# Goddard’s world

The blog Watts up with That is famous for its attempt to reinvent climate physics on Earth, and now they want to reinvent astrophysics as well. Steve Goddard, a frequent poster at WUWT, now has a couple of posts (here and here) as well as write ups by Lubos Motl, about how the extreme temperatures on Venus are not caused by the greenhouse effect, but rather the 90 bar atmospheric pressure on Venus. It essentially argues that the “extra warming” by that we see on Venus is mostly due to the adiabatic lapse rate. As with many posts at WUWT, this radical idea which contradicts any textbook on planetary climate has been accepted without question by most commenters. Unfortunately for the massive attempt at a paradigm shift that is underway, they are all wrong.

For a very good overview of some of the basic features of the Venusian atmosphere, I suggest this 2001 paper by Bullock and Grinspoon who discuss the very important role of CO2, as well as the various competitive roles of H2O and SO2 in its climatic history. Goddard, Motl and many bloggers feel that this description is all wrong based on fundamental physical misunderstandings.

From high school mathematics, we know that a straight line can be represented by the equation y=mx + b, where m is the slope, and b is the intercept of the line. It’s important to see that the slope alone cannot tell you the absolute value at a single point, only the difference between two points. It is this that neither Motl nor Goddard can grasp. The lapse rate in an atmosphere without a condensable substance can be derived fairly easily. Starting from the first law of thermodynamics:

$dQ = c_{v}dT + pdV = 0$

and from taking the derivative of the ideal gas law for a mole of substance,

$VdP + pdV = RdT$

Note that R = cp – cv. Re-arranging, and for an adiabatic process,

$PdV = (c_{p}-c_{v})dT - Vdp$
$c_{p}dT=VdP$

Using the hydrostatic equation, where $dp = - \rho g dz$ and defining $C_{p}=c_{p}/mass$ we obtain

$dT/dz = - g/C_{p}$

This gives the rate of temperature change as one changes altitude. For a dry case on Earth, this equation corresponds to a temperature decline of nearly 10 K km-1. The actual value is somewhat less (closer to 6 K km-1) because the latent heat release of water vapor acts to slow the rate of temperature decline relative to the dry case. Alternatively, in pressure-space, one can use Poisson’s equation to determine the temperature T(p) at some level. For an atmosphere whose temperature profile is given by the dry adiabat, the surface temperature
is

$T_{sur}=T_{rad}(p_{s}/p_{rad})^{R/c_{p}}$

where the exponent is approximately 0.286 on Earth and about 0.23 in a pure CO2 atmosphere. The “rad” subscript designates a mean radiating level. On Venus, the ideal gas law does not hold well in the lower atmosphere, but nonetheless the lapse rate can be found to somewhat resemble that of Earth

Bullock and Grinspoon, Icarus 2001

What the IR heating from the greenhouse effect does is primarily to establish the intercept (e.g., the lower tropospheric temperature) while on Earth convection essentially keeps the atmosphere near some adiabat and thus establishes the vertical structure of the atmosphere. See below for example:

Viewed in terms of mass the tropopause isn’t deeper than on Earth. On both planets, some 90% of the mass is contained within the troposphere. There are two ways to get a very high tropopause though: a low lapse rate or very large optical thickness. The latter is what is important for pushing up the Venusian tropopause so high, well above Earth’s. This rate of change establishes the difference between two layers in the atmosphere. Strictly speaking, a greenhouse effect does not require a surface (a greenhouse effect still exists on the gaseous planets); one simply needs to extrapolate along the adiabat to some new level below prad and using Poisson’s equation one can find the temperature at that level. For Venus, one can see that prad is quite high, since transmission is related to the opacity of the atmosphere in a declining exponential function. This, among many other things, is Goddard’s misunderstanding within his perspective for why the greenhouse picture is “absurd” in his first post. Also, the Venusian surface does receive a relatively small trickle of solar radiation to the ground, and actually this small trickle is very important for establishing the temperature structure of the planet.

There are three lines depicted in the the second graph above. Line “A” and “B” represent the same sloped adiabat, which means that the difference in temperature between the surface and some point in the atmosphere is the same. The difference is of course where the line is shifted which cannot possibly be due to pressure, but rather through energy balance…either because the planet with line “B” has a higher solar constant, a lower planetary albedo, or a stronger greenhouse effect. This falls right from the basic energy balance equation, where

$\frac{S_{0}(1-\alpha)}{4} = \sigma T^{4}$

for a planet with a fully transmissive atmosphere (i.e., prad = psur). The greenhouse effect is thus defined as

G = $\sigma T_{sur}^{4} - OLR$

Since Venus radiates at the surface at several thousands of W m-2 and yet the outgoing radiation is even less than that of Earth, this shows how important a strongly opaque atmosphere influences the planetary surface.

Line “C” in the figure represents a planet with an even higher surface temperature, although with a more mild temperature gradient between two points on the line. This is actually how the negative lapse rate feedback works (which partially offsets the water vapor feedback), since the top of the atmosphere has now experienced a greater change in temperature than the surface, and can become a better emitter. In no case can the lapse rate be the reason for the very high surface temperature on Venus. It is simply not physical for the planet to be hotter than that value set by the incoming absorbed solar radiation in the absence of a greenhouse effect or some internal heat source

Note that pressure does matter for the greenhouse effect. Continuum opacity due to CO2 pressure-induced transitions in the Venusian atmosphere are very important. On a planet like Mars with little atmosphere, very little broadening from collisions makes CO2 a much less effective absorber over a variety of wavelengths. Bottom line is CO2 becomes a much better greenhouse gas at high pressures. In addition, scattering of infrared radiation, unlike Earth, plays an important role in the energy balance.

Another issue that has popped up in discussions is the logarithmic relation between CO2 concentration and radiative forcing. Some have even tried to extrapolate Venus concentrations using a logarithmic law to get some guesstimate for Earth-like sensitivity. This is really apples and oranges. One issue is that you lose the logarithmic relationship at high enough CO2 concentrations. If you put in too much CO2, you make the overall width of the principal absorption region wide enough that you get out into a different shape of envelope which would happen at a significant fraction of Earth’s atmosphere (say, 0.2 bars or so) let alone on Venus. On Venus, a lot of weak bands that are inconsequential on Earth become dominant in determining the changes in OLR. It’s also worth keeping in mind, that CO2 absorption at high wavenumbers which are important on Venus but not on Earth come into play. These are crucial for understanding the main absorption on Venus, and unfortunately, high wavenumber areas are not satisfactorily resolved by databases such as HITRAN.

From R.T. Pierrehumbert, Principles of Planetary Climate, in publishing progress

As a conclusion, the blogosphere has not revolutionized the field of planetary climate or the greenhouse effect. They have made elementary mistakes that would be resolved by taking a basic undergraduate course in atmospheric thermodynamics. Neither Motl nor Goddard understand what they are talking about, and this is just another fad that will be forgotten about soon.

### 152 responses to “Goddard’s world”

Chris, Excellent summary. Thanks for bring light to the lunacy. Ray

2. I’ve skimmed this but have other things to do . Perhaps you could cut to the chase .

By StefanBoltzmann & Kirchhoff the temperature of a gray body in Venus’s orbit is about 328k ( see http://cosy.com/Science/TemperatureOfGrayBalls.htm ) .

The spectrum of an object modifies that equilibrium temperature by the ration of the correlation of its spectrum with that of its sources and sinks , ie , the “greenhouse effect” .

All three heat flow equations I know of flow from hot to cold . Please show me the equations by which the interior of a sphere can come to a higher equilibrium temperature than that determined by SB & K for its externally observed spectrum other than by having an internal heat source .

I will interpret a lack of answer to this question an admission that it is not possible .

Response: See my post on the greenhouse effect— chris

3. Chris – nicely done. However, I think your statement:
“What the IR heating from the greenhouse effect does is primarily to establish the intercept (e.g., the lower tropospheric temperature)”

is not correct – you’ve already got the solution to what the greenhouse effect does – it determines T_rad and p_rad in your notation. The lapse rate to the surface then determines the surface temperature. With no GHG’s p_rad = surface pressure and T_surface = T_rad. With high GHG’s p_rad is quite low, and the surface temperature is raised.

This is close to the sort of simple explanation I’ve been trying to find, but a little elaboration on how p_rad is determined would be nice…

4. carrot eater

Bob Armstong:

You still have the basic error of assuming everything is a grey body. Emissivity and absorptivity are allowed to be functions of wavelength. The only requirement is that they be equal to each other, at any given wavelength.

For example, snow can have an emissivity near 1 in the IR wavelengths, and yet reflect a good deal of light in the visible wavelengths. This can be measured, and is entirely uncontroversial.

Likewise, the gases in the atmosphere absorb weakly in some wavelengths, and have strong absorption bands at another wavelengths.

So long as you insist on making everything a grey body, you won’t get anywhere.

5. jon

Nice post. I am really impressed by Goddard’s posts and I hope my undergraduate students understand the concept of adiabatic lapse rate better than Goddard does.

A simple response to Bob Armstrong.

Read ANY introductory book on atmospheric science (Wallace and Hobbs is nice, Hartmann is also nice, there are other classics such as Salby, Houghton and so on). Once you read and understand ONE of those introductory books, you will not pose preposterous questions in blogs, questions which only show your ignorance. Are you seriously pretending that the lack of a detailed answer means that we (Chris and any other sensible atmospheric scientist) admit your position? No, sir, you simply know nothing about the game and people are simply too busy to discuss with you about absurd questions. Have you thought on the fact that shortwave radiation is not all absorbed in the atmosphere? Have you actually solved the equations of energy balance for the cases $a_s a_l$?

This is hard to believe …

6. Carrot eater , did you miss my second paragraph ?

“”The spectrum of an object modifies that equilibrium temperature by the ration of the correlation of its spectrum with that of its sources and sinks , ie , the “greenhouse effect” .

