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| Misconception: Without greenhouse gasses the Earth would be a frozen snowball at -18 C. |
What would thus the temperature be if the atmosphere was fully transparent without the so-called greenhouse gasses water vapour and CO2, thus without effects of (infrared) radiation? This would be like an Earth with no atmosphere.
And the other way around: What would the temperature be if the atmosphere was fully opaque?
These questions connect to earlier posts such as this, and these if you want to browse. For a revelation of the mystery of black body radiation, see the web site Computational Black Body Radiation.
The standard answer propagated by global warming alarmism is that the greenhouse effect is +33 C. It is claimed that with a fully transparent (or no) atmosphere, the Earth would be a frozen ball at -18 C instead of the observed +15 C with a difference of 33 C.
The presence of greenhouse gasses is thus what makes the Earth livable. The message is that the greenhouse effect is big = 33 C and as such will lead to dangerous global warming of 3 C upon a small increase of CO2 as the "best estimate" of IPCC, as 1/10 of an estimated big greenhouse effect.
This is the very basis of climate alarmism demanding a stop to emission of CO2 to prevent the Earth + atmosphere passing a tipping point into a run-away greenhouse effect approaching the surface temperature of Venus with its atmosphere filled with CO2, that is a roaring Hell at 462 C.
This is the apocalypse waiting unless we cut down CO2 emissions from human activity to zero and form a fossil free world following the lead of Sweden now transforming into the first fossil free welfare state as required by the New Swedish Climate Law.
To check the alarm signal of 33 C let us recall Stephan-Boltzmanns radiation law for a grey body:
- $Q = \epsilon\sigma T^4$
connecting radiance $Q$ at temperature $T$ in Kelvin K into a background at 0 K, through Stephan Boltzmann's constant $\sigma$ with $0\le\epsilon\le 1$ a coefficient of emissivity with a black body characterised by $\epsilon =1$.
Assuming absorptivity=emissivity (according to Kirchhoff's law), we can use the SB-law to compute the temperature $T$ of a grey body at a certain distance $D$ from the Sun knowing that the temperature $T_S$ at the emitting surface of the Sun (acting like a black body) is 5778 Kelvin K. What is needed is the ratio $q=R/D$ with $R$ the radius of the Sun, with $q^2$ the dilution effect depending on distance/area. All grey bodies at the same distance from the Sun would then have the same surface temperature (compare with discussion here).
Assuming absorptivity=emissivity (according to Kirchhoff's law), we can use the SB-law to compute the temperature $T$ of a grey body at a certain distance $D$ from the Sun knowing that the temperature $T_S$ at the emitting surface of the Sun (acting like a black body) is 5778 Kelvin K. What is needed is the ratio $q=R/D$ with $R$ the radius of the Sun, with $q^2$ the dilution effect depending on distance/area. All grey bodies at the same distance from the Sun would then have the same surface temperature (compare with discussion here).
For the Earth $q =0.00465047$ which gives the surface temperature $T_E$ through the following formula resulting from the above SB-law:
- $T_E = (0.25*q^2)^{0.25}*5778 = 279$ K
with the first factor $0.25$ the ratio between projected surface to total surface of a sphere. The temperature of the Earth as a grey body with fully transparent (or no) atmosphere, would thus be 279 K or +6 C.
For Mars with a very thin almost transparent atmosphere and with a distance to the Sun equal to 1.524 astronomical units, the formula gives 225 K, to be compared with observed about 228 K with then a small 3 C greenhouse effect. The two small moons Phobos and Deimos of Mars are reported to have about the same temperature of 233 K.
For Mercury essentially without atmosphere with a distance to the Sun of 0.4 au, the formula gives 440 K, just as observed with zero greenhouse effect.
The mean value of max and min temperatures of Ganymede, the largest of Jupiter's moons with a very thin atmosphere of Oxygen, is -125 C, which fits well with the formula with a distance of 5.2 au, again with zero greenhouse effect.
For Titan the largest moon of Saturn at a distance of 9.6 au the formula gives 90 K to be compared with an observed surface temperature of 94 K, thus with a very modest greenhouse effect of 4 K from an atmosphere somewhat denser than that of the Earth consisting mainly of nitrogen.
For Pluto with average au = 40 the formula gives 44 K = - 229 C in agreement with observed temperature varying between - 223 and - 233 C.
For the Moon (without atmosphere) rotating once every month, it is more natural to use the formula with the factor 0.25 replaced by 1 representing maximal (instead of mean) temperature to get +121 C fitting fairly well with observed maximal temperature +127 C.
We thus see that the formula works (surprisingly or not) very well for Mercury, Mars, Ganymede, Pluto and the Moon essentially without atmospheres, and so we may expect it to serve also for an Earth without atmosphere:
For Mars with a very thin almost transparent atmosphere and with a distance to the Sun equal to 1.524 astronomical units, the formula gives 225 K, to be compared with observed about 228 K with then a small 3 C greenhouse effect. The two small moons Phobos and Deimos of Mars are reported to have about the same temperature of 233 K.
