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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January 22, 2012 12:02 pm

Thanks for posting this Anthony. The debate on the correct grey-body temperature of the Moon and Earth will be interesting!

Steve (Paris)
January 22, 2012 12:04 pm

Above my paid grade. In fact out of my orbit. But I love it.

Editor
January 22, 2012 12:06 pm

Thanks, Anthony, particularly for the caution at the top. I see one typo? in the work:

Comparing the final form of Eq. (5) with Eq. (3) shows that in accordance with Hölder’s inequality.

Seems like part of the sentence is missing.
All the best,
w.
REPLY: Thanks, fixed. Many instances like that where copy/paste from the original document outright fails for some reason. – Anthony

Alvin
January 22, 2012 12:11 pm

Thank you. This will take several readings, and several cups of hot tea to fully digest.

Ceri Phipps
January 22, 2012 12:14 pm

I am trying to get my head around all this, but that’s a good thing. Real science rather than polar Bears fall ing off icebergs. I love it!

Joel Shore
January 22, 2012 12:16 pm

I’ll post here what I communicated to Ned Nikolov directly:
Thanks for the reply.
First, I would like to say what I think is good about this reply: It is well-written and clearly explains your thinking and I thank you for that. Furthermore, the actual calculations that you do appear to be correct.
However, I believe there are some quite erroneous statements and interpretations as discussed below that will presumably set the stage for major incorrect conclusions in the second part of the reply.
Here is a brief discussion of the most major errors that I found:
page 3:

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 (alpha_0) 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 ˜ 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 alpha_0 = 0.12 in Eq. (3) produces 269.6K, which translates into an ATE of only 18K (i.e. 287.6 – 269.6 = 18K).

This is mainly a semantic issue: The conventional assumption of a constant albedo is used in order to derive the temperature rise attributable to the greenhouse effect alone. It is true that clouds also have another effect, in that they change the planet’s albedo, but that is a separate issue. Furthermore, if you do want to imagine removing the albedo due to clouds, you really also have to then ask how the surface albedo of the colder planet will change due to the increase of snow and ice. (But then…Does the planet still have water on it to form snow and ice? It depends by what magic we got rid of the IR-absorbing gases in the atmosphere! We thus get into various hypotheticals!) I think the best “clean” statement that we have is that if we imagine somehow turning off the radiative greenhouse effect without changing anything else, then our planet would be about 33 K colder.
page 5:

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.

No…You guys have failed to fully understand the implications of Holder’s Inequality. First of all, there is no condition that says that additional thermal energy can’t be added to the planet: It is receiving energy from the sun! Hence, conservation of energy has been misapplied here.
The correct way to apply energy conservation is to note that what is required is that the planet will be near (and always pushed toward) a state of radiative balance where it is radiating back into space the same amount of radiation as it receives from the sun. What Holder’s Inequality shows is that there are lots of different temperature distributions having lots of different average temperatures that satisfy the criterion that the average radiative emission is, say, 240 W/m^2. A planet could thus hypothetically have any average temperature compatible with Holder’s Inequality, which means any average temperature lower than the average temperature for a perfectly uniform temperature distribution (which is ~255 K for a blackbody earth); the actual one that occurs will depend not only on the distribution of insolation (in space and time) but also on the convective, advective, and conductive transport mechanisms in the atmosphere, oceans, and (to a lesser degree) solid surface.
page 9:

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)!

It is not really a matter of one being more accurate than another. It depends very sensitively on how you define the average temperature of the surface, e.g., is the surface the first millimeter of the planet’s surface or is it the first meter? The data from Diviner presumably is measuring the temperature right at the surface. However, if you go several centimeters down, you find the temperature remains remarkably uniform between day and night and you get the average value of around 240 K. This is explained, for example, here: http://www.asi.org/adb/m/03/05/average-temperatures.html
Hence, average surface temperature of an airless planet is not a very well-defined quantity because of the large temperature swings right at the surface. This is fundamentally because lots of different average temperatures (corresponding to different temperature distributions) lead to the same amount of total power emitted by the planet’s surface.
page 11:

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.

Not really…What the results, correctly interpreted, show is that it is probably not very useful to talk about an average temperature for an airless planet where the temperature distribution is so non-uniform. That is why scientists usually talk about the “average temperature” determined by averaging T^4 and taking the 4th root.

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?

First of all, you can’t really appeal to intuition here…What is a reasonable thermal enhancement due to the IR-absorbing gases? How is one to know?
More importantly, nobody is claiming that the IR-gases are responsible for such a large thermal enhancement. They are responsible for raising the average emissions of the surface from ~240 W/m^2 to ~390 W/m^2, which for any sort of reasonably uniform temperature distributions (such as that which occurs on the Earth presently) corresponds to a temperature rise of ~33 C.

Hence, the evidence suggests that the lower troposphere contains much more kinetic energy than radiative transfer alone can account for!

The limitation imposed by the physics is not directly on the amount of kinetic energy that the lower troposphere can contain. It is receiving huge amounts of energy from the sun! The limit is instead set by radiative balance…i.e., that the Earth must radiate back out into space the same amount as it receives from the sun (or else it will warm up or cool down).

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).

The “among other things” is alas the rub. The Ideal Gas Law alone does not uniquely constrain the surface temperature.

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) What process is isobaric?
(2) Try as one might, one is not going to explain the fact that the Earth’s surface emits an average of ~390 W/m^2 while the Earth + atmosphere absorb an average of ~240 W/m^2 without acknowledging that the atmosphere must absorb some of the terrestrial radiation emitted, i.e., that there is a radiative greenhouse effect.
Sincerely yours,
Joel

Stephen Wilde
January 22, 2012 12:17 pm

Taking Ned’s ATE as a starting point various implications for climate would follow. In that context I had this exchange with Joel on one of the other threads.
“For a planet without a greenhouse effect, said height is necessarily the surface of the planet”
Irrelevant because energy still gets into the air via conduction and convection and back to the surface via convection and conduction for then radiating out. So there will still be a lapse rate and it must match ATE otherwise the system is unstable. If it doesn’t match ATE then the atmosphere will accumulate energy via conduction and convection indefinitely until it boiled away or lose energy to the ground via conduction and convection indefinitely until it congealed on the surface.
The true Perpetuum Mobile is the concept of a planet with an atmosphere that is not precisely in equilibrium with its ATE.
Any disequilibrium will either boil off the atmosphere or congeal it on the ground.Once congealed on the ground it would be lost via sublimation.
The radiative GHG theory is itself a Perpetuum Mobile because it proposes that changing the composition changes ATE. Thus a bit more human GHGs are amplified by more water vapour and that gives more GHGs which amplifies again ad infinitum.
The atmosphere and oceans would get hotter and hotter until they boiled away.
We would see lots more planet sized bodies with no atmospheres at all because a little change in atmospheric composition would have been enough to destabilise it.
IIf we were to reduce GHGs so that they changed the ATE lapse rate the other way then the Earth would get steadily colder until the oceans and air congealed on the ground.
The system won’t allow it. Any change in composition that might introduce a disequilibrium with ATE is neutralised by a reconfiguring of the circulation pattern.
If you could find one planet where the Gas Laws do not apply then you would have me. Where is it ?”

January 22, 2012 12:22 pm

Willis Eschenbach says:
January 22, 2012 at 12:06 pm
Thanks, Anthony, particularly for the caution at the top. I see one typo? in the work:
Comparing the final form of Eq. (5) with Eq. (3) shows that in accordance with Hölder’s inequality.
Seems like part of the sentence is missing.

I notice the sentence following contains the key to understanding the reason for the inclusion of one of the terms which has been referred to as a ‘free parameter’ by some critics.
To make the above expression physically more realistic, we add a small constant 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: eq 6

January 22, 2012 12:29 pm

Willis Eschenbach says:
January 22, 2012 at 12:06 pm
Thanks, Anthony, particularly for the caution at the top. I see one typo? in the work:
Comparing the final form of Eq. (5) with Eq. (3) shows that in accordance with Hölder’s inequality.

It’s an error in the formatting. The .pdf reads:
Comparing the final form of Eq. (5) with Eq. (3) shows that Tgb << Te in accordance with Hölder’s inequality.

Joel Shore
January 22, 2012 12:30 pm

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.

It is probably useful to go into a little more detail on this claim than I did above by providing a little more discussion as to why Nikolov and Zeller’s claim is wrong here: Let’s suppose they were right, so we take a non-spinning planet and start a planet spinning and the temperature distribution becomes more uniform without changing the average surface temperature. Is this a reasonable steady-state?
Initially, before spinning, the planet was in radiative balance. Now that it has a more uniform temperature distribution but the same average temperature, the average value of T^4 will necessarily be lower. Hence, it will now be emitting less power back into space than it absorbs from the sun. What we such a planet do? It will warm up until such point as it is now emitting the same amount of power as it is absorbing from the sun. Hence we see how energy conservation, correctly applied, constrains not the average temperature but the emitted power.

Stephen Richards
January 22, 2012 12:35 pm

In principle it looks fine but relys heavily on the grey body definitions and calculations, obviously. Quite a few typos and missing words I think but so what. I know how difficult ( and time consumming) that would have been for you Anthhony.

Stephen Wilde
January 22, 2012 12:37 pm

“Try as one might, one is not going to explain the fact that the Earth’s surface emits an average of ~390 W/m^2 while the Earth + atmosphere absorb an average of ~240 W/m^2 without acknowledging that the atmosphere must absorb some of the terrestrial radiation emitted, i.e., that there is a radiative greenhouse effect.”
The surface of the Earth exchanges 150W/m2 by a dynamic conductive exchange between surface and atmosphere.
The Earth also exchanges 240W/m2 with the incoming solar energy from the top of the atmosphere.
Joel says the atmosphere gets its energy solely from downward radiation from GHGs. The truth is that it is conduction and convection.
Wiki is quite clear on the issue:
http://en.wikipedia.org/wiki/Lapse_rate
“Because the atmosphere is warmed by conduction from Earth’s surface, this lapse or reduction in temperature normal with increasing distance from the conductive source.”
And it all applies to:
“any gravitationally supported ball of gas.”
without any breach of the Laws of Thermodynamics.

A physicist
January 22, 2012 12:37 pm

Nikolov and Zeller asked: “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? … The answer obviously is …”

With respect, Ned and Karl, didn’t NASA ask-and-answer this question back in the 1960s?
NASA asked the question in an engineering context: “Can a huge (> 80%) improvement in thermal insulation [in a rocket stage] be the result of a few grams of reflective plastic foil, that collectively amount to less than 0.0001% of total cryogenic booster mass? …”
NASA’s answer (which obvious in retrospect) was “tiny thickness of foil create a huge increase in insulation.”
That is why (it seems to me) that your entire last paragraph should be dropped, for the simple reason that engineers familiar with radiative heat transfer will immediately appreciate that its reasoning is just plain wrong.

January 22, 2012 12:39 pm

We have over 260 comments on the thread we have been running at the Talkshop about this paper since Jan 17th. It’s a hot topic!

Editor
January 22, 2012 12:43 pm

Ned (or anyone), I don’t understand this claim:

However, we will point out that increasing the mean equilibrium temperature of a physical body always requires a net input of extra energy.

The mean equilibrium temperature, as Holder’s inequality establishes, cannot exceed what you are calling the “effective emission temperature” (Equation 3).
However, for a given input of solar energy, the mean equilibrium planetary temperature can take up a variety of values as long as it is less than the effective emission temperature. For a planet the value is mainly set by the length of the day (speed of rotation) and the presence/absence of an atmosphere.
Again for a given input of solar energy, the rule is “the bigger the temperature swings, the lower the temperature average”.
So we can have two identical planets whose only difference is rotation speed. Both are receiving the identical amount of solar energy. The planet that is rotating faster will have a higher mean equilibrium temperature without any “net input of extra energy”.
Which seems to contradict your statement that I quoted. What am I missing?
All the best,
w.

Kev-in-UK
January 22, 2012 12:46 pm

Personally, I don’t see anything wrong in their explanation (a couple of minor typos though) as a basic description of an assessment of direct evidence (measurement) of a comparable irradiated object (the Moon) and its theoretical SB derived temperature. As I see it, they have then ‘reverse engineered’ to dare to suggest that the SB derived temperature for the earth (and the subsequent assumed GHE) is potentially in error.
In my opinion, there is definate merit in detailed independent analysis – especially concerning the math around Holders inequality (!). The thinking is indeed somewhat ‘out of the box’ – but it should not be discounted ‘out of hand’ and if the actual measurements from the moon are not meeting ‘theoretical’ expectation they are absolutely correct in raising this…their actual theory may be somewhat wrong or overstated, but the big question, as I see it – is that the accepted theory (SB derived temp) may well not be correct either!

G. Karst
January 22, 2012 1:01 pm

Very elegant and convincing. I keep looking for some huge hole in the logic (it must be there – mustn’t it?). It will be a tough row to hoe because it makes most of the climate community look so foolish. Why do we so easily and often get the cart in front of the horse??
Right or wrong, this will be one interesting subject to follow, in the coming future. Hopefully it will shake out new ideas and approaches. GK

Joel Shore
January 22, 2012 1:15 pm

tallbloke says:

I notice the sentence following contains the key to understanding the reason for the inclusion of one of the terms which has been referred to as a ‘free parameter’ by some critics.
To make the above expression physically more realistic, we add a small constant 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: eq 6

Nope…Except at the very earliest stages of comments of their original paper, I have not been counting this value as a free parameter because I have recognized that it is not really free. The four parameters that are really free are those in Equation (7) of their original paper.

Bob Shapiro
January 22, 2012 1:22 pm

Nice presentation, except that it ignores conduction transfer of energy from earth’s surface to the atmosphere.
Imagine a ball of nitrogen & oxygen gas, bound up in a zero thickness, purely transparent shell, in deep space. With the “local environment” existing at barely warmer than absolute zero, it is reasonable to assume that our ball of non-greenhouse gases would obtain a comparable temperature.
Now, move that ball to the vicinity of the moon, with the sun’s radiation pouring into it. Since the gases are nearly (not quite 100%) transparent to the sun’s range of emitted radiation, as well as to the moon’s re-radiation wavelengths, our nitrogen & oxygen still would remain only slightly warmer than in deep space.
If instead, our ball were filled with CO2 and other “greenhouse gases,” then our new ball would warm up significantly, since those gases do indeed absorb the moon’s re-radiation.
This is a distinction which exists only in outer space, and this usually is used to justify the distinction here on earth. I believe this is a flawed interpretation of how the system works. Our environment here on earth is quite different from outer space. Here, other forms of energy transfer are at play – conduction and convection.
Let us imagine once more. This time, we have an earthlike planet, covered by an atmosphere made up entirely of nitrogen and oxygen – with no so-called greenhouse gases. As in deep space, our sun’s radiation essentially passes through our atmosphere, with negligible absorption.
When that radiation reaches earth’s surface, some of that energy is reflected, while most is absorbed by the land and seas. (So long as we’re imagining, let’s assume water that doesn’t evaporate into the atmosphere.)
This incident radiation striking the earth warms up the surface. While some of this heat is reradiated back into space, once again passing through our IR transparent gases, much of the heat is conducted to the atmosphere. (This article admits that, on the moon, “Surface temperatures at low latitudes soar to 390K (+117C) around noon…”)
The air that is touching the earth’s surface, warms, becomes less dense than the cooler air above it, and rises, leaving cooler air now in contact with the surface. Convection currents work even if there are no greenhouse gases.
Since the sun beats down on only half our earthly disk at a time, the sun facing side warms, through radiation from the sun, compared to the dark side. Planetary differences between surface temperatures cause planet scale convection patterns in our nitrogen-oxygen atmosphere.
These convection patterns will warm even the night time earth above a bare earth temperature. Yes, this nitrogen-oxygen atmosphere will lose energy radiation to space, but it also will radiate partly back onto the earth from which it originally received its energy through conduction. Though the nitrogen and oxygen have warmed through conduction rather than radiation absorption, from that point forward they essentially are greenhouse gases.
Our “non-greenhouse gas” earth system still will warm considerably compared to an atmosphere free planet, and this warming will be well above the SB black body equilibrium of an airless body in space. How much warming our earth would experience, I will leave for the physicists to determine, but ignoring conduction and convection energy transfer mechanisms is not acceptable.
One consequence of our considering only radiative absorption as the way our atmosphere can warm, is that we ascribe the entire difference in earth’s temperature, between its current level and what it would be without any air at all, to greenhouse gases. We are conceding ground to the CAGW activists by default.
Our non-greenhouse gas system still acts as a greenhouse; our non-greenhouse gases indeed are greenhouse gases. The whole distinction based on how the various gases receive their energy, some of which then is radiated partly out to space and partly back to earth, is false.
We cannot determine how much CO2 and the IR absorbing gases raise earth’s temperature without including conduction absorbing gases.

January 22, 2012 1:23 pm

Willis says:
For a planet the value is mainly set by the length of the day (speed of rotation) and the presence/absence of an atmosphere.
Again for a given input of solar energy, the rule is “the bigger the temperature swings, the lower the temperature average”.

Sounds about right to me. So the Moon is likely considerably colder on average than the long accepted calcs say it is. But since Earth spins more rapidly (once a day rather than once a month) then even without an atmosphere it would be warmer. It’s axial tilt has to be accounted for too.

Joel Shore
January 22, 2012 1:30 pm

Stephen Wilde says:

Joel says the atmosphere gets its energy solely from downward radiation from GHGs. The truth is that it is conduction and convection.

No…You are attributing things to me that I have not said. What I have said is that the fact that conduction and convection exists doesn’t get you around the fact that the Earth with its current surface temperature is emitting ~390 W/m^2 and hence that, in the absence of an IR-absorbing atmosphere, the Earth system would be emitting ~150 W/m^2 more than it is absorbed from the sun and would rapidly cool down.

bair polaire
January 22, 2012 1:35 pm

Formatting problem in Table 1: last column should read 143 instead of 14 for the south pole ATE. The cursor seems to hide the last figure…
Very interesting discussion! (No fishing lines this time… 😉 )

January 22, 2012 1:42 pm

Consider two bodies, A is fast rotating and therefore has a near uniform temperature, and B is rotating slowly and consequently has huge temperature differences.
From the non-linearity of eq. (1) you can see that if A and B have the same mean temperature then B will emit more radiation than A. To understand that, you can for example say that A is uniformly at 300K and B has one half at 200K and the other at 400K, and do the math.
If A and B are in the same distance from the sun and have the same albedo, they will receive the same incoming radiation.
Remember that the outgoing radiation has to equal incoming radiation over time, ie. the outgoing radiation from A has to equal the outgoing radiation from B. Therefore the mean temperature of A has to be higher than B. This proves that equation (6) is wrong for a rotating body.

Jordan Yankov
January 22, 2012 1:45 pm

Mr Nikolov, I will sugest to read this paper by acad. Sorohtin:
http://fiz.1september.ru/articlef.php?ID=200501111
This is in Russian but propbably you will not have problems. I think this is quite close to your theory.

January 22, 2012 1:50 pm

// sarc on //
God, I hate this stupid denier site. No science ever goes on here!
// sarc off //
Great job here. It’s above my pay grade, as they say, but I’m learning tons.

James Reid (from Arding)
January 22, 2012 1:51 pm

This on the surface at least looks to me a systematic approach to the problem that is at least accessible to the lay person.
errata: just before equation 2 should it not be So not Sa for TOA solar irradiance?
It raises some questions that I have had about the orthodox approach which may already be covered in the parts I haven’t yet read… will ask them later if not.

gnarf
January 22, 2012 1:53 pm

There is a big problem in the integral when they make the substitution u=cos(theta).
if u=cos(theta) you have to express dtheta using du to make the substitution.
dtheta=-du/sqrt(1-u2)
So after substitution you have something with u^0.25/sqrt(1-u^2) to integrate, and certainly not u^0.25 only!!!
http://www.sosmath.com/calculus/integration/substitution/substitution.html
Sorry, but this is plain terrible.

Kev-in-UK
January 22, 2012 1:53 pm

Roger says:
Sounds about right to me. So the Moon is likely considerably colder on average than the long accepted calcs say it is. But since Earth spins more rapidly (once a day rather than once a month) then even without an atmosphere it would be warmer. It’s axial tilt has to be accounted for too.
I don’t follow how rotation/spinning increases temp – logically (and ignoring any internal heat generation etc) the sum of the irradiance must equal the sum of the re-radiance i.e. energy in = energy out ? – in the no atmosphere scenario, why would spinning make the temp warmer? The diurnal change (or whatever frequency of change one has!) doesn’t change the total amount of solar radiation received nor the total amount of reradiatione emitted!

John Blake
January 22, 2012 1:55 pm

In summary, may we conclude that circulating atmospheric pressure is a hitherto neglected factor that in proper SB context both empirically and theoretically multiplies Earth’s previously hypothesized global equilibrium temperature by ~1.86, nearly doubling previous estimates?
On this basis, given requisite variations on such an enhanced figure, it seems that Gaia’s standard range of temperature fluctuations encompasses virtually all geohistorical extremes from steaming Jurassic jungles to pre-Cambrian ice ages. This suggests that not global atmospheric/oceanic circulation but plate-tectonic dispositions have been the root cause of Earth’s non-cyclic but periodic warming-cooling phases from the planet’s earliest times.
As geophysical phenomena, Pleistocene Era glaciations averaging 102,000 years interspersed with interglacial epochs of median 12,250 years, are accordingly symptoms of a more fundamental “fluid dynamic,” that is, of Earth’s essential thermodynamic planetary composition/structure. If so, any and all AGW hypotheses are iatrogenic artifacts not of empirical investigation but of scholastic preconceptions– not objective, rational conclusions but mere semantic exercises, words.

Robertvdl
January 22, 2012 2:00 pm

“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”
This includes ‘air’ (atmosphere) pressure?