Do you disagree with it ? Do you understand it ? You are saying the same thing in non-quantitative language . We are about 9c warmer than a gray body in our orbit . Since I don’t have time to spend on it , I am now offering $300 on my website to any student who extends the array language implementation for gray spheres to full spectra and computes actual equilibrium temperatures for all the relevant spectra . That should reduce the unknown variance for the earth’s temperature down to a few degrees at most . But that has essentially nothing to do with the extreme temperature of the surface of Venus since by the ubiquitous “frozen earth” amateurish computation seen , eg , on Wikipedia , Venus , having the highest albedo of all inner planets , should have the temperature most below its SB&K gray body temperature . So some other mechanism must be being called upon . It’s a simple , basic thermodynamics question . How does the interior of a sphere can come to a higher equilibrium temperature than that determined by SB & K for its externally observed spectrum ? I’m not looking for any “detailed answer” , just the couple of relevant equations that get you from the SB&K value for the surface to the higher interior temperature . Perhaps if you have one of those text books you refer to you could scan the couple of pages with the critical equations , if not transcribe them , and post them somewhere on the web to convert those of us who are stopped cold by this apparent violation of thermodynamics and common sense into true believers . Until then , I will continue to consider the level of understanding of the physics apparent on both sides of this debate pathetic . Response: But where is the incoming radiation balanced by the OLR via S-B? Hint: It isn’t at the surface– chris 7. “ration” should be “ratio” of course . 8. Anonymous Re Bob Armstrong: “All three heat flow equations I know of flow from hot to cold” As well they should be. But: 1. while convection and conduction carry heat from one location through the next via all points in between, radiation can carry heat across distances without necessarily depositing and then extracting that heat into and from intervening material. Radiative heat transfer among non-photons is from point of emission to point of absorption. Alternatively, one can say that heat can be transfered from other matter to photons or vice versa, but if the photons can move some distance without interacting with matter, then the heat the photons carry can be transported some nonzero distance with the photons independently of the temperature gradients in the matter. 2. Consider what happens at thermodynamic equilibrium. All activity does not cease, but what happens is that energy and matter are in an equilibrium distribution. Energy and matter tend to spontaneously spread (with some amount of inhibition from kinetic barriers) outward from whereever they are, to different places and forms. In an equilibrium distribution, energy and matter of any form and place are spreading into other forms and places at the same rate that energy and matter from other states are spreading into the first form and place. Molecules of any given type are diffusing in opposite directions at the same rate, going into and out of different physical states at the same rate; chemical reactions are occuring in the forward and reverse directions at the same rate; molecular energy (internal energy) is being transported in opposite directions at the same rate (via random molecular collisions and motions), photon energy is being emitted and absorbed at the same rate and photons are propagating in opposite directions at the same rate. When there is some concentration of something into a place or phase or form that is greater than the equilibrium amount, relative to how much is found in other places, phases, or forms, then more of that something tends to spread out than is replaced. That is causes a flow of heat from higher to lower temperature. It isn’t that all the relevant forms of energy or matter flows in the opposite direction cease, but that flux is just less then the flux in the opposite direction. Because of the way radiation is emitted and absorbed and propagates over distances, it is convenient to seperately consider the two opposite fluxes of radiant energy. Hence we may refer to the ‘backradiation’ from the generally colder atmosphere to the generally warmer surface. This doesn’t violate the second law of thermodynamics. What would violate the second law is if this backradiation were sustained somehow while the radiant fluxes emitted by the warmer surface and absorbed in the colder atmosphere, or for that matter, emitted from warmer parts of the system and absorbed by colder parts of the system, were less than those in the opposite direction. What the backradiation does is reduce the the net flux of radiant energy – the flux of radiant heat from warmer to colder can be reduced by changing optical properties, as can convection and conduction be reduced by changing other properites; for any such change, some changes in temperature would be required to cause an equal and opposite change to bring the heat fluxes back to what they were. This is how the greenhouse effect works, this is how a glass building greenhouse works, this is how a winter coat works; at the energy of some other form (solar energy, metabolism of food energy) goes into the system one way. Energy builds up unless and until it can leave the system at the same rate that it is supplied, but the system is not able to lose the energy at the same rate by the same way (because it is too cold to emit radiation at the same wavelengths, because it doesn’t make food), and so, energy builds up unless it can be lost by some other route (emission of photons at other frequencies, convection, conduction, perspiration), and because those flows of energy are in a different form or along a different channel then the energy supplied, the flow of those forms of energy can be regulated (via greenhouse gases, a glass enclosure, thermal insulation that also impedes convection and radiation) at least somewhat independently of what regulates the inflow of the energy supply. 9. Chris , I don’t understand the import of your question . There’s no argument that within 10s of meters of the earth’s sunlit surface the outgoing IR jiggles the CO2 and H2O molecules in particular and they warm the rest of the more transparent gasses by conduction . At night , the process reverses and the warmed air radiates its heat back via the IR active gases thus having an enormous effect on our diurnal temperature variance which I have never seen discussed . It’s a very different thing to change a mean than to change a variance . Anonymous , first , I’ve got to interject that it was shown as early as 1909 that the overwhelming effect of glass green houses comes from the stopping of convection . Further , putting a winter coat on a corpse won’t warm it . Yes , back-radiation will change equilibrium temperature by the difference of the 4th powers of the temperatures . For instance , if the cosmic background radiation was about 100k , we would be about 1.2c warmer . If it were 200k , we would be about 17c warmer . But , the bottom line , whatever your argument is : Show me the equations ! Otherwise , it’s just technobabble . 10. Patrick 027 (PS accidentally posted as “anonymous” earlier) “Viewed in terms of mass the tropopause isn’t deeper than on Earth. On both planets, some 90% of the mass is contained within the troposphere.” Actually, I would tend to think that the absolute mass (per unit area) above the tropopause is a relevant value to compare, since that (for a given composition, line broadenning and line strength effects) adds a greenhouse effect that can warm the tropopause level (on Earth, the tropopause is colder than the equilibrium surface temperature absent a greenhouse effect; this needn’t always be the case, particularly if enough of the flux to space is emitted from above the tropopause). By that measure, the tropopause is deeper into the atmosphere on Venus then on Earth, right? 11. Your implicit opinion that only greenhouse (IR-absorbing and IR-emitting) gases in the atmosphere determine all the temperatures is bizarre. Do you doubt that there would be a temperature gradient (cooling with altitude) even if the atmosphere were fully composed out of nitrogen or other non-greehouse gases? Now, imagine that the thickness of such atmosphere would be 40 km and the lapse rate would remain close to 10 deg C per km. Do you agree that at the top of the atmosphere, the temperature would still have to be above 0 Kelvin because it can’t go negative? Do you agree that it implies that the temperature at the bottom would have to be hundreds of degrees? Your opinion that only the optical thickness matters for the temperatures can’t be right. • jon Lubos, using your own words (slightly changed, of course). Your implicit opinion that only vertical temperature gradient in the atmosphere determines surface temperatures is bizarre. The fact is that: You take for granted that upper tropospheric temperature is Tu. So, as the gradient is the observed one (you are not explaining why it is that way) gives you: Tsurf=Tu+grad*Tthick with Tthick the troposphere thickness. You are translating the problem from surface to upper troposphere. How do you maintain the temperature of the upper troposphere at Tu without convection, without surface heating, without radiative energy fluxes? Just wondering about it. • jon Just a single answer to another part of your post, Lubos. You ask: “Do you doubt that there would be a temperature gradient (cooling with altitude) even if the atmosphere were fully composed out of nitrogen or other non-greehouse gases?” Well … it depends on the kind of absorption. You don’t need to go to Venus. If the absorption is in the shortwave part of the spectrum, the vertical temperature gradient will be POSITIVE … as it is the stratosphere, due to the absorption of shortwave radiation by ozone. Look at the standard vertical temperature profile of the atmosphere, for instance. No model here, just observations. The fancy part of this response (which shows the problem of blind extrapolation) is that the way you get a surface temperature following your “extrapolative scheme” clearly depends on the part of the atmosphere that you use to define the vertical temperature gradient. If you take stratospheric values and a very thick atmosphere … you get negative Kelvin temperatures at the surface. Boltzmann would be interested in hearing about this. 12. By the way, it is easy to pick some basic data about the atmosphere of Venus, e.g. http://en.wikipedia.org/wiki/Atmosphere_of_Venus#Troposphere The graph in the section of the page linked above shows that the Venus temperature drops by a large lapse rate resembling the Earth’s lapse rate up to the height of 80-100 km above the Venus’ surface. Only above 100 km of altitude, decrease begins (as the circulation disappears and the adiabatic assumption is not applicable). Around 50 km above the surface, both pressure and temperature pretty much match the Earth’s surface: 1 atmosphere, 20 deg Celsius, a nice place for future (flying) colonies. So there’s 50 additional kilometers of gas beneath this Earth-like point. What a surprise that the surface has hundreds of degrees Celsius, just by the lapse rate. This extra gas is there and it irrelevant whether it absorbs in the infrared. The graph of the temperature also shows the sulphuric clouds between the altitudes 50 km and 80 km, and the sulphur haze between 30 km and 50 km above the surface. The clouds clearly can radiate, too. One doesn’t need CO2 at all, to keep the temperature at these layers substantially above 0 Kelvin, determined by the balance of incoming and outgoing thermal radiation, and the high surface temperature is then given purely by the adiabatic lapse rate. The difference is more important than the “intercept” for the surface temperature, anyway, because the minimum temperature in the Venus’ atmosphere, 100 km above the surface, is much much smaller than the surface temperature – effectively close to 0 Kelvin. So both the slope and the intercept are dictated by the lapse rate and presence of things such as sulphur clouds, so you can see that the greenhouse gases are irrelevant for the qualitative explanation of the high temperature near the surface, up to at most a dozen of percent of accuracy. The ability of CO2 (as the major atmospheric gas) to absorb and emit doesn’t play any role for the explanation of the “majority” of the temperatures and temperature differences on the graph above. The huge mass or mechanical thickness of the atmosphere together with the existence of anything that absorbs and emits thermal radiation at the relevant frequency – sulphur clouds are enough – is sufficient to explain all qualitative features of the Venus temperature profiles. The greenhouse effect is just a small correction, relatively speaking, just like it is a small correction on the Earth. Cheers, LM 13. just me @Lubos yeah, in an atmosphere with a 40 km high tropopause and the lapse rate of 10K/km you have a damn pretty high surface temperature. But, what do you need to create such an atmosphere with these parameter values? What could cause this combination of parameters? As I understood, Chris said small lapse rates and/or optically thick atmospheres create high tropopauses. Are there more factors? What causes the tropopause anyway? Sorry for my ignorance, I am just a layman, I really do not exactly know. I am trying to learn 🙂 Response: The tropopause is, in loose terms, the height of convection. On Earth, a less steep lapse rate in the tropics (since you are on a moist adiabatic curve) is what causes a higher tropopause relative to the mid-latitudes or poles. The optical thickness of the atmosphere plays a decidedly secondary role. Venus, on the other hand, has a tropopause height governed largely by the optical thickness. In the optically thin case, it is even possible for the atmosphere to be nearly isothermal in the absence of convection (the lapse rate is extremely mild in the Martian winter for example, as it would be on a snowball Earth), where the surface temperature is much warmer than the overlying air. Once convection kicks in, the layer will grow in depth until the temperature at the top of the mixed layer matches the skin temperature. The adjustment of the temperature profile to follow an adiabat does not have an impact on the ground temperature in this case since the atmosphere is having no influence on the OLR.– chris 14. jg Chris, thanks for covering this topic. Jon, thanks for the references. I found these titles: Houghton, The Physics of Atmospheres Salby, Dmowska (editor), Pielke Sr. (editor): Fundamentals of Atmospheric Physics, Vol 61 Wallace, Hobbs, Atmospheric Science, 92 Are they the ones you recommended, and are they comparable so that I can consider price in my decision? thanks, John G 15. MarkB Goddard appears to be backtracking pretty quickly. 1. Goddard last week: This is very close to what we see on Venus. The high temperatures there can be almost completely explained by atmospheric pressure – not composition. How did such bad science become “common knowledge?” The greenhouse effect can not be the cause of the high temperatures on Venus. “Group Think” at it’s worst, and I am embarrassed to admit that I blindly accepted it for decades. 2. Goddard more recently: “The whole point of these articles is to demonstrate that Earth could not become like Venus, unless all the limestones dissociated. ” —– The goalposts have been almost literally moved from one planet to another. Goddard is reduced to furthering his attack on Carl Sagan, implying that Sagan thought Earth will become like Venus. As much of a red herring this is (Sagan, deceased more than a decade ago, is obviously not an active climate scientist) I’ve read enough Carl Sagan and a biography about him to know Goddard’s assertions are disingenuous and misleading. While Sagan wasn’t afraid to boldly speculate (he also turned out to be correct regarding the extreme heat on Venus), he was often careful to use the appropriate context and nuance. From Pale Blue Dot (1994): “Those who are skeptical about carbon dioxide greenhouse warming might profitably note the massive greenhouse effect on Venus. No one proposes that Venus’s greenhouse effect derives from imprudent Venusians who burned too much coal, drove fuel-inefficient autos, and cut down their forests. My point is different. The climatological history of our planetary neighbor, an otherwise Earthlike planet on which the surface became hot enough to melt tin or lead, is worth considering — especially by those who say that the increasing greenhouse effect on Earth will be self-correcting, that we don’t really have to worry about it, or (you can see this in the publications of some groups that call themselves conservative) that the greenhouse effect is a “hoax”. ” Partial content listing of Chapter 11 of his book “Billions and Billions”: http://books.google.com/books?id=bPWy5ta4lM0C&pg=PA117&lpg=PA117&dq=%22The+Warming+of+the+World%22+sagan&source=bl&ots=2Z5cJd99cL&sig=gvdsV483bnBEgrrHq7hR7YArkCs&hl=en&ei=z0PsS8iRNo3gtgPOqejDDw&sa=X&oi=book_result&ct=result&resnum=3&ved=0CCIQ6AEwAg#v=onepage&q=prov&f=false 16. Hi, I hope that many people on “both sides” will agree with this conclusion which may be viewed as a synthesis. The convection would almost always go to the very highest portions of the atmosphere, because of the fluctuating temperatures (diurnial cycles and seasons, mechanics) so a very massive atmosphere would have a high tropopause. However, if there were a very low concentration of substances absorbing the thermal radiation so that the surface would be too hot to match the incoming solar radiation, then the whole atmosphere would be cooling down because the outgoing energy (from a hot surface, going through a transparent atmosphere) would exceed the incoming one. By cooling down, one or several of the following things would be happening: 1) the cooler atmosphere would have a smaller volume, i.e. smaller height 2) the cooler atmosphere would be closer to the boiling point of the major gases, which means that the dry lapse rate would be reduced to a “wet” or saturated adiabatic lapse rate which is smaller because the work is also spent for latent heat 3) if needed, a part of the atmosphere would condense into the oceans, reducing the amount of gas in the atmosphere All these things would reduce the product of the height of the atmosphere and the lapse rate, and they would therefore reduce the predicted surface temperature. Eventually, a value would be found that would correspond to a stationary state where the outgoing radiative energy matches the incoming one. So can the warm surface of Venus be considered a consequence of the greenhouse effect? It’s a question similar to what was first, hens or eggs. It’s because the “same size” atmosphere without any greenhouse substances cannot exist at all. This non-existence may be described in two ways: either the greenhouse effect is a consequence of the large height of the Venusian atmosphere – the atmosphere simply finds its stuff that takes care of the balancing of the radiation energy, or the thickness of the atmosphere is a consequence of the greenhouse substances. At any rate, the life without the greenhouse gases couldn’t differ “by the surface temperature” only. The structure and height of the atmosphere would be completely different – and there’s no canonical “non-greenhouse counterpart” of the Venusian atmosphere that we could compare Venus with because many quantities have to be changed simultaneously for us to find another stationary solution. Do you agree with that? Cheers LM • Patrick 027 From what you’ve written, you seem to have admitted that the greenhouse effect is important. Any absorption, or even scattering, of LW radiation (radiation emitted at lower temperatures, as opposed to SW radiation, which is dominated by solar radiation), produces a greenhouse effect. The huge mass or mechanical thickness of the atmosphere together with the existence of anything that absorbs and emits thermal radiation at the relevant frequency – sulphur clouds are enough – is sufficient to explain all qualitative features of the Venus temperature profiles 1. Aside from whether the clouds are enough, the second part of that statement refers to the greenhouse effect. 2. Yes, for a given lapse rate, in terms of change in T per unit p, you need more weight in the troposphere to have a higher surface temperature for a given tropopause or tropospheric-average temperature. However: a. the weight of the troposphere as a fraction of the whole atmosphere is not a fixed value dependent only on the weight of the atmosphere. b. the temperature at the tropopause is not a fixed value dependent only on the weight of the atmosphere. c. while not the case for a solid or liquid, there is a further important point regarding a gaseous layer – the adiabatic lapse rate is a function of T and p, and all adiabats to T=0 at p=0. So even if the atmosphere had 1/10 the weight of the Earth’s, if you could get the tropopause up to near p=0, then you can have the surface temperature become very large with even a cold tropopause. Modification of tropospheric lapse rates by the larger-horizontal-scale circulation and it’s causes, and in particular by latent heating in convection, or the departure from an ideal gas at larger pressures, will modify this picture but doesn’t put a clear upper limit on things that I know of. d. OF course, the pressure and temperature affect the optical properties of materials (such as via line-broadenning (also affected by composition) and line-strength). That effect aside, a more massive atmosphere of the same composition will have greater optical thickness. But composition can be varied. If the mass of the atmosphere affects temperatures by providing or enhancing a greenhouse effect, that doesn’t take away the fact that there’s a greenhouse effect. If the conditions caused by the mass of the atmosphere influence the composition, that doesn’t take away the fact that composition is important. What sustains the convection in a troposphere? Convection (plus conduction and diffusion of sensible and latent heat from the surface into the air) is (in a time-averaged sense) continually transporting heat upward. Where does that heat come from and where does it go? In order to sustain this, there must be a net radiative heating at the surface and net radiative cooling aloft. When an atmosphere is very opaque to LW radiation and when some of that opacity comes from absorption, then the photons travel short distances from point of emission to point of absorption. This means that the radiation coming from any direction is being emitted from nearby, where the temperature is similar to where it is at the point of view. This means that radiant fluxes in opposite directions are more similar. This means that the net flux of radiant energy is small. If there is no LW opacity at the surface, than LW radiation emitted by the surface escapes directly to space, and there is no other source of LW radiation, which means that the emission to space from the surface is (in equilibrium) equal to the solar heating of the whole system. If the atmosphere absorbed any solar radiation, then the lapse rate would have to be negative so that heat could flow by conduction/diffusion downward to the surface so that it could then be radiated to space. On the other hand, if the opacity is very large, their will be very little net LW cooling at the surface, so even a small amount of solar heating of the surface would sustain some convection from the surface. In general, in order for convection to be sustained with a tropospheric-type lapse rate, a pure radiative equilibrium would have to produce a lapse rate that is unstable, and that depends on optical properties. • jon Dear Lubos. I don’t see this part of your text. “However, if there were a very low concentration of substances absorbing the thermal radiation so that the surface would be too hot to match the incoming solar radiation” I don’t agree with this. You can solve a simple system. Suppose a gray atmosphere (absorptivity for longwave radiation constant, al and absorptivity of shortwave radiation constant, too, as). The atmosphere is a single layer, no more. You can easily prove that the equilibrium temperature of the surface is Ts=sqrt((2-as)/(2-al)) Te with Te the equilibrium temperature of a point planet placed at that orbital place (sqrt(sqrt(irrad*(1-albedo)/4/sigma))). You can easily check the differences in the temperatures of the surface for the cases as>al and al>as. Well, the example is naive and highly idealized, but I think it illustrates the point. If al is VERY low, as you suggest (the lowest value is al=0), then, the surface temperature is exactly the equilibrium temperature corresponding to a point (Te) or lower than that, depending on the value of as. Ts<Te if as higher than 0. Ts=Te if the atmosphere does not exist, in radiative terms (it does not absorb neither shortwave nor longwave…). So, I don't see the point above in your argument. By the way, it is a very interesting exercise to compute the derivative of Ts with respect to al. It is positive 😉 17. gallopingcamel My understanding is that the apparent temperature of Venus viewed from space is ~288 K which corresponds to the temperatures found at an altitude of ~55 km. This would approximate to the level at which incoming radiation (after adjusting for the planet’s albedo) is in balance with the energy radiated into space. With a measured adiabatic lapse rate in the range 7-8 degrees Kelvin/km one would expect a surface temperature in the range 673-728 Kelvin. That seems close enough for government work so why is Chris getting his knickers in a twist? • Steve Yes, absent greenhouse gasses, *if* Venus’ atmosphere were still 288K at 55km with a 7-8 lapse rate, then the surface would still be 673-728 Kelvin. The point is that this is a very big *if*. Without greenhouse gasses, 288K would occur at a much lower altitude, if at all. 18. Pingback: Niche Modeling » Weekly Roundup 19. Steve I’m interested in your point about Poisson’s equation. I can imagine if you had two measurements of temperature and pressure (P0,T0) and (P1,T1), let’s say at different altitudes, then you could use Poisson’s equation to calculate T(P) for any P between P0 and P1. But you seem to be saying you can use Poisson’s equation with only one boundary value given, e.g. (P1,T1). Also, if Poisson’s equation is \Delta u = f, what f do you use? 20. Lubos – suppose Venus atmosphere was replaced with pure helium. There would be no condensing out (your 2 and 3). There would be no IR absorption. What you’ve left out of your consideration is that as the surface cools, convection also reduces. As the temperature gradient drops below the lapse rate, vertical mixing ceases and the atmosphere becomes stratified, as in Earth’s stratosphere. As convection gets slower, the motion of gas becomes less adiabatic because there is more time for energy and molecule exchange with surrounding regions. Venus with a helium atmosphere would have a cool surface and a cool atmosphere with very little temperature gradient from surface to space. • Dear Arthur, stratosphere is stratified because the temperature *increases* with the altitude. That’s why the warmer air – which is lighter – doesn’t ever want to move lower. It stays where it is, the air stays in the layers, and the part of the atmosphere is stratified, therefore “stratosphere”. The inverse relationships for the temperature occur because of the absorption of the solar radiation: the UV heat is absorbed mostly at the top of the stratosphere which is why it’s warmer over there. Now, for your thought experiment, it’s important to figure out how deeply the direct heating of the atmosphere from the Sun goes. If the atmosphere is sufficiently transparent for sunlight, the light will get up to the surface and the vertical temperature gradient will be more affected by the heat conduction etc. from the surface, and you won’t get the inversion and you won’t get stratification. Of course, we would have to study how much of which gas you have etc. But it’s surely not the case that you always get stratification without greenhouse gases. After all, it also fails in the opposite way: ozone is also a greenhouse gas, yet it is a key gas that is responsible for stratification because it absorbs UV radiation in the stratosphere. Best wishes Lubos • Patrick 027 1. if you are isolating the effect of LW opacity (the greenhouse effect), then you have to mathematically seperate the LW effects from the SW effects of any material. The greenhouse effect of ozone is the greenhouse effect of ozone, it doesn’t include the effect of ozone on solar heating. 2. if there is no greenhouse effect, the atmosphere cannot emit any radiation to space. Thus, all solar heating of the whole system has to be balanced by radiation to space from the surface. Thus, if there is any solar heating of the atmosphere, the atmosphere has to have a negative lapse rate so that heat can flow downward to the surface. 3. A stratosphere can have a positive or zero lapse rate. It just has to be smaller than the tropospheric lapse rate. • Lubos, you claim “it’s surely not the case that you always get stratification without greenhouse gases” and then present ozone as an argument in the other direction – but that’s a classical fallacy – Improper Transposition. I think the best way to understand the issue from a physical standpoint is as a cascade of energy from low entropy and high temperature (the sun) through the various planetary systems and to increasingly higher entropy and lower temperatures (ultimately into space). If the atmosphere is completely transparent to radiation, all the interaction with sun and space is at the surface and the atmosphere is merely a bystander (i.e. essentially isothermal, stratified, aside from surface winds due to diurnal and latitudinal temperature differences). If the atmosphere partially absorbs incoming sunlight (like ozone) but is still transparent in the infrared, then some of the incoming energy flow goes through the atmosphere, but it cannot leave through the atmosphere (no infrared emission) so it has to go to the surface: the atmosphere must be warmer than the surface, and have a positive temperature gradient so that energy can flow down. If the atmosphere has any components (greenhouse gases, clouds, dust etc.) that absorb and emit in the thermal infrared region, then energy can flow outward directly from the atmosphere. If the atmosphere is completely transparent to incoming sunlight, then the energy flow has to be from the surface into the atmosphere – the atmosphere must be colder than the surface, negative temperature gradient. Natural convection only starts when the negative temperature gradient is more negative than the appropriate adiabatic lapse rate (dry or moist). So you still could have stratification even with GHG’s, if the absorption level was small. But once you get a high enough temperature gradient, convection then pins that gradient at the lapse rate and you get the usual scenario. 21. See Arthur Smith’s comment in response to Lubos for a quick scenario on Venus with no greenhouse effect. It would not be possible to achieve a surface temperature in any significant excess of what blackbody radiative balance from incoming sunlight allows, and Steve’s comment is essentially correct that the level which I defined as p_rad in my post would essentially be at the surface. Also keep in mind if you could keep the pressure the same (say, 90 bars of an atmosphere composed of diatomic molecules that did not condense) it would still be possible to generate a diatomic-molecule greenhouse effect from continuum effects. Such is seen on Titan or Jupiter for example, but not really important on Earth- chris 22. stupmy I recall scientist used to think Venus had temperature similar to earths, funny how things change over time and hypothesis changes. I understand that the poles show no temperature difference to the equator, and there is no temperature change at night. How is this explained with the greenhouse effect? I also recall other scientists in the past have argued that the surface of Venus is hot due to thermal venting from the relatively hot surface, this would explain the fairly constant temperature – how would this geothermal heat factor into the dicussion? My view is we dont fully understand our own planets climate, and the greenhouse effect is by no means a complete equation – and we certainly are no where near understanding the Venus climate! Both arguments are nothing more than hypothesis and very difficult to ever prove. Essentially no one is “correct” or “wrong”. Whilst the greenhouse effect is thought to be a well established fact, it would not suprise me if it turned out we are all wrong, nature has a way of doing that! Sacred cows can die – it has happened many times in the past! Response: Oh please. You argument is essentially that we don’t know everything, therefore we know nothing (or, therefore we can reinvent the laws of physics just to make everyone happy with their personal opinions) and it is bogus. As much as it offends people’s democratic sensibility, not everyone’s random thought is equally valid. We know well enough to say that the greenhouse effect is very important on Earth, and even moreso on Venus. Further, we know well enough to say that the proposal by Goddard is simply not plausible, unless much of what we know about physics (e.g., energy conservation) breaks down on our neighbor planet. I don’t buy it– chris • Patrick 027 The atmosphere has some heat capacity. It moves around, from equator to pole and back, around from day to night and back, so it can have similar temperatures over the area of Venus at a given vertical level. As the atmosphere moves around, it can transfer heat from the low-latitude midday region and bring it to the night time side. Via the greenhouse effect, at a given time and place, the effect of the strong solar heating cycle and gradient is moderated by the (in this case, much larger) radiant fluxes to and from the surface and layers of atmosphere (as the heat from the sun passes back and forth by photons or other means to different layers, heat capacity filters the temporal cycle at each step). 23. gallopingcamel Arthur Smith, Your question about a Helium atmosphere is very interesting. With dry Helium the lapse rate would be 1.9 degrees Kelvin/km so temperature would fall much more slowly with altitude. If there were no other gasses present most of the long wavelength radiation would pass thorough the atmosphere and into space. The surface temperature would therefore ~288 Kelvin if the planet were Venus. As there would be no clouds the albedo of the surface of the planet would come into play and that might be a significant effect. Let’s suppose sulphuric acid is present in the atmosphere at concentrations similar to measurements discussed by David Grinspoon (http://www.funkyscience.net/). The sulphuric acid would form clouds and vapour layers which would absorb outgoing long wavelength radiation, thereby allowing the “Greenhouse Effect” to operate. However, the cloud layers would be at ~600 km instead of ~50 km owing to the low density of Helium. These clouds would be radiating to outer space at a temperature of ~277 Kelvin (owing to the increased radiative diameter of the planet). With a Helium/sulphuric acid atmosphere I would expect the surface temperature of Venus to be in the range 1,100 to 1,400 degrees Kelvin. You need absorbing gasses and the adiabatic lapse rate to make the “Greenhouse Effect” work. This is a “back of the envelope” calculation so I hope someone can offer a more precise prediction. 24. Dear Arthur, I agree that the example of ozone proved the failure of the opposite implication, not the original one, but it’s still true that stratification may be avoided in the absence of greenhouse gases. If there were no vapor or CO2 in the lower atmosphere – imagine (dry) Sahara as a good approximate example – there would still be warming and cooling of the surface, and heating of the low troposphere from the surface. Colliding air would still lead to vertical circulation as well. So even in Sahara, the temperature decreases with the altitude, at least near the surface. The layers where you can get the stratosphere-like behavior are closer to the surface than elsewhere, but they’re not immediately adjacent to the surface, despite the de facto absence of greenhouse gases (water vapor is the key here). We’ve had exactly this debate with CapitalistImperialistPig, an alarmist. Subadiabatic lapse rates can be stable for a while or locally but eventually, circulation always kicks in from somewhere, and it drives the atmosphere towards the universal adiabatic rate(s) wherever the circulation can get. So a daily, annual, and/or global average for the lapse rates will show numbers that are close to the adiabatic values. Cheers LM • Patrick 027 1. There is CO2 in the air over the Sahara 2. “Subadiabatic lapse rates can be stable for a while or locally” They can be stable in the sense that they can be sustained by sufficient heat fluxes along with some impediment or resistance to convection (viscosity, a boundary to motion, a phase change with negative Clapeyron slope (the later being important in the mantle)). But by definition they are unstable to convection. Just wanted to clarify that point. 