Unified Theory of Climate: Reply to Comments

Foreword – I’ve had this document since January 17th, and it has taken some time to get it properly reproduced here in full due to formatting issues. Some equations have to be converted to images, and I have to double check every superscript, subscript, and symbol for accuracy, then re-insert/re-format many manually since they often don’t reproduce properly in WordPress. WordPress doesn’t manage copy/paste of complex documents well. I hope that I have everything correctly reproduced, if not, please leave a note. A PDF of the original is here: UTC_Blog_Reply_Part1 This is a contentious issue, and while it would be a wonderful revelation if it were proven to be true, I personally cannot see any way it can surmount the law of conservation of energy. That view is shared by others, noted in the opening paragraph below. However, I’m providing this for the educational value it may bring to those who can take it all in and discuss it rationally, with a caution – because this issue is so contentious, I ask readers to self-moderate so that the WUWT moderation team does not have to be heavy handed. I invite you take it all in, and to come to your own conclusion. Thank you for your consideration. – Anthony

Part 1: Magnitude of the Natural ‘Greenhouse’ Effect

Ned Nikolov, Ph.D. and Karl Zeller, Ph.D.

  1. Introduction

Our recent paper “Unified Theory of Climate: Expanding the Concept of Atmospheric Greenhouse Effect Using Thermodynamic Principles. Implications for Predicting Future Climate Change” spurred intense discussions at WUWT and Tallbloke’s Talkshop websites. Many important questions were raised by bloggers and two online articles by Dr. Ira Glickstein (here) and Dr. Roy Spencer (here). After reading through most responses, it became clear to us that that an expanded explanation is needed. We present our reply in two separate articles that address blog debate foci as well as key aspects of the new paradigm.

Please, consider that understanding this new theory requires a shift in perception! As Albert Einstein once noted, a new paradigm cannot be grasped within the context of an existing mindset; hence, we are constrained by the episteme we are living in. In that light, our concept requires new definitions that may or may not have exact counterparts in the current Greenhouse theory. For example, it is crucial for us to introduce and use the term Atmospheric Thermal Effect (ATE) because: (a) The term Greenhouse Effect (GE) is inherently misleading due to the fact that the free atmosphere, imposing no restriction on convective cooling, does not really work as a closed greenhouse; (b) ATE accurately coveys the physical essence of the phenomenon, which is the temperature boost at the surface due to the presence of atmosphere; (c) Reasoning in terms of ATE vs. GE helps broaden the discussion beyond radiative transfer; and (d) Unlike GE, the term Atmospheric Thermal Effect implies no underlying physical mechanism(s).

We start with the undisputable fact that the atmosphere provides extra warmth to the surface of Earth compared to an airless environment such as on the Moon. This prompts two basic questions: (1) What is the magnitude of this extra warmth, i.e. the size of ATE ? and (2) How does the atmosphere produce it, i.e. what is the physical mechanism of ATE ? In this reply we address the first question, since it appears to be the crux of most people’s difficulty and needs a resolution before proceeding with the rest of the theory (see, for example, Lord Monckton’s WUWT post).

  1. Magnitude of Earth’s Atmospheric Thermal Effect

We maintain that in order to properly evaluate ATE one must compare Earth’s average near-surface temperature to the temperature of a spherical celestial body with no atmosphere at the same distance from the Sun. Note that, we are not presently concerned with the composition or infrared opacity of the atmosphere. Instead, we are simply trying to quantify the overall effect of our atmosphere on the surface thermal environment; hence the comparison with a similarly illuminated airless planet. We will hereafter refer to such planet as an equivalent Planetary Gray Body (PGB).

Since temperature is proportional (linearly related) to the internal kinetic energy of a system, it is theoretically perfectly justifiable to use meanglobal surface temperatures to quantify the ATE. There are two possible indices one could employ for this:

  1. The absolute difference between Earth’s mean temperature (Ts) and that of an equivalent PGB (Tgb), i.e. ATE = TsTgb; or
  1. The ratio of Ts to Tgb. The latter index is particularly attractive, since it normalizes (standardizes) ATE with respect to the top-of-atmosphere (TOA) solar irradiance (So), thus enabling a comparison of ATEs among planets that orbit at various distances from the Sun and receive different amounts of solar radiation. We call this non-dimensional temperature ratio a Near-surface Thermal Enhancement (ATEn) and denote it by NTE = Ts / Tgb. In theory, therefore, NTE should be equal or greater than 1.0 (NTE ≥ 1.0). Please, note that ATEn is a measure of ATE.

It is important to point out that the current GE theory measures ATE not by temperature, but by the amount of absorbed infrared (IR) radiation. Although textbooks often mention that Earth’s surface is 18K-33K warmer than the Moon thanks to the ‘greenhouse effect’ of our atmosphere, in the scientific literature, the actual effect is measured via the amount of outgoing infrared radiation absorbed by the atmosphere (e.g. Stephens et al. 1993; Inamdar & Ramanathan 1997; Ramanathan & Inamdar 2006; Houghton 2009). It is usually calculated as a difference (occasionally a ratio) between the total average infrared flux emanating at the surface and that at the top of the atmosphere. Defined in this way, the average atmospheric GE, according to satellite observations, is between 157 and 161 W m-2 (Ramanathan & Inamdar 2006; Lin et al. 2008; Trenberth et al. 2009). In other words, the current theory uses radiative flux units instead of temperature units to quantify ATE. This approach is based on the preconceived notion that GE works by reducing the rate of surface infrared cooling to space. However, measuring a phenomenon with its presumed cause instead by its manifest effect can be a source of major confusion and error as demonstrated in our study. Hence, we claim that the proper assessment of ATE depends on an accurate estimate of the mean surface temperature of an equivalent PGB (Tgb).

  1. Estimating the Mean Temperature of an Equivalent Planetary Gray Body

There are two approaches to estimate Tgb – a theoretical one based on known physical relationships between temperature and radiation, and an empirical one relying on observations of the Moon as the closest natural gray body to Earth.

