. And the opacity inequality between the two bounding conditions: Free Space / Solid Earth indicate whether it will be mostly going down or going up, or equal.
Convective processes manage to transport 78W/m^2.
Thermal radiation back down (so not the entire radiative loss) is 324W/m^2.
324>>78
Even considering net flow, the difference isn’t much, though it reverses (66 vs 78).
And increasing evaporation increases the opacity, increasing the downward flux by increasing the feedback proportion and increasing the temperature (ratev aries as T^4). There is a reduction in the earth outflow by radiation, but if the earth cools then evaporation doesn’t happen so quick, so it still has to be a net positive, just not as positive as it would have been without evaporation).”
All the NET surface fluxes absorbed into the atmosphere are thermalised. So the correct number for the surface energy transported in the atmosphere up to the levels where it is radiated to Space is 128W/m^2 (78W/m^2 evaporation, 26W/m^2 net radiation and 24W/m^2 conduction).
I hope this answers Mark’s points above, which I must confess I found difficult to follow.
“3. 26W/m^2 (One fifth of the surface energy absorbed by the atmosphere) from NET radiation from the surface, which enters the atmosphere with an unknown height profile, probably weighted towards the surface.”
66 NET.
Not 26.
[We are using the numbers from Kiehl &Trenberth 1997, see IPCC AR4 WG1 Chap1]
Mark has correctly calculated the NET Surface Radiation (= Surface Radiation – Back Radiation = 390-324 = 66 W/m^2) but not the NET Sureface Radiation Absorbed By the Atmosphere (=Surface Radiation – Back Radiation – Surface Radiation through the window to Space = 390-324-40 = 26 W/m^2)
26W/m^2 is the correct number
Mark asked for some references.
There is an interesting paper on watervapour/relative humidity and evaporation: “Water Vapour and the DYnamics of Climate Change” By Schneider, O’Gorman and Levine, AGU, 2009. This paper exposes the Surface energy balance problem and derives numbers for evaporation.
On the other hand, see “How much more rain will Global Warming bring?” , by Wentz, Ricciardulli, Huilburn and Mears, Science, 2007, which suggests, based on measurements, that evaporation is underestimated in the models. They found increased wind over the tropical oceans in the period 1987-2006, the opposite of what the models predicted, with a similar story for precipitation/evaporation.
I think that this is unresolved science.
If it’s unresolved, why the certainty from you that it will happen one way?
Colin:
“I think that this is unresolved science.”
Mark:
“If it’s unresolved, why the certainty from you that it will happen one way?”
A fair question. My position is that, from this really simple view of the energy fluxes, the rate of evaporation change is a critical number – the Surface Air Temperature change is very sensitive to evaporation. It seems from the literature that there is not a great agreement on what the rate of change of evaporation with temperature is, with some recent measurements indicating that it is at the high end, rather than the low end. What then is the scientific basis for the low rate of change assumed by the modellers?
“– the Surface Air Temperature change is very sensitive to evaporation.”
And therefore..?
“It seems from the literature that there is not a great agreement on what the rate of change of evaporation with temperature is,”
But it can’t cool the earth more than the increased temperature from the evaporation.
“What then is the scientific basis for the low rate of change assumed by the modellers?”
There IS NO “assumed rate of change” in the models. The models are big enough to not assume any such thing.
Pop along to the GISS code of the modelE and check for yourself:
http://www.giss.nasa.gov/tools/modelE/
I think one of the problems denial of AGW and the climate science has is that people (like Colin) hear that there’s something being assumed and, rather than check (which in the case of GCMs requires finding the code and understanding it) the tale, just take it as true (since that’s a lot less work).
Therefore here we have “why did they assume such a low evaporation input to the models?”.
In other places, there’s “why don’t they release the code?!?!?” when if they looked, they could see GCM code, though not necessarily the one they’re asking for. Which is a pointless distinction if the GCM code you can get a hold of gives pretty much identical results.
In others “why do they hide the data” when it’s trivially possible to get the same data from another source or a large subset that still gives the same result.
People seem to assume that the proclamations about what’s being hidden or assumed or an input to the climate science is true and never think to check (or if they do check, they don’t ask, therefore a self-limiting selection: only those who don’t check ask).
If you don’t know, don’t make such assertions. *ASK*, yes. ASK if they make that assumption, or if you can get a CGM codeset or the records for the last 150 years. But DO NOT state as truth that the assumption is made, the code not available or the records destroyed.
I thank Mark for his most generous response. I particularly agree with his characteriation of me as someone who unquestioningly picks up hearsay and runs with it.
I had been feeling very guilty about this thread, as I seemed to have taken it over. Many of my final posts were necessarily trying to unconfuse Mark, and I hope I succeeded in that.
In particular, I hope I managed to correct his mathematics and unconfuse the physics in the cases of the calculation of Net Surface Radiation absorbed by the Atmosphere, the amount of heat convected in the atmosphere, and the fall in Net Surface Radiation absorbed by the Atmosphere as the Greenhouse tightens.
It seems from his post at 0823 30Jul, that Mark has been arguing that evaporation is just a derived term within the model, not a control knob with a specific setting.
That may be true, but does not invalidate what I have been saying – The very simple KT model of the atmosphere implies that the surface temperature is very sensitive to evaporation – you can’t MAINTAIN a large temperature change without having a weak change in evaporation.
So I will take Mark’s advice and take a peek at the model.
Finally, I must thank Science of Doom, both for the quality of his headline posts and his forebearance and tolerance of my own poor offerings.
“In particular, I hope I managed to correct his mathematics and unconfuse the physics in the cases of the calculation of Net Surface Radiation absorbed by the Atmosphere”
No, you haven’t corrected and increased the confusion.
Among other things, such as bring along hearsay to a discussion about knowledge.
“The very simple KT model of the atmosphere implies that the surface temperature is very sensitive to evaporation ”
No, you haven’t managed to show this, you’ve only maintained that this “must” be so.
Despite radiative processes being 50x greater than evaporative ones.
“you can’t MAINTAIN a large temperature change without having a weak change in evaporation”
And here is one last example of how you’ve been adding to confusion: large temperature change IN WHAT?
[…] 31, 2010 by scienceofdoom In Part One we took a look at what data was available for “back radiation”, better known as Downward […]
I hope Science_of_Doom will further indulge me, as I respond to Mark’s post above.
I will start at the beginning, which is the very strange world depicted in the energy flux diagram by Kiehl & Trenberth 1997, and reproduced in IPCC AR4, WG1, Chapter 1. There are many implied assumptions in this diagram, one of which is that it depicts a world in energy equilibrium.
This diagram shows a mean surface temperature of 15DegC. From the diagram, the Surface Energy Flux Balance can be written as:
(Forcings) = (Responses), or,
Sunlight_Absorbed_By_The_Surface + IR_Back_Radiation =
IR_Radiation_from_the_Surface + Evaporated Water + Conduction (shown as thermals in the KT diagram).
IPCC AR4 identifies the most likely change in temperature for a doubling of CO2 to be about 3DegC. It does not however show a new KT diagram for this new higher temperature. Note that I am assuming an Equilibrium KT world. I have assumed:
1. No change in the Conduction term, on the basis that any change would imply a different balance in temperature between the Surface and the Boundary layer.
2. No change in the Sunlight Absorbed by the Surface.
3. Minimum change in evaporation rate =0, maximum is the Clausius-Clapeyron limit of 6.5% per degC.
If I do this, then I get:
Change in IR_Back_Radiation = Change in Surface IR Radiation + Change in Evaporated water
= (4sT^3 + 78y)dT, where s=5.67×10^-8, T=Temp in DegK, y= change rate of evaporation, dT = change in temperature in DegC.
ie d(Back Radiation) = 5.4dT(minimum evaporation) to 10.5dT (max evaporation), giving a steady state mean average KT diagram temperature change of 0.1 to 0.2 DegC/W/m^2 change in Back Radiation.
For an 18DegC KT world, the Back Radiation increase required to MAINTAIN the new Surface Temperature is
16W/m^2 (zero evaporation change) to 32W/m^2 (maximum evaporation change).
I think Mark agrees to here. If not, perhaps he should identify which assumption is questionable.
It is difficult to see in detail where the additional forcing comes from in both cases. In the zero evaporation change case we have to assume that the atmosphere warmed by 3DegC. ie the 4W/m^2 imbalance created by doubled CO2 at the Trpopause translates to 16W/m^2 at the Surface. In the maxiomum evaporation case we have to assume that an additional 16W/m^2 is somehow generated by additional cloud reflection, without a corresponding decrease in the Sunlight reaching the Surface.
It is my view that balance can only be achieved if a low figure for evaporation rate change is used. But perhaps Mark can assign numbers to the various factors pertaining in the EQUILIBRIUM 18DegC KT world which show me to be wrong.
