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:

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.

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

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.

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.

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

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.

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:

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).

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).

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.

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.

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)

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

2. Steve (Paris) says:

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

3. Willis Eschenbach says:

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

4. Alvin says:

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

5. Ceri Phipps says:

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!

6. Joel Shore says:

I’ll post here what I communicated to Ned Nikolov directly:

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

7. Stephen Wilde says:

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 ?”

8. 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

9. 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.

10. Joel Shore says:

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.

11. Stephen Richards says:

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.

12. Stephen Wilde 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.”

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.

13. A physicist says:

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.

15. Willis Eschenbach says:

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.

16. Kev-in-UK says:

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!

17. G. Karst says:

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

18. Joel Shore says:

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.

19. Bob Shapiro says:

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.

20. 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.

21. Joel Shore says:

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.

22. bair polaire says:

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… ;-) )

23. Jan Kjetil Andersen says:

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.

24. // 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.

25. James Reid (from Arding) says:

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.

26. gnarf says:

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.

27. Kev-in-UK says:

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!

28. John Blake says:

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.

29. Robertvdl says:

“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?

30. PeterGeorge says:

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.

31. don penman says:

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 .

32. Zac says:

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.

33. Bryan says:

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.

34. 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.

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.

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]

35. gnarf says:

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

36. George E. Smith; says:

“”””” 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” ?

37. Bill Hunter 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.”

Clean? Clean in science is no missing stuff, not painting over the smudges.

38. George E. Smith; says:

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

39. gnarf says:

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 ??

40. HLx says:

@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.

41. Greg Elliott says:

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.

42. R. Gates says:

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.

43. Bill Hunter says:

“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.

44. DB-UK says:

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

45. “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.

46. PeterF says:

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.

47. Joel Shore says:

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.

48. wayne says:

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.

49. 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.

50. 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. 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.

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.

52. Julien says:

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.

53. Bill K says:

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.

54. 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.