According to the Stefan-Boltzmann (SB) law, any physical object with a temperature (T, oK) above the absolute zero emits radiation with an intensity (I, W m-2) that is proportional to the 4th power of the object’s absolute temperature:

image

where ϵ is the object’s thermal emissivity/absorptivity (0 ≤ ϵ ≤ 1 ), and σ = 5.6704×10-8 W m-2 K-4 is the SB constant. A theoretical blackbody has ϵ = 1.0, while real solid objects such as rocks usually have ϵ ≈ 0.95. In principle, Eq. (1) allows for an accurate calculation of an object’s equilibrium temperature given the amount of absorbed radiation by the object, i.e.

image

The spatially averaged amount of solar radiation absorbed by the Earth-Atmosphere system (Sα ̅̅̅, W m-2) can be accurately computed from TOA solar irradiance (Sα ̅̅̅, W m-2) and planetary albedo (αp) as

image

where the TOA shortwave flux (W m-2) incident on a plane perpendicular to the solar rays. The factor ¼ serves to distribute the solar flux incident on a flat surface to a sphere. It arises from the fact that the surface area of a sphere (4πR2) is 4 times larger than the surface area of a disk (πR2) of the same radius (R). Hence, it appears logical that one could estimate Earth’s average temperature in the absence of ATE from using the SB law. i.e.

image

Here (TeK) is known as the effective emission temperature of Earth. Employing typical values for S0 =W m-2 and αp = 0.3 and assuming, ϵ  = 1.0 Eq. (3) yields 254.6K. This is the basis for the widely quoted 255K (-18C) mean surface temperature of Earth in the absence of a ‘greenhouse effect’, i.e. if the atmosphere were missing or ‘completely transparent’ to IR radiation. This temperature is also used to define the so-called effective emission height in the troposphere (at about 5 km altitude), where the bulk of Earth’s outgoing long-wave radiation to space is assumed to emanate from. Since Earth’s mean surface temperature is 287.6K (+14.4C), the present theory estimates the size of ATE to be 287.6K – 254.6K = 33K. However, as pointed out by other studies, this approach suffers from a serious logical error. Removing the atmosphere (or even just the water vapor in it) would result in a much lower planetary albedo, since clouds are responsible for most of Earth’s shortwave reflectance. Hence, one must use a different albedo (αp) in Eq. (3) that only quantifies the actual surface reflectance. A recent analysis of Earth’s global energy budget by Trenberth et al. (2009) using satellite observations suggests αp≈ 0.12. Serendipitously, this value is quite similar to the Moon bond albedo of 0.11 (see Table 1 in our original paper), thus allowing evaluation of Earth’s ATE using our natural satellite as a suitable PGB proxy. Inserting= 0.12 in Eq. (3) produces Te = 269.6K, which translates into an ATE of only 18K (i.e. 287.6 – 269.6 = 18K).

In summary, the current GE theory employs a simple form of the SB law to estimate the magnitude of Earth’s ATE between 18K and 33K. The theory further asserts that the Moon average temperature is 250K to 255K despite the fact that using the correct lunar albedo (0.11) in Eq. (3) produces ≈270K, i.e. a15K to 20K higher temperature! Furthermore, the application of Eq. (3) to calculate the mean temperature of a sphere runs into a fundamental mathematical problem caused by Hölder’s inequality between non-linear integrals (e.g. Kuptsov 2001). What does this mean? Hölder’s inequality applies to certain non-linear functions and states that, in such functions, the use of an arithmetic average for the independent (input) variable will not produce a correct mean value of the dependent (output) variable. Hence, due to a non-linear relationship between temperature and radiative flux in the SB law (Eq. 3) and the variation of absorbed radiation with latitude on a spherical surface, one cannot correctly calculate the mean temperature of a unidirectionally illuminated planet from the amount of spatially averaged absorbed radiation defined by Eq. (2). According to Hölder’s inequality, the temperature calculated from Eq. (3) will always be significantly higher than the actual mean temperature of an airless planet. We can illustrate this effect with a simple example.

Let’s consider two points on the surface of a PGB, P1 and P2, located at the exact same latitude (say 45oN) but at opposite longitudes so that, when P1 is fully illuminated, P2 is completely shaded and vice versa (see Fig. 1). If the PGB is orbiting at the same distance from the Sun as Earth and solar rays were the only source of heat to it, then the equilibrium temperature at the illuminated point would be (assuming a solar zenith angle θ = 45o), while the temperature at the shaded point would be T2 = 0 (since it receives no radiation due to cosθ < 0). The mean temperature between the two points is then Tm = (T1 + T2)/2 = 174.8K. However, if we try using the average radiation absorbed by the two points W m-2 to calculate a mean temperature, we obtain = 234.2K. Clearly, Te is much greater than Tm (TeTm), which is a result of Hölder’s inequality.

image

Figure 1. Illustration of the effect of Hölder’s inequality on calculating the mean surface temperature of an airless planet. See text for details.

The take-home lesson from the above example is that calculating the actual mean temperature of an airless planet requires explicit integration of the SB law over the planet surface. This implies first taking the 4th root of the absorbed radiative flux at each point on the surface and then averaging the resulting temperature field rather than trying to calculate a mean temperature from a spatially averaged flux as done in Eq. (3).

Thus, we need a new model that is capable of predicting Tgb more robustly than Eq. (3). To derive it, we adopt the following reasoning. The equilibrium temperature at any point on the surface of an airless planet is determined by the incident solar flux, and can be approximated (assuming uniform albedo and ignoring the small heat contributions from tidal forces and interior radioactive decay) as

image

where is the solar zenith angle (radian) at point , which is the angle between solar rays and the axis normal to the surface at that point (see Fig. 1). Upon substituting , the planet’s mean temperature () is thus given by the spherical integral of , i.e.

image

Comparing the final form of Eq. (5) with Eq. (3) shows that Tgb << Te in accordance with Hölder’s inequality. To make the above expression physically more realistic, we add a small constant Cs =0.0001325 W m-2 to So, so that when So = 0.0, Eq. (5) yields Tgb = 2.72K (the irreducible temperature of Deep Space), i.e:

image

In a recent analytical study, Smith (2008) argued that Eq. (5) only describes the mean temperature of a non-rotating planet and that, if axial rotation and thermal capacity of the surface are explicitly accounted for, the average temperature of an airless planet would approach the effective emission temperature. It is beyond the scope of the current article to mathematically prove the fallacy of this argument. However, we will point out that increasing the mean equilibrium temperature of a physical body always requires a net input of extra energy. Adding axial rotation to a stationary planet residing in a vacuum, where there is no friction with the external environment does not provide any additional heat energy to the planet surface. Faster rotation and/or higher thermal inertia of the ground would only facilitate a more efficient spatial distribution of the absorbed solar energy, thus increasing the uniformity of the resulting temperature field across the planet surface, but could not affect the average surface temperature. Hence, Eq. (6) correctly describe (within the assumption of albedo uniformity) the global mean temperature of any airless planet, be it rotating or non-rotating.