You ask: “he should identify which assumption is questionable.”
This one:
“This diagram shows a mean surface temperature of 15DegC. From the diagram, the Surface Energy Flux Balance can be written as:”
It’s an a priori assumption that requires total flux, not net to enact. Then assumes total flux doesn’t make a difference and should be ignored.
“IPCC AR4 identifies the most likely change in temperature for a doubling of CO2 to be about 3DegC. ”
This is THE FIRST TIME you’ve said why you keep banging on about 3C.
“It is difficult to see in detail where the additional forcing comes from in both cases.”
Try here:
http://www.ipcc.ch
“16W/m^2 (zero evaporation change) to 32W/m^2 (maximum evaporation change).”
This is also incorrect.
With evaporation, the radiative outgoing component is reduced therefore the backradiation doesn’t have to increase by as much to cause a warming and new balance point.
This implies your figures are the wrong way around: it should be a higher backradiation with zero evaporation.
“It is my view that balance can only be achieved if a low figure for evaporation rate change is used.”
See above for the fatal mistake marker.
Mark raises several issues in his post. I will deal with them one at a time.
Mark wrote:
“You ask: “he should identify which assumption is questionable.”
This one:
“This diagram shows a mean surface temperature of 15DegC. From the diagram, the Surface Energy Flux Balance can be written as:”
It’s an a priori assumption that requires total flux, not net to enact. Then assumes total flux doesn’t make a difference and should be ignored.”
It is difficult to discuss matters with Mark. His point seems illogical. Does he disagree wiith the equation which is directly derived from the KT diagram?:
(Forcings) = (Responses), or,
Sunlight_Absorbed_By_The_Surface + IR_Back_Radiation =
IR_Radiation_from_the_Surface + Evaporated Water + Conduction (shown as thermals in the KT diagram).
If he disagrees, perhaps he would care to explain why he thinks it is invalid to copy down the fluxes across the surface/atmosphere boundary, fluxes which (if he adds it all up) balance exactly in the manner described in the equation.
Mark also wrote (I am ignoring the sniping):
““16W/m^2 (zero evaporation change) to 32W/m^2 (maximum evaporation change).”
This is also incorrect.
With evaporation, the radiative outgoing component is reduced therefore the backradiation doesn’t have to increase by as much to cause a warming and new balance point.
This implies your figures are the wrong way around: it should be a higher backradiation with zero evaporation.
“It is my view that balance can only be achieved if a low figure for evaporation rate change is used.”
See above for the fatal mistake marker.”
In my post at 0043, 1Aug, I derived the change in Back Radiation as 5.4dT(zero evaporation) and 10.5dT(Clausius-Clapeyron evaporation).
For a 3 DegC temperature rise, we then get 16W/m^2 and 32W/m^2 respectively for the change in Back Radiation required to maintain the new temperature.
The formula I used was derived from the Surface Flux Balance Equation and is:
Change in IR_Back_Radiation = Change in Surface IR Radiation + Change in Evaporated water
= (4sT^3 + 78y)dT, where s=5.67×10^-8, T=Temp in DegK, y= change rate of evaporation, dT = change in temperature in DegC.
Mark disagrees with this equation, but has offered zero argument as to what is wrong with it. Perhaps he can derive another equation which explains his point.
Then again, perhaps he can’t.
[His argument appears to be that if evaporation occurs, then the surface doesn’t warm as much. But we have stipulated a 3DegC temperature rise in all cases – and the system has to work a whole lot harder to maintain such a change if evaporation also increases.]
Mark said:
”
“IPCC AR4 identifies the most likely change in temperature for a doubling of CO2 to be about 3DegC. ”
This is THE FIRST TIME you’ve said why you keep banging on about 3C.”
I refer Mark to my post at 0037, 23Jul.
Mark has accused me of bringing Hearsay to this discussion about knowledge. I’ll bite. I invite him to be more specific.
And I quote you:
“It is difficult to see in detail where the additional forcing comes from in both cases.”
In his latest post, Mark kindly wrote:
““you can’t MAINTAIN a large temperature change without having a weak change in evaporation”
And here is one last example of how you’ve been adding to confusion: large temperature change IN WHAT?”
I have in previous posts stated an assumption that the Surface Air Temperature is tightly coupled to the Surface Temperature, and a second assumption that the relationship between these two is maintained in a KT world where the Surface Temperature increases.
So in answer to Mark’s question, both the Surface Temperature and the Surface Air Temperature are the quantities affected.
A further point. Mark failed to include the whole context in his question. My original post read:
“The very simple KT model of the atmosphere implies that the surface temperature is very sensitive to evaporation – you can’t MAINTAIN a large temperature change without having a weak change in evaporation.”
By failing to include the first phrase, Mark implied that I was being unspecific.
“you can’t MAINTAIN a large temperature change without having a weak change in evaporation”
And here is one last example of how you’ve been adding to confusion: large temperature change IN WHAT?”
But it wasn’t me confusing the issue, it was his inadvertent omission which added to the confusion. In any case, I hope my post above has unconfused the situation.
“you can’t MAINTAIN a large temperature change without having a weak change in evaporation.”
A large temperature change IN WHAT?
“I have in previous posts stated an assumption that the Surface Air Temperature is tightly coupled to the Surface Temperature”
And this is another hearsay production.
Your assumption is incorrect.
Mark has either impugned my honour (ie is accusing me of lying) or doesn’t understand the meaning of “hearsay”.
Either way, I think he needs to rephrase his post.
As to the correctness or otherwise of the assumption, Mark is entitled to have an opinion. But I would prefer a more reasoned argument other than flat out denial.
No,l all your work there is hearsay.
“I believe” “In my opinion”
backed up with half-baked maths and misapplied physics.
Mark wrote:
“The very simple KT model of the atmosphere implies that the surface temperature is very sensitive to evaporation ”
No, you haven’t managed to show this, you’ve only maintained that this “must” be so.
Despite radiative processes being 50x greater than evaporative ones.”
This last statement of Mark’s implies a misunderstanding of the physics on two levels:
1. The statement is mathematically incorrect. In the KT diagram, the Surface Radiation Flux is 390W/m^2 and the Evaporation Flux is 78W/m^2. on this simplistic (and wrong!) basis, the Surface Radiation is only 5 times the Evaporative Flux.
2. Fluxes only influence the world by their NET values. There are 3 NET fluxes to consider:
A. Surface Radiation which escapes directly to Space = 40W/m^2
B. Surface Radiation absorbed by the Atmosphere = 26W/m^2
C. Evaporated Water which Condenses in the Atmosphere = 78W/m^2
The evaporative flux absorbed by the atmosphere is 3 times the NET Surface Radiation absorbed by the Atmosphere.
Mark has indicated throughout this exchange that he doesn’t agree with the calculation of NET Radiation. I would ask him to describe the flux in a world which has a completely dry surface and a 1mm vaccuum magically interposed between the atmosphere and the Surface.
In my opinion, the flux from the Surface into the atmosphere which actually affects the atmosphere would be the NET: Suface IR – Surface IR escaping to space – Back radiation.
Now get rid of the magic vaccuum. The Surface flux into the atmosphere becomes the NET surface radiation into the atmosphere (calculated as above) + Conduction.
Finally add water. Surface flux into the atmosphere = Net Surface Radiation into the Atmosphere + Conduction + Evaporation.
The effect of Radiation is always the effect of the NET Radiation not the GROSS. (Otherwise it would truly be Gross!)
For some reason some get grossed out by the large absolute numbers (390W/m^2) rather than the small NET (26W/m^2 + 40W/m^2). Only the NET has any effect in a thermodynamically correct world. In that world the Surface Radiation has only one fifth of the effect of the Surface on the Atmosphere, and Evaporation has three fifths.
“2. Fluxes only influence the world by their NET values. There are 3 NET fluxes to consider:”
No, they don’t.
The temperature of the earth’s surface is increased by the total fluxes, not the net flux.
Else, since the space is colder than the earth, the earth WOULD NOT heat up, only cool.
I’m afraid YOU are the one who is getting the extremely basic physics wrong.
“The evaporative flux absorbed by the atmosphere is 3 times the NET Surface Radiation absorbed by the Atmosphere.”
And that is both wrong and pointless, for reasons stated above.
Your case would break the second law (and many others). It would also give the atmosphere a temperature that is hugely underrepresentative of the actual temperature.
Blind insistence on net is why you’re failing miserably at the physics.
Just because the car isn’t moving, doesn’t mean you’re not pushing it as hard as you can.
“In my opinion, ”
That doesn’t make it true.
“the flux from the Surface into the atmosphere which actually affects the atmosphere would be the NET: Suface IR – Surface IR escaping to space – Back radiation.”
Which would give it a temperature of what? 147K.
Please explain why the air is actually 150C warmer than that.
