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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HLx
January 23, 2012 10:05 am

Tilo Reber:
The moon surface has a very low density. It acts as a insulator. Imagine a planet which is all covered by a 5o cm thick layer of insulation. The surface would not allow much heat down – so I agree with you :)..
But, this would probably also be a reason why the temperature of the moon is much lover on the dark side (because of low density and also low heat capacity)?
My question is, is it fair to compare the dense surface of earth with the low density moon surface? :-/…

kdk33
January 23, 2012 10:09 am

It cannot be a pressure phenomenon. If the atmosphere was not picking up LW radiation and eventually reradiating it, it would gently cool and cool until the admosphere was very cold and very dense (only warmed by surface conduction).
Is that really what would happen? Maybe this would happen: conduction would warm air near the surface, that warm air would then begin to convect. As convection begins, the rising and falling gasses have to follow the lapse rate. Right?
That probably means something.

January 23, 2012 10:11 am

The extra warmth or ATE observed at the surface is NOT a result of slowing down (or reducing) of the surface cooling by the atmosphere, but is due to an extra energy produced by pressure in combination with solar heating. The free atmosphere looses heat as fast as it can and cannot retain much energy due to its very low heat capacity.
Which, in the case of a non-GHG containing atmosphere is “not very fast at all” because it doesn’t cool at the top due to radiation. The problem is very simple. Without GHGs, there is no average thermal gradient to the atmosphere. Heat is convected up and once there, it stops. Where is it going to go? Down?
The problem is, I suspect, that you are buying into Jelbring’s idea that a static thermally isolated column of fluid can maintain a thermal lapse rate. It cannot. You have to have convective transport, and in order to have convection you have to have differential heating. If the top of the atmosphere never cools by radiation, it just traps heat until there is no vertical convection to speak of. That’s what happens in the stratosphere, which not at all coincidentally sits right above the point where the GHGs lose all of their heat in a steady state of flow up from the surface.
So your explanation appears to be partly correct, but is more wrong than it is right, because without GHGs it doesn’t work at all.
rgb

Editor
January 23, 2012 10:11 am

Joel Shore says:
January 22, 2012 at 7:28 pm

Willis says:

But they are not integrating over the dark half, as near as I can tell. What am I missing? You can’t just ignore half of the planet like that.

They did integrate over the dark half…and the value they get is zero because that is what the insolation is over that half. (They then add something back in to account for the fact that the temperature on the dark side would not really be 0 K but the 3 K background. I haven’t really paid attention to whether they did that correctly because the power due to the 3 K background is so ridiculously small as to be inconsequential.)

They are describing the surface by two angles, the horizon angle starting at the north phi, and zenith angle theta. To integrate over the entire the surface, phi goes from 0 to 2 Pi, and theta goes from – Pi/2 to Pi/2.
The dark half is where the sun is below the horizon. Given mu=cos(theta), that means when mu is less than zero.
But their integral only goes from 0 to 1, not from -1 to 1 as it would if they covered the dark side.
Where is the part of the integral where the sun is below the horizon?

Oh, and Joel, what was your opinion of them substituting
mu = cos(theta)
into equation 7, and then integrating over mu? That seems like an incorrect procedure to me.

It is fine. The integral of the polar coordinate for a function f over a spherical surface is integral of f*sin(theta)*d(theta) but sin(theta)*d(theta) = -d(cos(theta)) = -d(mu) where mu = cos(theta). [The negative sign is accounted for by switching the limits of integration, i.e., 0 deg to 90 deg becomes mu = 0 to mu = 1.]
Like I said in my first post, as near as I can see, their mathematical calculations are fine. Their errors here are conceptual ones.

My point is a little different, let me see if I can explain it …
Suppose y = cos(theta). The integral of cos(theta) from 0 to 1 is 0.707.
Now replace cos(theta) with mu. The integral of mu from 0 to 1 is 0.5.
This is why I was taught, at least, that you cannot simply substitute at random before integration and expect to get the right answer. You can only substitute if your substitution is linear w.r.t. what is substituted for. Otherwise, your integral (as in the example above) will be incorrect.
Similarly, I get different values when I integrated above for theta, and when I replaced cos(theta) with mu.
If what you say is correct, shouldn’t I get the same answer whether I integrate over theta or over mu?
Regards,
w.