PeterGeorge
January 22, 2012 2:00 pm

As complex as this is, it seems to me to be missing something.
If the air at the surface is still (no convection), and the resulting surface temp is T0, then what happens when we add wind (i.e. convection)?
The windy air will convect energy away from the surface, so less energy can be radiated (from the surface), and its temperature must be T1 < T0.
To me is seems implausible that all else being equal, the surface temperature of a planet with little near surface atmospheric turbulence could be the same as that of a planet with lots of turbulence.

don penman
January 22, 2012 2:01 pm

the difference in temperature between the moon and the Earth cannot all be down to temperature swings and length of day ,the difference in temperature is still there when we compare the poles of the moon with the poles of the earth.I think it is unreasonable that our atmosphere does not insulate us from the cold of space not just ghg.I think thermodynamic equilibrium is an unreasonable assumption if the Earth and the moon are warmer than space then heat will flow from the moon and the Earth to space .

Zac
January 22, 2012 2:03 pm

I am not surprised that Anthony does not support this paper given the prominance and support he gave to Willis Eschenbach’s outburst. But to deny the laws of physics seems a daft tad to me.

January 22, 2012 2:13 pm

Joel Shore says
” I think the best “clean” statement that we have is that if we imagine somehow turning off the radiative greenhouse effect without changing anything else, then our planet would be about 33 K colder.”
Notice the without changing anything else logic.
Here’s Joel’s logic applied to a typical small car
Average speed = 45miles per hour
Fuel consumption in a year 5000 litres.
The engine is now removed without changing anything else.
Fuel consumption drops to zero, excellent carbon footprint numbers, average speed slightly increased because of reduced weight.

Editor
January 22, 2012 2:14 pm

Ned (or anyone) I think there is an error between equation 4 and equation 5.
In equation 4 you have correctly indicated that when the sun is below the horizon, the value of Ti is zero.
However, in equation 5 you have only integrated that over half of the surface of the sphere, the sunlit half. This is indicated by mu (the cosine of the zenith angle) varying from 0 to 1, rather than -1 to 1.
Integrating over the full surface gives a different answer than you have. Here’s Mathematica on the subject:

Note that this is just 1/2 of your answer, as we would expect when we include the unlit side.
In addition, to do the integral, in the derivation of equation 5 you have substituted mu for the cosine(theta) term in Equation 4. I’m not sure you can do that when you are going to integrate. My suspicion is supported by the fact that integrating over cos(theta) from -Pi/2 to Pi/2 gives a very different answer involving the Euler gamma function. Mathematica again:

This gives a value of 0.337 in place of the “2/5” at the left of your answer.
Comments?
w.
[EDITED TO ADD: I have corrected an error, I’d incorrectly integrated from -Pi to Pi, it is from -Pi/2 to Pi/2]

gnarf
January 22, 2012 2:15 pm

The integral of u^0.25/sqrt(1-u^2) from 0 to 1 gives a little bit less than 0.5 (quick numerical approximation).
So what you finally get with a correct substitution (and dividing the integral by 2PI the surface of half a sphere as the integral is made on half a sphere!) is not Tgb=2/5 […]^0.25 but Tgb~1/2[…]^0.25 that is to say 1.6 times your result.
For your example with earth it gives Tgb~1.6*154.7=247.52 K

George E. Smith;
January 22, 2012 2:17 pm

“”””” 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 “””””
This is a rather gross simplification and leads to erroneous calculations. For a start, there is ne spectral emissivity data, and published spectral emissivity data for many types of common rocks give quite different numbers. It is true that many common rocks have an emissivity around 95% at wavelengths longer than about 12 microns, and also below 8 microns, but in that crucial window from 8 to 11 microns, common rocks have considerably lower emissivities, often below 80%.
And this region; 8-12 microns is just the range where the 288 mean earth LWIR radiation spectrum peaks. So the peak emittance from most rocks is omewhat less than the above would suggest, and this is also where the atmospheric window is that allows much of this radiation to escape.
As for “Holder’s Inequality” (with an umlaut), is this some peer reviewed Nobel Prize winning fundamental discovery of Physics; or is it some self evident high school mathematical conclusion. I submit it is “climate scientists” trying to sound important by attaching some glorious title to trivial mathematics.
And any hobbyist playing around with simple elctronics; Radio Shack style is well aware of the fact the RMS value of a sinusoidal waveform is quite different from the average value (which is zero), and also from the average value of the rectified sine waveform. So stop trying to make a big deal out of trivia. If it was Eric Holder, the current Attorney General of the USA, it would make more sense; it’s about on par with his intelligence.
By the way; the very same reference text book, where rock emissivity measured data can be found, also shows BLACK BODY LIKE SPECTRA for the sky radiation as seen from the earth surface; and that spectrum (observed) has the shape expected for the ATMOSPHERIC TEMPERATURE. This despite the well known FACT among climate scientists that ordinary diatomic or monoatomic neutral gases do not emit thermal radiation, as do all other materials at Temperatures abou zero Kelvins.
I see a lot of thrashing around in this paper, including some interesting stuff, but I think it is a bit self congratulatory to describe it as “a unified theory of climate”. In what way is it “unified” ?

Bill Hunter
January 22, 2012 2:20 pm

“I think the best “clean” statement that we have is that if we imagine somehow turning off the radiative greenhouse effect without changing anything else, then our planet would be about 33 K colder.”
Clean? Clean in science is no missing stuff, not painting over the smudges.

George E. Smith;
January 22, 2012 2:24 pm

Also in the above analysis of this “unified theory” where is the part about atmospheric H2O and CO2 absorbing part of the incoming solar radiation spectrum, as between 0.7 and 4.0 microns; which itself will warm the atmosphere. Both CO2 and H2O have overlapping (but not coincident) IR bands at 2.7 microns, where a significant solar energy component still exists.
Just what is it that you suppose converts an extra-terrestrial 1362 W/m^2, into a ground level number more like 962 W/m^2

gnarf
January 22, 2012 2:27 pm

Yes there are two big big errors in this integral:
1) they divide the result by 4pi which is the surface of the entire sphere but the integral is made on half the sphere (cos theta varying from 1 to -1 covers the north hemisphere)
2) the substitution u=cos(theta) is a big big failure…it does not give u^0.25du but it gives
u^0.25/((1-u^2)^0.5)du
And the final, correct result is approximately 1.6 times the result presented here->247K. Is it some pro AGW people trying to show we are stupid to accept terrible mistakes if they tend to show things differently ??

HLx
January 22, 2012 2:27 pm

@Joel Shore:
“I think the best “clean” statement that we have is that if we imagine somehow turning off the radiative greenhouse effect without changing anything else, then our planet would be about 33 K colder.”
But this is an issue they are disputing, and it is a key point they are making. It is almost as you are saying “I do not want to look at step F, I’m contempt with things as they are” while their argument is step A, step B, step C etc. through step F all the way to a final conclusion. It is a faulty dismissal.
I’m inclined to think they have missed something obvious in all their steps, but I still accept that they must be allowed to have their logic heard.
And also, I for one think it is entirely fair to set up a Case Scenario where earth has no atmosphere and albedo at 0.12. It is the current scenario they are discussing. Not yesterday or tomorrows scenario with more or less snow, or whatever.

Greg Elliott
January 22, 2012 2:29 pm

Bob Shapiro says:
January 22, 2012 at 1:22 pm
Our non-greenhouse gas system still acts as a greenhouse; our non-greenhouse gases indeed are greenhouse gases. The whole distinction based on how the various gases receive their energy, some of which then is radiated partly out to space and partly back to earth, is false.
I’ve written an analysis to that effect. Reviewers welcome.
http://tallbloke.wordpress.com/2012/01/20/greg-elliott-use-of-flow-diagrams-in-understanding-energy-balance/
The paper also identifies the assumption of constant albedo as a source of error in climate models.

R. Gates
January 22, 2012 2:30 pm

The entire conclusion of Nikolov and Zeller rests on the assumption that the Moon is a gray-body equivalent to the Earth, and this assumption is quite erroneous. The Earth, because of having a far more dense and thermally active interior, generates far more longwave radiation than the Moon, leading to of course the fact that the surface of the Earth emits more net energy than it absorbs through sunlight. This energy of course leaves the ground and is abosrbed and re-emitted by the atmosphere in proportion to the concentration of GH gases, which leads to higher lor lower atmospheric temperatures, depending of concentration levels. The geologically dead and rather inert Moon on the other hand emits a tiny amount of LW naturally, beyond the solar energy absorbed, but it is at a tiny faction of what the Earth naturally emits beyond solar energy absorbed. This greater amount of energy generated by the surface of the Earth versus the Moon is accounted for in Trenberth’s energy balance etimates in upward directed LW, and leads to the fact that the conclusion by Nikolov and Zeller, that the ATE is ‘3 to 7 times higher’ than currently estimated, is also quite erroneous. In short, because of the very geologically and thermally active nature of the Earth (the core could be as high as 5000C or so), the heat generated by the surface of the Earth itself is not inconsequential, and makes the rather inert and thermally dead Moon a very poor and highly inaccurate grey-body equivalent to the Earth.

Bill Hunter
January 22, 2012 2:35 pm

“I see a lot of thrashing around in this paper, including some interesting stuff, but I think it is a bit self congratulatory to describe it as “a unified theory of climate”. In what way is it “unified” ?”
Considering the average dogma indoctrination rate of climate scientists its probably more than worthwhile to point out its far from divinely-inspired perfection.

DB-UK
January 22, 2012 2:41 pm

Since there is no such a thing as “global annual temperature”, but rather huge number of local temperature systems, this is all lot of theoretical arguments about some virtual planet that does not exist! Please comment on Essex et al 2007, Kramm and Dlugi 2011 and Pell et al (2007) on Koppen-Geiger climate classification in Pell et al 2007.
DB-uk

January 22, 2012 2:50 pm

“According to Eq. (2), our atmosphere boosts Earth’s surface temperature not by 18K—33K as currently assumed, but by 133K! This raises the question: Can a handful of trace gases which amount to less than 0.5% of atmospheric mass trap enough radiant heat to cause such a huge thermal enhancement at the surface? Thermodynamics tells us that this not possible.”
This is really nuts, and is being repeated. No-one claims that the atmosphere boosts surface temperature by 18K-33K. Some say that the GHG fraction of the atmosphere has that effect, relative to an atmosphere with no GHG.
But to then say that the 133K is due to a handful of trace gases when you’ve removed the whole atmosphere????
The GHE is that difference between air with and without GHG. Your ATE is something else.

PeterF
January 22, 2012 2:53 pm

I cannot accept your reasoning for being allowed to ignore rotation of a planetary body.
Imagine an infinitely fast rotating body. It would be the equivalent of incoming radiation falling on the whole surface of the planet, and not just half of it. This would result in the “shaded” side having a higher, and the “lit” side a lower temperature. And because of the 4th power law and Hölder’s inequality you would get a higher average temperature.If your further conclusions rest on this averaging, they cannot be trusted.
Proof me wrong by doing the integration over time, or show explicitely the “fallacy” of doing it.

Joel Shore
January 22, 2012 2:56 pm

Willis Eschenbach says:

Ned (or anyone) I think there is an error between equation 4 and equation 5.
In equation 4 you have correctly indicated that when the sun is below the horizon, the value of Ti is zero.
However, in equation 5 you have only integrated that over half of the surface of the sphere, the sunlit half. This is indicated by mu (the cosine of the zenith angle) varying from 0 to 1, rather than -1 to 1.

No, Willis. They have just used the fact that the integral of an integrand that is zero (as it is over the dark half) is zero. I think for the approximation that they are making (i.e., that the local temperature is determined by radiative balance with the local insolation), their calculation is correct.

wayne
January 22, 2012 3:02 pm

Ned & Karl:
I do see one possible slip in the text. It states:“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: …”; but 0.0001325 is closer to 5 K than 2.72 K stated. Maybe the text should be altered to state that 5.07K is the CMB plus mean star-shine that all interstellar bodies would also receive at all times. That is what I assumed that tiny difference represented.

January 22, 2012 3:27 pm

Anthony Watts.
Having just read the foreword, I paused before getting in to the main post because the only thing going through my mind was “what a gentleman Anthony is.
He is obviously hard working and intelligent, but he is also polite, courteous and considerate of others, hence the foreword.”
Anthony, you’re a credit to yourself and your family. I’d like to meet you one day, maybe the next time you come to Australia. I’d like to shake your hand.

R. Gates
January 22, 2012 3:27 pm

Just to add to the notion that the Moon is a very poor gray-body body proxy for Earth– if you took away Earth’s atmosphere and ocean and measured the radiation curve of the Earth’s surface, it would look very different than the Moon’s. This difference is due to the Earth’s internal energy, some of which is emitted at the surface, and is not insubstantial. The outer core of the Earth is quite dense and is around 5000C, and certainly some this heat reaches the surface. The moon is far less thermally active on the interior and thus far less heat reaches the surface and so the two bodies would have very different radiation curves.

Lady Life Grows
January 22, 2012 3:30 pm

I just skimmed the article, because it appears that it still takes zero cognizance of the fact that some component of the Earth’s surface temperature is due to internal radiation from U238 and other radioactive elements.
It does not appear to me that the author’s efforts to compare Earth to the moon will prove valid, as the moon has fewer radioactive elements, as far as I know.

Julien
January 22, 2012 3:51 pm

It’s a great contribution you’ve done there. To improve things, it’d be nice to include the fact that Earth is a terrestrial planet (it has volcanic activity and nuclear fusion/fission is occurring at the centre of our planet, which releases energy as well) and how it affects the ATE, as well as the earth’s rotation influence (which I think is void over a complete yearly solar revolution anyway, but nice to consider). So IMO because of volcanic activity you can’t compare the moon and earth directly, but it’s still a good push forward that you’ve done. Good job.

Bill K
January 22, 2012 3:57 pm

This is very interesting. I am not qualified to pass judgement and have great respect for many individuals with different opinions about the issues being discussed. I look forward to part 2.
For those interested, Konrad Hartmann has posted on an interesting experiment over at Tallbloke’s Talkshop. In comments Lucy Skywalker refers to a very interesting experiment by R. W. Graeff.
A little more confusing in this context but still quite interesting with apparently compelling logic is an article by John O’Sullivan that posts an essay by Dr. Pierre R Latour on the 33 deg greenhouse effect.
There are likely many processes involved in determining earth’s climates, but the one thing that seems least likely is that human contributions to atmospheric CO2 will lead to harmful warming of the climates or even a significant proportion of them.
Thanks to all contributors.

January 22, 2012 3:58 pm

R. Gates,
I often agree with you, but not here: “This difference is due to the Earth’s internal energy, some of which is emitted at the surface, and is not insubstantial. ”
The internal energy is mostly “insubstantial”. The flow of geothermal energy is typically estimated as ~ 0.1 W/m^2. Even if this was as high as 1 W/m^2, this is fairly small compared to the 240 W/m^2 from the sun. The moon would be 240 W/m^2, while earth would be 241 W/m^2, which would be a small difference (assuming both have the same albedo). Different albedos between the two would have a much bigger effect than any geothermal energy flows.

Ian H
January 22, 2012 4:02 pm

As a rule of thumb things that should not be added should not be averaged. “It was 293K at my house and 290K at yours for a total temperature of 583K”. Meaningful? No! So the concept of average temperature should be looked on with a great deal of suspicion, especially if you plan to use it in calculation rather than simply as a general measure of middle. On the other hand the radiation emitted per unit area is proportional to T^4, and the total radiation emitted by our two houses is a very meaningful idea. Hence it is much reasonable to average T^4. That is why it is much better to use units of energy rather than units of temperature to describe what is going on.
You hint at awareness of this problem when talking about Holder’s inequality. But then as far as I can see you then try to compute a mean temperature. Why not simply integrate $ \alpha T^4 over the surface of the planet and compare directly to the incident flux.

January 22, 2012 4:15 pm

Nick Stokes says:
January 22, 2012 at 2:50 pm
Your ATE is something else.

Wotcher Nick.
Good puzzle isn’t it?

January 22, 2012 4:16 pm

Lady Life Grows said:
January 22, 2012 at 3:30 pm
I just skimmed the article, because it appears that it still takes zero cognizance of the fact that some component of the Earth’s surface temperature is due to internal radiation from U238 and other radioactive elements.
———————————
Others, maybe – but U238 is barely radioactive.

Bill Illis
January 22, 2012 4:19 pm

I think there must be an error in the derivation of equations 5 and 6.
The justification for the new derivation is “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 (Te ≫ Tm), which is a result of Hölder’s inequality”
But the temperature versus forcing/radiation is logarithmic to the fourth power. You cannot average two radiation levels to surmise the temperature equation is off. You can only average the temperatures to do this. (Climate science actually does something similar in that they take shortcuts which average forcings over a wide range to derive the climate sensivity. See Hansen and the ice ages).
I think the 2/5 constant in equations 5 and 6 is not supposed to be outside of the equation which is then taken to the power of 1/4 or 0.25. Its equivalent is supposed to be inside. Its equivalent is supposed to apply to the Solar Forcing only, not to the SB constant as well.
Temp = (Forcing(radiation) / SB constant)^.25
Of course, the Forcing and the SB constant denominator can be further refined to reflect actual situations but the 2/5 is not a constant outside the equation.

Billy
January 22, 2012 4:22 pm

This is not so much a comment on the paper but a comment on process.
Anthony Watts wrote:

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.

Posts like this may explain why wattsupwiththat.com appears to get many more views than RealClimate. Watts provides a friendly forum to views he disagrees with in the hope that informed discussion of a position he believes to be wrong may improve understanding.
I think practices like this boost his credibility and the credibility of other posters on the blog. In contrast, RealClimate filters out comments. I had one of mine filtered out which i am pretty sure was correct and not even particularly politically incorrect. But, it disappeared. That disappearance made me distrust RealClimate. I suspect that if Stephen Hawking posted a devastating and correct critique of one of their posts, it would vanish. (Well, not if he signed it Hawking but it would if he signed it S.H.)
Believing that sound critiques are deleted at RealClimate, I see little point in reading their postings unless I am willing to do the work necessary to check their results. I’m quite confident that my skills and ability would allow me to do so, but it could take a lot of time. So, I look at RealClimate from time to time, but more for amusement than information.
I see material here from time to time that appears—mmmm—shall we say questionable. But, such material often gets strongly questioned so i can relax as i read it—figuring that flawed material will be critiqued and that the critiques will not be deleted.
Billy

Jose Suro
January 22, 2012 4:31 pm

He is absolutely correct about Holder’s Inequality. Read up on that instead on concentrating on the First Law, which by the by, is constrained be the Second Law, and no one seems to be taking that into account. The Second Law is an overriding factor as well.
Best,
J.

jorgekafkazar
January 22, 2012 4:33 pm

Typo? “…then the equilibrium temperature at the illuminated point would be (assuming a solar zenith angle θ = 45°), {something missing here?} while the temperature at the shaded point would be T2 = 0…”

Titixxxx
January 22, 2012 4:36 pm

That is why I love WUWT, the possibility for open debate!
My word of caution: I have not read all about the Unified theory, but as I see it is taking “importance”, I am getting curious about it.
As a consequence my question may have been answered before, sorry for that if it is the case, a link to the response will be sufficient.
“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.”
Is it undisputable? I always wondered about the impact of the oceans.

KevinK
January 22, 2012 4:39 pm

As formatted by Anthony; Ned Nikolov, Ph.D. and Karl Zeller, Ph.D. wrote;
“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?”
(Note; 0.5% of the atmosphere has less thermal capacity than the Oceans by orders of magnitude, between at least 4-6 orders depending on how deep you consider the Oceans to be for your modeling purposes).
And then;
“The answer obviously is that the observed temperature boost near the surface cannot be possibly due to that atmospheric IR absorption!”
In engineering speak I would translate this into;
Do you really expect me to believe that the miniscule thermal capacity of the “GHGs” are really pulling the massive thermal capacity of the Oceans into thermal equilibrium with said gases ?
Or, in technician speak;
Are you sure that throwing a burning rag onto that ice dam is going to melt it and stop the flooding ?
This is one of those engineering sanity checks that I and others have suggested for several years.
This comment is not a statement of the correctness or incorrectness of the “N&Z” paper. That will become apparent going forward.
This comment is just an observation about the usefulness of sanity checks, also called “gut feel” (i.e. every other time we tried it that way it eventually failed, so we stopped doing it that way), and also called the “TLAR” analysis method. TLAR stands for; “That Looks About Right” and is actually very useful when winding your way through a complex system trying to find out why it is not doing what you expect (i.e. the Earth is not warming as predicted and actually appears to be cooling). It becomes a sort of innate engineer thing, often times a skilled engineer can take one look at a design from a less skilled engineer and immediately point out the problem by seeing the things that don’t look ABOUT right.
No disrespect intended towards scientists, engineers or technicians, we all just think differently, all three professions are necessary.
Cheers, Kevin.

jorgekafkazar
January 22, 2012 4:40 pm

R. Gates says: “Just to add to the notion that the Moon is a very poor gray-body body proxy for Earth– if you took away Earth’s atmosphere and ocean and measured the radiation curve of the Earth’s surface, it would look very different than the Moon’s. This difference is due to the Earth’s internal energy, some of which is emitted at the surface, and is not insubstantial…”
But it is insubstantial. This confirms my suspicion that you don’t read all of these threads. The previous threads had several comments regarding this.

BenAW
January 22, 2012 4:44 pm

I made this comment @Tallblokes
Imo it is absolutely WRONG to use the blackbody approach for waterplanet earth.
It’s base temperature is 275K, not 0K.
Reason of course being the oceans, and no, I’m not assuming any heat exchange between the hot core and the oceans, just radiative balance for planet earth with incoming solar, so no temperature change for the whole system, just internal distribution of heat.
(oceancurrents, windpatterns etc.etc)
The 275K temp of the oceans is probably a left over from higher temps long ago.
I also have some problems with figure 3 in the paper.
The slope in the lines for 0, 60 and 75 latitude during “nighttime” suggests some heat storage capapcity for the moons surface, making it a non-perfect grey body.
The line for “latitude 89 winter” shows imo the effect of “earthshine” on this specific day.
If correct it’s magnitude is far greater than the 2,7K deep space temp you do compensate for.
Needs some elaboration imo.