“Only the NET has any effect in a thermodynamically correct world.”
In that case, since there is only net loss at night, the earth would be negative temperature, since only fluxes going away work in your world.
In this real one, the actual fluxes count. A mirror being shone on is different from one being left in the dark, despite the net radiation being the same.
Mark’s comments are not correct.
Consider a 1m^2 surface area sphere in an empty universe. The sphere is a blackbody, and its temperature will change by 1 degree per 100 Joules of energy added or lost. We start the sphere at a temperature of 16.9DegC, and allow it to radiate for 1 second.
The sphere loses 400J of energy by radiation. The energy for radiation comes from the kinetic energy of the material of the sphere, which falls. We measure a new sphere temperature of 12.9DegC.
We place the sphere, magically now at 0DegK, inside a Surface separated by a minute but finite space radiating as a blackbody with a temperature of minus 3.3DegC. After 1 second the sphere has gained 300J of energy by radiation and is at a temperature of 3DegK.
We now magically heat the sphere to 16.9DegC, keeping it enclosed by the very close spherical surface at minus 3.3DegC. What happens in the first second? The Sphere gains 300J from the Surface, but loses 400J to the Surface. It cools by 1DegC.
Note it is not the magnitude of the fluxes, but the Difference – the NET flux, which affects the temperature of the sphere.
The sphere does not heat up by 3DegC. It does not cool by 4DegC. It is not influenced by the magnitude of the fluxes but by the Difference in the fluxes.
I hope Mark agrees with all of this, which I know Science_of_Doom has also covered in the past.
This post negates the entirety of his post, which is 100% in error. Mark accuses me of “getting the extremely basic physics wrong.” Apart from the immoderacy of that statement, it is, unfortunately for Mark, incorrect.
“Mark’s comments are not correct.”
I’m afraid they are Col.
“Note it is not the magnitude of the fluxes, but the Difference – the NET flux, which affects the temperature of the sphere.”
No, it was the “magically at 0K” that broke the physics there.
“The sphere loses 400J of energy by radiation.”
Why? You don’t use this phenomena later.
Gish galloping?
“We place the sphere, magically now at 0DegK, inside a Surface separated by a minute but finite space radiating as a blackbody with a temperature of minus 3.3DegC. After 1 second the sphere has gained 300J of energy by radiation and is at a temperature of 3DegK.”
But what was it doing when it rose in temperature?
It was radiating. Not at 300W but the loss is unaccounted for in your unphysical model.
“Note it is not the magnitude of the fluxes, but the Difference – the NET flux, which affects the temperature of the sphere.”
No, the 300W flux was -3.3C. the net flux to maintain that is 300W. The net flux is zero when both are at thermal equilibrium. But their temperatures are not zero.
The net flux affects the rate at which heating occurs.
But not the temperature.
Mark seems unfamiliar with the First Law of Thermodynamics.
My point in my example was to restate that law:
If a body receives 300J of energy but loses 400J of energy, it will have lost 100J net energy.
The same would be true if the numbers were 1000J and 1100J. The effect on the body WOULD BE EXACTLY THE SAME.
Unless, as Mark does, you invoke a new law. Then you can have anything you like: The MAGNITUDE of the transfers affects the result, not the NET transfer.
The real point is that Mark doesn’t like the answers which result from my calculations. He thinks they are wrong, but in seeking to invalidate them has violated the First Law.
Let me give an example. In his post at 0800, 2Aug, Mark stated that a same world (same surface, same temperature, same atmosphere) with a higher evaporation rate would need a lower back-radiation. That statement is a violation of the First Law of Thermodynamics – If Mark does the sums he will find that his numbers don’t add up.
I restate the central question here: The radiative effect on a body is the NET Radiation (incoming minus outgoing), NOT the Gross. Otherwise the First Law of Thermodynamics is being violated (energy is being created or destroyed).
“If a body receives 300J of energy but loses 400J of energy, it will have lost 100J net energy.”
Never disagreed.
“The same would be true if the numbers were 1000J and 1100J. The effect on the body WOULD BE EXACTLY THE SAME.”
Yes, 100J would have been lost.
“Let me give an example.”
Would be nice. Just make it accurate.
“Mark stated that a same world (same surface, same temperature, same atmosphere) with a higher evaporation rate would need a lower back-radiation.”
Indeed it would.
Since radiative loss would be less from the earth.
Or doesn’t boiling water to steam require energy any more?
“That statement is a violation of the First Law of Thermodynamics”
No it doesn’t.
“If Mark does the sums he will find that his numbers don’t add up.”
Why don’t you do the sums and show this.
“The radiative effect on a body is the NET Radiation (incoming minus outgoing), NOT the Gross.”
you’ve never said that before. You’ve said that the net radiation tells you the temperature. Making this:
“I restate the central question here: ”
Except you never stated the central question you state “here” before.
“NOT the Gross. Otherwise the First Law of Thermodynamics is being violated (energy is being created or destroyed).”
Except you’re incorrect.
If the net radiation were the only effect you should consider, then the earth is far too hot to radiate that and there’s no source for that heat.
This would violate the second law because:
1) Earth radiates 66W/m^2
2) get a metal plate at 250K and a surface area 1m^2. It radiates at 221W.
3) If it radiates at 221W then the earth radiating at 66W underneath it will have a net input of 156W.
4) therefore the earth (290K) is being warmed by the plate (250K).
5) therefore the second law is being violated.
I find Mark’s posts to be confusing and confused. For example in his post at 1449, 2Aug he states:
”
[Colin]“Note it is not the magnitude of the fluxes, but the Difference – the NET flux, which affects the temperature of the sphere.”
[Mark] No, the 300W flux was -3.3C. the net flux to maintain that is 300W. The net flux is zero when both are at thermal equilibrium. But their temperatures are not zero.
”
What does the first phrase, “the 300W flux was -3.3C.” mean? Perhaps Mark can clarify what he meant to say.
“What does the first phrase, “the 300W flux was -3.3C.” mean? Perhaps Mark can clarify what he meant to say.”
P=sigmaT^4.
-3.3K, plug it in, P=300W.
It was YOUR model, not mine, why didn’t you know what your model was?
Your inability to understand your own model is confusing.
Should have been -3.3C. You can’t get negative absolute temps…
“I hope Mark agrees with all of this, which I know Science_of_Doom has also covered in the past.”
Yes, and he didn’t get your answer either.
We both disagree with you.
“This post negates the entirety of his post, ”
Except it manages that only by telling great big whoppers.
I assume merely because you’re confused about the energy flux and the radiative equations and the mechanisms of temperature.
“Except it manages that only by telling great big whoppers.”
Mark has again accused me of deliberately lying. I hope he will withdraw this remark.
I’ve accused you of telling whoppers. Is what you’ve said that I’ve called a whopper true?
No.
Also, technically, if you want to be pedantic (which would be how to get “you’re deliberately telling lies”), the post told whoppers to refute the back radiation/net flux dichotomy you hold to.
You may have forgotten radiative losses in a vacuum sealed object…
You didn’t get to the end of the message, did you Col.
“I assume merely because you’re confused about the energy flux and the radiative equations and the mechanisms of temperature.”
I have debated with myself as to whether I should continue what seems at present to be a pointless debate with Mark. I have decided to continue with one last attempt.
I am seeking the truth – that is my sole motivation, and I expect other posters to be similarly motivated. I have disageed with many other posters, and had meaningful and polite exchanges of views with them. I have learnt something from these exchanges of views, and I hope they have also benefited.
My problem with Mark is getting to the level of engagement. I find in general his posts very hard to follow for the reasons below:
1. His style is aggressive. He liberally sprays ad-hominem and untruthful accustions, and does not retract when called to account. But such should never be part of the corpus when trying to explain a point of view – it is simply distracting noise.
2. He selectively/out-of-context quotes, then uses this as a point of attack. This leaves me in the position of having to restate what has already been made clear in the original. Again all this does is clutter the thread – it doesn’t communicate, or clarify, just confuses.
3. His prose is unclear and imprecise, making it almost impossible to discern his line of argument.
It may be that Mark is also a seeker after the truth. I have decided, against the evidence of these last 50 posts, that that may be the case. I am therefore offering, in my next post, an opportunity for him to clearly state his position – if possible without the sort of distractions and confusions noted above.
In this thread, my posts have been almost exclusively based on the [unreal] world of Kiehl & Trenberth 1997 (see IPCC AR4, WG1, Chapter 1 for a copy of their figure).
That is an AVERAGE planet in EQUILIBRIUM (all transient processes moving the planet to a different state are completed.) with a Surface temperature of 15DegC.
The IPCC claims a temperature increase of 3DegC for a doubling of CO2. I have attempted to describe what some of a resultant KT diagram at 18DegC would be like – in particular what the new Surface Flux balance would be.