January 23, 2012 10:12 am

Look at this paper for the measured surface temperatures at the Apollo 15 site (we refer this in our paper):
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)
Both direct and indirect (satellite) measurements show that the MEAN temperature the lunar equator is about 210K. Like we stated in our article – there isn’t a Moon latitude. where the average diurnal surface temperature is anywhere close to 250K …

January 23, 2012 10:18 am

Willis,
Drop the subject of the integral! You are only embarrassing yourself more and more … Just stick to what you know (and do) best – construction management! Please!

Editor
January 23, 2012 10:23 am

Ned Nikolov says:
January 22, 2012 at 8:17 pm

Fellows,
I’d like to make a point to those of you, who have no formal training in physical science. You should be careful about how strongly you feel about certain science aspects of this new theory. For instance, if you know your calculus skills are weak, don’t try to figure out what’s wrong with our solution to the integral in Eq. 5, and then share your opinion on the blog as if your were making a contribution. You are NOT! Instead, you are only muddying the discussion. For example, Willis Eschenbach (whose comments I have followed for 2 weeks now) relentlessly tries to find problems with the math in our papers (to no avail, of course), while demonstrating at the same time a remarkable lack of skills in math. He actually admits his total lack of science credentials and science education in this 2010 video :

Ned, I have made two very clear objections to your math. Waving your hands and saying I’m wrong is a joke when you are replying to mathematical objections.
I may be wrong, and I acknowledged that when I asked. I asked what I was missing.
So … either answer my mathematical objections and tell me what I was missing, or not as you wish.
But don’t just come out here and complain that I’m finding what I think are errors. Perhaps they are not, but gnarf at least thinks I am right that you can’t just make the substitution of cos(theta) = mu, and you whining about people finding fault with your math doesn’t explain why you think you can make that substitution.
If your math is correct, then why do I get different answers when I integrate using theta and using mu? I put my Mathematica formulas up there, surely you are not claiming that Mathematica is wrong. So what is wrong with my Mathematica formulas?
I’ve shown exactly, specifically, and precisely where I think you are wrong. If you wish to ignore my valid mathematical objections and reply with an attack on my math skills, that’s your choice … but that’s not science, and it will just make people point at you and laugh.
Phil Jones tried your same bogus “how dare you doubt me, you’re just a commoner” answer to my questions, Ned, and he didn’t come off so well. You might try a real answer to my math questions … or not.
w.

Schrodinger's Cat
January 23, 2012 10:27 am

Consider our atmosphere, particularly the air nearest the ground (or sea) because it is this air that is highly compressed. It is compressed because of the force acting on it as a consequence of the mass of air above and the acceleration due to gravity.
Now let us switch on a source of energy. The sun is convenient. The short wave radiation heats up the surface of the planet. This in turn, heats up the air in contact with the surface. This is the same compressed air we discussed a moment ago. Consider the very high density of highly enegetic air molecules in this air. The kinetic energy equates to the air temperature in accordance with the Gas Laws and is a consequence of the high pressure (air mass) and solar energy. Of course as we consider increasing altitude, the force (air mass) decreases and so does the temperature.

Editor
January 23, 2012 10:29 am

Björn says:
January 22, 2012 at 8:28 pm

Willis , Joel is right about the integration and the result given is the correct one as he explains …

Many thanks, Bjorn, and you and Joel may certainly be right, but if that is the case then why does Mathematica give a different answer for the two integrals?
Also, you say:

… and the lower bound on the inner integral becomes +1 ( theta = cos(0)) and the upper bound becomes -1 …

but as you will note, their inner integral only goes from 0 to 1 … so where is the part where the sun is below the horizon?
w.