Inserting typical values for Earth and Moon into Eq. (6), i.e. So = 1,362 W m-2, αo = 0.11, and ϵ = 0.955, produces Tgb = 154.7K. This estimate is about 100K lower than the conventional black-body temperature derived from Eq. (3) implying that Earth’s ATE (i.e. the GE) is several times larger than currently believed! Such a result, although mathematically justified, requires independent empirical verification due to its profound implications for the current GE theory. As noted earlier, the Moon constitutes an ideal proxy PGB in terms of its location, albedo, and airless environment, against which the thermal effect of Earth’s atmosphere could be accurately assessed. Hence, we now turn our attention to the latest temperature observations of the Moon.

  1. NASA’s Diviner Lunar Radiometer Experiment

In June 2009, NASA launched its Lunar Reconnaissance Orbiter (LRO), which carries (among other instruments) a Radiometer called Diviner. The purpose of Diviner is to map the temperature of the Moon surface in unprecedented detail employing measurements in 7 IR channels that span wavelengths from 7.6 to 400 μm. Diviner is the first instrument designed to measure the full range of lunar surface temperatures, from the hottest to the coldest. It also includes two solar channels that measure the intensity of reflected solar radiation enabling a mapping of the lunar shortwave albedo as well (for details, see the Diviner Official Website at http://www.diviner.ucla.edu/).

Although the Diviner Experiment is still in progress, most thermal mapping of the Moon surface has been completed and data are available online. Due to time constraints of this article, we did not have a chance to analyze Diviner’s temperature data ourselves. Instead, we elected to rely on information reported by the Diviner Science Team in peer-reviewed publications and at the Diviner website.

Data obtained during the LRO commissioning phase reveal that the Moon has one of the most extreme thermal environments in the solar system. Surface temperatures at low latitudes soar to 390K (+117C) around noon while plummeting to 90-95K (-181C), i.e. almost to the boiling point of liquid oxygen, during the long lunar night (Fig. 2). Remotely sensed temperatures in the equatorial region agree very well with direct measurement conducted on the lunar surface at 26.1o N by the Apollo 15 mission in early 1970s (see Huang 2008). In the polar regions, within permanently shadowed areas of large impact craters, Diviner has measured some of the coldest temperatures ever observed on a celestial body, i.e. down to 25K-35K (-238C to -248C). It is important to note that planetary scientists have developed detailed process-based models of the surface temperatures of Moon and Mercury some 13 years ago (e.g. Vasavada et al. 1999). These models are now being successfully validated against Diviner measurements (Paige et al. 2010b; Dr. M. Siegler at UCLA, personal communication).

What is most interesting to our discussion, however, are the mean temperatures at various lunar latitudes, for these could be compared to temperatures in similar regions on Earth to evaluate the size of ATE and to verify our calculations. Figure 3 depicts typical diurnal courses of surface temperature on the Moon at four latitudes (adopted from Paige et. al 2010a).

image

Figure 2. Thermal maps of the Moon surface based on NASA’s Diviner infrared measurements showing daytime maximum and nighttime minimum temperature fields (Source: Diviner Web Site).

image

Figure 3. Typical diurnal variations of the Moon surface temperature at various latitudes. Local time is expressed in lunar hours which correspond to 1/24 of a lunar month. At 89◦ latitude, diurnal temperature variations are shown at summer and winter solstices (adopted from Paige et al. 2010a). Dashed lines indicate annual means at the lunar equator and at the poles.

image

image

Figure 4. Temperature maps of the South Pole of the Moon and Earth: (A) Daytime temperature field at peak illumination on the Moon; (B) Nighttime temperature field on the Moon; (C) Mean summer temperatures over Antarctica; (D) Mean winter temperatures over Antarctica. Numbers shown in bold on panels (C) and (D) are temperatures in oK. Panels (A) and (B) are produced by the Diviner Lunar Radiometer Experiment (Paige et al. 2010b). Antarctica maps are from Wikipedia (http://en.wikipedia.org/wiki/Antarctic_climate). Comparison of surface temperatures between Moon’s South Pole and Antarctica suggests a thermal enhancement by the Earth atmosphere (i.e. a ‘Greenhouse Effect’) of about 107K in the summer and 178K in the winter for this part of the Globe.

Figures 4A & 4B display temperature maps of the Moon South Pole during daytime peak illumination and at night (Paige et. al 2010b). Since the Moon has a small obliquity (axial tilt) of only 1.54o and a slow rotation, the average diurnal temperatures are similar to seasonal temperature means. These data along with information posted at the Diviner Science webpage indicate that mean temperature at the lunar-surface ranges from 98K (-175C) at the poles to 206K (-67C) at the equator. This encompasses pretty well our theoretical estimate of 154.7K for the Moon mean global temperature produced by Eq. (6). In the coming months, we will attempt to calculate more precisely Moon’s actual mean temperature from Diviner measurements. Meanwhile, data published by NASA planetary scientists clearly show that the value 250K-255K adopted by the current GE theory as Moon’s average global temperature is grossly exaggerated, since such high temperature means do not occur at any lunar latitude! Even the Moon equator is 44K – 49K cooler than that estimate. This value is inaccurate, because it is the result of an improper application of the SB law to a sphere while assuming the wrong albedo (see discussion in Section 2.1 above)!