George E. Smith;
January 22, 2012 4:47 pm

“”””” wayne says:
January 22, 2012 at 3:02 pm
Ned & Karl:
I do see one possible slip in the text. It states:“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: …”; but 0.0001325 is closer to 5 K than 2.72 K stated. Maybe the text should be altered to state that 5.07K is the CMB plus mean star-shine that all interstellar bodies would also receive at all times. That is what I assumed that tiny difference represented. “””””
So what is the basis for claiming that deep space has a Temperature of 2.72 K ?
For starters the 2.72 K number is the effective black body Temperature corresponding to the microwave background radiation that is detected from all directions.. That doesn’t mean there is any material at 2.72 K that emitted that radiation. And let’s not forget the climatists dictum that ordinary gases like interstellar hydrogen cannot emit thermal radiation spectra. Maybe it has more to do with expansion of the universe, than emission from a 2.72 K body of matter.

R. Gates
January 22, 2012 4:49 pm

Tim Folkerts says:
January 22, 2012 at 3:58 pm
R. Gates,
I often agree with you, but not here: “This difference is due to the Earth’s internal energy, some of which is emitted at the surface, and is not insubstantial. ”
The internal energy is mostly “insubstantial”. The flow of geothermal energy is typically estimated as ~ 0.1 W/m^2. Even if this was as high as 1 W/m^2, this is fairly small compared to the 240 W/m^2 from the sun. The moon would be 240 W/m^2, while earth would be 241 W/m^2, which would be a small difference (assuming both have the same albedo). Different albedos between the two would have a much bigger effect than any geothermal energy flows.
_____
Thanks for the feedback. I am wondering how even that 0.1 – 1 W/m2 would change the radiation profile considering the Moon is no doubt much less than this (as it has less internal heat and no convection). I would like to see the actual estimation for Earth’s radiation curve (sans atmosphere and ocean) versus the Moon, as their being very close is a foundational requirement for some of Nikolov and Zeller’s contentions. With the Earth’s outer core at 5000C (with a lot of convection to the surface) and the Moon’s maybe around 850C (with very little or no convection to the surface), I just find it hard to believe that the Moon is a good gray-body proxy for the Earth. And this doesn’t even begin to take into consideration albedo, which, as you point out, could be an even bigger factor in comparing the two as gray-bodies.

PeterGeorge
January 22, 2012 4:54 pm

A modest proprosal for an alternative approach.
1. Suppose that of all the IR radiated to space from all components of the climate system, the amount of increased absorbtion due to increased CO2 is 1%.
2. Now, partly because of the ease of energy transport among elements of the system, but mostly because of humility – it’s too damn complex to do all the accounting right, we assume that EVERY radiating component of the system increases its radiation by 1% to compensate for the absorbtion.
3. In particular, the near surface atmosphere will increase its radiation by 1%.
4. If T0 is the near surface temperature before, and T1 is the temp after, then:
T1^4 / T0^4 = 1.01
T1 = (1.01 * T0^4) ^ 1/4
5. If T0 is about 288K, then T1 = 288.72K
6. Therefore, the increase in near surface air temperature will be 0.72K, if the increase in absorbtion is 1% of the outbound radiation.
7. The total outbound radiation is 240 w/m^2. 1% of that is 2.4 w/m^2. 1% is just a ballpark number. Does anyone know of a reliable calculation of the increase in absorbtion that should result from a doubling of CO2 (i.e. the correct percentage of outbound radiation)?

KevinK
January 22, 2012 4:57 pm

A Physicist wrote;
“NASA asked the question in an engineering context: “Can a huge (> 80%) improvement in thermal insulation [in a rocket stage] be the result of a few grams of reflective plastic foil, that collectively amount to less than 0.0001% of total cryogenic booster mass? …”
I do wonder why they didn’t just wrap the Space Shuttle Fuel tank with a few grams of reflective plastic foil ? Seems they could have avoided that whole foam falling off problem.
The reflective plastic foil (aka MLI; Multi Layer Insulation) only works in a vacuum (sans conduction and convection). That’s why you can’t buy it at your typical neighborhood hardware store. Yes, they do sell some air cell stuff with foil facing, but the foil facing has almost no effect here on Earth. Does anybody remember when fiberglass insulation for your house had an aluminum foil facing (back in the 70’s and 80’s) ? I don’t think they stopped that because of the high price of aluminum foil. They stopped it because; it doesn’t achieve much insulation effectiveness, and if you happen to staple it down and contact a live electrical wire you create a fire and/or electrocution hazard. I seem to remember some recent problems in Australia regarding that.
Cheers, Kevin.

Ned Nikolov
January 22, 2012 5:00 pm

There appears to be a lot of typos and missing little pieces in the HTML version. This is inevitable when one tries to convert such a complex document containing so many math symbols from MS Word into HTML. Hence, we encourage the readers to use the PDF version for a smooth narrative:
http://wattsupwiththat.files.wordpress.com/2012/01/utc_blog_reply_part1-1.pdf
Thank you
Author

Latitude
January 22, 2012 5:01 pm

Thanks for all your hard work putting this together Anthony.

R. Gates
January 22, 2012 5:09 pm

jorgekafkazar says:
January 22, 2012 at 4:40 pm
R. Gates says: “Just to add to the notion that the Moon is a very poor gray-body body proxy for Earth– if you took away Earth’s atmosphere and ocean and measured the radiation curve of the Earth’s surface, it would look very different than the Moon’s. This difference is due to the Earth’s internal energy, some of which is emitted at the surface, and is not insubstantial…”
But it is insubstantial. This confirms my suspicion that you don’t read all of these threads. The previous threads had several comments regarding this.
_____
I would like to see this quantified, as “insubstantial” was admittedly a poor choice of words on my part. Would it make a difference in the kinds radiation profile that each had? Then, combining whatever difference that is, with the differences in albedo, is it still accurate to say the Moon is a good gray-body proxy for Earth? If find it very difficult to believe that the Earth, with an outer core temperature of at least 5000C and lots of convection from there to the surface, wouldn’t have a different enough radiation curve (if measured simply as rocks in space without oceans and atmosphere) from the Moon, to say that the Moon is not a very good gray-body proxy for the Earth. Finally, the 0.1 W/m2 of “average” geothermal energy over the Earth’s surface is quite misleading, as some areas, such as around volcanos and plate boundaries will have much more than this, and considering the Moon has none of these, there is a lot of heat coming from the interior of the Earth and ending up at the surface. The difference in their overall radiation profiles between an essentially dead and inert Moon and a geologically and thermally active Earth would seem to me to be likely something more than insubstantial.

David A
January 22, 2012 5:10 pm

R. Gates says:
January 22, 2012 at 2:30 pm
“The entire conclusion of Nikolov and Zeller rests on the assumption that the Moon is a gray-body equivalent to the Earth, and this assumption is quite erroneous.”
———————————————————————————–
R Gates, not as I read it. It rests on the assumption that the measured mean T of the moon is far less then previousely thought, and then gives reasons why the MEASURED T is different then the predicted. Focus on the measured moon T. Beyond that I think they are saying the earth, by virtue of its 3d atmosphere, has two albedos, and all the TSI which bypasses the clouds, falls on a new albedo that is closer to the moons, so you cannot figure the earths albedo correctly from the total albedo. ( I could well be wrong, just what I got from my read.)
Joel Shore says:
January 22, 2012 at 1:30 pm
Stephen Wilde says:
Joel says the atmosphere gets its energy solely from downward radiation from GHGs. The truth is that it is conduction and convection.
‘No…You are attributing things to me that I have not said. What I have said is that the fact that conduction and convection exists doesn’t get you around the fact that the Earth with its current surface temperature is emitting ~390 W/m^2 and hence that, in the absence of an IR-absorbing atmosphere, the Earth system would be emitting ~150 W/m^2 more than it is absorbed from the sun and would rapidly cool down.”\
——————————————————
Joel, can you think of no way for surface T to conduct to non GHGs, and back conduct energy to the surface? Can surface energy conduct to non GHG stay in the atmosphere longer then radiated surface energy, 50% plus of which leaves at the speed of light? So you have some DWR which by passes non GHG heating the surface, which via conduction then flows to insulating (incapable of radiating to space) non GHGs, staying in the atmosphere longer then radiating GHGs, warming the atmosphere, decreasing the gradient from the ground. This warmed non GHG, in the atmosphere may conduct to another molecue of non GHG, staying in the atmosphere far longer then if it conducted to a GHG where over 50% of the conducted heat can now speed from the atmosphere at light speed?

markus
January 22, 2012 5:18 pm

“The 275K temp of the oceans is probably a left over from higher temps long ago”.
There are 5 independent independent states of matter on this planet.
Flux, Thermodynamic, Dynamic, Potential, Radiative.
They relate to the stratification of the whole of the Earth & Atmosphere in this way;
Stratosphere, Atmosphere, Oceans, Earth, Deep Earth.
It is the opacity of matter that reflects light, the potentiality of it, reflects radiation. It is the ozone of atmosphere that energises the rest.
Two things to know.
1. Ancient energy at the centre of the earth re-rediates upon itself.
2. The potential energy of the earths crust prevents emission of this ancient energy.
So, no there are no leftovers past deep earth, but sometimes it does get re-emitted to space, by release, caused by geological changes.

David A
January 22, 2012 5:20 pm

Tim Folkerts says: January 22, 2012 at 3:58 pm
R. Gates,
I often agree with you, but not here: “This difference is due to the Earth’s internal energy, some of which is emitted at the surface, and is not insubstantial. ”
The internal energy is mostly “insubstantial”. The flow of geothermal energy is typically estimated as ~ 0.1 W/m^2. Even if this was as high as 1 W/m^2, this is fairly small compared to the 240 W/m^2 from the sun. The moon would be 240 W/m^2, while earth would be 241 W/m^2, which would be a small difference (assuming both have the same albedo). Different albedos between the two would have a much bigger effect than any geothermal energy flows.
====================================================
Thanks Tim, is that flow based on land borehole data? Does it include the thinner ocean crust? Does it include volcanic and active geo thermal flows? Does it include oceanic volcanic and geo thermal flows? Does it include the residence time of heat in the ocean depths which may be centuries? How much heat is in the oceans from 500 years of continues flow into the oceans from geo thermal energy?

jorgekafkazar
January 22, 2012 5:23 pm

Nick Stokes says: “…The GHE is that difference between air with and without GHG. Your ATE is something else.”
In my solar system astronomy course in the sixties, we talked about the GHE as due to the presence of an atmosphere containing CO2. The calculation was not based on removing the atmosphere, not just taking out the CO2. But if you really want to slice your definition that thin, how about looking at the AGHE, the amount of warming due to only the greenhouse gases of anthropogenic origin?

Bill Hunter
January 22, 2012 5:28 pm

Nick Stokes says:
January 22, 2012 at 2:50 pm
“This is really nuts, and is being repeated. No-one claims that the atmosphere boosts surface temperature by 18K-33K. Some say that the GHG fraction of the atmosphere has that effect, relative to an atmosphere with no GHG.
But to then say that the 133K is due to a handful of trace gases when you’ve removed the whole atmosphere????”
Technically you are correct at least about what has been claimed by the current climate science establishment. But here comes what has not been said.
This is important information because the initial calculation appears based upon comparison to the moon without an atmosphere. Since establishment climate science can’t seem to relate the earth to Venus except in terms of some completely vague, opaque, and unquantified, lacking a mechanism runaway greenhouse effect; (OKA quackery) that would seem to highly elevate this approach.
So it is a “net” figure! I don’t know what there next step is but as I see it, it should have to have something to do with heat gain and loss calculations using primarily emissivity numbers. Since the temperature of a poorly emissive atmosphere is going to be a lot higher (if it has access to pulse heat of a sufficient magnitude) than a highly emissive PGB. . . .just like you find in building energy loss calculations. In other words the 240watts/m2 now used only applies to a blackbody with no atmosphere, no storage capacity, and with uniform incoming radiation. Our planet has none of those conditions. They have stated and quantified the problems with trying to average temperatures using radiation that has a 4th power relationship to temperature already.
The figure I suspect they will be working with will be the full 1365 watts, possibly modified down in minor ways by some factors (like the conduction vs convection ratio which I think is only about 5%) then applied to greybody properties. All this is common sense to a passive solar designer, assuming I am right of course. Guess we will have to wait and see.

Alan Millar
January 22, 2012 5:38 pm

R. Gates says:
January 22, 2012 at 4:49 pm
Thanks for the feedback. I am wondering how even that 0.1 – 1 W/m2 would change the radiation profile considering the Moon is no doubt much less than this (as it has less internal heat and no convection).
The Moon’s heat flow per sq/m is less than a quarter of the Earth’s.
However, because heat flow, from the interior, is such a small component of the total energy budget, I wouldn’t concern yourself with it.
Alan

January 22, 2012 5:43 pm

Big improvement Ned – now I can understand this part of the N&Z theory.
I agree that what N&Z call the “Atmospheric Thermal Effect” (ATE) should encompass ALL the effects that make a barren planet colder than a similar planet with an Earth-like Atmosphere and Earth-like oceans and rivers and water-based clouds and snow and ice.
However, as I explain below, the N&Z estimate that ATE = 133K is faulty. But, neither do I agree that the 33K of the conventional accounting for the so-called “Greenhouse” Effect (GHE) accounts for the full warming effect of an Earth-like Atmosphere with so-called “Greenhouse” Gases (GHG). The value of the N&Z theory, and of Anthony’s decision to publish it, is that it made it clear to me and I hope others that the conventional 33K accounting is not the whole story of the contribution of an Earth-like Atmosphere to warming.
I think the true ATE lies somewhere between the N&Z 133K and the conventional 33K.
By the conventional accounting (as I wrote here the GHG of 33K is based on comparison of our actual Earth with a totally imaginary Earth that has a non-GHG atmosphere of the same mass as the Atmosphere of the Earth. To achieve a non-GHG condition, that imaginary Earth must lack water, and thus have no clouds or ice. Therefore the surface must be painted to raise the albedo from 0.11 to the 0.3 of our actual Earth. For that imaginary situation, and that alone, 33K may be a good estimate. (However, it may be off by some amount due to the absence on the imaginary Earth of normal atmospheric effects such as precipitation, winds, and storms.)
The N&Z posting illustrates the ATE in their Table 1 by comparison of various latitudes of the Earth and the Moon. They subtract mean temperatures, getting values of 93K, 143K and 158K which average out to 131K which is almost identical to their original 133K.
However, as N&Z acknowledge but fail to properly account for, the Moon rotates at a small fraction (1/28 = 3.6%) of the rate of the Earth. As PeterF noted in his comment above:

…I cannot accept your reasoning for being allowed to ignore rotation of a planetary body. Imagine an infinitely fast rotating body. It would be the equivalent of incoming radiation falling on the whole surface of the planet, and not just half of it. This would result in the “shaded” side having a higher, and the “lit” side a lower temperature. And because of the 4th power law and Hölder’s inequality you would get a higher average temperature….

A barren Earth rotating at the same rate as our actual Earth would have a considerably higher mean temperature than the Moon, because, on the day side, it would absorb and, due to heat capacity, store much of the incoming Solar energy, and release it on the night-side. As a result, the daytime temperatures would be less than on a body rotating at 3.6% of the rate, and the nightime temperatures would be greater, and by the T^4 averaging, the net mean would be higher. Thus, the N&Z Table 1 values are considerably higher than the actual ATE.
I understand that the above N&Z posting is only the first of two parts, and that it does not get into the “gravitational” aspects of the original N&Z paper, which is the most unconventional aspect. I eagerly await the second part.

Bill Hunter
January 22, 2012 5:44 pm

The 1365 watts is the relevant figure to work with.
Note they mention the moon surface soars to 390K. For a .89 emissive PBG thats the equivalent of 1365 watts solar with about 50 watts still be absorbed by storage.

richard verney
January 22, 2012 5:47 pm

Kev-in-UK says:
January 22, 2012 at 1:53 pm
/////////////////////////
Doesn’t it depend upon whether the ‘dark’ side has sufficient time to comduct and/or radiate away the heat that it gained when on the daylight side?
Imagine a planet rotating once every 100 years and then consider what it would be like if instead it rotated at 100rpm. In the latter case, the surface never gets time to cool when on the ‘dark’ side so its average temperature is nearly equivalent to that of the ‘daylight’ side. Very different to the slowly rotating planet.

Bob Fernley-Jones
January 22, 2012 5:48 pm

Anthony,
In your introduction, you said in part:

…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 does seem surprising that two PhD’s in physics could not be aware of Conservation of Energy or between them not have noticed a red flag, whilst at the same time probably having awareness that the establishment would try hard to tear it to shreds.
I seem to remember that their fundamental claim is NOT that gravitational pressure CAUSES heating, but that pressure ENHANCES energy input. As far as I can see, they have not claimed to surmount Conservation of Energy as you claim, but that the surface temperature will be higher as a consequence of the enhancement effect of P, for the same amount of energy input. Others have elaborated this including Richard Courtney.
It will be interesting to see if some queries above on the maths will be resolved, and what Part 2 will say.
BTW, Ira’s analogy using two pressure vessels looks rather faulty. See a detailed more balanced 1-off experiment by Konrad Hartmann that gives some initial empirical data, supporting N&Z here:
http://tallbloke.wordpress.com/2012/01/22/konrad-hartmann-experiment-to-determine-the-effect-of-pressure-on-temperature-in-earths-atmosphere/#more-4431
A work in progress I trust.

A physicist
January 22, 2012 5:59 pm

A Physicist wrote “NASA asked the question in an engineering context: “Can a huge (> 80%) improvement in thermal insulation [in a rocket stage] be the result of a few grams of reflective plastic foil, that collectively amount to less than 0.0001% of total cryogenic booster mass? …”

KevinK asks: I do wonder why they didn’t just wrap the Space Shuttle Fuel tank with a few grams of reflective plastic foil ? Seems they could have avoided that whole foam falling off problem.

Kevin, NASA’s multilayer reflective foil trick works only if there is a vacuum between the sheets, and it just wasn’t practical to provide vacuum insulation for the whole external tank. Here’s a picture of a NASA tank that *does* use multilayer reflective foil: the 650 liter dewar of Gravity Probe B, which kept its liquid helium cold in-orbit for more a year.
Whether the insulating layers are made of metal foil, or whether they’re made of CO2, that multilayer insulation trick works *really* well!

markus
January 22, 2012 6:04 pm

Gentlemen,
Please refrain from the term GHGs, and get back to reality.
The gases known by their names on the periodical table, PLEASE.

jorgekafkazar
January 22, 2012 6:05 pm

R. Gates says: January 22, 2012 at 5:09 pm
I would like to see this [internal energy] quantified, as “insubstantial” was admittedly a poor choice of words on my part.
Then I’ll try to answer your comment. I believe the previous thread had several numbers. I’d track down the links, but I’m short of time. I did look for my early calculations on conducted internal heat, but can’t find them. Sorry.
Would it make a difference in the kinds [of] radiation profile that each had?
Doubtful. The core heat would be combined with geothermal/radioactive heat AND the solar heating, which is orders of magnitude larger than the former. The emission spectrum would follow the surface characteristic emissivity vs. wavelength for the resulting temperature of the rocks, or whatever. Agreed?
Then, combining whatever difference that is, with the differences in albedo, is it still accurate to say the Moon is a good gray-body proxy for Earth? If find it very difficult to believe that the Earth, with an outer core temperature of at least 5000C and lots of convection from there to the surface, wouldn’t have a different enough radiation curve (if measured simply as rocks in space without oceans and atmosphere) from the Moon, to say that the Moon is not a very good gray-body proxy for the Earth.
You may be right, but I think the fairly close numbers used in the post are representative. You’d still be looking at differences in albedo of +/- 0.02 or less. As you well know, this topic of this post is intended to present a simple model of a complex planet, so the numbers need not be nailed down any further than necessary.
Finally, the 0.1 W/m2 of “average” geothermal energy over the Earth’s surface is quite misleading, as some areas, such as around volcanos and plate boundaries will have much more than this, and considering the Moon has none of these, there is a lot of heat coming from the interior of the Earth and ending up at the surface.
Well, yes, volcanic and tectonic areas abound, and you might be able to detect eruptions from space in the IR. There are about 3 million subsea volcanoes and a few of them may present a distinguishable image, though the nature of the signal might be hard to identify. Globally speaking, these are not huge areas. I doubt if they’re extensive enough to affect our estimates of the with-atmosphere albedo.
The difference in their overall radiation profiles between an essentially dead and inert Moon and a geologically and thermally active Earth would seem to me to be likely something more than insubstantial.
More like negligible, but again the differences would be more due to surface type (boulders, schist, mountain sides, dirt, gravel, sand, etc.) than heat source, assuming you mean profile and not map. The major difference is the lack of an ocean. That said, I must agree that we don’t know everything about the earth that we’d like to, and a better estimate of the airless Earth albedo might be helpful.
Best regards, Jorge..

wayne
January 22, 2012 6:13 pm

BenAW, as to your remarks about the ‘latitude 89 winter’, this may help; the very coldest spot at Hermite Crater at 25 K has a radiative equivalent to a mere ~0.022 W/m2 which is not too far off from what one half of the mean earthshine value is at the lunar surface. The 90 K is 3.7 W/m2 and you seem right there, seems to show about 3.5 W/m2 of thermal inertia.
Just realized this, if you have a surface receiving just 1 W/m2 the equivalent temperature would be right at 65 K. That sure highlights the fourth power effects.

richard verney
January 22, 2012 6:13 pm

@R. Gates says:
January 22, 2012 at 5:09 pm
////////////////////////////
I am still thinking about this but although I accept that the mantle temperature (for the sake of a better expression) does not contibute much in radiative terms, I am not convinced that the fact that the Earth is still geologically warm is not an important factor and explaims in part why the Earth and moon are different..
On the moon there is a large diurnal range partly because the moon is geologically dead. When the sun rises, it initially has to heat up the surface from a very cold starting point. However, if the moon was geologically active like the Earth, the dark side would not have been quite so cold and the solar irradiance would not have had to have done so much heavy lifting.
Of course much depends upon the latent heat capacity of the surface in question and its arbsorptive and emissitivity characterisrics. This effect may come to the fore in large areas of damp soil, eg vegetated areas. In these areas there is sufficient conductivity to effectively allow geothernal heat to contribute signicantly to the surface temperature. In some densely packed forest areas, the tree canopy is such that the surface may see all but no sunlight. However, the ground does not freeze.

gnomish
January 22, 2012 6:13 pm

it looks like the point of the exercise is simply to replace an obnoxious meme with anything else at all. if so, fine – let the propaganda wars continue – but if it’s about getting a paradigm shift toward improved conceptualization, then Bart is already at the goal line.
the atmosphere is to be considered a refrigeration system and all the engineering principles that are well known and used apply. that’s the proper identification of its nature.