2. “but eventually, circulation always kicks in from somewhere” Somewhere is a key word. If you remove the greenhouse effect from one place, you can still have vertical motions sustaining an adiabatic lapse rate. Even if you remove all differential heating -driven overturning, a mixed layer can be forced by kinetic energy supplied from elsewhere (differential heating-driven overturning in another location, or gravitational tides (which can’t do much, especially on Venus), etc.). Setting things like tides aside as an unecessary complication to the basic aspects, it is hard to get the differential heating necessary to sustain an adiabatic lapse rate within an atmosphere with no greenhouse effect (or even with a greenhouse effect if the greenhouse effect is based only on LW scattering). If the air is not able to absorb and emit LW radiation, then heat from the surface that goes into the atmosphere has no where else to go but back to the surface. If there is horizontally differential heating, then some overturning can be sustained, but the colder sinking air has to lose heat to the underlying even colder surface by conduction, so the lapse rate has to be negative near the surface in regions of subsidence. The requirement that cooling occurs by conduction constrains the vertical thickness of such overturning. (And if the vertical thickness of such a layer is such that conduction is effective in transporting heat through it, then the air in rising regions might be heated from below faster than convection, tending to reduce the lapse rate from adiabatic, if the motion is slow enough). But aside from all such complexities, there is a very simple point: If there is no radiation going out to space emitted by the atmosphere, and if there is no radiation emitted by the surface that does not go out to space, then there can’t be any LW radiation downward at the surface, and the radiation emitted by the surface must be, if at equilibrium, equal to the solar heating (plus geothermal heating and tidal heating, which are small enough to be ignored in this Earth/Venus context) of the surface and atmosphere (because all solar heat deposited in the atmosphere must exit by way of the surface – hence, except for effects of horizontal differential heating, the need for a negative lapse rate if the atmosphere absorbs any solar radiation). If the surface is a blackbody for the LW part of the spectrum, then the average of the fourth power of the surface temperature Ts will be equal to the Te^4, where sigma * Te^4 = S*(1-albedo)/4, S being the solar flux per unit area facing the sun in space at the distance of the planet. • I think I agree with all this stuff and have said the nearly same thing a few times, too. To look back at all the points, it’s important to realize that the circulation is as important for the climate as the IR absorption, and it can be ignited not only by the IR absorption but also by the heating and cooling surface – directly from the Sun – and heat conduction etc. • Patrick 027 Would you agree, then, that with no greenhouse effect, the equilibrium average surface temperature can not be more than Te, where sigma ~= 5.67 E-8 W/(m2 K4) and epsilon is the LW surface emissivity (effective for the spectrum at the temperature Te) and epsilon * sigma * Te^4 = external heat input = S*(1-albedo)/4 + geothermal heating (very small) + tidal heating (very small) + etc. (very very very small) • Patrick 027 Reworded/Redone sigma * Te^4 = external heat input = S*(1-albedo)/4 + geothermal heating (very small) + tidal heating (very small) + etc. (very very very small) where sigma ~= 5.67 E-8 W/(m2 K4) and epsilon is the LW surface emissivity (effective for the blackbody spectrum at the surface temperature) Would you agree, then, that with no greenhouse effect, the equilibrium average surface temperature Ts can not be more than Te/epsilon^(1/4), if epsilon is constant, or more generally, that the equilibrium global average of (epsilon*Ts^4) must be equal to Te^4 ? • Patrick 027 Example: Consider a planet that has 1/3 of it’s area absorbing 600 W/m2 of solar radiation and the other 2/3 of it’s area are in night. (of course only 1/2 would be in night, but the dayside solar heating would vary from near zero to higher than 600 W/m2 on this planet; this is an illustrative simplication.) And the surface of the planet is a blackbody for LW radiation. Global average solar heating: 200 W/m2 Te = 243.7 K Te1 (‘day’ area Te for 600 W/m2) = 320.7 K Te2 (‘night’ area Te for 0 W/m2) = 0 K Ts1 = surface temperature in ‘day’ Ts2 = surface temperature in ‘night’ For equilibrium: Given Ts1, Ts2 is required to be such that the global average emitted flux is 200 W/m2 (Te1^4 – Ts1^4)*sigma is a nonradiative cooling at the surface in ‘day’ and (Ts2^4 -Te2^4)*sigma is a nonradiative heating at the surface in ‘night’ Multiplied by their respective areas, the two resulting fluxes are equal and the value is the net heat flux that must be conducted into the air in ‘day’ and then convected, perhaps vertically, and necessarily horizontally, to ‘night’, and then conducted into the surface in ‘night’. Assuming that the thickness of the air through which heat must conduct out of the surface in ‘day’ is thin relative to the layer dominated by convection, assume an average lapse rate G1 = 10 K/km = 0.01 K/m, up to some height zeq. The lapse rate at the surface in ‘night’, G2s, is equal to the the negative of the nonradiative heating / thermal conductivity. Allow thermal conductivity to be 0.2 W/(m K) (about 8 times that of ‘air’ – see http://en.wikipedia.org/wiki/Thermal_conductivity) – no particular reason for choosing that specific value. Allowing the lapse rate in ‘night’ to be G2s up to a height zeq. Set the temperature over the globe to be the same Teq at zeq. Then zeq = (Ts1-Ts2)/(G1-G2s) A look at what we’re dealing with Ts1 (K), Ts2 (K), non radiative surface cooling in ‘day’ (W/m2) (divide by 2 to get non radiative heating of surface in ‘night’), G2s (K/m) (note units), zeq (m), Teq (K), global average Ts Te1, 0, 0, 0, ~ 32000, 0, 107 320, 83, 5.5, -13.6, 17.3, 319.8, 162 319, 103, 13, -32.1, 6.71, 318.9, 175 315, 139, 42, -104, 1.69, 315, 197 310, 161, 76, -191, 0.78, 310, 211 300, 188, 141, -352, 0.32, 300, 225 280, 217, 251, -629, 0.10, 280, 238 Te, Te, 400, -1000, 0, Te, Te Some qualifying notes to follow… • Patrick 027 What this means: 1. in this crude illustrative example, it is implied that zeq would be the tropopause level. It is the thickness of a layer that can be cooled in ‘night’ so as to allow convection in ‘day’. In the limit that the surface temperatures are in radiative equilibrium with solar heating within each region, the air in ‘night’ needn’t have much of a negative lapse rate to transport heat downard, so it can approach 0 K up to great heights, and this allows convection in ‘day’ to reach great heights. HOWEVER, the limit is physically inconsistent; if the night-cooled air is transported into the day and vice-versa, then there would be some heat flux that would raise the night air’s temperature. To actually reach the limit that each region is seperately in radiative equilibrium, there would be no horizontal circulation between them, and thus no heat sink for the day region’s convection from the surface. If mixing were externally forced, then their could be layer with an adiabatic lapse rate, but it wouldn’t change the equilibrium surface temperature. Absent such mixing, the atmosphere would over time tend to become isothermal. The limit of a globally isothermal surface is also impossible to achieve as an equilibrium, because it would require a large heat flux through a thin (in this calculation, zero thickness) layer between two surfaces of the same temperature. • Patrick 027 … Also, zeq needn’t be at the same geometric or pressure height everywhere. First, the gas (if it even remains a gas) in ‘night’ will be on average colder through the overturning layer, and especially so near the surface. So the pressure levels in the upper portion of this layer slope downward into night. Of course, this is the way that horizontal differential heating drives such overturning; the pressure in some lower portion of the layer exihibits the opposite tendency. In terms of mass, then, the layer should actually be thicker in the night region. On the other hand, a portion of the layer in the dayside could overshoot zeq, and the air in some upper region might sink downward with increasing temperature adiabatically while approaching or while within the night region, before reaching a point where conduction maintains a negative lapse rate. Kinetic energy is generated by thermally-direct motion (warmer rising, colder sinking; at least for constant composition and setting aside various counterexamples like fresh water below 4 deg C). Kinetic energy can then drive thermally-indirect motion (warmer sinking) and mixing of stably-stratified layers. Ultimately kinetic energy is dissipated and converted back to heat, so the propagation of kinetic energy can be viewed as part of a convective heat flux that doesn’t follow all the same behavior. Anyway, this energy allows motions to overshoot equilibria, and could entrain air from above to increase the thickness of the overturning layer; it might also enhance downward heat transport in the night region, reducing the necessary temperature gradient and thus allowing a thicker layer to overturn. Aside from that, the assumption of constant lapse rate on the night side supporting conduction with constant thermal conductivity implies that heat is only transported into the region at the very top of the layer and the heat sink is transported out just at the bottom; an even distribution of heat loss would result in (for conduction, assuming constant thermal conductivity) a parabolic temperature distribution, with the magnitude of the lapse rate decreasing with height. zeq would then be larger. Or what if the heat transported into the region is proportional to the horizontal temperature difference – in that case, the lapse rate (where much larger in magnitude than the day region’s) would decay exponentially with height; zeq would be an e-folding height. Note that the convective heat flux upward from the surface in the day region should decrease with height. Conceivably it might never reach zero. However, if there were any solar heating of the atmosphere, then there must be a downard flux to the surface. The component associated with daytime heating of the atmosphere would be downward and outward (spreading horizontally), so that in the dayside the flux that balances atmospheric solar heating above any given height would decrease toward the surface. At some point above the dayside surface, the total vertical nonradiative heat flux would be zero; a purely thermally-direct circulation would be trapped beneath such a level (I think), though kinetic energy could entrain air from above and thus pull heat downward. Of course, if some heating were not from the sun but from from some externally-powered mechanical stirring mechanism, for example, gravitational tides, then it would go into the system initially as gravitational potential energy, then (at least some portion – or would it be all?) converted to kinetic energy, which could drive mixing while being converted to heat. If strong enough, that could force the existence of a mixed, nearly adiabatic layer, and help pull heat downward from solar heating or in the night region. However, notice that while these factors can affect the way atmospheric temperature varies with height, none of it has any effect on the temperature in one region of the surface if the other region’s temperature is given. It would affect the surface temperature only by affecting the heat transport from one part of the surface to another. A more isothermal surface can have a higher average surface temperature to emit the same LW radiation flux. But the average temperature cannot be raised beyond Te (for a LW blackbody surface; or otherwise Te divided by a power of surface emissivity if surface emissivity is constant, or in more complex cases … etc.). Given the global average solar (+geothermal, tidal, etc.) heating (including atmosphere), if the surface temperature and surface optical properties are given for all other times and locations, then, for a climatic equilibrium, the remaining location and time’s surface temperature is set (given it’s optical properties), by the requirement of outgoing LW radiation balancing the heating. • Patrick 027 The limit of a globally isothermal surface is also impossible to achieve as an equilibrium, because it would require a large heat flux through a thin (in this calculation, zero thickness) layer between two surfaces of the same temperature. … well, the zero thickness of the layer that is supposed to transport the heat is a problem, but pressure variation at the surface would allow air from one region to adiabatically reach a lower temperature as it goes to another region. Of course, there still wouldn’t be a mechanism to sustain such flow (at least setting aside other factors besides differential solar heating). • Patrick 027 Another bottom line – with zero greenhouse effect (or with only LW scattering), there is no vertically differential heating to drive convection locally in the absence of export of heat horizontally. With that condition, and without horizontal differential heating, any stirring would have to be externally driven. And for constant surface optical properties, the equilibrium average surface temperature would be set regardless of any such mixing (for a LW blackbody surface, it would be Te). • Patrick 027 … any more continued at end of thread • Patrick 027 … except: In the case of zero horizontal heat flux in the example given, the calculated tropopause temperature in day was 0 K. For other reasons given above, the situation is unphysical for equilibrium; however, it is also true that an adiabatic process cannot actually get all the way to 0 K when p is not zero … at least for an ideal gas, 0 K is approched in the limit of zero pressure. A constant adiabatic lapse rate is often an approximation that may fail outside of some range. 25. BigWaveDave The troposphere as a whole loses heat by radiation. The radiative equilibrium temperature (Te) is perceived to be at the top of the troposphere, which is the upper limit of the turbulent zone. The surface is heated to greater than Te, and the heat is transferred upward by conduction and convection, which creates the turbulence The lapse rate explains the difference between the surface temperature and the radiative equilibrium temperature. • Patrick 027 Yes, although – depending on how you meant your comment or which context you intended for it, this is either a correction of your statement or simply an extension of your statement that you might have already agreed with – In order for the tropopause to lose heat by radiation, there must be gases or clouds or something within it that can emit (and therefore, absorb, assuming quasi-LTE or something not very far from quasi-LTE or at least something not too exotic) radiation in some of the relevant frequencies for the temperatures involved (mostly LW frequencies, as opposed to SW (solar-dominated) frequencies, for temperatures that are not too high). Aside from horizontal differential heating, in order to support convection through some layer, the vertical temperature gradient must be large enough and of the right sign so that the air is not stable to convection, while there must be net non-convective heating below and net-nonconvective cooling above to balance the convective heat flux through the layer; if the temperature below is too high and/or the temperature above is too low, then net cooling below and net heating above by LW radiation (or conduction/diffusion, if/when that is ever important – tends not to be within a gaseous atmosphere on large scales but is important at the surface) would be too great to support convection. Changing the adiabatic (dry/moist/etc.) affects the first condition; a smaller adiabatic lapse rate is more favorable to convection, other things being equal; Changing optical properties (or where important, other properties such as thermal conductivity or molecular diffusivity) adjust the second condition; setting other processes aside, zero LW optical thickness (from absorption) removes all LW cooling or heating within the atmosphere, so there is no heat radiatively-supported heat sink within the air, and also, if there is zero LW optical thickness (absorption or scattering), then there will be no backradiation from the atmosphere to the surface; adding LW optical thickness generally** increases backradiation from the atmosphere, so that the net LW cooling of the surface is reduced for the same temperature, allowing the surface temperature to be larger for the same net LW cooling; while adding some LW optical thickness with some fraction of that from absorption allows a net LW heat sink to exist within the air; increasing that optical thickness reduces the net LW fluxes while concentrating the region that has signficiant net LW cooling into a smaller mass at the top of the atmosphere. Thus increased LW optical thickness, with at least some fraction of that from absorption, allows a larger temperature variation to exist with the same net radiant fluxes, which is more favorable to convection, and also generally more favorable to convection up to higher levels of the atmosphere. Note that Te can be warmer than the tropopause temperature. For example, if the stratosphere is optically thin for LW radiation, then it absorbs a small fraction of the upward LW radiation from below (whether from the surface or troposphere); because it emits (if some opacity is from absorption) LW radiation both upward and downward, it will tend to become colder than the source of the radiation that it absorbs, but because it allows most radiation from below through, the effective T for the radiation from below the stratosphere will not be much larger than Te, so the stratosphere can be colder than Te, approaching (in the limit of near zero LW opacity) a lower T that is the skin temperature. (If the stratosphere’s optical properties don’t vary over wavelength over the LW portion of the spectrum, then there is a simple formula for the skin temperature: Te/(2^(1/4)).) If their is direct solar heating of the stratosphere, then the stratosphere will be warmer, but if this direct solar heating is not too large, and not too large within the lower stratosphere, then the lower stratosphere can still be colder than Te. If the tropopause were at Te and the air immediately above were significantly colder, then convection would adjust this arrangement. The tropopause temperature can be lower than Te. On the other hand, if the stratosphere is optically thick for LW radiation (at least in the case where a sufficient fraction of that is via absorption), then the tropopause could be warmer than Te. 26. Dear Patrick, yes, I agree with the bound, but your assumption is so unrealistic that it makes the conclusion worthless. With no greenhouse effect, the inequalities are true. But just a tiny portion of IR absorption is enough to change the situation qualitatively, much of the warming may be attributed to the lapse rate resulting from circulation, and the dependence of the warming on any further increase of greenhouse gases may be negligible. For example, it’s likely to continue to be logarithmic because all the major effects – vertical shifts of the troposphere, widening of the spectral lines – lead to the logarithmic effect. For Venus, it means that removing 90% of the CO2 and replacing it e.g. with Nitrogen would just move you ln(10)/ln(2) = 3.3 doublings, out of 18 doublings Venus has above us. So the warming of the surface would only change by 1/6 or so. Steve Goddard was careful to keep a fraction of the CO2. Cheers Lubos • Patrick 027 1. Whether or not the complete absence of a greenhouse effect is realistic (although I think it is for the Earth’s Moon or Ceres, for example) doesn’t matter to what the effect of the greenhouse effect is, even if just a little bit of it. 2. There are some complexities that must be kept in mind when comparing the radiative forcing and climate sensitivity of a small change (doubling of CO2) to the totals (total greenhouse effect, removal of all CO2, etc.) Climate sensitivity might vary over a large range of forcings (and will be different if, for example, there is no water; gravity would also have an effect); the logarithmic proportional of radiative forcing to CO2 is an approximation that works well within certain limits but fails outside of those bounds (very low or very high amounts); and very very important: radiative forcing is a function of the temperature distribution. 27. Phil. Our stratosphere is caused by solar absorption by a non-greenhouse gas, O2, since it is a non-greenhouse gas it’s not capable of radiating out to space so you might expect a continually increasing temperature! However in the process of absorbing the incoming radiation it undergoes photochemical reaction and produces O3 which is a GHG and therefore confers stability. A planet with an oxygen atmosphere would be interesting. By the way the Earth’s polar regions have temperature inversions which last for months on end due to high rates of radiational cooling from the surface through dry air. Response: It’s the O3 doing the UV absorbing in the stratosphere.– chris • Patrick 027 Increasing temperature with height is generally in equilibrium with solar heating of the air, but a stratosphere can include isothermal regions or even regions of decreasing temperature with height, so long as that decrease is not so large as to allow significant convection. With no solar heating of the air, a stratosphere would exist that could be isothermal or have some decrease of temperature with height, in the region above the tropopause level (what would be harder to get without direct solar heating of the upper atmosphere is the stratosphere-mesosphere-thermosphere layering that characterizes the Earth’s upper atmosphere). In the event that there is no greenhouse effect, and specifically, no LW emission … (a greenhouse effect theoretically could be based entirely on LW scattering (LW scattering could become important in a snowball climate with dry ice clouds), or for that matter, fluorescence, etc. (but that would be a rather exotic atmosphere)) … then the upper atmosphere would heat up to temperatures sufficient for conduction (plux any convective mixing that is forced by something other then the vertically differential heating, as that cannot drive convection in this case) to transport heat downward to the surface, or otherwise become so hot as to emit radiation in otherwise solar-dominated frequencies (SW). 28. Patrick 027 … To sum up, for equilibrium, with zero greenhouse effect, the equilibrium global average of (epsilon*Ts^4) must be equal to Te^4, regardless of atmospheric temperature and circulation. Now, with some greenhouse effect, with pure scattering, atmosphere temperature continues to only matter indirectly; the only LW cooling occurs at the surface, but the atmosphere only allows some of the surface’s emitted flux to space, the rest is absorbed at the surface. More applicable to Earth, Venus, and most planets so far as I know – if there is a greenhouse effect and some of that is from absorption, then the temperature of the atmosphere, and thus, the thickness and lapse rate of the tropopause, do have a direct effect on radiative heating and cooling. They **begin** to matter when some greenhouse effect is added. If only 10 % of the LW flux emitted by the surface is blocked before reaching space, then the effect of the atmospheric temperature on the total LW flux to space is not as large as it would be with a stonger greenhouse effect, and the backradiation from the atmosphere to the surface also has much room to increase. Furthermore, when climate equilibrium is attained, by adding such a greenhouse effect, the radiative heating and cooling as a function of temperature generally (some exceptions if the greenhouse agents are concentrated in some layers, especially if they are lower lying layers) becomes more favorable to supporting a thermally-direct troposphere with a higher tropopause with significant upward convective heat fluxes. Even when the effect at the surface is nearly saturated (nearly zero net LW flux), there can be room for changes in radiative effects at higher levels that continue to increase the difference in equilibrium temperature from the surface to the coldest optically-significant part of the atmosphere. Note that radiative forcing of a change in optical properties is a function of climate – of the temperature distribution in particular. Higher temperatures lead to larger LW fluxes. Increased vertical temperature variation can lead to larger radiative forcings for the same change in optical properties. If we segregate radiative forcings and feedbacks and also include the effect of climate itself, then, allowing climate to approach equilibrium and with no hysteresis, removing all CO2 and then adding it back will return the climate to it’s initial state, but the magnitude of the forcing and climate sensitivity are different in the forward and reverse directions (it’s the product that would be the same, assuming no hysteresis). • BigWaveDave The greenhouse effect in a greenhouse is due to its prevention of convection from carrying heat away. The same thing happens with earth’s ghe, Earth’s gravity prevents convection above the troposphere, where heat escapes by radiation. The only prerequisite for radiation is that there is a temperature difference. There is no one radiative equilibrium temperature for Earth. Just look at IR photo from about two million miles away: http://photojournal.jpl.nasa.gov/catalog/PIA00558 http://photojournal.jpl.nasa.gov/catalog/PIA00559 Earth’s gravity also creates the lapse rate that keeps surface temperatures higher than top of troposphere temperatures. • Patrick 027 There is no one radiative equilibrium temperature for Earth. Yes and no. Te = [S*(1-albedo)/4]^(1/4) is the radiative equilibrium temperature corresponding to the radiative flux that an isothermal blackbody surface of the same size would have to have to be in equilibrium. Of course not every unit area at every time emits a flux to space (or within the atmosphere or at the surface, has an upward and downward LW flux) that is the same – the brightness temperature varies, both over area, and over frequency (and – pertaining to the greenhouse effect – over vertical distance). But a necessary condition for climatic equilibrium is that, over area and time (long enough to characterize a climatic state), fluxes have to average and add up to the same total to balance the distribution of other energy fluxes – or for the balance at the top of the atmosphere, the outward LW flux has to be the same as that of an isothermal blackbody at Te. (PS the picture in the first link – note that, since this is seen from one viewpoint, the brightness temperatures are for the radiant intensities – quite understandably, in addition to being low at higher latitudes (and where there are high cloud tops, though that might not be well-resolved in that picture), they are also lower near the limb (edge of the disk), because of a greater thickness of air in each layer along such a line of sight.) Earth’s gravity prevents convection above the troposphere, where heat escapes by radiation. Just to be clear, gravity tends to impede convection generally everywhere, by increasing the weight of each unit of mass, thereby increasing the pressure decline over mass and thereby increasing the adiabatic lapse rate relative to mass, thus requiring radiative (setting aside conduction, etc.) fluxes on their own to sustain a larger lapse rate if thermally-direct convection by vertically differential heating is to occur. (For equilibrium, the convergences and divergences of the fluxes have to balance; if convection doesn’t occur, radiative fluxes (setting aside conduction, etc.) must be balanced by themselves. With convection and any other heat fluxes, radiation must be responsible for supplying the heat to where convection removes it and removing the heat from where convection puts it. For a given distribution of solar heating, If the temperatures required for the LW fluxes to sustain the lower level heating and higher level cooling require a lapse rate that is stable to convection, then the lapse rate falls to one that is sustained by radiation (setting aside alternative drivers of convection that could pull heat downwards, etc.). The optical thickness is of great importance in shaping the amount of convection and the thickness of the layer in which it is signficant. • Patrick 027 PS forget a “” up there, and “Just to be clear, gravity tends to impede convection generally everywhere, by increasing the weight of each unit of mass, thereby increasing the pressure decline over mass and thereby increasing the adiabatic lapse rate relative to mass, thus requiring radiative (setting aside conduction, etc.) fluxes on their own to sustain a larger lapse rate if thermally-direct convection by vertically differential heating is to occur.” …Refering to the lapse rate relative to mass, which will tend to be relative to optical thickness for a given composition, though optical properties are themselves affected by pressure and temperature even for the same composition and physical state. Of course, gravity also, via pressure, compacts the mass into a smaller vertical distance, but it will do the same to optical thickness and the temperature distribution over mass and over optical thickness. 29. gallopingcamel Dear Dr. (edit– Mr. :-))Colose, Suppose there was planet just like Venus but the CO2 was replaced by enough helium to create the same 90 bar pressure at the surface. I would estimate that the replacement of CO2, a so called “greenhouse gas”, by helium would actually raise the surface temperature of the planet. Here is my reasoning. There would still be 100% cloud cover mainly consisting of sulphuric acid. This cloud layer would effectively be the radiative layer with a temperature of ~288 Kelvin, essentially what is observed on the real Venus. While the adiabatic lapse rate for helium on Venus is only 1.9 Kelvin/km, the depth of the atmosphere would be increased in the ratio 48/4 or 12:1. The radiative layer would be at a height of ~600 km (12 times the present 50 km) and the surface temperature would be: Ts = 288 – 600*1.9 = 1,428 Kelvin Today Venus has a radiative radius of 6,100 km but a helium atmosphere would increase that to 6,650 km. Using the Stefan-Boltzmann equation the new radiative temperature would be 282 Kelvin. I do not know the “wet” adiabatic lapse rate for helium and sulphuric acid so I hope you will not object to a guess at this point. Suppose that the “wet” lapse rate is 1.0 Kelvin/km. The surface temperature of a “Venus” with a helium based atmosphere should therefore be in the range: Ts = 882 to 1,420 Kelvin In a nutshell, you don’t need an absorbent gas like CO2 to make Venus hot if you have clouds made of complex molecules like water or sulphuric acid. • Patrick 027 There would still be 100% cloud cover mainly consisting of sulphuric acid. This cloud layer would effectively be the radiative layer with a temperature of ~288 Kelvin, essentially what is observed on the real Venus. I haven’t actually gotten around to seeing whether the sulphuric acid clouds are mainly LW absorbers/emitters or if they significantly scatter LW – either process would contribute to the greenhouse effect but the mathematics is a bit different. But bottom line, H2SO4 clouds absorbing or scattering LW radiation is still a greenhouse effect. If it were/is the case that the upward LW flux from the top of the cloud layer (whether coming from the clouds or if some is penetrating from below) with a flux per unit area corresponding to 288 K blackbody radiation – that is too much for the solar heating; cooling will result – unless there is something (such as CO2) above the cloud layer that keeps the cloud layer or whatever is supplying the radiation coming up from it as warm as it is via additional greenhouse effect while reducing the flux that escapes to higher levels and space. While the adiabatic lapse rate for helium on Venus is only 1.9 Kelvin/km, the depth of the atmosphere would be increased in the ratio 48/4 or 12:1. PS adiabatic lapse rate in terms of p for an ideal gas is actually proportional to T and is larger at smaller p. At smaller p, the air is less dense (but T also affects this), so dp/dz is smaller, so the adiabatic lapse rate can decrease with height … I just thought that deserves mentioning. To better isolate the effect of LW opacity, we should assume same lapse rate and density of the gas, but okay… Today Venus has a radiative radius of 6,100 km but a helium atmosphere would increase that to 6,650 km. Using the Stefan-Boltzmann equation the new radiative temperature would be 282 Kelvin. A larger atmosphere could intercept more solar radiation as well (do the H2SO4 clouds absorb any solar radiation (my assumption: yes of course, but how much)? How much is absorbed there, vs at ground level or in between?) • No, a single layer of infrared absorption high in the atmosphere could not maintain such a high surface temperature. Look at the energy balance for the surface: less than 100 W/m2 from the sun, and at most another couple of hundred from the high clouds. But you’re proposing a surface that would emit something like 5000 W/m2; such a high surface temperature simply could not be sustained. There is not enough energy trapped in this atmosphere to maintain the flows needed for hundreds of miles of vertical adiabatic convection. With a pure helium atmosphere and infrared-absorbing high clouds as suggested here, the surface temperature on Venus would be similar to Earth, perhaps as high as 300 K. Most of the atmosphere would have a very low lapse rate due to the greenhouse effect being too low to drive full convective overturning on that scale. 30. Patrick 027 …|dp/dz| is smaller; dp/dz < 0, of course… 31. BigWaveDave ain response to: There is no one radiative equilibrium temperature for Earth. Patrick 027 said “Yes and no. Te = [S*(1-albedo)/4]^(1/4) is the radiative equilibrium temperature corresponding to the radiative flux that an isothermal blackbody surface of the same size would have to have to be in equilibrium.” So, yes, there is no one temperature that Earth radiates to space. If one assumes Earth loses energy to space (sink) at the same rate Earth gains energy from the Sun (source), one can calculate the surface temperature of a blackbody. under similar conditions, So, what? Earth does not resemble a blackbody, It more resembles a mobile patchwork of shades with a variable fuzzy coating. There is no fixed set of paths between incoming and outgoing energy, and radiation is not the primary mode of heat exchange between the points where the energy is received from the radiation source and where the energy is transmitted by radiation to the sink. Response: What on Earth does this mean? Essentially all of the energy input and output of the planet is done radiatively. The calculation shown by Patrick is a blackbody calculation which would closely resemble the Earth’s surface in the absence of an atmosphere. The isothermal approximation is not a bad one for Earth, and becomes useless only in situations where you have substantial diurnal gradients (e.g., Mercury or the moon).– chris • That formula for Te above needs to be modified to ( absorptivity / emissivity ) because its only the ratio of those two values , which in turn depend on the correlations between a sphere and its radiant sources and sinks , which makes the temperature any different than that of a gray ( flat spectrum ) ball . I think of “albedo” as only having meaning with respect to flat spectra . This is the computation the implementation on my http://CoSy.com handles allowing me to state the mean temperature of a gray ball ( think of a shell around Venus and its atmosphere ) in Venus’s orbit will be about 328 kelvin . Even a ball , black facing the sun , and totally reflective on the night side will only be around 390k . Note that I’m offering a prize to any student who extends that implementation to full spectra and calculates precise mean temperatures for any particular spectrum , that is , “greenhouse effect” . Dissing of other discussion groups is stupid . • Patrick 027 Typically, absorptivity = emissivity for the same frequency, direction (at least for the absorption of radiation from one direction compared to emission back toward that direction), and polarization. It’s the variation of optical properties over wavelength that’s of great importance here, particularly between the SW (solar dominated) and LW (dominated by emissions from the surface and atmosphere for a typical planet) portions of the spectrum. At night , the process reverses and the warmed air radiates its heat back via the IR active gases thus having an enormous effect on our diurnal temperature variance which I have never seen discussed . It’s a very different thing to change a mean than to change a variance . The backradiation exists day and night; because of the relatively small diurnal temperature range of the atmosphere, the backradiation tends to not have much of a diurnal cycle (though locally it can vary with the weather, of course). The diurnal average of the backradiation is not zero. Anonymous , first , I’ve got to interject that it was shown as early as 1909 that the overwhelming effect of glass green houses comes from the stopping of convection. But in doing so, it requires the temperature inside to increase to some higher value before reaching equilibrium, which is in a general sense analogous with a radiative greenhouse effect. Further , putting a winter coat on a corpse won’t warm it . In what way would this contradict anything I’ve said? (I was that Anonymous; I was in a hurry at that time and missed the fact that “Name” was blank.) But , the bottom line , whatever your argument is : Show me the equations ! It’s actually quite easy to understand the effects of optical properties in a qualitative sense – including the effects of scattering. The mathematics of scattering is potentially much much much more complicated, so I’ll stick to purely absorbing and emitting greenhouse agents and a blackbody surface. Also, setting polarization and refraction aside: For a given frequency, assuming quasi-LTE (relevent to radiation) (a good approximation for most of the mass of the atmosphere and at the surface), blackbody radiant intensity (units W/m2 per unit frequency of the spectrum per steradian (a unit of solid angle)) is given by a version of the Planck function (another version gives the value in terms of per unit wavelength of the spectrum, etc; either one works as long as you are consistent.). The fraction of that value which is emitted along a line of sight, from a small interval dx, is equal to the emissivity in that direction of dx. This is also equal to the fraction of radiant intensity from that direction which is absorbed by dx (absorptivity). The emissivity and absorptivity of dx are equal to the emission cross section per unit volume * dx and the absorption cross section per unit volume * dx. These cross section densities (let’s call them ecsv and acsv, for emission and absorption cross section per unit volume) are proportional to the densities of the particles or molecules which cause them via their absorption and emission of radiation. Assuming quasi-LTE, they are equal to each other. They are also equal to the optical thickness per unit x, so their products with dx are equal to the contribution of optical thickness made by dx. Because of absorption along a line of sight between an inteval dx and some point a distance x away, the fraction of blackbody intensity emitted by dx that reaches a distance x is equal to ecsv * dx * exp(-acsv*x), which is equal to ecsv*exp(-ecsv*x)*dx; the exponential can be derived from integration over x. In case ecsv is not constant over x, it is convient to use a distance o which is measured in units of optical thickness; in that case, the fraction of blackbody intensity (based on the temperature at the P1) emitted by an interval do at point P1 that reaches a distance o to point P2 is exp(-o)*do. This is equal to the fraction of radiant intensity in the line of sight at P2 that reaches a distance o to P1 and is absorbed within the interval do. (This fraction is actually the amount of the weighting function for P2 that is found in do at P1, in case that term comes up again later). Sum the contributions from all intervals along a line of sight to find the total radiant intensity at P2 in that direction. Integrate the product of radiant intensity and cos(q) over solid angle to get a flux per unit horizontal area (W/m2 per unit of spectrum), where q is the angle from vertical. Sum over the spectrum to get the total flux per unit area at P2 (noting that the distances in terms of optical thickness are not the same at different frequencies). Do this for the set of directions crossing the unit area from above and from below, subtract to get a net flux per unit area . The variation of the net flux per unit area over distance is equal to a net radiant cooling (or heating) rate per unit volume. For equilibrium, the heating rates of all SW frequencies, LW frequencies, and convection/conduction/diffusion, must sum to zero for a time-average for a climatically-relevant time period. • Patrick , you have lots of words , but I see no computable sequence of equations . I have few words but refer to a precise and succinct implementation on my website of the essential equations for non-uniformly radiantly heated , non-uniform gray balls and provide experimentally testable predictions . You seem to be lost in the forest of details without understanding it’s first order constraints . Perhaps you ought to spend a little time examining exactly what I’ve implemented so that we have a common point of understanding . As I outline , the extension of the algorithm to full , rather than flat , spectra will provide a rather precise mean temperature for a ball in Venus’s orbit with the observed spectrum of the lumped planet and its atmosphere . I find it highly unlikely that that value will be higher than the 390k of a ball black facing the sun , and totally reflective facing away . Then we can address again how the interior of the sphere can maintain a higher interior temperature than that with no more of an internal source of heat than a corpse . The argument that a static pressure differential cannot explain a maintained temperature differential is , of course , correct . But I have seen no other , quantitative , experimentally demonstrable , explanation for this asserted violation of thermodynamics and common sense , either . Response: You need to read a text on the physics of planetary climate. I suggest Houghton’s “Physics of Atmospheres,” Dennis Hartmann’s “Global Physical Climatology,” or Ray Pierrehumbert’s upcoming text on “Principles of Planetary Climate.” Even simple slab models (which I don’t really like as they don’t capture the way the greenhouse effect truly works) can at least verify to you quantitatively that an IR absorbing atmosphere can in fact warm the surface temperature well beyond that of radiative equilibrium with the sun. Beyond that, your understanding of Kirchoff’s law, and the calculations done for no-atmosphere planets is severely misplaced. For instance, the implication on your page that absorptivity=emissivity and thus the terms cancel in the radiative balance calculation is wrong, since the Earth radiates in a much different part of the spectrum than it receives from the sun– chris • Patrick 027 this asserted violation of thermodynamics and common sense No violation of thermodynamics or common sense to be found. q=zenith angle a=angle around a vertical axis v=frequency s=distance along line of sight note that ds = dz/cos(q) dIv(s,q,a) = ecsv*(Bv(T)-I)*ds Iv(P,q,a)) = integral of dIv over s along direction Q up to point P dFv(P) = Iv(P,q,a) * cos(q) * sin(q) * 2*pi*dq*da Fvup(P) = integral of dFv(P) from q = pi/2 to q=pi Fvdown(P) = integral of dFv(P) from q = 0 to q = pi/2 Fvupnet(P) = Fvup(P) – Fvdown(P) Fupnet(P) = integral of Fvupnet(P) over v (for a range of v pertaining to either SW or LW radiation or to all radiation; FupnetLW(P) would be the integral over the LW portion of the spectrum, FupnetSW(P) would be the integral over the SW portion of the spectrum; Fupnet(P) = FupnetSW(P) + FupnetLW(P)) RCRV = radiative cooling rate per unit volume = d[Fupnet(P)]/dz RCRVLW = d[FupnetLW(P)]/dz RCRVSW = d[FupnetSW(P)]/dz RCRV = RCRVLW + RCRVSW For equilibrium, this must be balanced by convection, etc, (or be zero). • Ok Patrick , do it . First , what is your computation for the temperature of a shell ( ball ) with the lumped spectrum of Venus and its atmosphere in Venus’s orbit ? How does it compare to my gray body approximation ? Second , what is the temperature you derive for Venus’s surface given whatever additional parameters , eg , atmospheric pressure gradient , your argument requires . Please abstract as succinctly as you can the step at by which the interior temperature becomes greater than that calculated for the exterior . • Patrick 027 rewrite two lines: dIv(P,s,q,a) = ecsv(v) * [Bv(T)-Iv(P,s,q,a)] * ds (this is written assuming s increases going towards P; it might be more convenient to assume s increases from 0 going away from P, in which case: dIv(P,s,q,a) = -ecsv(v) * [Bv(T)-Iv(P,s,q,a)] * ds ) Iv(P,q,a)) = integral of dIv over s along direction Q=q,a up to point P where Bv(T) = Planck function in terms of frequency ecsv(v) is a function of v replace frequency with wavelength everywhere and it works just the same, so long as Bv(T) is replaced with B_wavelength(T), the Planck function in terms of wavelength Where a line of sight intersects a perfect blackbody surface, ecsv(v)*ds can be treated as equal to 1 at that location, or alternatively, Iv(P,s,q,a) = Bv(T) when the surface is a distance s from P in the direction considered. • Chris , I don’t know how much clearer I can make it that my first order implementation is for GRAY , ie , FLAT SPECTRUM , objects for which Absorptivity = Emissivity for any source or sink spectra . For colored objects the effective ratio between Absorptivity from sources to Emissivity towards sinks is given by the ratio of correlations of the object’s spectrum with its sources and sinks . Is there any quarrel with that statement – which essentially defines the “greenhouse effect” ? That requires a couple of additional lines to my implementation , but also a fair amount of work acquiring and applying the algorithm to various spectra of interest . That’s why I think it far more worth my time to offer a$300 “prize” to any student who reproduces my gray body implementation and extends it to get the results for various spectra of interest . The implementation has been translated to the freely available language J , and I’m sure a free copy of Dyalog’s APL can be obtained for any student interested in tackling the project .