Similarly, the mean global temperatures of Mercury (440K) and Mars (210K) reported on the NASA Planetary Fact Sheet are also incorrect, since they have been calculated from the same Eq. (3) used to produce the 255K temperature for the Moon. We urge the reader to verify this claim by applying Eq. (3) with data for solar irradiance (So) and bond albedo (αo) listed on the fact sheet of each planet while setting ϵ = 1. This is the reason that, in our original paper, we used 248.2K for Mercury, since that temperature was obtained from the theoretically correct Eq. (6). For Mars, we adopted means calculated from regional data of near-surface temperature and pressure retrieved by the Radio Science Team at Stanford University employing remote observations by the Mars Global Surveyor spacecraft. It is odd to say the least that the author of NASA’s Planetary Fact Sheets, Dr. David R. Williams, has chosen Eq. (3) to calculate Mars’ average surface temperature while ignoring the large body of high-quality direct measurements available for the Red Planet!?

So, what is the real magnitude of Earth’s Atmospheric Thermal Effect?

Table 1. Estimated Atmospheric Thermal Effect for equator and the poles based on observed surface temperatures on Earth and the Moon and using the lunar surface as a proxy for Earth’s theoretical gray body. Data obtained from Diviner’s Science webpage, Paige at al. (2010b), Figure 4, and Wikipedia:Oymyakon.

image

Figure 5. Earth’s mean annual near-surface temperature according to Wikipedia (Geographic Zones: http://en.wikipedia.org/wiki/Geographical_zone).

Table 1 shows observed mean and record-low surface temperatures at similar latitudes on Earth and on the Moon. The ATE is calculated as a difference between Earth and Moon temperatures assuming that the Moon represents a perfect PGB proxy for Earth. Figure 5 displays a global map of Earth’s mean annual surface temperatures to help the reader visually verify some of the values listed in Table 1. The results of the comparison can be summarized as follows:

The Atmospheric Thermal Effect, presently known as the natural Greenhouse Effect, varies from 93K at the equator to about 150K at the poles (the latter number represents an average between North- and South- Pole ATE mean values, i.e. (158+143)/2 =150.5. This range encompasses quite well our theoretical estimate of 133K for the Earth’s overall ATE derived from Eq. (6), i.e. 287.6K – 154.7K = 132.9K.

Of course, further analysis of the Diviner data is needed to derive a more precise estimate of Moon’s mean surface temperature and verify our model prediction. However, given the published Moon measurements, it is clear that the widely quoted value of 33K for Earth’s mean ATE (GE) is profoundly misleading and wrong!

  1. Conclusion

We have shown that the SB Law relating radiation intensity to temperature (Eq. 1 & 3) has been incorrectly applied in the past to predict mean surface temperatures of celestial bodies including Mars, Mercury, and the Moon. Due to Hölder’s inequality between non-linear integrals, the effective emission temperature computed from Eq. (3) is always significantly higher than the actual (arithmetic) mean temperature of an airless planet. This makes the planetary emission temperature Te produced by Eq. (3) physically incompatible with any real measured temperatures on Earth’s surface or in the atmosphere. By using a proper integration of the SB Law over a sphere, we derived a new formula (Eq. 6) for estimating the average temperature of a planetary gray body (subject to some assumptions). We then compared the Moon mean temperature predicted by this formula to recent thermal observations and detailed energy budget calculation of the lunar surface conducted by the NASA Diviner Radiometer Experiment. Results indicate that Moon’s average temperature is likely very close to the estimate produced by our Eq. (6). At the same time, Moon measurements also show that the current estimate of 255K for the lunar average surface temperature widely used in climate science is unrealistically high; hence, further demonstrating the inadequacy of Eq. (3). The main result from the Earth-Moon comparison (assuming the Moon is a perfect gray-body proxy of Earth) is that the Earth’s ATE, also known as natural Greenhouse Effect, is 3 to 7 times larger than currently assumed. In other words, the current GE theory underestimates the extra atmospheric warmth by about 100K! In terms of relative thermal enhancement, the ATE translates into NTE = 287.6/154.7 = 1.86.

This finding invites the question: How could such a huge (> 80%) thermal enhancement be the result of a handful of IR-absorbing gases that collectively amount to less than 0.5% of total atmospheric mass? We recall from our earlier discussion that, according to observations, the atmosphere only absorbs 157 – 161 W m-2 long-wave radiation from the surface. Can this small flux increase the temperature of the lower troposphere by more than 100K compared to an airless environment? The answer obviously is that the observed temperature boost near the surface cannot be possibly due to that atmospheric IR absorption! Hence, the evidence suggests that the lower troposphere contains much more kinetic energy than radiative transfer alone can account for! The thermodynamics of the atmosphere is governed by the Gas Law, which states that the internal kinetic energy and temperature of a gas mixture is also a function of pressure (among other things, of course). In the case of an isobaric process, where pressure is constant and independent of temperature such as the one operating at the Earth surface, it is the physical force of atmospheric pressure that can only fully explain the observed near-surface thermal enhancement (NTE). But that is the topic of our next paper… Stay tuned!

  1. References

Inamdar, A.K. and V. Ramanathan (1997) On monitoring the atmospheric greenhouse effect from space. Tellus 49B, 216-230.

Houghton, J.T. (2009). Global Warming: The Complete Briefing (4th Edition). Cambridge University Press, 456 pp.

Huang, S. (2008). Surface temperatures at the nearside of the Moon as a record of the radiation budget of Earth’s climate system. Advances in Space Research 41:1853–1860 (http://www.geo.lsa.umich.edu/~shaopeng/Huang07ASR.pdf)

Kuptsov, L. P. (2001) Hölder inequality. In: Encyclopedia of Mathematics, Hazewinkel and Michiel, Springer, ISBN 978-1556080104.

Lin, B., P. W. Stackhouse Jr., P. Minnis, B. A. Wielicki, Y. Hu, W. Sun, Tai-Fang Fan, and L. M. Hinkelman (2008). Assessment of global annual atmospheric energy balance from satellite observations. J. Geoph. Res. Vol. 113, p. D16114.