Editor
January 22, 2012 6:24 pm

Joel Shore says:
January 22, 2012 at 2:56 pm

Willis Eschenbach says:

Ned (or anyone) I think there is an error between equation 4 and equation 5.
In equation 4 you have correctly indicated that when the sun is below the horizon, the value of Ti is zero.
However, in equation 5 you have only integrated that over half of the surface of the sphere, the sunlit half. This is indicated by mu (the cosine of the zenith angle) varying from 0 to 1, rather than -1 to 1.

No, Willis. They have just used the fact that the integral of an integrand that is zero (as it is over the dark half) is zero. I think for the approximation that they are making (i.e., that the local temperature is determined by radiative balance with the local insolation), their calculation is correct.

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.
w.

Editor
January 22, 2012 6:34 pm

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.
w.

Gregory Ludvigsen
January 22, 2012 6:49 pm

One question that puzzles me. The magnetic field and core differ greatly between the earth and the moon. The earth’s is much stonger. It would seem that the earth would generate more energy and heat internally than the moon. This, it would seem would heat the surface of the earth more that the moon. How has this been factored in in calculating the gray body temperature comparison between the eath and the moon?

January 22, 2012 7:07 pm

gnarf:
I had the same difficulty with the integral initially as you’re having, but I may have figured out a way for it to make sense.
Turn the earth on its end so that the sun is shining directly onto the North Pole, placing the whole Southern Hemisphere in niight. If phi is latitude, theta is longitude, and we define mu = sin phi, the radiation intensity at any location in the Northern Hemisphere is S_0 (1 – alpha_0) sin phi = (1 – alpha_0) S_0 mu, and the equivalent temperature is the fourth root of that value divided by epsilon sigma. To get the area-average temperature, integrate the product of that temperature and differential area over the Northern Hemisphere
The differential area is a latitudewise arc R d phi swept through a longitudewise arc R cos phi d theta, where R is the earth’s radius. No loss of generality for present purposes results if we assign R a value of unity, so lose the Rs.
Now convert the integration variable from phi to mu = sin phi: d phi = d mu / sqrt(1-mu^2) and cos phi = sqrt(1 – mu^2). With those substitutions, you simply end up with a constant times mu^(1/4) as the integrand. Unless I’ve made a further mistake myself, that should make it straightforward, with the appropriate integration-limit changes, to reach the result at the end of Equation 5.

Bob Fernley-Jones
January 22, 2012 7:14 pm

jorgekafkazar @ January 22, 6:05 pm
Jorge, you wrote in part in your interchange with R. Gates:

R.Gates: Finally, the 0.1 W/m2 of “average” geothermal energy over the Earth’s surface is quite misleading, as some areas, such as around volcanos and plate boundaries will have much more than this, and considering the Moon has none of these, there is a lot of heat coming from the interior of the Earth and ending up at the surface.
Jorge: Well, yes, volcanic and tectonic areas abound, and you might be able to detect eruptions from space in the IR. There are about 3 million subsea volcanoes and a few of them may present a distinguishable image, though the nature of the signal might be hard to identify. Globally speaking, these are not huge areas. I doubt if they’re extensive enough to affect our estimates of the with-atmosphere albedo.

I hypothesize briefly that the most important and underestimated geothermal consideration is that the oceanic crust is generally much thinner than the continental crust, and since the ocean is a massive very dynamic highly conductive heat sink, a much more rapid heat transfer is undetectable, over ~70% of the Earth’s surface. Not only is the continental crust much thicker but it arguably has lower thermal conductivity because of much layering including limestones and sandstones that contain micro-macro conductive interfaces.

Joel Shore
January 22, 2012 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.)

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.

Edim
January 22, 2012 7:33 pm

Joel Shore says:
“Try as one might, one is not going to explain the fact that the Earth’s surface emits an average of 390 W/m^2 while the Earth + atmosphere absorb an average of 240 W/m^2 without acknowledging that the atmosphere must absorb some of the terrestrial radiation emitted, i.e., that there is a radiative greenhouse effect.”
Here’s the Trenberth’s global energy balance:
http://theinconvenientskeptic.com/wp-content/uploads/2010/11/FT08-Raw.png
Earth’s surface emits in average ~396 W/m2, but it receives ~333 W/m2 from the atmosphere and ~161 W/m2 solar. Net radiative heat transfer at the surface is:
396 – 333 – 161 = -98 W/m2 (downwelling).
There’s no radiative balance, just like there shouldn’t be! Only in outer space, where the only heat transfer is radiative, but not in the atmosphere. To complete the energy balance, non-radiative heat transfer (outgoing) is added (17 + 80 = 97). So, it balances out.
It’s wrong to claim that Earth’s surface emits an average of 390 W/m2, implying that it’s the heat flux of 390 W/m2. Earth’s surface net radiative loss is:
396 – 333 = 63 W/m2 (if the Trenberth’s numbers are correct).

January 22, 2012 7:39 pm

Baa Humbug said @ January 22, 2012 at 3:27 pm

Anthony Watts.
Having just read the foreword, I paused before getting in to the main post because the only thing going through my mind was “what a gentleman Anthony is.
He is obviously hard working and intelligent, but he is also polite, courteous and considerate of others, hence the foreword.”
Anthony, you’re a credit to yourself and your family. I’d like to meet you one day, maybe the next time you come to Australia. I’d like to shake your hand.

Beat you to the handshake Baa 🙂 Anthony even came to deepest, darkest Tasmania, a place few bother to visit unpaid. I would go further and say that Anthony is a credit to the human race.

January 22, 2012 7:47 pm

Your new terminology of ATE itself needs clarifying. In fact, most of what you are discussing seems to actually be GERE (global energy redistribution effect).
Futhermore, GERE is composed of sub-effects — notably:
* GERE-A (global energy redistribution effect due to the Atmosphere),
* GERE-O (global energy redistribution effect due to the Oceans), and
* GERE-R (global energy redistribution effect due to Rotation of the earth).
These are all distinct from GHE, which I will indeed call the Greenhouse Effect. I (and I think most others) consider the GHE to be due only to IR radiation to/from certain gases.
So off the bat, the “Atmospheric Thermal Effect” as defined in the paper includes the non-atmospheric effects of the oceans and rotation, and thus is poorly named and potentially misleading.
——————————————————————
If all the forms of GERE are removed, along with removing the GHE, then we get your average temperature of ~ 155 K for a non-rotating world (assuming the integration is correct — and it seems in the right ballpark). The “effective temperature” that you call T_gb would still be ~ 255 K. So far so good.
Anything that redistributes energy around the world from warm areas to cooler area will lead to a higher average temperature (but a constant T_gb)
The atmosphere achieves this by large-scale convection (eg Hadley Cells)
The oceans achieve this by large-scale currents (eg the Gulf Stream)
Rotation achieves this by turning the cold night side into the light, and turning the warm day side into the darkness.
In principle, GERE could, by itself, raise the surface to an average of 255 K if these effects were strong enough and fast enough to create a uniform global temperature. Of course, in reality, there would always be some temperature differences (the poles will always be colder than the equator; the night side will always be colder than the day side). So in reality, GERE will only get us CLOSE to 255 K.
On the earth, GERE seems to be fairly effective. Rather than the temperature range of about 360 K (2.7 K to 360 K on the hypothetical planet in the paper), we have a range of only ~ 120 K (from ~ -70C to ~ 50 C). And most of the earth is in a much narrower range.
But GERE will never get the planet above an average temperature of T_gb = 255 K. Getting above this temperature requires something other than redistributing energy around the surface of the globe.
One well-known and well-established possibility is the GHE.

January 22, 2012 7:47 pm

Albedo has been mentioned for a body without atmosphere.
I have a problem with albedo because my understanding and the usage by climatic folks seems to differ.
My understanding is albedo has no effect on the temperature assumed by a body in a flux field isolated in space. (excluding zero, time constants etc.). The reason is reciprocity, in and out are the same (same resistance to heat in and to heat out). The effect is the albedo value cancels to unity.
Note: we can have a body painted half white, half black but planets don’t do that, they spin on their axis averaging.
Where am I wrong?

January 22, 2012 7:49 pm

KevinK said @ January 22, 2012 at 4:57 pm

The reflective plastic foil (aka MLI; Multi Layer Insulation) only works in a vacuum (sans conduction and convection). That’s why you can’t buy it at your typical neighborhood hardware store. Yes, they do sell some air cell stuff with foil facing, but the foil facing has almost no effect here on Earth. Does anybody remember when fiberglass insulation for your house had an aluminum foil facing (back in the 70’s and 80’s) ? I don’t think they stopped that because of the high price of aluminum foil. They stopped it because; it doesn’t achieve much insulation effectiveness, and if you happen to staple it down and contact a live electrical wire you create a fire and/or electrocution hazard. I seem to remember some recent problems in Australia regarding that.
Cheers, Kevin.

The problem with foil batts in Australia was down to multilayer foil batts — not foil-faced fibreglass batts. The foil batts consist of two, or three layers of aluminium foil 25mm apart with paper webbing to hold the layers apart. No vacuum reuired. They are widely used in the tropics and sub-tropics. And yes, untrained workers installing them were electrocuted and some caused housefires. All in the name of government mandated Saving the Planet of course.

wayne
January 22, 2012 7:59 pm

gnarf says:
January 22, 2012 at 2:15 pm
The integral of u^0.25/sqrt(1-u^2) from 0 to 1 gives a little bit less than 0.5 (quick numerical approximation).
So what you finally get with a correct substitution (and dividing the integral by 2PI the surface of half a sphere as the integral is made on half a sphere!) is not Tgb=2/5 […]^0.25 but Tgb~1/2[…]^0.25 that is to say 1.6 times your result.
For your example with earth it gives Tgb~1.6*154.7=247.52 K
gnarf says:
January 22, 2012 at 1:53 pm
There is a big problem in the integral when they make the substitution u=cos(theta).
if u=cos(theta) you have to express dtheta using du to make the substitution.
dtheta=-du/sqrt(1-u2)
So after substitution you have something with u^0.25/sqrt(1-u^2) to integrate, and certainly not u^0.25 only!!!
http://www.sosmath.com/calculus/integration/substitution/substitution.html
Sorry, but this is plain terrible.
— — —
Terrible? In fact, the math is correct, no errors there. This has been checked by numerical integration using two different geometries. Maybe the explicit meaning of mu under the radial threw you.

Joel Shore
January 22, 2012 8:03 pm

@Edim ( http://wattsupwiththat.com/2012/01/22/unified-theory-of-climate-reply-to-comments/#comment-872929 ):
Nothing that you show from Trenberth’s diagram contradicts what I said. Trenberth’s diagram is for the actual Earth’s atmosphere where some of the terrestrial radiation is absorbed by the atmosphere (and said atmosphere also radiates). If this were not the case and the Earth’s surface still emitted 390 W/m^2 of radiation from the surface, then all of that radiation would escape to space and the energy balance at the top of the atmosphere would be 240 W/m^2 of solar radiation coming in (and being absorbed, as opposed to the part that is reflected) with 390 W/m^2 of terrestrial radiation going out.
The problem is not getting those numbers to balance with the radiative greenhouse effect…The problem is getting them to balance without the radiative greenhouse effect.

January 22, 2012 8:05 pm

My take on this subject, is that an atmosphere (and ocean) transfers energy from the hotter portion of the surface (the tropics in our case) to the colder portions of the surface, simultaneously cooling the hottest portion and warming the cooler portion. This results in an average surface temperature that is much higher than a planet without the atmosphere physically distributing the energy, primarily due to the T^4 law.
Radiation, especially from the hotter surface portions, is a net loss and does not effectively transport the energy to the colder portions of the globe.
Just a glance at the atmospheric and ocean circulatory patterns demonstrate the energy pathways and mechanisms.

Ned Nikolov
January 22, 2012 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 :
http://www.youtube.com/watch?v=IrSjS0IYZ2M
Science is like any other field – it takes a formal education and many years of practice to reach proficiency. Just like I do not pretend to know farming or accounting, and would not express strong opinions on these topics, folks with no training in physical sciences and math, should have a bit more trust and respect in those, who make a living on it. Also, keep in mind that the only way for us to get out of the current AGW confusion is through solid and clear science, not by creating more confusion!
Please, do not take this post of a sign of arrogance on my part. It is not. I’m only trying to steer the discussion into a more productive mode … Thank you for your understanding!
Respectfully,
– Ned Nikolov

Björn
January 22, 2012 8:28 pm

Willis , Joel is right about the integration and the result given is the correct one as he explains , and the substitution from muy = cos(theta) is both eminently legal and a standard simplification used to get the double integral ( around the equator therfore from 0 to 2*pi for the phi , and then from pole to pole for theta i.e from 0 to Pi) of the area element dA = r^2 * sin(theta)d(theta)d(phi) to find an area of a sphere with radius r, r is a constant that does not affect the calculation here, it can therefore be set to 1 ( earth radii ) and as d(muy)/d(theta) = -sin(theta) it follows that
d(theta) = -d(muy)/sin(theta) and after the substitution and with r=1 the area element becomes
dA = – d(muy)d(phi) , and the lower bound on the inner integral becomes +1 ( theta = cos(0)) and the upper bound becomes -1 (= theta = cos(Pi) ), switching them gets rid of the minus in the -d(muy)d(phi) and the you have the integrand as Ti d(muy)d(theta) , and as the total area of a sphere with radius 1 is 4pi , you divide the total temperature integral with that to get the mean temp per area element. And as Joel pointed out you only have to integrate the from zero to one in the inner integral as Ti is zero when theta lies in the interval from [pi/2,Pi] i.e when muy is in [0,-1].
In other words the math used for getting from equation 4 to eqution 5 ( and equ 6 ) , is solid, and the result of course valid iff equation 4 can stand on its feet all by it self. The argument for it looks sensible but, there are some holes in my knowlegde which have to be mended before I can form passable opinion as to of its validity.
And Gnarf if you read this you see that the missing division by sqrt( 1 – muy^2) ( == sin(theta)) you were going on about really canceled out as a result of the a carefully chosen substitution. I suggest you consult the “Sphere” article in Wikipedia especially the paragraph about how formula for the surface of a sphere can be found using a spherical double integral.

kzeller
January 22, 2012 8:31 pm

The equations we have given you bloggers are simple and they work. 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? The question is how can we possibly have done it – there is no question that our equations work – if you haven’t verified that it works, why haven’t you? What are you all afraid of: the realization that the Earth could be 100% nitrogen or 100% CO2 or 100% naughty vapours of some sort, and using the same surface pressure, would provide for the same average global surface temperature? Why are you all trying to include so-called GH gases; ocean modulations; re-radiations; crusts, your grandma’s bad breath and so on ad nauseam? These are not part of our theory. These parameters & ideas have absolutely nothing to do with the long term average global surface temperatures we are addressing and we’ve proved it with actual data for crying out loud. This is the miracle of our theory and why we called it the UTC? Why aren’t you thinking: “hmmmm, N&Z have given us an equation that lo-and-behold when we plug in the measured pressures and calculate Tgb as they suggest, gives us a calculated Ts that also matches measured values! You can’t disprove the equation? So maybe we are cooking the data books somehow, but how?

David
January 22, 2012 8:33 pm

David A says:
January 22, 2012 at 5:20 pm
Tim Folkerts says: January 22, 2012 at 3:58 pm
R. Gates,
I often agree with you, but not here: “This difference is due to the Earth’s internal energy, some of which is emitted at the surface, and is not insubstantial. ”
The internal energy is mostly “insubstantial”. The flow of geothermal energy is typically estimated as ~ 0.1 W/m^2. Even if this was as high as 1 W/m^2, this is fairly small compared to the 240 W/m^2 from the sun. The moon would be 240 W/m^2, while earth would be 241 W/m^2, which would be a small difference (assuming both have the same albedo). Different albedos between the two would have a much bigger effect than any geothermal energy flows.
====================================================
Thanks Tim, is that flow based on land borehole data? Does it include the thinner ocean crust? Does it include volcanic and active geo thermal flows? Does it include oceanic volcanic and geo thermal flows? Does it include the residence time of heat in the ocean depths which may be centuries? How much heat is in the oceans from 500 years of continues flow into the oceans from geo thermal energy?

David
January 22, 2012 8:44 pm

Joel Shore says:
January 22, 2012 at 1:30 pm
Stephen Wilde says:
Joel says the atmosphere gets its energy solely from downward radiation from GHGs. The truth is that it is conduction and convection.
——————————————————–
‘No…You are attributing things to me that I have not said. What I have said is that the fact that conduction and convection exists doesn’t get you around the fact that the Earth with its current surface temperature is emitting ~390 W/m^2 and hence that, in the absence of an IR-absorbing atmosphere, the Earth system would be emitting ~150 W/m^2 more than it is absorbed from the sun and would rapidly cool down.”\
——————————————————
Joel, can you think of no way for surface T to conduct to non GHGs, and back conduct energy to the surface? Can surface energy conducting to non GHG stay in the atmosphere longer then radiated surface energy, 50% plus of which leaves at the speed of light? So you have some DWR which by passes non GHG, heating the surface, which via conduction then flows to insulating (incapable of radiating to space) non GHGs, staying in the atmosphere longer then radiating GHGs, warming the atmosphere, decreasing the gradient from the ground. This warmed non GHG in the atmosphere may conduct to another molecue of non GHG, staying in the atmosphere far longer then if it conducted to a GHG where over 50% of the conducted heat can now speed from the atmosphere at light speed? It appears logical that for conducted surface T non GHGs create more warming and are a larger heat sink then GHGs.
How is the 390 W/m2 measured? It must be hard to measure the actual surface, verses the air just above the surface. Any answers to my questions are helpful.

R. Gates
January 22, 2012 9:08 pm

Bob Fernley Jones said:
“I hypothesize briefly that the most important and underestimated geothermal consideration is that the oceanic crust is generally much thinner than the continental crust, and since the ocean is a massive very dynamic highly conductive heat sink, a much more rapid heat transfer is undetectable, over ~70% of the Earth’s surface. Not only is the continental crust much thicker but it arguably has lower thermal conductivity because of much layering including limestones and sandstones that contain micro-macro conductive interfaces.”
____
Very interesting idea. Considering how many undersea volcanoes there are, and how much heat could be transferred directly to the deep ocean water through direct heating, which through the THC, would eventually be brought to the surface and atmosphere, it is at least worth thinking about.
My basic point still seems valid in that the much higher internal heat of the Earth versus the Moon,(over 5000C versus 850C) and the much higher convection rates for Earth (or direct conduction of heat in the case of deep ocean water and volcanoes!), makes the Earth and Moon incompatible as gray-body equals. Even without an atmosphere (or ocean) if you did an infrared scan of the Earth and the Moon, you’d see two very different radiation graphs.

January 22, 2012 9:15 pm

I’m getting a lot out of this discussion, mostly sitting back and taking this thread in but wanted to make a few points:
1. No, the moon has a lot of drawbacks as a grey body model of the earth with no atmosphere, but that doesn’t make it useless. The moon does not exhibit the temperature profile one would expect from a straight SB Law calculation, and understanding why is of value.
2. The criticism that the moon has a 28 day cycle of day/night while the earth has a 24 hour cycle of day/night isn’t really complete. The poles for example, have a “night” and a “day” of several MONTHS each. Despite which, they don’t get either as cold or as warm as SB Law would suggest. Obviously air and ocean transports move a lot of energy from tropics to poles, but I’d suggest that effects from GHE would be neglible. While the tropics are busy blasting out 500 w/m2 of radiance, the poles are well under 200 w/m2. So there’s a sparcity of earth radiance to absorb and re-emitt in the first place. On top of that, there is a sparcity of GHG’s. Water vapour declines to nearly nothing at temperatures below freezing, and then there’s that famous ozone “hole”… so whatever moderates temperatures at the poles has a lot less to do with GHE than the rest of the planet.
3. I’ve mentioned heat capacity in several other threads and BenAW has also mentioned this. There is a fair bit of difference between heating something up from scratch and keeping something that is already warm at the same temperature. For example, at the putative 255K, radiance is 240 w/m2, which,over a 24 hour period, equates to 2.88 Kwh. So, at the risk of using the very “averages” I’ve argued against so strenuously, but in the interests of simplicity, if a surface at 255K were to receive 360 w/m2 for 8 hours, and 0 for 16 hours, it would remain in equilibrium because that would be 360*8=2.88 Kwh. But, the average insolation of that exact same surface would be, over the 24 hour time period, only 120 w/m2. This is part of what allows the temperature of the earth surface to be maintained at a temprature higher than would be anticipated by a calculation of average insolation.
dmh

January 22, 2012 9:29 pm

Ned Nikolov;
Science is like any other field – it takes a formal education and many years of practice to reach proficiency. Just like I do not pretend to know farming or accounting, and would not express strong opinions on these topics, folks with no training in physical sciences and math, should have a bit more trust and respect in those, who make a living on it. >>>
Ned, for those of us who have been following the debate in depth for any length of time, that statement is a non starter. Don’t take that the wrong way please, if you’ve been following the various threads you should know that I am one of your biggest chear leaders. But the sad fact is that climate science and the CAGW meme have reached the abysmal state of affairs that they are in today on the backs of horribly flawed and often completely misrepresented “science” churned out in truck loads by legions of researchers with those three letters… PhD… behind their name. We’ve got Keith Briffa PhD publishing 1,000 year temperature reconstructions of earth with 50% of the data coming from a single tree. We’ve got Michael Mann PhD with papers using the Tiljander data series literaly backwards from the rest of the data, not to mention his hockey stick graph drawn by a computer program that draws the same rough graph from almost any mix of data. Then there’s Phil Jones PhD who professes to be unable to draw a graph in Excel, and sees no harm in deleting decades of tree ring data that failed to show the temperature rise he was trying to prove. Kevin Trenberth PhD brags of getting an apology and resignation from the editor of an academic journal for the sin of publishing a paper that used observational data that conflicted with the results of computer models, following which he published a paper claiming that his “missing heat” was being sequestered in the ocean depths despite no instrumentation recording its passing from the surface on down.
We’ve been inundated with PhD’s pawning deeply flawed and sometimes frau)ulent work and demanding it be taken at face value because, well, they’re trained scientists and we’re not.
But thanks for expanding on your original article. Your work is a tour de force and while I expect that there will be plenty of people from both sides of the debate doing their best to pull it down, it will stand for one reason and one reason only.
You guys nailed it!