I browsed Pierrehumbert’s online draft and found no equation by equation quantitative derivations . It looked more like a “Physics for Poets” treatment .

32. Patrick 027

“note that ds = dz/cos(q)”
with a possible positive or negative sign depending on perspective; if s is increasing toward P then cos(q) 0 and cos(q) > 0 when ds/dz < 0, so a negative sign must be added to the equation. If s increases away from P in the direction that intensity is coming from, then the equation works as originally stated.

• Patrick 027

“note that ds = dz/cos(q)”
Correction: if the angles refer to the direction in which the radiation is going (toward P) and s is increasing toward P, then
ds = dz/cos(q)
but if one of those conditions is reversed
ds = -dz/cos(q)
but if both are reversed,
ds = dz/cos(q)

33. BigWaveDave

Patrick 027 said

In stating:: “Just to be clear, gravity tends to impede convection generally everywhere, by increasing the weight of each unit of mass, thereby increasing the pressure decline over mass and thereby increasing the adiabatic lapse rate relative to mass,”, you misstate the direction of gravity’s influence on convection, but almost seem to get the influence of gravity on lapse rate, And then, by concluding: “thus requiring radiative (setting aside conduction, etc.) fluxes on their own to sustain a larger lapse rate if thermally-direct convection by vertically differential heating is to occur.”

Gravity or its corollary, buoyancy, is a driver of convection. When the surface conducts heat to air molecules, the heated air will be less dense, and will rise. For dry air over dry surfaces, this effect is small. For dry air over wet surfaces,when evaporated surface water mixes with surface air, the molecular weight of the mixture is reduced, and the air will rise, even if the temperature remains unchanged.. For moist air, the effect of convection is very large. The latent heat of evaporation is on the order of four thousand times the specific heat required to raise the temperature of dry air by one degree.

Rising air is either cooled locally as it nears the tropopause, or, as in the case of Hadley circulation, changes from vertical to poleward direction at the tropopause

In either the dry or wet case, as air molecules move up, gravity forces other molecules to replace them. In addition, as higher molecules cool, they become more dense and gravity forces them toward the surface.

Patrick 027 said:

“For equilibrium, [RCRV = RCRVLW + RCRVSW] must be balanced by convection, etc, (or be zero).”

When considering the balance, how do you handle transfer of heat by convection?

• Patrick 027

Gravity is an organizer and ‘co-driver’ of thermally-direct convection. But it increases the adiabatic lapse rate, requiring radiative forcing on it’s own to sustain a greater lapse rate if thermally direct convection is to occur, and also tending to impede overturning forced by other means if such a lapse rate cannot be sustained except by mixing.

• Patrick 027

Although, once mixed, it is easier to keep mixing, but it must be done at a sufficient rate to overcome any heat fluxes that would stabilize the layering.

• Patrick 027

In the simplest (1 dimensional column, time average) approach, one can start by assuming that the potential for convection will prevent a lapse rate from getting larger than adiabatic. Then, use the radiative heating/cooling to compute a temperature change at each location for each time step, and keep doing iterations until the temperature distribution approaches an equilibrium, while specifying that convective (and conductive) heating and cooling must be zero wherever the lapse rate is smaller than adiabatic and allowing it to be nonzero elsewhere, but with the sum of convective+conductive heating over the whole column being zero (convective+conductive heating at one location must be balanced by convective+conductive cooling at another location). The convective heating and cooling necessary to maintain this condition determines the net radiative heating/cooling distribution within the layer that is convection. Outside that layer, the net radiative heating/cooling must be zero at each location when equilibrium is reached.

• Patrick 027

In this condition, the net radiative heating must sum to zero for the whole of the convecting layer and surface, but can be nonzero at locations within this layer (and at the surface). This means that at the boundary of such a layer, the net radiative flux (solar + LW) out must be zero.

