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All IPCC definitions taken from Climate Change 2007: The Physical Science Basis. Working Group I Contribution to the Fourth Assessment Report of the Intergovernmental Panel on Climate Change, Annex I, Glossary, pp. 941-954. Cambridge University Press.

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Sun & climate: moving in opposite directions

What the science says...

Select a level... Basic Intermediate Advanced

The sun's energy has decreased since the 1980s but the Earth keeps warming faster than before.

Climate Myth...

It's the sun

"Over the past few hundred years, there has been a steady increase in the numbers of sunspots, at the time when the Earth has been getting warmer. The data suggests solar activity is influencing the global climate causing the world to get warmer." (BBC)

Over the last 35 years the sun has shown a cooling trend. However global temperatures continue to increase. If the sun's energy is decreasing while the Earth is warming, then the sun can't be the main control of the temperature.

Figure 1 shows the trend in global temperature compared to changes in the amount of solar energy that hits the Earth. The sun's energy fluctuates on a cycle that's about 11 years long. The energy changes by about 0.1% on each cycle. If the Earth's temperature was controlled mainly by the sun, then it should have cooled between 2000 and 2008. 

TSI vs. T
Figure 1: Annual global temperature change (thin light red) with 11 year moving average of temperature (thick dark red). Temperature from NASA GISS. Annual Total Solar Irradiance (thin light blue) with 11 year moving average of TSI (thick dark blue). TSI from 1880 to 1978 from Krivova et al 2007. TSI from 1979 to 2015 from the World Radiation Center (see their PMOD index page for data updates). Plots of the most recent solar irradiance can be found at the Laboratory for Atmospheric and Space Physics LISIRD site.

 

The solar fluctuations since 1870 have contributed a maximum of 0.1 °C to temperature changes. In recent times the biggest solar fluctuation happened around 1960. But the fastest global warming started in 1980.

Figure 2 shows how much different factors have contributed recent warming. It compares the contributions from the sun, volcanoes, El Niño and greenhouse gases. The sun adds 0.02 to 0.1 °C. Volcanoes cool the Earth by 0.1-0.2 °C. Natural variability (like El Niño) heats or cools by about 0.1-0.2 °C. Greenhouse gases have heated the climate by over 0.8 °C.

Contribution to T, AR5 FigFAQ5.1

Figure 2 Global surface temperature anomalies from 1870 to 2010, and the natural (solar, volcanic, and internal) and anthropogenic factors that influence them. (a) Global surface temperature record (1870–2010) relative to the average global surface temperature for 1961–1990 (black line). A model of global surface temperature change (a: red line) produced using the sum of the impacts on temperature of natural (b, c, d) and anthropogenic factors (e). (b) Estimated temperature response to solar forcing. (c) Estimated temperature response to volcanic eruptions. (d) Estimated temperature variability due to internal variability, here related to the El Niño-Southern Oscillation. (e) Estimated temperature response to anthropogenic forcing, consisting of a warming component from greenhouse gases, and a cooling component from most aerosols. (IPCC AR5, Chap 5)

Some people try to blame the sun for the current rise in temperatures by cherry picking the data. They only show data from periods when sun and climate data track together. They draw a false conclusion by ignoring the last few decades when the data shows the opposite result.

 

Basic rebuttal written by Larry M, updated by Sarah


Update July 2015:

Here is a related lecture-video from Denial101x - Making Sense of Climate Science Denial

 

This rebuttal was updated by Kyle Pressler in 2021 to replace broken links. The updates are a result of our call for help published in May 2021.

Last updated on 2 April 2017 by Sarah. View Archives

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Further viewing

Related video from Peter Sinclair's "Climate Denial Crock of the Week" series:

Further viewing

This video created by Andy Redwood in May 2020 is an interesting and creative interpretation of this rebuttal:

Myth Deconstruction

Related resource: Myth Deconstruction as animated GIF

MD Sun

Please check the related blog post for background information about this graphics resource.

Comments

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Comments 451 to 475 out of 1289:

  1. Gord - Let's get this straight - Considering the approx. 33 K difference between the approx. 255 K surface temperature with no greenhouse effect and the approx. 288 K surface temperature with the greenhouse effect about as much as it is (or recently has been): 1. I correctly note that this 33 K warming is from the greenhouse effect, without any change in albedo, and I agree with you (as if I didn't know before) that albedo would change if the atmosphere in it's entirety is removed. 2. NASA makes a mistake in their explanation, implying that removal of the atmosphere would leave the albedo unchanged, or that the 33 K warming comes from the greenhouse effect minus the cooling from the albedo increase, which is of course not true. I know that NO atmosphere means NO atmosphere, but I, assuming I am not more knowledgable than NASA, give them the benifit of the doubt, and choose to think that they did not actually intend what they mistakenly wrote. That should be quite clear to you, and you should have no reason to think I don't understand words. 3. You seem to hold NASA's error against me or against climatology in general - as if it were up to me (or Kiehl and Trenberth (okay, maybe more likely), or...) what NASA says - and at the same time, tell me I have delusions of grandeur, when I said that I assumed that I was not more knowledgable on the subject than the people at NASA, and so chose to assume that they actually know that the 33 K warming is from the greenhouse effect with no change in albedo and that they were just sloppy in their writing. (I still do not think I am smarter or more knowledgable than the people at NASA in general - however I do think I much more knowledgable than you - you keep giving me evidence for that.) 4. For NASA's mistake, you seem to suggest that I am wrong about the greenhouse effect. I have offered over the last several pages of comments and references therein a very good description of the greenhouse effect, but you look for erroneous descriptions from other sources to argue against me, to argue against things - things that I also agree are erroneous. 5. Your entire point seems to be that there is no greenhouse effect. You were arguing against Kiehl and Trenberth's energy budget, and they do not make any mistake in assuming that the atmosphere contributes to albedo. I have made no mistake in that regard either. You also seem to be aware that there is some significant albedo. Yet you offer a calculation of temperature with zero albedo as support for your argument that there is no need to use the greenhouse effect to explain the temperature of the surface of the Earth. BOTTOM LINE: The albedo of the Earth, including atmospheric effects - is about 0.3. With that albedo, but in the absence of the greenhouse effect, the average temperature of the surface would be about 255 K, or possibly slightly colder. The average temperature of the surface is actually near 288 K. Meanwhile, the emission of radiation to space can be observed and it is less than the radiation that the surface would emit to space if it were completely exposed. The difference is especially associated with high cold cloud tops, and at wavelengths where absorption by atmospheric gases - H2O and CO2 in particular, is expected to be sizable based on known optical properties.
  2. Gord - Let's get this straight - Considering the approx. 33 K difference between the approx. 255 K surface temperature with no greenhouse effect and the approx. 288 K surface temperature with the greenhouse effect about as much as it is (or recently has been): 1. I correctly note that this 33 K warming is from the greenhouse effect, without any change in albedo, and I agree with you (as if I didn't know before) that albedo would change if the atmosphere in it's entirety is removed. 2. NASA makes a mistake in their explanation, implying that removal of the atmosphere would leave the albedo unchanged, or that the 33 K warming comes from the greenhouse effect minus the cooling from the albedo increase, which is of course not true. I know that NO atmosphere means NO atmosphere, but I, assuming I am not more knowledgable than NASA, give them the benifit of the doubt, and choose to think that they did not actually intend what they mistakenly wrote. That should be quite clear to you, and you should have no reason to think I don't understand words. 3. You seem to hold NASA's error against me or against climatology in general - as if it were up to me (or Kiehl and Trenberth (okay, maybe more likely), or...) what NASA says - and at the same time, tell me I have delusions of grandeur, when I said that I assumed that I was not more knowledgable on the subject than the people at NASA, and so chose to assume that they actually know that the 33 K warming is from the greenhouse effect with no change in albedo and that they were just sloppy in their writing. (I still do not think I am smarter or more knowledgable than the people at NASA in general - however I do think I much more knowledgable than you - you keep giving me evidence for that.) 4. For NASA's mistake, you seem to suggest that I am wrong about the greenhouse effect. I have offered over the last several pages of comments and references therein a very good description of the greenhouse effect, but you look for erroneous descriptions from other sources to argue against me, to argue against things - things that I also agree are erroneous. 5. Your entire point seems to be that there is no greenhouse effect. You were arguing against Kiehl and Trenberth's energy budget, and they do not make any mistake in assuming that the atmosphere contributes to albedo. I have made no mistake in that regard either. You also seem to be aware that there is some significant albedo. Yet you offer a calculation of temperature with zero albedo as support for your argument that there is no need to use the greenhouse effect to explain the temperature of the surface of the Earth. BOTTOM LINE: The albedo of the Earth, including atmospheric effects - is about 0.3. With that albedo, but in the absence of the greenhouse effect, the average temperature of the surface would be about 255 K, or possibly slightly colder. The average temperature of the surface is actually near 288 K. Meanwhile, the emission of radiation to space can be observed and it is less than the radiation that the surface would emit to space if it were completely exposed. The difference is especially associated with high cold cloud tops, and at wavelengths where absorption by atmospheric gases - H2O and CO2 in particular, is expected to be sizable based on known optical properties. ---- "It is very, very tedious explaining things to you and backing up my statements with links." Ha! I (perhaps quite unwisely) took the time to go over your mistaken ignorant babbling more or less point by point; I've used as pedantic an argument I ever would and then used another, I've explained the microscopic basis of macroscopic phenomena such as the second law of thermodynamics. I've explained here and offered references to other comments explaining just how the greenhouse effect and radiative energy transfer actually work (in a way that is easy to visualize, hence my occasional anthropomorphism, as in 'what the surface can see'), at times using the examples you offered. I've gone over the math of blackbody radiation. I might as well be your science tutor. And for the most part, all you ever do is repeat a bunch of ill-thought out, ill-informed assumptions that most serious physicists would laugh at.
  3. " I might as well be your science tutor." By the way, I QUIT.
  4. Patrick - Re: your posts #461 and #472: "2. Heat Radiation between hot and colder objects" "P = e*BC*A(T^4 - Tc^4)" When both the hot object and colder object have emissivities e and ec that could be less than 1, and there is a layer between with transmissivity T: P = (e*ec*T)*BC*A(T^4 - Tc^4)" ---------------------- e*BC*A(T^4 - Tc^4) does not equal (e*ec*T)*BC*A(T^4 - Tc^4) PROOF: e does NOT equal e*ec*T and the units of P (watts) have now been changed...by Patrick !!! All justified by Patrick's delusional OPINION. ------------ Mindless drivel that ANY algebra student could OBVIOUSLY see. Only a total "incompentent" and/or someone with "delusions of grandeur" would consider "modifying" the Stefan-Boltzmann Law and the units of Watts. Absolutely Hilarious! Adios, Patrick....you have "QUIT"...finally!
  5. Mizimi - RE: Your Post #467 Your "simplified" model violated two fundamental Laws of Science and actual measurements! Enough said.