55. Ian H says:

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. 56. Nick Stokes says: January 22, 2012 at 2:50 pm Your ATE is something else. Wotcher Nick. Good puzzle isn’t it? 57. 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. 58. Bill Illis says: 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. 59. Billy says: 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 60. 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. 61. jorgekafkazar says: 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…” 62. Titixxxx says: 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. 63. KevinK says: 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. 64. jorgekafkazar says: 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. 65. BenAW says: 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. 66. George E. Smith; says: “”””” 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. 67. R. Gates says: 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. 68. PeterGeorge says: 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)? 69. KevinK says: 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. 70. Latitude says: Thanks for all your hard work putting this together Anthony. 71. R. Gates says: 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. 72. David A says: 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? 73. markus says: “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. 74. David A says: 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? 75. jorgekafkazar says: 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? 76. Bill Hunter says: 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. 77. Alan Millar says: 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 78. 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. 79. Bill Hunter says: 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. 80. richard verney says: 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. 81. Bob Fernley-Jones says: 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. 82. A physicist says: 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! 83. markus says: Gentlemen, Please refrain from the term GHGs, and get back to reality. The gases known by their names on the periodical table, PLEASE. 84. jorgekafkazar says: 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.. 85. wayne says: 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. 86. richard verney says: @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. 87. gnomish says: 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. 88. Willis Eschenbach says: 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. 89. Willis Eschenbach says: 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. 90. Gregory Ludvigsen says: 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? 91. Joe Born says: 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. 92. Bob Fernley-Jones says: 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. 93. Joel Shore says: 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. 94. Edim says: 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: 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). 95. 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. 96. 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. 97. tchannon says: 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? 98. 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. 99. wayne says: 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. 100. Joel Shore says: 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. 101. Genghis says: 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. 102. 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 103. Björn says: 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. 104. 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? 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? 105. David says: 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? 106. David says: 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. 107. R. Gates says: 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. 108. 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 109. 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! 110. R. Gates says: 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. 111. 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 … 112. Dan in Nevada says: 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. 113. Bob Fernley-Jones says: 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. 114. George E. Smith; says: 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. 115. 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. 116. OOOPS! I bungled the math in my point 3 above. Too many beers. Night night, I’ll re do it in the AM. 117. 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! 118. R. Gates says: 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. 119. George E. Smith; says: 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. 120. George E. Smith: To answer your question about the cause of recent warming see my post above: https://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: 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). 121. Jenn Oates says: I’m poleaxed, and wondering however will I manage to condense this to elevator length for my students. :). The math, it burns! 122. Bob Fernley-Jones says: 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. 123. 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 ;-) 124. 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! 125. jimmi_the_dalek says: I think this thread needs a bit of humour, so I applaud kzeller’s nicely sarcastic parody at 8:21pm 126. Kasuha says: 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). 127. Surfer Dave says: 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. 128. Werner Brozek says: “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. 129. markus says: Hi Ned, Love your work. Is their a application for The Unified Theory of Climate, in a pressured enclosure above photovoltaic cells? Regards, 130. Rosco says: 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. 131. markus says: Ned, Could a three dimensional matrix, of all indices, based on incoming watts, explain your theory 132. Greg Elliott says: 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. 133. BenAW says: 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 134. wayne says: @ 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. 135. BenAW says: 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. 136. 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. 137. gnarf says: 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. 138. Willis Eschenbach says: 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. 139. Dr Burns says: 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 ? 140. 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. 141. markus says: Mr Eschenbach, I think it’s coming down to E-Mc2. Markus. 142. David Blake says: 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. 143. markus says: 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. 144. Steve Richards says: 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 145. A physicist says: 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). 146. John Marshall says: 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. 147. Schrodinger's Cat says: 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. 148. gbaikie says: “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. 149. Gras Albert says: 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 150. 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? 151. markus says: 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. 152. A. C. Osborn says: 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. 153. AusieDan says: 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. 154. AusieDan says: 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. 155. Stephen Wilde says: 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). 156. David says: 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. 157. AusieDan says: 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. 158. JinYu says: 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. 159. David says: 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. 160. richard verney says: @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? 161. Bill Illis says: 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. 162. JJThoms says: 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 163. 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. 164. dlb says: 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. 165. Joel Shore says: 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. 166. Sparks says: 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) 167. Edim says: 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. 168. Joel Shore says: 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. 169. Joe Born says: 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. 170. Joel Shore says: 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. 171. Joel Shore says: 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! 172. 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. 173. Kev-in-Uk says: 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? 174. John Marshall says: 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. 175. Mike Monce says: 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. 176. Richard M says: 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. 177. Joules Verne says: 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. 178. Spector says: 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. 179. Gary Swift says: 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. 180. 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 181. JJThoms says: 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 182. A. C. Osborn says: 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? 183. A. C. Osborn says: Or are you saynig that they are warmer than they would be? Which of course can’t be proved one way or the other. 184. Tim Folkerts says: “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. 185. A. C. Osborn says: 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 186. Ralph says: >>>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. . 187. PeterGeorge says: 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. 188. kwik says: 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…. 189. 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. 190. Bill Hunter says: 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! 191. Willis Eschenbach says: 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. 192. 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! 193. Spector says: 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. 194. Tilo Reber says: 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. 195. 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. 196. 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 197. 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’ … 198. HLx says: 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? :-/… 199. kdk33 says: 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. 200. 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 201. Willis Eschenbach says: 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. 202. 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 … 203. 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! 204. Willis Eschenbach says: 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. 205. Schrodinger's Cat says: 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. 206. Willis Eschenbach says: 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. 207. 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. 208. Willis Eschenbach says: 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? 209. 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 210. A. C. Osborn says: 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? 211. Willis Eschenbach says: 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. 212. A. C. Osborn says: 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? 213. Willis Eschenbach says: 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. 214. “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? 215. “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? 216. 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 … :-) 217. JPeden says: 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!” 218. 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. 219. Willis Eschenbach says: 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. 220. 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 … :-) 221. JJThoms says: 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! 222. A physicist says: 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. 223. Willis Eschenbach says: A. C. Osborn says: January 23, 2012 at 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? Since they are presumably averaging the entire surface, the effect seems obvious. It appears that they’ve left out half the planet, which is why I asked what I’m missing. w. 224. JJThoms says: A. C. Osborn says: January 23, 2012 at 10:43 am Ned Nikolov says: Can you offer the mechanism for Clouds keeping it warmer at night? ========== The air is more compresssed under the heavy cloud????! 225. Elevator Pitch for N&Z by davidmhoffer Respectfully, not bad at all, but not really N&Z as far as I can tell. Also you really mean to say net absorber of heat through radiation (solar input) and net loser of heat through radiation (outgoing thermal radiation integrated over all wavelengths) when you refer to the tropics vs temperate and arctic zones, right? Because as long as the Earth is approximately static in temperature, no zone is a net absorber or loser of heat from all mechanisms, correct? To put it another way, yes, heat is moved around a lot between where it is absorbed from the sun before it is radiated back to space, and there is some bulk transport of heat from the tropics towards the poles. The problem with the argument you give, however, is this: 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. This isn’t an argument, it is stating the desired conclusion of your argument (that changes in GHG concentrations aren’t important) and hence begging the question. If changing the GHG concentrations actually do alter the amount reradiated in the tropics given, as you put it, equal absorption in the unblocked visible band, they can certainly affect the net amount of heat left over to be transported to the poles. The more heat that goes back out with radiation in the tropics, the less there is to go back out in heat at the poles. The question then returns to whether or not altered GHG concentrations do alter the amount of heat reradiated in the tropics before it can take a trip poleward. And that isn’t answered by begging the question, it can only be addressed by indicating how it is that altering GHG concentrations do not alter the amount of heat radiated at the tropics, or for that matter, the poles. An argument for this might be “the temperature of the tropopause (and or emission height) does not change rapidly with either the GHG concentration or the underlying surface temperature because of vertical and lateral thermal mixing and the boundary condition imposed by the stratosphere”. As long as changes in the mean height of the tropopause from all causes remain small compared to its height and/or fluctuations that have nothing to do with GHG concentration, both are to be expected. But none of this is N&Z’s argument either, is it? As far as the currency argument goes, sure. Except that you mustn’t forget to take the fourth power of the currency conversion ratio somewhere in there. The actual radiative heat loss rate at the higher temperatures of the tropics is much higher than the rate at the much colder poles, much larger (or at the very least more complicated, with inversions and everything) than the temperature difference would suggest. This nevertheless does suggest net warming relative to at least a Tin-Foil Earth solar model because increased lateral heat transport makes at least part of the radiated energy more isotropic, which is more like the perfectly isotropic radiator, which is warmer than tin-foil Earth on average. That still doesn’t explain how to get from the superconducting Earth isotropic radiator up to GHG-enhanced temperatures. You need the GHE — split radiative loss between surface at one temperature and upper troposphere at another — to get the average surface temperatures up to where they are observed to be. That is by no means intended to suggest that the GHE is sensitive to CO_2 concentration — I doubt that it is — but this is a separate question from the question “is the GHE responsible for the bulk of the warming from Tin-Foil Earth”, or worse, Tin-Foil Earth with an albedo. Or better, from Earth with oceans and heat capacity and various forms of short and long timescale transport structures and variable albedo and a sometimes eccentric orbit and a lot of axial tilt and continents and… rgb 226. Ned Nikolov; Re-distribution of energy CANNOT create additional energy.>>> I never said it did. I said that re-distribution of energy from the warmer tropics to the cooler high latitudes creates more DEGREES. 227. JJThoms says (January 23, 2012 at 11:08 am): Have you read the article that is the topic of this thread? Did you not get the fundamental idea that we are talking about the temperature of a gray body with NO atmosphere, and that our integration is over an AIRLESS sphere ? What nighttime IR radiation from the atmosphere are you talking about?? Ok fellows, I need to to some real work now, so you wont see me on the blog for a while … 228. Kev-in-UK says: A. C. Osborn says: January 23, 2012 at 10:43 am I would have thought reduced vertical conduction and convection would be a possible mechanism? 229. Richard M says: Robert Brown says: January 23, 2012 at 10:04 am Dr Brown, thank you for describing in a vastly superior way essentially what I have been trying to say for the last couple of weeks. 230. HLx says: As Willis says, could someone do a step by step integration, with substitutions along with an explanation and post here? :).. 231. markus says: Willis Eschenbach says: January 23, 2012 at 11:11 am “”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 …”” Nobody will be using that term it to the future. Do not insult me, or I will treat you, as a newbie, in the reasoning of things. That is, a child. 232. mkelly says: A physicist says: January 22, 2012 at 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. ============================= Being somewhat familiar with radiative heat transfer, I find it odd you compare a solid to a gas. Since, radiation is a surface phenomena the thin slice is all that is needed. Do it this way, how many atoms/molecules are in a square meter of the foil used versus how many molecules of CO2 in a square meter of the atmosphere? Then you can see how bad your comparison is. By the way if I did my sums correctly there are 113800 molecules CO2 in a square meter of air. 233. R. Gates says: Sorry if this was answered, but I didn’t see the answer after a quick scan of the thread, so here it is again: 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? Bottom line: Downwelling LW at night is measured all over the world and it varies most directly with: Cloud cover Water Vapor present (in clear sky conditions) Concentrations of all greenhouse gases present Cloud cover is often inversely related to atmospheric pressure, in that rising air and lower pressure often accompany cloud cover, and this cloud cover is a strong source of downwelling LW. How are these bits of data assimilated into your new conjecture that the “GH effect is a pressure phenomenon”? 234. JPeden said @ January 23, 2012 at 11:02 am 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!” “a century’s worth of [CO2 = CAGW] physics”? Really? Do you have a reference for that? IIRC we were being exhorted to be very afraid of an impending ice age unless we liberally coated the polar ice with black carbon in the late 60s/early 70s. See: Nigel Calder’s The Weather Machine and the Threat of Ice. I rather think CAGW started in the 80s. 235. Joel Shore says: Ned Nikolov 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! 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. Ned: Frankly, I am getting very angry that you keep repeating nonsense like this that you can’t defend. I have told you many times in many forums that the reason convection completely offsets the radiative effect is that you have put convection in wrong. It is obvious that you have put convection in wrong because you tell us that you have put it in wrong when you say that it forces the temperatures T_a and T_s to be equal. Everybody knows that in the real atmosphere, convection does not drive the troposphere to an isothermal temperature distribution with height; it drives it to some rough compromise between the dry and saturated adiabatic lapse rates! Your claims that the models are treating convection incorrectly are bogus…It is you who are treating it incorrectly. And, it is well known that an atmosphere with an isothermal temperature distribution with height will not have a radiative greenhouse effect…Hence, it is no surprise that your error has led you to an incorrect conclusion. If you continue to repeat things that you can’t defend and that have been shown to be nonsense then one has to question why you do so!!! 236. Ned Nikolov said @ January 23, 2012 at 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 … :-) How long has temperature been energy? Or am I misreading Ned’s comment? 237. Ned Nikolov said @ January 23, 2012 at 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 … :-) So why do you not answer Willis’s question? Enquiring minds would like to know… 238. George E. Smith; says: “”””” erl happ says: January 23, 2012 at 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. “”””” Well Erl, I DID read the Paper first. Now you mention paper(s) plural, implying there are more than the one paper which I got to by clicking on the author’s link above. So what OTHER papers are there that the link didn’t show me. I read THE paper in considerable detail, and as I said, I didn’t check the equations and integrals for accuracy. I do give the authors credit for checking their math and ensuring their results are correct. So my aprehensions are with the underlying Physics; not the equations nor the integrals which are plain mechanical exercises. Now I DO understand how the WORK DONE in compressing a gas (increasing the pressure) will result in heating the gas (increasing the Temperature), but that is purely a temporary condition. Given that the total mass (and composition) of the gas remains fixed (an assumption), the principle changes in the system are the pressure increase, and the volume decrease, which is the direct cause of the pressure increase. The transient Temperature increase is a consequence of the work done during the compression; and any real body of gas will subsequently cool back to the Temperature that the previous condition permits. There will be no permanent increase in the Temperature following a pressure increase; uinless that cooling is somehow prohibited, which is what happens when stars collapse to the hydrogen ignition point, and its outer envelope is opaque to the internally generated thermal radiation. But I will re-read THE paper to see if I can find what I’ve missed so far. 239. 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. Nobody should be claiming that Anthony is in denial of anything. I think a confusion arose from Ned’s terminology back in the first paper which gave Anthony a concern that energy conservation was being broken. Ned said earlier that: “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.” I’ve double checked with Ned, and he agrees this would be better formulated as: “The extra warmth or ATE observed at the surface of Earth or Venus compared to an airless planet is NOT a result of slowing down (or reducing) of the surface cooling by the atmosphere. It is because there is a bigger proportion of the total atmospheric energy nearer the surface due to the gravitationally induced pressure gradient in combination with solar heating” So, no magicked-up-out-of-nowhere “extra energy”, just an arrangement of mass brought about by ordinary physical laws of nature which concentrates the atmospheric warmth nearer the surface. 240. A physicist says: 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 the form: “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 is obvious in retrospect) was “a tiny thickness of foil can accomplish a huge insulating improvement.” 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. mkelly asks: Being somewhat familiar with radiative heat transfer, I find it odd you compare a solid to a gas. Since, radiation is a surface phenomena the thin slice is all that is needed. Do it this way, how many atoms/molecules are in a square meter of the foil used versus how many molecules of CO2 in a square meter of the atmosphere? Mkelly, that is a good question to ask! Supposing that 0.5% of the atmosphere (by mass) is composed of greenhouse gases (mainly H20 and Co2), if we “magically” condensed those two gases down to a surface layer of water-soaked mylar (which is what it would chemically resemble), that surface layer would be (very roughly) one inch thick. Which is equal to the thickness of twenty-five thousand of NASA’s ultra-thin superinsulating mylar sheets. That’s why it is a fatal mistake for any climate theory to ignore those “trace” greenhouse gases — no matter how careful the subsequent calculations. 241. Phil. says: Willis Eschenbach says: January 23, 2012 at 11:11 am A. C. Osborn says: January 23, 2012 at 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?” Since they are presumably averaging the entire surface, the effect seems obvious. It appears that they’ve left out half the planet, which is why I asked what I’m missing. Willis I asked this earlier, basically they integrate over the sunlit side and assume that the darkside is everywhere at absolute zero (~3K), therefore all they have to do is halve the integral of the sunlit side. Essentially they are assuming a surface with negligible heat capacity, not even true for the moon. 242. Joules Verne says: Ned Nikolov says: January 23, 2012 at 11:02 am ” 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.” In a closed system this is true but as far as I know none of the planets or moons in our solar system are closed systems. You warned about thought experiments but ideal grey bodies are thought experiments in and of themselves since the ideal does not exist in nature anymore than closed systems exist in nature. Regardless, where you go completely off the rails is a wanton disregard for different forms of energy and you can certainly change the amount of kinetic energy in a closed system by converting it some other form of energy. In a gravitational field you can most do it by “moving the components around” because you’re trading off gravitational potential energy for kinetic energy by moving mass from one elevation to another. Moving a warm parcel of air from the surface to some altitude removes kinectic energy and adds gravitational potential energy with no change in total energy contained by the parcel. If you can’t understand how a non-greenhouse atmosphere reaches equilibrium in an isoenergitic state by kinectic energy reduction and gravitational potential energy increasing commensurately with altitude then you should stick with forestry and give up on theorectical physics. This would violate the First Law of Thermodynamics. Re-distribution of energy CANNOT create additional energy. 243. Konrad says: @Joel Shore “poorly conceived and constructed”? “century of well understood physics”? I think I have heard this song before. Remember when you were claiming that LWIR had the same effect over the oceans as it did over land? You had no contrary empirical evidence that time either. 244. kwik says: Funny thing, it all boils down to; When we measure radiation, do we measure radiation that causes a temperature increase…….or….do we measure the radiation from something that is already heated up….. 245. Joules Verne says: tallbloke “It is because there is a bigger proportion of the total atmospheric energy nearer the surface due to the gravitationally induced pressure gradient in combination with solar heating”” This is pure bollocks. The atmosphere farther away from the surface has as much gravitational potential energy as the atmosphere nearer the surface the has more kinetic energy. The result is a wash and each layer of the atmosphere regardless of altitude has the same total amount of energy. What part of that do you not understand? 246. @ One of the George Smith’s > So unless the authors can show a “cause and effect” > linkage between the atmospheric pressure, and the > average surface Temperature ; the “Unified” stature > of their theory remains unconvincing. How about the Ideal Gas Law, for starters: PV=kNT (http://en.wikipedia.org/wiki/Ideal_gas_law), which expresses a linkage between pressure and temperature. I think that’s where N&K start their theory. If you can regard the Earth as an isolated system in equilibrium, and its atmosphere as being sufficiently ‘ideal’ then that seems to be a reasonable starting point for the theory. I eagerly await Part II and further elaboration of the details of this premise. :-| 247. George E. Smith; says: January 23, 2012 at 12:05 pm There will be no permanent increase in the Temperature following a pressure increase; uinless that cooling is somehow prohibited… Hi George; that was Ira’s argument, but it isn’t initial compression and consequent transient heating we are talking about here. It’s simply the way nature has compressible gases and gravity arranged in a pressure gradient as an ongoing condition which causes there to be lots more warmth near the surface when illuminated by a star. Simply put, there are lots more molecules per cc to hold kinetic energy (and therefore heat) near the surface than at high altitude. And the nearest star warms them up. 248. Lars P. says: I think talking of “earth surface temperature” we are still missing the elephant in the room: the oceans. – water has an absorption band that allows for the visible light to penetrate deep – up to 200 m and dissipating the energy over a 200m column of water which is totally different to heating the rocks at the surface. This is why water is not heating up to 60-70°C at the surface – less radiance – water has a heat capacity that trumps almost any other element, rocks or atmosphere – the latent heat needed to freeze, unfreeze and enthalphy – the formation of ice at the surface – ice and snow are very bad for heat transfer – so when the oceans are frozen they don’t lose heat through radiation. This reduces the radiating area of the oceans, ice acts like a thermal skin there where the oceans would be else losing too much heat. – with 1000 Wm-2 solar insulation at equator & tropics there is enough heat to unfreeze these parts of the oceans which will not freeze during the night – heat loss only at the very superficial skin through radiation is not enough to freeze the oceans here, no IR radiation from below as water is opaque to IR All this makes the ocean play a relevant part of the climate not only through the redistribution of the heat, which also plays the major role in reducing differences between max and min temperatures (Hölders inequality). 249. George E. Smith; says: “”””” Ned Nikolov says: January 22, 2012 at 10:06 pm George E. Smith: To answer your question about the cause of recent warming see my post above: https://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! “”””” Well Dr Ned, perhaps you have not yet learned to recognize the tongue in cheek. Of course I don’t subscribe to the “accelerating warming” scare mongering; and I agree with your view of what seems to be happening. It was an attempt on my part to elicit a response as to whether your theory would explain what Temperature rise seems to be valid as the earth recovers from the last glaciation; which apparently per your theory, would imply a correponding pressure increase to go along with the Temperature increase. And I DID read the above linked paper in great detail, and as I have reported elsewhere, I did credit you chaps with having properly checked your equations and did your integrals correctly; so I was not going to waste my time, and Anthony’s bandwidth redoing your integrals, and equation checking, which I consider, just mechanical boilerplate, that authors have to go through. My apprehensions are with the Physics; not the mathematics. And I am also not happy to see the packs of pit bulls running around unsupervised; your work doesn’t deserve that. One aspect of the posts here at WUWT, is that much straight forward Physics, keeps constantly being misapplied. People throw around the Stefan-Boltzmann “law” as if it is some theory, yet to be validated; specially about the moon. In fact it is a fundamental constant of Physics, with a value in terms of other physical constants, and the newly introduced (at the time) Planck’s constant. Sigma = (pi)^2k^4 / 60h^3c^2 with a value 5.67051E-8 W / m^2K^4, and it is simply the definite integral from zero to infinity wavelength, of the Planck Radiation Law. Planck’s radiation law, and the Stefan-Boltzmann constant, apply ONLY to the electromagnetic radiation emitted by a BLACK BODY, which itself is a non-existent ideal theoretical object whose only property, is to totally absorb any and all electromagnetic radiation that falls on it; period. Neither the sun, nor the earth, nor the moon is a black body, and in the case of the earth and moon, they aren’t even close. When people talk about the “black body Temperature” of the sun, they often cite one of two; or both Temperatures. One cited value, is that Temperature of a black body, which gives the correct total EM radiation energy emitted by the sun, so as to yield an earth TSI value of 1362 W/m^2, which NASA recently said was the latest preferred value. It is fashionable to use a similar definition for the earth and even the moon, to get the bandied about numbers of 255K for the earth, and 240 K for the moon. None of these numbers should be used, either for the sun, or the earth, or the moon. The correct “black body” Temperature, to assign to each of these three bodies, is that Temperature, whose black body spectral radiant emission spectrum most closely matches the shape of the emission spectrum of the body in question; not the one that simply has the same area under the curve. The discrepancy between the Total Radiant Emittance of a real object, and that for the corresponding “black body” is properly handled via the “emissivity” of the real object, or more correctly the “spectral emissivity”. The space observed spectrum of the emission from a cloud free night time area of the earth, is either the shape of a black body at the correct surface Temperature; 288 K or 300 K or whatever the surface Temperature is (with the addition of holes for resonance absorption lines due to atmospheric gas components such as CO2 , O3, H2O) or else it is something else; such as the often stated 255 K effective Temperature of the upper atmosphere; plus the GHG holes. I don’t see how it can possibly be the latter, UNLESS one is prepared to abandon the most cherished belief of climate science; namely the insisted belief that gases DO NOT RADIATE THERMAL PLANCKIAN SPECTRA, ACCORDING TO THE GAS TEMPERATURE. Can’t have it both ways; the only way the earth from space EM radiation spectrum can have a 255 K BB spectrum shape (plus GHG holes) is if that cooler upper atmospheric gas IS RADIATING THAT THERMAL SPECTRUM. 250. Joules Verne says: January 23, 2012 at 12:41 pm tallbloke “It is because there is a bigger proportion of the total atmospheric energy nearer the surface due to the gravitationally induced pressure gradient in combination with solar heating”” This is pure bollocks. The atmosphere farther away from the surface has as much gravitational potential energy as the atmosphere nearer the surface the has more kinetic energy. The result is a wash and each layer of the atmosphere regardless of altitude has the same total amount of energy. What part of that do you not understand? Sure I understand, as you know I do from the perpetuum mobile thread. Where we were discussing the Jelbring Hypothesis. Which part of compressible gas being denser at the squashed end of a pressure gradient do you not understand? 251. scf says: Willis Eschenbach is right. “2. Can they simply substitute mu for cos(theta) and then integrate over mu?” The answer is no. The math, as written, from equation 4 to 5, is wrong. You cannot integrate with a double integral when the two variables in question are not independent. In this case, with mu = cos(theta), the two variables are obviously not independent variables. Therefore eliminating one variable with the inner integral is simply incorrect, you will get the wrong answer. Nikolov says “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. ” It doesn’t take many years to understand multivariable integral calculus. In a double integral, you cannot eliminate one of two variables with the inner integral, if the two variables are not independent. Double integrals deal with functions of two independent variables. Certainly I make no judgement myself about the rest of the article, however, to me it is clear that the math, as written, from equation 4 to equation 5, is wrong. 