The constraints are:
1. The planet is at a new EQUILIBRIUM (all transient processes moving the planet to a different state are completed.)
2. Sunlight absorbed by the Surface is unchanged.
3. Conduction to the atmosphere from the Surface is unchanged.
I have previously posted my picture of the new surface flux balance at 12am, 30Jul:
“At a surface temperature of 15DegC, the numbers from Kiehl & Trenberth 1997 (all figures are W/m^2)are:
Surface Radiation : 390
Back Radiation: 324
Surface Radiation through the window: 40
Thus NET SR absorbed by atmosphere: 26
Conduction: 24
Evaporation: 78
At 18DegC, and assuming a modest 6W/m^2 increase in evaporation, the numbers are:
Surface Radiation : 406
Back Radiation: (calculated so that energy balance is maintained): 366
Surface Radiation through the window: 40
Thus NET SR absorbed by atmosphere: 20
Conduction: 24
Evaporation: 84”
Mark is invited to provide his numbers for the stated quantities, and to justify each one – as I have done.
“In this thread, my posts have been almost exclusively based on the [unreal] world of Kiehl & Trenberth 1997”
They aren’t the source of the unreality.
Averages and equilibrium are very real ways of answering the world. Your average speed doesn’t exist because it varies slightly. Even f=ma is a result of QM where each term is taken as the average of each quantum entity in the system.
“The IPCC claims a temperature increase of 3DegC for a doubling of CO2. ”
Indeed they do.
“3. Conduction to the atmosphere from the Surface is unchanged.”
Why would that be? It’s 3C warmer.
“At 18DegC, and assuming a modest 6W/m^2 increase in evaporation, the numbers are:
…
Mark is invited to provide his numbers for the stated quantities, and to justify each one – as I have done.”
You’ve not justified the 18C and 6W/m^2 evaporation.
You’ve just said they are such.
Instead we have 3C warming. 1C from CO2 reduction at TOA and 2C from all feedbacks. Some of which are reflection reduction by reduction of sea ice.
Such reduction is part of the 30W/m^2 reflected by the surface.
You can find the actual calculations and work your own result using the REAL physics here:
http://www.giss.nasa.gov/tools/modelE/
Or you can go and read the papers here:
http://www.ipcc.ch
Mark said:
”
[Colin]“3. Conduction to the atmosphere from the Surface is unchanged.”
[Mark]Why would that be? It’s 3C warmer.”
In previous posts I have stated that this is the conservative assumption – it implies that the average temperature difference between the boundary layer air and the surface temperature is the same for all absolute values of surface temperature. In other words the temperature of the boundary air heats by the same amount as the surface.
Unless there is evidence to the contrary, this is the only safe assumption to make.
Mark said :
“You’ve not justified the 18C and 6W/m^2 evaporation.
You’ve just said they are such.”
The 3 DegC (=15+3) is what the IPCC says, as Mark has acknowledged. I don’t understand why Mark is challenging this. As explained in the original post, all we are trying to do is to see what the IPCC increase implies for the KT diagram.
The 6W/m^2 comes from the figure of 2.5% perDegC increase in evaporation rate from the paper by Schneider, O’Gorman and Levine (“Water Vapour and the Dynamics of Climate Changes”, AGU 2009). While I regard this as low, it is the latest I am aware of in the peer reviewed literature. Other authors have stated that real world measurements suggest that a 5%perDegC change is the actual case. (I have actually explained all this before in previous posts, including the verey large uncertainty in this important quantity.)
So 6W/m^2 seems a conservative number.
Mark was invited to provide his own numbers for a KT diagram which is 3DegC warmer.
Instead, ducking this, but attacking me for doing so, Mark said:
“Instead we have 3C warming. 1C from CO2 reduction at TOA and 2C from all feedbacks. Some of which are reflection reduction by reduction of sea ice.
Such reduction is part of the 30W/m^2 reflected by the surface.
You can find the actual calculations and work your own result using the REAL physics here:
http://www.giss.nasa.gov/tools/modelE/
Or you can go and read the papers here:
http://www.ipcc.ch”
Well. So sorry. Mark doesn’t like my numbers but won’t provide his own. Probably he either can’t, or they don’t give him the answer he wants. Either way this episode shows that he is not engaging but just making argument and selling herrings.
“Thus NET SR absorbed by atmosphere: 26”
But this is irrelvant when you’re talking about fluxes.
The flux absorbed by the atmosphere is
350W/m^2 from the earth. 67W/m^2 by the sun.
Compared to that, your 6W/m^2 increase is tiny, therefore your assertion that it is a major flux is false.
You could get the same flux by increasing upwelling absorption by 2%. Such could easily happen by increasing the opacity of the atmosphere to LWR by 2%.
Before: 324W/m^2.
After: 330W/m^2.
Before: T=274.94K
After: 276.21K
dT=1.27K.
Adiabatic lapse rate of 10C/km means that the absorbing layer is now 127m lower.
This isn’t what is going on, mind, but an illustration of why you NEED the total flux.
As does the experiment given earlier that shows that you need the total flux or you will break the laws of thermodynamics:
1) Earth radiates 66W/m^2
2) get a metal plate at 250K and a surface area 1m^2. It radiates at 221W.
3) If it radiates at 221W then the earth radiating at 66W underneath it will have a net input of 156W.
4) therefore the earth (290K) is being warmed by the plate (250K).
5) therefore the second law is being violated.
The whole of the above post is irrelevant. Mark was invited to provide numbers for a new KT diagram 3DegC warmer.
He hasn’t because he cannot or because he doesn’t like the numbers.
He is trying to deflect the discussion. I am concluding that he doesn’t want to engage, is not seeking the truth.
In addition, Mark’s inability to see the relevance of NET surface radiation is truly astounding. I doubt whether there are any people engaged in this debate who would subscribe to his views.
Mark is correct – the Heat flows into the Atmosphere are:
1. Absorbed sunlight(67)
2. Surface Fluxes:
A. Evaporation (78)
B. Conduction (24)
C. Net Radiation (26)
These are balanced by Atmospheric IR radiation to Space (195).
All figures in W/m^2.
Note that, in accordance with the Second Law of Thermodynamics, all the above fluxes are NET, and are flows from the hotter to the colder body.
If there was no atmosphere, the NET radiation would be very large. But the addition of an atmosphere with IR active gases means that the atmosphere is warm, so the NET Radiation drops, and at present is very small indicating that the Greenhouse is nearly perfect.
Dr Spencer has a long and interesting post on Miskolczi at http://www.drroyspencer.com/2010/08/comments-on-miskolczi%e2%80%99s-2010-controversial-greenhouse-theory/
This caught my eye:
“Even the (controversial and often maligned) K&T energy diagram shows the convective heat loss by the surface to the atmosphere (102 Watts per sq. meter) is about 4 times larger that the rate of IR loss by the surface to the atmosphere (26 Watts per sq. meter). ”
[By “convective heat loss” it is clear that he is talking about Conduction + Evaporation.]
Which is exactly what I have been saying.
I wonder whether “K&T is unreal” is the new meme for those trying to avoid science, now that even Roy Spencer and Pielke have shown that back radiation and the presence of a body at nonzero temperature is actually scientifically correct.
It does seem to have turned up several times just recently.
Meme spreading and infection?
In past posts I have made clear that the Kiehl & Trenberth diagram represents a strange “average” planet, with no day/night, no weather, average cloud, average latitude, average season, average everything, and is at Equilibrium (no dynamic processes). None of these things are “real world”.
The diagram has been widely attacked as a model for exactly that reason. I understand those criticisms and agree with them in that the most accurate models will include all dynamic processes to the highest possible resolution.
However those highly complex models are also justifiably in question – Can a chaotic system be sufficiently accurately described so that a meaningful prediction can be made – more meaningful than the simple average model provided by the KT diagram, or indeed the simple dynamic model used by Dr Spencer?
I think the KT diagram has merit, but it needs to be used as presented – as a static equilibrium, not a dynamic model, and its limitations need to be remembered.
Hence the reminder that it is “unreal” [an entirely accurate and succinct adjective].
Or if you want to move to more widespread realms:
http://www.ipcc.ch
http://www.realclimate.org
http://www.aip.org/history/climate/index.htm
realclimate.org does not have a moderation regime which would allow these sorts of discussions. (Science of Doom is again to be congratulated on his strict application of his policy).
ipcc.ch is of no help in constructing an 18DegC KT world.
and neither is the otherwise informative aip site.
The problem area is the Evaporation. How much increase will there be?
At present the only 18DegC KT model to hand is the one I have presented, Mark having declined.