  6. Patrick 027 464 Earth’s climate can be evaluated as a dynamic system for which the science of Control Theory applies. That means that all of the minutia of climate and weather, all of that stuff; forcings, Climate Science type feedbacks, GHG spectral absorption, ocean turnover, plant respiration, atmospheric carbon dioxide level, methane level, etc. etc. everything that Climatologists attend to (including all that you have described in excruciating detail) and even those things that haven’t been discovered yet; everything gets lumped together in a block in the Control Theory block diagram called ‘all internal factors that can alter average global temperature’. The only requirement is that none of them can add any energy to the block. The output from this block is average global temperature (agt). In Control Theory, feedback is a dimensionless number that is proportional to a change in output. In global climate it is a measure of the effect that change to agt has on how effective change to the energy entering the global climate system is at influencing agt. Feedback is a number that is added to 1.000 and the sum is multiplied by the input to the global climate system. Thus a feedback of 0.01 means that the input to the global climate system is multiplied by 1.01. If the feedback is -0.01 the input to the global climate system is multiplied by 0.99, etc. Note that feedback in Control Theory is dimensionless while feedbacks in Climate Science have units. Temperature data are readily available for the last and previous glacial periods (see, e.g. the few numerical data at 414 and graphs in my pdf linked from http://climaterealists.com/index.php?tid=145&linkbox=true . Since the agt trend changes the direction of slope from down to up and vice versa, Control Theory determines that the temperature can not be controlled by temperature feedback. A familiar example of a feedback system is the cruise control for a car. When working properly, it controls on speed and the speed of the vehicle is held fairly constant. Now if the speed changes substantially one must conclude that the unit is not controlling on speed. That is, there is no feedback from speed. In this analogy, agt is analogous to vehicle speed. Since earth’s temperature has changed as shown by the accepted temperature data from the last glacial period, it shows that there can be no significant positive (Control Theory) feedback from temperature in earth’s climate. Since there can be no significant Control Theory feedback, there can be no significant net Climate Science feedback either because that would influence temperature. Without Climate Science net feedback none of the twenty or so Global Climate Models that the IPCC uses predict significant global warming.
  7. Important correction to comment: http://www.skepticalscience.com/argument.php?p=19&t=474&&a=18#3338 I used the same symbol for two different variables (transmissivity (unitless value) and temperature (units K or some other unit on an absolute scale), which caused confusion: P = (e*ec*T)*BC*A(T^4 - Tc^4) What variable shall we use for transmissivity? How about N: P = (e*ec*N)*BC*A(T^4 - Tc^4) Then let U = e*ec*N*BC*A The rest was correct: P = U * (T^4 - Tc^4) This can be rewritten as: P = U*T^4 - U*Tc^4 Original point still valid. Units are correct. Inteligent people sometimes make simple slip-ups like that. This had the significance of a typo - I think I had meant to use a different variable and must have absent-mindedly used called transmissivity T. However, anyone reading that could have figured out what I was trying to communicate and made the correction for their self. And yes, I have been known to substitute 'their' for 'there', as might be noticed in prior comments.
  8. Patrick 027 468 Thermalized means that the energy is mostly conducted to adjacent non-GHG molecules that do not significantly re-radiate. "When a radiant flux is absorbed, it is assumed to be mostly thermalized" Where then does all that radiation back to the surface come from? If the K&E graphic was not misleading there would be no need for all the clarification/explanation/rationalization.
  9. Dan - "A familiar example of a feedback system is the cruise control for a car. When working properly, it controls on speed and the speed of the vehicle is held fairly constant. Now if the speed changes substantially one must conclude that the unit is not controlling on speed. That is, there is no feedback from speed." That doesn't quite make sense - what if there was positive feedback - then the speed could certainly change. ---- If the temperature goes up and down without forcing (which does happen - look at the all the wiggles about the longer-term trend of the last several decades), that doesn't mean that there are no positive feedbacks. (There are fluctuations in cloud cover patterns, humidity patterns, and evaporation and latent heating and temperature patterns associated with variability in circulation patterns in the atmospheres (which can also contribute to surface albedo fluctuations), so there are both thermal and mechanical (momentum) feedbacks through which circulation patterns cause changes in circulation patterns - but over time there tends to be a predictable statistical description of all that (which is the longer-term climate)... Heat can also be added to or removed from the surface environment by vertical transports in the ocean.) As I was explaining in the last comment to you, I think I saw where a major miscommunication may lie. In equilibrium (for the longer-term average over internal variability), the radiant energy out equals the radiant energy in. A solar TSI increase is a positive forcing, increasing the radiant energy input. An increase in CO2 is a positive forcing, decreasing the radiant energy output. In climatology, the equilibrium response without any feedbacks is generally described as the temperature increase required to change, directly through blackbody emission's dependence on temperature, the radiant energy output, so as to restore balance. However, I suspect that in Control theory, that is itself considered feeback. Remove that feedback, and what happens? The rate of net energy storage is constant - if the heat capacity doesn't change, temperature increases linearly over time to infinity. That is the climate response without any feedbacks, including the feedback of blackbody radiation as a function of temperature (let's call it the blackbody radiation feedback). The approx. 1 K increase due to a doubling of CO2 includes that feedback. And in the absence of forcings, this feedback is a negative feedback that must be stronger than the net positive feedback from all other feedbacks if the climate is to be stable. Consider the water vapor feedback again. Using the same values as before (and using a linear approximation to responses): in response to a positive radiant forcing, the temperature rises. Without any feedbacks including the blackbody radiation feedback, the temperature rises indefinitely. But with the blackbody radiation feedback, their is an equilibrium temperature increase of 1 K, and the difference between the equilibrium temperature and the temperature decays exponentially at a rate determined by the climate sensitivity and the heat capacity. If the water vapor feedback increases the equilibrium temperature change from 1 K to 2 K, then this implies that the water vapor response to a 2 K temperature change is sufficient to force a 1 K temperature increase - it would cause indefinite warming if the blackbody radiation feedback did not stop it. It might help to use some time-dependent formulas: Let T be the change in temperature from some reference value. Let R be the radiant forcing difference from some reference level. Let the blackbody radiant feedback be B, where B = b*T, and assume b is negative. Let F be the net of all other radiant feedbacks to the change in temperature, where F = f*T. Let the heat capacity (per unit area) be C. Time is t. Conservation of energy: dT/dt = (R+F+B)/C dT/dt = [R + (f+b)*T]/C dT/dt = R/C + T*(f+b)/C ------------------------------- For R = constant, Let T = Teq - A*exp(-Z*t) A*exp(-Z*t) = Teq - T then dT/dt = Z*[A*exp(-Z*t)] dT/dt = Z*(Teq - T) IF the climate is stable, Z is greater than 0. --- dT/dt = Z*Teq - Z*T and dT/dt = R/C + T*(f+b)/C --- therefore Z*Teq = R/C Z*T = -T*(f+b)/C --- Z = -(f+b)/C Z = R/(C*Teq) e-folding time of (Teq-T) = 1/Z = C*Teq/R = C * climate sensitivity -(f+b)/C = R/(C*Teq) -(f+b) = R/Teq climate sensitivity = Teq/R = -1/(f+b) Teq = -R/(f+b) ------------------------------- IN SUMMARY (where R, T, and Teq are defined as relative to reference values): Whatever has happened up to time t, if R is set to a constant value, then: If the climate is stable, there is a constant equilibrium T = Teq, of the same sign as R, that T tends to approach over time through the exponential decay of (Teq - T). This exponential decay has a time scale (a time constant, or e-folding time) of 1/Z = C * Teq/R, which is the product of the heat capacity (per unit area) and the climate sensitivity. Teq = -R/(f+b) 1/Z = C * Teq/R = -C/(f+b) ------ (Teq-T) = exp[-Z*t] = exp[ -t / (C * Teq/R) ] = exp[ t*(f+b)/C ] ------ The climate is stable if the climate sensitivity is either zero or positive and finite. The climate sensivity, defined as R/Teq, is equal to -1/(f+b), where B = b*T is the blackbody radiant feedback and b is negative, and F = f*T is the net effect of all other feedbacks. ------------------------------- An increase in R from an initial value to a new value shifts Teq. If (Teq-T) was zero before the change in R, it will not be immediately zero after the change; (Teq-T) will then exponentially decay to zero. Notice that what happens to (Teq-T) is the same if, instead of changing the external radiative forcing R, there is an unforced change in T. An unforced change in T from Teq will tend to exponentially decay over time if the climate is stable. A stable climate is compatible with a positive value of f, provided that f is not greater than the absolute value of b. When climatologists say the net feedback is positive, they are refering to f; when they state the climate sensitivity Teq/R without feedbacks, they mean with f = 0; they still include the effect of a negative b. The climate sensitivity with f = 0 is -1/b ---------- Revisiting an earlier example, considering water vapor as the only contributor to f: Let -1/b = 0.25 K / (W/m2) (which means that a 0.25 K increase in temperature increases the outgoing radiation by 1 W/m2 if f=0; b = -1/0.25 (W/m2)/K = -4 (W/m2)/K) Teq = 0 K when R = 0 W/m2 (these are both relative to a reference or baseline state). Then imposing a forcing R of 4 W/m2 increases the Teq by 1 K if f=0. If T = Teq before R was changed from 0 W/m2 to 4 W/m2, (Teq-T) will equal 1 K before T responds. As (Teq-T) decays to zero from 1 K, T rises from 0 to 1 K, at a rate depending on C and R/Teq for f=0. Before T has changed, C*T is increasing at a rate R = 4 W/m2. But for each 0.1 K increase in T, the radiative feedback -B increases 0.4 W/m2,But when T has increased 0.5 K, that rate has dropped to R+B = R+b*T = (4 - 2) W/m2 = 2 W/m2. For each halving of (Teq-T), the rate of energy gain is halved. Hence, if C is constant, (Teq-T) decays exponentially. The time for each halving of (Teq-T) is proportional to C and proportional to Teq/R. T (K), B (W/m2), R+B (W/m2), Teq-T (K), 0.0, -0.0, 4.0, 1.0 0.1, -0.4, 3.6, 0.9 0.2, -0.8, 3.2, 0.8 0.3, -1.2, 2.8, 0.7 0.4, -1.6, 2.4, 0.6 0.5, -2.0, 2.0, 0.5 0.6, -2.4, 1.6, 0.4 0.7, -2.8, 1.2, 0.3 0.8, -3.2, 0.8, 0.2 0.9, -3.6, 0.4, 0.1 1.0, -4.0, 0.0, 0.0 ----- Let f = 0.5 * (-b) = 2 (W/m2)/K. Thus, as T increases 0.1 K, there is a net radiative feedback (except for B) of F = 0.2 W/m2. F+B for each 0.1 K increase is half of B alone, F+B = -0.2 W/m2. Because f+b has been halved, Teq will double to 2 K for R = 4 W/m2. Because F and B are both 0 when T = 0 K, the initial rate of T increase is the same regardless of what f and b are; C*T increases initially at a rate R = 4 W/m2. Since halving the sum f+b doubles Teq/R, T has to change twice as much, so it makes sense that the climate response time will now be twice as long (PS this makes it hard to determine climate sensitivity based on the earliest state of a climate response to a change in R). T (K), B (W/m2), B+F (W/m2), R+B+F (W/m2), Teq-T (K), 0.0, -0.0, -0.0, 4.0, 2.0 0.1, -0.4, -0.2, 3.8, 1.9 0.2, -0.8, -0.4, 3.6, 1.8 0.3, -1.2, -0.6, 3.4, 1.7 0.4, -1.6, -0.8, 3.2, 1.6 0.5, -2.0, -1.0, 3.0, 1.5 0.6, -2.4, -1.2, 2.8, 1.4 0.7, -2.8, -1.4, 2.6, 1.3 0.8, -3.2, -1.6, 2.4, 1.2 0.9, -3.6, -1.8, 2.2, 1.1 1.0, -4.0, -2.0, 2.0, 1.0 1.1, -4.4, -2.2, 1.8, 0.9 1.2, -4.8, -2.4, 1.6, 0.8 1.3, -5.2, -2.6, 1.4, 0.7 1.4, -5.6, -2.8, 1.2, 0.6 1.5, -6.0, -3.0, 1.0, 0.5 1.6, -6.4, -3.2, 0.8, 0.4 1.7, -6.8, -3.4, 0.6, 0.3 1.8, -7.2, -3.6, 0.4, 0.2 1.9, -7.6, -3.8, 0.2, 0.1 2.0, -8.0, -4.0, 0.0, 0.0 The rate of change remains proportional to Teq-T, but per unit Teq-T is half what it was when b+f had twice its value and Teq was only 1 K. _____________________________
  10. "Patrick 027 468 Thermalized means that the energy is mostly conducted to adjacent non-GHG molecules that do not significantly re-radiate. "When a radiant flux is absorbed, it is assumed to be mostly thermalized" Where then does all that radiation back to the surface come from?" I explained that before when I explained what thermalization is. When a GHG molecule absorbs a photon, in most cases, it exchanges energy with other molecules before it would radiate a photon. This keeps the different gaseous substances in the same unit volume of air at about the same temperature in spite of different optical properties. But GHG molecules also radiate photons. They radiate photons, in each wavelength interval in each direction, at a temperature-dependent rate that statistically follows blackbody radiation as a function of temperature, multiplied by the emission cross section of each molecule. After a GHG molecule radiates a photon, it has less energy, but through molecular interactions, energy tends to be redistributed to keep all the gaseous substances near the same temperature in each unit volume. Hence, the air as a whole loses energy when the GHG molecules (and clouds, etc.) it contains emit photons. The net change in the enthalpy of the air (enthalpy is better to use with air because it expands as it gains heat, so that some of the energy actually does work, which increases the heat capacity (at constant pressure) relative to a constant volume value.) is caused by the absorbed radiant energy minus the emitted radiant energy, plus latent heat gain (if the latent heat present in water vapor is not counted toward the enthalpy of the air), and if within the first mm or so of the surface, conduction from the surface (and diffusion of water vapor from a wet surface, if water vapor latent heat is counted toward the enthalpy of the air). "If the K&E graphic was not misleading there would be no need for all the clarification/explanation/rationalization." Well, maybe it was intended for other scientists who would recognize what it means without much need to figure it out.
  11. "If the K&E graphic was not misleading there would be no need for all the clarification/explanation/rationalization." Actually, I think that's a bit like saying that this is misleading: that rain and snow, and other precipitation, comes from clouds which form when water vapor condenses (and sometimes freezes); the water vapor comes from evaporation from wet surfaces. Well, are dew and frost considered 'precipitation'? Sometimes clouds and precipitation evaporate before reaching the surface. Etc. Some plants get moisture directly from fog that is not actually precipitating. Aside from all that, though, there are a few missing links in the water vapor -> cloud -> precipiation chain: How does condensation take place without a surface (a small droplet has a high internal pressure because the surface tension of the spherical surface is squeezing it; this raises the equilibrium vapor pressure so that it tends to evaporate; this makes it very hard to form a droplet from pure gas, because it has to start out very small)? How do cloud droplets and ice crystals gather into particles large enough to fall out at significant rates (it is hard to grow large enough from continual condensation on particles from the vapor phase in the short time periods that are observed)? (Answers: generally aerosols. Some aerosols are more or less effective at nucleating cloud droplets. (If hydrophilic, they provide a surface so that the initial droplet can have less internal pressure from surface tension; perhaps much more important, if soluble, they provide a solute that can decrease the equilibrium vapor pressure, and more so at high concentrations - such as occurs when the smallest amount of water vapor condenses on a large enough aerosol. Some particles(such as salt) are considered hygroscopic because they pull moisture out of the air even when below 100% relative humidity. Moist particles go through a haze particle phase in which increasing size increases the relative