R. Gates
January 22, 2012 9:30 pm

kzeller said:
What are you all afraid of: the realization that the Earth could be 100% nitrogen or 100% CO2 or 100% naughty vapours of some sort, and using the same surface pressure, would provide for the same average global surface temperature?
____
Don’t know anything about “naughty vapours”, but the basic equations for the absorption and re-emission of LW radiation by greenhouse gases are pretty robust and have faced many decades of very intense scrutiny. Downwelling LW is measured daily all around the world, in all kinds of sky conditions, at a wide variety of latitudes, and the effects are quite quantifiable. Greenhouse gases warm the surface through the absorption and re-emission of LW radiation. Now, might they (and non-greenhouse gases) also warm the surface through surface pressure? Possibly– but then the question become one of quantifying the relative contribution of each. One thing is certain though– the “greenhouse” behavior of the “greenhouse” gases is essential to maintaining a fairly critical amount of the surface temperature, such that, if you were to take the “minor trace gas” of CO2 out of the atmosphere, and replace it with the exact same amount of a non-greenhouse gas, nitrogen for example, the Earth returns to the snowball state in a few decades– and surface pressure won’t change that. You could barely measure the additional nitrogen you’d have added to replace the CO2, but the effects would be quite noticeable. Bottom line: the atmospheric pressure contribution of CO2 to surface pressure is barely measurable, but the effects on surface temperature is quite so because the surface temperature is more a function of LW absorption and re-emission than surface pressure.

Ned Nikolov
January 22, 2012 9:31 pm

One more comment: Please, stop going in circles with this radiative transfer! Contemplate on this (which is a major conclusion from our analysis of observed planetary data): the long-wave (LW) radiation in the atmosphere is a RESULT (a BYPRODUCT if you will) of the atmospheric temperature, NOT a cause for the latter! The atmospheric temperature, in turn, is a function of solar heating and pressure!
The so-called GH effect is a pressure phenomenon, not a radiative phenomenon! That’s because no back radiation can rise the Earth’s surface temperature some 133K above the corresponding no-atmosphere (gray body) temperature. AND yes, the thermal effect of our atmosphere is well over 100K as proven by NASA’s recent observations of Moon surface temperatures.
Also, consider this: the Earth’s albedo is 0.3, while Moon’s albedo is only 0.11. This means that the Moon absorbs 27.1% more solar radiation than Earth (see Eq. 2 in this paper about how to calculate the planet’s mean absorbed radiation). Yet, Earth’s surface is 133 degrees warmer on average that the Moon surface!! Where is that huge enhancement coming from? It comes from pressure through its physical characteristic called FORCE! The kinetic energy of the air is given by the product PV (Pressure x Gas Volume), which is the same as Force Per Unit Area x Gas Volume. Without the force of pressure, there will be no kinetic energy of the atmosphere and no atmospheric temperature (T). On a planetary scale, pressure is a FORCE that is independent of solar heating since it is only a function of the atmospheric mass and gravity. Atmospheric volume, on the other hand, is a function of solar heating, so that the ratio T/V is constant on average! Solar heating can change the temperature through changing gas volume, while pressure can change the temperature through its physical force!
We are NOT claiming that the warming observed over the past 110 years is due to changes in pressure. Not at ALL. This warming, which incidentally began in 1650 is due to a different mechanism – a reduction in global cloud cover by about 1.2%. Earth’s total cloud cover is about 65%. These small changes in cloud cover are due to variation in solar magnetic activity, and they cannot exceed 1.1% – 1.4%. … Let me know if you have any questions …

Dan in Nevada
January 22, 2012 9:33 pm

The Diviner data, which I’d like to hear more about, seems to be rocking the boat. Now, all of a sudden, a GHG-free atmosphere can raise planetary temperature without violating the laws of thermodynamics, but only by 100 degrees. You still need GHGs, though, to get that extra 33 degrees. OR, the earth is different enough from the moon that calculating the theoretical S-B blackbody temperature absolutely can’t be done the same for both, whereas it was earlier explained that most known planetary bodies are similar enough that it doesn’t matter. OR, the earth generates enough geothermal heat to make all the difference. Like Steve McIntyre would say, you really have to keep your eye on the pea.
I’m just being facetious, but I agree with kzeller that hardly anybody is addressing their central point. Those knowledgable enough to do so appear to be saying they are right (so far). Can’t wait for part 2.

Bob Fernley-Jones
January 22, 2012 9:34 pm

wayne @ January 22, 7:59 pm,
Wayne, in your exchange with Gnarf, in part:

[Gnarf:] …Sorry, but this [N&Z integration] is plain terrible.
[ Wayne:] Terrible? In fact, the math is correct, no errors there. This has been checked by numerical integration using two different geometries. Maybe the explicit meaning of mu under the radial threw you…

Apparently in desperation, or plain cussedness, Gnarf and others, (including the font of total wisdom; Willis), have been arguing against the mathematical skills of two PhD’s in physics. I’m confident that quite apart from their own skills, N&Z are aware that their hypothesis is controversial, and that they have had access to second opinions on their maths.
I’m rusty on that stuff after decades away from it, but it seems to me that just because some do not understand the N&Z maths, it does not mean that they are wrong.

George E. Smith;
January 22, 2012 9:39 pm

So if the Temperature of the earth is a consequence of the atmospheric pressure, why does it aapear, and is claimed, that the earth Temperature is steadily increasing; and at an increasingly accelerating (and alarming) rate.
Do the authors have observational data showing that earth’s atmospheric pressure rises and falls with the rise and fall of the observed earth Temperature.
I’m not going to question their equations; nor am I going to question the accuracy of their integrations (the actual mechanics thereof); but any agreement between their calculations, and experimental observations is no proof of causation; and that is the case no matter how closely their calculated results match any observational results. In any case, do we even know what the earth mean Temperature is, in the absence of a sampled data set, that is even close to satisfying the Nyquist sampling theorem for sampled data systems.
And in case anyone thinks that an “accurate” agreement between observation and “theory” is such justification; I’m aware of an agreement between experiment and theory that was within 1/3rd of the standard deviation of the very best experimental measurement; and we are talking of agreement to 8 significant digits for a fundamental Physical Constant; the Fine Structure Constant.
The theory that calculated the fine structure constant to eight significant digits was totally bogus; just simply messing about with numbers; the theory contained absolutely no input at all from the physical universe. Subsequent investigation came up with a list of about 12 numbers derived from simple formulas of the same generic kind; all of which agreed with the fine structure constant to eight significant digits.
So unless the authors can show a “cause and effect” linkage bewteen the atmospheric pressure, and the average surface Temperature; the “Unified” stature of their theory remains unconvincing.
And by the way, I do have the necessities to check their calculations and their integrals; but there is no incentive to do so, since I doubt that their thesis rises or falls on the basis of some simple mathematical mistakes.
So I’ll leave it for the young lions to look for boo-boos; I doubt that mathematics will be the origin of any deficiencies.

January 22, 2012 9:50 pm

Ned Nikolov said @ 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 :

Science is like any other field – it takes a formal education and many years of practice to reach proficiency. Just like I do not pretend to know farming or accounting, and would not express strong opinions on these topics, folks with no training in physical sciences and math, should have a bit more trust and respect in those, who make a living on it. Also, keep in mind that the only way for us to get out of the current AGW confusion is through solid and clear science, not by creating more confusion!
Please, do not take this post of a sign of arrogance on my part. It is not. I’m only trying to steer the discussion into a more productive mode … Thank you for your understanding!
Respectfully,
– Ned Nikolov

I find it somewhat odd that you criticise Willis for being an autodidact and despite trying “to find problems with the math in our papers” not succeeding. Do I take this to mean that there are problems with the mathematics that Willis lacks the skill to find?
You ignore the contributions to the discussion of Robert Brown, De Witt Payne, Joel Shore and others despite that they appear to have the credentials that Willis lacks. You have provided no significant input to the discussion to clear the confusion other than ask for our trust based on your credentials.
You might want to learn a little humility.

January 22, 2012 9:52 pm

OOOPS! I bungled the math in my point 3 above. Too many beers. Night night, I’ll re do it in the AM.

Ned Nikolov
January 22, 2012 9:53 pm

Davidmhoffer (January 22, 2012 at 9:29 pm),
Your point about PhDs is well taken, and I totally agree with you that there have been a number of ‘bad apples’ labeled with those 3 letters. However, since the issue still needs a robust (real) science to be resolved, we cannot expect that such solutions will come from amateurs like Willis who struggle with basic math … That was my point.
Thank you for the high mark you gave our work!

R. Gates
January 22, 2012 10:02 pm

Ned Nikolov says:
January 22, 2012 at 9:31 pm
“The so-called GH effect is a pressure phenomenon, not a radiative phenomenon!”
______
Explain then please, how the higher amounts of downwelling LW, as measured over the Arctic occurs on cloudy nights, if there is no GH effect from the water vapor in the clouds? This is not to say some warming might not also occur from atmospheric pressure itself, but it seems that the actual measurable downwelling LW is being left out– and this can’t be coming from the non-greenhouse gases, so WUWT? Take away all greenhouse gases and replace them with more nitrogen and oxygen, (even though these would be very small additions to the current total of these gases in the atmosphere) and the surface temperature of Earth has a very different (and much lower profile). Surface pressure alone would not prevent another snowball Earth episode. Replace all the CO2 in Venus’ atmosphere with Nitrogen and Venus cools off quickly.

George E. Smith;
January 22, 2012 10:02 pm

The authors claim early in their thesis that the LWIR downward emission from the atmosphere is a consequence of the atmospheric Temperature which in turn per their new theory, is a function of the atmospheric pressure.
So how do the author’s stand on the claim that the atmospheric gases (sans GHGs) do NOT radiate thermal radiation esponsive to the gas Temperature. They seem to be arguing that those ordinary atmospheric gases are and must radiate a themal spectrum. I don’t disagree with that assertion, but I fail to see how the ideal gas law applies to an open system where the volume, the Temperature, and the pressure are all varying quantities.
It seems to me that so long as the total mass (number of molecules) in the earth atmosphere remains fixed, the average pressure is also constant, and Temperature and volume would vary together, as the atmosphere rises and falls due to heating.

Ned Nikolov
January 22, 2012 10:06 pm

George E. Smith:
To answer your question about the cause of recent warming see my post above:
http://wattsupwiththat.com/2012/01/22/unified-theory-of-climate-reply-to-comments/#comment-873000
Also, read our original paper linked the article of this blog.
In regard to the ‘accelerating warming’ claim, it is baseless! All global temperature records (surface and satellite alike) show that the global temperature stopped rising in 1998 – 2000. The temperature trend has been flat for about 12-13 years now. Over the same period, temperatures over the Continental US have been dropping at a rate of -0.42C/decade! So, we have large continental masses cooling for 11 years now. That’s what the actual data show, see:
http://www.climate4you.com/
Yes, we know what’s global temperature is with a fairly high precision (+14.4C). We have a network of surface stations supplemented with satellite observations (which cover the World from wall to wall).

Jenn Oates
January 22, 2012 10:09 pm

I’m poleaxed, and wondering however will I manage to condense this to elevator length for my students. :). The math, it burns!

Bob Fernley-Jones
January 22, 2012 10:10 pm

Edim @ January 22, 7:33 pm
In your exchange with Joel Shore, I’ll comment on your final lines:

…It’s wrong to claim that Earth’s surface emits an average of 390 W/m2, implying that it’s the heat flux of 390 W/m2. Earth’s surface net radiative loss is:
396 – 333 = 63 W/m2 (if the Trenberth’s numbers are correct).

Well yes! For a start, EMR (radiation) is a different form of energy to HEAT. Furthermore, the 396 EMR from the surface is omnidirectional, (= equally in all directions hemispherically), and in an absorptive atmosphere, it cannot all escape to space, or magically get changed to travel only upwards. Most of it approaches the horizontal and is self-cancelling. See my article on this in latest draft @: http://bobfjones.wordpress.com/2011/10/16/studying-the-trenberth-et-al-earths-energy-budget-diagram/
No one has dismantled it yet, on a very similar version at WUWT, attracting 639 comments.

January 22, 2012 10:13 pm

George E. Smith; said @ January 22, 2012 at 9:39 pm

So if the Temperature of the earth is a consequence of the atmospheric pressure, why does it aapear, and is claimed, that the earth Temperature is steadily increasing; and at an increasingly accelerating (and alarming) rate.
Do the authors have observational data showing that earth’s atmospheric pressure rises and falls with the rise and fall of the observed earth Temperature.

One early commenter remarked that the Pterosaurs (flying “dinosaurs”) could only have flown in a much denser atmosphere than that of the present day and that N&Z’s hypothesis explained this higher temperature/higher density atmospheric relation. Warren Carey of expanding Earth fame claimed that gravity was very much less in those days and that explained their ability to fly. Since he was a fully qualified geologist and the founder of the school of geology I attended in 2003, perhaps I should take Ned Nikolov’s advice and defer to his fully qualified judgement 😉

Ned Nikolov
January 22, 2012 10:15 pm

George E. Smith:
We call our Theory ‘Unified‘ not because it explains all temperature changes with pressure, but because it proposes a hierarchical framework of climate drives. Some drivers are identified as a result of our research while others have been studied by other scientists. Look at Figure 10 in our original paper here:
http://tallbloke.files.wordpress.com/2011/12/unified_theory_of_climate_poster_nikolov_zeller.pdf
It displays the hierarchy of climate drivers and their time scale of operation according to our Unified Theory. Our theory really builds on a large body of previous research!

jimmi_the_dalek
January 22, 2012 10:18 pm

I think this thread needs a bit of humour, so I applaud kzeller’s nicely sarcastic parody at 8:21pm

Kasuha
January 22, 2012 10:27 pm

The very first thing I dislike on this matter is the name, “Unified Theory of Climate”. Sorry but it’s not unified, it doesn’t qualify as theory (at least not yet) and it’s not about climate. It’s just reassessment of effect of atmosphere on surface temperatures.
Further on there are things just placed there, such as claim about “undisputable fact that the atmosphere provides extra warmth to the surface of Earth compared to an airless environment such as on the Moon” without defining what’s meant under “warmth”. Surface of Moon directly facing Sun is way warmer than corresponding surface of Earth so without proper definition what you mean by it, there’s a lot to dispute.
Also, Moon is far from ideal “gray body” as its temperature is also affected by thermal capacity of its surface material (otherwise the night side’s temperature would be 3K) and I don’t think it’s correct to assume thermal capacity of watery surface on Earth can be taken equal to thermal capacity of Moon rock.
It’s hard to read such paper as it resembles crackpot theories too much from the very beginning and one must believe the fact that these imperfections won’t have negative effect on the result.
And if there is one thing I’m missing, then it’s the assessment of “runaway state”. I can accept the atmospheric mass can affect temperature gradient near surface and Venus is hot because it’s got way heavier atmosphere, but if you imagine our oceans boiled away, that’d be an awful lot of mass that’d go to the atmosphere. It’d be nice to see, with your own methods, what surface temperature would that cause – particularly whether that’d be more than 100 C (stable state) or less (unstable state).

Surfer Dave
January 22, 2012 10:31 pm

The moon is geothermally cold isn’t it? No vulcanism?
The earth is geothermally hot, isn’t it? What is the heat flux? Why is there a linear relationship between depth and temperature that seems to be constant around the planet below the surface layer of 10m to 20m?
I don’t think an atmosphereless Earth with volcanic raging core can be compared to the volcanically cold Moon.
The SMU site says continental USA averages a geothermal flux of 250mW-2 with many areas well over that, and it is not an even distribution so the mean may be completely misleading. The Rockies have vast areas of well over 1Wm-2 with the outliers being upto 15Wm-2.
Sure, ignore that. We have no idea what the distribution of the global geothermal heat flux is nor how it has varied over time. We don’t even really know the magnitudes of the many processes that contribute to that heat. It *is* the 800-pound gorilla.

Werner Brozek
January 22, 2012 10:36 pm

“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.”
I just have some comments with regards to this point. While it is true that even in the best of cases, 99% of the energy may be successfully transformed to different types of energy, then 1% is still lost as waste heat. But in this case, there really is no waste heat in that sense since all “waste heat” becomes part of the so called ATE.
Are there other sources of energy that amount to anything? As we presume, compressing a gas causes it to heat up, but then the heat dissipates. However what about the time within 3 or 4 days of a full moon? We know the moon affects tides on earth. So is it possible the full moon very slightly compresses gases on its opposite side while the sun heats up the slightly compressed gases and thereby creates some additional heat? And is it also possible that the sun very slightly compresses the gases on the side of the full moon and that some heat from the reflected light of the full moon causes some heating on the side away from the sun? My guess is that these affects are totally negligible, but I am not sure.

markus
January 22, 2012 10:52 pm

Hi Ned,
Love your work.
Is their a application for The Unified Theory of Climate, in a pressured enclosure above photovoltaic cells?
Regards,

Rosco
January 22, 2012 11:05 pm

Who cares about averages ?? There is no demonstrated mechanism whereby the Sun’s energy illuminating a body the size of a planet can result in some meaningful average – it is either receiving more net energy during the day or losing more during the night.
The more of these theoretical analyses I see the more ludicrous they appear – to me at least.
The biggest hole in all this averaging nonsense is the unarguable fact that the daytime temperature on the airless Moon hits ~396 K (123 C) while the maximum temperature recorded on Earth is of the order of 323 K (50 C) yet both are subjected to a similar solar radiative flux albeit they have a different albedo.
In the light of this our atmosphere is obviously NOT enhancing the solar radiation but is doing the opposite by convecting the energy away from the surface and allowing it to be distributed over the planet and ultimately escape to space – the Earth’s surface never approaches its “blackbody” temperature for a radiative flux of 70 % of the Solar constant = ~360 K, whereas the moon reaches its “blackbody” temperature.
At night the atmosphere cools – moderated by the thermal mass of the oceans. If the nighttime of the Earth was longer than it is the Earth could cool to much lower levels.
The atmosphere would not have sufficient retained energy to prevent serious cooling during nights of the length of lunar nights.
I do not believe the atmosphere enhances the temperature on Earth during the day – I believe it is the opposite. At night the atmosphere and the oceans moderate the heat loss until the cycle begins over again.
Gases or water/water vapour do not generate energy – to imply they do is silly as it could easily be demonstrated by experiment.

markus
January 23, 2012 12:20 am

Ned,
Could a three dimensional matrix, of all indices, based on incoming watts, explain your theory

Greg Elliott
January 23, 2012 12:20 am

What we are taught in schools and universities is based on one side of a 150 year old dispute that has never been experimentally resolved. We learn one side, but not the other.
It was not known at the time of Loschmidt, Boltzmann and Maxwell that gravity acted on both mass and energy. Newton’s Law only applied to mass. Thus there was no mechanism to explain how gravity could separate matter by temperature.
However, Einstein showed that gravity applies to mass, energy and time. Thus the bending of light near stars, the prediction of black holes and the confirmation of time dilation.
There is no good reason to believe that gravity doesn’t apply to temperature. It most certainly will affect IR radiation from GHG. That is clearly demonstrated by light, so why assume that gravity cannot affect heat and thereby temperature?
For example, hotter molecules have higher energy levels and thus could well have greater gravitational attraction than colder molecules, according to e=mc2. A force of this size would have been near impossible to detect with the crude instrumentation of 150 years ago.
Gravity remains the least understood force in nature. We don’t know the cause or the force carrier. We don’t even have a good handle on the speed. About the only thing that gravity doesn’t affect is gravity, which suggest it may not be a force at all.