Paige, D.A., Foote, M.C., Greenhagen, B.T., Schofield, J.T., Calcutt, S., Vasavada, A.R., Preston, D.J., Taylor, F.W., Allen, C.C., Snook, K.J., Jakosky, B.M., Murray, B.C., Soderblom, L.A., Jau, B., Loring, S., Bulharowski J., Bowles, N.E., Thomas, I.R., Sullivan, M.T., Avis, C., De Jong, E.M., Hartford, W., McCleese, D.J. (2010a). The Lunar Reconnaissance Orbiter Diviner Lunar Radiometer Experiment. Space Science Reviews, Vol 150, Num 1-4, p125-16 (http://www.diviner.ucla.edu/docs/fulltext.pdf)

Paige, D.A., Siegler, M.A., Zhang, J.A., Hayne, P.O., Foote, E.J., Bennett, K.A., Vasavada, A.R., Greenhagen, B.T, Schofield, J.T., McCleese, D.J., Foote, M.C., De Jong, E.M., Bills, B.G., Hartford, W., Murray, B.C., Allen, C.C., Snook, K.J., Soderblom, L.A., Calcutt, S., Taylor, F.W., Bowles, N.E., Bandfield, J.L., Elphic, R.C., Ghent, R.R., Glotch, T.D., Wyatt, M.B., Lucey, P.G. (2010b). Diviner Lunar Radiometer Observations of Cold Traps in the Moon’s South Polar Region. Science, Vol 330, p479-482. (http://www.diviner.ucla.edu/docs/paige_2010.pdf)

Ramanathan, V. and A. Inamdar (2006). The Radiative Forcing due to Clouds and Water Vapor. In: Frontiers of Climate Modeling, J. T. Kiehl and V. Ramanthan, Editors, (Cambridge University Press 2006), pp. 119-151.

Smith, A. 2008. Proof of the atmospheric greenhouse effect. Atmos. Oceanic Phys. arXiv:0802.4324v1 [physics.ao-ph] (http://arxiv.org/PS_cache/arxiv/pdf/0802/0802.4324v1.pdf ).

Stephens, G.L., A. Slingo, and M. Webb (1993) On measuring the greenhouse effect of Earth. NATO ASI Series, Vol. 19, 395-417.

Trenberth, K.E., J.T. Fasullo, and J. Kiehl (2009). Earth’s global energy budget. BAMS, March:311-323

Vasavada, A. R., D. A. Paige and S. E. Wood (1999). Near-surface temperatures on Mercury and the Moon and the stability of polar ice deposits. Icarus 141:179–193 (http://www.gps.caltech.edu/classes/ge151/references/vasavada_et_al_1999.pdf)

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Mike Monce
January 23, 2012 7:13 pm

Willis,
They are indeed integrating over the whole sphere. The theta integral runs from 0 to pi, the phi integral runs from 0 to 2pi. These are the standard limits to cover an entire sphere. To prove they are correct, do the integral:
R^2 INT sin(theta)d(theta)d(phi) which yields 4piR^2 which of course is the surface area of a sphere. Or easier, just check a mulitvariable calculus book 🙂

hotrod (larry L)
January 23, 2012 7:15 pm

George E. Smith; says:
January 23, 2012 at 6:30 pm

And of course with very little moisture in the tropical deserts, the atmosphere never had it so good in passing most of the surface LWIR radiation energy through to space.

George I do not think that is correct.
If measured as relative humidy, yes the deserts have very little water (compared to what they could hold), but their absolute water content is actually higher than in the arctic due to two effects.
First hot air can hold enormous amounts of water, cold air even at 100% humidity can hold very little.
Second the tropopause is much higher in the equatorial regions and adjacent deserts so the atmospheric layer that can hold lots of water vapor is also thicker.
If you look at the map on page 789 of this pdf you will see that the Sahara has 2x-3x the mean precipitable water content as the polar region.
http://docs.lib.noaa.gov/rescue/mwr/096/mwr-096-11-0785.pdf

“It has often been said that the air over deserts is not
really dry and these areas lack precipitation primarily
because of their high saturation deficits and stable
atmospheric conditions. This is confirmed by the annual
map of precipitable water. The Sahara, for example, has
nearly as much precipitable water as the cooler regions
of northern Europe and the northern United States.
The deserts do have less precipitable water than other
areas in the same latitudes, however.”

Larry

kdk33
January 23, 2012 7:21 pm

So, WUWT readers are supposedly a technically literate, science oriented bunch. I’m assuming there are a few out ttherre who have a bit of a handle on engineering thermodynamics. Can someone please help me out with the following:
The dry adiabatic lapse rate can be derived assuming an isentropic ideal gas atmosphere with a gravity imposed pressure gradient (I can provide the math; I’m hoping it’s not necessary). Furthermore, the T,P relationship for an ideal gas is T2/T1 = P2/P1^(R/Cp). R is the gas constant and, for and ideal gas, Cp can be taken as 5/2R so T2/t1 = P2/P1^0.4.
If I solve for T2 setting P1 and T1 as the conditions at the tippy tippy top of the atmosphere and P2 as the pressure at the planet surface I calculate an enormous surface temperature, T2.
My questions are as follows:
1) is the near surface atmsophere generally considered isentropic? If not, why does the DALR follow from this assumption?
2) If the atmosphere were isentropic all the way up, am I correctly calculating the surface temperature? It not, why not?
3) If the near surface atmosphere is isentropic, but the entire atmosphere is not, where does the isentropic assumption break down? Why does it break down?
4) (this one is for bonus points) Is the isentropic assumption and equilibrium criteria or a steady state condition?
Thanks!

Bill Hunter
January 23, 2012 7:43 pm

Here is a thought experiment that might help:
In Jelbring’s world with its energy impenetrable shell and surface, heat is trapped inside of the system.
So You then turn off gravity.
Clearly then two things will happen.
1) The gases will go isothermic;
2) No heat will be lost in the system.
From that you should conclude that gravity is the cause of the “surface temperature” or more correctly heat distribution but not the cause of any energy in the system.
Thus the perpetual motion arguments really don’t apply to this world.

hotrod (larry L)
January 23, 2012 7:47 pm

The lapse rate only applies to the tropsphere (60,000-80,000 ft altitude). This is the area that normal convective mixing takes place, At higher levels the atmosphere stratifies, into several distinct layers some of which get hotter with altitude as other mechanisms modify the air temperature. Such as ozone heating.
http://apollo.lsc.vsc.edu/classes/met130/notes/chapter1/vert_temp_all.html
Larry