34. Leonard Weinstein

You have your physics basically correct above but totally miss the point that Steve and Lubos made. They AGREE that the greenhouse effect is necessary to move the tropopause to the upper edge of the atmosphere of Venus. However, even if the composition was only .01 times as much CO2, and the rest an equal density non greenhouse gas (e.g., Argon), the surface temperature would only change a few degrees C. The point is that it is not a runaway greenhouse gas effect, only that a small but sufficient amount of some greenhouse gas is needed to move the location of the tropopause, but going to a lot more CO2 has only a small additional effect. This is not how a runaway effect is defined. In fact if the CO2 on Venus were only .001 times the present value (with the replacement Argon), it still would only have a small effect of difference in ground temperature. The same for Earth. Going from present CO2 level to 2 times the present level would only directly change the surface temperature a little over 1 degree C if there were no feedback effect from water vapor. The question of water vapor feedback is not settled for Earth, but indications are that it may be negative, although that is not an issue for Venus.

Response: There is absolutely no support for your assertion that Venus would only change little if you removed most of the CO2 and replaced it with a non-GHG. Certainly no text in planetary climate or anything in the refereed literature will support this and it’s something you basically made up. This falls right from the absorptive properties of GHG’s and their influence on the transmission [exp(-tau)] of the atmosphere. Once again, I recommend the Bullock and Grinspoon link.

It is true that a very thick N2 (or some other non GHG) atmosphere with just a bit of CO2 can still generate a strong greenhouse effect. In fact, in the Sagan and Muller 1972 paper on the faint young sun, the authors believed that adding one bar of N2 to the GHG atmosphere that got them above freezing gave an unrealistically hot situation. Though, you cannot get to Venus with just trace amounts of CO2. This is largely a broadening effect and also because the other gases influence the thermodynamic properties of the air (heat capacity, etc)

I’m also aware that the runaway is not defined by high pCO2 and I didn’t claim otherwise. It’s not really appropriate to say that Venus is in a runaway as much as it was in a runaway and is now in a state where it can’t get rid of the CO2 atmosphere and won’t ever have liquid water again.

As for water vapor, there is absolutely no evidence for a negative feedback effect, and the WV feedback is now rather well understood. See the review paper by Sherwood/Dessler, the work of Soden et al., etc. There is really no question now that water vapor is a strong positive feedback and you’re trying to over-inflate the uncertainty by baseless assertion. The issue is clouds, but even there, not much evidence for a negative feedback.– chris

35. Leonard Weinstein

I was partially wrong in my previous response stating that your physics was basically correct. On re-reading it more carefully it is clear that you started out with basically correct physics, but then lost it in the following material. The lapse rate basic analysis was correct. Much of the rest is partially to totally wrong. I would be glad to go into any desired detail why. One of the many points is that there is no need for any direct Sunlight to reach the ground (convection and back radiation do just fine alone).

• Patrick 027

Absent geothermal or tidal heating from below (these things do exist but are very small), if there is no SW (solar) heating below some level, then the net convective + LW radiative + conductive/diffusive heat fluxes upward from that level must be zero. Since they tend to be in the same direction, they must individually tend to be zero. What temperature lapse rate would give rise to that situation? Sufficient LW opacity can bring the net LW flux to near zero.

The lapse rate one gets is that which allows the fluxes to balance. In low viscosity fluids, thermally-direct convection can prevent the lapse rate from becoming significantly superadiabatic on large scales and take up any slack from radiation, etc, that are required by the cap on the lapse rate, but if there is no such slack, thermally-direct convection cannot be sustained; if the non-convective non-SW heat transport occurs too easily for the same SW heat fluxes, the lapse rate will be too small to support thermally-direct convection.

36. Leonard Weinstein

Chris,
As you well know, even a relatively small amount of greenhouse gas blocks most direct radiation from surface to space (notice that at the higher surface temperature, different long wave wavelengths come into play than for the Earth, but I have not examined which bands are now most important. However I still assume effective blockage for Venus). The CO2 on Venus is 230000 times as much as Earth. At .001 times the Venus level, but the same atmospheric mass, the CO2 would still be 230 times that of Earth. The amount on Earth (along with water vapor) is sufficient to be a real greenhouse effect, i.e., already blocking the outgoing radiation in the lower atmosphere. Adding the clouds on Venus makes the block even more complete. In the case of Earth and Venus, the fact of direct radiation block from the surface to space results in almost all of the transfer of energy from the surface upwards coming from atmospheric convection, not radiation (the upwards and back down radiations almost balance at lower levels in the atmospheres). This continues up to a high enough region so that radiation can escape directly to space. The only change more greenhouse gas has once the minimum amount to produce reasonable lower atmosphere blocking occurs is to slightly raise the level of the location radiating to space. However, the raising of tropopause level would only affect a small fraction of the mass of the atmosphere (that above the tropopause). The effective tropopause level determines (locks) the location of input and outgoing balance. The lapse rate due to -g/Cp does the rest of the heating going downwards.

• Patrick 027

Could we clear something up. I think the use of dT/dz = -g/cp as an adiabatic lapse rate is not preferable to Poisson’s equation:

However, the raising of tropopause level would only affect a small fraction of the mass of the atmosphere (that above the tropopause).

But what fraction of that fraction is affected? I don’t really understand why we use -g/cp for an adiabatic lapse rate anyway because that only applies if … well maybe I’ll let someone else explain that ** – but the actual adiabatic change in temperature following fluid motion is given by cv*dT = dq – p*d(alpha) when dq = 0 (where alpha = 1/density), which, using p*alpha = R*T (for an ideal gas), alpha*dp + p*d(alpha) = R*dT; substitution into the first equation. I think the issue is state vs environmental variables:

(using ‘D’ instead of ‘d’ to emphasize derivatives following a parcel of material (Lagrangian instead of Eulerian)):

First law of thermodynamics
Dq = Du + Dw = cv*DT + p*Dα

where α = 1/ρ = 1/density,
equation of state for an ideal gas:
p*α = R*T
therefore
p*Dα + α*Dp = R*DT
p*Dα = R*DT – α*Dp

substitute into first law of thermodynamics:

(cv+R)*DT – α*Dp = Dq
note that if Dp = 0, Dq = cp*DT, therefore cp = cv+R;

also, Dh = Du + D(p*α) = cv*dT + p*Dα + α*Dp = Dq + α*Dp

Dh = cp*DT = Dq + α*Dp

For the dry adiabatic lapse rate, set Dq to zero, substitute R*T/p for α (equation of state for ideal gas), and integrate:

for dq = 0:
cp*DT = (R*T/p) * dp
cp/T * DT = R/p * Dp
integration:
cp*ln(T/T0) = R*ln(p/p0)
ln(T/T0) = ln[(p/p0)^(R/cp)]
T = T0 * (p/p0)^(R/cp)

which describes the temperature as a function of pressure in adiabatic pressure changes. Note that this implies the lapse rate is proportional to temperature, and also that T = 0 when p = 0, and a fractional change in p0 produces a fractional change in T for the same p and T0. Surface T would actually approach infinity if the tropopause approaches p=0, for the same tropopause temperature.

The lapse rage -g/cp is derived from:

cp*DT = α*Dp

using the hydrostatic relationship ∂p/∂z = -g/α. If a parcel maintains an environmental value of p – which is approximately what typically happens (sound waves are not a major aspect of atmospheric dynamics), then p(parcel) = p(environment) = p, and Dp/Dt = ∂p/∂t + Dz/Dt * ∂p/∂z during vertical motion Dz/Dt, and for a sufficiently small parcel, p(x,y,z) can be treated as a constant, so that ∂p/∂t = 0, so that Dp/Dt = Dz/Dt * ∂p/∂z; dividing both sides by Dz/Dt, Dp/Dz = ∂p/∂z, and
cp*DT = α*Dp = α*(∂p/∂z)*Dz = -g*Dz

Wait! Try again:
cp*DT = α_parcel*Dp = α_parcel*(∂p/∂z)*Dz
= -[α_parcel / α_environment ] * g*Dz

(see the important distinction?)

But consider the situation where the environtal lapse rate is dry adiabatic, the two α values should be the same, and we’re back to -g*Dz…

… okay, but there must be something wrong with that, because it simply isn’t true that raising a parcel of air dry adiabatically actually brings it to 0 K at some finite height with nonzero p, and in the hydrostatic relationship, p decays exponentially (varying with temperature)…

(Chris, what did I miss here? Every time I’ve seen the -g/cp lapse rate, it’s bothered me because it doesn’t square with the other adiabatic lapse rate, and I know the other lapse rate is the ‘real one’…)

——-

Anyway,

The effective tropopause level determines (locks) the location of input and outgoing balance.
Yes and no; the tropopause level can be but isn’t generally the effective emitting level (approximately, the centroid of a weighting function) for the flux going to space. On Earth, a majority of LW radiation to space comes from within the troposphere and some still comes from the surface. If I’m not mistaken, on Venus, the effective emitting level is somewhere within the stratosphere (?).

The lapse rate due to -g/Cp does the rest of the heating going downwards.

I will give you the benifit of the doubt and assume you mean that the adiabatic lapse rate determines, in combination with the location of the boundaries of a convecting layer and the temperature at at least one location within it, the temperature at any other point within the convecting layer.

The lapse rate is not itself a source or sink of heat, though it shapes those things.

There is significant LW cooling of the Earth’s surface. Some of this goes directly to space, some is a net heat transfer to the atmosphere, which, combined with convection and direct solar heating of the atmosphere, is balanced by atmospheric heat loss to space. LW heat loss to space is not distributed the same way as net LW cooling, because some LW cooling is associated with emission that is absorbed in another layer of the system.

• Patrick 027

But consider the situation where the environtal lapse rate is dry adiabatic, the two α values should be the same, and we’re back to -g*Dz…
… for the case where a parcel comes from within the same dry-adiabatic layer.

• Patrick 027

Oops, I just realized what I missed…

… okay, but there must be something wrong with that, because it simply isn’t true that raising a parcel of air dry adiabatically actually brings it to 0 K at some finite height with nonzero p, and in the hydrostatic relationship, p decays exponentially (varying with temperature)…

(Chris, what did I miss here? Every time I’ve seen the -g/cp lapse rate, it’s bothered me because it doesn’t square with the other adiabatic lapse rate, and I know the other lapse rate is the ‘real one’…)

I was wrong to say ‘real one’, though it is the more generally applicable relationship.

It’s because T as decreases with z along a dry adiabat for an adiabatic layer, p decays ‘exponentially’ at faster and faster rates – not really exponentially, then – it could go to zero.

Of course, if we are changing the T and preserving mass, then relative to mass, the lapse rate will change…

• Patrick 027

This line:
Dh = cp*DT = Dq + α*Dp
was incorrect (take the middle out, I think)
though not pertinent to the rest…

• Patrick 027

… no, now I’m not sure about that last bit (the equation for enthalpy Dh related to cp*DT); need to review it again…

• Patrick 027

Du = cv*DT

Dh = Du + D(α*p)
= Du + p*Dα + α*Dp
= Dq + α*Dp

Dq = Du + Dw
= Du + p*Dα
= Du + D(α*p) – α*Dp
= Dh – α*Dp

Dh – α*Dp = Du + p*Dα

For an ideal gas, α*p = R*T, so D(α*p) = R*DT, thus
Dh = Du + R*DT = (cv+R)*DT
and
Dq = (cv+R)*DT – α*Dp
when Dp = 0, Dq = cp*DT = (cv+R)*DT
therefore (allowing nonzero Dp),
Dq = cp*DT – α*Dp
Dh = cp*DT

But for other materials, α*p = ?, Dq = ?? – α*Dp, cp = ???, Dh = ????…
——————–
An attempt to find relationships:
where (∂a/∂b)c is a partial derviative with c held constant

α = α(p,T)
Dα = D[α(p,T)] = (∂α/∂p)T * Dp + (∂α/∂T)p * DT,
p*Dα = [p*(∂α/∂T)p]*DT + [p*(∂α/∂p)T]*Dp

For clarity, let (these are nonstandard variables)
r = p*(∂α/∂T)p
A = -p*(∂α/∂p)T
so that
p*Dα = r*DT – A*Dp

substitution into first law of thermodynamics
Dq
= cv*DT + p*Dα
= (cv + r)*DT – A*Dp

Since Dq = cp*DT for Dp = 0, cp = cv + r, and so

Dq = cp*DT – A*Dp

—-

substitute p*Dα = r*DT – A*Dp
into equation for Dh:

Dh = Du + D(α*p)
= cv*DT + p*Dα + α*Dp
= cv*DT + (r*DT – A*Dp) + α*Dp
= (cv+r)*DT + (α-A)*Dp
= cp*DT + (α-A)*Dp

for at least an ideal gas, A = α, so Dh = cp*DT

Hope I didn’t make any algebraic errors.

37. Any attempt to explain the temperature of Venus needs to take into account some factors that are commonly missed.

1. The slow rotation of the planet.

If we view things in terms of strata-and-heat-budgets and not in terms of watts-per-square meter we see that this becomes important since it leads to less disruption between the strata.

2. Transfer of electrical energy. Via the solar wind and Birkeland currents. We also want to know the effect of air pressure on the conversion of this electrical energy to thermal energy.

3. Air pressure as mentioned. Its air pressure that would make greenhouse effective. Since the typical greenhouse molecule will merely scatter its absorption region, where the air pressure is low, and not absorb this energy. This is because to create the transfer to thermal energy, as opposed to mere scattering, you would presumably need the molecule to be disturbed be a second molecule, in the limited time before it would merely re-release the “photon” at a different direction. (Not that I believe in photons but thats the simple model we are working with.) It is air pressure which gives us the probability of one molecule being disturbed by another in that limited time.

Therefore we must infer that greenhouse properties are pretty useless without sufficient air pressure.

4. The super-rotation of the clouds. Surely this is direct evidence for a sort of convection-oven arrangement. This directly implies powerful overturning that can be used to recycle thermal energy. Just like having a ceiling fan helps with the effectiveness of your heater.

People who concentrate on these little Watts-Per-Square-metre equations are denying the physicality of the situation.

Response: No one is downplaying the effects of fluid dynamics on climate but the fact is that rotation rate does not set the global mean temperature to be several hundred degrees K in excess of what incoming sunlight allows. And obviously the GHE strength depends on the pressure, no one said otherwise– chris

38. Patrick– I suppose it’s worth noting a few things, since I can’t quite pick up on where you are (seem?) confused, so maybe something here will help:

— Poisson’s Equation is a means to determine how T changes with pressure for an adiabatic process (and potential temperature is conserved for a dry adiabatic process), while g/cp tells us how the temperature of a moving parcel changes with height

— The dry adiabatic lapse rate (g/cp) is an approximation because it assumes a particular relationship between changes in pressure and changes in height. It’s also possible to introduce a small error if hydrostatic balance is invalid or T(environment) =/= T(parcel), but these are not large issues. I agree Poisson’s equation is in general a better form because pressure is often much better than height as a preferred vertical coordinate in atmospheric science.

— The environmental lapse rate (elr) is not, in general, the same as the dry adiabatic lapse rate (dlr). In fact, a measure of stability is whether elr > dlr (unstable), elr < dlr (stable), or elr = dlr (neutral). In potential temperature thinking, the stable case is the same as saying ∂(theta)/∂z > 0 where theta is the potential temperature.

–Note I’m not thinking about moisture at all here and this breaks down above the tropopause where there is no convection

• Patrick 027

Thank you Chris. I actually realized the mistake (and was a bit embarrassed) I was making not long after I posted that (see my follow-up) … I got fixated on the idea that p never actually goes to zero at some finite height, forgetting that it actually will go to zero if α doesn’t go to zero, which is possible if T does go to zero. In other words, I just needed to find p(z) for the situation where T = -g/cp * z + T0; then I would have found agreement between the two lapse rates (as long as I’m here, might as well do this):

T_poisson = T0 * (p/p0)^(R/cp)
T_adiabaticlayer = T0 – g/cp * z
where p(z=0) = p0

Let T_adiabaticlayer = T_poisson = T
T0 * (p/p0)^(R/cp) = T0 – g/cp * z
solve for p(z)
p^(R/cp) = p0^(R/cp) * [ 1 – g/cp * z/T0 ]
p = p0 * [ 1 – g/cp * z/T0 ]^(cp/R)
find ∂p/∂z
∂p/∂z
= p0 * (cp/R) * [ 1 – g/cp * z/T0 ]^(cp/R – 1) * [-g/(cp*T0)]
= -g * p0/(R*T0) * [ 1 – g/cp * z/T0 ]^(cp/R – 1)
substitute z = (T0-T)*cp/g
∂p/∂z
= -g * p0/(R*T0) * [ 1 – g/cp * [(T0-T)*cp/g]/T0 ]^(cp/R – 1)
= -g * p0/(R*T0) * [ 1 – (T0-T)/T0 ]^(cp/R – 1)
= -g * p0/(R*T0) * [ T/T0 ]^(cp/R – 1)
= -g * p0/(R*T0) * [ T/T0 ]^(cp/R) * T0/T
= -g * p0/(R*T) * [ T/T0 ]^(cp/R)
substitute Poisson’s equation (T/T0)^(cp/R) = p/p0 :
∂p/∂z
= -g * p0/(R*T) * p/p0
= -g * p/(R*T)
= -g/α
which is the hydrostatic relationship, as expected.

So to wrap up, for a dry adiabatic process with an ideal gas,
T = T0 * (p/p0)^(R/cp), where, if p0 is a standard reference pressure level, T0 = θ = potential temperature, which is conserved following dry adiabatic processes, so
DT/Dp
= θ/p0 * R/cp * (p/p0)^(R/cp-1)
= α0/cp * (p/p0)^(R/cp-1) (that’s a form you don’t see everyday)
= R/cp * θ * (p/p0)^(R/cp) / p
= R*T/p * 1/cp
= α/cp

When a layer of ideal gas has constant θ and is in hydrostatic balance, the environmental lapse rate = a dry adiabatic lapse rate, and the way p varies with z allows a linear temperature decline with height, ∂T/∂z = -g/cp, and any dry adiabatic inviscid motions within a dry-adiabatic layer that don’t disrupt hydrostatic balance (so that DT/Dz = ∂T/∂z ) will involve an exchange between enthalpy and gravitational potential energy: Dh = cp*DT = -g*Dz, with no production or consumption of kinetic energy.

———
What’s interesting to point out with regards to massive changes in climate is that pressure may be approximately conserved at the surface but p(z) changes if T(z) changes; if surface T increases then the density decreases, so the same mass of air occupies a larger volume, so one has to go higher for to reach the same p, so the same ∂T/∂z will extend over a larger range of z for the same range of p, so the temperature difference between the bottom and top of such a layer will increase for the same layer mass.

(Of course, with feedbacks such as water vapor (think early Venus), some mass will be added to the atmosphere, while expansion of mass outward decreases mass per unit area and decreases g, thus decreasing the pressure for the same mass (typically a minor effect, and with upper atmospheric cooling for GHG forcing, reversed for the mass of the upper atmosphere).

• Patrick 027

… or maybe α does go to infinity, but not fast enough to prevent p from going to zero at finite z, …. etc.

• Patrick 027

forgetting that it actually will go to zero if α doesn’t go to zero,
I meant infinity, and of course, it could (and based on the math above, does) go to infinity, but only as p approaches zero at finite z; in agreement with α, ∂p/∂z goes to zero at that same point (at the top of an adiabatic layer that runs through an entire ideal gas hydrostratic atmosphere).

The freakish corollary is that they’re also questioning the very idea of black body radiation – so they want us to go back way before Planck … before Kelvin, maybe?

Not since Velikovsky has easy-to-prove science been so challenged so confidently.

Velikovsky’s first crank paper once he ventured out of psychoanalysis, which he studied in college, was about the atmosphere. A central claim was that the gases sorted out like thin layer chromatography, instead of mixing. That’s where i feel we are with this.

All of this was hashed out in the 1950s. Real theories, including ones with elements of what Goddard is claiming, and those involving particulates, and so on, vied with crank theories like Velikovsky’s electrostatic pool table solar system. Hansen was a latecomer to the runaway greenhouse theory, so he had to work on one of the other lines, because Sagan and colleagues owned it by then.

Also, putting a winter coat on a freshly dead corpse will slow its rate of cooling. And the atmospheric argument about cold to hot is like the bellboy and the missing dollar riddle – misdirection.

There’s no need to talk about heat flowing at all. Lower frequency radiates from the earth. It heats up CO2 molecules. They warm up then reradiate in all directions. About half those directions are back towards the Earth. In practice, it’s a chain both ways, of countless molecules of CO2, H20, CH3, differently transparent to some frequencies of radiant energy. It is what it is.

