Late Prof. Richard Schwartz was an astrophysicist, and had his article entitled “An Astrophysicist Looks at Global Warming” published posthumously by the Geophysical Society of America in

This question does deserve an answer. Let’s answer the question by addressing what Schwartz considered the scientific basis of greenhouse gas warming was.

*GSA Today*,**22**(1), 44-45 (January 2012). Schwartz demanded in the article “most important, contrarians*must*show why the scientific basis of greenhouse gas warming is incorrect.”This question does deserve an answer. Let’s answer the question by addressing what Schwartz considered the scientific basis of greenhouse gas warming was.

First of all, Schwartz employed the planetary mean temperature for the Venus, the Mars and the Earth to interpret the greenhouse gas warming effect; the total atmospheric greenhouse gas warming raised the temperature by 33 °C for the Earth, 6°C for the Mars and 460°C for the Venus.

To explain why this interpretation is incorrect, we need to examine how the 33°C greenhouse effect for the Earth is obtained. 33°C = 15°C – (-18°C). The -18°C is obtained by radiative equilibrium between incoming absorbing radiant flux from the Sun and outgoing emitting radiant flux from the Earth:

(1) p

*r*^{2}(1-a)*S*_{0}= 4 p*r*2 ε σ*T*^{4}where,

*r*, is radius of the Earth, and*T*earth’s mean surface temperature; α is albedo and*S*_{0}is the solar constant representing the incoming solar radiation energy per unit area and unit time with its value being around 1368 W/m^{2};*σ*is the Stefan-Boltzmann constant equal to 5.670373 x 10^{-8}(W/m^{2}K^{4}), and*ε*is the emissivity of the earth surface.In current climate research,

*ε*is either missing in the equation or is assumed to be unit. Inserting the value of*α*= 0.3 and*ε*= 1 into and rearranging Eq. (1) leads to:(2)

*T*= 254.9 (K) @ 255 (K) @ -18°CHowever, by adopting

*ε*= 1, one has assumed that the earth surface is a black-body surface, which of course can not be true. If*ε*is not 1, but 0.9, 0.8, 0.7 and 0.6,*T*would be -11.5°C, -3.6°C, 5.5°C or 16.5°C respectively. This -18°C is simply a result of technical error.On the other hand, the Earth’s mean near-surface air temperature, as measured by global weather stations, is around 15°C (@ 288K). Another widely spread technical error is to use this 15°C to subtract the -18°C. To explain why, it is essential to decode highly symbolised notions “surface” and “surface temperature

*T*” of the Stefan-Boltzmann law to extract true physical meanings for the case of earth-atmosphere system.If there is no atmosphere, the surface means the land and water ground surface of the Earth, and

*T*represents the mean temperature of the ground surface. If there is atmosphere that are all of nitrogen and oxygen, the surface is still the ground surface, and*T*still the mean temperature of the ground surface, regardless what the temperature of nitrogen and oxygen may be. This is because nitrogen and oxygen are non-radiative (literally*ε*= 0 for transparent and white bodies). 0 multiplying anything leads to 0.The real earth-atmosphere system consists of the ground surface, non-radiative gases as well as radiative gases such as water vapour and carbon dioxide. In this case the earth surface is not straightforward any more: over the absorption bands of water vapour and carbon dioxide (e.g. the absorption band 15 μm for CO

_{2}), the surface is a layer of atmosphere starting from the top of atmosphere (TOA) with thickness equal to absorption depth, and the temperature is the mean temperature of CO_{2}in this air layer,*T*_{co2}(h). One can similarly find out the surface and surface temperature for any other absorbing bands of radiative gases. For the rest of infrared bands, the surface and surface temperature are the ground surface and its mean temperature,*T*_{GSurf}. What*T*stands for in Equation (1) is the mean temperature averaged over all the infrared bands. Figure 1 shows that over the 15 μm infrared band, the earth surface and surface temperature are the top layer of the atmosphere and the temperature of CO_{2}in this layer respectively.Figure 1 An illustration showing what surface and surface temperature should be over the CO

_{2}absorption band 15 μm. For infrared bands transparent to the atmosphere, the surface and surface temperature are the ground surface and ground surface temperature,*T*_{GSurf}The global mean surface temperature 15°C is measured by weather stations using thermometers, and can be denoted

*T*_{air}(h). Obviously,*T*_{air}(h) is the mean temperature of*T*_{n2o2}(h),*T*_{co2}(h) and temperature of water vapour etc., averaged in terms of heat capacity. As such the 15°C is largely of the temperature of nitrogen and oxygen gases that consist of 99% dry air. Therefore, it is not physically meaningful to subtract -18°C from this 15°C.In calculation of Mars planetary mean temperature using the Stefan-Boltzmann equation, climate scientists made the same error as they did in calculation of the earth mean temperature, i.e. they falsely assumed the Mars has a black-body surface. If this error is corrected, one obtains the mean surface temperature for Mars -47.13°C, which is in agreement with measurements -47°C. http://www.ucar.edu/learn/1_1_2_1t.htm

When there is a net heating source on a planet, we’ll not be able to calculate its planetary mean temperature any more from radiative equilibrium. For example, the incoming radiation energy for the Sun is almost 0, while its outgoing radiation energy is still εσ

*T*^{4}, with*ε*@ 1 and*T*= 5778 K, which is a result of its energy generation due to nuclear fusion of hydrogen nuclei into helium.The Venus has almost same temperature day and night in spite of the fact that a venusian day is as long as 243 earthly days. This is an indication that there are heat generating sources, most likely magma covering generated by volcanoes on the Venus.