humidity (relative to a flat surface of pure water) necessary to keep it in equilibrium (because the size increase dilutes the solute); once enough particles per unit volume reach some size where the necessary supersaturation for equilibrium decreases with increasing droplet size (due to the decrease in internal pressure from the surface tenstion), those particles continue to grow as cloud droplets, bringing the relative humidity back to about 100 %, and causing the evaporation of remaining haze particles. See also "Kohler curve".) Some are more or less effective at nucleating ice crystals. When the temperature gets down toward -40 deg C, homogeneous nucleation of ice within liquid water becomes more likely; the rate of homogeneous nucleation is a per unit volume rate, so larger droplets will freeze faster at higher temperatures than smaller droplets will. Larger droplets may also be more likely to contain an effect ice nucleus (?). When droplets hit ice particles, ice crystals may be nucleated. After some ice is nucleated, the surface of a droplet may freeze first, leaving liquid water within it, which can break the ice particle into smaller pieces when it freezes and expands. Different size particles have different terminal velocities and fall at different rates, so they may run into each other and sometimes combine. (Turbulence might have a local centrifuge effect that would amplify the differential motion of particles, though I'm not sure this is significant). Ice surfaces have a lower saturation vapor pressure at a give temperature than liquid water surfaces, so in the presence of a few ice particles, many small liquid droplets tend to evaporate as water vapor diffuses and is deposited to form a few larger ice particles that can then fall at a higher speed.)
  12. Correction: ... Larger droplets may also be more likely to contain an effectIVE ice nucleus (?)...
  13. Patrick - Re: 476 Original Equation (Stefan-Boltzmann Law) P = e*BC*A(T^4 - Tc^4) which when expanded and giving ec as the emissivity of Tc gives: P = e*BC*A*T^4 - ec*BC*A*Tc^4 If T = temp of the Earth and Tc = temp of the atmosphere, this is a clear subtraction of power produced by the Earth and the power produced by the atmosphere. Clearly, the emissivity of the atmosphere does not have any affect on the power produced by the Earth and vice-versa. ---------- Here is the "Patrick's Law" equation pulled out of his imagination with no logical development shown: What variable shall we use for transmissivity? How about N: P = (e*ec*N)*BC*A(T^4 - Tc^4) Patrick said..."When both the hot object and colder object have emissivities e and ec that could be less than 1, and there is a layer between with transmissivity T:" (Patrick later changed T to N.) ---------- Notice that "Patrick's Law" does not require a temperature for the layer between T and Tc! He just has N = transmissivity for this layer....THAT'S ALL! emissivity = 1 - Transmissivity - reflectivity http://www.optotherm.com/emiss-physics.htm Thus, the emissivity of this new layer of can be expressed as en = (1 - Tn - Rn) Substituting En into Patrick's Law gives: P = (e*ec*en)*BC*A(T^4 - Tc^4) and expanding gives: P = e*ec*en*BC*A*T^4 - e*ec*en*BC*A*Tc^4 Now the power produced by the Earth is changed by the new atmospheric layer emissivity and the higher atmospheric layer emissivity! A clear violation of the Stefan-Boltzmann Law....and ABSOLUTELY HILARIOUS! Only a total "incompentent" and/or someone with "delusions of grandeur" would consider "modifying" the Stefan-Boltzmann Law. What utter drivel.
  14. Patrick 027 478 "That doesn't quite make sense - what if there was positive feedback - then the speed could certainly change." Maybe it doesn't make sense because you have not grasped that feedback as used in Control Theory is different from feedback as used in Climate Science. See 475.
  15. Patrick 027 479 So what fraction of the energy radiated from the surface and absorbed by the GHGs is thermalized?
  16. Dan - "Maybe it doesn't make sense because you have not grasped that feedback as used in Control Theory is different from feedback as used in Climate Science. See 475." Well, maybe, ... but what do you think of the feedback in climate science now? (Point being that you can't say that feedback by one definition is not positive because feedback by another definition is not positive - When climatologists say that the warming from a doubling of CO2 would be a bit over 1 K without feedbacks, they mean without feedbacks except for the increased thermal radiation to space as a function of temperature; adding positive feedback to make the warming 3 K may be entirely compatable with a negative feedback in Control theory. ---------- "So what fraction of the energy radiated from the surface and absorbed by the GHGs is thermalized?" Essentially all of it.
  17. To whomever may wish to know: Regarding: ----------------------------- P = (e*ec*N)*BC*A(T^4 - Tc^4) Then let U = e*ec*N*BC*A The rest was correct: P = U * (T^4 - Tc^4) This can be rewritten as: P = U*T^4 - U*Tc^4 ----------------------------- This refers to the net radiant power transfer between two layers, that is emitted and absorbed by those layers. It does not include emissions and absorptions in other layers. It assumes the layers are thin enough for each to be nearly isothermal within itself vertically (horizontal temperature variations are generally far to spread out compared to optical thickness to matter much, though there could be occasional exceptions on a small scale). Relative to perfect blackbody layers with entirely transparent space in between, the amoung of radiant power emitted by one layer and absorbed in another is proportional to the emissivity e1 of the emitting layer and the absorptivity of the absorbing layer, a2, and the transmissivity of the intervening space, N. The radiant power emitted by the second layer and absorbed in the first is proportional to e2 * a1 * N. At local thermodynamic equilibrium, emmissivity = absorptivity (at each location and time, direction, and wavelength). It is generally a very safe assumption that each small unit volume (small enough to be nearly isothermal) is approximately in local thermodynamic equilibrium. (Fluorescence is an example of the type of radiation that can be emitted when a system is not in local thermodynamic equilibrium - the energy involved has not been thermalized.) Hence, a2 = e2 and a1 = e1. So, for the radiant power exchanged by two layers by emission and absorption, the radiant power in both directions, and thus, the net radiant power, are all proportional to e1*e2*N. In the above equations, e1 = e and e2 = ec, but more generally: Pjk = net P per unit area from layer j to layer k = ej*ek*Njk*BC* ( Tj^4 - Tk^4 ) ------ Of course, this formulation is for the most simple case where ej, ek, and Njk are not wavelength dependent, and also requires Njk to be the transmissivity for the total flux, integrated over direction (it also assumes the refractive index of each layer is close to 1). Still keeping the approximation of index of refraction = 1 (a very safe approximation for the atmosphere): For a single direction and wavelength, if the intervening space is made of layers l=j+1 to k-1 (assuming k is greater than j, otherwise ... you get the idea), and Nm is the transmissivity of layer m, then Njk = Nj+1 * Nj+2 * Nj+3 * ... * Nk-1 [where j+n,k+n are all subscripts], and in the absence of scattering or reflection, Nm = 1 - em. Identifying the direction by it's angle from verticle, q, and identifying the wavelength by L, and for each wavelength, measuring the layer thicknesses out so each layer has the same e and thus the same N, Ijk(L,q) = e'j*e'k*[(1 - e'j+1)*(1 - e'j+2)* ... *(1 - e'k-1)]*BC*[ Ibb(Tj) - Ibb(Tk) ] = e'^2 * (1-e')^(k-j-1) * BC * [ Ibb(Tj,L) - Ibb(Tk,L) ] Where Ijk(L,q) is the spectral intensity (monochromatic intensity per unit interval of the spectrum) of the radiation - the radiant power per unit solid angle per unit area normal to the direction of radiation, per unit interval of wavelength (it can also be formulated as per unit interval of frequency); Ibb(T,L) is the blackbody intensity at wavelength L at temperature T. **In this case, e' is the emissivity along a path at angle q through a layer, so it will be different for different q. What is e' if e is the emissivity along a vertical path through a single layer? Well, if N is the transmissity along a vertical path, and N' is the transmissivity in the direction q, then: N' = 1 - e' N = 1 - e and N' = N^sec(q), because sec(q) = 1/cos(q) = the