January 23, 2012 10:30 am

Elevator Pitch for N&Z by davidmhoffer
I’m a salesman. Elevator pitches are what I do. Further, I’ve been selling bleeding edge technology for 30 years. Taking a known technology and explaining it to a highly technical decision maker whose technical expertise is not necessarily relevant to the technology I am selling (but which solves a problem for them) is what I do. I have been struggling with coming up with an elevator pitch to explain N&Z but, given the myriad of issues that they raise, I couldn’t find a way to pull them all together. Thanks to my debating of the issues with many, but mostly Joel Shore and Willis Eschenbach, I think I’m finally ready to speak to the issue in an elevator pitch style. I shall actually cheat a bit and break it up into two pieces.
N&Z Summarized.
The earth is a net absorber of energy in the tropics. The earth is a net loser of energy in the temperate and arctic zones. Energy is moved from the tropics to the temperate and arctic zones via a variety of mechanisms including conduction, convection, back radiation, oceanic currents and more. Thermal equilibrium is achieved when the temperate and arctic zones warm sufficiently that they radiate at a net loss to space that exactly equals the net gain in the tropics. Since the concentration of GHG’s does nothing to change the amount of net absorption in the tropics, how much they absorb and re-radiate is immaterial. The amount of energy that must be moved from tropics to temperate and arctic zones remains identical. Only the means by which it moves changes. The various mechanisms are locked together via feedback loops such that an increase in the amount of energy moved by one mechanism by default causes a reduction in the amount of energy moved by other mechanisms until thermal equilibrium is once again restored (and vice versa).
Temperature and Currency
To understand “warming” in any context, we must also understand it in the context of “currency conversion”. Suppose that the Canadian “dollar” is worth 0.90 of an American “dollar”. Suppose that Willis Eschenbach has 100 dollars (he lives in the States) and I have 100 dollars (but I live in Canada). On the surface, between the two of us, we have an “average” of 100 dollars per person. Now suppose that Willis sends his 100 dollars to me (thankyou Willis, I was running a bit short this month). I can’t deposit Willis’ generous donation to my bank account until I convert it to Canadian. Between the two of us, Willis and I now have a total of 211 dollars, 11 more than we started with, and an average of 105.50 per person. Moving energy from the tropics to the temperate zones works exactly the same way, ony the currency is “degrees”. When the tropics cool by moving energy to the temperate and arctic zones, they are moving “degrees” which to not convert one to one when they land at higher latitudes. For the simple reason that P(w/m2) varies with T(degrees) raised to the power of four, when the tropics cool by one degree, thermal equilibrium cannot be achieved by the temperate and arctic zones warming by one degree. One degree at -40C is worth 2.9 w/m2 but one degree at +40 is worth 7.0 w/m2. By failing to account for the currency conversion, the average of T across the planet surfave gives the illusion of a higher “average” temperature than a simple blackbody calculation that ignores currency conversion would suggest.

Editor
January 23, 2012 10:36 am

kzeller says:
January 22, 2012 at 8:31 pm

… Why aren’t you all trying to disprove our MIRACLE equation rather than banging your heads against walls trying to prove or disprove who knows what and exclaiming you have problems with this or that? …

Ummm … because that’s how science works? People are going to very carefully examine every one of your arguments, and the logical steps among them, despite the fact that you might think they are 100% bulletproof … who knew?
w.
PS—Which one of the equations is the MIRACLE equation, and just what miracles can I expect it to perform when I find it?

January 23, 2012 10:37 am

Suppose y = cos(theta). The integral of cos(theta) from 0 to 1 is 0.707.
Now replace cos(theta) with mu. The integral of mu from 0 to 1 is 0.5.

Or to be a bit more precise:
\int_-\pi/2^\pi/2 \cos(\theta) d\theta = \sin(\theta)_-\pi/2^\pi/2 = 2
or (letting x = \sin\theta, so that dx = d(\sin(\theta) = \cos(\theta) d\theta)
\int_-\pi/2^\pi/2 \cos(\theta) d\theta = \int_-1^1 dx = x|_-1^1 = 2
If you want to change variables to do an integral, that’s fine, but you have to change all of the variables and the limits of integration when you do. Sometimes that is algebraically helpful (a short and easy way to do an integral). Sometimes not.
rgb

A. C. Osborn
January 23, 2012 10:43 am

Ned Nikolov says:
January 23, 2012 at 9:49 am says “The implications from our analysis of planetary data as well as that of the current structure of climate models (see our first paper) clearly suggests that the effect of so-called GH gases on surface temperature is ZERO!”
Can you offer the mechanism for Clouds keeping it warmer at night?

Editor
January 23, 2012 10:43 am

markus says:
January 23, 2012 at 2:05 am

Mr Eschenbach,

“If you were to mention which laws of physics you think Anthony is denying it would assist us all.”