BenAW
January 23, 2012 12:26 am

markus says:
January 22, 2012 at 5:18 pm
“The 275K temp of the oceans is probably a left over from higher temps long ago”.
I think it’s safe to assume they have been warmer in the distant past.
(since their creation, or since the last major meteor impact)
Presently they have a certain temperature, and it isn’t changing much. (275K)
By just assuming thermal balance with the hot core through the crust
(no heat flow from core to oceans) and balance with the atmosphere they will
maintain that temperature.
The discussed ATE and the GHE use blackbody calculations to arrive at a radiative temp.
for earth of 154K and 255K resp. and then try to explain why the earth is actually at 288K.
The 154K and 255K is caused by solar radiation already!!
Now enter the oceans, 275K WITHOUT solar influence.
The difference between 275K and 288K can be explained by solar radiation, ATE, and GHE.
Imo solar radiation is the major player here

wayne
January 23, 2012 12:33 am

@ Bob Fernley-Jones: January 22, 2012 at 9:34 pm
Hi Bob. Oh how so true you are. This last few weeks has really opened some eyes, I just can’t believe what has been going on, I see some colors have changed.
Partially my mind it on your last post… wanting to get a bit deeper in the geometry of radiation but right now I can’t help but keep digging on deeper into N&Z’s absolutely earth shaking theory. I mean, how does nature do that you see in their Figure 7? I’m reserving a place in the base for your horizontal aspect. I think that partially kicked started this. Your post became troll city till you shut it down. You can tell every time how close you are to something.
Got a minute? Let me see if I can lay out what we know so far, the core parameters.
It definitely has to do with pressure. The pressure is governed by the weight of the mean atoms or molecules, how many there are in the atmosphere, that defines the mass of the atmosphere. Then you have the mass of the body and its radius, which sets the gravitational acceleration at the surface. Now knowing those two, the mass and acceleration, you can calculate the surface pressure. At this point you know the specific heat capacity from the gas components and properties. Next you need the TSI which can come from the solar effective temperature, the suns radius, and the body’s orbits semi-major axis. Standard stuff.
From here on there is some speculation. Pressure is a constant but the density is not and looking at the U.S. Standard Atmosphere and dividing the pressure at each level with the density at that level you will notice this ratio P/μ goes smaller with every increase in altitude. But, if you take that ratio and multiply it by the ratio of the molecular mass per mole by the gas constant (M/R), poof, you have the temperature at each and every level. See, the last ratio is a constant so P/μ strictly sets the temperature and that is a lapse rate, the environmental one. That also means the density at the surface is controlled by the surface temperature, or, the surface temperature is controlled by the density, or, each could affect the other.
So what are we missing? Radiation. All energy enters and by it and this is where your horizontal aspect might have a place though it will be buried in a term named optical thickness or depth.
Both Ned Nikolov and Karl Zeller are probably shaking their heads. Why in the world did not one single person who can spin out some 6000 comments notice the ratio 287.6/154.7 = 1.86… I couldn’t believe it when I saw it. Could this be somehow be related physically to the Miskolczi tau? N&K is the temperature ratio: without atmosphere to one with an atmosphere. Miskolczi is the ratio of radiation at the surface (without atm) to how thick the atmosphere is to retard radiation through it. N&Z are temperature ratios, Miskolczi is speaking of radiation ratios when taken as an exponent of the negative. Surely a coincidence but I’m not letting go quite yet.
His figure is just a tiny bit off, that figure is tau, the LW optical thickness and you can get some meaning from it my transforming it into the fractional portion that can actually pass more or less vertically from the surface and on into space. exp(-1.86) gives 0.1557. I don’t remember Miskolczi’s tau but I remember the fraction –ln(0.1546) = 1.867. Working Miskolczi’s backward you get a surface temperature, evidently an upper limit of 289.1 K. Suspicious isn’t it? Haven’t been able to nail down it’s reality (if it is).
Oh, the ~0.1547, that’s Trenberth’s 396 * 0.1547 or 61 W/m2, the NET that leaves the surface. Sort’a look familiar?
Maybe these two papers are mates? This just keeps getting better and better. I have even more but it’s getting long, if I left something out so far, stick it in or correct it.

BenAW
January 23, 2012 12:35 am

Quick reality check for the calculation of the blackbody temp for earth:
1364 W/m^2 incoming, spread over half a sphere, 30% reflection:
1364*0,7/2 = 477W/m^2 SB> 303K
Other half at 3K makes the total average temp 153K. Close enough imo.

January 23, 2012 12:45 am

Joel Shore says:
“Trenberth’s diagram is for the actual Earth’s atmosphere where some of the terrestrial radiation is absorbed by the atmosphere (and said atmosphere also radiates). If this were not the case and the Earth’s surface still emitted 390 W/m^2 of radiation from the surface, then all of that radiation would escape to space and the energy balance at the top of the atmosphere would be 240 W/m^2 of solar radiation coming in (and being absorbed, as opposed to the part that is reflected) with 390 W/m^2 of terrestrial radiation going out.
The problem is not getting those numbers to balance with the radiative greenhouse effect…The problem is getting them to balance without the radiative greenhouse effect.”

Joel, I understand it’s hard for you to get your head around this, but consider Ned’s statement that:
“the long-wave (LW) radiation in the atmosphere is a RESULT (a BYPRODUCT if you will) of the atmospheric temperature, NOT a cause for the latter! The atmospheric temperature, in turn, is a function of solar heating and pressure!
The so-called GH effect is a pressure phenomenon, not a radiative phenomenon! That’s because no back radiation can rise the Earth’s surface temperature some 133K above the corresponding no-atmosphere (gray body) temperature. AND yes, the thermal effect of our atmosphere is well over 100K as proven by NASA’s recent observations of Moon surface temperatures.”

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!
The radiation measured by AERI and other such devices is the radiation buzzing around between the molecules in the air. The air is denser near the surface, which is why we see 390 squiggles per square metre whizzing about just above the surface there. Up at 7km or thereabouts on average, where the air is less dense and there are fewer molecules per cubic cm, we see around 240 squiggles per square metre whizzing around. This fails to surprise me.
As we all know, there is plenty of convection and evaporation and condensation leading to latent heat release going on in the troposphere, such that radiation is not required as a shifter of heat there. It just buzzes around doing its buzzy thing. Above the troposphere, those radiatively hyper-active water vapour and co2 molecules do a sterling job shifting heat back into space for us so we stay cool here on the surface. Has there really been a change in the effective radiating height in the last 40 years? Got any data on that?
Empirical data from pyrheliometers gathered and studied by Doug Hoyt and others show no overall change in the opacity of the atmosphere for 70 years or so during the C20th. It’s a real result which has been ignored for too long IMO.
Stay cool Joel.

gnarf
January 23, 2012 12:53 am

I was wrong your integral calculation is OK.
Could you please explain why you consider Ti=0 on the dark side of the earth, while it is measured at 90-100K on the moon?
You consider all energy received is immediately radiated? Nothing stored? It does not fit with the moon, why would it fit with the earth?
It seems you state that heat capacity does not change the average temperature of a planet. That is not true. Introducing heat capacity does not change the overall radiative flow there is no added energy, but it strongly changes the average temperature. Less radiation is emitted on the bright side as a part of incoming radiation is stored as non-radiated heat…and more radiation is emitted on the dark side as this heat is progressively released as radiation . As the temperature is a power 1/4 of radiation, the average changes strongly because of this. Increasing heat capacity increases average temperature.
You definitely have to introduce the heat capacity of ground…it makes the result vary from 157K to 250K and above.

Editor
January 23, 2012 12:53 am

Zac says:
January 22, 2012 at 2:03 pm

I am not surprised that Anthony does not support this paper given the prominance and support he gave to Willis Eschenbach’s outburst. But to deny the laws of physics seems a daft tad to me.

Thanks, Zac. If you were to mention which laws of physics you think Anthony is denying it would assist us all in evaluating your claim.
w.

Dr Burns
January 23, 2012 1:24 am

Ned,
For those without neither the time, inclination nor background to wade through your math, as an elevator presentation, what do you suggest is the relative warming effect of the Earth’s IR-absorbing gases, compared to your atmospheric thermal effect ?

January 23, 2012 1:34 am

Joe Born says:
January 22, 2012 at 7:07 pm
gnarf:
I had the same difficulty with the integral initially as you’re having, but I may have figured out a way for it to make sense.
Turn the earth on its end so that the sun is shining directly onto the North Pole, placing the whole Southern Hemisphere in night. If phi is latitude, theta is longitude, and we define mu = sin phi, the radiation intensity at any location in the Northern Hemisphere is S_0 (1 – alpha_0) sin phi = (1 – alpha_0) S_0 mu, and the equivalent temperature is the fourth root of that value divided by epsilon sigma. To get the area-average temperature, integrate the product of that temperature and differential area over the Northern Hemisphere
The differential area is a latitudewise arc R d phi swept through a longitudewise arc R cos phi d theta, where R is the earth’s radius. No loss of generality for present purposes results if we assign R a value of unity, so lose the Rs.
Now convert the integration variable from phi to mu = sin phi: d phi = d mu / sqrt(1-mu^2) and cos phi = sqrt(1 – mu^2). With those substitutions, you simply end up with a constant times mu^(1/4) as the integrand. Unless I’ve made a further mistake myself, that should make it straightforward, with the appropriate integration-limit changes, to reach the result at the end of Equation 5.

That looks like an efficient and smart way to restate and resolve the issue to me Joe. Thanks for helping me understand the way the substitution works in the integral. I need to be able to visualise those kinds of maths puzzles in order to understand them post-accident, and your explanation did it for me. Geometry rules!
Thanks again.
TB.

markus
January 23, 2012 1:40 am

Mr Eschenbach,
I think it’s coming down to E-Mc2.
Markus.

David Blake
January 23, 2012 2:04 am

Some fascinating stuff in the paper, which I am not qualified to comment on. The moon temperature data seems to be a smoking gun of some sorts.
Anthony; a suggestion regarding HTML formatting. The document looks to have been written in LaTex. This is a popular typestting format in academia as it gives superior math output. Next time ask the author to provide the LaTex (or TEX) document (rather than the .pdf output), and you can then translate the LaTex to HTML with one of these tools : http://enc.com.au/docs/latexhtml/
It’ll save you a lot of typesetting problems.
Looking forward to the next discussion on it.

markus
January 23, 2012 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.

Steve Richards
January 23, 2012 2:09 am

Greg Elliott says:
January 23, 2012 at 12:20 am
What we are taught in schools and universities is based on one side of a 150 year old dispute that has never been experimentally resolved. We learn one side, but not the other.
Here is a paper that documents an experiment that appears to ‘prove’ the Nikolov & Karl Zeller theory:
http://tallbloke.files.wordpress.com/2012/01/graeff1.pdf

A physicist
January 23, 2012 2:18 am

Zac concludes:  … to deny the laws of physics seems a daft tad to me.

That what I’ve been saying all along: the assertions of the concluding section of Nikolov and Zeller “Part I” are daft in precisely the sense that Zac describes.
As for detailed comparisons of earth-versus-moon surface temperature … it’s daft too to conceive a simple model that describes the moon’s surface temperature (no-atmosphere and no-surface-water and no-icecaps and a long 655-hour sidereal rotation period) and then expect that same model to describe the earth (thick-and-flowing atmosphere, thick-and-flowing oceans, icecaps, and a short 24-hour sidereal rotation period).

John Marshall
January 23, 2012 2:27 am

Many thanks for the expansion.
I dislike the GHG theory because:-
1. The forecast troposphere heat anomaly is not there.
2. The reradiated heat has to come from a cooler area than the surface which violates the 2nd law.
3. GHG’s are said to ‘store’ heat which again violates 2nd law.
Your theory is basically that of adiabatic compressive heating which is an accepted principle and can be experienced. It certainly does not conflict with the 1st law.
Convective cooling can also increase heat at the surface as any body of rising air must displace upper air which must descend to the surface. The rising air will be saturated air and cool at the SALR of 5C/Km rise whilst the descending air will warm up at the DALR of 9.5C/Km descent, rather like a vertically looped Foehn wind.
My question to GHGers is always ‘why does Jupiter with its H2/He atmosphere emit more energy than it receives?’ Without compressive heat this would not happen. I am told that this does not have any baring on the question of Earth which is complete rubbish since all planetary atmospheres must obey the same laws of physics.

Schrodinger's Cat
January 23, 2012 2:29 am

The bottom line is that this work claims (1) that the SB law has been incorrectly applied to the moon and the earth and as a consequence (2) the current GHG effect underestimates the extra atmospheric warmth of the earth by about 100K. This is what we should be debating.
Many commenters are introducing all sorts of alternative sources of warming and cooling which are not relevant.
If the authors are correct about issues (1) and (2) then that seriously questions the credibility of current climate science and that is before we hear about part 2 of their theory.
So far there seems to be emerging agreement that their calculations are correct.

gbaikie
January 23, 2012 2:34 am

“wayne says:
January 22, 2012 at 6:13 pm
BenAW, as to your remarks about the ‘latitude 89 winter’, this may help; the very coldest spot at Hermite Crater at 25 K has a radiative equivalent to a mere ~0.022 W/m2 which is not too far off from what one half of the mean earthshine value is at the lunar surface. ”
This crater never receives sunlight or earthshine- hasn’t for millions of years.
I would guess a large portion of ~0.022 W/m2 is internal heat from the Moon
“The 90 K is 3.7 W/m2 and you seem right there, seems to show about 3.5 W/m2 of thermal inertia.
Just realized this, if you have a surface receiving just 1 W/m2 the equivalent temperature would be right at 65 K. That sure highlights the fourth power effects.”
That is interesting.
If a airless planet’s internal heat was extremely high, say 5.67 watts per square meter, surface temperature would be 100 K.
If planet was at earth distance in which it received receiving 400 watts per square meter.
Would we divide 400 watts per square meter by 4 giving average of 100 watts square meter?
Giving us average temperature of 5.67 K
And what would the surface temperature of the planet be if it had internal heat of 5.67 watts per square meter and had solar flux 400 watts per square meter?
How much energy would this planet receive from sunlight.
And measure the amount energy coming from this planet, what energy budget look like?
Would be average amount energy leaving the planet be 100 watts per square meter?
Indicating that planet was absorbing 100 watts per meter?
Let’s move the planet nearer the sun.
Now it’s receiving solar flux of 800 watts per square meter. The average would 200 watts per square meter. Which gives average temperature of 90.7 K
So it would be said that the planet is absorbing average of 200 per square meter square
and was emitting an average of 200 watts per square meter.
The internal heat is making planet 100 K.
The sun without such an internally warmed planet would apparently make the planet have average temperature of 90.7 K.
But since it is such internally warmed planet, what would it’s temperature be?
Any alarm bells ringing in anyone’s head?
Yet.

Gras Albert
January 23, 2012 2:36 am

NikolovHence, the evidence suggests that the lower troposphere contains much more kinetic energy than radiative transfer alone can account for!
Indeed!, some of that kinetic energy is derived because the planet is rotating, consider a line drawn perpendicular to the surface and extended through the stratosphere. At any instant molecule A at 50km, molecule B at 10km and molecule C at the surface sit on that line, but the surface is rotating at 7.272 x 10-5 radians/sec (0.278 miles/second, 1000mph), does the atmosphere at 50km rotate with the same angular velocity? No it does not, just as convection triggered thermal bubbles never attain the horizontal velocity of the air mass in which they rise so there is an angular velocity gradient with increasing altitude. Where is the consideration in the energy budget of the effect on the lower troposphere of the constant movement of continental mountain ranges through an atmosphere which rotates at some fraction of the surface angular velocity?
There is another atmospheric energy transfer system in regular operation which is never mentioned in Consensus Climate Theory, atmospheric (lee) waves, lee waves reach from the surface to the upper stratosphere, they extend hundreds of kilometres from their source and have been measured with vertical velocities exceeding 20m/s (40kts, 4000ft/min). Their capacity to do work dwarfs that of thunderstorms, with appropriate atmospheric conditions they operate 24/7 sometimes for several days continuously, the Maori label for the islands of New Zealand, the land of the long white cloud is highly appropriate, consider the work done to create a lenticular cloud system 1000km long and several km high!
I applaud Nikolov et al for thinking outside the box, I challenge Joel Shore to start the same process

HLx
January 23, 2012 2:55 am

Refraction of thermal(LW) radiation.
Hi! Can someone answer a question from me.
For radio waves we get a refraction as a result of the different densities (different n) in the atmosphere. Waves traveling normal to the ground, do not refract. But with ever increasing angle to the normal, a larger refraction is observed. (This refraction is also observable for visible light, as our sky is blue).
A point on a body radiates heat in all directions (0 to 90 degrees of the normal). One could believe the higher the angle, the higher the thermal refraction.This should help keep a noticeable amount of LW radiation, thus increasing the temperature, of the ground (and atmosphere).
My question:
1. –> Does anyone take this into account?
(this would make a planet with atmosphere hotter)
Also:
2. –> How much does this effect account for?

markus
January 23, 2012 3:24 am

And for the other views out there;
Heat is a manifestation of kinetic energy which cannot penetrate the mass of earth.
We cannot add extra heat to our planet by any direction of radiation.
Markus.

A. C. Osborn
January 23, 2012 3:35 am

A physicist says: January 22, 2012 at 5:59 pm says “Whether the insulating layers are made of metal foil, or whether they’re made of CO2, that multilayer insulation trick works *really* well!”
There is agreat deal of difference between a Mirror reflecting radiation away, it is using an Albedo of nearly 100% and CO2 molecules absorbing a very small percentage of outgoing radiation.

AusieDan
January 23, 2012 3:57 am

I think that it would helpful at this stage to take up Willis’ challange and attempt a brief elevator summary.
Start:
What N&Z are claiming is that maximum annual temperature at the surface is a function of both distance from the sun (irradiance) plus atmospheric pressure at the surface.
End of summary.
That is what we should be debating.
All the detail about maths and so forth can be left until this essential mattter is settled.
Unlike many here, I have not been delving too much into mathematical theory or physics.
Instead, I have been doing lots and lots of calculating and charting.
I urge you all to do likewise.
What you will find is that N&Z theory is robust.
You can make all the changes that you like, but you will find that Tgb remains a function of irradiance, Nte canbe expressed as a function of atmospheric pressure that works for at least eight solar bodies and Ts (equal to Tgb * Nte) is almost exactly equal to the observed temperature.
OK, now lets start debating the Unified Theory.

AusieDan
January 23, 2012 4:02 am

Perhaps I should have added that there is a body of mostly informal experiments (but some published ones as well) that demonstrate quite clealy that all gases can be heated.
Some of these make quite plainly explicit that, given equal pressure, the increase in temperature of CO2 rises to exactly the same temperature as normal air, when acted on by an equal heat source.
So let’s once and for all, forget this nonesense about greenhouse gases.

Stephen Wilde
January 23, 2012 4:08 am

The consensus seems to be developing that N & Z are correct and the radiative theory of the Greenhouse Effect is unravelling.
Quite right too.
It was always an unproved and unlikely speculation in light of the 150 year old known physics of the Gas Laws.
The next step is to decide what happens to the extra energy in the air from GHGs INSTEAD of raising the system temperature.
That leads to a true Unification Theory of Climate and I have been banging on about it here and elsewhere for four years now.
Sinply put, there is just a miniscule surface air pressure redistribution.
Climate change and weather is simply the negative response of the climate system to ANY factor that seeks to disrupt N & Zs ATE (formerly known as the Adiabatic Lapse Rate).

David
January 23, 2012 4:12 am

Other half at 3K makes the total average temp 153K. Close enough imo.
tallbloke says:
January 23, 2012 at 12:45 am
Joel Shore says:
“Trenberth’s diagram is for the actual ……
Joel, I understand it’s hard for you to get your head around this, but consider Ned’s statement that:
“the long-wave (LW) radiation in the atmosphere is a RESULT (a BYPRODUCT if you will) of the atmospheric temperature, NOT a cause for the latter! The atmospheric temperature, in turn, is a function of solar heating and pressure!
The so-called GH effect is a pressure phenomenon, not a radiative phenomenon! That’s because no back radiation can rise the Earth’s surface temperature some 133K above the corresponding no-atmosphere (gray body) temperature. AND yes, the thermal effect of our atmosphere is well over 100K as proven by NASA’s recent observations of Moon surface temperatures.”
————————
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!
The radiation measured by AERI and other such devices is the radiation buzzing around between the molecules in the air. The air is denser near the surface, which is why we see 390 squiggles per square metre whizzing about just above the surface there. Up at 7km or thereabouts on average, where the air is less dense and there are fewer molecules per cubic cm, we see around 240 squiggles per square metre whizzing around. This fails to surprise me.
===========================
Tallbloke, this “The radiation measured by AERI and other such devices is the radiation buzzing around between the molecules in the air. The air is denser near the surface, which is why we see 390 squiggles per square metre whizzing about just above the surface there.””
I have asked about this many times, with no real answers. At what point are we measuring the radiation and T from the earth ocean mass, and the land, verses the atmospheric radiation just above it. ( And, if there is heat in non GHGs, just above the surface, we are missing that conducted heat when we only quantify T via LWIR measurements) It makes perfect sense to me that the denser the gas, due to either more atmosphere, of more gravitational force, there will be a larger heat pool capable of holding more energy per m2. It does not make sense to me that all atmospheres, regardless of composition will have the same T as long as pressure, gravity, TSI and albedo are the same. Yet I also see non GHGs as insulators, where the conducted energy is held far longer then it would be by a GHG.
I asked Joel Shore the following, but alas, he apparently did not deam my questions worthy of response. Joel, can you think of no way for surface T to conduct to non GHGs, and back conduct energy to the surface? Can surface energy conducting to non GHG stay in the atmosphere longer then radiated surface energy, 50% plus of which leaves at the speed of light? So you have some DWR which by passes non GHG, heating the surface, which via conduction then flows to insulating (incapable of radiating to space) non GHGs, staying in the atmosphere longer then radiating GHGs, warming the atmosphere, decreasing the gradient from the ground. This warmed non GHG in the atmosphere may conduct to another molecue of non GHG, staying in the atmosphere far longer then if it conducted to a GHG where over 50% of the conducted heat can now speed from the atmosphere at light speed? It appears logical that for conducted surface energy, non GHGs create more warming and are a larger heat sink then GHGs.

AusieDan
January 23, 2012 4:19 am

Having time to consider the above, I realise that I might not have made myself quite clear.
Perhaps I should have expressed myself thus:
Within reasonably wide tolerances, no matter how you change the equation for Tgb to account for your view of the applicable physics and maths, there is always a function of pressure (Nte) which when multiplied with your version of Tgb, which will come very close to the actual observed annual temperatures of at least eight solar bodies.
In other words, temperature is a function of irradiance and pressure, or at least can be calculated as such.

JinYu
January 23, 2012 4:23 am

Well, I have carefully read this and I also read some elevator speak before, which I do not think was relevant enough to take this theory down. This is not about conservation of energy. Here is another elevator try with another angle.
Consider the previous elevator speak about the gas chamber and resulting in a isothermal gas. This means the gas should have the same temperature throughout. Well, then the atmosphere should also have the same temperature all the way up, which it do not have. So that elevator speak is wrong. The gas is not isothermal, well in a smaller container it might well be but not when unlimited and restrained only by gravity.
To add to this discussion, we have the theory of the creation of galaxies. The clouds created by gravity grew bigger and bigger and temperature raises. Where do that temperature come from if not from gravity. And then there is not even a sun or a planet in the beginning, still the sun’s is lightened up because of the temperature caused from gravity alone. This is because the mass is so great and this is the reason all the shining stars is so huge compared to our planet. So gravity cause temperature to raise, simple as that.