path length through the layer in the direction q relative to a vertical path length, and the trasmissivity through a path is equal to the product of the transmissivities of subunits of the path. hence, e' = 1-N' = 1 - N^sec(q) = 1 - (1-e)^sec(q) --- Solid angle (symbol w used below) is measured in steradians (symbol sr), which is the projection (from the center of a sphere) of a solid angle onto a spherical surface area, divided by the square of the radius, and thus, like radians, steradians can be treated as unitless. The spectral power (monochromatic power per unit of spectrum) per unit horizontal area = Pjk(L) = (integral from q = 0 to q = pi/2)[ Ijk(L,q) * cos(q) * 2*pi*sin(q)*dq ]. The integrand has a factor of cos(q) because Ijk(L,q) is a value per unit area normal to the direction q; that unit area is a fraction cos(q) of its projection onto a horizontal surface. The factor 2*pi*sin(q)*dq is the the solid angle dwencompassed by the interval of directions dq (it is a ring centered around q=0, of angular width dq from the center and angular length 2*pi (full circle) around the center. cos(q) * 2*pi*sin(q) = pi * sin(2*q) *** pi/2 is a measure in radians, in case anyone was confused by that; pi ~=3.1415927 and pi/2 is a 90 deg angle. Integrating from q = 0 to q = pi/2 encompasses all the contributions from a hemisphere of solid angle. It is not necessary to integrate over the whole sphere of directions because the quantity being integrated is a net value that is the sum of radiant intensities in opposing directions. The same result would be obtained by integrating the contributions to power per unit horizontal area from the intensity from each direction, over the whole sphere of solid angle. PS: For isotropic radiation I(q) = I, the radiant flux per unit area F = (integral from q = 0 to q = pi/2)[ I * cos(q) * 2*pi*sin(q)*dq ]. = I * (integral from q = 0 to q = pi/2)[ pi*sin(2*q)*dq ] = I * pi * [ -cos(pi) + cos(0) ]/2 = pi * I The solid angle in the hemisphere is: (integral from q = 0 to q = pi/2)[ 2*pi*sin(q)*dq ] = 2*pi*[-cos(pi/2) + cos(0) ] = 2*pi There are 2*pi steradians in a hemisphere. --- There is a less clunky way to figure this stuff out - we can let the layer thickness as measured by optical thickness become infinitesimally thin, and use calculus: still at a given wavelength: In a unit volume, there is a density of emission cross section, the emission cross section per unit volume = ecsv. This is equal to the absorption cross section per unit volume if at local thermodynamic equilibrium. ecsv from one substance is equal to the product of the emission cross section of that substance per unit mass times the mass of that substance per unit volume. The exv of different substances and additional amounts of substances add linearly to give the total. The exv is the effective blackbody surface area facing a given direction - each molecule contributes to ecsv with a cross-section. It is possible for ecsv to be direction-dependent (analogous to the molecules having elongated optically-defined shapes with some non-random alignment) but it is generally the case in gases that exv is isotropic (analogous to each molecule being spherical or to random alignments of molecules, so that on average, each molecule looks like a spherical blackbody that presents the same cross-sectional area in each direction). (Of course, the image of a molecule having some well-defined shape with a clear edge is just a mathematical equivalent to what it contributes, on average, to bulk optical properties.) (It is possible to imagine some macroscopic analogue to molecular cross-sections - imagine how radiation propagates through a room filled with blackbody spheres - except, there will be some diffraction around those spheres, which corresponds to a contribution to the scattering cross section on the molecular or small particle scale. Scattering on the molecular scale has a macroscopic analogue of mirrored and/or clear glass spheres or other shapes. The total (extinction) cross section (emission + scattering = extinction) can be different in different directions but must be the same in a pair of opposing directions; however, it would be possible to construct a macroscopic analogue in which the scattering and emission cross sections vary between pairs of opposing directions - for example, spheres that are mirrored on one side and perfect blackbodies on the other - but I think this might not have a molecular analogue, at least for radiation with wavelengths longer than the molecular scale (??).) A unit volume is equal to a unit path length with a given area. Thus, the escv, the emission cross section per unit volume is equal to the emission cross section (an area) per unit area per unit path length, and is thus a fraction of area per unit path length. We can consider an infinitesimally thin layer, of thickness dx, so that the fraction of area per unit path length, escv*dx, is nearly zero. Since the fraction of area of the whole layer dx is nearly zero, the fraction of area of any sublayer must be nearly zero, so the fraction of any sublayer cannot significantly block the fraction of another sublayer (the molecules that contribute to the emissivity are spread out enough so as to be very unlikely to align on top of each other and block each other from view in the direction x). Thus, essentially the entire emission cross section within such a thin layer dx is visible in the direction x. Thus, the emissivity of that layer in the direction x is equal to the fraction escv*dx. And the emissivity of the layer 2dx will be approximately 2*escv*dx, but actually just slightly less than that; the emissivity approaches 1 and cannot increase farther. The transmissivity (in the absence of scattering, and with the absorption cross section = emission cross section) is 1 - escv*dx for the layer dx in the x direction. The transmissivity through a layer n*dx is approximately 1 - n*escv*dx if n is not large, but it is exactly (1-escv*dx)^n; The transmissivity decreases by a fraction escv*dx for each dx. Where the transmissivity is N, dN/N = -escv*dx N = 1 when x = 0. Integrating: ln(N) - ln(1) = -escv *(x-0) ln(N) - 0 = -escv *x N = exp[-escv *x] For a radiant intensity at x, I(x), going in the x direction, I(x) = I(0)*N(x) = I(0) * exp[-escv*x], which I believe is refered to as Beer's Law - this applies to the portion of the radiation that was present at position 0 that reaches x, and does not include radiation emitted along the path from 0 to x. ------------ For the spectral intensity Ixs emitted from a dx at position x that reaches position x = s (going in the positive x direction): Ixs = Ibb(T(x),L) * escv * dx * exp[-escv *(s-x)] Let Is is the intensity of radiation at s, in the x direction, emitted from the layer between x=0 and x=s: Is = (integral from x = 0 to x = s)[ Ibb(T(x),L) * escv * exp[-escv *(s-x)] * dx ] Which can be numerically integrated for any T(x) function, but if we assume T(x) is a constant T between 0 and s: Is = (integral from x = 0 to x = s)[ Ibb(T,L) * escv * exp[-escv *(s-x)] * dx ] Is = Ibb(T,L) * escv * exp[-escv *s] * (integral from x = 0 to x = s)[ exp[-escv *(-x)] * dx ] Is = Ibb(T,L) * escv * exp[-escv *s] * (1/escv) * [ exp[escv *s] - exp[escv *0] ] Is = Ibb(T,L) * exp[-escv *s] * [ exp[escv *s] - exp[escv *0] ] Is = Ibb(T,L) * [ exp[escv *(s-s)] - exp[escv *(0-s)] ] Is = Ibb(T,L) * [ exp[0] - exp[-escv *s] ] Is = Ibb(T,L) * [ 1 - exp[-escv *s] ] = Ibb(T,L) * [ 1 - Ns ] where Ns is the transmissivity of the layer from 0 to s in the x direction. Not surprisingly, this implies the emissivity of the layer, es, is 1 - Ns. Of course, the absorptivity of the layer is equal to the emissivity of the layer (local thermodynamic equilibrium) ------ So the net radiative transfer in the x direction, from emissivity to absorptivity, between two layers g and h, seperated by layer s (where g, h, and s are the thicknesses of the layers in the x direction), defined as positive in the direction from g to h, is: Igsh(L) = eg * ah * Ns * Ibb(Tg,L) - eh * ag * Ns * Ibb(Th,L) where ej, aj, and Nj are the emissivity, absorptivity, and transmissivity of the layer j. Since ej = aj (local thermodynamic equilibrium): Igsh(L) = eg*eh*Ns * [ Ibb(Tg,L) - Ibb(Th,L) ] eg = 1 - exp[-ecsv *g] eh = 1 - exp[-ecsv *h] Ns = exp[-ecsv *s] --- What if we want to know the contribution of this radiant transfer (which is per unit area normal to the x direction) to the radiant transfer PGSH(L) per unit horizontal area, where G, S, and H are the layer thicknesses measured in the vertical direction? Let