For, matter without potential energy, there is none, kinetic energy cannot be potential energy, radiation is the enhancement of potential energy to the state of geomagnetism. Energy from our Sun cannot penetrate the potential energy of Earth, unless we have irreversibly entered its magnet fields. Kinetic energy from our Sun cannot obtain the properties of potential energy, why, because it has no mass.
Energy doesn’t equal mass. It is the energy retained by mass, from creation of the universe, that cannot be penetrated by the remnants of that creation.
E=mc2.
Markus.

Riiiiight … your claim is that Anthony is denying that E=mc2 … riiiight …
w.

A. C. Osborn
January 23, 2012 10:52 am

Willis Eschenbach says:
January 23, 2012 at 10:29 am says “but as you will note, their inner integral only goes from 0 to 1 … so where is the part where the sun is below the horizon?”
What affect does that have on an Atmospherless body?

Editor
January 23, 2012 10:56 am

Ned Nikolov says:
January 23, 2012 at 10:18 am

Willis,
Drop the subject of the integral! You are only embarrassing yourself more and more … Just stick to what you know (and do) best – construction management! Please!

Thanks, Ned, but until you answer a question it doesn’t go away, and neither do I, as Phil Jones found out to his cost. And I am never embarrassed by asking questions. That’s how people learn. So no, I won’t drop the integral question.
Why not? Because I want to learn something. I want to find out where I’m wrong. I want to advance my knowledge, to answer the question I asked to begin this whole discussion—what am I missing?
No one has explained why my two Mathematica equations don’t give the same answer, and now you advise that I just walk away?
Not happening, my friend. Answer my questions or not, that’s up to you. But I’ll keep asking until I learn where I’m wrong … or where you are.
w.

January 23, 2012 10:58 am

“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).”
If you tie this in with Nicola Scafetta’s recent paper ( http://www.appinsys.com/GlobalWarming/SixtyYearCycle.htm ) on the 60 year interplanetary influence, you then have planetary gravitational forces exerting changes in atmospheric pressure, which alters climate on earth.
Changes in Gravitational Pull = Changes in Climate?

January 23, 2012 11:00 am

“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).”
If you tie this in with Nicola Scafetta’s recent paper ( http://www.appinsys.com/GlobalWarming/SixtyYearCycle.htm ) on the 60 year interplanetary influence, you then have planetary gravitational forces exerting changes in our atmospheric pressure, which alters climate on earth.
Changes in Gravitational Pull = Changes in Climate?

January 23, 2012 11:02 am

davidmhoffer says
“One degree at -40C is worth 2.9 w/m2 but one degree at +40 is worth 7.0 w/m2. By failing to account for the currency conversion, the average of T across the planet surfave gives the illusion of a higher “average” temperature than a simple blackbody calculation that ignores currency conversion would suggest.”
David, unfortunately your description is physically TOTALLY wrong, and we do not subscribe to it. In the physical world, one cannot increase the total amount of kinetic energy in a system by simply moving internal components of the system around. This would violate the First Law of Thermodynamics. Re-distribution of energy CANNOT create additional energy. In fact, in the process of re-distribution, some of the energy inevitable gets lost as heat outside the system. That’s what makes it impossible to create an engine of a 100% efficiency. This is governed by the Second Law of Thermodynamics regarding ever increasing entropy! Sorry for the bad news … 🙂

JPeden
January 23, 2012 11:02 am

Joel Shore says:
January 23, 2012 at 6:13 am
Sorry…but one poorly conceived and carried out experiment does not overturn more than a century’s worth of physics…It is really bizarre what you guys seem to think constitutes evidence!
No, Joel, what’s really bizarre is how “a century’s worth of [CO2 = CAGW] physics” has been shown to count for next to nothing here in the real world as approached by real science – which is inherently sceptical science – but non-sceptical mainstream Climate Science just keeps right on claiming that we should all do something really stupid in order to appease its alleged findings, “before it’s too late!”

January 23, 2012 11:06 am

Tim Folkerts said January 23, 2012 at 8:40 am

The “new” science is that the authors reject the radiative effects of the atmosphere. While ERE’s are certainly important and are responsible for much of the rise in temperature from ~ 155 K to ~ 288 K, they cannot do this by themselves. However, the RGE (radiative greenhouse effect) can raise the ground temperature above 255 K by well-known, well-accepted methods.