For Mercury essentially without atmosphere with a distance to the Sun of 0.4 au, the formula gives 440 K, just as observed with zero greenhouse effect.
The mean value of max and min temperatures of Ganymede, the largest of Jupiter's moons with a very thin atmosphere of Oxygen, is -125 C, which fits well with the formula with a distance of 5.2 au, again with zero greenhouse effect.
For Titan the largest moon of Saturn at a distance of 9.6 au the formula gives 90 K to be compared with an observed surface temperature of 94 K, thus with a very modest greenhouse effect of 4 K from an atmosphere somewhat denser than that of the Earth consisting mainly of nitrogen.
For Pluto with average au = 40 the formula gives 44 K = - 229 C in agreement with observed temperature varying between - 223 and - 233 C.
For the Moon (without atmosphere) rotating once every month, it is more natural to use the formula with the factor 0.25 replaced by 1 representing maximal (instead of mean) temperature to get +121 C fitting fairly well with observed maximal temperature +127 C.
We thus see that the formula works (surprisingly or not) very well for Mercury, Mars, Ganymede, Pluto and the Moon essentially without atmospheres, and so we may expect it to serve also for an Earth without atmosphere:
The recorded mean temperature on the Earth surface is 288 C with gives a total atmosphere effect of +9 C, from fully transparent (or no) 279 K to observed 288 K with greenhouses gasses present into a semi-opaque atmosphere.
The total greenhouse effect is thus at most 9 C, instead of the 33 C as the corner stone of global warming alarmism.
The observed greenhouse effect of 9 C would then represent an observation of the total effect of the atmosphere on surface temperature, including both radiation and thermodynamics with gravitational lapse rate. Observation and not speculation.
Of course, the assumption that for the Earth without atmosphere emissivity=absorptivity, can be debated, since absorption and emission occurs at vastly different light frequencies, but yet may serve to get a rough estimate of the greenhouse effect (with Mercury, Mars, Ganymede, Pluto and the Moon essentially without atmospheres as observational support of the formula).
The temperature 255 K (-18 C) behind 33 C comes from an application of the SB-law assuming absorptivity = 0.7 and emissivity = 1 with questionable logic.
We can go one step further and predict what the temperature $T_E$ would be with a fully opaque
atmosphere by extrapolation from the present observed situation with the "infrared atmospheric window" acting as fully transparent atmosphere letting through 1/6 of the total emitted (infrared) radiation from the Earth surface directly into outer space. Closing the window from 5/6 to fully shut into a fully opaque atmosphere could then have an effect of $9/5$ C, less than 2 C. This is the observed variation of temperature after the last ice age.
The ultimate effect of making the atmosphere fully opaque would thus be less than 2 C and so the possible effect from more CO2 would thus be much smaller.
This argument thus supports an idea that climate sensitivity as the temperature increase upon doubling of CO2 from preindustrial level, is less than 1 C. This is based on observation of temperature 288 K (15 C) and atmospheric window 5/6 shut combined with the SB-law. Pretty basic and undisputable.
One can argue that the observations used in the argument include "feed-back" (from convection and evaporation). This is to be compared with another common argument based on (invented) "radiative forcing without feed-back" as 1 C, which is inflated to 3 C by free invention of thermodynamic feed-back.
One can argue that the observations used in the argument include "feed-back" (from convection and evaporation). This is to be compared with another common argument based on (invented) "radiative forcing without feed-back" as 1 C, which is inflated to 3 C by free invention of thermodynamic feed-back.
We can see the reduction of the basic greenhouse effect from 33 C to 9 C with a factor of 3-4, as a
reduction of the "best prediction" of climate sensitivity by IPCC of alarming 3 C into non-alarming 1 C. It may be as simple as that, to give the hope back to the people of the world.
reduction of the "best prediction" of climate sensitivity by IPCC of alarming 3 C into non-alarming 1 C. It may be as simple as that, to give the hope back to the people of the world.
PS1 In recent work by Nikolov and Zeller (referring to work by Volokin and ReLlez) the greenhouse effect is claimed to be whopping +90 C. A coming post will explain the origin of this utterly alarming (misleading) number. Nikolov and Zeller do not start out very promising: Thermal enhancement of 90 K creates a logical conundrum...appears inexplicable..Stay tuned...
PS2 The infrared atmospheric window is indicated in blue in the following picture:
PS3 The thick CO2 atmosphere of Venus is fully opaque, while the very high surface temperature of +462 C is a thermodynamic effect of high pressure from gravitation and not any "greenhouse effect" from CO2. For a Venus without atmosphere the grey body formula gives +60 C.