252. Jordan Yankov says: January 22, 2012 at 1:45 pm Paper in Russian but Google Translate is your friend. Eureka! 2005 paper The adiabatic theory of greenhouse effect. Thanks Yankov. More vindication for Nikolov & Zeller by the looks. Joel, get thee to some more maths! 253. Joules: And if density equals out due to the thermal gradient, then it’s still the case that the total kinetic energy is greater near the surface, as you point out, which is what I should have said rather than just total energy, so sorry about that. 254. tallbloke says: January 23, 2012 at 12:43 pm George E. Smith; says: January 23, 2012 at 12:05 pm There will be no permanent increase in the Temperature following a pressure increase; unless that cooling is somehow prohibited… Hi George; that was Ira’s argument, but it isn’t initial compression and consequent transient heating we are talking about here. It’s simply the way nature has compressible gases and gravity arranged in a pressure gradient as an ongoing condition which causes there to be lots more warmth near the surface when illuminated by a star. Simply put, there are lots more molecules per cc to hold kinetic energy (and therefore heat) near the surface than at high altitude. And the nearest star warms them up. TB is correct. George is somehow hung up on the notion of ‘heat generated by compression’. But no change in pressure is need to establish the temperature T of a system. The kinetic energy of the gas particles and the force they exert on each other is sufficient to establish T. What TB is saying is essentially a restatement of the Ideal Gas Law. Look the derivation ‘from statistical mechanics’ here: http://en.wikipedia.org/wiki/Ideal_gas_law#Derivations Note the first equation, which shows that the concept of a temperature T can be derived from Newton’s laws of motion and the Equipartition Theorem (ET), which states that this kinetic energy is shared equally among the degrees of freedom (x,y,z etc). Note also that ET also supports vibrational and rotational modes of gas molecules in the sharing of energy. I will be particularly interested in how Ned&Karl treat the absorption of heat energy by GH gases in this respect. I’m guessing that it somehow contributes to temperature, but that the effect of H2O dominates over CO2. :-| 255. Joe Born says: HLx : “As Willis says, could someone do a step by step integration, with substitutions along with an explanation and post here? :)..” I did that above, but I’ll repeat it here if you missed it: 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 (since that’s the only non-zero part) and divide the result by the earth’s surface area, 4 pi R, where R is the earth’s radius 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. The phi integration limits are 0 to pi/2, and the phi integration limits are 0 to 2 pi. 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). The mu integration limits are now 0 to 1. With those substitutions, you simply end up with a constant times mu^(1/4) as the integrand. That should make it straightforward to reach the result at the end of Equation 5. (My opinion is that they got Equation 6 wrong, but numerically it doesn’t matter much.0 256. Willis Eschenbach says: January 23, 2012 at 11:11 am “It appears that they’ve left out half the planet, which is why I asked what I’m missing.” I think there are two errors in your Mathematica. In the first, you integrated Phi from 0 to Pi instead of 0 to 2*Pi. That’s why you got half their answer. In the second, you substituted cos[theta] for mu in the integrand, but you have to substitute properly also in dμ = d(cos(θ)) = – sin(θ)θ. 257. Should be dμ = d(cos(θ)) = – sin(θ) dθ. 258. OzWizard says: Willis, Re your observation {January 22, 2012 at 2:14 pm} about the integration limits in equations 4 and 5: I read them as covering the top (Northern?) hemisphere only. The assumption is that the average for the top and bottom hemispheres will be identical, and I agree with that assumption. The inner (mu) limits for the integration steps leading to equation 5 run along a meridian through the N pole, i.e. from the “zenith point” (theta = 0) to the point opposite zenith on the dark side (theta = pi). Thus the “mu” limits are from cos (0) = 1 to cos (pi) = -1. The outer (psi) limits are from 0 to 2 pi, around the equator (zenith back to zenith) closing the hemispherical area of their integration. This new “grey body model” is looking good to me and is what I earlier called the game-changer. 259. George E. Smith; says: “”””” tallbloke says: January 23, 2012 at 12:43 pm George E. Smith; says: January 23, 2012 at 12:05 pm There will be no permanent increase in the Temperature following a pressure increase; uinless that cooling is somehow prohibited… Hi George; that was Ira’s argument, but it isn’t initial compression and consequent transient heating we are talking about here. It’s simply the way nature has compressible gases and gravity arranged in a pressure gradient as an ongoing condition which causes there to be lots more warmth near the surface when illuminated by a star. Simply put, there are lots more molecules per cc to hold kinetic energy (and therefore heat) near the surface than at high altitude. And the nearest star warms them up. “”””” Well tallbloke, I undertand perfectlythat the pressure at the surface consists of the weight of all the gas above, so that they higher you go, the less gas remaining above so the lower the pressure, so yes I undertand how gravity creates a pressure gradient. I also understand that it is generally colder, at higher altitudes, so there is also a Temperature gradient, neglecting for the moment the stratospheric regions, where the Temperature goes up again, So of course we should have the exact same conditions in the earth’s oceans; perhaps our two PhDs, can apply their equations and integrals to the earth ocean to calculate how much hotter it is down at the bottom, compared to the surface Temperature. Now as you say, the bottom of the atmosphere IS warmer than at higher altitudes; but that is simply a consequence of the atmosphere being partially transparent to the solar spectrum TSI, so that most of that energy reaches the surface and warms the surface. That atmosphere is also partially opaque to the solar spectrum, which means the atmosphere also absorbs some of that TSI energy, which warms the atmosphere. BUT, the atmospherically absorbed solar energy, is directly proportional to the local density of that atmosphere, so the energy per unit mass is approximately constant so the Temperature increase is independent of altitude. Actually, it’s a bit more complex than that, since the upper atmosphere attenuates the portions of the TSI, that the lower layers can absorb, so the lower layers see a smaller TSI value, so they absorb less than they otherwise would have. So actually, the direct solar heating of the atmosphere has a reverse Temperature gradient; the upper layers of the atmosphere are heated MORE by the sun, than the lower atmospheric layers. But the majority of the TSI reaches the surface, which then in turn heats the lower atmosphere by conduction and LWIR radiation and any other operating thermal processes. Convection of course directly conveys warmer air to higher altitudes. But the resulting Temperature gradient, from the warmer surface to the colder high altitudes, is simply a consequence of the outward flow of HEAT ENERGY, which requires a Temperature gradient, falling T with altitude. The atmospheric pressure has nothing to do with creating the Temperature difference. Turn off the sun, and the Temperaturte gradient will drop significantly, because the ground will get colder. 260. Joe Born says: scf: “In this case, with mu = cos(theta), the two variables are obviously not independent variables. ” Among the maddening things about the paper is that they didn’t make that equation more user-friendly even though the issue had been raised before. What you find is that mu is not rally cos(theta). See my immediately previous post for the detailed integration. 261. Rosco says: Is anybody really suggesting GHGs are responsible for all radiation in the atmosphere ? I understand they are the prime absorbers of IR though O2’s role is downplayed. But emissions of IR must come from ALL constituents of the atmosphere else that makes a few gases that constitue ~98 % of the atmosphere scientific miracles – they either do not absorb heat by any means which is amazing if true as they should fall out of the sky as solids OR upon becoming heated they do no radiate- a proposal I find ludicrous. I think 98% of something at ambient temperature is likely to completely overshadow a few percent at most or in the case of CO2 less than 0.06% by mass. At least water has the capacity to store large amounts of energy without increasing in temperature through a very high latent heat – CO2 increases in temperature by the injection of small amounts of energy – if it were a “potent” GHG the effects claimed for such a small amount would be easily reproducible by experiment – and in 150 years of trying the only results have been spectacular farce and failure. I once saw an experiment meant to show the power of CO2 where 2 glass bottles were heated – one contained air and the other 100% CO2 – yes you heard it right 100%. The 100% CO2 bottle was claimed to have reached a temperature a “few” degrees higher than the air bottle. The significance of this to me was twofold : 1. I am supposed to be concerned by a few degrees caused by 100% CO2 ?? — With current percentages at 0.04% by volume I don’t think I care at all. 2. Despite the glass “trapping” IR both bottles cooled down – the 100% CO2 bottle did not trap any heat for any appreciable length of time. Some people with qualifications argue increasing CO2 levels will cool the Earth by providing an increased pathway for heat to space and not the reverse – who knows – I certainly don’t but I do know that 100% CO2 bottle cooled at a rate similar to the air bottle. 262. Tilo Reber says: HLx: “My question is, is it fair to compare the dense surface of earth with the low density moon surface?” I’m not really doing that. I’m just disputing Joel’s claim that after a few centimeters you get a subsurface temperature that is much warmer than the space immediately above the surface. In order to maintain such a temperature gradient there would have to be an internal source that was generating heat within the moon. Even insulators conduct heat – even if it is much slower. So there is no way that Joel’s supposition that the subsurface temperature could be much warmer than the space immediately above the surface. I think that Nikolev and Zeller have the emperical evidence that agrees with their computation. So if one want’s to claim that their computation is breaking the laws of thermodynamics, one needs to explain why they come up with the right answer. Also, I think people should wait for part 2 before attacking them on the basis of the physical explanations that they will cover in part 2. Also, their priciple seems sound to me. Radiation cannot transfer heat to empty space. That is why empty space is close to zero K. Before radiation can transter heat it must have something to transfer it to. In the case of an atmosphere, the more pressure the atmosphere has, the more densly packed the molecules of that atmosphere are. The more densly packed the molecules are, the more energy per cubic meter that will be transfered to them by solar radiation. At some point, the absorption and reradiation will reach equilibrium. But it is obvious to me that the equilibrium is higher at a higher molecular density, and that density is directly proportional to the pressure. So I cannot understand why people object to the statement that temperature is proportional to pressure in the presence of a constant radiation source. It strikes me as intuitive. 263. gnarf says: >>Gnarf and others, (including the font of total wisdom; Willis), have been arguing against the mathematical skills of two PhD’s in physics. Yeah yeah… we all have big diplomae… aren’t the two PhD’s in physics mentionned here explaining that many many other PhD’s are 100K off? You want to belive, I want to understand…. I made a mistake the integral is right I did not realise they make two substitutions… My problem with this result (150K) is that last year I took a little time to make some simple numerical simulations to find the average temperature of earth without atmosphere. And the results were nowhere as low as 150K (only in one situation…see later). That’s why I first thought something was wrong with the integral. But now I remember I considered each point of earth has a heat capacity different from zero. It stores a part of the received radiations as internal heat, and as a result continues radiating when on the dark side. It does not change the amount of energy radiated….but it changes the timing. Less is radiated on the bright side than received, more than zero on the dark side. And this has a huge impact on the average temperature. Depending how big the heat capacity is, It moves the average temperature up to 250K and more. Here the heat capacity of earth is considered null, temperature falls to 3K as soon as the soil is on the dark side. That is not realistic at all….as they display on their moon temperature charts, the moon soil has a non null heat capacity and it’s temperature does not fall below 90K during night. So they create a result which applies only to bodies with a null heat capacity, and later they say it works nice with the moon, but the charts show the moon does not respect at all the hypothesis!! The result (5) is obtained using over-simplified hypothesis that even the moon does not respect…using more realistic hypothesis gives results up to 100K higher. 264. Stephen Wilde says: “whether your theory would explain what Temperature rise seems to be valid as the earth recovers from the last glaciation; which apparently per your theory, would imply a correponding pressure increase to go along with the Temperature increase” Not necessarily, the changes could be a result of more solar energy getting into the oceans from a reduced global albedo. N & Z do say that their ATE also depends on the radiative energy coming in to the system as well as pressure. Changes in the amount of energy getting past the clouds and into the oceans would achieve warming and cooling without a change in pressure. There would however be a change in surface pressure distribution with the climate zones shifting latitudinally (that is my novel contribution to this debate). Even that need not involve any significant temperature changes with height globally. It would only be a reorganisation of the structure of the atmosphere to maintain the pressure and solar driven lapse rate. For all planets to show the same relationship between solar input and pressure requires that the atmosphere of whatever composition must configure itself around the adiabatic lapse rate. Indeed that would be the only stable configuration. If the atmosphere were to allow a divergence then the outcome over time would be the atmosphere boiling away or congealing on the surface. 265. Joel Shore says: Robert Brown: 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. Robert: The case of the Earth is quite a bit different because the liquid oceans provide much faster heat transport than the almost purely conductive transport one is talking about on the moon. That said, you might be right that my statement about the temperature 1m down on the moon is not obtained just by averaging the insolation received over time…It may be more appropriate to average the temperature. (But, then why do they seem to find the average lunar temperature is so much warmer when they average over depth than when they look right at the surface layer?) I’ll have to think about this some more. 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. Robert: I would strongly suggest you read more about how the radiative-convective transport problem is solved before you hook yourself to the “CO2 effect is saturated” wagon! This is a point that no serious climate scientist believes…Roy Spencer and Richard Lindzen both acknowledge that the radiative forcing due to doubling CO2 levels is about 4 W/m^2. Why CO2 continues to be important is well-understood and the lack of saturation emerges in everything from simple “toy” models to the most sophisticated radiative-convective calculations. The basic things to recognize are these: (1) It is a multiple absorption-emission problem, not an issue of whether or not the CO2 can absorb the terrestrial emission once. (2) Even if the center of the absorption band is so saturated that the effective radiating level is into the stratosphere, the wings of the absorption band are not saturated in this manner. 266. Legatus says: 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? Hence, the evidence suggests that the lower troposphere contains much more kinetic energy than radiative transfer alone can account for! This shows a major conceptual problem, you are assuming that the entire claimed 100K difference between what you estimate the difference between our earth as a airless body and our earth as having our atmosphere is comes entirely from radiative transfer. Of course it does not, it has already been pointed out that there is conduction and convection, to which I add evaporation and friction caused by all that and by the temperature differences for various reasons in different areas resulting in movement of wind and water. Another possible conclusion, therefore, is that all the above non radiative gasses related ways to warm the air amount to 100K, with 18-33K difference from the GHG’s alone. Therefore, your idea that you need another way to add the missing 100K you claim is there may not be needed. Until the heating of air can be measured or at least accurately estimated (preferably the former) from conduction, convection, evaporation, friction, and any other ways heat is added to the atmosphere other than by radiative transfer, you cannot tell us that this 100K additional you say is there cannot be made up by radiative transfer of 0.5% of gasses alone. Assuming your 133K figure is right, the only way you can tell how much is from radiative transfer to the atmosphere is by subtracting all of the non radiative ways the atmosphere is heated. Your two statements above strongly suggest that you have not done that. Your reasoning appears to go like so: 1) The earth without an atmosphere would be 133K cooler than now. 2) This is 100K more than the amount claimed for GHG gasses. 3) the only way the atmosphere can be heated, or the majority of the way, is by radiative transfer. 4) 100K is far more than GHG’s alone can account for. 5) Therefore, something must be adding 100K. 6) Hence, your new theory about how pressure does this, to make up the 100K. Assuming for the moment that the 100K figure above is correct (just assuming), there is an obvious flaw in the above logic, and that is premise 3, that the only way the atmosphere can be heated is by radiative transfer. Since that premise can be shown to be false, we can therefore say that the number 100K which is claimed must be lower than 100K. If it is very much lower, than the ways we already know of that add heat to the atmosphere, conduction, convection, evaporation, friction etc, may be quite sufficient to make up the difference, and therefore your idea of pressure making up the difference may be unneeded. However, I must congratulate you on this new fangled way you have discovered to find out the temperature of a gray body, that of actually gathering data of an actual real world gray body, rather than the present method used throughout climate science today of using models, often the latest method known as “models all the way down”. Perhaps you could introduce this newfangled idea to the climate “scientists”. However…there may be a danger with this new idea of basing at least some of your science on fact, rather than fiction, and that is that fact can be directly measured by others as well, and they may find that you have done it wrong (as some are trying to do in this thread), since they can actually check it out for themselves (replicate the experiment). I suggest that, instead, you switch back to the presently excepted method, which is to base your ideas entirely upon fiction, with careful attention to covering up the fact that it is fiction (hold off all FOI requests). This way you can avoid a lot of possible embarrassment, since you can never be proven wrong, since it will be impossible to check out whether you are right or wrong since the “experiment” cannot be replicated. If anyone then questions your idea, just add the word, “robust’ and it’s all good. Oh, and claim “I am an actual scientist and you are not”, and “I have a PhD and you do not”, that always works (except with Willis, ‘cuz as we all know, he’s a heretic). Come to think of it, though, people are now calling you a heretic, to which I can only say, welcome to the club. I think, however, that you (and all others on this site) need to tell yourself the most important statement any scientist can make, and the central statement that underlies the whole of this newfangled (especially to climate science) idea called “The Scientific Method”, that statement is “I may be wrong”. If you and everyone else here can accept that, maybe we can actually get to some truth here. 267. Joel Shore says: Ned Nikolov 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! 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’. This is completely wrong. It has been demonstrated that the reason you got the greenhouse effect to go away by adding convection is because you added it in wrongly. And, we know you added it in wrongly because you have told us so in your original paper: “Equation (4) dramatically alters the solution to Eq. (3) by collapsing the difference between Ts, Ta and Te and virtually erasing the GHE (Fig. 3).” In other words, you have told us that you added convection into the model in such a way that it gave the unphysical result of an atmosphere with an isothermal vertical profile. And, any elementary book on climate science will tell you that the effective radiating layer being at a lower temperature than the surface is a necessary condition for there to be a radiative greenhouse effect. For example, Ray Pierrehumbert’s book (p. 148) says: “The key insight to be taken from this discussion is that the greenhouse effect only works to the extent that the atmosphere is colder at the radiating level than it is at the ground.” Nobody has contested this point, including you, and I don’t blame you for not contesting it because your own paper essentially points out the error that you have made. However, I do blame you for continuing to post things like the above paragraph that fails to acknowledge this fundamental error and instead pretends that the reason for the discrepancy between what you get and the rest of the scientific community gets is some obscure numerical detail of how they include convection. 268. jjthoms says: R. Gates says: January 23, 2012 at 11:48 am Sorry if this was answered, but I didn’t see the answer after a quick scan of the thread, so here it is again ==== Ned Nikolov claims this thread is about airless planetoids and will not answer my question about how: day=400w/m^2 night = 260+ w/m^2 if ghg are not important then these measurements are these not similar levels to his pressure induced heating As to LW IR in the arctic (North Slope of Alaska (NSA) in Barrow) there is this document : http://www.slf.ch/ueber/mitarbeiter/homepages/marty/publications/Marty2003_IPASRCII_JGR.pdf 140W/m^2 night 150W/m^2 day For Southern Great Plains in Oklahoma http://www.patarnott.com/atms749/pdf/LongWaveIrradianceMeas.pdf Day=260W/m^2 night=400W/m^2 In both these documents this LW IR is in good agreement with Modtran Models based soley on GHGs These measured values show that averaging of GHG heating is happening over the dark and light side – it is not heating on/off with sunlight. 269. Björn says: January 22, 2012 at 8:28 pm “…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…” Hi Björn – sorry,but I must disagree with you here. The surface of Earth is a 3-D fractal object and its surface area depends on the scale of the device used to measure it. For example, if conduction effects at the boundary between solid land and atmosphere are being considered, measurement needs to be done at the scale of the size of an average air molecule and this will make a huge difference to the results obtained as the area will be vastly greater. Applying simple maths to complicated natural structures/systems is always a recipe for ambiguous results and lots of ensuing confusion. 270. Stephen says: I have not been able to read all of the comments so I don’t know if this has been addressed before: Have you considered the atmosphere’s and ocean’s effect in evening out temperatures across the Earth? Because heat-emission varies with T^4, the planet would radiate away heat more slowly as variation across the surface drops (without a change in the mean). The math behind this is the same as the math behind the error in calculating the moon’s temperature. While I haven’t done the math, the data in Table 1 suggests that the temperature across the Earth is much more even than across the moon so the drop in emissions could be important. 271. I’d have a lot more respect for Mr Nikolov and his theory if he would stop with the insults and just answer Willis’s question. It’s a perfectly reasonable question to ask, and it deserves an answer. 272. Peter Spear says: I think the integral in Eq.5 is wrong. If you wish to substitute mu = cos(theta) then you need to correctly substitute for the d(theta) term also: mu = cos(theta) d(mu)/d(theta) = -sin(theta) = – sqrt(1-cos^2(theta)) = – sqrt(1-mu^2) Therefore the correct substitutions should have been cos(theta) = mu d(theta) = d(mu)/sqrt(1-mu^2) It looks like the limits in the integration are screwed up too. We should be integrating mu from -1 to 1 and phi from 0 to Pi since we are trying to skip the night side rather than the southern hemisphere. 273. Joules Verne says: tallbloke Here’s the bottom line, first principle from which you must start to understand this. -a gravity field cannot maintain an energy gradient at equilibrium The presense of an energy gradient means work can be extracted from it and if gravity restores the gradient you have then just violated conservation of energy and have an infinite energy source because in the process of extracting useful work from the gradient you do nothing to diminish the force of gravity. This is the knee-jerk reaction to this theory from anyone with an intuitive understanding of thermodynamic principles. They know there cannot be an energy gradient maintained by gravity in equilibrium. That’s just SO perpetual motion. What most people cannot do, including Duke physicist Robert Brown, is describe how the energy is equally distributed. Brown says the system must be isothermal at equilibrium. That’s his gut talking but this will not happen in a gravity induced pressure gradient. What happens in the gravitational field with a compressible gas. Kinetic energy density is concentrated in the lower regions while gravitational potential energy is concentrated in the higher regions at equilbrium. Willis actually described this to Brown then let Brown somehow convince him he wasn’t right. Willis shouldn’t have rolled over. In the end there is NO mechanism introduced by gravity to increase the total energy capacity of the column nor to give any horizontal layer a greater capacity than any other. Any gain in kinetic energy capacity in a lower layer at equilibrium has an opposing higher gravitational energy capacity in a higher layer resulting in an isoenergetic system at equilibrium. This requirement for isoenergetic system at equilibrium is not negotiable as it would be a blatant violation of first principles in thermodynamics and, quite frankly, the kind of thing upon which crank science is built. Mark my words, this IS crank science and it WILL NOT get anywhere even near as much traction as P&F cold fusion. It’s much easier to show in this case to show what the authors propose is simply not possible. 274. Lars P. says: BenAW says: January 22, 2012 at 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.” Ben it is not only redistribution of heat. The oceans work like a heat pump. Incomming sun radiation by day of 800-1000 W/m2 and radiating 300-400 W/m2 in the night will not cool enough the oceans to freeze at the equator and tropics. As I specified above the oceans are not warmed only at the surface to radiate the heat like solid rock but to the depth – much less escaping radiation. The warmth accumulated this way will be redistributed to the north and south areas above where the sun insulation would be less. At the extremes Nordpole and Southpole where the oceans freeze at the surface, they lose only minimal heat as water is almost as cold as ice and ice does not transmit heat. So ice above will be much cooler then the ocean and radiate very little and not really heat from the ocean. This all functions like a heat pump as I explained above. It is the oceans that drive the temperature of the atmosphere and not vice versa. I further think that the radiation from “greenhouse” is wrongly computed as net radiation. This is Prevost heat exchange and should be substracted from earth or ground radiation causing a net heat transfer from ground to the atmosphere. Only the sun can be computed as net heat transfer as the heat exchange from earth to sun through radiation can be ignored. Here a nice example of solar versus greenhouse: http://climaterealists.com/index.php?id=9004 The oceans are the heat pump. It is the water in the bath that heat the air and not the other way around. 275. Ralph says: >>>>kdk33 says: January 23, 2012 at 10:09 am KDK, the warming of the atmosphere cannot be solely due to conduction and convection, and we know this because when it is cloudy overnight, even a thin layer of cloud can ‘warm’ the atmosphere by 10 – 15 oc. The cloud layer is acting like GHGs, and reraditing the LW back to the surface, reducing the heat-loss to space, making the surface warmer than it would be otherwise. If clouds can do this, and we know they can, then so can GHGs, because we know that (like clouds), they can absob and reemit LW radiation. We don’t notice the GHGs doing this, because they do it all the time. But the intermittent appearance of clouds and their huge effect on night temperatures proves that an ‘insulator’ (an absorber and reemitter) can reduce surface and atmospheric heat loss. Note also that clouds can do this at a very low level. An ‘insulating’ cloud layer at just 1,500 ft can keep the surfsce much warmer, and so this is not simply a matter of increasing the emmision height in the atmosphere. . 276. kdk33 says: Hey Joules Verne (luv that name), What is the proper constraint to apply to the near surface atmosphere. It isn’t isothermal. It isn’t isobaric. Is it at equilbirium? Is it isentropic? If so, how high into the atmosphere does the assumption hold. 277. Kev-in-UK says: Joules Verne says: January 23, 2012 at 3:11 pm Joules – Ok, but why is it that the denser atmosphere cannot provide an increased insulative ‘effect’ and thereby increased temperature? – by the action of simply having more material around at a given atmospheric depth, the kinetic energy is logically ‘more concentrated’ and ergo, the apparent temperature is higher? So, to my (admittedly) limited logic, in the gas phase, a denser gas has more kinetic energy than a lighter gas at the same volume. I don’t see from the N&Z theory that they are insisting on extra energy being introduced – just that, if you like, the energy thats around is more concentrated within the denser gas zone (and thus temps are higher). 278. Dan in Nevada says: tallbloke says: January 23, 2012 at 12:43 pm George E. Smith; says: January 23, 2012 at 12:05 pm There will be no permanent increase in the Temperature following a pressure increase; uinless that cooling is somehow prohibited… My understanding of what N&Z are saying is way simpler than what most people seem to be assuming. It’s quite possible that I’m wrong and if so, please point out where I am. The best I can come up with is a thought experiment regarding a wandering planet that is coated with a layer of frozen N2. The planet gets captured by a sun and assumes a stable circular orbit a fixed distance from the sun. Over time, the planet absorbs radiation from the sun, heats up, and eventually achieves thermal equilibrium. At some point in that process, the N2 becomes gaseous and an atmosphere forms. Conditions are such that none of this atmosphere escapes. Because of the ideal gas law (PV=nRT), a pressure P will develop and remain constant because the planet’s gravity is constant and n does not change. Using eighth grade algebra (third grade in the U.K.), we can show that the relationship between pressure, volume, and temperature will be P = T / V OK, now run the same scenario except this time the wandering planet has enough extra frozen nitrogen such that P will be exactly double the first scenario. This will be slightly more than 2n since the planet is spherical. The planet gets captured by the sun, etc. and achieves thermal equilibrium. Now since we’ve doubled P, we can show the effect of the extra atmosphere compared to the first scenario by 2P = 2T / V meaning that if the atmosphere expands to take up the exact same volume as in the first case, then the average temperature (K) will be twice as high. Arguably, the volume would be larger than V, but then T would have to be higher by the same factor, implying that this process is non-linear with respect to T vs P. That’s my elevator version and it doesn’t involve compressing gas. Whether I’m representing N&Z or not, and if so whether they are right or not, is the question. 279. Stephen Wilde says: “-a gravity field cannot maintain an energy gradient at equilibrium” You are conflating energy and temperature. All molecules of a gas have energy so the more densely they are packed the more energy they carry per unit volume. However one unit of volume can have the same temperature as another unit of volume containing less molecules. Gravity therefore DOES maintain an energy gradient by causing more dense molecular packing at the base of a column. It does not in itself maintain a temperature gradient as far as I know but some have been arguing that it might do that as well to a small degree. The biggie is then what happens if an external energy source is played onto a column with greater density at the bottom. Obviously it will generate more heat in the denser gases. There is 150 years of science to that effect and I cannot understand the intransigence here of those who really should know that to be true. 280. George E. Smith; says: January 23, 2012 at 1:52 pm “”””” tallbloke says: January 23, 2012 at 12:43 pm George E. Smith; says: January 23, 2012 at 12:05 pm yes I undertand how gravity creates a pressure gradient. I also understand that it is generally colder, at higher altitudes, so there is also a Temperature gradient, neglecting for the moment the stratospheric regions, where the Temperature goes up again, So of course we should have the exact same conditions in the earth’s oceans; perhaps our two PhDs, can apply their equations and integrals to the earth ocean to calculate how much hotter it is down at the bottom, compared to the surface Temperature. George: Water is not compressible. Air is. 281. hotrod (larry L) says: Eureka! 2005 paper The adiabatic theory of greenhouse effect. And then there were three independent papers drawing the same essential conclusion! Larry 282. Joules Verne says: tallbloke One might next ask what is different about CO2 from, say, nitrogen. How does CO2 raise the equibrium temperature where N2 cannot when everything else is equal. CO2 in effect lowers the albedo of the surface (land-only). The ground soaks up shortwave energy from the sun, reemits it as longwave energy, CO2 absorbs a portion of the LW energy in it’s absorption band, and reflects about half of it back at the ground. In effect this forces the ground to absorb more energy which in turn raises its equilibrium temperature. There’s a another big misconception floating around here as well in this regard. It’s not possible to attain a higher equilibrium temperature than the S-B temperature for an ideal blackbody. All greenhouse gases can do is raise surface temperature closer to ideal blackbody S-B temperature. There is no way to attain a higher temperature than can be obtained by a perfect absorber. The perfect absorber by definition is, well, perfect and will attain the highest possible temperature for the given energy input – anything less than perfect can only result in a lower temperature. 283. Joules, maybe you didn’t see my followup comment. We are in violent agreement. 284. Mike Monce says: Peter, George, Willis, et al: The integral is correct. The average is taken over the surface of a sphere. i.e. they are averaging over the solid angle 4pi, hence the reason they divide by 4pi outside the integral. What everyone is missing is that the integral is NOT being done in cartesian coordinates but in SPHERICAL coordinates. The integral is done as follows: Set the z-axis of the system pointing towards the sun at the equator, thus their theta angle corresponds to the usual theta in the spherical corrdinate system. At a constant radius, the element of solid angle d(omega) is given by sin(theta)d(theta)d(phi), the integral we are thus calcuating is INT cos(theta)^.25 sin(theta)d(thetad(phi) from zero to pi mu = cos(theta) then d(mu) = -sin(theta) d(theta) subbing in gives: INT mu^.25 (-d(mu)) d(phi) = – 2pi INT mu^.25 d(mu) = -2pi (4/5) mu^5/4 = -2pi (4/5) cos(theta)^5/4 evaluated from zero to pi The limit evaluation just gives a factor of -1 so the net result is 2pi (4/5) Now divide by 4pi and you get exaclty their value of 2/5. I’m surprised no other physicists has caught this before (rgb??) It’s standard physics from the junior year in college. 