x be an angle q from vertical. Consider the solid angle dw. Note that h*cos(q) = H, so h = H*sec(q), and g = G*sec(q), and s = S*sec(q) d(PGSH(L)) = Igsh(L) * cos(q) * dw = eg*eh*Ns * [ Ibb(Tg,L) - Ibb(Th,L) ] * cos(q) * dw = [ Ibb(Tg,L) - Ibb(Th,L) ] * (1 - exp[-ecsv *g]) * (1 - exp[-ecsv *h]) * exp[-ecsv *s] * cos(q) * dw = [ Ibb(Tg,L) - Ibb(Th,L) ] * (1 - exp[-ecsv *G*sec(q)]) * (1 - exp[-ecsv *H*sec(q)]) * exp[-ecsv *S*sec(q)] * cos(q) * dw Assuming layers G and H are isothermal over horizontal distance (generally a good approximation for the spatial scales involved in radiative energy transfers, except in some conditions), then the same value applies to all directions at an angle q from vertical, so dw can be replaced with the ring-shaped solid angle 2*pi*sin(q) * dq, and integration can be done over q from 0 to pi/2 to find PGSH(L). Of course, ecsv will vary over vertical distance (And anywhere the path runs into a cloud, or surface, etc.) Within the atmosphere, ecsv will tend to be proportional to air density at wavelengths dominated by well-mixed gases (in the absence of clouds), but at many wavelengths (in between absorption lines), decrease faster with height, while at some wavelengths (near line centers), decrease more slowly. Water vapor is concentrated downward, and there are clouds, etc. --- The mathmatics can be simplied by using optical depth or optical thickness-based coordinates. A unit 1 optical thickness is the distance x over which ecsv*x = 1, so that transmissivity is 1/exp(1). Various different vertical coordinates are used in atmospheric science, including geometric height z (as in x,y,z), pressure (x,y,p), potential temperature (x,y,theta), and sigma coordinates (x,y,sigma), where sigma is set to one value at the surface regardless of surface pressure, and one value at infinite height (p=0), and varies linearly with pressure in between. Any of those vertical coordinates can be mapped onto optical thickness coordinates and vice-versa based on the optical properties at each position. Different mappings will be necessary for different wavelengths, however, so integration over the spectrum to find total fluxes will require converting back to one of the other vertical coordinates. Pressure coordinates will tend to have nearly-constant heat capacity per unit vertical distance, which is convenient for converting an energy flux convergence (in the absence of horizontal net fluxes, -1 * the vertical derivative of the net upward energy flux (power) per unit area) to a rate of temperature increase. Redefining (to conserve on variables) G, H, and S to vertical layer thicknesses in optical thickness coordinates: eg = 1 - exp[-G*sec(q)] eh = 1 - exp[-H*sec(q)] Ns = exp[-S*sec(q)] d(PGSH(L)) = eg*eh*Ns * [ Ibb(Tg,L) - Ibb(Th,L) ] * cos(q) * dw = [ Ibb(Tg,L) - Ibb(Th,L) ] * (1 - exp[-G*sec(q)]) * (1 - exp[-H*sec(q)]) * exp[-S*sec(q)] * cos(q) * dw --- For dw = 2*pi*sin(q) * dq, d(PGSH(L)) = [ Ibb(Tg,L) - Ibb(Th,L) ] * (1 - exp[-G*sec(q)]) * (1 - exp[-H*sec(q)]) * exp[-S*sec(q)] * pi*sin(2*q) * dq Using optical thickness coordinates is convenient; the surface and space can be regarded as layers with infinite thicknesses, over which T is (for x,y, and time) constant. To figure out the total net radiant energy transfer emitted from one layer G and absorbed in all layers above Hj, one must integrate over the different Hj layers. If one wants to know the total net radiant energy absorbed from layer by all layers above and below, one must sum the net fluxes from the other layers. If one wants to know the net upward flux at a given position, one must integrate the net upward flux between each pair of layers across that position. Of course, one must also integrate over solid angle and over the spectrum (or the portion of the spectrum being considered). Fortunately, the qualitative effects are easy to visualize. One can see some fraction of perfect blackbody radiation radiation intensity from a layer according to that layer's optical thickness, the optical thickness of the space between you and that layer, and the angle; and putting yourself in the place of another layer at the viewing location, the fraction absorbed depends on your optical thickness in that direction - that, multiplied by the other factors, is how much 'you would see' of the other layer from that direction, and that other layer would see the same amount of 'you'. With radiation, 'what you see' (pretend you can see at any wavelength) is what you get. --- What about scattering? For the extinction cross section per unit volume xcsv and the scattering cross section per unit volume scsv, xcsv = ecsv + scsv. The transmissivity over distance s without scattering Ns = exp[-xcsv * s] There will, however, be a glow of diffuse scattered radiation, which will add to the radiant intensity in other directions; some scattered radiation still reaches some forward distance after scattering. ... it's complicated (but qualitatively easy to understand - scattering, and reflection in general, decrease transmission between layers going in any one direction but replace what they block from view with radiation from the same side of the scattering/reflection as the viewing position; for a mix of absorption and scattering - any blocking of radiation from behind will be replaced with a combination of radiation from the blocking layer and radiation from the same side of the blocking layer as the viewing position, but with varying intensities depending on the temperature of the emitting sources); to be continued if/when I get to it...
  18. Patrick 027 485 There can be no significant NET Climate Science type positive feedback. So there could be at least as much positive as there is negative from the temperature increase. I expect that additional negative Climate Science type feedback will be discovered that is associated with clouds. I modeled the K&E graph except I included thermalization as an unknown. (first model April 29, updated today) I used average cloud temperature 258 K, cloud emissivity .5, cloud coverage 50% (see Zero-dimensional models at http://en.wikipedia.org/wiki/Climate_model) and varied the fraction of cloud radiation that reaches the ground from 50% to 90%. (I assumed cloud radiation to be graybody type from the water/ice particles) Thermalization varied from 16% for 50% cloud-to-ground to 24% for 90% cloud-to-ground. Considering all of the GHGs that the cloud radiation needs to penetrate to reach the ground, the fraction that makes it is probably closer to 50% than 90%. That which is not thermalized is needed as back radiation to the surface. I was surprised to see such a low number for thermalization but can't find anything wrong. Using their values for 'absorbed by atmosphere', thermals and evapotranspiration and correctly combining numbers I get acceptable agreement with what they got.
  19. Patrick - In your Post #461 you said... "Surface absorbs RSe from the sun and Rae from the atmosphere. Surface loses heat by Rea to the atmosphere, Res to space, and Cea (convection) to the atmosphere. The atmosphere absorbs RSa from the sun and Rea from the surface, and gains Cea from the surface, and loses energy by Rae to the surface and Ras to space. THESE EQUATIONS ARE BASED ON THE CONSERVATION OF ENERGY: Rate of energy gain by the surface = RSe + Rae - Res - Rea - Cea Rate of energy gain by the atmosphere = RSa + Rea + Cea - Ras In climatic equilibrium, the average of each rate of energy gain is zero. Where is energy being created or destroyed?" ---- Answer: + Rae = energy absorbed by the Earth's surface. Rae came from the Atmosphere that got it's energy from the Earth's radiation. The Earth's radiation cannot cause it's own radiation to increase. It's called Energy Creation....a violation of Conservation of Energy. Further, Rae IS GREATER THAN Rse and Rse is the energy absorbed by Earth's surface from the SUN ...THE ONLY ENERGY SOURCE!..another Energy Creation and violation of Conservation of Energy. ------------------- Your equations ARE BASED ON A VIOLATION OF CONSERVATION OF ENERGY. Like I have said...With Patrick, 1 + 1 = anything but 2. What great Comedy!
  20. CORRECTION TO http://www.skepticalscience.com/argument.php?p=19&t=488&&a=18#3338 The Rae term was mistakenly missing in the atmospheric energy budget; here is the correction: THESE EQUATIONS ARE BASED ON THE CONSERVATION OF ENERGY: Rate of energy gain by the surface = RSe + Rae - Res - Rea - Cea Rate of energy gain by the atmosphere = RSa + Rea + Cea - Rae - Ras