[…] at the earth’s surface corresponds to the missing sections at the top of the atmosphere. See The Amazing Case of “Back-Radiation” and The Amazing Case of “Back Radiation” – […]
[…] much more about the downward radiation from the atmosphere – The Amazing Case of “Back-Radiation”, Part Two, […]
[…] 11, 2010 by scienceofdoom In the The Amazing Case of “Back-Radiation” series, which included Part Two and Part Three, someone commented that it would have been good […]
[…] so I offer the series, The Amazing Case of “Back Radiation” as proof, especially Part Three. Result of Part Three was – “well, that can’t […]
[…] the surface we can measure this downward radiation from the atmosphere. See The Amazing Case of “Back Radiation” -Part One and the following two parts for more discussion of […]
[…] the complete picture of the energy transfer – the atmosphere also radiates to the ocean (see The Amazing Case of “Back Radiation” -Part One and the following parts). The issue that was probably implied in the original question was […]
[…] The Amazing Case of “Back Radiation” -Part One – for measurements of radiation from the atmosphere […]
[…] The Science of Doom, The amazing case of “back-radiation”, 27 July […]
[…] […]
[…] […]
What seems to be missing here is a discussion on the real reason the atmosphere has been heated.. It is not by absorption of long wave radiation from the earth. it is by convection, the abstraction of heat by conduction, enhanced by turbulence and a phased transmission to upper parts of the atmosphere. The whole of the atmosphere transmits the heat, not just the “greenhouse gases” , whose presence is unnecessary.
This atmosphere, heated by convection radiates part of the heat it has abstracted from the earth to outer space,from each level and transmits half of it as “back radiation”, back to earth.
The article’s premise is:
The question about why the atmosphere is at the temperature it is is a completely different question, not posed by this article.
But seeing as you raise it, the transfer of energy from the surface to the atmosphere is a result of all of the heat transfer mechanisms which includes absorption of longwave radiation as well as convection.
Your statement implies you believe that the atmosphere does not absorb any longwave radiation from the surface of the earth.
If you think this, why does the spectrum of TOA radiation have those notches in that you see in, for example, the first figure of Theory and Experiment – Atmospheric Radiation?
If you don’t think this, why did you write: “It is not by absorption of long wave radiation from the earth“
Yes, I admit there might be some absorption of outgoing long wave radiation by greenhouse gases, but almost exclusively water vapour, and probably negligible. At least there is no evidence that it has affected the climate.
Not much point explaining anything to you then.
If anyone else wonders whether Vincent is onto something take a look at actual spectral measurements of TOA radiation and explain them, with reference to spectroscopic published papers and textbooks on radiation theory.
The “big dip” around 15μm is the bit that needs explaining:
On the face of it, it is evidence of atmospheric absorption, by water vapour.. However, I would like to have information on how accurate it is and how it varies over time or over different parts of the earth. The radiation emitted by the ocean is surely different from land. By night there is a different average temperature and a different spectrum..
I may have exaggerated the importance of convection, but there is no doubt its importance has been grossly downplayed by the greenhouse enthusiasts.
Whatever the reasons for the heating of the atmosphere, “back radiation”: from the atmosphere undoubtedly exists, whatever its temperature, and whatever itrs cause.. It should be seen as returning to the earth, by radiation, one half of the heat that was removed, either by convection, or atmospheric absorption, or by any other cause, modified, of course by albedo..
There still remains the absence of realisation that any “global warming” may be due to the efforts of humans to reduce convection and evaporation
Additional information is given by Bob Fernley Jones at
http://wattsupwiththat.com/2011/10/26/does-the-trenberth-et-al-%E2%80%9Cearth%E2%80%99s-energy-budget-diagram%E2%80%9D-contain-a-paradox/
He shows that the use by Trenberth et al to use W/msq instead of heat causes an error which fails to give sufficient attention to convection.
This is a nice little applet http://phet.colorado.edu/sims/blackbody-spectrum/blackbody-spectrum_en.html that allows you to see the relative proportions of the power intensities from the sun and the earth. [An image at the top of this scienceofdoom page says “sun (scaled by a factor 10^-6)”, which some might not realize.]
The correct scaling factor for solar radiation viewed at the Earth’s orbit is 2.16E-05 rather than 1E-06. But then you reduce that by a factor of 4 to get the average intensity at the TOA over 24 hours or 5.4E-06. So a scaling factor of 1E-06 isn’t unreasonable, especially considering that logarithmic wavelength scale makes the area under the solar curve look larger than it actually is.
Here is my mistake: I was appealing in my mind to the intense radiation given off by the sun, not realizing that for the purposes of seeing the impact on earth we are going to be talking about a scaled down effect. Thanks for crunching (or looking up) some of those numbers and bringing up the point. D’oh.
What proportion of backradiation is attributable to the 390-280=110 ppm of CO2 mankind has added ?
Punksta,
The important value in the energy balance of the climate system is the change in outgoing radiation to space (not the backradiation at the surface).
The absorbed solar radiation is a function of the solar radiation received at top of atmosphere and the albedo (reflectivity) of the climate system to solar radiation.
You can see values in The Earth’s Energy Budget – Part Four – Albedo.
This needs to match outgoing emitted radiation to space or there will be an imbalance in the earth’s energy budget and therefore an effect on temperature.
The outgoing radiation is affected by the change in CO2 by about 2W/m2, globally annually averaged. This value is calculated in W.D. Collins et al (2006), Radiative forcing by well-mixed greenhouse gases: Estimates from climate models in the Intergovernmental Panel on Climate Change
(IPCC) Fourth Assessment Report (AR4).
The reason that backradiation at the surface is not as important is explained in Understanding Atmospheric Radiation and the “Greenhouse” Effect – Part Four, where a simple model demonstrates how surface radiation can be constant with increases in a trace gas, yet outgoing radiation to space can reduce.
Thanks for the general commentary.
No direct answer to my specific question though, the (fair enough) reason being that, even though backradiation exists, it is not considered to be a major factor in the energy budget ?
Punksta,
Correct, kind of..
The TOA budget is what determines the final surface temperature. (note 1)
Back radiation is “a major factor” in the surface budget, but not in the whole climate budget – in effect it is “dragged along” by the TOA budget.
Note 1 – And now the caveat that the linkage between the TOA budget and the surface temperature is the “lapse rate” – which is how the temperature drops with altitude. This is not a simple subject. On the subject in question (back radiation), the lapse rate is not a function of back radiation
Can I ask a question? In your GEBA table of DLR values….. the flux quantities worry me….
If the surface is emitting LR at around 350 W/m^2, surely the DLR readings should only be a fraction of this? The quantity coming back from GHG in the atmosphere must be less than half of what went up, surely? So I wil thinking in the order of 150 W/m^2 or thereabouts.
So why are the average annual readings for DLR all in the 300s? How can that be? Have I missed something fundamental?
The surface emits a radiative flux of about 390 W/m^2. The averages for DLR are in the 300s W/m^2 because there are multiple sources of downward LW from the atmosphere to the surface boundary besides just that which is emitted from the surface.
TheBigYinJames,
The values are correct. And observation and theory match.
The theory you have implies something different so of course you expect a different result.
What theory leads to your statement: “..The quantity coming back from GHG in the atmosphere must be less than half of what went up, surely..” ?
A surface of water at 15’C (288K) will emit at about 374 W/m2. (Emissivity = 0.95).
Why must the atmosphere emit “less than half” of this value?
In fact, the atmosphere will emit according to its temperature and its emissivity. This is the critical point to understand. This is the fundamental of emission of thermal radiation
You tell me:
– What do you think is the temperature of the atmosphere?
– What do you think is the emissivity of the atmosphere?
Steve, thanks for the reply.
I realised why I made the mistake – I was thinking of flux in terms of a quantity of energy – forgot that W/m^2 already contains a term removing the area aspect. Flux is flux, it’s the same in all directions. Since the atmosphere near the ground is at a similar temperature as the ground, the flux will be similar, I get it. Silly me. Confusing work per area with Joules.
Incidentally, I wasn’t criticising GH theory in my original post. I class myself as lukewarmer, and spend a great deal of time explaining back radiation to people on Bishophill myself – so I like to have my ducks in a row, and those flux values unnerved me for a while. It was a genuine mistake, though.
Having read a lot of comments here, I do understand why you knee-jerked, though. But I really was asking for my own information.
O/T TheBigYinJames : thething is if people are to debate then they must be on the same planet, and some people are off in their own Lalaland
Interesting. Can you be a bit more specific? I’m not doubting you, just interested in what other LW emitters are contributing to the atmospheric downward LW apart from the surface? What else is emitting in the correct wavelengths?
“What else is emitting in the correct wavelengths?”
Post albedo solar power absorbed in the atmosphere and non-radiative flux moved from the surface into the atmosphere (convection and latent heat primarily).
Quote
“Notice that DLR does not drop significantly overnight. This is because of the heat capacity of the atmosphere – it cools down, but not as quickly as the ground.”
While this statement is obviously correct for land is it true over the sea ?