David
January 23, 2012 4:30 am

Schrodinger’s Cat says:
January 23, 2012 at 2:29 am
The bottom line is that this work claims (1) that the SB law has been incorrectly applied to the moon and the earth and as a consequence (2) the current GHG effect underestimates the extra atmospheric warmth of the earth by about 100K. This is what we should be debating.
——————————————
Yes, I agree. But the authors also appear to be implying that all atmospheres, regardless of composition, will have the same average T as long as pressure, gravity, TSI and albedo are the same, regardless of the type of gas, oceans, water vapor (beyond albedo affects), rate of rotation, etc. They claim observational data of other planets confirm this.

richard verney
January 23, 2012 4:37 am

@Edim says:
January 22, 2012 at 7:33 pm
//////////////////////////////////////////////////////////////////
Further to the point that Edim makes, The Trenberth diagram suggests that the Earth itself (in which I include the oceans) net absorbs 0.9 W per sqm.
Edim states: “….396 – 333 – 161 = -98 W/m2 (downwelling)… To complete the energy balance, non-radiative heat transfer (outgoing) is added (17 + 80 = 97)…..”
The difference between these figures is the net absorption by the Earth itself of 0.9 W per sqm.
What is the effect of this net absorption over time?
How has this affected current temperatures?
If this net absorption has been occurring since the last ice age, surely that amounts to quite some significant effect?

Bill Illis
January 23, 2012 5:01 am

We should incorporate time into all these equations.
Let’s take the moon’s equator. Just before the Sun rises, it is 95K, as if it was radiating at 4.67 joules/m2/second of energy.
The sun comes up. Now the surface rapidly warms up. But does it really?
The solar energy is coming in at 9:00 am (or in 3 days moon time) = (1362 joules/m2/second / 2 * (1-0.11)) = 606 joules/m2/second. Yet the temperature is only increasing at 0.004 joules/m2/second. (a number not much different than Earth’s equivalent).
The Sun is beaming in and getting absorbed but the rocks are only warming up as if they are absorbing a tiny fraction of that. Where is it going. Warming the surface immediately below? Being re-emitted? (No, not unless the instruments measuring the radiation are missing it). There is just a tiny increase in the emission/temperature/energy rate per second.
By noon, (7 days moon time), the surface is now up to 390K or emitting as if were 1311 joules/m2/second of energy. This is actually higher than the solar forcing now which is = (1362*(1-0.11) = 1211 joules/m2/second.
In accumulating 0.002 joules/m2/second, it is now hotter than just the solar energy coming. As the sun starts to get lower in the sky, the surface starts losing energy at the same 0.002 to 0.004 joules/m2/second and the temperature gets down to 95K just after the sun sets.
Different picture.

JJThoms
January 23, 2012 5:09 am

Please look at the spectra shown in slide 9 of:
http://www.patarnott.com/atms749/powerpoint/ch6_GP.ppt
This shows grond and TOA spectra GHG bands mising from TOA and present in upward looking ground spectra.
This shows that GHGs are changing the radiation paterns exiting the planet.
Now Please look at
http://www.patarnott.com/atms749/pdf/LongWaveIrradianceMeas.pdf
Figure 3
This shows a few days of MEASURED longwave downwards irradiance. Note that there is significant downward radiation measured at night gtreater than 260W/sq metre.
GHGs can only be the source of this since there is insignificant LW IR from N2 and O2
How can conduction and no significant GHG radiation explain these measured values.
Forget the guff about black body/grey body/etc forget lapse rate there is measured 360W/sq metre during the day and 260W/sq metre during the night hitting Southern Great Plains in Oklahoma,
This is radiation not conduction. If you suggest that conduction is the source of the overtemperature above non atmosphered BB then there is a problem with total radiation received.
Please help my confusion

January 23, 2012 5:09 am

tallbloke says:
January 23, 2012 at 12:45 am …
Doug Hoyt has just reminded me that his pyrheliometry doesn’t measure near IR but visible spectrum solar radiation.

dlb
January 23, 2012 5:13 am

N & Z are way out with their lunar surface temperatures. The Apollo programme measured lunar temperatures a metre down and found them to be fairly constant over a lunar day in the order of 253 to 255K depending on the landing site. See (Nagihara et al 2010). Which is close enough for me to the blackbody figure for a nongreenhouse planet.
You can do the same on earth, bury a thermometer a meter down and it will average out diurnal variations, bury it even deeper and it will average out seasonal differences.

Joel Shore
January 23, 2012 5:18 am

Ned Nikolov says:

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.

Science is like any other field – it takes a formal education and many years of practice to reach proficiency.

Ned: However, by the same token, I would say that if you are going to try to change the paradigm in a field, it is imperative that you first take the time to understand the field and the current paradigm first. Even someone as brilliant as Einstein did this before embarking on his own paradigm-changing work.
By contrast, it is quite clear that you guys haven’t done this and this and as a result you have made several dramatic errors:
(1) Proposing a “theory” that can’t even be shown to satisfy conservation of energy.
(2) Misapplying conservation of energy to make the incorrect claim that the average temperature of a planet won’t change if, for example, you go from a non-rotating to a rotating case.
(3) Putting convection into a simple radiative model of the greenhouse effect incorrectly (so that the temperature profile with altitude is driven to an isothermal profile when the actual profile should be an adiabatic lapse rate profile) and then marveling that this gets rid of the radiative greenhouse effect, a fact that you could have read about in many textbooks on the subject.
You criticize Willis Eschenbach for his errors; however, his belief that you might have a mistake in your mathematical calculation is a relatively modest error compared to the large conceptual errors that you have made and that Willis has correctly pointed out to you.

January 23, 2012 5:21 am

If you understand that the Stefan-Boltzmann (SB) law was originally derived from other known laws in 1879 on the basis of experimental measurements made by John Tyndall and was derived from theoretical considerations, using thermodynamics, by Ludwig Boltzmann (1844-1906) in 1884. as well as ohm’s law (See; Nonlinear correction to Ohm’s law derived from Boltzmann’s equation.), and other references, (which, in my opinion ohm’s law is one of the best diagnostic tools there is, if not the best).
The (SB) law seems incomplete and doesn’t provide any explanation about the distribution of energy as a function of frequency and leaves a theoretical infinitive within the equation. And therefore the “black body” theoretical object that absorbs 100% of the radiation that hits it seems to me as being too far from reality, It can be compared to absolute zero, at which entropy reaches its minimum value, too augmented from reality to be used as a practical tool, even tho it appears theoretically sound.
I say Avoid the rabbit holes and work on a reference frame, this is one of the reasons I like the Ned Nikolov and Karl Zeller paper “Unified Theory of Climate” that, and the bonus of the introduction of proper terminology “Atmospheric Thermal Effect (ATE)” in place of the term “Greenhouse Effect” (GE) as it is “(b) 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”…
Just for fun, how would an analogy of “Unified Theory of Climate” compared to a circuit look?
here’s what I think it would look like using ohm’s law. (needs a bit of work). 🙂
Governing:
(g) Gravity, (M) Mass, (t) Time, (D) Distance
Primary:
(P) Wattage or power P = Temperature, energy from the sun i.e transferred one Joule per second is one Watt.
(E or V) Voltage or volts = Potential difference of energy between the SUN and a planets surface through a body of Resistance (atmosphere, oceans etc…)
(I) Amperage or current I = pressure, (force per unit area) pressure exerted anywhere.
(R) Resistivity or ohms Ω = composition, material, volume resistivity, (including the reciprocal, conductivity, sigma “mho”?)
Secondary:
R= impedance, impedance (Z) varies with frequency! Four electrical quantities determine the impedance of a circuit: resistance (R), capacitance (C), inductance (L) and frequency (f). Weather, Life, geographic (erosion etc..), chemical ?
* Resistance R (the part which is constant regardless of frequency)
* Reactance X (the part which varies with frequency due to capacitance and inductance)

Edim
January 23, 2012 5:33 am

George E. Smith; says:
“I don’t disagree with that assertion, but I fail to see how the ideal gas law applies to an open system where the volume, the Temperature, and the pressure are all varying quantities.”
George, ideal gas law is simply an equation of state for ideal gases. If the IGL is not accurate enough for the gas in question, there are equations of state for real gases. They all look like this:
F(p,ρ,T) = 0.
If any 2 of the 3 state variables is fixed (known), the third one is fixed too and can only have a value according to the equation of state. The IGL argument is not a process nor an energy balance argument – it’s simply a state variables argument. The equation of state applies to any “point” of the gas. If you know the pressure and density at some point (M(x,y,z,t)), you can calculate the temperature. That’s all, I think.

Joel Shore
January 23, 2012 5:35 am

kzeller says:

The equations we have given you bloggers are simple and they work. 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? The question is how can we possibly have done it – there is no question that our equations work – if you haven’t verified that it works, why haven’t you?

It is not a “miracle” to fit 8 pieces of data with a functional form that you chose that has 4 free parameters (plus some possible additional freedom from your choice of how to compute T_gb and your choice of which data to believe for surface temperatures and pressures on the 8 celestial bodies, especially since the average temperature for an airless body is a poorly-defined quantity that depends strongly on how thick a surface layer you look at).
And, I have verified that it works. However, I have also found that I can change the surface temperature data points for the 3 bodies that have a significant radiative greenhouse effect by replacing their average temperature by the traditionally-determined value of T_sb instead (e.g., 255 K for Earth), hence essentially removing the radiative greenhouse effect entirely. The data is then “squirrellier” since Venus now has a small value of N_TE than Earth or Titan, but nonetheless your functional form does a good job of fitting it! I imagine I might be able to do even better if I “shopped around” for another functional form.
So, in other words, your functional form can fit data for 8 bodies that have no significant radiative greenhouse effect at all (5 of them because they never did and 3 of them because I have removed the effect). Why can it do this?
(1) Because it has lots of free parameters.
(2) Because there are various reasons why we do expect a general positive correlation between pressure and average temperature. The most important of these (given your definition of T_sb) is simply that a planet with more atmosphere will have a more uniform temperature distribution and this alone explains most of the rise in average temperature. Other lesser factors are that the radiative greenhouse effect also tends to be positively correlated with surface pressure both because an atmosphere has to have at least some significant pressure to have a significant quantity of greenhouse gases and another being that pressure causes broadening of the absorption bands of the greenhouse gases.
It would be nice if you stopped claiming that we haven’t addressed this issue of your “miracle” fit and actually addressed our explanations of it.

January 23, 2012 5:44 am

tallbloke: “Thanks for helping me understand the way the substitution works in the integral. I need to be able to visualise those kinds of maths puzzles in order to understand them post-accident, and your explanation did it for me. Geometry rules!”
You’re welcome. And I can identify with the math-skills impairment, although age is the culprit in my case.
Just to be clear: Although I (belatedly) recognize that Equation 5 works, I don’t think Equation 6 does; unlike solar radiation, background radiation is isotropic. Numerically this is de minimis, but the authors’ repeating that equation without explanation in this write-up after having had the problem brought to their attention in connection with the poster makes me question how much care they apply to their work.
Among the other serious reservations I have, however, this is just a minor point. For reasons among which I gave some at your site, I think that these authors are not the ones to explain why, as I suspect but can come nowhere near to proving, the greenhouse effect is virtually insensitive to greenhouse-gas concentration beyond a low level.

Joel Shore
January 23, 2012 6:09 am

tallbloke says:

The radiation measured by AERI and other such devices is the radiation buzzing around between the molecules in the air. The air is denser near the surface, which is why we see 390 squiggles per square metre whizzing about just above the surface there. Up at 7km or thereabouts on average, where the air is less dense and there are fewer molecules per cubic cm, we see around 240 squiggles per square metre whizzing around. This fails to surprise me.

A shorter statement would have been for you to simply say, “I have no idea what radiation is.” Electromagnetic radiation exists even in a vacuum…It has nothing to do with the kinetic energy of the molecules of air where the radiation happens to be passing through.

Joel Shore
January 23, 2012 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!

January 23, 2012 6:17 am

George Smith. says “So unless the authors can show a “cause and effect” linkage bewteen the atmospheric pressure, and the average surface Temperature; the “Unified” stature of their theory remains unconvincing.”
Like most of the posters here George, a well educated and erudite fellow misses the point by a country mile.
You must read the papers first. If you do so you will see that the relationship between pressure and temperature enables the authors to calculate planetary surface temperature from atmospheric pressure and solar irradiance alone and it does not seem to matter what the atmosphere is composed of. That is something that Joel and Willis must consider.
Temperature also changes at the surface (without change in pressure) due to change in albedo, i.e. cloud cover. The authors have told us that they think that it is this process that accounts for recent climate change on Earth.
Posters on this thread have been quick to condemn the theory without reading the work, without understanding the maths and without considering the fact that, the denser the atmosphere, the more energy it will absorb, store and transmit (to a thermometer).
Take the case to its logical extreme. Conduction is the chief means of surface cooling when a heated object is contacted by a dry object. Not until you touch something do you begin to perceive just how hot it is. Can you touch something with a vacuum? The rate of transfer of energy to an object and the amount of energy stored by the touching object is dependent in the first instance on its density, the number of molecules in a given volume.
Its really that simple.The atmosphere is the touching object. The planetary surface is the object being touched. At Tallblokes Konrad Hartmann describes an experiment that confirms that simple observation. And, if its a gas, it does not matter what the touching object is composed of.
The treatment that the authors have received on this thread is one of the saddest things that I have witnessed on this forum.
There is only so much that you can teach a parrot. And a parrot with an ego is a very poor learner.
I had to smile when I read this from KZeller: “Why are you all trying to include so-called GH gases; ocean modulations; re-radiations; crusts, your grandma’s bad breath and so on ad nauseam? These are not part of our theory.”
This is a very careful and methodical piece of work based on a good understanding of the simplest physical principles.

Kev-in-Uk
January 23, 2012 6:22 am

richard verney says:
January 22, 2012 at 5:47 pm
……..
The calculations are based on the whole surface area, yes? so, if a rotating planet gets an even smattering of incident radiation, it then radiates evenly too. A static body having all its incident radiation on one side will simply be hotter and radiating more (on that side) to compensate. Like I said, I cannot see how a rotating planet can be ‘warmer’ as Roger suggested. Incident radiation is X, and outgoing radiation is Y. Once equilibrium is reached, either on a static or rotating body – the equivalent average temps should be the same IMO? Thats why I queried Tallblokes comment – on simple energy conservation rules (and the no atmosphere criteria), they must be equal? Yes, the ‘appearance’ of one constantly heated side will be hotter than the non heated side, but the averaging of the surface temp for the sphere should be the same, should it not?

John Marshall
January 23, 2012 6:34 am

Schrodinger’s cat claims that we should debate the problems with the SB law as applied to Moon/Earth not anything else.
Well this can be discussed. I do not think that the SB formulae is relevant to a planet with so many daily variables driving the heat transfer system. It works well in the laboratory. But if we are to lay claims to the problems with the GHG theory we must propose another mechanism for the extra heat, even if the BB/GB figures are wrong it is how much. The GHG theory was formulated over 100 years ago well before detailed radiation measurements were possible so the theory is open to debate before my objections above.(it should have been obvious back then, we did have a good understanding of the laws of thermodynamics then.).
And to continue the consideration of a 1Km high enclosed column of air is getting over the top. this imaginary container suffers the same problems as the garden greenhouse in explaining the GHG theory, there is no convective mixing with the outside atmosphere as reality dictates.

Mike Monce
January 23, 2012 6:36 am

I haven’t quite made up my mind on the N&Z theory. However, the authors do one thing the crtitics here have failed to do: compare their calculated results with empirical data. If the critics are to dismantle the N&Z theory, then they need to do calculations based on their ideas and then compare their results to the data. So, for Joel, Willis, etc, to start, show how you would calculate the average temperature of the Moon, get a numerical result, and then compare to the actual data. At that point, the rest of us can then decide which is the better physical theory. All these wordy criticisms are not convincing; show your own calculations!
If the Moon temperature can be properly calculated, then the next step is to try a planet with an atmosphere. What N&Z have done is to show a roadmap to the proper steps to building up a physical description of planetary temperatures. While we may all disagree with their details, their basic “attack plan” seems sound and is the way physics has always progressed: understand the simple system first, then work in steps to the more complex system.

Richard M
January 23, 2012 6:41 am

Ned Nikolov says:
January 22, 2012 at 9:31 pm
One more comment: Please, stop going in circles with this radiative transfer! Contemplate on this (which is a major conclusion from our analysis of observed planetary data): the long-wave (LW) radiation in the atmosphere is a RESULT (a BYPRODUCT if you will) of the atmospheric temperature, NOT a cause for the latter! The atmospheric temperature, in turn, is a function of solar heating and pressure!
The so-called GH effect is a pressure phenomenon, not a radiative phenomenon!

Or maybe it is both. Right now I don’t see the second part of your theory holding up. I’m still of the belief that you have, in fact, changed the paradigm. The so-called GHE is actually determined by the irradiance and atmospheric pressure. However, that does not preclude a mechanism for heating the air.
The GHE is often defined as if there was one layer back-radiating to the surface. A more valid view, imho, is GHGs, at all levels in the atmosphere, absorb radiation (from both the surface and other radiating items in the atmosphere) and convert the energy into kinetic energy. That is, they warm the atmosphere. Because of density reductions as one moves up in a gravity field, this effect occurs less often. Eventually radiation escapes to space. The net result is the lapse rate and an effective radiation height defined by the pressure. However, this effect is NOT possible without having GHGs in the atmosphere.
The GHGs act as a resistor to the movement of energy out of the system. The gravity and mass of the atmosphere sets of the physical nature of this resistor. The only way to change the overall GHG effect is to change that physical nature.
This description explains why we are not warming at the present. We only add .005% mass to the atmosphere when we add 100 ppm of CO2 by burning fossil fuels. It also explains your findings of similar warming profiles. It just requires understanding that overall radiation is a flow of energy and not two separate and distinct flows forwards and backwards.

Joules Verne
January 23, 2012 6:49 am

Joel Shore says:
January 22, 2012 at 12:30 pm
Nikolov writes: “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.”
Nikolov et al display their poor grasp of energy transfer here for sure.
Shore writes: “It is probably useful to go into a little more detail on this claim than I did above by providing a little more discussion as to why Nikolov and Zeller’s claim is wrong here: Let’s suppose they were right, so we take a non-spinning planet and start a planet spinning and the temperature distribution becomes more uniform without changing the average surface temperature. Is this a reasonable steady-state?”
“Initially, before spinning, the planet was in radiative balance. Now that it has a more uniform temperature distribution but the same average temperature, the average value of T^4 will necessarily be lower. Hence, it will now be emitting less power back into space than it absorbs from the sun. What we such a planet do? It will warm up until such point as it is now emitting the same amount of power as it is absorbing from the sun. Hence we see how energy conservation, correctly applied, constrains not the average temperature but the emitted power.”
Shore is correct. As the spin rate slows on a real planet its average surface temperature falls because the hotter the hot side is when the lights go out the faster the temperature drops. This can be seen in an equatorial diurnal temperature graph for the moon which is not, as Nikilov’s hypothesis requires, a sine wave. The daytime warming is a lovely positive half of a sine wave but the cold side is almost a square wave. As one moves towards higher latitudes on the moon and the diurnal temperature swing lessens the graph becomes more and more shaped like a sine wave.
This is critically important. Spinning spheres of rock do not behave like massless superconducting grey bodies. For one thing absorption and emission coefficients are not equal in rocks and these parameters also vary with temperature. Figuring out what the temperature of the moon should be according to theory is not as easy as applying simple formula. It requires engineering knowledge of the properties of regolith – things such as emissivity and absorption coefficients, thermal conductivity, and so forth. This is why we went to moon and determined these things empirically through experimental packages. Teh actual measured average temperature of the moon taken about a meter deep in the regolith where the temperature is constant is 250K. The moon spins slower than earth and when taking the spin rate difference into account, given the earth and the moon have the same crustal rock composition, the earth should be slightly warmer from the faster spin which means the commonly given 255K number is strongly supported by empirical evidence.
Where Joel Shore goes wrong is that the earth isn’t a rock world like the moon. It is a water world and until one understands teh truly vast difference betweent the physical properties of rocks and water one will never begin to understand the earth’s climate. The so-called greenhouse effect is a consequence of a liquid ocean which, like a greenhouse gas, is transparent to visible light from the sun and is completely opaque to long wave infrared. The same properties that distinguish greenhouse gases from non-greenhouse gases are even more pronounced in liquid water. The albedo of water is also much lower than rocks and its ability to sequester energy against diurnal, latitudinal, and even seasonal temperature swings distributes heat almost as good as a far faster spin rate would accomplish and the better temperature distribution also works to raise the average temperature just as the moon would become warmer if it spun faster.
So long as the earth presents a surface that is largely liquid water non-condensing greenhouse gases play only tiny role in average temperature modulation and that small role is largely confined to land surfaces where evaporation isn’t the big Kahuna in cooling mechanisms. One physical fact of life that most people just don’t grok is that you cannot heat water with long wave infrared radiation. All you can do to water with long wave infrared is make it evaporate faster – it simply cannot be heated that way nor, because the GHG mechanism is radiative, can it be insulated that way as all downwelling LWIR will accomplish is immediate rejection of the energy in latent heat of vaporization. As many people do know but often neglect we humans, like thermometers, cannot feel latent heat. That’s why it’s called latent heat. So GHGs have very little surface warming effect over the ocean. Over land it’s a different story because rocks, simply put, don’t evaporate in response to DLWIR. Rocks absorb DLWIR and that in turn raises the equilibrium temperature and we, as well as thermometers, do feel the effect of warmer rocks under our feet.