Bob Fernley-Jones
January 23, 2012 8:09 pm

Joel Shore January 23, 5:11 pm
Thank you for your typical style of response. I compare your response to part of mine:

[Bob_FJ:] Another way of looking at it is that the consensus community is quite happy to accept that under the GHE, (including any feedbacks), it is currently argued that there is ~150 W/m^2 more energy at the surface than leaves at TOA. So, is this extra energy, offending conservation of energy? Clearly, it cannot be. Somewhat similarly, N&Z are proposing that the surface of a planet will be warmer with an atmosphere, than with an airless one, by virtue of a different mechanism. They are not claiming that the air pressure creates energy, but that the pressure enhances the uptake of the energy source near the surface
[Joel:] You are still not getting it. How is it possible to have 150 W/m^2 leaving the surface as radiation than leaving the TOA unless some of that radiation is getting absorbed (or reflected), i.e., unless there is a greenhouse effect? Air pressure does not absorb electromagnetic radiation….You are simply talking nonsense.

You seem to have misunderstood what I wrote; particularly see the part where I’ve herewith added bold emphasis. Oh and BTW, it is not me doing the theorising, but what I believe that N&Z are proposing, which many here do not seem to understand, including as a lecturer, yourself.
As Wayne points out below, you will not find it in Raymond Pierrehumbert’s book or the like, so you may find it hard to even consider it.
Concerning N&Z’s qualifications, sorry, I thought they were introduced somewhere in one of the threads as belonging to your discipline.

Nikolov does not have a PhD in physics…His PhD is in Forest Ecology ( http://www.fs.fed.us/rm/analytics/staff/nikolov.html ) Zeller does not have a PhD in physics…His PhD is in fluid mechanics and wind engineering ( http://www.fs.fed.us/rm/analytics/staff/zeller.html ). In fact, neither of them have any sort of degree (e.g., even a B.S.) in physics.

Of course physics is quite a wide field and according to your bio, you have been mostly active in non CAGW stuff. Recently, he has also developed an interest in global climate change.. Also, according to your bio it seems that you are no more qualified than they in this area, and they have certainly been more active in diverse constructive publications.
http://www.rit.edu/~w-physic/Faculty_Shore.html
Once again your elitism and dogma comes to the fore in these exchanges

KevinK
January 23, 2012 8:23 pm

A physicist wrote;
“Kevin, NASA’s multilayer reflective foil trick works only if there is a vacuum between the sheets”
Previously, KevinK wrote;
“The reflective plastic foil (aka MLI; Multi Layer Insulation) only works in a vacuum (sans conduction and convection).”
DID I MISS SOMETHING HERE ??? Yes, both statements are in the English language, but they sure seem to express the same conclusion to to me ???
So, the external fuel tank of the Space Shuttle spent almost all of its operational life outside of a vacuum environment. I.E. from Cape Kennedy, filled with cryogenic fuel, then launched, followed up by burning up during reentry to the Earth’s surface. All of these environments seem to be lacking any significant vacuum ?
Last time I visited Cape Kennedy, I do not seem to remember any significant presence of a vacuum. But I was on vacation, so I may have missed it.
So, is your position that the “GHE” only occurs when a vacuum is present? If so, how many molecules of GHGs are present in your vacuum, most definitions of a vacuum seem to exclude the presence of gas molecules.
By the way, that “NASA TRICK” of applying MLIs was in fact figured out by the folks that originally figured out how to operate “spy” satellites (look up the CORONA system) before the NASA folks borrowed it.
Cheers, Kevin.

Bill Hunter
January 23, 2012 8:52 pm

In Jelbring’s world with its energy impenetrable shell and surface, heat is trapped inside of the system.
So You then turn off gravity.
Clearly then two things will happen.
1) The gases will go isothermic;
2) No heat will be lost in the system.”
More stuff, add a radiative surface with storage capability.
3) The surface will cool and equilibriate with the atmosphere via drawing stored heat from the soil.
4) Surface radiation will reduce.
5) The total energy in the larger system has not changed.
Conclusion: Surface radiation is a very poor indicator of the GHE (or better stated as ATE) Its total energy and its distribution between various materials with various emissivities that determine how much heat can be stored in the system and gravity only determines its distribution.
With zero storage capacity there would be a zero greenhouse effect and the GHE at the surface is caused by gravity, not backradiation.

January 23, 2012 9:18 pm

kdk33 says: January 23, 2012 at 7:21 pm
“My questions are as follows:”

I think your use of isentropic is unorthodox. It applies to a process or a flow, but not usually to a continuum. What would make sense is to speak of a region where you can move air from one level to another isentropically (your P2/P1 formula) and it arrives at the ambient temperature. But that, in conjunction with hydrostatic pressure, just means that the lapse rate is the DALR.
So the answers are:
1. Doesn’t mean anything. The DALR is just a number from gas and gravity properties. It doesn’t follow from an assumption about the state of any one atmosphere.
2. Just asks (as above) if the actual lapse rate was equal to the DALR. Answer, yes, your calc is then correct. But usually that lapse rate is less than DALR.
3. Again, you’re just asking whether the lapse rate is the DALR, and if not, why not. It’s usually less, and the commonly quoted reason is condensation of water, which changes the effective heat capacity. My view is that another reason is that work has to be done to maintain the lapse rate (overcoming Fourier Law type conductive leakage), with energy provided by the wind. Radiative transfer is powerful leakage, and I think causes the lapse rate to fall short of DALR.
4. Again, check the meaning of isentropic.