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Hi, Professor Johnson
SvaraRaderaI enjoy your work. I've read your wave-only light theory book where you model light as a damped spring. I didn't fully absorb the math, but I think I got the main idea.
I hope you don't buy into the idea that GHGs warm anything.
The official radiation budget shows GHGs sending all the radiation it receives. Therefore GHGs can’t transfer their vibrational energy into translational energy, otherwise there would be no or less vibrational energy to generate IR – thus destroying the radiation balance.
By conserving radiation flows, there is no heating of the air by GHGs.
And if there is heating of the air by GHGs then the radiation balance is destroyed.
Do to the constraint placed by the energy budget, this leaves only the surface to be THEORETICALLY heated by CO2.
As you know any surface heating will be diffused into the atmosphere such that dT/dh = -g/Cp.
A 3.7 W/m^2 forcing at the surface will result in an addition of
3.7 W/m^2 / 11,000 m = 0.00034 W/m^3.
If we go outside the energy budget constraint and allow direct heating of the air by IR, then we get the same thing. Each cubic meter in a column will take 0.00034 W/m^3 from "Downwelling" IR and pass on 0.00034 W/m^3 to the cubic meter below, and so on.
What do you think?
I feature your work here:
https://phzoe.wordpress.com/2019/11/11/why-up-is-not-down/
-Zoe
* and pass on 0.00034 W/m^3 LESS to the cubic meter below
SvaraRaderaCould you please show your derivation of $T_E = (0.25*q^2)^{0.25}*T_S$ (where $T_S$ is the temperature at Sun surface)?
SvaraRaderaRemember that the radiation absorbed by the earth is $\epsilon_E*(0.25*q^2)*Q_s = Q_{Ein} = Q_{Eout} = \epsilon_E*\sigma*T_E^4$
RaderaThen the formulae given by professor Johnson follows directly.
What do you think of:
SvaraRaderahttps://phzoe.wordpress.com/2019/12/25/why-is-venus-so-hot/
and
https://phzoe.wordpress.com/2019/12/04/the-case-of-two-different-fluxes/
and
https://phzoe.wordpress.com/2019/12/06/measuring-geothermal-1/
?
Would love to hear your thoughts.
Your calculation is valid for a thin flat metal disc facing the sun, not for a massive rotating sphere with poor heat conduction properties. The earth with water and air must be compared to the moon without water or atmosphere, like Nikolov and Zeller have done, which is also consistent with Mulholland and Wilde https://wattsupwiththat.com/2019/06/27/return-to-earth/
SvaraRaderaThe adiabatic heating effect is essential.
No, the formula represents a mean temperature for a (quickly) rotating sphere. For the slowly rotating Moon, it is more relevant to look at maximal temperature, see PS2.
SvaraRaderaIncoming radiation hits only half the earth at any instant, whereas outgoing radiation takes place from all of the surface continuously.
RaderaThe factor 1/4 takes care of that. This is standard bringing incoming 1360 W/m2 down to mean 340 W/m2 over the globe.
RaderaThe mean flat-earth-cold-sun value of 340 W/m2 is unphysical and can not explain the processes taking place in the atmosphere, as Mulholland and Wilde explains https://wattsupwiththat.com/2019/06/27/return-to-earth/
RaderaMoved PS2 about the Moon into the text.
SvaraRaderaThe figure 255K comes from an earth without GH gases but still an albedo 0,7
SvaraRaderaThat is as believable as waterless clouds.
Maybe worth a though: The application of the SB law in any form describe radiation from a surface, but the earth-atmosphere system radiates to the outer space from a volume. Consequently, the "equivalent" temperature is without sense (whether it is 9 or 90 K). The most pressing question is not to investigate an earth with or without atmosphere (and possible different albedos), but an atmosphere with or without greenhouse gases. The greenhouse "theory" usually deny a thermal gradient due to gravity. Rather, it explains that the lapse rate is due to radiation and convection effects. In an otherwise isothermal atmosphere the greenhouse effect must be huge.
SvaraRaderaProfessor Ivan Kennedy has published multiple interesting papers. In one of them he calculated a theoretical lapse rate of 6,8 K/km, effectively modelling the earth-atmosphere system as a heat engine in quasi-equilibrium between gravitational work and radiation. Link to one of the papers: https://www.researchgate.net/profile/Ivan_Kennedy/publication/281525629_Computation_of_Planetary_Atmospheres_with_Action_Mechanics_Using_Temperature_Gradients_Consistent_with_the_Virial_Theorem/links/55ec8d3808ae65b6389f3c38/Computation-of-Planetary-Atmospheres-with-Action-Mechanics-Using-Temperature-Gradients-Consistent-with-the-Virial-Theorem.pdf
With these numbers we are in the margins of any effect at all (due to greenhouse gases), or a small cooling effect as well as warming.