285. Joules Verne says: hotrod (larry L) says: January 23, 2012 at 3:32 pm “And then there were three independent papers drawing the same essential conclusion!” Three wrongs don’t make a right but as PT Barnum said there’s a sucker born every minute. Lord only knows how many people contrived and believed in a carbuerator that would give their big block Chevy 100 miles per gallon fuel efficiency but the oil companies made sure none are for sale. Undoubtedly the next step for the junk science in the OP will be a claim that they can’t get it published in a reputable journal with a significant impact factor because of a similar conspiracy. Wait for it. In the meantime I’m sure it’s a great source of page views for blogs that dally and dance with crank science because they don’t have anyone on staff who can tell the wheat from the chaff so they print anything that looks remotely feasible to them. Nature prints all the science that’s fit to print. Watt’s Up With That prints everything else.l Joel Shore sure got one thing right – taking this seriously makes all skeptics of the ridiculous CAGW narrative look bad. There’s a boatload of reasons why CAGW boffins are nutters but them not acknowledging crank science that violates the most basic principles of thermodynamics is not one of those things. 286. BenAW says: Lars P. says: January 23, 2012 at 1:01 pm Hi Lars, seems nobody wants to see this elephant, I called it a 600 pound gorilla on another blog. See: https://wattsupwiththat.com/2012/01/22/unified-theory-of-climate-reply-to-comments/#comment-873088 I think that the oceans make the use of any blackbody approach to the temperature on waterplanet earth nonsense. Still it is the basic premise in the theory we are discussing, which starts with: “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.” The temp of the moon is arrived at by using a blackbody approach. The only thing the sun is heating directly is the ocean layer above the thermocline, just ~200m in the tropics, reducing to 0m around the polar circles, depending on the seasons. This is a direct GHE killer since the GHE is supposed to explain the difference between the blackbody 255K temp and our 288K. If this reasoning is totally wrong, let somebody at least explain why, but I would be amazed. 287. Ralph says: >>>Kev-in-UK says: January 23, 2012 at 3:23 pm >> – Ok, but why is it that the denser atmosphere cannot provide >>an increased insulative ‘effect’ and thereby increased temperature? … , >> in the gas phase, a denser gas has more kinetic energy than a lighter >>gas at the same volume. … the energy thats around is more concentrated >>within the denser gas zone (and thus temps are higher). Because without warming, such an atmosphere would continue to cool until you are left with puddles of liquid oxygen and nitrogen (and thus no adiabatic lapse rate whatsoever). (Depensing on the strength of the insolation and the rate of rotation of the planet.) To maintain its depth and pressure gradien an atmosphere needs energy, and you can only get that through surface conduction and LW absorption. Does the latter happen, as well as the former? Yes, of course it does, and we know this because a thin cloud layer can ‘insulate’ (absorb and reemit) the surface and lower atmosphere for some considerable time. . 288. R. Gates says: jjthoms, Thanks for your response…very helpful. Perhaps they only want to talk about airless planetoids, but for a “Unifed” Theory of Climate, for them to restrict it to bodies which have no climate, seems a bit…odd…and not much to “unify”. Seems a basic question to them: Downwelling LW from clouds at night…measured all over the world…without the need for atmospheric pressure theory. Basic greenhouse gas physics works quite well, so WUWT? 289. Joules Verne says: tallbloke says: January 23, 2012 at 3:35 pm “Joules, maybe you didn’t see my followup comment. We are in violent agreement.” My apologies. I had a hard time believing we were somehow not in agreement in the first place because as far as I know this would be first time in the couple of years since I first saw your writing I’d ever read anything you wrote without nodding my head and thinking “Hallelujah, at least one person has a clear understanding of the big picture”. I still think the most pithy comment I’ve seen is yours: “In the big picture the sun heats the ocean and the ocean heats the atmosphere”. If more people would start from that big picture and work backwards filling in the details there would be a lot less misunderstanding of what drives the climate. Climate boffins do it ass backwards starting from the rarefied gases in the top of the atmosphere and working down to the surface from there and the ocean then becomes a detail instead of the biggest single factor other than the sun. 290. paulhan says: January 23, 2012 at 3:03 pm I’d have a lot more respect for Mr Nikolov and his theory if he would stop with the insults Heh. Where have you been for the last fortnight? 291. Tilo Reber says: Joules: “Here’s the bottom line, first principle from which you must start to understand this. -a gravity field cannot maintain an energy gradient at equilibrium” But it can and does maintain a temperature gradient at equilibrium. The potential energy contained by a quantity of gas is not measured as a part of temperature – only the kinetic energy is. So how does the potential energy have any relevance in a discussion of temperature versus pressure? 292. wayne says: A few here seem to have a problem with how to perform this spherical integration… this may help. To be brief, look up at the top-posted article for any parameters you may not understand. A spherical integration of the per-point average temperature field: T.gb = ¼π ∫[0,2π] ∫[0,1] T.i dμ dφ or even clearer: T.gb = ½ * ½π ∫[0,2π] ∫[0,1] T.i dμ dφ As for μ = cos(θ.i), this will only need to be evaluated at two locations, 0 & π/2; first is when the sun is directly above and the second at the 360° terminator that is 90° from the first. So the ultimate results of these two evaluations are cos(0)=1 & cos(π/2)=0. T.gb = ½ * ½π * ∫[0,2π] ∫[0,1] root4( S.0 (1-α.0) μ / (εσ) ) dμ dφ T.gb = ½ * ½π * root4(S.0(1-α.0)/(εσ)) ∫[0,2π] ∫[0,1] μ^(1/4) dμ dφ First integrate ∫ [0,1] μ^(1/4) dμ: = 4/5 * ( cos(0)^(5/4) – cos(π/2)^(5/4) ) = 4/5 * ( 1^(5/4) – 0^(5/4) ) = 4/5 * 1^(5/4) = 4/5 Giving: T.gb = 4/5 * ½ * ½π * root4(S.0(1-α.0)/(εσ)) ∫[0,2π] 1 dφ T.gb = 4/5 * ½ * ½π * root4(S.0(1-α.0)/(εσ)) ∫[0,2π] dφ Next integrate ∫[0,2π] 1 dφ 2π*1 – 0*1 T.gb = 2π * 4/5 * ½ * ½π root4(S.0(1-α.0)/(εσ)) * 1 Simplifying: T.gb = 8π/20π root4(S.0(1-α.0)/(εσ)) T.gb = 2/5 root4(S.0(1-α.0)/(εσ)) or T.gb = 2/5 (S.0(1-α.0)/(εσ))^0.25 If any one out there disagrees with this derivation, lay out yours step-by-step and most here would like to see your expertise. I have also personally numerically integrated this using two separate geometries, one as stated above and the other by a front facing latitude band view… both verify the above math … in Earth’s case 154.7 K. I also further extended this numeric integration to address other configurations that Dr. Brown raised in a prior article here at WUWT. Those too have also been double-checked and all appear correct. 293. Bob Fernley-Jones says: Some here, including Anthony have asserted that N&Z must be wrong because their hypothesis surmounts conservation of energy. However, that may be through a misunderstanding of what N&Z have theorised. Tallbloke, has elaborated on that potential misunderstanding here: https://wattsupwiththat.com/2012/01/22/unified-theory-of-climate-reply-to-comments/#comment-873608 Another way of looking at it is that the consensus community is quite happy to accept that under the GHE, (including any feedbacks), it is currently argued that there is ~150 W/m^2 more energy at the surface than leaves at TOA. So, is this extra energy, offending conservation of energy? Clearly, it cannot be. Somewhat similarly, N&Z are proposing that the surface of a planet will be warmer with an atmosphere, than with an airless one, by virtue of a different mechanism. They are not claiming that the air pressure creates energy, but that the pressure enhances the uptake of the energy source near the surface. These two physics PhDs have probably heard about conservation of energy maybe? Maybe the critics here are putting the cart before the horse without understanding the hypothesis? REPLY: Read Willis latest on the front page of WUWT about equation 8 – Anthony 294. Joules Verne says: Stephen Wilde says: January 23, 2012 at 3:30 pm “-a gravity field cannot maintain an energy gradient at equilibrium” “You are conflating energy and temperature.” Absolutely not. Temperature is a measure of kinetic energy. There are other forms of energy. You are conflating temperature with all forms of energy. “All molecules of a gas have energy so the more densely they are packed the more energy they carry per unit volume.” All molecules of gas have mass and in a gravity well have potential energy such that when they move outboud they lose kinetic energy and gain gravitational potential energy and when the move inbound they lose gravitational potential energy and gain kinetic energy. That’s how a motionless rock dropped from a 10th floor window gains a lot of kinetic energy by the time it lands and hopefully it lands on a crank scientist’s head and knocks some sense into them so the energy isn’t wasted. Molecules of air are like little bitty rocks. Gravity sorts them out into a pile where the lowest ones have a lot of kinetic energy and the highest ones have a lot of potential energy – just like a rock. Capicse? 295. George E. Smith; says: “”””” tallbloke says: January 23, 2012 at 3:30 pm George E. Smith; says: January 23, 2012 at 1:52 pm “”””” tallbloke says: January 23, 2012 at 12:43 pm George E. Smith; says: January 23, 2012 at 12:05 pm yes I undertand how gravity creates a pressure gradient. I also understand that it is generally colder, at higher altitudes, so there is also a Temperature gradient, neglecting for the moment the stratospheric regions, where the Temperature goes up again, So of course we should have the exact same conditions in the earth’s oceans; perhaps our two PhDs, can apply their equations and integrals to the earth ocean to calculate how much hotter it is down at the bottom, compared to the surface Temperature. George: Water is not compressible. Air is. “”””” Gee ! I’ll have to check my Timoshenko, and get back to you on that. I thought that everything; even Neutron Stars are compressible; they squish down to black holes, if you squeeze hard enough. That’s two things I’ve learned here at WUWT. Gases can’t emit thermal radiation; but they are the only things you can squeeze; simply amazing ! 296. Joules Verne says: @Stephen Wilde “It does not in itself maintain a temperature gradient as far as I know but some have been arguing that it might do that as well to a small degree.” You still don’t get it. Gravity maintains TWO energy gradients. One kinetic and one potential. The kinetic gradient decreases with altitude and the potential gradient increases with altitude. The two opposing gradients cancel out and the column is isogenergetic. This is how you can have a perpetual temperature gradient yet not be able to extract any work from it for a perpetual motion machine – a temperature gradient can be nullified by an equal but opposite gradient of energy in a different form. You can’t connect the cold and hot sides of the atmosphere without climbing up in a gravity well and the useful energy represented by the change in temperature is exactly used up by the energy required to climb uphill against gravity. The books thus balance and conservation of energy is once again safe from the abuses of junk science. 297. Joel Shore says: scf says: The math, as written, from equation 4 to 5, is wrong. You cannot integrate with a double integral when the two variables in question are not independent. In this case, with mu = cos(theta), the two variables are obviously not independent variables. Therefore eliminating one variable with the inner integral is simply incorrect, you will get the wrong answer. …. It doesn’t take many years to understand multivariable integral calculus. In a double integral, you cannot eliminate one of two variables with the inner integral, if the two variables are not independent. Double integrals deal with functions of two independent variables. Certainly I make no judgement myself about the rest of the article, however, to me it is clear that the math, as written, from equation 4 to equation 5, is wrong. theta (or mu) and phi are independent variables. Come on guys…Nikolov and Zeller have made huge conceptual errors that render their conclusions completely incorrect, but let’s not bludgeon them about the things that they have done right. It just gives them the opportunity to show that you are wrong and thus to avoid discussing the fatal errors in their theory. You are playing into their hands. 298. Joules Verne says: Dan in Nevada says: January 23, 2012 at 3:25 pm P = T / V Yes but the atmosphere’s volume is not fixed. The pressure is fixed. If you heat or cool the atmosphere the pressure doesn’t change – the volume changes instead. If temperature doesn’t control pressure then it follows that pressure doesn’t control temperature. The author’s contention that pressure sets temperature is thus disproven. QED 299. hotrod (larry L) says: KDK, the warming of the atmosphere cannot be solely due to conduction and convection, and we know this because when it is cloudy overnight, even a thin layer of cloud can ‘warm’ the atmosphere by 10 – 15 oc. The cloud layer is acting like GHGs, and reraditing the LW back to the surface, reducing the heat-loss to space, making the surface warmer than it would be otherwise. If clouds can do this, and we know they can, then so can GHGs, because we know that (like clouds), they can absob and reemit LW radiation. Are you sure about that? You could have the cause and effect inverted. Perhaps the clouds exist because the ground is warmer and the atmospheric moisture profile allows the clouds to form. If you had a dry atmosphere you would have the air cooling at the dry adiabatic lapse rate as you go up in altitude. The dry adiabatic lapse rate is one degree Celsius of cooling for every 100 meters (1°C/100m, 10°C/kilometer or 5.5°F/1000 feet How ever if the air column is not dry, than you can have any lapse rate between the dry adiabatic lapse rate and the wet adiabatic lapse rate. wet adiabatic lapse rate is 0.5°C/100 m (5°C/kilometer or 3.3°F/1000 feet). The saturated adiabatic lapse rate does vary with temperature. In conditions that allow clouds to form, the lower air below the cloud deck will cool at some rate between the dry adiabatic lapse rate and the wet adiabatic lapse rate of 3.3 degF per 1000 ft until the air reaches the lifted condensation level (LCL) {if it is mechanically lifted} where it becomes saturated and the water vapor condenses liberating all its latent heat of evaporation, warming the air it is mixed with. Other wise condensation will release this latent heat at the convective condensation level (CCL). This can cap off convection if the warmed air does not reach the conditions to begin free convection and is too dense to continue rising buoyantly. This can cap off the convective lifting as that layer of air warmed by the condensing water vapor is warmer and less dense than the air below it. But once it is shed of its water load, and dry it is not hot enough and too dense to continue rising through the dryer layers above. You have created a thermal inversion layer that cuts off convection cooling to high altitude. You can see this on a skew T chart under conditions of a cloud deck forming. http://www.theweatherprediction.com/thermo/qanda/ http://geog-www.sbs.ohio-state.edu/courses/G620/jrogers/L12ThermoDiag.pdf It is this liberated energy and the fact that in the cases of low level clouds, indicates that convection is capped that are actually causing the warm nights. The clouds are the symptom not the cause of the warming. The cause of the warm night and lack of cooling is the lack of convective transport to higher altitudes and liberated latent heat of evaporation of water contained in the local air mass. Larry 300. thepompousgit; How long has temperature been energy? Or am I misreading Ned’s comment>>> Methinks he is simply responding to poor wording on my part. I said that a “degree” is worth more “watts” in the tropics than in the arctic. That’s true, but it conveys the suggestion that by moving a “degree” from the tropics to the arctic I’m creating watts. That wasn’t my intention. Better would have been to not that if I take a given number of w/m2 from the tropics, and via conduction and convection, move it to high latitudes where it is then radiated away, I do not create any additional energy, but I do create additional “degrees”. For example, if conduction/convection/whatever cools the tropics from 40C to 39C, that’s a reduction of about 7 w/m2. If that 7 w/m2 gets moved to the arctic where it is -40C, it will increase the surface temperature of one square meter of arctic by 2.4 degrees. If we just consider the one square meter in each region, +40 and -40 have the same total radiance as +39 and -37.6 which averages to 0.7 degrees instead of 0 degrees. The more energy is moved from tropics to poles, the higher the “average” temperature is. How efficiently energy gets moved from tropics to poles is going to be governed by a variety of factors, but two factors will dominate. 1. The higher the surface pressure, the more efficient the tansport will be, the more efficient the transport, the more uniform the temperature, and the more uniform the temperature, the higher the “average” temperature. Other processes such as GHG’s get to play, but they are dominated by surface pressure to the point of being negligible. 2. Re-distribution of energy from tropics to poles must increase until thermal equilibrium is reached. Once the high latitudes have warmed sufficiently that their net loss to space equals the net gain in the low latitudes, the planet is in thermal equilibrium. Low pressure means inefficient movement and hence a very hot tropics and not so much warmer high latititudes. High pressure means more efficient movement which means cooler tropics and warmer high lats. Both have the same net radiance to space, but the average T of the latter is higher. 301. George E. Smith; says: “”””” Ralph says: January 23, 2012 at 3:21 pm >>>>kdk33 says: January 23, 2012 at 10:09 am KDK, the warming of the atmosphere cannot be solely due to conduction and convection, and we know this because when it is cloudy overnight, even a thin layer of cloud can ‘warm’ the atmosphere by 10 – 15 oc. The cloud layer is acting like GHGs, and reraditing the LW back to the surface, reducing the heat-loss to space, making the surface warmer than it would be otherwise. If clouds can do this, and we know they can, then so can GHGs, because we know that (like clouds), they can absob and reemit LW radiation. We don’t notice the GHGs doing this, because they do it all the time. But the intermittent appearance of clouds and their huge effect on night temperatures proves that an ‘insulator’ (an absorber and reemitter) can reduce surface and atmospheric heat loss. Note also that clouds can do this at a very low level. An ‘insulating’ cloud layer at just 1,500 ft can keep the surfsce much warmer, and so this is not simply a matter of increasing the emmision height in the atmosphere. “”””” I’d like a dollar for every time somebody says that clouds warm the surface or keep the surface warmer, and the higher the cloud, the warmer they keep the surface. I’ve been watching clouds for the better part of a century, and I’ve never yet seen one that warmed me up. A cloud gets between me and the sun, and I get colder, instantaneously, no matter what species of cloud it is, and no matter how high or low it is. As a practical matter; as far as “reflection”, or “re-emission” from a cloud, of the surface emitting (right under my feet (nearly)) LWIR, the reflection / re-emission from the cloud; back to my feet (nearly) diminishes as the fourth power of the cloud altitude; in fact worse, because the cloud also emits isotropically, so half goes up, and half comes back down, and the lower moisture density of the higher cloud absorbs less LWIR energy. Now the interesting thing is that the surface is mostly lower in Temperature than my body is at, and I can’t even sense the LWIR from the surface (near my feet) , so how could I possibly detect the miniscule portion of that undetectable radiation that finally makes it back down from the cloud. But I can detect the near IR radiation from the sun, that is strongly absorbed in the water in my skin, so it heats my skin, and often that solar radiation also evaporates moisture and heats the ground and the air, so that it feels hotter than usual, since my skin can’t sweat into that high humidity warm air. And when the sun finally does go down or set, the warm moist air which is constantly rising moves into a cooler upper air layer, until the Temperature drops to the dew point and that magic “warming” cloud appears. Funny thing is, despite the appearance of that cloud, the surface continues to cool after sunset; it never warms up again; and the hotter it was during the day, the higher that moist air has to go for the normal Temperature lapse rate drops it to the dew point, so the cloud forms higher in altitude. And if the amount of moisture was lower during the day, the relative humidity will be lower, and the dew point will be even colder, so the clouds form higher still; but they still don’t warm the surface; the SUN did that during the day. 302. Joel Shore says: Bob Fernley-Jones says: Another way of looking at it is that the consensus community is quite happy to accept that under the GHE, (including any feedbacks), it is currently argued that there is ~150 W/m^2 more energy at the surface than leaves at TOA. So, is this extra energy, offending conservation of energy? Clearly, it cannot be. Somewhat similarly, N&Z are proposing that the surface of a planet will be warmer with an atmosphere, than with an airless one, by virtue of a different mechanism. They are not claiming that the air pressure creates energy, but that the pressure enhances the uptake of the energy source near the surface. You are still not getting it. How is it possible to have 150 W/m^2 leaving the surface as radiation than leaving the TOA unless some of that radiation is getting absorbed (or reflected), i.e., unless there is a greenhouse effect? Air pressure does not absorb electromagnetic radiation….You are simply talking nonsense. These two physics PhDs have probably heard about conservation of energy maybe? Nikolov does not have a PhD in physics…His PhD is in Forest Ecology ( http://www.fs.fed.us/rm/analytics/staff/nikolov.html ) Zeller does not have a PhD in physics…His PhD is in fluid mechanics and wind engineering ( http://www.fs.fed.us/rm/analytics/staff/zeller.html ). In fact, neither of them have any sort of degree (e.g., even a B.S.) in physics. Robert Brown and I have PhD’s in physics. That doesn’t make us right but what does make us right is that we have simple compelling arguments and nobody has been able to refute our arguments using correct physics principles. 303. Willis Eschenbach says: Nick Stokes says: January 23, 2012 at 1:47 pm Willis Eschenbach says: January 23, 2012 at 11:11 am “It appears that they’ve left out half the planet, which is why I asked what I’m missing.” I think there are two errors in your Mathematica. In the first, you integrated Phi from 0 to Pi instead of 0 to 2*Pi. That’s why you got half their answer. Indeed, you are correct. My bad, moving too fast. The problem remains, let me illustrate it correctly. To integrate over the entire surface, mu has to go from -1 to 1. N&Z only integrate from 0 to 1. Shouldn’t the answer for the whole surface be the average of those two? In the second, you substituted cos[theta] for mu in the integrand, but you have to substitute properly also in dμ = d(cos(θ)) = – sin(θ)θ. My point exactly. N&Z didn’t do that. They made one conversion but not the other. They started out with a function in theta, equation 4. Theta is the zenith angle. That is the function they are integrating for all possible angles of theta and phi. So when they replace it with mu, as you point out, they have to make the corresponding change in d(theta) d(phi). But they haven’t. They are integrating straight d(mu) d(phi). Hope that is clearer. Regards, w. 304. Willis Eschenbach says: OzWizard says: January 23, 2012 at 1:50 pm Willis, Re your observation {January 22, 2012 at 2:14 pm} about the integration limits in equations 4 and 5: I read them as covering the top (Northern?) hemisphere only. The assumption is that the average for the top and bottom hemispheres will be identical, and I agree with that assumption. Can’t be, because theta is the zenith angle of the sun (the angle the sun makes with the point overhead. So it’s from sun overhead to sun at the horizon. w. 305. Joel Shore says: Willis Eschenbach says: They started out with a function in theta, equation 4. Theta is the zenith angle. That is the function they are integrating for all possible angles of theta and phi. So when they replace it with mu, as you point out, they have to make the corresponding change in d(theta) d(phi). But they haven’t. They are integrating straight d(mu) d(phi). Willis: What you are missing is what I explained to you before. To integrate a function f(theta,phi) over a sphere, you don’t integrate f(theta,phi)*d(theta)*d(phi). You integrate f(theta,phi)*sin(theta)*d(theta)*d(phi), which can be rewritten as -f(mu,phi)*d(mu)*d(phi) where mu = cos(theta). N&Z have done the math correctly. It is the physics that is completely wrong. 306. George E. Smith; says: Joel Shore says: January 23, 2012 at 2:56 pm Ned Nikolov says: …………………………….. And, any elementary book on climate science will tell you that the effective radiating layer being at a lower temperature than the surface is a necessary condition for there to be a radiative greenhouse effect. For example, Ray Pierrehumbert’s book (p. 148) says: “The key insight to be taken from this discussion is that the greenhouse effect only works to the extent that the atmosphere is colder at the radiating level than it is at the ground.” Joel, As I understand what you just excerpted from Peter Humbug’s book, “the radiating level”, is the apparent level in the atmosphere, that is ACTUALLY EMITTING the LWIR radiation that finally cools the planet, and the Temperature at that level in the atmosphere is LOWER THAN THE SURFACE TEMPERATURE which was the ORIGINAL source of the LWIR emission. The molecular absorption bands of the common GHG species are functions of the electron structure of the emitting molecules; they are NOT first order functions of Temperature. Therefore any radiation being emitted from GHG molecules rather than the surface (ground or ocean), must have a Temperature independent emission spectrum. The only sort of radiation which has a primary dependence on the Temperature of the emitting medium, is thermal Planckian style radiation, which has a spectrum given by the Planck formula, and a total radiant emittance that depends on the emissivity of the emitting material. And we have it beaten into us every day, that this source cannot be the atmosphere principal gases, N2, O2, and Ar, sincer they can’t emit black body style radiation, having no electric dipole moment. So how can the upper atmosphere be emitting a Temperature dependant emission spectrum, since neither the main atmospheric gases, nor the GHGs are capable of doing that (so they say) The ground (solid or liquid) is the only thing capable of emitting a BB like spectrum; so the outgoing LWIR from the atmosphere, should have a spectrum depending on the ground Temperature, less the holes carved by the GHG trace gases. 307. Dan in Nevada says: Joules Verne says: January 23, 2012 at 4:33 pm “Dan in Nevada says: January 23, 2012 at 3:25 pm P = T / V Yes but the atmosphere’s volume is not fixed. The pressure is fixed. If you heat or cool the atmosphere the pressure doesn’t change – the volume changes instead. If temperature doesn’t control pressure then it follows that pressure doesn’t control temperature. The author’s contention that pressure sets temperature is thus disproven. QED” Joules, You didn’t read very carefully. I stipulated that pressure is fixed. That’s what I meant by “a pressure P will develop and remain constant because the planet’s gravity is constant and n does not change”. “Constant” and “fixed” are synonymous for purposes of my example. That being the case, only V and T can vary AND according to the ideal gas law they will vary proportionate to each other, i.e. if V doubles then T must double. For you to say that is wrong, you have to be saying that a heated atmosphere will neither be warmer (have a higher temperature) or occupy more volume than the same atmosphere at absolute zero. If you argue that the volume WILL be greater, then the ideal gas law apparently requires the temperature will be greater also. Dan 308. Joel Shore; You are still not getting it. How is it possible to have 150 W/m^2 leaving the surface as radiation than leaving the TOA unless some of that radiation is getting absorbed (or reflected), i.e., unless there is a greenhouse effect?>>> It isn’t and it doesn’t. You keep averaging things that should not be averaged and then drawing false conclusions from them. Here is the ERBE image of net radiance of the earth: Note that the tropics are net absorbers of energy, and hence radiating at a temperature well below their SB Law equilibrium temperature for the amount of radiance they receive. Note that the high latitudes are net losers of energy, and hence radiating at a temperature well above their SB Law equilibrium temperature for the amount of radiance they receive. Note that for the planet to be in thermal equilibrium, the net loss from the high latitudes must equal exactly the net gain from the low latitudes. Note that there are multiple mechanisms by which energy is moved from the tropics to the high latitudes, and note further that it matters not in the slightest what they are, only that thermal equilibrium is achieved. Now for the hard part that requires that you STOP averaging things that should not be averaged. Note that the increase in the number of degrees above SB Law of the high latitudes is in excess of the decrease in the number of degrees of the low latitudes despite. Despite the amount of energy being moved FROM the tropics being exactly equal to the amount of energy being moved TO the high latitudes, the TEMPERATURE of the high latitudes increases MORE than the temperature that the low latitudes decrease by. Hence, the “average” temperature of the earth with an atmosphere is higher than the average temperature of the earth with no atmosphere because the atmosphere enables the movement of energy from tropics to high latitudes, and while doing so does NOT change the total amount of energy radiated to space, it DOES increase the average temperature of the earth. This is not to say that GHG’s do not also alter surface temperature, but it does show that the surface temperature can be raised, and raised significantly, simply through conduction and convection moving energy from the hottest part of the planet to the colder parts. Every watt moved from tropics to artic and then radiated out raises the temperature of the arctic more than it cools the tropics. That is why if you ignore SB Law which states that P varies with T^4 and instead average just P and compare to averae of just T, you get wrong answers. 309. Willis, “My point exactly. N&Z didn’t do that.” I think they just skipped explanation. The area element that you need for sphere with θ, φ surface integration is sin(θ) dθ dφ. They have gone straight to the μ form, but I think correctly. In this context, if you start getting gamma functions of fractional arguments, something has gone wrong. 310. Willis, I should have added that their changing the range of integration from (-1,1) to (0,1) is OK, because their integrand as defined is zero on (-1,0). 311. Joel Shore says: davidmhoffer says: [A bunch of repetitive stuff that doesn’t at all address the issue.] Dave: We all understand the difference between averaging T^4 and T. That doesn’t get you around having to explain how the Earth’s surface can emit an average of 390 W/m^2 at the surface and 240 W/m^2 at the top of the atmosphere. The only explanation of that is that there is a greenhouse effect. (For those incapable of understanding how averages work, not to mention any names, you can multiply all my numbers by the surface area of the earth and then you just get power.) 312. George E. Smith; said @ January 23, 2012 at 5:37 pm So how can the upper atmosphere be emitting a Temperature dependant emission spectrum, since neither the main atmospheric gases, nor the GHGs are capable of doing that (so they say) I rather thought GHGs are capable of emitting IR and do so in the boundary layer (rather than the upper atmosphere). My reference is Oke’s Boundary Layer Climates (1987) rather than Humbert. I have been feverishly rereading Oke rather than attempting to follow most of the er… convoluted arguments here recently. 313. George E. Smith; says: “”””” Joel Shore Robert Brown ……………………….. Robert: I would strongly suggest you read more about how the radiative-convective transport problem is solved before you hook yourself to the “CO2 effect is saturated” wagon! This is a point that no serious climate scientist believes…Roy Spencer and Richard Lindzen both acknowledge that the radiative forcing due to doubling CO2 levels is about 4 W/m^2. Why CO2 continues to be important is well-understood and the lack of saturation emerges in everything from simple “toy” models to the most sophisticated radiative-convective calculations. The basic things to recognize are these: (1) It is a multiple absorption-emission problem, not an issue of whether or not the CO2 can absorb the terrestrial emission once…….””””” Gotta agree with you there Joel; “CO2 is saturated” is not a good hill to die on. Throughout the bulk of the atmosphere, as Phil has pointed out on many occasions, the mean time between molecular collisions is a lot less, than the mean lifetime of the CO2 excited states (in the 15 micron band), so upon capturing a 15 micron photon, the CO2 quickly loses the enrgy to collisions with other gas moleculesd, N2, O2, and Ar. The CO2 molecules think they are one of a kind, and have no knowledge of a kindred molecule since the nearest one is on average about 13 molecular layers removed. So the captured LWIR energy is quickly thermalized, and the CO2 is reset for its next victim. I keep thinking that some terraflop computer jockey must have rigorously simulated this cascade absorption / thermalization /absorption process for CO2; maybe Peter Humbug; and if he hasn’t, why hasn’t he done it? He reported on a simulation where he took ALL of the H2O out of the atmosphere, and he said he got it all back in just three months (I believe him). I wish he had taken ALL the CO2 out as well. I believe he would at least get all the H2O back in about three months; might have to wait a good while for his CO2. Time to hit the books again Robert . 314. wayne says: @ Bob_FJ Bob, don’t let Joel Shore throw you, he’s trying to be his dishonest self, but don’t let him do that. He shows with every word how little he knows of physics in the real world. He lives in books. If his books never say it, to him it does not exist. He says: You are still not getting it. How is it possible to have 150 W/m^2 leaving the surface as radiation than leaving the TOA unless some of that radiation is getting absorbed (or reflected), i.e., unless there is a greenhouse effect? Air pressure does not absorb electromagnetic radiation….You are simply talking nonsense. BS! Bob, you are not talking nonsense, Joel is. Some of it is absorbed, a lot, depending on the pressure and density. It is this horizontal resonance, the passing of the same energy back and forth horizontally, that it what keeps us warm at the surface. In a way you can call IT radiation trapping. Not specifically GHGs but any gas depending on it’s molar mass and the total mass as I explained last night. Thick gases of any type absorb radiation. This absorption varies from gas to gas; even which multiple gas species are present in the composition, for you have gas-gas interactions, which even modify that. Think once again of your ‘horizontal radiation’ thread. Great insight. Look at some ground based astronomy sites and read how they must always allow for this absorption (mass extinction coefficient), especially at any angle except straight up at the zenith where the minimal occurs. The thicker the gas, the greater the density and pressure and the greater this absorption occurs, any frequency. Bob, you still have never acknowledged that you understand this real world aspect. Please let me know if that occurs so I can stop repeating this to you and start repeating it to some others. You are the closest that shows the insight to ‘see’ it. That is key to really understanding how N&Z’s physically manifests itself in real parameters. 