That’s also reminiscent of the cheap 2nd law arguments by ID proponents.

42. Bob_FJ

Chris,
Do you agree that there must be convection on Venus, stronger than on Earth? If so, do you agree that there would be mechanical work involved, and that the constantly descending gases will be compressed, thus raising their temperature?

Marion Delgado, you wrote:

“…About half those directions [of re-radiation] are back towards the Earth…”

Not true; the majority of re-radiation is sideways which importantly, in an homogenous parcel has no heat transfer effect. See this distribution illustration:

43. Patrick 027

Bob FJ – for material with isotropic radiative properties, for a sufficiently small volume such that it may be approximated as isothermal, the emitted radiation intensity is approximately isotropic; if that volume is spherical, the flux will be isotropic; In terms of photons, the same number would be emitted in each direction. While most of those directions are not purely up or down, most are not purely horizontal. It can truly be said that half are emitted upward and half downward, though generally over a range of angles, and that of course affects how far up or down they can go before they are absorbed.

I don’t know the details here but it seems likely that solar heating at the surface and in the lower troposphere of Venus is significantly less than that on Earth. There is ~ 60 W/m2 (from memory) of net LW cooling at the Earth’s surface. This leaves ~ 100 W/m2 available to be transported from the surface by convection. How much is available on Venus? Even a small amount could be enough to sustain a nearly-adiabatic lapse rate. An adiabatic layer is not necessarily being ‘vigorously’ mixed. And, in an equilbrium time average, the mass of gas going downward and increasing in temperature adiabatically must be balanced by a mass going upward and decreasing in temperature adiabatically. Actually, the thermally-direct process removes heat, converting it to kinetic energy, though ultimately that goes back into heat again via viscosity if not via thermally-indirect motion; the net process is a transport of energy within the system, which (in equilibrium) will be balanced by non-convective (ie radiative, and conduction/diffusion at the surface, though in the context of climate and weather, that is often lumped together with convection) heat flux, which can only leave an imbalance that allows/drives convection if the temperature variations allow; in conditions where temperature variation that would leave an imbalance necessary to drive convection would be stable to convection, convection (at least of the thermally-direct sort) would not tend to occur.

• Patrick 027

Correction:
for material with isotropic radiative properties, for a sufficiently small volume such that it may be approximated as isothermal, the emitted radiation intensity is approximately isotropic
If the volume is spherical, etc…

A volume that is spherical, isothermal, homogeneous and with isotropic properties, in quasi-LTE, will emit the same intensities and flux in all directions. Intensities is plural because the highest intensities will come from paths going through the center of the sphere, with the intensity going to zero for paths just grazing the sphere, but the same combination of different intensities for different paths can be found in every direction.

Of course, if we talk about optically-thin layers, the largest intensities will be nearly horizontal. However, some of what is emitted is absorbed in the same layer, and the intensity emitted by the layer that emerges from the layer only approaches and doesn’t exceed the blackbody value for that layer’s temperature. There will be some such radiation that emerges from a layer at a grazing angle – nearly horizontal, but still coming from the top or bottom and thus exiting the layer upward or downward.

Accounting for emissions from other layers, some of which is transmitted by the closer layers, there are total intensities generally in all directions (except in space – actually could consider there are intensities there in all directions but some are equal to zero – or actually, approximately zero), but half of those directions are downward and half upward. The intensity (as measured or seen from some location) from a direction is the volume integral of the product of the emission weighting function (which itself has a volume integral of 1) and the blackbody intensity (in the absence of scattering or reflection, the weighting function is compressed onto a single path and so the volume integral is equal to a line integral along that path). Specifically for constant optical properties over direction (isotropic) and space (homogeneous composition (physical and chemical)), there is a greater difference between intensities in opposite directions when the directions are more closely aligned with the temperature gradient, so that the emission weighting function-weighted blackbody intensities are more different. Hence the net intensities tend toward zero along isothermal paths, which tend to be nearly horizontal. Nonetheless, it is more elegant to note that there are intensities in all directions and then consider the effects of temperature distribution on that, rather than to suggest that there are two different populations of photons and one has no effect whereas the other does.

Also, it can’t just be said that a majority of radiation is emitted sideways. Of the photons emitted by a layer, the intensity is generally largest for photons emitted along the plane of the layer; however, that intensity is a value per unit area normal to the direction, and for a layer of given dimensions, that area is largest for directions perpendicular to the plane of the layer. Generally (for isotropic optical properties), the same number of photons (per unit direction (and frequency), but totalled over area, and path length, etc.) are emitted in each direction, regardless of the shape of the space within which the emission occurs. Of photons that are able to leave a thin layer (not emitted and then absorbed, but emitted and then escaping), generally, the greatest number of photons will escape closer to normal to the plane of the layer, as those are the least likely to be absorbed within the same layer.

44. Bob_FJ

Patrick 027, Reur June 9, 2010 @ 11:28 pm and 12:48 pm

Thanks for your lengthy response, but my comment was, as a late visitor to this thread, a bit quick and flippant. I want to go over some other points first with Chris, to explain my issues in a more logical sequence

45. Bob_FJ

Chris, you wrote in your lead article:

Since Venus radiates at the surface at several thousands of W m-2 and yet the outgoing radiation is even less than that of Earth, this shows how important a strongly opaque atmosphere influences the planetary surface.

I guess you have applied S-B based on the surface temperature, but please consider that this approach is not correctly applicable in the case of immersion in a very opaque fluid, as demonstrated in the following first principles argument:

a] Consider a block of concrete that is covered with fully opaque paint; There are no photon emissions from the concrete, but only from the paint surface.

b] Consider a block of wet water-ice at zero degrees; There are zero or almost no photon emissions from the ice, but only from the water surface. (a water skin is very opaque to EMR photons that might otherwise emanate at low temperatures at such long wavelengths…. See footnote).

c] Consider the rock surface of Venus; Apart from small windows around 1 or 2 microns wavelength, it seems probable that almost all free path lengths for photons would be blocked, and the surface will lose HEAT via conduction, and thus convection will carry this HEAT upwards, (to less opaque regions), accelerating that process.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
There is also a paradoxical consideration as follows:

For any given surface temperature, S-B provides calculation of EMR loss rate. However, if the body is immersed in a fluid, even if it is transparent, there will also be HEAT loss via conduction. If it is a gas, the rate of conduction, and consequent convection, increases with temperature and/or pressure, which is very relevant on Venus. So; is this extra or substitutional energy?

Footnote; Visible light penetrates liquid water effectively to about 100m, but infrared absorptivity is about 7 orders of magnitude higher.

Response: Sure, convection is important, but sigma(Ts^4 – Te^4) is a good first-order measure of the greenhouse effect– chris

• Patrick 027

I had come across a comment somewhere on RC stating that the actual emission and absorption of photons is complicated within such opaque materisls that the average paths are on the order of a wavelength or less, so … maybe the concrete surface is not emitting many photons (though would the net flux still be as it would in the ‘normal’ formulation as a difference in fluxes from different emission weighting functions with different temperatures determined by optical thicknesses…?). However, if you removed the paint, there would be emission of photons according to the optical properties of the concrete surface; aside from the effect mentioned in the first sentence, the concrete would be emitting photons just the same, they would just be absorbed by the opaque paint (which would be emitting photons that are absorbed by the concrete) or backscattered (and then absorbed by the concrete) by the opaque paint. Either way, the paint affects the flow of radiation, which is the whole point, really. If the paint and concrete surfaces had the same emissivities and the paint had a different temperature, then the paint covering would change the radiative flux up from the whole system.

“So; is this extra or substitutional energy?”
In equilibrium the energy fluxes – all of them – must balance. If a change is made to LW radiation and SW fluxes are held constant, then the conduction and convection would change to balance the change in LW radiation. But bear in mind that a change in optical properties that changes LW radiation will cause accumulations and/or depletions of heat, changing the temperature, until some combination of changes in potentially both LW fluxes and convection, etc, restores balance.

46. Bob_FJ

Chris, Reur response on my June 15, 2010 @ 8:16 pm

Response: Sure, convection is important, but sigma(Ts^4 – Te^4) is a good first-order measure of the greenhouse effect– chris

OK, let’s look at this another way:
On Earth, there is a significant greenhouse effect caused by a mere 0.038% of CO2. However, on Venus, CO2 approaches 97% concentration, or is about 2,500 times greater in percentage composition. Additionally, the atmospheric pressure at about 92 times greater than on Earth is a large multiplier in quantity of CO2. What with increased density and absorption band broadening, it seems reasonable to conclude that its absorptivity in the infrared is extremely high; perhaps on a par with liquid water. My argument above shows that a body immersed in water, which is opaque in the infrared, cannot significantly emit EMR, (photons), if at all. (just as in the analogy; concrete cannot emit EMR when coated with opaque paint). HEAT transfer when immersed in a fully opaque fluid can only take place via conduction, which is accelerated by convection and advection.

Thus, the conclusion from the first principles discussed, is that there is probably no or negligible greenhouse effect at the surface on Venus. That is to say that there is not significant CO2 photon radiation to cause greenhouse effect until the ascending air reaches less opaque regions.

Chris, can you logically dispute my points a] b] & c] above etc?
Do you know of any references concerning S-B application for bodies immersed in fluids?

Response: But one of the key aspects of the greenhouse effect is to consider what is happening higher up where it is colder and less optically thick. If all the CO2 on Earth were squished to a thin layer near the surface you would not get any significant greenhouse effect, regardless of the layers optical thickness. I personally can’t make much sense out of the terminology “greenhouse effect at the surface…” since the surface temperature is just dragged along with the IR trapping which requires consideration of the whole atmosphere.– chris

47. Patrick 027

Re Bob FJ –

1. I’m not sure that covering up a surface of concrete with paint actually eliminates photon emissions from the concrete surface – if it doesn’t, what it does do is hide those photons, either absorbing them (and emitting some back to the concrete) or backscattering them into the concrete.

2. Either way, IT HAS AN EFFECT (emphasizing, not shouting) on the radiative fluxes. Saying that covering a surface with some material eliminates or reduces or blocks emission from the surface and then arguing that therefore there is no greenhouse effect at the surface is self-contradictory. Unless what you mean is that the greenhouse effect has been saturated at the surface. Which could be true, except, the totality of the greenhouse effect where convection occurs depends on the effect of optical properties on radiative heating/cooling over the extent of convection (which can be expressed in part in terms of changes in net fluxes at the boundaries of such a layer, such as at the tropopause), because convection will tend to respond to changes in radiative heating or cooling in such a way as to spread out the changes. For example, if the cooling at the top of a convecting layer is reduced, then the convective cooling of the lower part of that layer will tend to be reduced.

48. Bob_FJ

Chris, Reur response on my June 16, 2010 @ 8:48 pm

RESPONSE: [1]But one of the key aspects of the greenhouse effect is to consider what is happening higher up where it is colder and less optically thick. [2]If all the CO2 on Earth were squished to a thin layer near the surface you would not get any significant greenhouse effect, regardless of the layers optical thickness. [3]I personally can’t make much sense out of the terminology “greenhouse effect at the surface…” since the surface temperature is just dragged along with the IR trapping which requires consideration of the whole atmosphere.– chris

[1] Yes, but on Venus most of the HEAT transfer from the surface is via conduction. (or if you like, that EMR emissions are negligible). Conjecture then arises concerning at what altitude is the opacity reduced sufficiently to enable significant emissions that would result in absorption and back radiation, and to what effect. Whatever, importantly, the application of S-B to determine the rate of energy loss from the surface and then to equate that to a greenhouse effect is flawed. An analogy would be to attempt to apply S-B to the hot interior core of the Earth, and then describe the insulating effect up to the crust as a greenhouse effect rather than a thermal conductivity effect. The rate of heat loss from the Venus surface is predominantly related to the thermal conductivity of CO2 & convection speeds, and would be rather tricky to calculate given the lack of data. (however, there are some negligible EMR losses, that are almost off the end of the Planck curve, at around 1 or 2 microns)

[2] Resorting again to first principles, considering EMR alone, a major factor is that the surface radiates hemispherically such that 3 dimensionally, most radiation is “sideways”. Therefore, even if the free path lengths are long, many initial absorptions by GHG’s are sensibly rather close to the ground. The GHG’s then firstly re-radiate spherically from these many low heights, but at reduced intensity, because it is not just the GHG molecules that get hotter. Most of their photon increased KE is not lost by re-emission but in collisions that warm the hugely more numerous non GHG molecules. As the upward components of the EMR re-emissions progress, the air becomes thinner and colder, so that free path lengths increase, whilst the intensity of re-emission reduces by virtue of the lower T’s.

Thus, despite the depictions in the Trenberth et al energy budget diagrams, showing massive back radiation from the clouds all the way to the surface, most of the action is rather close to the surface.

[3] The way the greenhouse effect* on Earth is quantified in terms of AGW increase, is partly by considering the 150/180 – year surface T records. The greenhouse effect, on Earth is at its greatest at or rather close to the surface. That is what I meant by GHE at the surface, and that effect reduces with altitude. Venus is very different.

*(which by definition = when surface originating infrared EMR is absorbed by GHG’s)

49. Bob_FJ

Patrick 027 Reur June 17, 2010 @ 4:13 pm (and partly to your two earlier comments)
Thanks your lengthy comments but you have lost me a bit in some interesting contemplations and divergent thinking. If I can quickly pick-out a few points, by your item numbers in your latest:

1) In the analogy of concrete coated with opaque paint, I mean paint that it is opaque to EMR, which in context is in the infrared region. By opaque, I mean that infrared photons are unable to pass through it, and instead, HEAT transfer is via a process of conduction, by which it traverses through the paint before an EMR emissible surface atop the paint is found.
It is not hard to find opaque paints in the infrared region, and a comparison can be found in the reverse direction with the higher energy UV. Plastics that can be degraded by UV can be fully protected by suitable UV resistant (opaque) paints.

2) The opaque layer of paint may have an effect on radiative flux versus the unpainted concrete, depending on its thermal conductivity or insulating effect, but it is I would think trivial. As for the rest of what you say, perhaps you should reword it so that I can better understand it.

Footnote: EMR is a different form of energy to HEAT, although they are commonly confused amongst climate scientists.

50. Patrick 027

Footnote: EMR is a different form of energy to HEAT, although they are commonly confused amongst climate scientists.

We’ve been over that before. No confusion. if EMR is radiated by processes that take up heat, then EMR emission is a flux of heat out from material. If EMR is absorbed by a process that adds heat to a material, then the EMR absorption is a flux of heat into the material. EMR itself propogates with some (group) velocity and has some energy density and also typically has some entropy and a finite brightness temperature.

• Bob_FJ

Hi Patrick,
The only difficulty I have with your comment is the assertion: No confusion Unfortunately I’ve not archived some stuff I’ve seen…. perhaps it was worse about 5 years ago though?

51. Leonard Weinstein

Here is a simple thought experiment: Consider Venus with it’s existing atmosphere, and put a totally opaque enclosure (to incoming Solar radiation) around the entire planet at the average location of present outgoing long wave radiation. Use a surface with the same albedo as present Venus for the enclosure. What would happen to the surface temperature over a reasonably long time? For this case, NO Solar incoming radiation reaches the surface. I contend that the surface temperature will be the same as present. Now assume the initial gas temperature and distribution are different from present, but are not condensed. I contend the temperature will go to the present distribution and level.

• DeWitt Payne

For a temperature gradient to exist, there must be heat flow. The temperature gradient in the Earth’s atmosphere exists because SW solar radiation reaches the surface and must then leave the TOA as LW radiation. There is resistance to heat transmission through the atmosphere so a temperature gradient forms. If there were an opaque shell at any altitude at constant temperature, there would be no heat flow inside the shell at equilibrium and the interior will eventually become isothermal at the temperature of the shell. If that were not the case, then one could not make a near blackbody by taking a box with walls at constant temperature and cutting a small hole in it.

• Bob_FJ

Hi, DeWitt,
But you are asserting that the radiative processes are all-embracing, when they are not. There is strong speculation together with supporting evidence that on Venus there is powerful convection from the surface to a great height. You should not ASSUME that the descending aspect of convection is the same thermally as the ascending. (with no net thermal effect between the surface and the upper levels of the clouds). I’ll elaborate on this somewhere soon below, when I have time.

• Patrick 027

BobFJ, in the absence of forced convection (ie work done to force heat downward), DeWitt is correct. IF there is a level beneath with there is no solar heating, then (approximating geothermal and tidal heat fluxes as zero) the net upward LW flux + upward convective flux at that point must be zero at equilibrium. There is no need for a temperature gradient within such a layer to allow convection or drive a net LW flux.

Of course, LW fluxes can be transmitted across intervening layers, so depending on optical properties, an isothermal layer could have some net LW flux coming out of it, so to keep the fluxes at zero, the temperature might have to increase with height up to the level where solar heating occurs or else remain isothermal past that level, …. etc. – I haven’t thought about that situation in so much detail yet but the more basic point of the first paragraph still stands.

• Bob_FJ

Hi Leonard,
Certainly an interesting thought starter, but rather speculative with the deletion of a small amount of sunlight reaching the surface, and what might happen to convection.

• Bob_FJ

Leonard Weinstein Reur June 19, 2010 @ 5:14 pm
And comments following from me,
Dewitt, and Patrick

Catching up; I did have a convective argument for your thought experiment until I realized that it was self destructive, if your opaque “shell” was a fluid, because it would result in its mixing.
If the shell were to be a solid then it needs to be defined somewhat.
With such a shell, even if there is a small amount of geothermal heat release, it would be reasonable to predict that over a long time period it would become rather hot beneath the shell. (just as the surface of the Earth would do so, if its HEAT could not escape as it does now) There is also speculation from the ESA that Venus volcanism is “recent”, and she may be active currently.
Also, if the shell were opaque to sunlight it is difficult to conceive that it would not also be opaque to infrared. Thus, the near-infrared that currently escapes directly to space at around 1 or 2 microns would also cause heating below the shell.

52. Patrick 027

… wherein *the* flow of radiant heat is from higher to lower (brightness) temperature, then net EMR flux such as described is a heat flux through space, net EMR emission is a heat flux from non-photons to photons, net EMR absorption is a heat flux from photons to non-photons, and net EMR flux (from emission to absorption) between two volumes is a heat flux from one to the other.

Work can be used to produce EMR and such EMR can have very little entropy – for example, coherent radio waves, lasers – the energy can be converted back to work or it can be converted to heat …

(my understanding is that brightness temperature increases when the same radiant energy is concentrated into a smaller space or time (greater flux per unit area) and/or into a smaller solid angle (range of directions) (greater intensity), but also, into a smaller range of frequencies (greater spectral intensity), and also, I think, into a smaller range of polarizations, and also into a smaller range of phases – the idea being that a perfect blackbody emits radiation with some flux per unit area, which is isotropic (constant intensity over all directions), has a particular distribution over the spectrum (so there is a particular flux/area and intensity at some frequency), is unpolarized, and is incoherent; the entropy of the radiation is the energy/brightness temperature; entropy is reduced if the same energy is concentrated in time (greater power) or space (per unit area) or in other ways; selecting a smaller subset of directions, polarizations, frequencies, and phases, their is less radiant power from the same blackbody emitter with the same temperature. Presumably, brightness temperature should correspond to the temperature of the blackbody that would emit the same power in whatever subset of polarizations and phases is considered. Of course, upon absorption, if the absorption process cannot distinguish among phases, then there will be some minimum increase in entropy during the absorption of coherent radiation as the result would be indistinguishable from absorption of the same power of incoherent radiation – etc. for absorption processes that don’t distinguish among polarizations, directions, or frequencies.) …

EMR with a higher brightness temperature relative to an available heat sink has some potential to do work (photosynthesis, photovoltaic cells, solar-mechanical heat engine, in addition to production of electric current within an antenna recieving radio waves). This is limited by the entropy of the EMR and the entropy of a heat flux at the lower temperature of the heat sink; at best they can be equal, which requires some fraction of radiant energy to be absorbed as heat (in addition to that, the process that absorbs EMR will also be able to emit EMR, so there is some trade-off between higher temperature (for efficiency of conversion of power input) and higher power input – gaining one sacrifices some of the other).

53. Bob_FJ

Further to discussion above on the high importance of convection on Venus, the following data are relevant:
Thermal conductivity of 100% CO2, approximated to Venus, is considerably higher than the air on Earth, which means that heat transfer into and through the gases is more rapid, and hence convection should be faster and have greater impetus.