Dan in Nevada
January 23, 2012 9:23 pm

Joel Shore says:
January 23, 2012 at 6:13 am
“tallbloke says:
Now, The practical demonstration by Konrad Hartmann in the recent post on my site (linked above in an earlier comment) shows that higher pressure does indeed enhance the sensible atmospheric heat generated by the passage solar radiation. This is an empirical result. No conservation law is harmed during the process. Empirical reality cannot break laws of nature!
Sorry…but one poorly conceived and carried out experiment does not overturn more than a century’s worth of physics even when it tells you what you want to believe. Konrad hasn’t even tried to figure out how his data, even if correct, could be compatible with well-understood physics.
It is really bizarre what you guys seem to think constitutes evidence!”
I’ve been doing a little self-educating via Google – got to love the inter-webs! What Konrad has been investigating has been known as Gay-Lussac’s law since the early 1800’s. According to Wikipedia, it should really be attributed to Guillaume Amontons who published it around 1700~1702. So, Joel Shore believes Konrad is attempting to overturn a century’s worth of physics when he’s actually confirming 3 centuries worth of physics. It’s not worth pointing out that this could conceivably be considered “compatible with well-understood physics”, but I will anyway. Bizarre or not, I’ll consider three centuries’ worth of accepted physics as evidence.
The only reason Konrad is taking on these experimental chores is to try to rebut the notion that “settled team science” somehow trumps physical laws that have been accepted for literally centuries and have never been refuted. I’m beginning to see what N&Z are up against.
N&Z are claiming, I believe, that Charles’s law, a corollary to Gay-Lussac’s law, is what governs the surface temperature of planetary bodies with atmospheres. That’s harder to test since you have to allow the gas volume to vary, but what Konrad is doing is completely valid since everything is based on the IGL (PV=nRT). This is a very simple algebraic expression that even I can wrap my old brain around. Why is it so hard for actual physicists?

don penman
January 23, 2012 9:23 pm

I don’t think that the length of day will effect the average temperature of a planet without an atmosphere ,it will just get hotter during the day and colder at night .Some are trying to confuse by including the effects of an atmosphere which would be affected by the length of day.The effect of having an atmosphere even one without ghg is like the true greenhouse or wearing an extra item of clothing ,you could argue that the air trapped by the glass of the greenhouse will become the same temperature but that temperature will still be warmer.

Editor
January 23, 2012 9:25 pm

Mike Monce says:
January 23, 2012 at 7:13 pm

Willis,
They are indeed integrating over the whole sphere. The theta integral runs from 0 to pi, the phi integral runs from 0 to 2pi.

They are integrating over mu = cos(theta). For theta running from 0 to Pi as you say, this gives a corresponding interval for mu from 1 to -1.
However, they are only integrating over half of that:

Joel says it makes no difference, because the temperature is zero on the night side. I can’t see how that wouldn’t affect the integral, but Joel’s science-fu is strong.
I just don’t understand why they’re only integrating from 0 to 1, and not -1 to 1.
Thanks,
w.

Bill Hunter
January 23, 2012 9:30 pm

The conclusion above was poorly stated. The last sentence was OK though.
I like to use a flow diagram to demonstrate how the system really works. Tallbloke’s talkshop as a nice discussion on flow diagrams and radiation that worth reading.

January 23, 2012 9:39 pm

Willis,
An integral is like a summation of many really small segments of a function. So, if you have all ZEROS (T = 0 over the whole night side of the planet), and you add them, what do you get?
You get 0 + 0 + 0 + … + 0 = 0.0. That’s why the night-side portion of the integral, where mu varies from 0 to -1 can be ignored when calculating the average temperature … Do you get it now?

January 23, 2012 9:42 pm

Willis,
Further clarification regarding the integral – night-time temperatures (being ZEROS) are ignored in the summation (integration), but the overall sum (integral) is divided by the ENTIRE surface area which is 4*pi …

Ralph
January 23, 2012 9:45 pm

<<<<>>>>>>
Nonsense.
A cloud layer can move in from a totally different weather system hundrds of miles away, and the effect will be the same. Whether the night air is still convective (rare), or whether the airmass is completely stable (more usual), a cloud layer will ALWAYS result in higher nighttime temperatures – because the cloud is ‘insulating’ the LW radiation from the surface.
And I have been watching the weather for 30 years.
.

George E. Smith;
January 23, 2012 10:00 pm

“”””” hotrod (larry L) says:
January 23, 2012 at 7:15 pm
George E. Smith; says:
January 23, 2012 at 6:30 pm

And of course with very little moisture in the tropical deserts, the atmosphere never had it so good in passing most of the surface LWIR radiation energy through to space.
George I do not think that is correct.
If measured as relative humidy, yes the deserts have very little water (compared to what they could hold), but their absolute water content is actually higher than in the arctic due to two effects. “””””
Dang !! Rod, I think you done got me there. I suppose if it is hot enough and the supply of atmospheric water is limited by distance from bulk supplies, then the relative humidity can get so low, that there just never is much chance of getting down to the dew point.
Gosh I learn something new every day, specially at WUWT.
However I stick by my point that because of the surface Temperature (+60 deg C or more), the Wien displacement moves the LWIR spectral peak right into the center of that atmospheric window, which as I recall, is a hole in the H2O absorption spectrum The peak even moves below the 9.6 micron Ozone band.
Incidently, although Total radiant emittance goes as T^4 for BB like emitters, the peak spectral radiant emittance goes as T^5, so the peak of the desert floor emittance spectrum is way higher (over 2x) the value for the global mean Temperature.
Thanks for the insight and correction hotrod.

Ralph
January 23, 2012 10:06 pm

<<<<<<<>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
George, you are wrong on so many points, it is difficult to know where to start.
Firstly I did not mention the Sun, because I was talking about night temperatures. I’ll give you a clue here, there is no Sun at night.
Secondly, I took great pains to not mention ‘warming’ and when I did use the term once I placed it in inverted commas, because the ‘warming’ is only reletive to the temperature you would obtain without nighttime clouds. ie, a night with clouds will ALWAYS be warmer than one without, or in effect you will percieve a distinct ‘warming’ sensation in comparison to a cloudless night.
Thirdly, although you may rail against the term ‘warming’ and claim that that clouds do not warm you (or the surface of the Earth) well technically they do. There is radiative transfer from you to the cloud (and to the atmosphere in general), which cools you down (try standing naked in the desert at night – that is not all conductive cooling, as an infra-red camera will easily demonstrate). But there is also radiative transfer from the cloud to you, warming you up (and again, a sensitive infa-red camera will pick up the warmer clouds in comparison to the colder cloudless sky). Now just because there is more transfer from you to the atmosphere, than there is from the clouds to you — and so the net effect you feel is cooling — does not mean that the clouds are not trying to warm you up.