315. scf says: I believe the correct way to integrate equation 4 over the sphere is: T.gb = ∫[-½π,½π] ∫[-½π,½π] root4( S.0 (1-α.0) / (εσ) ) root4 (cos(x)) root4 (cos(φ)) dx dφ with x and φ being the angle of incidence in both directions, or more simply: T.gb = C ∫[-½π,½π] ∫[-½π,½π] root4 (cos(x)) root4 (cos(φ)) dx dφ where C is the constant root4( S.0 (1-α.0) / (εσ) ). Anyway, that’s the way I see it according to the argument in the article, as the incident solar flux striking the sphere. 316. George E. Smith; says: Why is it that people keep explaining that the sun is heating the ground and the ocean during the day, and then during the night everything reverts to radiating energy away to cool things. The earth COOLS much faster during the day, than it does during the night. In the hottest tropical deserts, the earth cools during the day at over twice the global average cooling rate. If it didn’t then the earth would be very much hotter than it is. On the other hand the cooling rate at the poles is more than ten times lower than it is during the day in the tropical deserts; so the polar regions serve very little to cool this planet. In addition the surface emission in the tropical deserts moves even further into the atmospheric window, getting further away from the 15 micron CO2 absorption band, and even dodging much of the Ozone block. And of course with very little moisture in the tropical deserts, the atmosphere never had it so good in passing most of the surface LWIR radiation energy through to space. 317. Willis Eschenbach says: Mike Monce says: January 23, 2012 at 3:44 pm Peter, George, Willis, et al: The integral is correct. The average is taken over the surface of a sphere. i.e. they are averaging over the solid angle 4pi, hence the reason they divide by 4pi outside the integral. What everyone is missing is that the integral is NOT being done in cartesian coordinates but in SPHERICAL coordinates. … [good stuff snipped] I’m surprised no other physicists has caught this before (rgb??) It’s standard physics from the junior year in college. Finally, someone explains the question. You are right, I was 100% wrong. I just ran through it and it makes perfect sense now. I had neglected the Sin[theta] term necessary for the spherical integration. Many, many thanks for clearing that up for me. However, I’m still perplexed about them only integrating over half the sphere. In equation (5) they only integrate mu for the interval zero to one … what about the dark half of the globe? Shouldn’t the final integral be the average of the two halves? Thanks again, w. 318. Joel Shore says: Tilo Reber says: 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. The gradient is not always in the same direction…When it is daytime, the surface is warmer than the subsurface. When it is nighttime, the surface is colder than the subsurface. The subsurface averages out the variations that occur at the surface. 319. Joel Shore says: Willis Eschenbach says: Finally, someone explains the question. I thought I explained that last night: https://wattsupwiththat.com/2012/01/22/unified-theory-of-climate-reply-to-comments/#comment-872924 However, I’m still perplexed about them only integrating over half the sphere. In equation (5) they only integrate mu for the interval zero to one … what about the dark half of the globe? Shouldn’t the final integral be the average of the two halves? To get the average temperature, they are integrating the temperature over the surface area of the sphere (i.e., the integral of T*dA where dA is the surface area element for a surface of radius 1) and dividing by the surface area (i.e., the integral of dA). The surface area is 4*pi, thus the 1/(4*pi) factor. The integral of T*dA consists of the integral over the day side plus the integral over the night side. HOWEVER, THE INTEGRAL OVER THE NIGHTSIDE IS ZERO (BECAUSE THE INTEGRAND IS ZERO) AND HENCE IT DOESN’T CONTRIBUTE. 320. dlb says: Tallbloke, here is the link to one of the Nagihara papers detailing the surface and subsurface temps for the Apollos 15 and 17 landing sites http://adsabs.harvard.edu/abs/2010LPICo1530.3008N The subsurface temps 0.5 to 1.0 metre down are very close to 255K Regards dlb 321. gbaikie says: “davidmhoffer says: January 23, 2012 at 11:18 am Ned Nikolov; Re-distribution of energy CANNOT create additional energy.>>> I never said it did. I said that re-distribution of energy from the warmer tropics to the cooler high latitudes creates more DEGREES.” You are saying the joules of energy from tropics gives more increase the cooler temperate zones. You sort of saying it’s harder for temperate zone to get warmer from every watt from the sun. Or if dial up the watts- the tropics get 1 watt per watt and temperature get 1/2 watt per watt- not necessarily that ratio. I would say something slightly different, the tropics has excessive amount money and if gives some away it can get more energy easier than temperate zones. So not only is it’s money worth more, but tropic as much higher income, and gets substantial tax break for charity. 322. Mike Monce says: Willis, They are indeed integrating over the whole sphere. The theta integral runs from 0 to pi, the phi integral runs from 0 to 2pi. These are the standard limits to cover an entire sphere. To prove they are correct, do the integral: R^2 INT sin(theta)d(theta)d(phi) which yields 4piR^2 which of course is the surface area of a sphere. Or easier, just check a mulitvariable calculus book :-) 323. hotrod (larry L) says: George E. Smith; says: January 23, 2012 at 6:30 pm And of course with very little moisture in the tropical deserts, the atmosphere never had it so good in passing most of the surface LWIR radiation energy through to space. George I do not think that is correct. If measured as relative humidy, yes the deserts have very little water (compared to what they could hold), but their absolute water content is actually higher than in the arctic due to two effects. First hot air can hold enormous amounts of water, cold air even at 100% humidity can hold very little. Second the tropopause is much higher in the equatorial regions and adjacent deserts so the atmospheric layer that can hold lots of water vapor is also thicker. If you look at the map on page 789 of this pdf you will see that the Sahara has 2x-3x the mean precipitable water content as the polar region. http://docs.lib.noaa.gov/rescue/mwr/096/mwr-096-11-0785.pdf “It has often been said that the air over deserts is not really dry and these areas lack precipitation primarily because of their high saturation deficits and stable atmospheric conditions. This is confirmed by the annual map of precipitable water. The Sahara, for example, has nearly as much precipitable water as the cooler regions of northern Europe and the northern United States. The deserts do have less precipitable water than other areas in the same latitudes, however.” Larry 324. kdk33 says: So, WUWT readers are supposedly a technically literate, science oriented bunch. I’m assuming there are a few out ttherre who have a bit of a handle on engineering thermodynamics. Can someone please help me out with the following: The dry adiabatic lapse rate can be derived assuming an isentropic ideal gas atmosphere with a gravity imposed pressure gradient (I can provide the math; I’m hoping it’s not necessary). Furthermore, the T,P relationship for an ideal gas is T2/T1 = P2/P1^(R/Cp). R is the gas constant and, for and ideal gas, Cp can be taken as 5/2R so T2/t1 = P2/P1^0.4. If I solve for T2 setting P1 and T1 as the conditions at the tippy tippy top of the atmosphere and P2 as the pressure at the planet surface I calculate an enormous surface temperature, T2. My questions are as follows: 1) is the near surface atmsophere generally considered isentropic? If not, why does the DALR follow from this assumption? 2) If the atmosphere were isentropic all the way up, am I correctly calculating the surface temperature? It not, why not? 3) If the near surface atmosphere is isentropic, but the entire atmosphere is not, where does the isentropic assumption break down? Why does it break down? 4) (this one is for bonus points) Is the isentropic assumption and equilibrium criteria or a steady state condition? Thanks! 325. Bill Hunter says: Here is a thought experiment that might help: In Jelbring’s world with its energy impenetrable shell and surface, heat is trapped inside of the system. So You then turn off gravity. Clearly then two things will happen. 1) The gases will go isothermic; 2) No heat will be lost in the system. From that you should conclude that gravity is the cause of the “surface temperature” or more correctly heat distribution but not the cause of any energy in the system. Thus the perpetual motion arguments really don’t apply to this world. 326. hotrod (larry L) says: The lapse rate only applies to the tropsphere (60,000-80,000 ft altitude). This is the area that normal convective mixing takes place, At higher levels the atmosphere stratifies, into several distinct layers some of which get hotter with altitude as other mechanisms modify the air temperature. Such as ozone heating. http://apollo.lsc.vsc.edu/classes/met130/notes/chapter1/vert_temp_all.html Larry 327. Bob Fernley-Jones says: Joel Shore @ January 23, 5:11 pm Thank you for your typical style of response. I compare your response to part of mine: [Bob_FJ:] Another way of looking at it is that the consensus community is quite happy to accept that under the GHE, (including any feedbacks), it is currently argued that there is ~150 W/m^2 more energy at the surface than leaves at TOA. So, is this extra energy, offending conservation of energy? Clearly, it cannot be. Somewhat similarly, N&Z are proposing that the surface of a planet will be warmer with an atmosphere, than with an airless one, by virtue of a different mechanism. They are not claiming that the air pressure creates energy, but that the pressure enhances the uptake of the energy source near the surface [Joel:] You are still not getting it. How is it possible to have 150 W/m^2 leaving the surface as radiation than leaving the TOA unless some of that radiation is getting absorbed (or reflected), i.e., unless there is a greenhouse effect? Air pressure does not absorb electromagnetic radiation….You are simply talking nonsense. You seem to have misunderstood what I wrote; particularly see the part where I’ve herewith added bold emphasis. Oh and BTW, it is not me doing the theorising, but what I believe that N&Z are proposing, which many here do not seem to understand, including as a lecturer, yourself. As Wayne points out below, you will not find it in Raymond Pierrehumbert’s book or the like, so you may find it hard to even consider it. Concerning N&Z’s qualifications, sorry, I thought they were introduced somewhere in one of the threads as belonging to your discipline. Nikolov does not have a PhD in physics…His PhD is in Forest Ecology ( http://www.fs.fed.us/rm/analytics/staff/nikolov.html ) Zeller does not have a PhD in physics…His PhD is in fluid mechanics and wind engineering ( http://www.fs.fed.us/rm/analytics/staff/zeller.html ). In fact, neither of them have any sort of degree (e.g., even a B.S.) in physics. Of course physics is quite a wide field and according to your bio, you have been mostly active in non CAGW stuff. Recently, he has also developed an interest in global climate change.. Also, according to your bio it seems that you are no more qualified than they in this area, and they have certainly been more active in diverse constructive publications. http://www.rit.edu/~w-physic/Faculty_Shore.html Once again your elitism and dogma comes to the fore in these exchanges 328. KevinK says: A physicist wrote; “Kevin, NASA’s multilayer reflective foil trick works only if there is a vacuum between the sheets” Previously, KevinK wrote; “The reflective plastic foil (aka MLI; Multi Layer Insulation) only works in a vacuum (sans conduction and convection).” DID I MISS SOMETHING HERE ??? Yes, both statements are in the English language, but they sure seem to express the same conclusion to to me ??? So, the external fuel tank of the Space Shuttle spent almost all of its operational life outside of a vacuum environment. I.E. from Cape Kennedy, filled with cryogenic fuel, then launched, followed up by burning up during reentry to the Earth’s surface. All of these environments seem to be lacking any significant vacuum ? Last time I visited Cape Kennedy, I do not seem to remember any significant presence of a vacuum. But I was on vacation, so I may have missed it. So, is your position that the “GHE” only occurs when a vacuum is present? If so, how many molecules of GHGs are present in your vacuum, most definitions of a vacuum seem to exclude the presence of gas molecules. By the way, that “NASA TRICK” of applying MLIs was in fact figured out by the folks that originally figured out how to operate “spy” satellites (look up the CORONA system) before the NASA folks borrowed it. Cheers, Kevin. 329. Bill Hunter says: In Jelbring’s world with its energy impenetrable shell and surface, heat is trapped inside of the system. So You then turn off gravity. Clearly then two things will happen. 1) The gases will go isothermic; 2) No heat will be lost in the system.” More stuff, add a radiative surface with storage capability. 3) The surface will cool and equilibriate with the atmosphere via drawing stored heat from the soil. 4) Surface radiation will reduce. 5) The total energy in the larger system has not changed. Conclusion: Surface radiation is a very poor indicator of the GHE (or better stated as ATE) Its total energy and its distribution between various materials with various emissivities that determine how much heat can be stored in the system and gravity only determines its distribution. With zero storage capacity there would be a zero greenhouse effect and the GHE at the surface is caused by gravity, not backradiation. 330. kdk33 says: January 23, 2012 at 7:21 pm “My questions are as follows:” I think your use of isentropic is unorthodox. It applies to a process or a flow, but not usually to a continuum. What would make sense is to speak of a region where you can move air from one level to another isentropically (your P2/P1 formula) and it arrives at the ambient temperature. But that, in conjunction with hydrostatic pressure, just means that the lapse rate is the DALR. So the answers are: 1. Doesn’t mean anything. The DALR is just a number from gas and gravity properties. It doesn’t follow from an assumption about the state of any one atmosphere. 2. Just asks (as above) if the actual lapse rate was equal to the DALR. Answer, yes, your calc is then correct. But usually that lapse rate is less than DALR. 3. Again, you’re just asking whether the lapse rate is the DALR, and if not, why not. It’s usually less, and the commonly quoted reason is condensation of water, which changes the effective heat capacity. My view is that another reason is that work has to be done to maintain the lapse rate (overcoming Fourier Law type conductive leakage), with energy provided by the wind. Radiative transfer is powerful leakage, and I think causes the lapse rate to fall short of DALR. 4. Again, check the meaning of isentropic. 331. Dan in Nevada says: Joel Shore says: January 23, 2012 at 6:13 am “tallbloke says: Now, The practical demonstration by Konrad Hartmann in the recent post on my site (linked above in an earlier comment) shows that higher pressure does indeed enhance the sensible atmospheric heat generated by the passage solar radiation. This is an empirical result. No conservation law is harmed during the process. Empirical reality cannot break laws of nature! Sorry…but one poorly conceived and carried out experiment does not overturn more than a century’s worth of physics even when it tells you what you want to believe. Konrad hasn’t even tried to figure out how his data, even if correct, could be compatible with well-understood physics. It is really bizarre what you guys seem to think constitutes evidence!” I’ve been doing a little self-educating via Google – got to love the inter-webs! What Konrad has been investigating has been known as Gay-Lussac’s law since the early 1800’s. According to Wikipedia, it should really be attributed to Guillaume Amontons who published it around 1700~1702. So, Joel Shore believes Konrad is attempting to overturn a century’s worth of physics when he’s actually confirming 3 centuries worth of physics. It’s not worth pointing out that this could conceivably be considered “compatible with well-understood physics”, but I will anyway. Bizarre or not, I’ll consider three centuries’ worth of accepted physics as evidence. The only reason Konrad is taking on these experimental chores is to try to rebut the notion that “settled team science” somehow trumps physical laws that have been accepted for literally centuries and have never been refuted. I’m beginning to see what N&Z are up against. N&Z are claiming, I believe, that Charles’s law, a corollary to Gay-Lussac’s law, is what governs the surface temperature of planetary bodies with atmospheres. That’s harder to test since you have to allow the gas volume to vary, but what Konrad is doing is completely valid since everything is based on the IGL (PV=nRT). This is a very simple algebraic expression that even I can wrap my old brain around. Why is it so hard for actual physicists? 332. don penman says: I don’t think that the length of day will effect the average temperature of a planet without an atmosphere ,it will just get hotter during the day and colder at night .Some are trying to confuse by including the effects of an atmosphere which would be affected by the length of day.The effect of having an atmosphere even one without ghg is like the true greenhouse or wearing an extra item of clothing ,you could argue that the air trapped by the glass of the greenhouse will become the same temperature but that temperature will still be warmer. 333. Willis Eschenbach says: Mike Monce says: January 23, 2012 at 7:13 pm Willis, They are indeed integrating over the whole sphere. The theta integral runs from 0 to pi, the phi integral runs from 0 to 2pi. They are integrating over mu = cos(theta). For theta running from 0 to Pi as you say, this gives a corresponding interval for mu from 1 to -1. However, they are only integrating over half of that: Joel says it makes no difference, because the temperature is zero on the night side. I can’t see how that wouldn’t affect the integral, but Joel’s science-fu is strong. I just don’t understand why they’re only integrating from 0 to 1, and not -1 to 1. Thanks, w. 334. Bill Hunter says: The conclusion above was poorly stated. The last sentence was OK though. I like to use a flow diagram to demonstrate how the system really works. Tallbloke’s talkshop as a nice discussion on flow diagrams and radiation that worth reading. 335. Willis, An integral is like a summation of many really small segments of a function. So, if you have all ZEROS (T = 0 over the whole night side of the planet), and you add them, what do you get? You get 0 + 0 + 0 + … + 0 = 0.0. That’s why the night-side portion of the integral, where mu varies from 0 to -1 can be ignored when calculating the average temperature … Do you get it now? 336. Willis, Further clarification regarding the integral – night-time temperatures (being ZEROS) are ignored in the summation (integration), but the overall sum (integral) is divided by the ENTIRE surface area which is 4*pi … 337. Ralph says: <<<<>>>>>> Nonsense. A cloud layer can move in from a totally different weather system hundrds of miles away, and the effect will be the same. Whether the night air is still convective (rare), or whether the airmass is completely stable (more usual), a cloud layer will ALWAYS result in higher nighttime temperatures – because the cloud is ‘insulating’ the LW radiation from the surface. And I have been watching the weather for 30 years. . 338. George E. Smith; says: “”””” hotrod (larry L) says: January 23, 2012 at 7:15 pm George E. Smith; says: January 23, 2012 at 6:30 pm And of course with very little moisture in the tropical deserts, the atmosphere never had it so good in passing most of the surface LWIR radiation energy through to space. George I do not think that is correct. If measured as relative humidy, yes the deserts have very little water (compared to what they could hold), but their absolute water content is actually higher than in the arctic due to two effects. “”””” Dang !! Rod, I think you done got me there. I suppose if it is hot enough and the supply of atmospheric water is limited by distance from bulk supplies, then the relative humidity can get so low, that there just never is much chance of getting down to the dew point. Gosh I learn something new every day, specially at WUWT. However I stick by my point that because of the surface Temperature (+60 deg C or more), the Wien displacement moves the LWIR spectral peak right into the center of that atmospheric window, which as I recall, is a hole in the H2O absorption spectrum The peak even moves below the 9.6 micron Ozone band. Incidently, although Total radiant emittance goes as T^4 for BB like emitters, the peak spectral radiant emittance goes as T^5, so the peak of the desert floor emittance spectrum is way higher (over 2x) the value for the global mean Temperature. Thanks for the insight and correction hotrod. 339. Ralph says: <<<<<<<>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> George, you are wrong on so many points, it is difficult to know where to start. Firstly I did not mention the Sun, because I was talking about night temperatures. I’ll give you a clue here, there is no Sun at night. Secondly, I took great pains to not mention ‘warming’ and when I did use the term once I placed it in inverted commas, because the ‘warming’ is only reletive to the temperature you would obtain without nighttime clouds. ie, a night with clouds will ALWAYS be warmer than one without, or in effect you will percieve a distinct ‘warming’ sensation in comparison to a cloudless night. Thirdly, although you may rail against the term ‘warming’ and claim that that clouds do not warm you (or the surface of the Earth) well technically they do. There is radiative transfer from you to the cloud (and to the atmosphere in general), which cools you down (try standing naked in the desert at night – that is not all conductive cooling, as an infra-red camera will easily demonstrate). But there is also radiative transfer from the cloud to you, warming you up (and again, a sensitive infa-red camera will pick up the warmer clouds in comparison to the colder cloudless sky). Now just because there is more transfer from you to the atmosphere, than there is from the clouds to you — and so the net effect you feel is cooling — does not mean that the clouds are not trying to warm you up. 340. Ralph says: Sorry mods, the first comment here was in response to Hotrod Larry, while the second was in response to George Smith. The brackets deleted the quotes !! . 341. George E. Smith; says: “”””” Ralph says: January 23, 2012 at 9:45 pm <<<>>>>> Nonsense. A cloud layer can move in from a totally different weather system hundrds of miles away, and the effect will be the same. Whether the night air is still convective (rare), or whether the airmass is completely stable (more usual), a cloud layer will ALWAYS result in higher nighttime temperatures – because the cloud is ‘insulating’ the LW radiation from the surface. And I have been watching the weather for 30 years. “”””” So I could have a cloudless day with surface Temperatures of say 40 deg F, and typical wintertinme low humidity, and after sundown a cloud layer at 20,000 feet comes in on some jet stream, and my surface Temperature will go up above 40 deg F ?? So the cloud might be -50 deg C, so just where does the energy come from to raise the surface Temperature; it’s after sundown, so it isn’t solar, and my cold air was going all the way up along with the lapse rate to where that 20,000 ft cloud came in, and my surface is getting even colder, as it radiates its 400 Watts per m^2 or more. I will grant you that a massive storm of air might sweep in from some hot place with clouds, and simply replace all the cold air I had during the day, and presumably I will be warmer; but it wasn’t radiation trapping from my surface that warmed me up; and notwithstanding that storm cloud, it will still cool down as the night progresses; it won’t continue to get hotter, unless even hotter air contines to come in from somewhere else. If you need to invoke a Santa Ana to make your case; then I submit, that you have no case; and i’ve been working at this for more than twice as long as you’ve been watching the weather. 342. Willis Eschenbach says: Ned Nikolov says: January 23, 2012 at 9:42 pm Willis, Further clarification regarding the integral – night-time temperatures (being ZEROS) are ignored in the summation (integration), but the overall sum (integral) is divided by the ENTIRE surface area which is 4*pi … Excellent, Ned, many thanks. That was the missing part in my understanding, that you are dividing through by the surface area. That being the case, it doesn’t matter that you have not integrated over the night side. Much appreciated, w. 343. Dan in Nevada says: re: Dan in Nevada says: January 23, 2012 at 3:25 pm A self-correction. Obviously, on reflection, my second scenario vs my first scenario should have been stated as 2P = 2nT / V If we are roughly doubling n (in order to double P), then if V2= V1, T2=T1. However, it seems likely that V2 would be significantly larger than V1, hence T2 would have to be proportionately larger than T1. So, less bang for the buck, but a bang nonetheless. 344. George E. Smith; says: “”””” Ralph says: January 23, 2012 at 10:06 pm <<<<<<>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> George, you are wrong on so many points, it is difficult to know where to start. Firstly I did not mention the Sun, because I was talking about night temperatures. I’ll give you a clue here, there is no Sun at night. “”””” Ralph, why don’t you start at “it is difficult to know where to start.” Yes I know you didn’t mention the sun; so let’s NOT mention it; the global warming positive cloud feedback people never do, so why should we. So we will have perpetual night without that unmentionable, and we can have jet stream clouds come and go, and absolutely NO external source of ground level energy from anywhere, since that is unmentionable; and the longer I wait, the colder I will get, and I will continue to get colder, cloud or no cloud. Now I can’t even detect the 10 micron radiation that is emitting from the ground around my feet, my six senses are completely insensitive to it. So after it leaves the ground and disperses according to the inverse square law of cloud height; not to mention the cosine^4 of the oblique angle to that cloud that just came by, and some small part of it, gets absorbed by the cloud, the part that hits the cloud that is, and then the cloud re-emits isotropically, so half of it is lost to space, and only half of the cloud emission heads towards the surface, and that too, has an inverse square law dispersion plus another cosine^4 obliquity factor, so I have a 1/height^4 attenuation plus a cos^8 obliquity angle attenuation, before any of that undetectable radiation from near my feet makes it back to make me feel warmer as you put it. But one thing is for sure, I will never get warmer with or without that cloud, and if morning twilight ever comes, it will be much colder than the evening twilight was, and I’ll be praying for some unmentionable source of energy to show up, because under your evening sky, cloud or no cloud, there was only a continual loss of energy from the status quo prior to the disappearance of that unmentionable source of energy. And since we are talking only night time; in the complete unmentionable absence of any external energy source, other than the ground around me, and the atmosphere, with or without cloud, then we can completely dispense with the usual explanation of the cloud positive feedback mechanism, which is that the higher the cloud, the less moisture it contains so it blocks less of that unmentionable external energy source, so more of that external energy reaches the ground; ergo, the higher the cloud the more the cloud warms the ground; that is the guts of THEIR explanation of cloud positive feedback. The fact that the same diminishing density with altitude has the exact same diminishing effect on the capture of outgoing LWIR by the cloud, seems to be entirely lost on those people. Well I’m so happy to be so wrong on so many poinst Ralph, that you don’t even know where to start. 345. Ralph says: >>>George Smith >>>So I could have a cloudless day with surface Temperatures >>>of say 40 deg F, and typical wintertinme low humidity, and after >>>sundown a cloud layer at 20,000 feet comes in on some jet stream, >>>and my surface Temperature will go up above 40 deg F ?? Who says the temperature goes up? You are distintly bad at reading, George, try again. I NEVER SAID THE TEMPERATURE GOES UP, is that clear enough for you?? Clouds reduce cooling, because of the Greenhouse Effect. Note the term here, George, REDUCE COOLING. This results in a warmer temperature than a night without clouds. Do you understand that this is not warming, George, it is reduced cooling. Warmer than without clouds, but still a net cooling effect (but slightly reduced). Clear?? . 346. Ralph says: >>>Willis Eschenbach says: January 23, 2012 at 10:38 pm Willis, just a quick question. Why cannot you just assume the Earth to be a flat disk with a sun-side and a night side? That is, after all the full amount of energy that the earth receives on its sun-side – a complete disk-full. Is this an approximation too far? . 347. gnarf says: To calculate this average surface temperature, you need to consider not only the surface element with null heat capacity, but the column below, and apply Fourier’s heat conduction law with heat capacity. This way, the temperature of a soil column around it’s one day trip is the solution of a differential equation Ti(theta, t, depth). This solution shows the heat stored below surface is responsible for the 100K on the dark side of the moon. T can be expressed using phi and angular rotation speed of earth. You get Ti(theta, phi, depth) to integrate over the sphere for depth=0. 348. gnarf says: January 23, 2012 at 11:26 pm To calculate this average surface temperature, you need to consider not only the surface element with null heat capacity, but the column below, and apply Fourier’s heat conduction law with heat capacity. This way, the temperature of a soil column around it’s one day trip is the solution of a differential equation Ti(theta, t, depth). This solution shows the heat stored below surface is responsible for the 100K on the dark side of the moon. T can be expressed using phi and angular rotation speed of earth. You get Ti(theta, phi, depth) to integrate over the sphere for depth=0. What you say is correct here. The question is, how much difference does it make? I don’t think it’s going to get you back up to 255K or anywhere near it. Do you? We have the Apollo results, from mid latitudes, which give readings over depths down to a metre or so, which should be enough. So let us know when you’ve done the calcs. I assume you now accept that Ned’s integral for the very surface is correct and you were wrong about that? 349. tallbloke says: January 24, 2012 at 12:36 am “What you say is correct here. The question is, how much difference does it make? I don’t think it’s going to get you back up to 255K or anywhere near it.” Yes, it does. A solid in contact with an oscillatory surface temperature has a decaying oscillation at depth, about a fairly constant (with depth) mean. And for the Moon that mean is about 255K. 350. Dr Burns says: Nothing is to be gained by Willis or Ned attempting to denigrate each other. Ned has some good ideas but I disagree with his conclusions, as obviously does Willis. My summarised thoughts are that there are two broad situations in a simple model of the Earth. a) the 30% of the Earth with no clouds and b) the 70% with clouds. a) If a packet of air containing IR absorbing gases is considered, it will exchange heat with its surroundings by evaporation/condensation, convection, conduction and radiation. The closer to the earth’s surface, the smaller the significance of radiative transfer, yet at high altitudes, radiation is the dominant mechanism. The IPCC view is that only radiative heat transfer takes place and clouds are effectively ignored. b) For the 70% of the Earth’s surface covered by clouds, radiation from the Earth’s surface is irrelevant. Heat is transferred primarily by evaporation/condensation and convection. Cloud temperature is mainly set by the lapse rate. Heat is absorbed from the sun and lost to space mainly by radiation, from the tops of clouds. IR absorbing gases have little impact on heat transfer. My view is that overall the impact of IR absorbing gases is far less than that claimed by IPCC but not zero as claimed by Ned. Looking forward to any constructive criticism. 351. OzWizard says: My mistake, Willis, regarding use of the word “Zenith”. Thanks for pointing that out (we can all learn from each other). Try this corrected version: The inner (μ) limits for the integration steps leading to equation 5 run along a meridian through the N pole, i.e. from the “sunlit equator on the left of Figure 1” (theta = 0) to the point diametrally opposite that point, on the equator on the dark side (theta = pi). Thus the “μ” limits are from μ1 {= cos (0) = 1} to μ2 {= cos (pi) = -1}. The outer (psi) limits are from 0 to 2 pi, i.e. one complete circuit around the equator, closing the hemispherical area of their integration. So I believe my primary statement stands: … I read them as covering the top (Northern?) hemisphere only. The assumption is that the average for the top and bottom hemispheres will be identical, and I agree with that assumption. Are we on the same page now? 352. gnarf says: >>What you say is correct here. The question is, how much difference does it make? I don’t think it’s going to get you back up to 255K or anywhere near it. Do you? Yes, that’s enough to change the average surface temperature a lot. In this article, the author calculates average temperature for a single point on moon surface, using different values for the heat capacity. He does not model the underground surface…he simply considers the surface element has the heat capacity of the entire soil column below. From 169K with null heat capacity it goes to 247K and more. Increasing heat capacity leads to increased average temperature. The last curve giving Taverage=247K uses a heat capacity of 5MJ/K/M2 which is lower than earth ground capacity (5MJ/K/M2 is the heat capacity of earth atmosphere in fact). http://scienceofdoom.com/2010/06/03/lunar-madness-and-physics-basics/ So yes, introducing heat capacity in these formulae gives you a temp in the 250K range. >>I assume you now accept that Ned’s integral for the very surface is correct and you were wrong about that? Yes, wrote it yesterday, Sir. >>So let us know when you’ve done the calcs. What do you think if people who built an entire theory based on a null heat capacity do the calcs? Afer all it’s their job and they are better at maths than me as shown here. 353. wayne says: scf says: January 23, 2012 at 6:29 pm I believe the correct way to integrate equation 4 over the sphere is: T.gb = ∫[-½π,½π] ∫[-½π,½π] root4( S.0 (1-α.0) / (εσ) ) root4 (cos(x)) root4 (cos(φ)) dx dφ with x and φ being the angle of incidence in both directions, or more simply: T.gb = C ∫[-½π,½π] ∫[-½π,½π] root4 (cos(x)) root4 (cos(φ)) dx dφ where C is the constant root4( S.0 (1-α.0) / (εσ) ). Anyway, that’s the way I see it according to the argument in the article, as the incident solar flux striking the sphere. — — Sure. I see that, the alternate geometry. Just look at this on 12/30: https://wattsupwiththat.com/2011/12/30/feedback-about-feedbacks-and-suchlike-fooleries/#comment-848611 . Is that not your geometry? Mine too. That is the first thing I had to verify, take the opposite geometry and quickly verify a point for point integration to satisfy myself. The only thing you should find with your direct integration equation above is you are missing one more integral to account for the decreased area as the latitudes decrease toward each pole (see the third cos term in the ‘c’ code? yours would be a sine). Still, either way, you should end up with: 2/5*(1362*(1-0.11)/(0.955*5.67e-08))^(1/4) = 154.723 on a perfect sphere or you slipped somewhere. 354. Bob Fernley-Jones says: Anthony, Thank you for your appended reply on my comment above: REPLY: Read Willis latest on the front page of WUWT about equation 8 – Anthony; link: https://wattsupwiththat.com/2012/01/22/unified-theory-of-climate-reply-to-comments/#comment-873764 If you have been following the maths issues on this thread, you will see that Willis has not demonstrated a great understanding of it here, sometimes one would think almost to the point of ego damage when he acknowledges corrections, for instance this latest link: https://wattsupwiththat.com/2012/01/22/unified-theory-of-climate-reply-to-comments/#comment-874022 Since you asked, I had a quick look at his new article, but have no interest in it, and anyway, the point I was making is independent of the maths. It will be interesting to see how it goes, and I can just follow the reports over at Tallbloke’s, without risking breaking out in hives visiting a Willis thread. 355. JJThoms says: George E. Smith; says: January 23, 2012 at 10:18 pm please look at my referenced documents about measuring LWE IR upwards. In both documents the night time downward LW IR is about 75% of the daytime IR. There is no sun at night! Where is this IR coming from? Add a cloud and there will be more re-emission of IR at the cloud level. 