Height (km) Ref: [a]
—–Temp. K Ref: [a]
————– Atmospheric pressure (bar) Ref: [a]
————————–Thermal conductivity (mW/M^2/K) Ref: [b]
0—- 735— 92.10— ~56
5—- 697— 66.65— ~52
10— 658— 47.39— ~47
15— 621— 33.04— ~43
20— 579— 22.52— ~42
25— 537— 14.93— ~36 Sulphuric haze & water vapour above Ref: [d]
30— 495— 9.851— ~33
35— 453— 5.917— ~29
40— 416— 3.501— ~25
45— 383— 1.979— ~23
……………………………… Somewhat Earth-like above 50 Km
50— 348— 1.066— ~21 Sulphuric clouds & water vapour above. T= 75C
55— 300— 0.531— ~17 Temperature, T = 23C
60— 263— 0.236— ~17 Sulphuric clouds & water vapour above. T= -10C
~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Compare with:
0—- 273— 1.032—- 14.7.…….. Earth Air at zero C Ref: [c]
[a] http://en.wikipedia.org/wiki/Atmosphere_of_Venus
[b] http://www.nist.gov/srd/PDFfiles/jpcrd723.pdf
[c] http://encyclopedia.airliquide.com/Encyclopedia.asp?GasID=26
[d] http://en.wikipedia.org/wiki/File:Venusatmosphere.svg

There is also speculation that there is a sulphurous/water precipitation and evaporation cycle in the clouds which would provide additional impetus for convection to high altitudes, e.g. from Harvard:

ABSTRACT: Mariner 10 occultation measurements have provided evidence of a dense cloud deck in the lower atmosphere of Venus with a peak liquid content of about 1 g/cu m. This, in conjunction with other measurements – such as turbulence, updrafts and the presence of aerosol – seem to favor the possibility of precipitation on Venus. Modeling of droplet growth in the Venusian environment shows that precipitation size drops can be formed over periods of only a few hours, similar to growth rates on earth. The precipitation region, if it exists, would extend from the cloud base at about 50 km to the 38-km level where most of the droplets will have evaporated. Precipitation regions can be detected with a variety of remote sensing radar and radio techniques.

There is also this from the ESA that talks of huge weather systems in the clouds that infer both convection and potential for precipitation:

EXTRACT: The clouds on Venus are very inhomogeneous in all directions. This is possibly due the formation of enormous cumulus-type clouds.
http://www.esa.int/esaMI/Venus_Express/SEMFPY808BE_0.html

Something that interests me is the very steady temperature gradient described for Venus at around 10.5 K/Km from the surface up to about 60 Km..

• Patrick 027

Roughly doubling the thermal conductivity will reduce the temperature difference between the surface and the air a very very very short distance above it. It will have very little effect on larger scales (> ~ 1mm) and will have very little effect on convection on on those larger scales unless the imbalances in net radiative heating/cooling are so small that a significant amount can be balanced by conduction through a tropospheric lapse rate (or if viscosity is quite a bit larger, such as in the mantle). This is not generally likely to occur.

Otherwise, I don’t see any points to argue over here.

54. Patrick 027

0. Heeding advice, this will be the last comment I make of anywhere near this length in response to you (Bob FJ) unless you ask some rather different questions than what you have been asking:

[1] Yes, but on Venus most of the HEAT transfer from the surface is via conduction. (or if you like, that EMR emissions are negligible).

1. No, there are emissions and absorptions of EMR at the surface and throughout the atmosphere. When net LW fluxes go to zero, conduction/diffusion and convection must balance solar heating. On Earth, aside from small-scale heat fluxes between particles and gases, conduction is only important within about 1 mm of the surface, where convection is relatively inhibited (air doesn’t just go in and out of the land or water in bulk); in the context of weather and climate, the conduction and diffusion of heat between the air and the surface is often lumped together with convection, with the entirety labelled ‘convection’, hence the convective flux from the surface to the troposphere is understood to include that initial conduction and diffusion of sensible and latent heat from the surface to the air right next to it.

Conjecture then arises concerning at what altitude is the opacity reduced sufficiently to enable significant emissions that would result in absorption and back radiation, and to what effect.

2. No. Backradiation reaching the surface is emitted from wherever it can reach the surface. With greater opacity, backradiation simply comes from air closer to the surface.

3. Emissions don’t result in absorption that results in backradiation in the way implied. Emission occurs as temperature and optical properties allow. Absorption occurs as radiation (wherever it comes from) and optical properties allow. Temperature declines if emission and divergence of convective fluxes are greater than absorption; temperature rises in the opposite case.

Whatever, importantly, the application of S-B to determine the rate of energy loss from the surface and then to equate that to a greenhouse effect is flawed.

4. True, which is why we don’t do that. (or do you mean net energy loss – well in that case, that can be considered part of the greenhouse effect, but it certainly isn’t *the* greenhouse effect (unless there’s no troposphere – although even then, one could still consider the effects of LW opacity on fluxes at other locations.)

An analogy would be to attempt to apply S-B to the hot interior core of the Earth, and then describe the insulating effect up to the crust as a greenhouse effect rather than a thermal conductivity effect.

5a. in the most general sense of a greenhouse effect (impeding the types of heat fluxes that can remove heat from a system), Planetary and stellar interiors have a greenhouse effect, via the limited thermal conductivity and mass diffusion, the physical limitations of convection (adiabatic lapse rate being a part of that), and some opaqueness to radiation in some range of frequencies. But considering ab optical greenhouse effect, do consider what would happen if the ocean, crust and upper mantle (say down to 660 km) and atmosphere were made transparent to radiation. Then emission from 660 km would escape to space. The cooling would reduce the lapse rate from that level upward, reducing convection and conduction upward from 660 km. Convection and conduction would initially increase from below, but the cooling would increase viscosity and … anyway, the equilibrium temperature at 660 km would be much reduced.

5b. The lack of radiative cooling of the Earth’s interior requires opaqueness, which, if reduced, would increase cooling of the interior. So there is an optical greenhouse effect. We don’t think of it that way very much simply because it’s taken for granted that there won’t be any significant direct radiative cooling of the interior of the Earth (much as it is for good reason taken for granted that conduction and molecular diffusion of heat is an insigificant contribution to heat fluxes in the (Earth’s, I’m guessing Venus’ and Mars’ too) atmosphere except to and from particles in the air (generally with little or no larger macroscopic-scale conductive/diffusive fluxes) and within a thin layer in immediate contact with the surface and between that and the surface itself.)

The rate of heat loss from the Venus surface is predominantly related to the thermal conductivity of CO2 & convection speeds, and would be rather tricky to calculate given the lack of data. (however, there are some negligible EMR losses, that are almost off the end of the Planck curve, at around 1 or 2 microns)

6. The graph shown above (in the original post) of the CO2 absorption spectrum shows a big gap between wavenumbers ~ 1100/cm and ~ 1800/cm, where CO2 absorption coefficient is less than 1E-6 m2/kg, meaning that 1 million kg/m2 of CO2 (approximately the entire atmosphere of Venus, in a vertical path) would provide an optical thickness less than 1. I haven’t been able to read the literature yet to find out if the gap is filled in at high pressures (and temperatures) such as those found near the Venusian surface. Even if it is, there might still be some nonzero LW flux through the air from or near the surface. And if that isn’t the case, that would itself be an effect of the optical properties of the Venusian atmopshere (an aspect of the greenhouse effect), and it would be because the backradiation absorbed at the surface is equal to the emission from the surface. At any vertical position, The net upward flux is the difference between the upward flux and the downward flux.

[2] Resorting again to first principles, considering EMR alone, a major factor is that the surface radiates hemispherically such that 3 dimensionally, most radiation is “sideways”. Therefore, even if the free path lengths are long, many initial absorptions by GHG’s are sensibly rather close to the ground.

8a. It’s not arbitrarily close to the ground just because of the distribution over directions. For isotropic emission from a horizontal surface, the average vertical distance a photon travels freely is 2/3 of that value for vertically-propagating photons (see 599
http://www.realclimate.org/index.php/archives/2010/05/solar/comment-page-12/#comment-174597 ). The transmission (without scattering) of some flux of photons which are isotropic and through a horizontal area at z=0, in a vertical direction z, initially decreases twice as fast over z, and only twice as fast, as it does for the photons travelling in the vertical direction (not at an angle from z); the transmission eventually decays less rapidly, approaching the rate of decay that applies to photons travelling vertically, because photons travelling at smaller angles from vertical come to dominate as the other photons are blocked. Transmission of radiation that is not in a ‘beam’ (approximately like direct sunlight at sufficient distance from the sun, or like a laser) can be expressed as a sum of exponential terms, some decaying faster than others.

9. (A small volume that is nearly transparent will have low emissivity as well as low absorptivity (setting aside scattering, reflection; for a particular frequency (and polarization)), so the flux of emitted photons will be small, but one can use an approximation that none of those photons are absorbed within the same volume. If optical properties are isotropic, then the same number of photons are emitted from and escape that volume in all directions. But if that volume is large enough that it can absorb some of the radiation that it emits, then the shape of the volume affects the spatial distribution of the photons which are emitted and escape. Over an isothermal path, intensity approaches a blackbody value as the optical thickness increases. Thus the intensity saturates as the thickness of an isothermal volume in a given direction increases. In the limit that all paths through some surface area (small enough to be considered flat, averaged over smaller-scale texture) have very large optical thicknesses contained within an isothermal region, the intensities escaping that surface area will approach the blackbody value and be isotropic – but the intensity is a photon (or photon energy) flux per unit solid angle per unit area perpendicular ot the direction of photon propagation; an surface area A presents an area A in the direction perpendicular to that surface, but presents a smaller area at other angles – approaching zero at grazing angles. Thus, though the radiation intensity is isotropic, the number of photons coming through such a surface area is not the same in each direction. Intensity is analogous to how bright something looks within some portion of a field of view; from the same distance, a surface area will occupy a smaller portion of a field of view when viewed at a more grazing angle.)

8b. The transmission of isotropic radiation in a direction is still of the same order of magnitude as the transmission of parallel rays in that direction.

10. At some wavelengths most photons are absorbed very close to the ground. At other wavelengths, in the absence of clouds or large concentrations of water vapor, for present-day and similar Earth-like conditions, many photons may be transmitted through the atmosphere, and for those that are absorbed, the absorption is only concentrated a little (relative to mass or optical thickness) in the lower atmosphere. The atmosphere transmits fewer and over shorter vertical distances those photons that are farther from vertical, but the effect of opaqueness on the transmission of all photons over some vertical distance is generally qualitatively similar and still of the same order of magnitude as that on photons in the vertical direction.

—————-
The GHG’s then firstly re-radiate spherically from these many low heights, but at reduced intensity, because it is not just the GHG molecules that get hotter. Most of their photon increased KE is not lost by re-emission but in collisions that warm the hugely more numerous non GHG molecules. As the upward components of the EMR re-emissions progress, the air becomes thinner and colder, so that free path lengths increase, whilst the intensity of re-emission reduces by virtue of the lower T’s.

11. It is in this context unecessarily complicated to think in terms of such iterations (first the GHGs absorb, then the lose energy to other molecules, then they emit, then they gain energy from other molecules, then those photons go here, then it happens again somewhere else, then…). It is much easier to take emission of photons from everywhere in whatever direction to whereever they are absorbed as a function of conditions; intensities and fluxes and net values of each are then a function of location and temperature (in the LTE approximation) and optical properites. (In the LTE approximation) emission is not a function of absorption at a given time; emission is a function of temperature, which will change through time if the LW, SW, convective, etc, net heating or cooling is not zero.

12. reduced intensity relative to what? Non-LTE conditions? Maybe, but it depends on the temperature of the air that contains the GHGs (and cloud particles, etc.). LTE could increase emission relative to non-LTE if the temperature is high enough relative to the flux being absorbed.

13. re-emission is not the best term to use when molecular collisions are tending to maintain LTE and most molecules exchange energy with other molecules in between absorbing a photon and emitting a photon.

14. Other than point 13, The last sentence you have there has a couple of correct points, but for clarity they should be seperated because they are not necessarily correlated.

The free path of photons is larger where optical thickness per unit distance is shorter. Note that, at some particular frequency (and polarization), at some particular time and place, it can be convient to use mass (per unit area) or optical thickness as a vertical coordinate. Optical thickness tends to be proportional to mass path to a first approximation, aside from concentrations of water vapor, clouds, or ozone, etc, but otherwise it does tend to be smaller per unit mass path higher in the atmosphere over at least a significant fraction of the spectrum. Using optical thickness as a measure of vertical distance, the vertical components of free paths would not vary with height. The variation of free path with height is not really of fundamental importance; it is how free path is related to the spatial scale of temperature variation that is truly important. Temperature and optical thickness, etc, can be mapped onto whatever vertical coordinate is convenient.

The temperature generally declines with height from the surface to the tropopause; that is very very important – see 11. above, 16-17. below.

Thus, despite the depictions in the Trenberth et al energy budget diagrams, showing massive back radiation from the clouds all the way to the surface, most of the action is rather close to the surface.

15. It’s a schematic diagram and shouldn’t be taken so literally.

16. The backradiation shown is just the downward LW radiation flux per unit area reaching and absorbed by (in the approximation of zero LW-albedo) the surface. It originates from various parts of the atmosphere; At some frequencies it is mostly near the surface; at others it is distributed over much of the atmosphere, or concentrated within humid or cloudy layers.

17. At a given location, for intensity at a given frequency (and polarization) and from a given direction, the source distribution (emission weighting function * blackbody intensity) is similar to the distribution of absorption of radiation (same frequency and polarization, back into the given direction, at the same location) (distribution equals the emission weighting function for radiation from that direction, etc.) – this is true at the surface, as it tends to be for emission to and absorption from any location. There are upward and downward LW fluxes at each vertical level, and the source region (emission weighting function for flux/area) for each shifts with shifting vertical level.

[3] The way the greenhouse effect* on Earth is quantified in terms of AGW increase, is partly by considering the 150/180 – year surface T records.

18. The greenhouse effect on Earth is not quantified in terms of AGW increase – AGW is described and predicted by the physics of the greenhouse effect and feedbacks, etc.; the temperature and other historical records are some of the data that backs-up theory and modelling results.

*(which by definition = when surface originating infrared EMR is absorbed by GHG’s)

Huh? What? Come again?

19. In totality the greenhouse effect can be considered to be the effect of LW optical properties on fluxes at all levels and the resulting effects on average temperature and temperature cycles and all other weather/climate variables at all places and times of day and year and etc. But a more narrow definition could be the effect of LW optical properties on global average surface temperature. Backradiation may concievably affect that via feedbacks from changes in convection; regional variations in circulation and precipitation and vegetation, etc, will have effects… but the equilibrium surface temperature more generally depends on the lapse rate of the tropoposphere and the radiative forcing of the troposphere. Temperature at any location can be affected by radiative forcing at other locations both through changes in LW radiation from the resulting temperature changes elsewhere, and very importantly within the troposphere, through communication of changes occuring elsewhere via convection itself.

The greenhouse effect, on Earth is at its greatest at or rather close to the surface. That is what I meant by GHE at the surface, and that effect reduces with altitude. Venus is very different.

20. In the sense that removal of LW opacity would increase the net upward LW radiation more near the surface than at least somewhere higher in the atmosphere, that is true for Earth – it should also be true for Venus (removal of all atmospheric LW opacity would leave the net upward LW flux everwhere equal to the emission from the surface; with the LW opacity, the net upward flux is reduced by reduction of upward LW flux from below and/or increase in downward LW flux froma above (depending on vertical position); if all SW heating were at the surface and there were no non-radiative fluxes, then the net upward LW flux would still be the same at all levels (in equilibrium) and so the change (by removing all LW opacity) would be the same at all levels – however, any SW heating within the atmosphere, and thermally-direct convection, would lead to increasing net upward LW flux with height (in equilibrium), meaning that removing all LW opacity increases the net upward flux more below than above generally).. The net LW flux upward from the surface is the LW flux from the surface minus the backradiation flux. When the greenhouse effect is as strong as it can get, the net LW flux is zero.

(Note that a change in radiative forcing is typically calculated as the change in flux (per unit area) before the temperature response.)

21. (PS And the net LW flux can never actually go to zero in equilibrium above some portion of solar heating outside the extent of convection; if there is any SW heating below the tropopause, in the approximation of zero convection above the tropopause, there must be a net upward LW flux at the tropopause, so there is generally always room for additional greenhouse effect.)

22. It’s different on Venus because there is so much CO2 that additional absorption bands are important. It’s different because the temperature in the lower atmosphere is so much higher that fluxes at shorter wavelengths have greater importance than otherwise. It’s different because the different atmospheric conditions alter the absorption spectra of gases. It’s different because the types and arrangements of clouds are different (so far as I know). It’s different because there is little methane or ozone (so far as I know), etc. It’s different because of the slightly smaller gravitational acceleration and the different specific heats of the air (affects adiabatic lapse rate). It’s different because of the lack of importance of latent heating over much of the lower atmosphere (so far as I know) (affects the tropospheric lapse rates). It’s different because of the different distance from the sun and the different albedo (combined effect: less solar heating). It’s different because the stratosphere itself has different optical thickness and the tropopause is not in the same place (so far as I know).

It’s not fundamentally different in the importance (and limitation of that importance) and effect of backradiation at the surface.

55. Bob_FJ

Hi Patrick,
Phew! I admire your determination and great volume of words, but you lost me with your first point and I went no further.

Please advise what part of the word ‘opaque’ do you not understand!

Consider this graphic (Fig 1) of the absorption spectra of liquid water, and elaborate just how far relevant infrared wavelengths could penetrate. (BTW; notice the log-log scale)
http://en.wikipedia.org/wiki/Electromagnetic_absorption_by_water

What makes you ASSERT that the Venus atmosphere near the surface cannot be compared with water?

You might also like to interpolate this Plancky collection between 800K & 600K, to see what significance the near infrared windows around 1 or 2 microns might have.

• DeWitt Payne

The emission spectrum of Venus from space looks like this:

Most of the emission is described by a black body temperature of 230 K, but there is a peak at about 4500 cm-1 and increasing emission starting at 5,000 cm-1. But it’s orders of magnitude less than the peak flux (log y axis scale). So while there is obviously some transparency near 2 micrometers, there isn’t much.

Response: Your link does not appear to work and I don’t know what the correct link should be so I cannot fix it– chris

• DeWitt Payne

Let’s see if this one works:

• Patrick 027

We both understand ‘opaque’. I never asserted that the Venusian air near the surface cannot be compared with water (technically, you have to have made a comparison in order to conclude that there is no comparison – that aside, of course, the gas is dense at those conditions even with the high temperature (but I don’t think it’s liquid ?); the absorption spectrum will be altered by that and I wouldn’t be surprised if it is significantly opaque over a broader range of wavelengths – but I haven’t actually found that information yet!

Besides which, given the nonlinear relationship of blackbody radiation to temperture, the very high temperatures near the Venusian surface will lead to greater net fluxes for the same temperature gradients and opaqueness. Whether that is merely a larger insignificant net flux instead of a smaller insignificant net flux, I don’t claim to know.

But you might better understand my point by figuring this out: Cover a surface with some layer of water with some temperature distribution in the water. What is the backradiation at the surface from the layer of water?

I could sum up a significant portion of what I’ve been saying this way:
Reducing the net LW flux doesn’t reduce the greenhouse effect; it is (how) the greenhouse effect (works).

56. Bob_FJ

Patrick 027 Reur June 21, 2010 @ 11:34 pm
Sorry, but you’ve lost me again, and I wonder if you checked the following as I suggested in my June 21, 2010 @ 3:04 am :

Consider this graphic (Fig 1) of the absorption spectra of liquid water, and elaborate just how far relevant infrared wavelengths could penetrate. (BTW; notice the log-log scale)
http://en.wikipedia.org/wiki/Electromagnetic_absorption_by_water

Hint: the extinction depth of the blue portion of visible light is around 100m, (or 100 * 10^3 mm), with red being rather less. Notice the Y axis scaling, and now estimate from the orders of magnitude, the probable range for infrared

Are you shocked to find that infrared extinction depth is apparently in but a mere skin of water?

If you don’t believe this graph, I’ve seen similar depictions elsewhere. It is not cherry picked.

• Patrick 027

I might look at the graph at some point. I haven’t yet for our discussion because I’ve seen a graph of the H2O liquid spectrum before and what you say about it sounds about right. It is not a graph of the CO2 spectrum, and not having established (amongst ourselves) how similar CO2 at ~ 90 bars and above 700 K is to liquid H2O, it seems a bit irrelevant to discuss in great detail.