Spector
January 23, 2012 7:12 am

The Stefan-Boltzmann temperature is a value determined by energy flow considerations, not temperature considerations. It would only be true on a body like a star with a more or less uniform surface temperature. As radiant energy is proportional to the fourth power of temperature at any point on the surface, it might be said to be a FRMFP (Fourth Root of the Mean Fourth Powers) average value.
This is analogous to the RMS (Root of the Mean Squares) average, used in electronics, which is based on the fact that potential electrical power is proportional to the square of the voltage or current. Thus 110 VAC does not indicate the average alternating voltage, which is zero, but the constant (DC) voltage required to produce the same average power (energy flow) across an identical standard resistor.
The FRMFP temperature only indicates the uniform, non-varying surface temperature required to produce the same power (energy flow) from the surface of a body as the average power that is actually flowing from that surface.
An FRMFP average of zero and 100 is about 84, and this average of 99 and 101 is about 100.015, thus the FRMFP average is close to the real average if the individual differences are small. Some climate scientists may justify their use of FRMFP averages to characterize greenhouse effects because they are only dealing with small changes around 287 degrees kelvin. It is easy to lose sight of this qualification when it no longer applies.

Gary Swift
January 23, 2012 7:23 am

I do not think pressure alone generates heat, but friction under pressure does. I’m not sure how much friction is generated at one atmosphere of pressure though. I’m guessing that it isn’t much, but it’s easy to demonstrate that forcing air through a small tube under pressure generates heat quite well, as does a fast moving object going through air.

January 23, 2012 7:52 am

dlb says:
January 23, 2012 at 5:13 am
N & Z are way out with their lunar surface temperatures. The Apollo programme measured lunar temperatures a metre down and found them to be fairly constant over a lunar day in the order of 253 to 255K depending on the landing site.

Where can I see the data?
Thanks

JJThoms
January 23, 2012 8:03 am

Mike Monce says: January 23, 2012 at 6:36 am
I haven’t quite made up my mind on the N&Z theory. However, the authors do one thing the crtitics here have failed to do: compare their calculated results with empirical data.
=========
this papere refered to above give you exactly what you require
Measured downward radiation comared to modtran model.
http://www.patarnott.com/atms749/pdf/LongWaveIrradianceMeas.pdf
perhaps Ned Nikolov, and Karl Zeller could explain where this 260W/sqm at night is coming from since only GHGs radiate

Zac
January 23, 2012 8:05 am

Harry Huffman has now responded to “The Unified Climare Theory” on his blog.
http://theendofthemystery.blogspot.com/2012/01/observation-on-unified-climate-theory.html
It’s well worth a read as is his original “Venus No Greenhouse effect” post
http://theendofthemystery.blogspot.com/2010/11/venus-no-greenhouse-effect.html

A. C. Osborn
January 23, 2012 8:39 am

Joules Verne says:
January 23, 2012 at 6:49 am Says “Over land it’s a different story because rocks, simply put, don’t evaporate in response to DLWIR. Rocks absorb DLWIR and that in turn raises the equilibrium temperature and we, as well as thermometers, do feel the effect of warmer rocks under our feet.”
This is certainly true during the day with sunshine, but are you sure that any surface feels warmer than the air during the night time?

A. C. Osborn
January 23, 2012 8:39 am

Or are you saynig that they are warmer than they would be?
Which of course can’t be proved one way or the other.

January 23, 2012 8:40 am

“So, for Joel, Willis, etc, to start, show how you would calculate the average temperature of the Moon, get a numerical result, and then compare to the actual data. ”
Here’s the thing — I doubt any of them have a problem with the calculation leading up to Equation 6. They do the proper calculation for a slowly rotating, airless, massless world. This is nothing new, and it is how I would do it (and presumably Joel or Willis or Dr Brown).
Where we disagree is what to do next. The Energy Redistribution Effect (which I will call ERE since it is fun to have acronyms) changes the distribution of energy, which in turn changes the distribution of temperatures, which in turn changes the distribution of radiation to space. By Specter’s FRMFP (aka Holders inequality), this can change the “average temperature” quite a bit while leaving the “effective temperature” the same.
Consider Equation 6 — it is extremely bad on the the night side (2.7 K vs 90K) and only reasonably good on the day side. Since there is no atmosphere on he moon, this cannot be attributed to “ATE”. The difference is due to the ERE — on the day side, some of the energy that hits the surface does NOT immediately get radiated, but rather gets transferred by conduction to lower, colder layers of rock. This is the ERE-C (ERE via Conduction) and is the main correction to Equation 6 for the moon.
On earth, we can add in ERE-O (ocean currents carrying energy from equator to poles). We can add in ERE-R (rotation carrying cold night-time areas to warm daytime areas & vice verse). We can add in ERE-A (atmosphere carrying energy from equator to pole, or between ocean & land (sea breezes), or even on a small scale by convection drawing air into (and then up from) spots like hot parking lots). And of course, the thicker the atmosphere, the more effectively it will redistribute energy, so in this sense ERE-A is enhanced, which would indeed raise the average temperature more than a thinner atmosphere.
RECAP
* Any ERE from warmer areas to cooler areas will, via FRMFP, raise the average temperature.
* The MOST that ERE can do is to raise the average temperature of the surface to the effective temperature (~ 255 K for earth; slightly higher on the moon with its lower albedo)
* “ATE” (as define by the authors) is only due in part to the atmosphere; it is a combination of all the ERE’s. It is a poorly named effect.
* ERE-A will depend on the pressure of the air, and hence the global average temperature should correlate to the atmospheric pressure.
* NONE OF THIS IS NEW SCIENCE.
************************************************
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.

A. C. Osborn
January 23, 2012 8:41 am

Actually they can, it is always warmer under a surface, take a car under a Car Port as an example.
So sorry I can’t agree about DWIR warming the earths surface

Ralph
January 23, 2012 8:49 am

>>>Talbloke
>>>The so-called GH effect is a pressure phenomenon, not a
>>>radiative phenomenon!
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). So you would end up with a surface of the planted that behaved much like the airless Moon, and a very cold atmosphere only 5,000 ft thick that took no part in the warming of the planet.
The fact that the atmosphere is warm and 100,000 ft thick demonstrates that its absoption of LW radiation is a critical element in its present form. And if it is reemitting LW, then that LW can be emitted downwards, warming the surface (or rather, not allowing the surface to cool so quickly – as fast as it tries to emit energy, some of it comes right back). The ‘steel atmosphere’ thought experiment on WUWT some years ago demonstrated the GHE is a reality, and can be achieved with steel, glass or even an atmosphere.
.

PeterGeorge
January 23, 2012 9:12 am

Can’t the question of pressure induced temperature be tested experimentally in a centrifuge?
Place an insulating tube with a suitable gas inside along the radial arm of a centrifuge and spin it up. The gas at the distal end will compress and heat up while gas at the proximal end will rarify and cool. That is almost certain. But will it stay that way? Thermometers at the distal and proximal end would answer that question in a relatively short time.
The discussion here proves one thing for sure. The theory is very complex and potentially subject to errors. Isn’t it anathema to science to theorize about something and not bother to check the result experimentally?
I think so, and I’ll wait for this or some other simple experiment to resolve the question.

kwik
January 23, 2012 9:25 am

erl happ says:
January 23, 2012 at 6:17 am
” Conduction is the chief means of surface cooling when a heated object is contacted by a dry object. Not until you touch something do you begin to perceive just how hot it is. Can you touch something with a vacuum? The rate of transfer of energy to an object and the amount of energy stored by the touching object is dependent in the first instance on its density, the number of molecules in a given volume.”
You got that right! And imagine when the temperature is only, say, 40-50 degrees….

January 23, 2012 9:32 am

Tim Folkerts says:
January 23, 2012 at 8:40 am
“So, for Joel, Willis, etc, to start, show how you would calculate the average temperature of the Moon, get a numerical result, and then compare to the actual data. ”
Here’s the thing — I doubt any of them have a problem with the calculation leading up to Equation 6.

Ahem. Reread this thread from the start Tim. I saw quite a few problems developing quite quickly.
ill Illis: “I think there must be an error in the derivation of equations 5 and 6.”
gnarf: “There is a big problem in the integral when they make the substitution u=cos(theta)… this is terrible”
Willis: “mu for the cosine(theta) term in Equation 4. I’m not sure you can do that when you are going to integrate. My suspicion is supported by the fact…”
Instant falsification was in the air last night, and the crowd smelled blood.
Today, not so much.

Bill Hunter
January 23, 2012 9:34 am

markus says:
January 22, 2012 at 6:04 pm
“Gentlemen,
Please refrain from the term GHGs, and get back to reality.
The gases known by their names on the periodical table, PLEASE.
A lot of interesting stuff is going to come out of this.”
Exactly! The GHE is a double edged sword. You can cause stuff to warm by radiation or by insulating something that radiates. Poorly radiating gases are great insulators when you isolate them from a conductive source, like in the atmosphere which is sort of a box with only one side . . . .the side you want an explanation for its warmth . . . .”Doh!”
Whats going to come out of the answer of the ATE being 3 to 7 times larger than believed (between 100 and 233 degrees?) is this is a net figure. It will include everything, positive and negative added together.
There could be a lot of interesting combinations.
For instance if you allow a zero ATE for 98.96% of the atmosphere (non-GHG) as is done in all the “popular” climate studies the ATE might calculate much closer to the extremely high figures.
But I think the most interesting information will come from an ATE being in the range of 100K-133K. is that it is going to have a huge impact on natural solar variation calculations modifying their effects by a multiple of 3 to 4.
Then maybe we can actually start accounting for natural variation rather than discarding it as being at least 3 times too small relative to other stuff and end the problem of having to blindly plug aerosol and ENSO explanations to fill the gaps.
As Colonel Smith says: I love it when a plan starts to come together!

Editor
January 23, 2012 9:48 am

markus says:
January 22, 2012 at 6:04 pm

Gentlemen,
Please refrain from the term GHGs, and get back to reality.
The gases known by their names on the periodical table, PLEASE.

Last I looked, methane and R12 were not on the periodic table, my friend …
Every specialty has its jargon and acronyms. In climate science the GHGs are generally agreed to be the following
water vapor
carbon dioxide
ozone
methane
nitrous oxide
various halocarbons (R12, R22, etc)
While it is possible to disagree in theory with the usage of “GHGs” to mean those gases, that meaning and usage is well established in the climate science field. If you insist that people not use the term, you’ll just get treated as a newbie. It’s one of those quirks that you’ll just have to get used to if you want to be involved in the field—no one’s going to stop using it just because some random anonymous poster requests it, even when the request is in capital letters …
w.

Ned Nikolov
January 23, 2012 9:49 am

To Dr Burns (January 23, 2012 at 1:24 am)

For those without neither the time, inclination nor background to wade through your math, as an elevator presentation, what do you suggest is the relative warming effect of the Earth’s IR-absorbing gases, compared to your atmospheric thermal effect ?

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! That is because convection completely offsets on average the thermal effect of down-welling IR radiation, and it must do so in order to conserve the system’s internal energy. Climate models, on the other hand, effective violate the law of energy conservation by allowing changing atmospheric composition to alter the kinetic energy of the near-surface air environment. Projected warming with rising CO2 is, therefore, a pure model artifact resulting entirely from the artificial separation (decoupling) between radiative transfer and convective cooling in these models. There are NO ‘greenhouse gases’ in terms of ability to cause a NET warming. Hence, the term is a total misnomer as is the term ‘Greenhouse Effect’.
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. The higher surface temperature (above that of a gray body) is ALL a manifestation of the pressure enhancement of solar energy. As soon as the amount of absorbed solar radiation changes (due to a change in cloud cover/albedo or orbital characteristics), temperature at the surface will change as well. The absolute magnitude of the surface thermal enhancement (measured as a temperature difference) varies with absorbed solar radiation, but the RELATIVE enhancement (NTE = Ts/Tgb) stays constant as long as pressure remains the same. NTE is only a function of pressure!

Spector
January 23, 2012 9:54 am

RE John Marshall: (January 23, 2012 at 2:27 am)
“2. The reradiated heat has to come from a cooler area than the surface which violates the 2nd law.”
There is no violation of the laws of thermodynamics unless *more* radiant energy flows from the cool region than is being absorbed from the warm region. In order for there to be a violation of the second law, there must be more net energy flowing from the cool region to the warm region than the reverse. Heat flow is the difference between the incoming and outgoing radiant energy flows. Do not confuse radiant energy transfer with ‘heat’ transfer. They are two different entities.
All the cool region does is return a portion of the outgoing energy back to the surface or to lower levels in the atmosphere, thus preventing the escape of that energy to outer space. The returned radiant energy allows the lower surface to be warmer than it would have been if that energy had not been returned. The greenhouse effect is all about the transfer of radiant energy and the consequential thermal effects.
Greenhouse gases are those gases that have the ability to impede or slow the flow of radiant energy to outer space. A non-greenhouse atmosphere, by definition, offers no such impedance. The sun only provides the Earth with enough power to support an average flow of about 240 W/m² to outer space. An Earth with no radiation impeding atmosphere above would be limited to this average surface emission rate.
“3. GHG’s are said to ‘store’ heat which again violates 2nd law.”
*All* materials store heat; their ability to do this is known as heat capacity. Stored heat is measurable in calories per cubic meter. Radiant energy flow (or power) is measurable in watts per square meter.

Tilo Reber
January 23, 2012 9:54 am

Joel Shore: “However, if you go several centimeters down, you find the temperature remains remarkably uniform between day and night and you get the average value of around 240 K.”
So, you claim that there is a large thermal gradient between the surface and a few centimeters below the surface. Why would such a thermal gradient persist? Why wouldn’t the subsurface heat conduct to the surface and then radiate away until the surface and “several centimeters down” become close to equal? I’m afraid I can’t buy your assertion.

John Day
January 23, 2012 9:57 am

tallbloke says:
January 23, 2012 at 7:52 am
dlb says:
January 23, 2012 at 5:13 am
N & Z are way out with their lunar surface temperatures. The Apollo programme measured lunar temperatures a metre down and found them to be fairly constant over a lunar day in the order of 253 to 255K depending on the landing site.
Where can I see the data?
Thanks

Looks like dlb’s recollection on Apollo-sampled lunar soil temps might be faulty:
http://www.lpi.usra.edu/publications/books/planetary_science/ (Ch 4, pg 116)
“The temperature on the lunar surface increases by about 47°K in the top 83 cm. The top surface (2-3 cm) is a loosely packed porous layer. Surface temperatures vary considerably. At the Apollo 17 site, the surface reaches a maximum of 384K (111C) and cools to 102K (-171C) at the end of the lunar night [2]. The near-surface temperature is 216K (-57C). At the Apollo 15 site, these temperatures are about 10°K lower. The agreement with previous estimates based on terrestrial observations was very close [8, 9]”
Those references at the end of the quote:
8. Saari, J. M. (1964) Icarus. 3: 161.
9. Mendell, W. W., and Low, F. J. (1970) JGR. 75: 3319.

January 23, 2012 10:04 am

I would like to leave a very few remarks on this paper. First of all, it is much clearer than their original paper — so much to the good, although it is the second half that they haven’t presented yet that I am more interested in. Second, the idea of what they are attempting is valid enough, although there appear to be a few errors on both sides of the discussion. The arithmetical mean temperature of a rotating body (straight up averaged over the surface) should indeed be lower than the mean temperature of an ideal heat superconducting black body, because of the T^4. Trying to compute exactly how large the effect is from first principles is going to be difficult, however, because “tinfoil world” — an airless planet made out of a thin layer of perfectly black foil with a very low heat capacity and very little lateral conduction of heat wrapped around a perfectly adiabatic interior — will have a very different temperature distribution and average cooling rate than “water world”, a world with a thick layer of a fluid perfectly transparent to incoming radiation, that perforce smooths the sharp thermal fluctuations of tinfoil world into a nearly smooth small amplitude sinusoid (ignoring things like evaporation, sorry JV but this is imaginary water not real water that is even MORE complex). The point is that the heat capacity and conductivity and thermal mixing of the active region of solar absorption make a large difference in thermal distribution and mean temperature simply as a parameter in the computation attempted by N&Z. I therefore agree with Joel Shore that it isn’t likely that they’ve done this precisely correctly or defensibly — it is a useful model result but not necessarily a predictive one.
On the other hand, I don’t agree with Joel that the temperature 1 meter down on the moon is at all relevant to the radiative balance problem, any more than it is on the Earth. One meter down on the Earth it is (as J.V. would fondly remind us, given the chance:-) it is 4 C over some 70% of its solid surface, whereas everywhere on the land it is more like 10C or even a bit higher. Also, why one meter? Because 1000 meters down, or 10000 meters down, the temperature is something else, and radiation from the surface of an opaque solid is produced by the atoms on the very surface of the solid only! As far as radiative balance is concerned, then, we cannot speak of the correct value of the N&Z integral for average temperature without knowing the actual surface temperature — not the temperature a meter down, and worse, one has to actually use the heat equation all the way down to some depth where the temperature is effectively constant to work out how the surface temperature changes as it radiatively cools. Tin-foil world heats and cools to local detailed balance instantly so that incoming irradiance equals outgoing irradiance. Water world, or real world, deposits some of the incoming heat into the underlying heat capacitive ground (or water) when the surface is hot (keeping it from warming as fast) and releases it back to the surface when it is cool (keeping it from cooling so fast). At some point the actual equilibrium temperature is likely determined more by the rate heat is conducted out of or into the interior than it is just by insisting on local radiative balance.
The actual problem is even more complex. The person I agree with the most (so far) is Richard Moore, with J.V.’s ocean getting an honorable mention as well. Let me give a synopsis of what I think the GHE “really” is, and why it is not terribly sensitive to CO_2 concentration.
* Sunlight differentially delivers heat to the Earth. I include all heat delivery to the surface, the water, and the land. All radiative heat in. Yes, it is modulated by albedo and all sorts of other things. Ignore that for the moment as that modulates the temperature in ways we can treat as a rule of thumb later, if we need to. Ultimately, though, some fraction of top of atmosphere incident radiation is “absorbed” by the Earth (including atmosphere) and “thermalized” — not directly or elastically scattered back out to space.
* The heat delivered to the Earth differentially warms it, with the bulk of it delivered (over most of the illuminated area) to the surface and not the atmosphere. The only reason this is important is that it makes the surface warmer than the atmosphere, at least during the day.
* Some fraction of the total heat budget ends up in the atmosphere. How it gets there is of little concern. Conduction, convection, and radiation all play a role. The point is that the Earth’s surface and the thermosphere are generally not in thermal equilibrium, and because the surface isn’t the same temperature as the air most of the time in most places, there is a lot of convection. Convection both vertically and laterally transports heat, establishing a lapse rate in the Earth’s atmosphere. We can be certain that vertical transport of heat is not predominantly due to radiation because the observed lapse rate is not particularly exponential, as one would expect, until one gets to the very top of the troposphere. The temperature of the top of the troposphere is thus determined by the adiabatic lapse rate up to within the optical depth of the various radiation channels that cool it. Heat (on average) in at the bottom is removed at the top by outgoing radiation from the cold gases in the upper troposphere, and free convection up and down maintain a non-exponential lapse rate otherwise.
* The total incoming heat absorbed (on average) must be balanced by outgoing radiation (on average) emitted from different places in the entire column from the ground on up, at rates that are at least partly determined by the buffering of heat moderating thermal swings and heat transport warming some parts and cooling others. The simplest “model” for this is a ground emitting radiation directly to space through the “water window” in the atmosphere plus GHGs in the upper atmosphere radiating at the temperatures of the upper troposphere. The total outward integrated radiative flux (over all bands and the surface) has to — on average — balance the average incoming flux. It does so in a very chaotic and variable way, but any significant or systematic imbalance would quickly heat or cool the whole system.
* This outgoing radiation has to be self-consistent with this requirement. The thermal radiation of the ground and troposphere have to together balance out the absorbed radiation. It makes no difference how heat is internally transported, trapped, or released within this statement — I make no reference to “upwelling IR” or “downwelling IR” because I don’t know how to balance that against conduction, convection, evaporation, and all of the other things that might affect surface temperatures or the distribution of atmospheric heat. The point is that the heat is distributed, and at the end of the day convection probably wins in the troposphere as a mechanism as the thermal profile of the troposphere seems to remain close to the DALR, at least where the surface remains much warmer than the troposphere.
* The DALR is more or less completely insensitive to CO_2 concentration. The height of the tropopause is almost completely insensitive to CO_2 concentration. The optical thickness in the e.g. CO_2 band of the upper troposphere is a weak function of CO_2 concentration. Finally, the temperature of the upper troposphere is both laterally and vertically mixed and is considerably less variable than the temperature at the surface — heat is laterally transported to minimize lateral thermal differences in an environment. Consequently the radiative rates from the upper troposphere aren’t strongly modulated by well-mixed CO_2 concentration. The temperature up there doesn’t change much, although modulating e.g. the water vapor content of the stratosphere can make a difference (because it is thermally inverted relative to the top of the troposphere).
I think that this provides a reasonable (if provisional) answer to Joel’s question about why the GHE appears to be so insensitive to GHG concentrations once one achieves an optically thick atmosphere. It isn’t that the GHE isn’t real — of course it is. It is just subject to a boundary condition at the top of the troposphere that is not a strong function of CO_2 concentration. The DALR and thickness of the troposphere are self-consistently determined by the GHE, to be sure, but once established and maintained by adequate thermal imbalance (ground warming) to maintain convection, the rate of loss from the upper atmosphere is almost independent of the rate of loss from the ground.
The ground then does what one would expect. It gets hotter and cools faster, or cooler and cools more slowly, according to all the usual mechanisms for heat transfer. None of those change the outgoing top of cloud radiation (by much) and consequently there is little long term modulation of the ground temperature — the lower atmosphere is rife with negative feedbacks and quickly moves heat around to where it can be lost more effectively than just waiting for it to radiate from where it was absorbed.
rgb

Ned Nikolov
January 23, 2012 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. Once the math is in, thought experiments should be replaced with actual calculations, since for the most part we are dealing with a highly non-linear system (see for example Fig. 5 in our original paper), which cannot be grasped well by a ‘thought experiment’ …

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.

Ned Nikolov
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 …

Ned Nikolov
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?

Ned Nikolov
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.

Ned Nikolov
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.