50 % will be returned to the earth. CO2 and H2) do not share all the same Spectra. Hence the rate of loss of radiation upwards will be less. There is ample proof that clouds keep the ground warmer at night. NOTE that they do not add energy to the night just slow its cooling. LW IR in the arctic (North Slope of Alaska (NSA) in Barrow) there is this document : http://www.slf.ch/ueber/mitarbeiter/homepages/marty/publications/Marty2003_IPASRCII_JGR.pdf 140W/m^2 night 150W/m^2 day For Southern Great Plains in Oklahoma http://www.patarnott.com/atms749/pdf/LongWaveIrradianceMeas.pdf Day=260W/m^2 night=400W/m^2 In both these documents this LW IR is in good agreement with Modtran Models based soley on GHGs Note this last sentence. Using theoretical models of GHGs the same values of radiation are calculated as are measured. Does this not show proof of GHG theory? Also see this: Please look at the spectra shown in slide 9 of: http://www.patarnott.com/atms749/powerpoint/ch6_GP.ppt This shows ground and TOA spectra GHG bands mising from TOA and present in upward looking ground spectra. What is not being lost to space is actually hitting the ground and some is being absorbed. Since this LWIR originated on the ground and was radiated upwards, but is now hitting the ground again, a pretty simple sum shows that the ground is loosing less energy and must therefore be warmer than withourt GHGs 356. Tilo Reber says: Joel Shore: “The subsurface averages out the variations that occur at the surface.” Then you can’t have a subsurface that averages out to 40C warmer than what the surface averages out to. Because if it did you would have an average temperature gradient that would need to be maintained by an internal heat source. And if that gradient was 40C over a few centimeters, then that would have to be one hell of an internal heat source. 357. Tilo Reber says: Joules Verne: “The pressure is fixed.” By what? Are you claiming that adding atmosphere would not change the pressure? Are you claiming that increasing the gravitational field would not increase the pressure? Are you claiming that more molecules in a smaller area would not capture more radiative energy than less molecules in the same area? 358. Fellows, I understand that the vast majority of you are not scientists, and do not know certain facts. For example, the debate about the effect of clouds on surface temperature was settled many years ago. The net effect of clouds (averaged over night and day) on surface temperature is COOLING, not warming! For those of you, who gets confused by the often sited ‘warming effect’ of clouds at night, understand the following: Clouds do NOT provide ‘insulation’ to the outgoing IR radiation in the sense that they act as a ‘blanket’ stopping the IR flux emitted. That’s a misconception! Clouds only increase the downward IR flux due to their higher long-wave emissivity compared to regular air. This causes the NET loss of thermal radiation by the surface to become less. In other word, the IR flux coming from the surface gets absorbed by clouds and passed on along the attitudinal temperature gradient. Clouds do NOT return or reflect IR radiation emitted from the ground! That’s a common misconception. that laymen have about LW radiative transfer. Objects of high LW emissivity/absorptivity do NOT reflect IR radiation. They absorb the IR flux coming to them with whatever original intensity, but emit a flux that’s only proportional to the 4th power of their own temperature. In order to reflect IR radiation in the true sense of the word (like a mirror reflecting sunlight), the object must have very LOW emissivity, because IR Reflectivity = 1 – IR emissivity. Clouds have high emissivity (>0.9), and hence low IR reflectivity. Materials of high IR reflectivity (that actually reflect LW radiation) are for example aluminum, polished silver, and polished brass. This is why specially processed aluminum folio has been used by NASA for over 40 years to construct high-efficiency thermal insulation for astronauts and equipment in space. It is known as ‘radiant barrier technology‘, which specifically insulates against thermal radiation losses. In other words, materials of high thermal emissivity such as water vapor and CO2 CANNOT provide thermal insulation, since they do not reflect IR radiation. Only materials of extremely low emissivity such as aluminum can do that! 359. JJThoms, No one argues that the atmosphere emits a substantial flux of IR radiation towards the surface. The question is – does this flux on average cause any increase (warming) of the surface temperature? There answer is NO, because this flux gets neutralized (cancelled out) completely by the convective cooling at the surface. This can be easily proven using an energy balance model that couples convection and radiative transfer and solves simultaneously for both. The confusion comes from climate models, which do not solve for those two modes of heat exchange simultaneously! Also, how do you explain this fact? Globally, the atmosphere emits on average 343 W m-2 of LW radiation towards the surface, while the total radiation absorbed from the Sun by the ENTIRE Earth-atmosphere system is only 238.3 W m-2, i.e. the atmosphere emits down 44% more radiation than the total amount provided by the Sun!! These are real measurements! Does this not tell us that there is MORE energy in the lower atmosphere that the Sun can account for? It does! And it raises the question where is that extra energy coming from? The answer is – from pressure! Pressure provides a thermal enhancement through its physical characteristic called FORCE. This enhancement is demonstrated in the empirical relationship we have found between the NTE factor (Ts/Tgb) and mean surface pressure across a wide range of planets in the solar system. See Eq. 7 and Fig. 5 in our original paper: Arguing against relationships derived from real empirical data is unscientific and silly! That’s because science is about developing theories based on actual OBSERVATIONS, not on ‘thought experiments’. 360. Lars P. says: An appeal to rationality Can somebody point at the process how ATE or GE warms the oceans? I would like to learn and understand. Over 90% of the Earth energy budget runs through the oceans. If the oceans average temperature changes with 0.1°C we have a new ice age or a new climate optimum. And yet the great majority of climate scientists ignore the oceans when discussing earth energy budget and talk only of the solid ground and atmosphere. It is that greenhouse gase which traps energy. That 99% of the captured energy is in the oceans is no problem for them. The oceans are opaque to downward LW IR. It took long until people realised that the weather is run by the oceans and we need to understand such phenomenon as La Nina and El Nino to understand how weather happens. And yet the climatologists keep ignoring the oceans and make energy budgets only above them. They keep looking at the tail and wonder why is it whacking. Any energy budget of the Earth which does not describe the oceans energy budget is doomed to fail. 361. Joel Shore says: Ned Nikolov says: No one argues that the atmosphere emits a substantial flux of IR radiation towards the surface. The question is – does this flux on average cause any increase (warming) of the surface temperature? There answer is NO, because this flux gets neutralized (cancelled out) completely by the convective cooling at the surface. This can be easily proven using an energy balance model that couples convection and radiative transfer and solves simultaneously for both. The confusion comes from climate models, which do not solve for those two modes of heat exchange simultaneously! Every time you say this, I am going to point out that you have been shown to be absolutely wrong on this: There is no ambiguity…You made a huge mistake here. You have not bothered to defend what you have done because there is no defense…You implicitly admitted that your way of adding in convection yields a result in contradiction to reality in your description of what happened. Admit it and come clean rather than continuing to deceive people on this point! 362. Joel Shore says: Tilo Reber says: Then you can’t have a subsurface that averages out to 40C warmer than what the surface averages out to. Because if it did you would have an average temperature gradient that would need to be maintained by an internal heat source. And if that gradient was 40C over a few centimeters, then that would have to be one hell of an internal heat source. Yeah…As I noted to Robert Brown last night https://wattsupwiththat.com/2012/01/22/unified-theory-of-climate-reply-to-comments/#comment-873702 , my statement about the subsurface having the same average T^4 rather than the same average T as the surface may have been wrong…although we are still left with the empirical mystery (at least claimed) that “the average temperature on the surface is about 40-45 C lower than it is just below the surface” ( http://www.asi.org/adb/m/03/05/average-temperatures.html ). I did notice that the thermal conductance of the lunar surface materials is a function of temperature and I think this could lead to an asymmetry whereby the temperature below the surface is not equal to the average at the surface. However, from what I have seen of these thermal conductances, I keep getting the effect going in the other direction. 363. See - owe to Rich says: Nikolov said above: “Arguing against relationships derived from real empirical data is unscientific and silly! That’s because science is about developing theories based on actual OBSERVATIONS, not on ‘thought experiments’.” Well, one of the greatest of all, Einstein, used both observations and thought experiments with amazing results. So I’m with Willis Eschenbach – a good thought experiment can tease out limits of scientific understanding. So please try this one for size, Ned. Imagine the Earth as it is except with no clouds. Its albedo will be lower, so it will be hotter, and your equations can show by how much. Now for the kicker, imagine that iall over the ever changing night side, and only there, a 1000 metre thick layer of cloud forms at altitude 1000 metres. Under the N-Z model, the mean temperature of the Earth will not be changed by this blanket of greenhouse clouds. But experience tells me that the night will not get so cold, and so the Earth will warm up – unless of course the clouds are radiating as much heat upwards as the ground would have through clear air (but I don’t think so, as the clouds are cooler). It’s not that I don’t believe that the N-Z theory gives a good first order estimate of the temperature of a planet or moon, but that greenhouse gases will give a non-negligible second order effect. I don’t think it’s huge, but I’d be surprised if it wasn’t worth at least 10K on Earth. Rich. 364. mkelly says: Ned Nikolov says: January 24, 2012 at 8:47 am “In order to reflect IR radiation in the true sense of the word (like a mirror reflecting sunlight), the object must have very LOW emissivity, because IR Reflectivity = 1 – IR emissivity.” Sir, I would amend this slightly (if my heat transfer prof was right) that higher energy state objects donot absorb lower energy radiation. ie it is reflected. Thanks for your interaction. 365. Joel Shore says: mkelly says: Sir, I would amend this slightly (if my heat transfer prof was right) that higher energy state objects donot absorb lower energy radiation. ie it is reflected. Either your heat transfer prof was severely incompetent or (more likely) you misunderstood him. 366. Phil. says: 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 Unfortunately N&Z used a model where they introduced an additional change, namely the assumption of negligible surface thermal capacity. Most of the large difference that they calculate results from this assumption (if they hadn’t made this assumption there would still be a discrepancy due to non-uniformity but it wouldn’t be as large), the remainder results from the radiative exchange characteristics of GHGs. 367. Bill Hunter says: Dr Burns says: January 24, 2012 at 1:14 am “My view is that overall the impact of IR absorbing gases is far less than that claimed by IPCC but not zero as claimed by Ned. Looking forward to any constructive criticism.” Seems to me that large values are indeed being ruled out by the models themselves. The amount of effort put into them to model backradiation as the primary means of the GHE appears to have failed. So have the efforts to minimize natural variation. Its in your face now and I assign a close to zero probability what we are currently seeing is anthropogenic in origin. Whats left in the absence of significant investment in non-anthropogenic causes being less than catastrophic is really too small to rule out either gentle warming or gentle cooling influences. I that range of possibilities zero seems as likely as any other explicit figure. I am not sure but there are a lot of calculations with radiation that go absolutely nowhere. Heat loss with conduction equals heat gain from conduction just as it does for radiation. It seems with either you have to postulate some unphysical backradiation or back conduction in order to get cold objects to warm warm objects. I look to the evolution of passive solar water heating as an answer to this conundrum. If you put a black can of water out in the sun on top of a shower enclosure you can get warm showers in the evening but not in the morning. Early solar water heaters put the cans in greenhouses and by the restrictions on convection got better performance to the point that patents were awarded I think in the late 19th century. The final evolution of the totally passive water heating system was to greenhouse the collector system and put them below the storage system to which insulation was added. Actually a potentially ideal analogy to our earth surface and poorly emissive atmosphere heat storage system where convection is the single imbalanced heat exchange medium brought to us 100% by gravity. Turn off gravity and the black painted can would work as well as the older patented systems and both of those would work a lot better than the final evolution of the system with collectors placed remote to and below the storage system (though a small pump would fix that specific problem, it would be the least of our problems with either the system or anything else). Climate models represent a huge waste of tax payer dollars. They are completely incompetent and probably never will be useful until we actually understand better how the surface warms, which will probably end up mostly entailing knowing more about solar variability. Before the models will be useful we need to gain some confidence we know the cause and have some idea of what drives the cause. What needs to be done are some more far more practical experiments with say a solar gas heating system built on the principles of the passive solar water heating system to demonstrate how the system actually operates with various levels of emissivity using say shiny radiators above and below the insulated storage tank to mimic various combinations of surface, atmospheric and cloud emissivities. Anybody who has a two story house knows a little about it and heating contractors building forced air systems know a bit more. One detect it in a single floor home but putting thermometers at various levels in a room. A lot of work is put into design efficiency to not waste heat from convection. Also, there actually is some work thats been done on that in the passive solar industry with underground and greenhouse convection driven systems using ordinary air. Put the whole apparatus in a large climate controlled room and apply a variable energy source from above that can mimic the diurnal cycle at various latitudes and seasons. Kind of like what the GCR advocates were directed to do and Cloud 9 came into being. 368. Bob Fernley-Jones says: Joel Shore @ January 24, 10:47 am In your rant against Ned Nikolov, concerning the GHE, you wrote in part: There is no ambiguity…You made a huge mistake here. You have not bothered to defend what you have done because there is no defense…You implicitly admitted that your way of adding in convection yields a result in contradiction to reality in your description of what happened. Admit it and come clean rather than continuing to deceive people on this point! Perhaps you should read more carefully what Ned wrote. I understand it to mean that he accepts that there is what is called a greenhouse effect, but that in context, it is neutralized by other effects. In the case of the Earth, and in the context of his hypothesis, even Trenberth says so in his 2009 Earth’s Energy Budget cartoon, giving these numbers for heat loss from the surface (W/m^2): • Evapotranspiration = 80 • Thermals = 17 • Radiative absorbed (GHE) = 23 (14%) • Radiative direct to space = 40 • Disappeared = 1 Yet again you have lowered your credibility, maybe because you are too ready to criticise, and lack in comprehension of what has been said. 369. Joel Shore says: Bob Ferney-Jones says: Perhaps you should read more carefully what Ned wrote. I understand it to mean that he accepts that there is what is called a greenhouse effect, but that in context, it is neutralized by other effects. In the case of the Earth, and in the context of his hypothesis, even Trenberth says so in his 2009 Earth’s Energy Budget cartoon, giving these numbers for heat loss from the surface (W/m^2): • Evapotranspiration = 80 • Thermals = 17 • Radiative absorbed (GHE) = 23 (14%) • Radiative direct to space = 40 • Disappeared = 1 No…It is you who have failed to read and comprehend what I have written on this subject in many previous comments, some even linked to in the comment of mine that you quote. You have also failed to read and comprehend Nikolov and Zeller’s own paper where they say: “Equation (4) dramatically alters the solution to Eq. (3) by collapsing the difference between Ts, Ta and Te and virtually erasing the GHE (Fig. 3).” In other words, they have added convection into the model in a way that drives the temperature distribution in the vertical to be isothermal. In case you don’t know, the troposphere actually has a significant lapse rate in the vertical, the reason being that convection can only drive the temperature distribution down as far as the adiabatic lapse rate because beyond that the atmosphere is stable and convection is suppressed. (And, as Nick Stokes has pointed out, to the extent that some forced motions cause a “heat pump” effect, this effect actually can drive the lapse rate back up to a steeper value if it is below the adiabatic lapse rate.) Yes, Trenberth et al. acknowledge a role for convection and, yes, convection reduces the radiative greenhouse effect from the magnitude it would have if convection were not present and lapse rates steeper than the adiabatic lapse rate were stable. (As I recall, a purely radiative calculation gives a greenhouse effect about twice as large as the value once convection in included.) ****However, convection can only partially cancel the radiative greenhouse effect precisely because convection cannot drive the lapse rate all the way down to zero, i.e., an isothermal distribution with height.**** Read that last sentence several times until you understand it. Yet again you have lowered your credibility, maybe because you are too ready to criticise, and lack in comprehension of what has been said. In fact, it is you (and Nikolov and Zeller) who have lowered your credibility by showing your lack in comprehension of what has been said. 370. I had a go at the puzzle of the airless planet mean temperature using a simple numerical simulation model to figure out the temperatures on the surface as it rotated and re-radiated heat and conducted it into the ground. It was a bit rough and ready, but here’s my results. First the changing temperatures throughout the day: And then the mean temperature of the planet over the day: 247 degrees K. 371. Ralph says: >>Ned Nikolov says: January 24, 2012 at 8:47 am I understand that the vast majority of you are not scientists, and do not know certain facts. For example, the debate about the effect of clouds on surface temperature was settled many years ago. The net effect of clouds (averaged over night and day) on surface temperature is COOLING, not warming! For those of you, who gets confused by the often sited ‘warming effect’ of clouds at night, understand the following. —————–—————————————————– Ned, it is you who is confused here. We all know the overall net effect of clouds is cooling, but we are not talking about net effect here. The reason we are talking about clouds ONLY at night, is because these night clouds PROVE that the Greenouse Effect is real (since you do not seem to believe in it). Forget the day. At night, clouds will always make the surface warmer, because of their absorption and reemision of LW. This is a known meteorological effect, look it up. Thus a LW absorber and emitter will reduce the cooling of the surface at night. Fact. Ok, so during the day, a LW absorber and emitter will similarly reduce the cooling of the surface. And if the incomming daytime SW radiation remains constant, the daytime effect will be a warmer surface temperature than if there was no LW absorber and emitter sitting above it. Thus gasses that provide this ‘service’, like H2O and CO2, will make the surface warmer than if they were not present. Perhaps we need a name for this effect. I know, let’s call it the Greenhouse Effect. . 372. jjthoms says: Ned Nikolov says: January 24, 2012 at 9:24 am “No one argues that the atmosphere emits a substantial flux of IR radiation towards the surface. The question is – does this flux on average cause any increase (warming) of the surface temperature? There answer is NO, because this flux gets neutralized (cancelled out) completely by the convective cooling at the surface. This can be easily proven using an energy balance model that couples convection and radiative transfer and solves simultaneously for both.” ================= Care to explain this? Why should convective cooling increase without the surface temperature increasing first? Makes no sense to me! ============== “Also, how do you explain this fact? Globally, the atmosphere emits on average 343 W m-2 of LW radiation towards the surface, while the total radiation absorbed from the Sun by the ENTIRE Earth-atmosphere system is only 238.3 W m-2, i.e. the atmosphere emits down 44% more radiation than the total amount provided by the Sun!! These are real measurements! Does this not tell us that there is MORE energy in the lower atmosphere that the Sun can account for? It does! And it raises the question where is that extra energy coming from? ” ============== I quoted referen ces to real measurements. The downward LW IR is coming from somethin. The only thing radiating at these wavelengths at night is GHGs. Note that the solar TSI is greater than these DLR levels (but in a different part of the spectrum. These geat the ground/water and then these radiate at a lower “temperature” some of which gets re-radiated back down (hence the difference between day and night. So, sorry but your explanatuion seems totally wrong to me. 373. George E. Smith; says: “”””” JJThoms says: January 24, 2012 at 5:01 am George E. Smith; says: January 23, 2012 at 10:18 pm please look at my referenced documents about measuring LWE IR upwards. In both documents the night time downward LW IR is about 75% of the daytime IR. There is no sun at night! Where is this IR coming from? “”””” “”””” George E. Smith; says: January 23, 2012 at 10:18 pm “”””” Ralph says: January 23, 2012 at 9:45 pm <<>>>> Nonsense. A cloud layer can move in from a totally different weather system hundrds of miles away, and the effect will be the same. Whether the night air is still convective (rare), or whether the airmass is completely stable (more usual), a cloud layer will ALWAYS result in higher nighttime temperatures – because the cloud is ‘insulating’ the LW radiation from the surface. And I have been watching the weather for 30 years. “”””” So I could have a cloudless day with surface Temperatures of say 40 deg F, and typical wintertinme low humidity, and after sundown a cloud layer at 20,000 feet comes in on some jet stream, and my surface Temperature will go up above 40 deg F ?? So the cloud might be -50 deg C, so just where does the energy come from to raise the surface Temperature; it’s after sundown, so it isn’t solar, and my cold air was going all the way up along with the lapse rate to where that 20,000 ft cloud came in, and my surface is getting even colder, as it radiates its 400 Watts per m^2 or more. I will grant you that a massive storm of air might sweep in from some hot place with clouds, and simply replace all the cold air I had during the day, and presumably I will be warmer; but it wasn’t radiation trapping from my surface that warmed me up; and notwithstanding that storm cloud, it will still cool down as the night progresses; it won’t continue to get hotter, unless even hotter air contines to come in from somewhere else. If you need to invoke a Santa Ana to make your case; then I submit, that you have no case; and i’ve been working at this for more than twice as long as you’ve been watching the weather. “”””” So II, I just reposted the entire post that I placed here on Jan 23 2012, at 10:18 pm So please tell us the line numbers where you found this: “”””” “”””” JJThoms says: January 24, 2012 at 5:01 am George E. Smith; says: January 23, 2012 at 10:18 pm please look at my referenced documents about measuring LWE IR upwards. In both documents the night time downward LW IR is about 75% of the daytime IR. There is no sun at night! Where is this IR coming from? “”””” I submit that on Jan 23 2012, at 10:18 pm, I made no such statement. 374. Bob Fernley-Jones says: Joel Shore @ January 24, 2:03 pm YAWN! Very quickly, I will simply repeat what I wrote before but with bold emphasis added to some key words, (although it is tempting to put it all in bold), in the hope of you achieving English language comprehension: Perhaps you should read more carefully what Ned wrote. I understand it to mean that he accepts that there is what is called a greenhouse effect, but that in context, it is neutralized by other effects. In the case of the Earth, and in the context of his hypothesis, even Trenberth says so in his 2009 Earth’s Energy Budget cartoon, giving these numbers for heat loss from the surface (W/m^2): • Evapotranspiration = 80 • Thermals = 17 • Radiative absorbed (GHE) = 23 (14%) • Radiative direct to space = 40 • Disappeared = 1 BTW, my native tongue is SE England born, but corrupted by Oz lingo since 1969. During the 1980’s I worked in the USA maybe 30% of the time and additionally worked with Americans in Oz and Italy. Whilst I and colleagues were sometimes confused by Americano lingo, we usually managed to translate it OK. (probably). With that experience, and more, I doubt that my expose above has any ambiguity to you, if you try really hard to understand it. But then you have difficulty emerging from the covers of your apparently favorite book by Raymond Pierrehumbert, despite that he is by popular definition* not a gentleman. * DEFINITION; Gentleman: A man who can play the accordion but doesn’t 375. George E. Smith; says: “”””” Ralph says: January 23, 2012 at 11:10 pm >>>George Smith >>>So I could have a cloudless day with surface Temperatures >>>of say 40 deg F, and typical wintertinme low humidity, and after >>>sundown a cloud layer at 20,000 feet comes in on some jet stream, >>>and my surface Temperature will go up above 40 deg F ?? Who says the temperature goes up? You are distintly bad at reading, George, try again. I NEVER SAID THE TEMPERATURE GOES UP, is that clear enough for you?? Clouds reduce cooling, because of the Greenhouse Effect. Note the term here, George, REDUCE COOLING. This results in a warmer temperature than a night without clouds. Do you understand that this is not warming, George, it is reduced cooling. Warmer than without clouds, but still a net cooling effect (but slightly reduced). Clear?? “”””” Yes Ralph it is perfectly clear; and you might want to critique your own reading prowess before criticising mine. Yes I totally agree with you: THE TEMPERATURE DOES NOT GO UP AT NIGHT; EITHER WITH CLOUDS OR WITHOUT CLOUDS. Ergo, and I do so hope we can agree on this; the previous day, THE TEMPERATURE MUST HAVE BEEN HIGHER, during the day THAN IT WAS THE FOLLOWING NIGHT. And if on the cloudy night it was a higher Temperature, than on the cloudless night, THE DAY TIME TEMPERATURE BEFORE THE CLOUDY NIGHT WAS HIGHER THAN THE DAYTIME TEMPERATURE ON THE CLOUDLESS NIGHT. Besides Ralph, what is in contention is the CLIMATE CONSEQUENCE of clouds; NOT last night’s weather. So the question is quite simple. If the long term average mean global cloud cover (whatever that means) is say 61%, as some NOAA/NASA source has been quoted as suggesting; and then that number increases to 62% for a period of time of relevence to climate; such as 30 years for example. Will the mean ghlobal Temperature rise as a result of the increased cloud cover or will it decrease, due to a net reduction of total solar energy that gets stored in the earth system; mostly the oceans. I fully agree the clouds DO NOT make the night time Temperature go up. The higher DAYTIME temperature was the cause of the clouds (along with some moisture), not a consequence of the clouds, and the higher the daytime temperature is, the higher will be the altitude of the subsequent clouds, other conditions being equal. The problem of a diffuse emitting surface irradiating a second distant receiving surface, is a very well researched problem in the electronics industry. We might call it the “Opto-coupler” problem. Well some people will call it the “Opto-isolator” problem. It’s a very common signal transmission device for transmitting electronic signals over an electrically isolated path. Typically,an LED emitting some red or near IR radiation, illuminates a silicon detecting chip, separated by some light conducting insulator that might be required to withstand 25,000 Volts electrical isolation; or a million for that matter. Modulating the current of the LED creates an anlog signal that is propagated over the isolated path. Modern ones actually transmit digital signals rather than analog, so that high signal fidelity can be achieved. As early as 1960, Tektonix Inc built a complete Oscilloscope plug in vertical amplifier, that was isolated from ground for many kV, and the amplified signals were passed on to the mainframe via optocouplers.They even powered the isolated circuits using an ordinary incandescent lamp in the plug in, and solar cells on the isolated component. For such products it was of course desirable to have the highest signal coupling efficiency commensurate with the required isolation Voltage capability. These devices were produced in the multimillions in the 1970s. Hewlett Packard was one manufacturer of a line of these products. As it happens, I worked for a competitive company making similar products. The various manufacturers collaborated on some industry standards for these products, so we did get together to discuss mutually interesting questions. A very competent engineer at HP, and I set out to calculate the theoretical optical signal coupling for a general geometry for such devices. I won’t name him, because he is no longer with us, and we were good friends. The problem involves a quadruple integral over two areas, sending and receiving, and two solid angles. It so happens that I selected a poor choice of order to do the four integrals, and generated a real birdsnest of mathematics, that I never did solve. The HP chap, lucked out and made a better choice of order, and he succeeded in getting an accurate closed form solution for the signal coupling for a fairly generic system design. He sent me the solution, and when I looked at it, I realized that the equation was not a good equation to use for calculation; especially in a computer program, because the equation contained a small difference of two large numbers, so large errors could arise, when part of a transparent design process, with nobody watching the details. I worked on it, and was able to transform the quation into an equivalent form which instead contained the sum of those two large numbers, instead of their difference, so it was computer tractable and would go haywire in some special case. So I sent him the transformed equation, which he ended up writing up in an HP application note. We celebrated later with lunch together. His equation can be used to calculate the total energy coupling between two parallel coaxial areas at any distance apart, such as an earth surface area and some cloud area, and in the surface cloud case, the reverse transmission problem can also be solved, and then the two combined, for the round trip. So there is no great mystery to calculating the return signal energy in the surface/cloud/surface round trip problem. Well you don’t have to do eight integrals since the same result works both ways, and since the solution already exists, you just need to use the formula. I’ll let all you hotshot math PhDs do the quadruple integral for yourselves. Now there is one aspect of this problem that seems to have been overlooked somewhat. Take away the clouds, and we already have a surface to atmosphere (GHG bearing) to surface round trip, that handles the specific GHG such as CO2 absorption band spectral energies. The apparent consequence of this (I’ll take their word for it), is that not too much of the GHG band energy makes it up to where higher clouds might be; it’s largely been absorbed by GHGs lower down and returned. So much of the cloud situation revolves around the rest of the surface thermal LWIR spectrum, that runs afoul of the very broad liquid water absorption in the 5-50 micron region. The interesting thing is that absent the clouds, and much water vapor, the CO2 round trip mechanism is fully operational; yet nobody seems to notice much effect from it. In the high dry deserts sans clouds and much water vapor; CO2 seems to be remarkably inefficient, in keeping the surface warm at night. So much for the power of CO2. 376. George E. Smith; says: By the way Ralph, My use of CAPS is not shouting; its one way of my indicating I want to emphasize something. Sometimes I might use (….) instead or “…….” but sometimes that is misconstrued as a quotation; I’ve tried ‘………..’ to demark something without it looking like a quote. To be sure when I am quoting someone else, I like to use “””””…………””””” which is supposed to be five double (‘)s, but sometimes I over or undershoot. So when you see all those (“””””) you can be fairly sure I am cut and pasting what someone else said. Now I do often erase part of what someone said, that isn’t germane to the point I want to make and I try to always put in a ………. to indicate I slashed. I do try to not alterthe meaning of what someone posted, so if my deletions ever change someone’s intent, by all means call me on it, because that is NOT my intent. And no, we don’t really need to shout here at WUWT. George 377. David says: George E Smith, some questions for you please. Joel Shore has ignored these questions. Please consider the following assertions to also be questions. A planet (earth, ocean atmosphere) can only cool by radiating to space. Non GHG store conductive specific heat, but cannot radiate it to space. (Insulation) They can however conduct specific heat to radiating GHGs. Conducted specific heat from a non GHG molecue to another non GHG molecue stays within the atmosphere (increased residence time) longer then if the non GHG molecue conducts that heat to a GHG molecue, where that energy can radiate to space. Adding more GHGs to an atmosphere accelerates the loss of conducted heat to space, while it slows the loss of radiated heat from the surface. (cooling and warming) So GHG reduce the TSI reaching the surface, (cooling) Intercept UWLWR from the surface and return a portion towards the surface, (warming) Recieve conducted specific heat from non GHGs (which insulate conducted heat ) and allow some of this to zip to space, (cooling) George could you outline some possible paths of a LWIR photon within the CO2 absorbtion band? How many times can that energy be redirected before leaving to space? How long does this take? Is the residence time of conducted energy in non GHGs far longer then it is with GHGs? How can adding more of the the only gases (GHGs) which can cool (radiate to space) warm the planet? 378. Joel Shore says: @Bob Fernley-Jones: Your reply doesn’t address anything I wrote in any way whatsoever. I’ll just assume that you implicitly concede that you are wrong but are too embarrassed to admit it. 379. Smokey says: Joel Shore, I’ll just assume that you implicitly concede that you are wrong but are too embarrassed to admit it. After all, that’s exactly what the planet is saying, too.☺ 380. George E. Smith; says: So here’s an experiment for all of you budding Meteorologist / Climatologist afficionados to perform; it’s very simple from the Meteorology 0.01 textbook. I first studied this problem while I was raising a New Zealand white rabbit (NZWR) for a 4-H club project (to see if NZWR tastes “just like chicken”). I would probably have done it in 8th grade science; but then we didn’t have “grades” in NZ. I was killing time in between sorting NZWR pellets; separating a pile of pellets into two classes. Fresh virgin pellets, and pellets that had already been once through the rabbit, which don’t taste as good. So we need to define some variables since the brain dead editor in this box is even less like an IBM 360 typewriter than Micro$oft WORD.