The idea is to give a new (simple) estimate of climate sensitivity based on observation and see that it is small.
SvaraRaderaCO2's 2D view factor of Earth's ~390 W/m^2 can't exceed 0.0004.
SvaraRadera0.0004 * 390 = 0.156 W/m^2
CO2 only absorbs 5-10% of the spectrum, and thus you get anywhere between
0.0078 and 0.0156 W/m^2
Am I wrong?
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SvaraRaderaThe temperature $T_o$ of a gray body object with absorptivity $\epsilon_{in}$, emissivity $\epsilon_{out}$, with projected area averaging factor $a$, located $d$ astronomical units away from the Sun with temperature $T_s$, will be given by the formulae $
SvaraRaderaT_o=0.06819 \left({\frac{a}{d^2} \frac{\epsilon_{in} }{\epsilon_{out} }} \: \right)^{\frac{1}{4}}T_s$
Interesting problem: If the initial setting was a total lack of greenhouse gases (no water) leaving only nitrogen, oxygen and argon; all of which, I am being told, are not radiatively active. So such a transparent atmosphere would not be warmed by direct solar insolation, but surely the atmosphere in contact with the warm Earth surface would warm via conduction. This would mean that the warm air would become bouyant and thus convection would leed to a warming atmosphere. The question would then be: How high a temperature could the atmosphere then reach? This question assumes that the gases (N2 O2, etc) would not be radiatively active, although I have heard voices saying that even these gases can radiate and absorb.
SvaraRaderaAnyhow, just fooling, because the fact remains that we are living on a very complex water planet, and it is actually the thermodynamically strong processes that switch between the three phases of water (gas, fluid and ice); in fact making a very efficient thermostat. This thermostat has kept our planet inhabitable ever since the primeval ocean formed and photo-synthesis led to atmospheric CO2 and H2O forming the basis for plantlife and liberating oxygen.
Hi Boris: With a fully transparent atmosphere the Earth surface temperature would be set by radiation to outer space at 0 K and thus would be around +6 C, while the temperature in the atmosphere would be determined by thermodynamics alone and as such could be motionless at +6 C as one possible state.
RaderaPretty basic and indisputable you say. But is the division between 2/3 thermodynamic and 1/3 radiation basic and indisputable? Where does this come from?
SvaraRaderaGoogle Earth Energy Balance/Budget and you will find that, e g:
SvaraRaderahttp://www.hydrogenambassadors.com/background/earths-energy-balance.php
https://www.weather.gov/jetstream/energy
RaderaHere it seems that 70% is longwave as seen from the top of the atmosphere.
And 80% seen from the surface....
RaderaHi Claes,
SvaraRaderaJust a question. I agree that the surface of earth would during nighttime loose heat, but what about the inert gases (N2, O2) that are warmed by conduction, via contact with surface warmed by the Sun and that are not radiative?. Convection would move this warm air to higher altitudes. This would also mean substantial winds or not?
Regards
Boris
A fully transparent atmosphere is just a thought experiment and one can assume motionless air without conduction for the discussion. More relevant is to compare with a fully opaque atmosphere including full thermodynamics as in the post.
RaderaDen här kommentaren har tagits bort av bloggadministratören.
SvaraRaderaLaws mandating "clean" CO2 free energy will only result in adoption of nuclear energy as it is the only economical form. Then the whole AGW idea arose to prominence after Three Mile Island nuclear accident and has overcome environmentalist opposition to nuclear energy.
SvaraRaderaYou say "The temperature 255 K (-18 C) behind 33 C comes from an application of the SB-law assuming absorptivity = 0.7 and emissivity = 1 with questionable logic." Well, the absorbtivity at 0,7 (or 0,71 to be more correct) is for the Sun's spectrum. Substances like snow, ice, and clouds do not absorb so much of the visible light, and therefor its reflectivity is quite high. The reason you can see the earth by the naked eye is due to its albedo.
SvaraRaderaWhat is absorbed, is reradiated as heat radiation where emissivity is far greater than for the visible spectrum. So there you go.
Even if Kirchhoff’s Law states that at thermal equilibrium, the emissivity of a surface equals its absorbtivity, Kirchhoff’s Law is one of the most misunderstood in thermodynamics. Keep therefor in mind that the Earth's absorbtivity at 0.71 for the Sun's radiation does not mean that there is a mismatch with the Earth'r emissivity at it's heat radiation spectrum. The absorbed and the emitted energy for any given body will be the same at thermal equilibrium.
Therefor, the Earth's albedo at 0,29 (or absorbtivity at 0,71) for short wavelengths is not in conflict with its emissivity at 1 (0,95 on average to be more correct) at longer wavelengths.
So if you do the math correct, and not least put in the relevant values of absorbtivity and emissivity for different wavelengths, you will see that the Earth's temperature with a fully transparent atmoshere, would be 255 k. Not 279 k.