Since the daily variation of sea surface temperature is normally less than 1 C, the DLR will vary even less. Over land on a clear cold dry night, the surface can have a radiative excess on the order of 100 W/m². If the surface is relatively dry, the surface will cool rapidly. That’s not going to happen over the ocean because the humidity near the surface is never low and the near surface layer of the ocean is well mixed so the effective heat capacity is much larger than the land surface.
[…] concept is supported by the laws of physics. A good three article explanation from scienceofdoom : "The Amazing Case of "Back Radiation" What DLR Measurements Exist? Hundreds, or maybe even thousands, of researchers over the […]
Copied from Temperature Profile in the Atmosphere..
SOD and LW
I must be missing something; for this and the previous article and comment all seem more complicated than need be.
The slope of the dry adiabat (the lapse rate) varies inversely with specific heat at constant pressure. H2O has a higher specific heat than dry air and the moist lapse rate is steeper than the dry one. CO2 has a lower specific heat than dry air and moves the adiabat back towards its original slope. Neither change in the slope of the adiabat affects the surface temperature. Why, then, do we expect increasing quantities of CO2 in the atmosphere to warm the surface?
A warmer surface requires an outwards shift of the adiabat. Adding CO2, the heaviest of the atmospheric gases, would do this. By contrast, adding the lighter-than-air H2O gas would shift the adiabat inwards and reduce surface temperature.
But change in atmospheric density is not among the explanations usually given for a warming surface. One of these is that adding CO2 forces atmospheric radiation to space higher and weaker resulting in the atmosphere losing energy to space at a lesser rate. Supplied with energy from below at an unchanged rate, the atmosphere thus stores energy and warms. A warmer atmosphere relative to the surface steepens the adiabat (reduces the lapse rate). It doesn’t outward shift the adiabat and warm the surface, as Leonard Weinstein’s explanation implies. The warming of the atmosphere carries a self-regulating negative feedback – its rate of energy loss to space increases and stored energy and temperature reduce. With reference to the adiabat, the addition of CO2 moves the radiating co-ordinate up the adiabat and rotates the adiabat clockwise about its origin at the surface; while the negative feedback reverses these movements. But wherever the adiabat and the radiating co-ordinate on it settle at the new equilibrium, the surface temperature remains unchanged.
Another explanation involves the warmer atmosphere radiating to the surface at greater intensity causing it to warm. This is the contentious “back- radiation” which follows Prevost’s theory of radiative exchanges (1792) but contravenes Clausius’s 2LoT (which forbids energy transport from colder to warmer regions- 1850). SOD explains away the apparent theoretical conflict with his credulity-stretching “imaginary vs real 2LoT”. LW on the other hand, though recognising back-radiation, disallows its direct surface warming quality, instead attributing to it a quality of resisting energy loss from the surface which translates, via an ill-defined mechanism, to a warming surface due to storage of solar energy.
Hoping that my interpretations of these analyses are reasonable, evidently both can’t be right and, I would submit, both are flawed. The models employed by both analysts ignore the ubiquitous presence of radiation and direct interaction with bodies it surrounds. Absorption and emission of radiation involving the surface or atmospheric GHGs would then result in changes in the energy density of the surrounding radiation medium, absorption reducing it, emission increasing it. This would obviate the general need for complicating view factors between bodies mutually exchanging radiation and for atmospheric back-radiation specifically. It would also obviate the cognitive disjunct between K&T’s assessment of the relative energy supply to the surface from the atmosphere and from the sun, 2:1 in favour of the atmosphere, and everyday experience.
SOD, if I might be so bold as to venture to suggest a theme for a future article it would be: Experimental provenance of mutually exclusive theories: radiative exchanges which prescribes atmospheric back-radiation and 2LoT which proscribes it.
Copied from Temperature Profile in the Atmosphere, from RW:
John,
“It would also obviate the cognitive disjunct between K&T’s assessment of the relative energy supply to the surface from the atmosphere and from the sun, 2:1 in favour of the atmosphere, and everyday experience.”
Are you aware that not all the direct radiative power from the atmosphere to the surface is acting to warm the surface or maintain the surface temperature? Much of the direct radiative power from the atmosphere to the surface is just replenishing non-radiative power leaving the surface but not returned, making it a net zero energy flux entering the surface.
RW, what you describe looks a bit like what I think of as a “useless loop” between the surface and atmosphere. Maybe “redundant” would be a better word. However, whereas you see a zero balance between downward radiative and upward non-radiative powers, my zero balance is purely radiative. Why does the upward non-radiative power need to be replenished?
Because the surface receives more direct radiative power from the atmosphere and Sun than it emits, so much of the direct radiative power from the atmosphere to the surface has to be replenishing non-radiative energy leaving the surface but not returned.
Don’t forget that the energy balance at the surface is the sum of a radiative fluxd and a non-radiative flux, and that sum cannot be greater or less than about 390 W/m^2 if the surface is directly radiating 390 W/m^2 as a result of its temperature (288K).
[…] […]
[…] radiation for clear-sky and cloudy conditions, but none has been adopted for generalized use. The Amazing Case of ?Back-Radiation? | The Science of Doom __________________ Quote of the Week:Bri Pat: "Plenty of history needs to be rewritten […]
[…] done much research on it… Here's one of the proponents websites trying to explain this… https://scienceofdoom.com/2010/07/17/…ack-radiation/ and another from the famous Roy Spencer a guy who believes in theory but just doesn't buy the […]
“satellites cannot measure the downward radiation at the surface” However as DL:r causes heating at the earths surface it is reflected in OLR.which can be measured by sattelite.
Well, DLR is not “reflected”. It’s absorbed and re-emitted. The temperature at the surface is a function of many variables.
OLR is a function of surface temperature and the concentration and absorption line strengths of the various absorbing molecules in the atmosphere.
So measuring OLR only allows a calculation of DLR based on many other parameters. In that case it would be better to measure surface temperature and infer the DLR. But convective heat transfer from the surface is the most difficult to calculate and is only approximated via parameterized equations – so with two important unknowns the surface temperature does not really allow us to measure DLR (note 1).
This is why DLR measurements are so important. If you knew the surface temperature, the incident solar radiation and the DLR (note 2) at each location vs time you would have a mechanism for accurately knowing convective heat loss from the surface.
Note 1: We can only infer it via models when we know things like the concentration of radiatively-active gases, the temperature profile of the atmosphere, and the wind speed.
Note 2: This also requires knowledge of the “material properties” of the surface – emissivity, absorptivity vs wavelength. These values need to be known to calculate reflection of solar radiation and emission and reflection of terrestrial (longwave) radiation.
[…] David Archer ha mandato apprezzamenti e critiche al paper "molto importante ma faticoso" di Jim Hansen et al., e al loro modello per il raddoppio della concentrazione atmosferica della CO2, conseguente fusione delle calotte polari e innalzamento del livello del mare; Andrew Revkin del New York Times segnala due articoli che dovrebbero contraddire l'evoluzione di tsunami e tempeste prodotta dal modello; Erik Stabenau, oceanologo del National Park Service per le Everglades, vorrebbe sapere "quali osservazioni attuali o entro un decennio verificherebbero gli assunti del modello"; Matthew Whipple obietta che non c'è consenso sul collasso del West Antarctic Ice Sheet durante l'Eemiano; il commento più bizzarro è di un certo Nabil Swedan che nega l'esistenza della back radiation: […]
Are you aware of any long term datasets of measured DLR showing that its actually increasing in line with expectations?
The post seems to concentrate on the fact its measurable and has been measured over short time frames.
Tim,
The value of DLR depends on the temperature of the atmosphere (actually the temperature profile) and the GHG concentrations (water vapor, CO2, etc).
Actually this is just fundamental physics and verified many times. (Simple examples, clear night, low water vapor content = low DLR; hot humid day = high DLR).
What are your expectations of its increase?
What drives the surface warming from GHG changes (all other things being equal) is from the changes in OLR (outgoing longwave radiation).
I’m expecting that a doubling of CO2 ought to cause about a doubling of DLR. Isn’t that the expectation?
Tim,
No.
You should read through the Visualizing Atmospheric Radiation series – it will probably explain a lot.
I’m not talking “full atmosphere” at the moment. I’m considering a very local effect. Say 1m off the surface. Is the answer still “No”
Perhaps I should clarify. If the GHGs double then would you expect the DLR from say 1m above the surface to then produce double the radiation back at the surface (ignoring any that would be captured over that 1m)?
There are, afterall, twice as many molecules capturing and radiating at the same atmospheric temperature.
Tim wrote: “If the GHGs double then would you expect the DLR from say 1m above the surface to then produce double the radiation back at the surface… There are, after all, twice as many molecules capturing and radiating at the same atmospheric temperature.”