TNn is the Temperature at any time t at Night, where t = 0 represents the time at which the last vestige of the sun upper limb disappears on the horizon, with NO clouds present, at t = 0.
TNc is the night time Temperature at time t when clouds ARE present, at t = 0.
TDn is the daytime Temperature at some time, t = – epsilon, where epsilon can be any positive number, when no clouds are present at t = 0.
TDc is the daytime Temperature for t = – epsilon , when clouds are present at t = 0.
Mn is the gradient of the night time Temperature decay when no clouds are present at t = 0.
Mc is the gradient of the Temperature decay when clouds are present at t = 0.
Both Mn and Mc are positive , and constant for any particular night, since in climatology, trends are ALWAYS straight line, by definition.

The following axioms apply from the meteoroclimatology 0.01 text book.

Axiom # 1 TNc > TNn Cloudy nights are always warmer than cloudless nights.
Axiom #2 Mc 0

TNn = TDn – Mn.t

TNc = TDc – Mc.t

So we have from axiom #1, TDc – Mc.t > TDn – Mn.t

This “miracle” inequality is universally true for 0 <= t TDn Quite Easily Done !

It is ALWAYS warmer on a cloudy night,than on a cloudless night, because it was warmer during the previous day time, and those warmer conditions at night and the clouds themselves are the result of that simple fact; the clouds are NOT the cause of the warmer night, they are the consequence of the warmer day.

And yes as I recall NZWR does taste just like chicken.
This version is correct

381. George E. Smith; says:

“”””” David says:

January 25, 2012 at 1:52 am

George E Smith, some questions for you please. Joel Shore has ignored these questions.
Please consider the following assertions to also be questions. “””””

David, I am NOT a chemist, and my formal chemistry was limited to one Unit, Chem-1 at University, but following five years at high school. I went to a fabulous high school. I haven’t followed too closely your discussion with Joel, but I generally treat his postings quite seriously, and I have never doubted his sincerity.

As for my Physics, that stopped just short of getting heavily into Quantum Mechanics to where I could dash off solutions to Schroedinger’s equation for common problems; but I’ll take a whack at your questions.
First it does seem self evident that EM radiation is the only substantial process for losing energy from earth. What is less certain is where that energy originates (the long wave IR) I presume that it starts at either the solid or liquid surface, so it is somewhat black body like; but I choose to not accept that it is an emission at some uniform isothermal grey body at the 288 K putative global mean surface Temperature. OK that number may actually be the lower troposphere Temperature (izzat two metres above the ground, or over the top of the Weber grill as the case may be?)
Now on any ordinary midsummer northern day, the earth surface Temperature (simultaneously) can be as low as -90 deg C at places like Vostok Station, which will be in winter midnight, or it can be as high as +60 deg C in the tropical deserts. Due to an argument by Galileo, every single Temperature between those Temperature extremes, can be found somewhere on earth, in fact there is an infinite number of such places.
So it is absurd to talk of the earth as a black or even grey body radiator, but any small area can be close to grey; they all have different characteristic Temperatures.

Secondly, I am not a believer of the notion that ONLY GHGs radiate EM radiation. Every assemblage of molecules that can interract thermally with each other (collisions), and every such molecule, can be said to have a Temperature, and it can and will radiate and absorb EM radiation as a consequence of those collisions, which is what Temperature is all about.
Isolated molecules that are in free flight, and don’t collide don’t radiate “Thermal” radiation. Well they also don’t have a Temperature, sans collisions.
The “excited states” of say CO2 that result in the absorption or emission of the 15 micron band of spectral lines have lifetimes, that are generally much longer than the time between collisions at STP conditions, so a molecule that has captured such a photon, seldom gets a chance to spontaneously re-emit such a photon; collisions terminate the condition, which transfers that photon energy to the regular N2, O2, Ar molecules of the atmosphere, as thermal energy.

As Phil has pointed out on several equations, in the higher cooler lower density atmosphere, the longer mean time between collisions is long enough to allow spontaneous re-emission, so the upper atmosphere would then radiate the characteristic 15 micrton CO2 resonance lines. These wavelengths would not be first order Temperature dependent; they are not a consequence of Temperature.

Some people believe that the way a heated atmosphere loses energy by radiation, is for a typical molecule , N2, O2, Ar for example, to collide with say a CO2 molecule, with sufficient energy exchange to set the CO2 ringing (bending) at the 15 micron resonance frequency, and then that molecule subsequently radiates a 15 micron photon, so the only emission from the atmosphere would be that 15 micron radiation. Assuming of course for the moment that CO2 was the only GHG present. Now the CO2 could also emit a 4.0 micron photon, as a result of having its assymmetric stretch mode of oscillation excited by the collision; but that would take a collisions with about 4 times the energy required for a 15 micron event.

Problem with that thesis, is that we already decided that the mean time between collisions is much shorter than the CO2 excited state lifetime, so that spontaneous re-emission is highly unlikely. Now suddenly, it becomes the primary mechanism of atmospheric cooling.

So CO2 is an intelligent molecue;it is capable of distinguishing between a molecular collision, and a surface emitted LWIR 15 micron photon that set it ringing .A surface captured photon is quickly dumped, but energy received from a collision is automatically retained until spontaneous re-emission occurs !!

So try pulling my other leg.

In any case, once an excited CO2 molecule collides with another air molecule, and loses its enhanced energy, it is now reloaded to grab another 15 micron photon from the surface, and repeat the process. That is why the “CO2 is saturated” theory simply doesn’t hold H2O. Continuous cascades of absorption, thermalization, absorption occur, and more CO2 simply means a thinner air layer is needed to absorb a given amount of 15 micron radiation. It is quickly unloaded ready to do the same thing again, and the process repeats for a slightly thicker layer above that one, and so on.

I hope some of that is understandable; it is the wee hours, and I need some sleep.

382. Lars P. says:

BenAW says:
January 23, 2012 at 3:51 pm
Hi Lars, seems nobody wants to see this elephant, I called it a 600 pound gorilla on another blog.

Hi Ben. Yes you are right. So I will take the popcorn and watch. Maybe a little post now and then…. I ask myself how long will it take until the oceans will be properly addressed in the effort of defining Earth equilibrium temperature? What are your expectations, do you think a new theory based on the oceans will come before summer 2012, after it or “never” (not in the next 10 years or so). Have you seen anywhere in the web something that would explain sufficiently this part yet?

383. George E. Smith; says:

“””””Something went ape and this got all garbled; let’s try to fix it up.

************

Axiom # 1 TNc > TNn Cloudy nights are always warmer than cloudless nights.
Axiom #2 Mc TDn – Mn.t

This “miracle” inequality is universally true for 0 <= t TDn Quite Easily Done !(QED, not quantum electrodynamics)

The Temperatures never cross, if the cloudy night is warmer than the cloudless night, at all times during the night, then the previous daytime Temperture must have been higher too

It is warmer on a cloudy night,than on a cloudless night, because it was warmer during the previous day time, and those warmer conditions at night and the clouds themselves are the result of that simple fact; the clouds are NOT the cause of the warmer night, they are the consequence of the warmer day.

And yes as I recall NZWR does taste just like chicken.
This version is correct

384. Bob Fernley-Jones says:

Joel Shore @ January 25, 3:59 am

@Bob Fernley-Jones: Your reply doesn’t address anything I wrote in any way whatsoever. I’ll just assume that you implicitly concede that you are wrong but are too embarrassed to admit it.

Ah well, I guess us mere mortals should understand that certain academics sure do make a lot of assumptions

385. R. Gates says:
January 22, 2012 at 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. … so the two bodies would have very different radiation curves.
———————————–
Not so much on the surface. Earth has a substantial temperature gradient under the surface for four different reasons. However, all four combined are about 0.04% of solar insolation.
1. Radioactivity, leading to a high crustal temperature. Rocks are an effective insulator, so the rate of energy transfer to the surface is in the range of milliwats/m^2.
2. Tidal forces. Since Earth has a liquid surface and core, this leads to frictional forces that are probably many times that of radioactivity. Geologists would probably have a better idea of the magnitude of tidal forces.
2a. Luna also has tidal forces, which are larger than Earth’s since Earth exerts more force per volume on Luna than the reverse. However, much of Luna’s tidal forces are tied into its rotational wobble. Since Luna shows only one side to Earth (wobble makes that close to 60%) the location of the Earth-Luna tidal bulge moves less than 18 degrees off its center. This should cause some heating, but in the absence of other internal heating and a GHG atmosphere, not enough to make Luna significantly geologically active or warm at night.
3. Electromagnetic heating. Since the core and the surface rotate at slightly different rates, among other things, and both the core and surface have iron and nickel, some of the electromagnetic forces will be dissipated as heat.
4. Pressure gradient. This should have minimal current effects; the heat form planet creation has long since dissipated, and the annual increase in Earth’s mass is small compared to Earth’s total mass.
However, we now know that Luna gets *very* cold in places where the sun doesn’t shine. Antarctica has liquid water under the ice, obviously from internal heating. So the latitudinal profile of Earth vs. Luna is very different. Going from 4K to 150K, Earth’s internal heat is significant, less so from 150K to 288K.

386. Stephen Wilde says:

If temperature is set by pressure and solar input then in the end the effects of other processes or events such as volcanoes and GHGs must be neutral unless they affect total atmospheric mass.

When either tries to alter the lapse rate set by pressure then the entire atmospheric circulation must adjust to neutralise the effects.

The lapse rate in any given atmosphere actually appears to be set by pressure alone such that the level of solar input is relevant only to atmospheric height.

No sun and the atmosphere would freeze on the surface and the atmospheric height would be zero. Add sunshine and the atmospheric height would increase commensurate with the level of solar input.

The lapse rate for a given planet would appear not to be any different at different levels of solar input.

We all know that severe volcanic outbreaks have an effect on atmospheric circulation don’t we?

Those changes in circulation constitute the inevitable negative system response to the volcanic event. In due course the system returns to the previous equilibrium.

The system responds similarly to rising or falling quantities of GHGs and to all other forcings such as variations in the rate of energy release from the oceans.

More total atmospheric mass or higher solar input are the only things that will raise the system equilibrium temperature but the lapse rate set by pressure must remain the same for any planet of a given mass.

Anything other than planetary mass that might seek to alter the basic lapse rate can only do so by altering the atmospheric height but what goes up must come down so for every location where the heights are pushed up by warmth rising from the surface there is another location where the heights are pushed down by cooled upper air descending to the surface for a zero net effect on heights globally.

The only exception would seem to be solar input which would raise absolute atmospheric height. Everything else seems only to affect relative atmospheric heights.

There would be a climate consequence though because a warmer world gives deeper surface low pressure cells and more intense surface high pressure cells.

However that climate effect from more GHGs would appear to be miniscule compared to similar such variability from sun and oceans interacting together in an ever changing dance.

To summarise:

i) The gravity field alone seems to determine the slope of the lapse rate on any given planet.

ii) Adding solar input appears to affect only the ABSOLUTE height of the atmosphere and NOT the slope of the lapse rate.

iiI) Anything other than solar input only affects the RELATIVE atmospheric heights.

iv) The consequence is that anything other than solar input will only affect the rate at which energy flows from surface to space and not the equilibrium temperature of the planet.

387. Stephen Wilde says:

“I ask myself how long will it take until the oceans will be properly addressed in the effort of defining Earth equilibrium temperature?”

http://climaterealists.com/attachments/ftp/TheSettingAndMaintainingOfEarth.pdf

“The Setting and Maintaining of Earth’s Equilibrium Temperature”

Basically atmospheric pressure sets and maintains ocean temperature by fixing the energy cost of evaporation and the oceans then control the air temperatures above.

388. Joel Shore says:

Stephen Wilde says:

More total atmospheric mass or higher solar input are the only things that will raise the system equilibrium temperature but the lapse rate set by pressure must remain the same for any planet of a given mass.

How many times do we have to explain this to you? The lapse rate alone does not determine the surface temperature. You also need to know the temperature at one height. The constraint from radiative balance of the Earth is that the effective radiating level has to be at an average temperature of 255 K. As the concentration of greenhouse gases increases, the effective radiating level rises and, hence, the surface temperature (determined by extrapolating from 255 K at the effective radiating level using the average lapse rate) rises.

Is real science so difficult to understand that you have to fall back on pseudoscience?

389. I wonder if Professor Robert G. Brown knows anything at all about computer science !
Well wait a minute, this thread is about Thermodynamics, not computer science.

Just for grins, google “rgb beowulf”. The hit count is way down because the old beowulf archives aren’t active much any more, but it is still pretty respectable.

The answer is “yes”. I teach computer science (usually in independent studies). I’m an ubercoder, defined to be somebody with over some absurdly large number (pick one) of lines or bytes of code they have written, lifetime. Not to brag or anything, but I’m better known for my work on beowulf style computing than I am for my papers in physics. But the bulk of my work in physics over the last 20 years was statistical thermal simulations in the context of magnetic critical phenomena.

None of which really matters in this argument. Anybody who thinks the second law of thermodynamics will be violated, in steady state, for a system as simple as this one is is simply wrong. No, it won’t. One doesn’t have to think twice about it. No, it won’t. Anybody who asserts otherwise simply hasn’t learned physics, at least not properly. They believe in magic, because a stable violation of the second law of thermodynamics in an otherwise ordinary gas in a gravitational field would be friggin’ magic.

rgb

390. What I think he should have based it on is a showing that a system exhibiting the specific lapse rate Jelbring argues for would not be in a maximum-entropy state.

Joe, I have such an algebraic demonstration and would (and may yet) post it, but it is a) boring. Of course isothermal is maximum entropy — otherwise you could violate the second law! b) how many people would understand it? I can’t even get them to understand the far simpler and entirely valid second law argument.