  21. Re 467 Mizimi - yes, basically, GHGs reduce the rate of energy loss to space from the surface for a given temperature profile. Additional amounts can also reduce the energy loss to space from the lower atmosphere when opacity has become high enough.
  22. Following radiation from one point along a path, scattering removes the radiation from the direction of the original radiation. Considering the radiation I(0) in one particular direction x that is present at x = 0, the amount I(x) found at any other x (that has continued unabsorbed and unscattered; not including radiation that has been emitted or scattered into the path between 0 and x) again follows Beer's Law: I(x) = I(0) * exp(-xcsv * x) where xcsv is the extinction coefficient, equal to the extinction cross section per unit volume (a unit volume as measued in the coordinate system being used). xcsv = ecsv + scsv; extinction cross section = absorption cross section (assumed to be equal to emission cross section) + scattering cross section. optical thickness of path length x = xcsv * x The contributions of scattering and absorption/emission to optical thickness add linearly. The ratio of the scattering cross section to the extinction cross section is called the 'single-scatter albedo' (don't remember if that is supposed to be hyphenated or not). The loss to scattering and absorption per unit x can be found by taking the rate of change over x in the reverse direction: -dI/dx = xcsv * I(0) * exp(-xcsv * x) = xcsv * I = ecsv * I + scsv * I where the last two terms are the absorption per unit x and the scattering per unit x; the single scatter albedo is the ratio of the scattered I to the total loss of I without scattering and emission into the path. Scattered radiation of course continues in a different direction, and adds to I along some other path. If the new direction is not perpendicular to the original, then there has been either forward scattering (the angle between the directions is less than 90 deg) or back scattering. Note that when refering to solar radiation that is back scattered to space, the term may apply a bit differently, because as the sun gets farther from directly overhead at any one location, a greater portion of radiation that is forward scattered relative to the direct rays of the sun is still redirected upward. The directional distribution of scattered radiation varies; the scattering of sunlight by air molecules is dominated by Raleigh scattering, where the scattered radiation intensity is distributed in a symmetric dipole pattern aligned with the incident radiation, with equal amounts forward scattered and back scattered, and the lowest scattering intensities are found in the ring of directions in between. In at least some other types of scattering, the forward scattering dominates. For scattering by particles, Raleigh scattering occurs for particles much smaller than wavelength; Mie scattering occurs over a range, including when the particle size and wavelength are similar; geometric optics dominates for particles much larger than a wavelength (see p. 307, "Atmospheric Science - An Introductory Survey", Wallace and Hobbs, 1977 (but there is at least one other more recent edition, I believe)).
  23. Refinement to 487: Assume an average emissivity for earth’s surface at 0.98 then the energy radiated from the surface at 288°K average is 382 W/m2. K&E show 40 W/m2 or 10.5% getting all the way to space. Barrett at http://www.warwickhughes.com/papers/barrett_ee05.pdf , assuming 50% cloud cover, calculates 11.2% which corroborates this estimate. Since radiation from the clouds to the surface is not half-blocked by cloud cover, it can be said with reasonable confidence that the fraction of radiation that leaves the clouds going towards the surface that actually gets all the way from the average cloud cover to the surface is about 21%. Plugging this value in to the model results in the determination that only about 37 W/m2 or 11% of the radiation that leaves the surface but doesn’t go through the ‘window’ getting thermalized.
  24. Some interesting things about scattering: For geometric optics, aside from the effects of diffraction, the scattered light from a mirrored sphere(specular (as opposed to diffuse) reflection: angle of incidence = angle of reflection) is redirected equally in all directions; it is isotropic. If my understanding is correct on this point: while one might expect that, when in the realm of geometric optics, the extinction cross section of a sphere (either by scattering by specular or diffuse reflection, scattering by refraction, or absorption) would be equal to its actual cross sectional area. But it is actually twice that, because of diffraction around the edge. I would think that diffraction contributes essentially to just forward scattering, and for relatively large objects compared to wavelengths, the scattered rays are mostly not at large angles to the incident radiation (so it is hard to observe over relatively short distances where slight changes in direction have little effect on location). ----- Depending on conditions, a photon may scatter many times (multiple scattering) between emission and absorption. When a given scattering cross section per unit volume produces mainly forward scattering within a narrow cone close to that of the incident rays, then multiple such scatterings might be approximated by using a wider scattering distribution with a smaller scattering cross section per unit volume (fewer scattering events with greater directional change per event). ---------- In clear sky, outside the disk of the sun itself, the highest intensity radiation may be observed (careful!) in a narrow cone of solid angle just around the sun. I think this cone is too narrow and focussed to be explained by Raleigh scattering by air molecules - I presume it is by aerosols (dust, etc.). Aside from that, at least in the middle of the day, the brightest parts of the sky (the highest intensities) are generally nearer the horizon - this is because one looks through more air when looking near the horizon than straight up; the 'glow' of scattered radiation comes from a greater scattering cross section per unit area because of the longer optical path. (For the same reason, the incandescent glow of LW radiation from the sky will tend to appear brighter near the horizon - in that case, there is greater emission cross section per unit area, both from the total atmosphere (between the surface and space) and from the warmer, lower layers (at wavelengths with higher opacity, low level inversions could make the sky appear dimmer near the horizon). - However, with scattering, there is the added complexity of the distribution of scattering relative to incident rays; it may be necessary for significant multiple scatterings (including (generally) diffuse reflection from the surface) to occur to explain the visible brightness near the horizon if the sun is overhead (???)) ---------- A haze or thin cloud may appear significantly brighter when viewed toward the sun (careful!) than when viewed looking away from the sun. This indicates that forward scattering dominates over backscattering. What happens in thick clouds? For a thick cloud, each layer of cloud adds another round of scattering for forward scattered radiation. Also, even if an initially parallel set of rays is spread out over a small solid angle for an individual scattering, successive forward scatterings continue to spread the rays out more and more, and some 'net backscattering' can occur from several forward scatterings. Whatever has not left the cloud is still subject to the high density of scattering cross section. The initial scattering is concentrated within the cloud toward the source of the radiation, and while successive scattering allows further penetration, it also spreads out the directions of the rays, so if the cloud is thick enough, the rays may be going in opposite directions nearly equally before getting more than halfway through the cloud, and so more escapes out the side it came in than escapes the other side if the cloud is thick enough.
  25. ". Plugging this value in to the model results in the determination that only about 37 W/m2 or 11% of the radiation that leaves the surface but doesn’t go through the ‘window’ getting thermalized. " Setting aside scattering (which is a minor issue for LW radiation), the radiation emitted by the surface that does not directly reach space must either be absorbed by clouds or absorbed by something else (gases, or aerosols) - either way, it is essentially all thermalized.

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