That’s the take home message for me. And acceptance of GHE which is well-supported empirically, does not imply acceptance of CAGW’s enhanced GHE which is not well-supported empirically.
I think of GHE as analogous to adding a salt such as NaCl to water. Pure water does not conduct electricity and is the analog of N2 & O2. Adding a tiny amount (X gm) of salt (analog of GHG) increases conductance Y dS/m. Adding 2X gm increases conductance by somewhat less than 2Y dS/m. Each addition of salt has less effect than the previous addition until the electrical resistance falls to zero ohms at the maximum possible conductance.
In this analog, Earth’s atmosphere is not yet at maximum conductance. Hansen’s “runaway greenhouse” seems to imply (in my admittedly feeble mind) the possibility of negative resistance/greater than infinite conductance.

Editor
January 23, 2012 11:07 am

Robert Brown says:
January 23, 2012 at 10:37 am

Suppose y = cos(theta). The integral of cos(theta) from 0 to 1 is 0.707.
Now replace cos(theta) with mu. The integral of mu from 0 to 1 is 0.5.

Or to be a bit more precise:
\int_-\pi/2^\pi/2 \cos(\theta) d\theta = \sin(\theta)_-\pi/2^\pi/2 = 2
or (letting x = \sin\theta, so that dx = d(\sin(\theta) = \cos(\theta) d\theta)
\int_-\pi/2^\pi/2 \cos(\theta) d\theta = \int_-1^1 dx = x|_-1^1 = 2
If you want to change variables to do an integral, that’s fine, but you have to change all of the variables and the limits of integration when you do. Sometimes that is algebraically helpful (a short and easy way to do an integral). Sometimes not.
rgb

Thanks, Robert. So … not sure what your conclusion is. Are my two objections valid or not?
1. In their equation 6, have they integrated over the entire surface of the sphere?
2. Can they simply substitute mu for cos(theta) and then integrate over mu?
and my allied question …
3. If the answer to (2) is “yes”, then why do my two Mathematica integrals above give different answers, if the substitution is valid?
w.

January 23, 2012 11:07 am

Willis,
I’m happy to see the change in your attitude … I fully agree that everyone deserves to know and an honest desire to learn is a truly noble quality. No problem with that whatsoever … 🙂

JJThoms
January 23, 2012 11:08 am

Ned Nikolov says: January 23, 2012 at 10:05 am
A note of caution if I may offer one … Stop relying too much on ‘thought experiments’, for those must always take a back set to actual math and a quantitative analyses
========
I have offered you twice the paper below
This shows that 260w/sqm emerges from the atmosphere and hits the earth at NIGHT. compared to 400w/sqm during the day. Note that this is LWIR.
http://www.patarnott.com/atms749/pdf/LongWaveIrradianceMeas.pdf Figure 3
1. there is 0.75 as much heat hitting the earth in the LWIR during the night as during the day therefore your integration of zero night to full day insolation is not correct
2. Radiation only occurs with GHGs so although all the atmosphere is contributing to the GHG gasses re-radiation (as is earth/sea generated LWIR), without the GHGs there would be 260w/sqm less heating at night. This is not to say that your non-ghgs are not conducting to earth just that the MEASURED 260W/sq metre is in addition to this.
Take away the ghgs and 400W/sqmetre day and 260W/sw metre disappear in a flash! PLEASE note that these are actual measurements!

A physicist
January 23, 2012 11:09 am

Ned Nikolov says:
January 23, 2012 at 10:05 am
A note of caution if I may offer one … Stop relying too much on ‘thought experiments’, for those must always take a back set to actual math and a quantitative analyses. Arguing against accurate math results with ‘thought experiments’ is silly and can be terribly misleading and counterproductive … Thought experiments are useful as a general guidance ONLY at the preliminary stages of an inquiry into an unknown phenomenon, when actual processes and precess responses are not well known in terms of their mathematical functions.

Ned, although your broad point has merits, please let me say that (in my opinion, and the opinion of many) Robert Brown’s physical arguments and thought experiments have exposed a considerable number of weak and/or irrelevant and/or mistaken starting assumptions in your theoretical framework.
IMHO, you and Karl should take Dr. Brown’s criticisms seriously, because no amount of ingenious mathematics can rescue a physical theory whose starting assumptions are weak, irrelevant, or just plain wrong.

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