You cannot have a nonzero thermal lapse because if you did, and connected the two ends of the gas with a heat engine, it would turn all of the heat energy in the gas into work (dropping the temperature of the gas to zero) as Joules correctly stated.

That’s why you can’t. That is a direct violation of both Kelvin-Planck and Clausius statements of the second law. It would be friggin’ magic, the air in the column growing colder and colder not only without outside work, but turning all of that heat directly into work!

Now here’s a very, very simple “maximum entropy” exercise, one that anybody can do. We’re removing heat from the gas at nonzero temperatures, so its entropy is strictly decreasing. The heat is showing up in the Universe as nice, reversible, work, \Delta S = 0, quite independent of the temperature of the room outside.

What is the net entropy change of the Universe while all of this is going on?

Now that’s a maximum entropy computation, and I didn’t even use stat mech.

rgb

391. DeWitt Payne says:

George E. Smith; says:
January 25, 2012 at 7:09 am

The “excited states” of say CO2 that result in the absorption or emission of the 15 micron band of spectral lines have lifetimes, that are generally much longer than the time between collisions at STP conditions, so a molecule that has captured such a photon, seldom gets a chance to spontaneously re-emit such a photon; collisions terminate the condition, which transfers that photon energy to the regular N2, O2, Ar molecules of the atmosphere, as thermal energy.

Close, but no cigar. The rate of emission is proportional to the number of molecules in the excited state only. The lifetime of an individual molecule before it is de-excited by collision makes no difference at all in the rate of emission. The proportionality constant is the Einstein A21 coefficient. That coefficient can be use to calculate a half life, t_12, for a free excited molecule, but an excited molecule can emit at any time. It just means that if you have, say a thousand excited molecules at t(0), the probability is that 500 of them will have emitted at t = t_1/2. But notice that they did emit at less than the half life time.

Kirchhoff’s Law that absorptivity equals emissivity only holds if Local Thermodynamic Equilibrium (LTE) exists. A requirement for LTE to exist is that the energy distribution of the molecules follows Maxwell-Boltzmann statistics. If collisional excitation/de-excitation is not the dominant form of energy exchange, MB statistics don’t exist, nor does LTE. MB statistics determine the fraction of the population that is excited. That fraction decreases with temperature. The effective height of emission at a given wavelength depends on the optical density which depends on the number density of molecules in a unit volume, i.e. the partial pressure, not the volumetric mixing ratio and the absorption coefficient. Emission to space peaks at an optical density of 1. For CO2, that happens quite high in the atmosphere where it’s much colder than the surface or the altitudes where water vapor optical density reaches 1. But at lower altitudes CO2 molecules still emit and absorb. However, the probability that any of those photons will escape to space becomes vanishingly small as the optical density (measured from infinity or wherever you pick as the top of the atmosphere) becomes much greater than 1.

392. Bob Fernley-Jones says:

Joel Shore @ January 23, 6:13 am
You wrote in response to Konrad’s low budget experiments, which gave evidence that the basic premise of N&Z may be correct, my bold added:

Sorry…[I Joel, of godlike infinite wisdom declareth]: 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!

Oh really? Were you there to observe the detail of the experiment? Given some resource restraints on Konrad, please do not abuse the use of a rubber hot water bottle as an air pressure bladder/regulator, and house-bricks atop a plywood pressure base as an invalid mass. To me as an engineer I think it is a highly valid and quite impressive low budget/capacity approach.
How about you elaborate as to why you assume it all to be wrong. Do you get what I say? Describe the errors that you elusively assert!

393. George E. Smith; says:

“”””” DeWitt Payne says:

January 26, 2012 at 5:05 pm

George E. Smith; says:
January 25, 2012 at 7:09 am

The “excited states” of say CO2 that result in the absorption or emission of the 15 micron band of spectral lines have lifetimes, that are generally much longer than the time between collisions at STP conditions, so a molecule that has captured such a photon, seldom gets a chance to spontaneously re-emit such a photon; collisions terminate the condition, which transfers that photon energy to the regular N2, O2, Ar molecules of the atmosphere, as thermal energy.

Close, but no cigar. The rate of emission is proportional to the number of molecules in the excited state only. The lifetime of an individual molecule before it is de-excited by collision makes no difference at all in the rate of emission. “””””

So Dewitt,

Could you extract or point to the specific words in the piece of my post that YOU
chose out of my longer post; which piece I have recopied above from your post, that you consider are contrary to either of the two statements that you then made; which I have also repeated above; to whit:-

“”””” The rate of emission is proportional to the number of molecules in the excited state only. “”””
“”””” The lifetime of an individual molecule before it is de-excited by collision makes no difference at all in the rate of emission. “””””

And I say TWO statements, because the two pieces, I have referenced immediately above are contradictory with each other.

You first state that the rate of emision (actually spontaneous emission) is proportional to the number of molecules in the excited state. I have no disagreement with that; nor does my post and particularly the piece you chose above refute that.

But then you claim that the lifetime of an individual molecule before it is de-excited by collision (which will remove it from the number of molecules that are in the excited state) makes no diference at all in the rate of emission.

Now you can’t have it both ways. Either removing a given molecule from the count of excited molecules reduces the rate of emission or it doesn’t.

The process of spontaneous emission from an excited molecule; let’s say CO2 (isolated from ANY OTHER MOLECULE) , is a property of that particular molecular species, and that excited state has some statistical “lifetime” , presumably calculable from quantum mechanics; somewhat akin to how radioactive decay has some statisticl “lifetime”, commonly expressed as a half life.
Moreover, the spontaneous emission lifetime of our excited molecule is absolutely Temperature independent; it HAS to be that way, because a molecule that is isolated from any other molecule clearly hasn’t undergone any collision with any such other molecule, and any such free flight molecule has NO Temperature anyway.

So the spontaneous radiative lifetime of some specific excited state of some GHG molecule is Temperature independent.
On the other hand, both the Temperature and pressure of the ambient atmosphere, in which our free flight GHG molecule exists effect the mean time between molecular collisions, according to the statistical mechanics of ordinary gases.

So we have on the one hand the possibility of a spontaneous emission of a GHG characteristic photon, at any time with the appropriate statistical probability for that species and excited state, and we also have the probability that the GHG molecule may undergo a collision with an ordinary gas molecule, which presumably will terminate the excited state, resulting in thermal redistribution of the energy, rather than a spontaneous emission event.

The collision induced de-excitation is dependent on the gas Temperature, the spontanous emission is not Temperature dependent; but depends on having molecues in the excited state.

Now ALL that I have said, was that Phil has pointed out that spontaneous emission may be common in the cold rarified stratospheric atmosphere BECAUSE of the much longer mean time between collisions but at the lower (2 metre) STP atmosphere conditions, the mean time between collisions is much shorter so that it is more likely that collisions will de-excite the GHG molecule, rather than spontaneous emissions, because collisions reduce the number of molecules in the excited state.

We are then asked to believe that since the thermal energy of the ordinary atmosphere gases can’t be radiated away by those ordinary gas molecules, it can ONLY be given up to a GHG molecule in a collision which presumably is sometimes able to excite one of the resonances of the GHG molecule; but that process is still dependent on the GHG molecule being abe to spontaneously decay, emitting it’s characteristic wavelength photon.
Now I agree that at higher Temperature and pressure, the number of collisions with enough energy transfer to excite the GHG molecule, will be higher, so there will be more molecules going to the excited state; and by inference, more of them in the excited state at any moment; so yes, the rate of spontanous emissions will increase with Temperature.
BUT the photon energies emitted will still be the Temperature independent spectral lines of the GHG molecule, not ones representative of the atmospheric Temperature.

So according to the prevailing belief that atmospheric gases do not radiate themal spectra, the radiant emission from the atmosphere, must consist ONLY of GHG characteristic lines and bands.
So any black body like radiation seen in the earth emission spectrum from space MUST have originated from the ground; not the atmosphere, and its specrum has to be characteristic of the surface Temperature; not some upper atmosphere Temperature.

394. George E. Smith; says:

By the way Dewitt,

I don’t smoke; so why not leave out the condescending “no cigar” bit.

395. George E. Smith; says:

Kirchoff’s law of equal spectral radiant emittance and spectral radiant absorptance requires thermal equilibrium, between the radiating/absorbing material and the EM radiation field.

396. DeWitt Payne says:

George E. Smith; says:
January 27, 2012 at 1:56 pm

Now you can’t have it both ways. Either removing a given molecule from the count of excited molecules reduces the rate of emission or it doesn’t.

Sure I can. The reason is that for every molecule that is de-excited by collision, another is excited, keeping the total number constant. That’s required by MB statistics and LTE.

Kirchoff’s law of equal spectral radiant emittance and spectral radiant absorptance requires thermal equilibrium, between the radiating/absorbing material and the EM radiation field.

Correct. The radiative temperature and the kinetic energy temperature must be the same. That, however, can only happen if MB statistics apply to the kinetic energy distribution and the temperature calculated from determining the fraction of molecules in a particular excited state by measuring the radiant emission from that excited state matches the temperature determined by measuring the average kinetic energy of the gas molecules.

I used to do plasma emission spectrometry before I retired. In a plasma, there are lots of ways to measure the temperature besides the average kinetic energy.

397. George E. Smith; says:

Just a note related to a post much higher up, wherein Ralph, who has been watching the weather for 30 years said I was wrong on so many levels, in saying that clouds don’t produce positive feedback. Basically I refuted the long held meteorologist argument that clouds at night warm the surface.

You know how it goes; the six PM weather geek on T&V tells you that there are some high clouds so it is going to be a warm muggy night. Everybody and his mother in law knows from anecdotal self evidence, that high clouds and warm nights go hand in hand.

Every now and then one of them explains that it is a bit more complicated.

High clouds at night warm the surface, and low clouds at night cool the surface, and intermediate clouds at night neither warm not cool the surface. It’s part of what they learn in Meteorology school.
On rare occasions, one of them may point you to some standard meteorology text book, as one of them did to me,so I could actually read the book on Amazon,com

So I looked at his reference, and sure enough there was the tell tale graph. Specifically, the graph had an X-axis from 0 to 100 % of cloud cover, and a Y-axis of Temperature increase or decrease, and sure enough there were three graphs there.

One graph for some intermediate altitude showed that the Temperature went neither up nor down as the cloud cover went from zero to 100% cover.
Above that was an upward sloping line that showed with more cloud cover you got more temperature rise; while below it was a graph for a lower altitude that showed you got more cooling with more cloud cover for low altitude clouds.

Oddly enough all three graphs did not have the same origin. So when you have ZERO cloud cover, the Temperature depends on whether you have ZERO cloud cover at a high altitude, or whether the ZERO cloud cover isn’t at an intermediate altitude, and finally with no cloud cover not being at some lower altitude you get a cooler Temperature.

Assuming that the physics of this is regular, one would expect that at altitudes in between these three test altitudes you would get intermediate results so that the lower the low cloud the cooler it would get, and the higher the high cloud the warmer it will get.

I don’t dispute these data, after all they are in a major meteorological text book.

Besides Ralph has been watching the weather for 30 years so we can take his word for it.

Now one other important thing that Ralph told us;and he did point it out to me specifically, there is NO SUN at night time; and he was only talking about the weather AT NIGHT

So we must remember THERE IS NO SUN OR SOLAR ENERGY IN THIS PICTURE.

So on the first night, I have no clouds, so I get the normal cooling, since there is no sun to stop that.
On the second night, I have a continuous intermediate cloud cover, and I get the same surface Temperature all night as the first night.
On the third night, I have a very low all sky cloud cover; just a few hundred feet, so there is absolutely no LWIR EM radiation going to make it through that cloud; so it all gets absorbed and then re-emitted half downward; and now I find that the Temperature gets much colder than it did the previous two nights; it’s like if you put the blanket on the bed, instead of on the ceiling it is going to cool you down.
On the fourth night, I have a continuous layer of high cloud at 20,000 feet, and now I find, just as predicted; excuse me, that’s projected, the temperature remains much warmer all night long than it was the night before with the low cloud cover.

Now you see how easy this is, you don’t have to watch the weather for 30 years to understand this; you just have to read the right text books.
The dense low layer of water bearing cloud stops virtually ALL of the LWIR emission from the surface, and sends about half od it back down to the ground to keep you colder all night.

A much higher more ethereal cloud layer with very little water content by comparison, will stop only a small portion of the LWIR radiation from the ground and return half of that so with less LWIR returned to the ground you will be much warmer at night.

I know this is very hard to understand, and it would be easier to understand if we introduced the sun into the situation but as Ralph made a point to tell me, there is no sun at night.

I also posted a derivation which got garbled and regarbled twice, so neither post has it correct, despite that both claim they do, but the gist of it was simply
Given axiom #1 that nights with clouds are warmer than nights without clouds, as everybody knows, and axiom #2 that on cloudy nights it cools down much slower than on nights without clouds, as everybody also knows; then on the following morning (twilight), the cloudy night started off warmer and cooled less, so it will be much warmer than the morning twilight after a cloudless night.
But given those two axioms, I showed, that the daytime Temperature before sunset, must be warmer, for the cloudy night case, than for the cloudless night case.

So the reason it is warmer on cloudy nights than cloudless nights, is because it was warmer the day before, and it was that warmer day particularly after rain that both made the high clouds, and also made it warmer on the cloudy night. The clouds themselves have nothing to do with it; they are the result of the warm day/night, and not the cause of it.

And I don’t care how many meteorology text books claim otherwise.

398. hotrod (larry L) says:

George E. Smith; says:
January 25, 2012 at 10:04 am

It is ALWAYS warmer on a cloudy night,than on a cloudless night, because it was warmer during the previous day time, and those warmer conditions at night and the clouds themselves are the result of that simple fact; the clouds are NOT the cause of the warmer night, they are the consequence of the warmer day.

There is another contributing factor in play as well. A layer of moist air close enough to saturation temperature (allowing the formation of clouds) also has more heat capacity that the same thickness of atmosphere at a lower humidity due to the high specific heat of the water vapor.
As the more moist air cools, it condenses out more and more water, which as it condenses releases its latent heat of vaporization, turning potential energy into thermal kinetic energy.

Just like a pan of water will hold at the boiling temperature of the water until all the water is boiled away, a parcel of moist air will hold at the dew point of saturation until all the moisture is condensed out to form fog/clouds. This puts a kink in the temperature curve as the air cools at night. It cools at the normal rate consistent with its specific heat until it begins to condense out water droplets, but even though it is losing heat energy the temperature will stabilize at the dew point temperature until most of the water vapor is condensed out to liquid.

It would be more correct to say that night air cools more slowly if it is moist enough to form clouds. The slow cooling is not due to the radiation characteristics of the clouds so much as due to the stored energy available in the moist air’s reservoir of latent heat of vaporization.

Larry

399. George E. Smith; says:

Note to Dewitt Payne,

Thanks for the heads up; it’s nice to know what people do/did

I know nowt about plasmas, other than the correct spelling. When I studied BB and like, thermal radiation, which was 55 years ago, the Kirchoff condition was derived in terms of a cavity filled with material and thermal radiation, and I can’t rightly say I remember LTE being involved.

I understand that at higher temps the collision rates and energies will be higher so earlier terminations of excited states, and also that the higher energies will lead to more collision induced excited states as distinct from photon capture events; but as to why those rates must be equal escapes me. I can see that absent any LWIR from somewhere else that collision induced excitations would increase with Temperature; but it seems that a photon induced (LWIR) excited state can be terminated by a low energy collision that would ot create a collision induced excited state.

In any case, I’m going to take your word for it, and then try to figure out for myself why that is so; that’s how I learn.

In any case it gets me when people apply Kirchoff as if it applies universally all the time to anything emitting or absorbing.
The oceans which give a pretty good imitation of a black body, pretty much absorbing all (98%) of sunlight; yet I don’t see the oceans emitting a bright white light all the time. Kichoff’s law must have been repealed without telling me.

400. Phil. says:

DeWitt Payne says:
January 27, 2012 at 2:18 pm
George E. Smith; says:
January 27, 2012 at 1:56 pm

“Now you can’t have it both ways. Either removing a given molecule from the count of excited molecules reduces the rate of emission or it doesn’t.”

Sure I can. The reason is that for every molecule that is de-excited by collision, another is excited, keeping the total number constant. That’s required by MB statistics and LTE.

Except in the case we’re dealing with we’re de-exciting a ro-vibrational state with a collision (there’s no requirement that a collision will produce the same ro-vibrational state as has been de-excited). If you collide an excited CO2 molecule with an N2 molecule, you’ll lose an excited CO2 and gain an excited N2 which won’t emit. Excite some molecules in a gas cell with a focussed laser beam, vary the pressure and you’ll find that the rate of emission will drop as you raise the pressure by adding a diluent gas (e.g. N2).

401. Phil. says:

George E. Smith; says:
January 27, 2012 at 3:28 pm
Note to Dewitt Payne,

Thanks for the heads up; it’s nice to know what people do/did

I know nowt about plasmas, other than the correct spelling. When I studied BB and like, thermal radiation, which was 55 years ago, the Kirchoff condition was derived in terms of a cavity filled with material and thermal radiation, and I can’t rightly say I remember LTE being involved.

I understand that at higher temps the collision rates and energies will be higher so earlier terminations of excited states, and also that the higher energies will lead to more collision induced excited states as distinct from photon capture events; but as to why those rates must be equal escapes me. I can see that absent any LWIR from somewhere else that collision induced excitations would increase with Temperature; but it seems that a photon induced (LWIR) excited state can be terminated by a low energy collision that would ot create a collision induced excited state.

In any case, I’m going to take your word for it, and then try to figure out for myself why that is so; that’s how I learn.

In any case it gets me when people apply Kirchoff as if it applies universally all the time to anything emitting or absorbing.
The oceans which give a pretty good imitation of a black body, pretty much absorbing all (98%) of sunlight; yet I don’t see the oceans emitting a bright white light all the time. Kichoff’s law must have been repealed without telling me.

George, Kirchoff’s Law says emissivity at a wavelength = absorptivity at the same wavelength.
So if seawater has an absorptivity of 0.98 at 500 nm then its emissivity at 500nm will be 0.98, trouble is that that would only come into play at a temperature around 5,000K!

402. Spector says:

Just for reference, here is the basic greenhouse theory calculation:

The radiant energy flow or power from the sun at the Earth’s orbit is known as the ‘solar constant.’ It has a typical value of 1361 W/m². According to the Stefan-Boltzmann formula, this is the characteristic power emitted (per square meter) from a surface having a temperature of 393.6 degrees K or 120.5 degrees C.

Next, an *assumed* optical reflection factor of about 29.9 percent reduces the power being absorbed across the disk of solar radiation blocked by the Earth to an average of 953.9 W/m², which has a characteristic emission temperature of 360.1 degrees K or 87.0 degrees Celsius. This reflection value comes from the 102/341 ratio on the left side of the 2009 version of the Kiehl-Trenberth, Energy Balance Diagram. (All values on that diagram are referenced to the Earth’s spherical surface area.) It must be assumed that this reflection coefficient may depend on temperature and water content on the Earth.

The next factor that is applied is the 25 percent ratio of the area of the solar radiation interception disk to the area of the spherical surface of the Earth. In this calculation, that reduces the required average power emission from the surface of the Earth to about 238.5 W/m², which has a characteristic emission temperature of 254.7 degrees K or minus 18.6 degrees Celsius. This is the equilibrium average power emission required for stable conditions on the surface of the Earth.

The complicated 2009 Kiehl-Trenberth diagram shows the Earth being warm enough to emit an average of 396 W/m², which has a characteristic emission temperature of 289.1 degrees K or 15.9 degrees Celsius. The point here is that the Earth’s surface is continually emitting an average of about 157.5 W/m² more power by thermal radiation *alone* than it is receiving from the sun–that is, ignoring the extra power also lost by convection and evaporation. A long-wave, infrared radiation blocking, atmosphere is required to prevent the escape of this excess radiated power to outer space. That means an atmosphere containing greenhouse gases or equivalent long-wave infrared radiation blocking agents. One might contest the exact values of some of these numbers, but the overall requirement remains the same.

Compression heating by gravity is a one-time event, but thermal radiation is continuous. The second law of thermodynamics does not prevent the escape of radiation to outer space, nor does it prevent back radiation returning from a cooler region, as long as there is more forward radiation being emitted from the warm region. Thermodynamic heat flow is the net result of both forward and back radiation.

I believe that these attempted assertions of the nonexistence of the greenhouse effect serve only to enshroud skeptics of the hypothetical severe danger presented by anthropogenic carbon dioxide in the atmosphere with a wrapper of obviously ignorant pseudo-scientific nonsense. The devil is in the details, not in the basic greenhouse effect theory itself.

It does appear, however, that there has been a tendency, when presenting the results of these energy balance relationships, to gloss-over the difference between a fixed temperature equivalent to an average outgoing radiated power level (W/m²) and an average temperature.

403. DeWitt Payne says:

Phil. says:
January 27, 2012 at 3:31 pm

Excite some molecules in a gas cell with a focussed laser beam, vary the pressure and you’ll find that the rate of emission will drop as you raise the pressure by adding a diluent gas (e.g. N2).

You’ve violated LTE by using a laser beam. The CO2 molecules will have a higher total energy than the nitrogen molecules locally so MB statistics don’t apply. The reverse is true for an analytical plasma torch. Many more photons are being emitted than are absorbed because the plasma is optically thin so the plasma is not in LTE either. I’m tempted to subscribe to Spectralcalc again and calculate some emission spectra for CO2 at constant temperature and partial pressure but different mixing ratios with air. I would expect to see line broadening at lower mixing ratios (higher pressure), but not a significant change in total emission.

404. Thanks for the fantastic posts. I have been following you for a while on my rss reader, thought i would make the effort to say THANK YOU these days.

405. Phil. says:

DeWitt Payne says:
January 28, 2012 at 1:04 pm
Phil. says:
January 27, 2012 at 3:31 pm

Excite some molecules in a gas cell with a focussed laser beam, vary the pressure and you’ll find that the rate of emission will drop as you raise the pressure by adding a diluent gas (e.g. N2).

You’ve violated LTE by using a laser beam. The CO2 molecules will have a higher total energy than the nitrogen molecules locally so MB statistics don’t apply.

As they do in the atmosphere, you will always have a population of excited CO2 molecules due to absorption of IR, what’s their fate? In the lower atmosphere to lose energy by collisions mostly. I find no persuasive argument that a deactivating collision will automatically promote another molecule to the same excited state as you assert, could you elaborate on how you think that would work?

406. Just for grins, google “rgb beowulf”. The hit count is way down because the old beowulf archives aren’t active much any more, but it is still pretty respectable.

You’re *that* rgb. Seems to me you could reproduce the vaunted climate supercomputers using some variation of a beowolf cluster. Of course, they’d have to supply source code, although I suspect the data (manipulation) is critical, judging by the HARRY READ ME.txt collection. Don’t see any partial differential equations, Mathematica, etc.

407. LdB says:

I had tears rolling down my eyes reading this … sorry you have a massive problem with this theory, at least as many as problems you claim the original has.

Start at the beginning with 1859 with Kirchhoff realizing that if you put a hole in the side of a box then it is a good absorber because only radiation entering the hole has very little chance of getting back out. He realized by reversal that if you placed a hot object in the box what came out of the hole must be the radiation of the temperature.

Kirchhoff challenged theorists and experimentalists to work the energy/frequency curve for this “cavity radiation”, being German he called it hohlraumstrahlung. (where hohlraum means hollow room or cavity, strahlung is radiation).

Stefan-Boltzmann law is a direct derivative of that analysis you can’t integral it or any of the other stupidities in this article. The S-B law is directly derived from a black box you want to make a gray box fine now deal with those complications properly. A gray box >> IS NOT < http://piers.org/piersonline/pdf/Vol1No6Page691to694.pdf

I have no problem you don’t like the black box approximation work done in climate science but please this garbage is as bad as what you claim you are trying to fix.