When CO2 is doubled, twice as many photons are emitted, but their mean free path before absorption is half as long. This means the same flux will reach the ground after a doubling – to a first approximation.
This approximation is true for strongly absorbed wavelengths. At a weakly absorbed wavelength where the mean free path is 2 km, doubling CO2 will reduce the MFP to 1 km. After doubling, there will be slightly more photons reaching the surface because the average photon reaching the ground is now emitted from lower where it is warmer. .
You can watch DLR from increasing CO2 saturate using MODTRAN at
http://climatemodels.uchicago.edu/modtran/
Zero out all the GHGs and choose to look up from 0 kilometers (at the DLR arrive at the surface). Notice the baseline isn’t flat. Add 1(!) ppm of CO2 and keep doubling. Notice that saturation prevents a doubling of DLR even between 1 and 2 ppm. The differences between 200, 400, and 800 are subtle.
Blackbody spectral intensity (the colored curves) are what is present when absorption and temperature-dependent emission are frequent enough to reach equilibrium with the absorbing molecules they are traveling past.
Thanks Frank, but my argument when it emerges, assuming I’ve not misunderstood anything, wont be “full atmosphere”. I fully appreciate the importance of the lapse rate in the standard argument. Mine is going to be at a much lower level and I dont mean altitude wise 😉
Oh and to build on what you’ve said here (although somewhat unrelated to where I am heading)
“When CO2 is doubled, twice as many photons are emitted, but their mean free path before absorption is half as long. This means the same flux will reach the ground after a doubling – to a first approximation.”
I expect there is an “altitude” that is quite low to the ground that is analogous to the ERL where most of the radiation does reach the surface. Would you agree?
TTTM wrote: “I expect there is an “altitude” that is quite low to the ground that is analogous to the ERL where most of the radiation does reach the surface. Would you agree?”
Sure. For the strongest line absorbed by 400 ppm of CO2, the ERL is allegedly 1 m. Most of the photons emitted downward from CO2 below 1 meter with reach 1 m^2 of surface. 1 m^3 of atmosphere with 400 ppm provides the DLR. At 800 ppm, the ERL will be at 0.5 m and only 0.5 m^3 of atmosphere will provide the DLR. Since there is no temperature difference between 0.5 and 1 m, there will be no change in the spectral intensity of DLR from doubling at this wavelength.
Radiation of blackbody intensity is what is emitted after absorption and (temperature-dependent) emission have come into equilibrium with each other and absorber they are passing through. Planck’s Law was derived by assuming such an equilibrium and a material composed of “quantized oscillators. Hohlraums were devised to create such an equilibrium. For some wavelengths passing through our atmosphere, effective equilibrium can be reached within a few 10’s of meters in the lower atmosphere and temperature doesn’t change much over short distances. Those wavelengths have BB spectral intensity. Doubling the concentration of a GHG has no effect when absorption and emission are in equilibrium. The band is “saturated”. At other wavelengths, the temperature changes faster than equilibrium can be reached. Then DLR arriving at the surface has less that BB intensity for Ts. In the atmospheric window, no equilibrium occurs and DLR at these wavelengths has the spectral intensity for the source above the atmosphere. That is the microwave radiation left over from the Big Bang that fills “empty” space.
Playing with MODTRAN is the best way to develop a reliable intuition about radiation transfer. Until you understand why CO2 alone increasing from 0 to 1 to 2 to 4 … to 1024 ppm changes DLR as it does, you shouldn’t rely on intuition. Equations can be interpreted. Numerical integration of the differential equations of radiative transfer is a black box. The blackbody curves on the MODTRAN plots tell you when radiation has the spectral intensity for a blackbody at 300, 280, 260, 240 or 220. When CO2 emission from a tropical atmosphere (Ts = 300 K) has blackbody intensity, the ERL must be near the surface.
…well a doubling of DLR caused by the CO2 part anyway. Obviously all the other GHGs aren’t impacted.
TTTM: Try MODTRAN. Wate,r vapor has nothing to do with the inability of doubled CO2 to double DLR.
Consider an 0.5 thick pane of glass or other material. It emits approximately like a blackbody, eoT4.. Add another 0.5 cm sheet of the same material. Does it emit twice blackbody intensity, 2eoT4? Of course not. Why do you expect twice as much CO2 to emit twice as much radiation when you don’t expect the same of our materials?
So what I meant to say was that a doubling of GHGs (rather than CO2 specifically) would mean a doubling of DLR.
Again, the series I referenced above is well worth working through. In brief, no, nothing in this field is linear. And, simplifying to “too simple by far”, there are really two relationships – the relationship between the surface temperature and the temperature profile of the troposphere; and the relationship between OLR, concentration of GHGs and temperature profile of the troposphere.
More CO2 => the upper atmosphere is more opaque to longwave radiation = causes OLR to reduce, because the higher up the radiation to space the colder the atmosphere.
This means OLR reduces but (all other things remaining equal) the absorbed solar radiation is the same so the climate system warms up.
Eventually, due to the warming, the OLR matches the absorbed solar radiation – and this means the surface temperature will be warmer.
Change in DLR is not really the mechanism.
Another one to read – The “Greenhouse” Effect Explained in Simple Terms.
please see my specific question above.
TimTheToolMan,
“So what I meant to say was that a doubling of GHGs (rather than CO2 specifically) would mean a doubling of DLR.”
No, not even close to doubling DLR (at the surface). DLR is around 300 W/m^2! The surface radiates only around 390 W/m^2.
Tim,
What SoD says. The effect is not only not linear, it’s logarithmic for doubling CO2. Most of the contribution CO2 (at pre-industrial levels) has on the GHE has already occurred, but doubling it still incrementally enhances it somewhat.
The magnitude of the enhancement so far as surface warming is debatable, but radiative transfer models estimate it to be approximately 3.7 W/m^2 before the system has adapted to the change or imbalance. That is -3.7 W/m^2 at the TOA, or a reduction in IR flux passing into space from around 240 W/m^2 to 236.3 W/m^2.
As you’ve not responded, I’ll try to simplify the question even further. If one GHG molecule radiates at a certain rate for a given atmospheric temperature, will two molecules radiate at twice that rate for the same atmospheric temperature? Is that what science would tells us?
From my point of view it would seem so given that radiation is basically a probability.
TTTM,
As SoD pointed out above, it’s not that simple. Sure, two molecules may, in principle, emit twice as much radiation as one. But there are problems. Thermodynamic temperature is not defined for a few molecules. They won’t follow the Boltzmann energy distribution, etc., etc.
In one cubic meter of air at standard temperature and pressure with a CO2 partial pressure of 0.0004 atmospheres, there are ~1E22 molecules of CO2. You can answer your own question by using MODTRAN or spectralcalc.com. I could do your homework for you by working out an example, but you’ll learn more if you do it yourself.
http://climatemodels.uchicago.edu/modtran/
Doubling CO2, even at fairly high climate sensitivity, won’t double the amount of water vapor, which is the primary source of DLR. So doubling the well mixed ghg’s other than water vapor won’t double DLR and won’t even double the DLR from the ghg’s. At high concentrations, the system isn’t linear, it’s logarithmic.
TTTM,
To put it another way, you can find the answer to your question in the articles linked, but you probably haven’t bothered to read them.
TimTheToolMan,
You are asking an excellent question. You deserve a proper answer rather than being told that you are asking the wrong question. I will give it a shot.
For simplicity, let us imagine a planet for which the only greenhouse gas is CO2; no water, clouds, etc. Now imagine doubling CO2, and keeping the temperature the same (it won’t stay the same, but I don’t think the extra complication matters for your question). Consider a layer in that atmosphere, say, between 100.0 m and 100.1 m. You will indeed get twice as much emission from that layer.
But you will also get twice as much absorption in that layer. As a result, a photon will travel half as far. So if before the CO2 increase a photon would travel on average 1.0 meter before being re-adsorbed, then after the increase it would travel only 0.5 m. Thus, although the total number of photons being emitted or absorbed in a given layer will be twice as large, the total passing *through* a particular level will be the same.
So if you are looking up from the surface, with the lower CO2 concentration you would “see” the lowest 1.0 meter of the atmosphere and with the higher concentration you would see only the lowest 0.5 meter. So in the latter case you are seeing photons from half the volume. Although there are twice as many photons emitted per unit volume, the total number you see is the same.
OK, examples:
MODTRAN Tropical atmosphere, observer at 0 km, looking up, default settings: Total energy from 100-1500 cm-1 = 347.912 W/m².
Same conditions except increase CO2 from 400 to 800 ppmv: 349.482 W/m², an increase of 1.57 W/m².
Same conditions except water vapor scale, methane, tropospheric and stratospheric ozone set to zero so we’re mostly looking at CO2:
400ppmv CO2 = 101.736 W/m²
800ppmv CO2 = 109.492 W/m²
Restoring the other ghg’s and setting CO2 to zero:
340.69 W/m²
Surface DLR in the tropics is mostly from water vapor.