Ralph
January 23, 2012 10:08 pm

Sorry mods, the first comment here was in response to Hotrod Larry, while the second was in response to George Smith. The brackets deleted the quotes !!
.

George E. Smith;
January 23, 2012 10:18 pm

“”””” Ralph says:
January 23, 2012 at 9:45 pm
<<<>>>>>
Nonsense.
A cloud layer can move in from a totally different weather system hundrds of miles away, and the effect will be the same. Whether the night air is still convective (rare), or whether the airmass is completely stable (more usual), a cloud layer will ALWAYS result in higher nighttime temperatures – because the cloud is ‘insulating’ the LW radiation from the surface.
And I have been watching the weather for 30 years. “””””
So I could have a cloudless day with surface Temperatures of say 40 deg F, and typical wintertinme low humidity, and after sundown a cloud layer at 20,000 feet comes in on some jet stream, and my surface Temperature will go up above 40 deg F ??
So the cloud might be -50 deg C, so just where does the energy come from to raise the surface Temperature; it’s after sundown, so it isn’t solar, and my cold air was going all the way up along with the lapse rate to where that 20,000 ft cloud came in, and my surface is getting even colder, as it radiates its 400 Watts per m^2 or more.
I will grant you that a massive storm of air might sweep in from some hot place with clouds, and simply replace all the cold air I had during the day, and presumably I will be warmer; but it wasn’t radiation trapping from my surface that warmed me up; and notwithstanding that storm cloud, it will still cool down as the night progresses; it won’t continue to get hotter, unless even hotter air contines to come in from somewhere else.
If you need to invoke a Santa Ana to make your case; then I submit, that you have no case; and i’ve been working at this for more than twice as long as you’ve been watching the weather.

Editor
January 23, 2012 10:38 pm

Ned Nikolov says:
January 23, 2012 at 9:42 pm

Willis,
Further clarification regarding the integral – night-time temperatures (being ZEROS) are ignored in the summation (integration), but the overall sum (integral) is divided by the ENTIRE surface area which is 4*pi …

Excellent, Ned, many thanks. That was the missing part in my understanding, that you are dividing through by the surface area. That being the case, it doesn’t matter that you have not integrated over the night side.
Much appreciated,
w.

Dan in Nevada
January 23, 2012 10:50 pm

re: Dan in Nevada says:
January 23, 2012 at 3:25 pm
A self-correction. Obviously, on reflection, my second scenario vs my first scenario should have been stated as
2P = 2nT / V
If we are roughly doubling n (in order to double P), then if V2= V1, T2=T1. However, it seems likely that V2 would be significantly larger than V1, hence T2 would have to be proportionately larger than T1. So, less bang for the buck, but a bang nonetheless.

George E. Smith;
January 23, 2012 10:56 pm

“”””” Ralph says:
January 23, 2012 at 10:06 pm
<<<<<<>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
George, you are wrong on so many points, it is difficult to know where to start.
Firstly I did not mention the Sun, because I was talking about night temperatures. I’ll give you a clue here, there is no Sun at night. “””””
Ralph, why don’t you start at “it is difficult to know where to start.”
Yes I know you didn’t mention the sun; so let’s NOT mention it; the global warming positive cloud feedback people never do, so why should we.
So we will have perpetual night without that unmentionable, and we can have jet stream clouds come and go, and absolutely NO external source of ground level energy from anywhere, since that is unmentionable; and the longer I wait, the colder I will get, and I will continue to get colder, cloud or no cloud.
Now I can’t even detect the 10 micron radiation that is emitting from the ground around my feet, my six senses are completely insensitive to it. So after it leaves the ground and disperses according to the inverse square law of cloud height; not to mention the cosine^4 of the oblique angle to that cloud that just came by, and some small part of it, gets absorbed by the cloud, the part that hits the cloud that is, and then the cloud re-emits isotropically, so half of it is lost to space, and only half of the cloud emission heads towards the surface, and that too, has an inverse square law dispersion plus another cosine^4 obliquity factor, so I have a 1/height^4 attenuation plus a cos^8 obliquity angle attenuation, before any of that undetectable radiation from near my feet makes it back to make me feel warmer as you put it. But one thing is for sure, I will never get warmer with or without that cloud, and if morning twilight ever comes, it will be much colder than the evening twilight was, and I’ll be praying for some unmentionable source of energy to show up, because under your evening sky, cloud or no cloud, there was only a continual loss of energy from the status quo prior to the disappearance of that unmentionable source of energy.
And since we are talking only night time; in the complete unmentionable absence of any external energy source, other than the ground around me, and the atmosphere, with or without cloud, then we can completely dispense with the usual explanation of the cloud positive feedback mechanism, which is that the higher the cloud, the less moisture it contains so it blocks less of that unmentionable external energy source, so more of that external energy reaches the ground; ergo, the higher the cloud the more the cloud warms the ground; that is the guts of THEIR explanation of cloud positive feedback. The fact that the same diminishing density with altitude has the exact same diminishing effect on the capture of outgoing LWIR by the cloud, seems to be entirely lost on those people.
Well I’m so happy to be so wrong on so many poinst Ralph, that you don’t even know where to start.

Ralph
January 23, 2012 11:10 pm

>>>George Smith
>>>So I could have a cloudless day with surface Temperatures
>>>of say 40 deg F, and typical wintertinme low humidity, and after
>>>sundown a cloud layer at 20,000 feet comes in on some jet stream,
>>>and my surface Temperature will go up above 40 deg F ??
Who says the temperature goes up? You are distintly bad at reading, George, try again. I NEVER SAID THE TEMPERATURE GOES UP, is that clear enough for you??
Clouds reduce cooling, because of the Greenhouse Effect. Note the term here, George, REDUCE COOLING. This results in a warmer temperature than a night without clouds. Do you understand that this is not warming, George, it is reduced cooling. Warmer than without clouds, but still a net cooling effect (but slightly reduced).
Clear??
.

Ralph
January 23, 2012 11:19 pm

>>>Willis Eschenbach says: January 23, 2012 at 10:38 pm
Willis, just a quick question. Why cannot you just assume the Earth to be a flat disk with a sun-side and a night side? That is, after all the full amount of energy that the earth receives on its sun-side – a complete disk-full.
Is this an approximation too far?
.

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