No matter, because whether there is a sizable, small, or insignificant net LW flux at the surface of Venus or not (I haven’t claimed to know) doesn’t change the more basic point (so long as it is smaller than the surface solar heating, so that convection can be sustained). The reason why the net LW flux would be less than what it would otherwise be is the very thing you are describing – a layer of relatively opaque material – and the mechanism by which this occurs is not reducing the upward emission but by emitting (or scattering if that were the case) a backradiation from a layer with a temperature that is not much colder. Backradiation is important at the surface. (Or else, if the opacity were so great that photon paths were on the scale of photon wavelengths or less – well then I’m not sure what happens exactly, but the effect on net radiation should be similar.)

Dealing with a case where absorption (and thus, given near-LTE, emission) is a sizable to dominant contributor to opaqueness, that layer of opaque material – let’s say liquid water, following your analogy – will not only reduce or nearly eliminate the net LW flux from below and block the LW emission from below from reaching higher, it will also emit a LW flux upward. Is there some material above this layer of water to absorb or scatter that radiation and emit some radiation back down at the top of this water layer? Etc.

57. Bob_FJ

Patrick 027 Reur June 21, 2010 @ 11:44 pm
Whoops, I’ve just found it “hiding” way above.

Convection is the consequence of heating a fluid from below, and thermal conductivity on Venus is the essential means by which HEAT is transferred from the surface and into the air. When you claim that this only affects about the first mm of air, then for a start you seem to have overlooked (at least) that it is a dynamic process. So, the first mm is heated via conduction and floats upwards to be replaced by a new 1mm of air, which is then heated, which… …which…

There is also some advection on Venus, and BTW the thermal conductivity at the surface (per 100% CO2), is about 3.8 times greater than on Earth

• Patrick 027

“about 3.8 times greater ”

I had read that air’s thermal conductivity was something like 0.026 W/(K*m). My point still stands. And that point was

“then for a start you seem to have overlooked (at least) that it is a dynamic process. So, the first mm is heated via conduction and floats upwards to be replaced by a new 1mm of air, which is then heated, which… …which…”

No, that’s exactly what I mean. For the same net radiative balance, the same non-radiative (convection + etc.) must occur in equilibrium. If convection is able to occur and does so with sufficient strength then it will tend to maintain an adiabatic lapse rate or something close to that. The amount of conduction is then determined by that lapse rate. It will still be exceedingly small, unless the thermal conductivity get’s much much larger. To maintain the same non-radiative heat flux then requires, to a very good approximation, maintaining the same convective flux, unless the convective flux is actually very very small. At the boundaries of a convecting layer (in regions of thickness depending on conductivity, viscosity, …), conduction is necessary to feed the convection if there is a net radiative flux at that point that needs to be balanced. Thus, it is (assuming significant convective flux) only here that changing the thermal conductivity would significantly affect the lapse rate, and thus, the temperature. That would have some feedback on radiation via the change in temperature; however, I think that would be a marginal effect for the ranges of conditions being considered here.

58. You accuse “Watts up with That” attempting to reinvent climate physics on Earth but no mention of the Hockey Team’s attempts to lose the Medievil Warm Period and the Little Ice Age. Global warming may have been happening during the 19th and 20th centuries but it looks as though it may be cooling now. That’s what nature does. Man has an insignificant impact upon global climates. Perhaps you’d like to have a read of “Rescue from the Climate Saviors Is the “Global Climate” really in Danger?” (http://www.scribd.com/doc/33181109/Rescue-from-the-Climate-Saviors-1-1) provided by “The Hockey Schtick” (http://hockeyschtick.blogspot.com/2010/06/rescue-from-climate-saviors.html) which concludes QUOTE:

* The terms “greenhouse effect” and “greenhouse gas” are deli-berate misnomers and obstruct understanding of the real world.
* Earth has a “cooling system”. If our planet gets warmer, it will automatically raise its cooling power (Fig. 28).
* An increase of earth temperatures is only achievable if the heating power is stepped up: first to “load” matter with more energy (i.e. to raise temperatures) and then (and that is our point) to compensate for the increasing cooling, which results from the increase of IR radiation into space.
* CO2 and other IR-active gases cannot supply any additional heating power to the earth. Therefore, they cannot be a cause of “global warming”. This fact alone disproves the greenhouse doctrine.
* The “natural greenhouse effect” (increase of earth temperatures by 33°C) is a myth.
* IR-active gases do not act “like a blanket” but rather “like a sunshade”. They keep a part of the solar energy away from the earth’s surface.
* IR-active gases cool the earth: 70% of the entire coo- ling power originates from these molecules. Without these gases in the air the surface and the air immedia- tely above the ground would heat up more.
* The notion that a concentration increase of IR-active gases would impede earth’s cooling is impossible given the true me-chanisms explained above.
* As a consequence the very foundation of the “Green Tower of Climate Dogma” crumbles. Computer models alleging to fore- cast warming based on “greenhouse effects” are worthless, and any speculation about the “impact of climate change” ac- cordingly dispensable.
* Since the greenhouse hypothesis has been disproven by the laws of physics, it is only a matter of time until the truth becomes public opinion
UNQUOTE.

It’s being sent to politicians all around the globe.

Best regards, Pete Ridley

Response: Do you people rehearse this stuff in front of a mirror before breakfast every morning? — chris

• Patrick 027

Pete Ridley, Thank you for providing such d#$@ing evidence of another website’s stupidity. • Bob_FJ Patrick 027 Reur June 23, 2010 @ 12:36 pm Pete Ridley, Thank you for providing such d#$@ing evidence of another website’s stupidity.

I haven’t visited the websites that Pete Ridley quotes, but on the surface they appear to be extreme, without knowing the context of the quotes.

However, it is an opportunity for me to congratulate Chris on running a truly excellent site where he allows sceptical comments to appear, including those that have been deleted at some other sites without explanation or trace. The most famous for such deletions are; RC, Joe Romm, and Tamino

For instance see here for one example on the open thread here:
Or, for a bunch of older ones, see here at ’rcrejects‘:
http://rcrejects.wordpress.com/2009/12/29/post-your-rejected-posts-here-3/#comment-451

It looks like me, or my 2 computer IP’s have been excommunicated at RC, and I have more recent examples if anyone is interested.

Thanks again Chris.

59. Bob_FJ

Patrick 027,
I have some catching-up to do with some of your posts, and I’ve started here, way-way up the page with the Leonard & DeWitt thought experiment at:
https://chriscolose.wordpress.com/2010/05/12/goddards-world/#comment-2445
Sorry for html tag error above.

60. Bill Illis

We do have a nearby example of an object with an atmosphere which contained virtually no greenhouse gases and had only 0.0003 Watts/m2 of incoming EM radiation which reached temperatures of 10,000,000K about 4.55 billion years ago.

That is a very unusual greenhouse effect. Do any standard atmospheric thermodynamic formulae explain that.

• Patrick 027

In anology to a planet, the sun is heated mainly from ‘geothermal’ energy. Original heat would also have come from gravitational potential energy.

(When the solar system formed, compression under gravity would increase kinetic energy, some **(most, I’d guess?) of which went into kinetic energy on the molecular scale. In addition, there was radioactivity from nuclei made in previous stars, including some very-short-lived isotopes from any recent supernova. The decay of the longer-lived radioactive isotopes continues to provide internal heat to the planets so that they don’t ‘cool off (including loss of latent heat of physical (or chemical) latent heat))’ as rapidly as they actually lose heat to space. The sun has nuclear fusion in it’s core. For at least the inner planets, the absorption of radiation from the sun is much much much greater than any geothermal heat losses or tidal heating; thus, to attain equilibrium climate, planetary OLR approximately balances solar heating for the inner planets.)

• Patrick 027

(OLR = outgoing LW radiation – this must balance the rate of planetary cooling plus internal heat sources plus external heat inputs. For equilibrium climate (atmosphere and near-surface conditions), the cooling is entirely in the interior and is slow, and is the difference between the heat flux out from within the planet and the nuclear (radioactivity if not fusion) and tidal heating and heating by compaction under gravity, and latent heat release (if that is not counted in the ‘cooling’ term). It is mainly LW radiation because of the temperatures found within the layers that can emit radiation to space; more generally, it would be OR – For the sun, the OR is mainly SW radiation because of the high temperatures found within the layers visible from space.

The (radiative) greenhouse effect alters the relationship between LW radiation fluxes and the temperature distribution.)

• Patrick 027

“tidal heating” – actually, on Earth, most of that takes place within the ocean and is thus not beneath the climate system but within it – however it is very very very small.

• Bob_FJ

Patrick,
You should be able to analyse from what Bill Illis actually wrote, that he is rather knowledgeable about the Sun. However, you may not be aware that he is also very knowledgeable and active in the climate change debate. If you might accept that, perhaps you could then turn your attention to his final line, and as to how it relates analogously to Chris‘s use of certain standard formulae for Venus:

Do any standard atmospheric thermodynamic formulae explain that.

61. Bill Illis

The point is that gravity is constanting compressing a gas (or any material) – applying a force – and that results in temperatures of:

10,000,000K in the Sun before nuclear fusion eventually started;

10,000K in Jupiter’s core;

400K of temperature on Venus which is not explained by a greenhouse effect; and,

a non-Zero number in Earth’s atmosphere and a significant number at Earth’s core – a non-Zero fraction of the lapse rate is, in fact, gravitational compression – (the warmest places on the surface of the Earth are those below sea level like the Dead Sea and Death Valley and the Afar Depression and the Turpan Depression at 42N which averages 40C in July – the ocean/water being the exception to this idea).

• Patrick 027

Each bit of gravitational compression only happens once (unless there is expansion for some time periods). Potential energy is converted to other forms such as heat. But that heat can be lost by conduction, convection, and radiation. You need a continual input of energy to balance the energy outflow to maintain a temperature.

Obvioulsy BobFJ…

62. Patrick 027

… never mind that last incomplete thought.

63. Bill Illis, actually, doesn’t know what he’s talking about. The greenhouse effect on the Sun is actually quite huge, except that when you’re talking about the sun, you call it “opacity.” Hydrogen is an excellent greenhouse gas, and part of the reason the core of the sun is as hot as it is, is because it’s extremely difficult for a photon to get out through all that hydrogen. What we see is actually “fossil light” that could be up to millions of years old.

Pressure does not in and of itself cause heat. Yes, I know about the ideal gas law. I also know that hot objects radiate, and if Venus’s atmosphere were hot because of a compression event, the heat would have radiated away in a matter of a few thousand years. It doesn’t because it’s in thermal equilibrium, with the high surface temperatures caused by the greenhouse effect.

64. BTW, does Bill really think the sun’s core heat is due to PRESSURE? Has he ever heard of “fusion?”

65. Bob_FJ

Barton Paul Levenson Reur July 22, 2010 @ 2:32 pm

BTW, does Bill really think the sun’s core heat is due to PRESSURE? Has he ever heard of “fusion?”

Quickly; yes, he unambiguously mentioned fusion! How about you read his ‘June 29, 2010 @ 8:31 pm’ etc, above a little more carefully, and BTW consider the possibility of of some sarcasm in there?
Oh, and furthermore, are not the convection cells on the Sun a tad violent, and the constantly descending bits not subject to recompression?

• Patrick 027

Re Bob FJ –

Where BPL wrote that Bill Illis doesn’t know what he’s talking about, he basically said what I was going to say back where I said Obvioulsy BobFJ…… never mind that last incomplete thought. – which would have been “Obviously BobFJ has misjudged Bill Illis’s level of knowledge”, note that you not only said he was knowledgable about the sun but also knowledgeable about climate change (or at least the ‘debate’ about it, though even there, I disagree strongly).

The Way was written

Illis’ first comment – “June 28, 2010 @ 9:44 am ” – mentions nothing of fusion, though that is not an error, since temperature did increase (from the conversion from gravitational potential energy) before fusion initiated. However the implied assertion of no greenhouse effect was incorrect. Being able to emit radiation, the material would lose heat to space, and so, without fusion reactions and continued gravitational contraction and the little input from radioactivity (I wouldn’t know how much there is in the Sun but presumably there is at least a little), the object would inevitably cool toward absolute zero – well not quite (some heating from radiation from space), but close. In that case, the combination of the greenhouse effect and the heat capacity would slow this cooling, at least at the center; heat would have to flow out from the center by diffusion/conduction to reach some point where it could escape to space, thus the cooling on the exterior would produce a temperature gradient that would allow heat to flow out from the center, allowing the center to cool, but the cooling rate depends on that temperature gradient. The role of the pressure gradient? It causes moving materials to expand and contract, moving outward and inward, thus the existence of an adiabatic lapse rate, which limits the ability of convection to reduce the temperature gradient below that lapse rate for a given amount of heat flux but also limits how large the temperature gradient gets.

His second comment mentions fusion but the implication of his comment doesn’t seem to account for it, as if it is assumed that the sun would stay as hot at it is indefinitely without fusion, while continuing to radiate heat.

Yes, descending gas/plasma will increase in temperature with compression. But if this is part of ongoing convection with no net change in the mass distribution, then there is no net loss of gravitational potential energy and the process doesn’t produce heat on a net basis (unless it is forced by some input of energy of another form, which is not what’s happening in this context). (Thermally-direct convection actually converts heat to kinetic energy, which is then generally converted back to heat by visosity, except for whatever is emitted by electromagnetic energy of the organized, non-blackbody type). IF the object is losing energy to space then the object has to be gaining energy (or getting heat energy from another form like mass or gravitational potential energy) in order to have no net gain or loss of heat. The greenhouse effect slows the flow of heat from interior portions to space or requires a warmer interior to support the same heat loss to space.

66. Michele

I am late, sorry, but I would like to specify some matters.

First. I think that natural processes are more and more simple than we imagine them and mostly we depict them.
An atmospheric process is a fluid dynamics and thermodynamics process, that must be studied by means of the three well known conservation laws (mass, linear momentum, energy) and, if the fluid is a gas, also using the ideal gas law. On steady state conditions, into gravitational field, the flow energy is given by Bernoulli’s equation for compressible fluid g*z + cp*T + 0.5*v^2 = const, and with zero speed, g*z + cp*T = const = cp*T0, that gives T = T0 – g*z/cp, i.e. how T changes as one changes altitude z into an atmosphere without motion, i.e. without convection.
Also, with zero speed, the linear momentum law simplifies to gravitational equilibrium dp/dz = -Ro*g = g*p/(R*T) = g*p/(R*(T0-g*z/cp)). Separating the variables and integrating one gets p/p0 = (T/T0)^(cp/R) = (T/T0)^(k/(k-1)), i.e. the pressure and T (and density Ro) vary each other as in adiabatic process.
The Bernoulli’s equation also gives g*z + cp*T = cp*T0 if the speed (KE) is constant, but the mass conservation requires that is Ro*v*A = const; so, if A = const (we relate to a vertical column with cross area = one square meter) and v = const, also Ro must be constant, i.e. there doesn’t no longer occur the adiabatic process.
That means that the vertical gradient -g/cp for the temperature only occurs into a purely conductive troposphere, and the convection only can occur on unsteady conditions.
Note that the heat flow by conductivity is about 0,025(W/mK)*0,01(K/m) = 0,00025 W/m2, which is only about one millionth of heat flow needed for Earth’s cooling. Besides the strongest convection won’t be able to increase the heat flow more than 10^6X. Thus both conduction and convection are wholly negligible in Earth’s cooling. For Venus the order of magnitude is full analogous.

Further. Goddard’s statement is completely wrong. I explain.
If Earth’s atmosphere increases 100X its mass, maintaining constant its concentration, the mass will increase 100X anywhere and also the pressure will increase 100X anywhere. If the ratio between tropospheric mass and atmospheric mass, then the ratio p/p0, and Ro/Ro0 and T/T0 (0 subscript indicates the ground values), whatever thermodynamic process relate them each other, will remain constant at any altitude. That means that T2/T02 = T1/T01 (1 and 2 subscripts indicate the values ante/post increase 100X). But there’s also T02 – T2 = T01- T1 for any altitude, because the Bernoulli’s equation is still valid. The trivial system gives T02 = T01, i.e. the ground temperature don’t change increasing 100X (or any other valueX) the ground pressure of a planet.

Third. The offhand use of Plank radiation law for any EMR active gas in the whole spectrum, as is usual in atmospheric physics, gives me hives.
Plank can be used if there’s thermodynamic equilibrium that’s more different from thermal equilibrium because it requires not only a constant temperature but also that the radiant energy flux that becomes thermal energy must be balanced by an identical thermal energy flux transformed into radiant energy. At planetary temperatures any gas has unfrozen only the translational and rotational grades of freedom, as confirmed by their molar thermal capacity at constant volume (3R/2 for monatomic gases, 5R/2 for diatomic, and 6R/2 for triatomic or polyatomic). The thermal energy can cause only translational and purely rotational acceleration and thus, if gas molecules are electric dipoles, thermal energy can become radiant energy only at translational and purely rotational frequencies, i.e. in a very little portion of the spectrum, where is rightful to use the Plank radiation law. Using Plank outside this portion is wholly wrong, because for the gases at planetary temperature there isn’t thermal excitation for vibrational frequencies or higher, that can be energized only by EMR and then thermalized or scattered.

67. Don’t beat around the bush.

Either you have some evidence, or some sort of mighty logical argument, for the global warming fraud …

Or you don’t.

And so far it looks like you don’t.

And so a retraction is in order.

So retract.

68. Michelle you are just talking gibber.

Are we in an ice age or not? Are we past the normal length of time of an interglacial or is that not the case?

Nothing you said is relevant here. We have fraudulent movement. Nothing you’ve said can breathe legitimacy into it.

69. “…… When the solar system formed ……..”

What the good god damned is this? Another creation story from science-commie-central?

You don’t know a thing about the evolution of the solar system. Don’t pretend that you do …… Or spell out your stupid creation theory in all its unrighteousness.

70. This really is a very simple story. The global warming fraud is a transparent fraud just like Piltdown man. But unlike Piltdown man there is an endless source of funds behind the global warming fraud.

You really don’t need to make things more complex than that.

We are talking about a movement so fraudulent that they are yet to prove an anomaly that we ought to take to be their alleged greenhouse affect. They haven’t proven that anomaly beyond a “god of gaps” basis. And no-one here has any clue so much as to be able to find an anomaly and to make it relevant to this trifling back-radiation theory.

71. Michele

The posts of graemebird have given me the occasion to reread what I had written on the matter and I have to correct me because what I affirmed is wrong.

The atmosphere of Venus, as also those of Earth and of Mars, radiates at its top the energy that comes to it from the ground, transported aloft by convection. The process is certainly adiabatic because the time scale of the convection is very very smaller than that of the conduction. This means that the thermal-geo-potential energy (CpT+gz) of the carrying particles stays constant during the climbing though there is a continuous transformation of the thermal energy to geo-potential energy. In other words we have

Cp*Tbottom + g*Zbottom = Cp*Ttop + g*Ztop

or rather

Tbottom = Ttop + g*H/Cp

where H = Ztop – Zbottom, is the height of the atmosphere.

All this is always verified however the temperature changes in rising, i.e., whatever is the lapse rate, because the variation of the energy (function of state) only relies on the outmost points and not on the path that joins the point of departure to that of arrival (a lapse rate almost linear means that the ascending flow is nearly uniform).
Thus, it is the height of the atmosphere H that determines Tbottom with the same Ttop, that is only affected by the amount of energy radiated at the top of the atmosphere. If H increases also the pressure at the ground increases but this means that H affects both Pbottom and Tbottom, not that the pressure determines Tbottom.

Chris, has you seen Postma’s take greenhouse eg http://www.ilovemycarbondioxide.com/pdf/Understanding_the_Atmosphere_Effect.pdf? I notice it going around the denialosphere big time but hadnt seen a comprehensive debunk. It appears to be the same argument as this.

73. Your calculations ought to verify Venus as a new planet. Because thats where the evidence seems to point.

74. Patrick 027

Re Michele, also appearing at :
http://www.realclimate.org/index.php/archives/2011/06/unforced-variations-june-2011/comment-page-5/#comment-208321
…(and re the last comment there, the height from which radiation reaching space is emitted is not a single level and not necessarily at the very top of the atmosphere – in pressure coordinates in can get clost to TOA but in geometric coordinates it should tend to stay below – there should tend to be definable layer of gas that is so thin in terms of amount of material that is is relatively transparent, yet in geometric coordinates it may extend effectively many km.)

75. Patrick 027

… also: http://www.realclimate.org/index.php/archives/2011/06/unforced-variations-june-2011/comment-page-5/#comment-208341
(and giving the lapse rate in terms of height like that (cp*DT = -g*Dz) only strictly applies to a layer that is (dry) adiabatic; the more general relationship is between DT and Dp, and applies to any (dry) adiabatic process, at least for an ideal gas of constant composition, regardless of what the environmental lapse rate is. Also, don’t forget latent heating[where applicable].)