As SoD pointed out, it’s not the increase in DLR at the surface, it’s the decrease in OLR to space.
Since MODTRAN doesn’t change the stratosphere in it’s calculations, radiative forcing is best observed from the tropopause, 17km for the Tropical Atmosphere looking down.
Default conditions, 400ppmv CO2 OLR (100-1500cm-1) = 290.136 W/m²
800 ppmv CO2: OLR = 285.489 W/m²
Nothing but CO2:
0 ppmv CO2: OLR = 414.166 W/m² 0km looking down 416.05 W/m²
400 ppmv CO2: OLR = 368.322 W/m²
800 ppmv CO2: OLR = 362.67 W/m²
Everything else but CO2: OLR = 325.304 W/m²
TimTheToolMan,
..catching up on the questions and answers from others..
Not much to add with the answers from others.
It’s always dangerous to read into a question what the questioner intended, this brief attempt of mine is only aimed at being helpful, not at being adversarial..
Perhaps you meant – can we confirm the change in “greenhouse” effect, due to changes in CO2, by our measurements of DLR? Is theory matched by experiment?
Answer – no. The change in “greenhouse” effect, due to increases in CO2, is not revealed by DLR changes.
TTTM: Why would you expect twice as much CO2 to emit twice as much DLR? All common objects emit eoT^4 W/m2 of thermal radiation. If you make the material twice as thick, you still get eoT^4 W/m2, not 2eoT^4. If you accept this behavior, then you should believe it is possible that an atmosphere with 2XCO2 won’t emit twice as much DLR as 1XCO2. You are completely correct in saying that sometimes twice as much material does emit twice as much radiation. Sometimes it doesn’t. So how do you decide which rule applies?
Take a sheet of any material that you expect to emit eoT^4 radiation even if its thickness were doubled. Make it half as thick. Still expect eoT^4? Keep cutting the thickness in half. At some point, the material will be thin enough to partially see thru. Apparently gold foil can be pounded into a thin enough layer to partially see through. Silver coated sunglasses are partially transparent. When material is thin enough to partially see through, it must be obvious it is no longer emitting eoT^4, so of the radiation you are see is clearly coming from behind. The rest of the radiation you see will vary with the thickness of the material, but you will need a cold background to measure it.
Reverse this process and reconstruct a thicker layer of material from partially transparent sheets. The reason the emitted radiation changes is because the radiation emitted by each layer is partially absorbed by the layer(s) in front, which are also emitting. Blackbody radiation is the result of many rounds of absorption and emission going on inside solid objects.
In the atmosphere near the surface, molecules are about 10X further apart in each direction than in a solid. It is fully transparent at some wavelengths of thermal infrared, partially transparent at others and nearly opaque at others. The 400 ppm of CO2 in 10 m of atmosphere is enough to block almost all radiation at some wavelengths, at others you can see the cosmic background microwave radiation left over from the Big Bang. (That isn’t thermal IR, of course.)
WIth everything, there is a transition point in thickness between seeing twice as much radiation emitted by twice as many molecules and seeing radiation of blackbody spectral intensity (when equilibrium between emission and absorption occurs before the radiation leaves the material). (Emissivity arises from reflection at the interface between the surface and air.) Our atmosphere is confusing because it spans both of these extremes.
When you hear about “optically-thick layers of atmosphere”, that is the code-phrase for saying that the layer is thick enough to emit oT^4. (I hate models with such layers, because they are misleading about our atmosphere.) When you hear about “optically-thin” layers, emission is proportional to the number of GHG molecules in the layer.
Modtran numerically integrates the emission and absorption by a many optically thin layers. You get to choose the composition of the atmosphere and a few other variables.
“What’s amazing about back-radiation is how many different ways people arrive at the conclusion it doesn’t exist or doesn’t have any effect on the temperature at the earth’s surface.”
I can explain, it is very simple. A pyrganeometer is a thermopile with a sensor recording the temperatur of the pyrganeometer. This is because the emission of the device has to be accounted for. A thermopile measures the transfer of heat from it’s surroundings, the net transfer, through changes in resistance due to the incoming heat flux. The device is shielded to measure heat flux from only one direction by measuring the difference in intensity at the surface of the thermopile and the air above. It has a filter that excludes nearly all solar radiation. What it measures is the gradient.
The basic function is only a thermopile, a thermometer that measures net heat transfer. It could be described as a directed thermometer.
The pyrgeometer measures a bit further up in the air than a thermometer, but not much, according to wikipedia 25 meters and in manuals I find no mention of it’s reach. The thing is, it is used based on the assumption that there is DLR in the amounts claimed in the greenhouse theory, but what it really does is measure the difference between the emission of the thermopile and the incident radiation. It is assumed that the incoming radiation is from a stack of air that is equal to the mean free path length of 25 meters.
What we have is a an equivalent to thermometer measuring the temperature above it. At best, it measures the lowest 25 meters but I think it measures what is directly above. That means that the values that are claimed to be downward flux actually show how fast the air cools in the first 25 meters, since it measure the difference in emitted intensity at the surface of the thermopile and the first 25 meters. This is done by measuring the resistance in the thermopile that change according to it’s temperature.
In this manual from kipp&zonen it is explained how it works:
https://www.google.se/url?sa=t&rct=j&q=&esrc=s&source=web&cd=2&cad=rja&uact=8&ved=0ahUKEwjIntXpj-DSAhViKcAKHYR6COcQFggrMAE&url=http%3A%2F%2Fwww.kippzonen.com%2FDownload%2F33%2FCG-4-Manual&usg=AFQjCNEYJUNXMUGRTl9j7bjI39QatAc9pw&bvm=bv.149760088,d.ZGg
It says that the output shows a negative value, coming from a heat transfer -from- the thermopile. They use the difference calculating the net transfer from the pyrgeometer to the air above, and here comes the strange thing, the two fluxes that give the net transfer are added together and claimed to be the DLR. Net downward radiation measured by a pyrgeometer is not a measure of how much heat that is transferred to the ground, even less it can be claimed to measure DLR since it actually measures temperature in a slightly different way over a possibly slightly larger depth in the atmosphere.
What a pyrganeometer actually shows is how much heat the thermopile transfer to the air directly above it. The calculation of what is called DLR is done by using the emission of the air that is implied by the transfer measured from the thermopile, and claiming to be a measured downward flux. The only thing that is really measured is how much heat, or longwave radiation, that is transferred from the surface upwards. The whole device is based on an unproved assumption that there is thermal energy transferred from the cold atmosphere to the surface, but in textbooks including heat transfer such an explanation is not found. Actually, in textbooks about heat transfer there is no descriptions of any energy being transferred from a cold body to a warmer body, that have any effect on temperature at all. If the authors didn´t include a chapter about the earth and the greenhouse effect. The theory of the greenhouse effect is the only place where you find this kind of heat transfer from cold to warm, and that is not a good thing, that should everyone agree of.
If a theory in physics is proven and clearly defined, as the theory of heat transfer where heat is calculated as the net energy transferred, and another theory about the temperature of the planet is the only one claiming heat can be transferred even though it is not included in the calculation of net transfer, we should be very skeptical of that theory. Because the theory of thermal radiation and the included transfer of heat shows us through countless experiments and also by application in countless devices in modern society, that heat is the net energy transfer from a warm to a cold body, and that only the net energy transferred is involved in the process of rising temperatures.
And the greenhouse theory tries to say that in the troposphere, most of the net energy transferred is from cold temperatures in the air to the hot surface.
It think it is very easy to come to the conclusion that there is no downward transfer of energy to the surface from the atmosphere that has an effect on temperature when dissecting the theory and the device. The device has the greenhouse theory built into it. And there are other problems with the device as well. Looking at spectrums this is confirmed, the up-looking spectrum has very little radiation in most parts where earth emits. The area under the curve is equal to the intensity of incident radiation, and as can be seen, co2 has a almost full coverage, it follows the curve of surface emission. But it is only a fraction of the width of the spectrum so it emits much less radiation in comparison. That is actually a sign that it emits at it´s full capacity in the temperaturerange we find at the surface, and that makes it a efficient transporter of energy to the parts that is colder. The colder parts in the surroundings is the other gases shown in the spectrum that emits with almost no intensity. The spectrum shows how co2 is a very efficient transporter of energy from the surface to the other greenhouse gases that swallows the heat entirely without any significant emission, which means without rising in temperature. That´s how we are used to see water interacting with heat, lots of energy has to be absorbed to increase temperature.
The upwards spectrum on the other hand, shows how co2 absorbs large amounts of heat from it´s surroundings at higher altitude without rising in temperature, cooling the higher altitudes efficiently. A natural AC-gas for the atmosphere.