Clearly, how to apply and interpret the Stefan-Boltzmann’s law is the problem that has led to a misunderstanding of the greenhouse gas warming effect.

In explaining the molecular mechanism of greenhouse gases warming, Schwartz stated:

“When a greenhouse molecule absorbs an infrared photon, the molecule rotates or vibrates faster and is said to be in an “excited” state. At low gas densities, an excited greenhouse gas molecule will spontaneously (by the rules of quantum mechanics) reradiate an infrared photon, which may escape the atmosphere into space and produce no net warming.

“At the higher densities of Earth’s atmosphere, the excited molecule will bump into (collide with) another molecule (any molecule in the atmosphere). In the collision, the energized greenhouse gas molecule loses its rotational energy, which is transferred to the kinetic energy of the molecule it collides with (this is called collisional de-excitation). The increased kinetic energies of the colliding molecules means that the molecules are moving faster than they were prior to the collision, and the increased velocities of such molecules represents a direct measure of increased atmospheric temperature.

“

**’Greenhouse gas’ warming occurs because the collisional de-excitation time for greenhouse molecules in Earth’s lower atmosphere is much shorter than the radiation lifetime of excited molecular states.**This is the basic science of greenhouse gas warming.”A paper by Prof. Pierrehumbert (Pierrehumbert, R.T., 2011, Infrared radiation and planetary temperature:

*Physics Today*, v.64, p.33–38.) specifies radiation lifetime ranging from a few milli-seconds to a few tenth of a second, and collisional time 10^{-7 }s, implying that the thermal transfer process between N_{2}O_{2}and CO_{2}/H_{2}O by molecular collisions is far faster than the heat loss/gain by radiation for CO_{2}/H_{2}O.Unfortunately, wrong physics has been employed to explain the thermal absorption and in particular emission phenomena of radiative gases. De-excitation occurs only after excitation; however, emission occurs 24/7 to a non-white/transparent object as long as temperature of the object is not 0 K (-273.15°C), regardless whether it absorbs or not.

The kinetics of heat transfer by radiation is determined by the equation εσ

*T*^{4}(or*I*=*a**I*_{0}) that measures the heat energy gained/lost per unit area and unit time for an object. Assuming the total surface area,*S*, mass,*M*, and specific heat capacity,*c*_{p}, for the object, one can readily convert absorption/emission energy to temperature rise/drop for the object. The emission rate will be 10^{12}times faster at 1000 K than that at 1 K. On the other hand, the collision time of 10^{-7}s, which is related to the mean free path of air molecules, does not mean that air homogenizes its temperature in 10^{-7}s time scale. In fact air is a thermal insulator because thermal transfer for air by molecular collision is slow. Industrial examples of taking advantage of air’s thermal insulation property include: double glass windows for trains and hollow synthetic fibres etc.Hopefully, this article is comprehensive enough and does answer late Schwartz’s question.

Hello again Dr Cao

ReplyDeleteI'm getting addicted to your posts lol

I'm having trouble fully understanding the last 2 paragraphs.

How is the emission rate difference between 1K and 1000K relevant? What is the emission rate at 288K compared to collisional de-exitation rate of 10^-7?

And how does the fact that the ground surface emits continuously tie in with all this?

May I impose on you to maybe explain all this in a different way so as a layman like me can understand it.

p.s. So far my understanding is that radiative emission happens too fast for CO2/H2O to be able to warm N2/O2.

regards

Oguzhan in Brisbane

You are welcome, Oguzhan in Brisbane.

DeleteThe emission rate difference between 1K and 1000K ((10^3)^4 = 10^12) was a question/reply to Prof. Pierrehumbert who specified radiation lifetime ranging from a few milli-seconds to a few tenth of a second (range 10^2).

Assuming a tiny spherical black body object with radius, r, density, D, and specific heat capacity, cp, one can easily calculate the rate of temperature drop (dT/dt) due to emission for the object.

4 pi r^2 σT^4 = 4/3 pi r^3 D cp dT/dt

Therefore: dt/dT = D cp r / (3σT^4)

where, 4 pi r^2 and 4/3 pi r^3 are the surface area and volume of the sphere respectively. dt/dT tells the time required to drop 1 K for the sphere.

If we put numbers into the equation:

σ = 5.67 x 10^-8 (J/s)/m^2K^4

T = 288 K

D = 1000 kg/m^3

cp = 1000 J/kgK

r = 1.0 nm = 10^-9 m

One obtains dt/dT = 8.54 x 10^-7 (s/K), i.e. to drop 1 K at 288 K, it takes 8.54 x 10^-7 second.

Yes, the ground surface continuously emits, so that the radiative gases continuously absorb. But equally, the gases continuously emit. This is how we determine the radiative equilibrium temperature for CO2 etc.

The earth ground surface radiation is weak, it can only warm up CO2 up to -80C. CO2 gains further energy from molecular collision with N2O2 to rise temperature to -50C. As a result of energy transfer, N2O2 lose energy and cause temperature drop. Therefore, CO2 and water vapour are the cooling agents for the atmosphere.

Maybe carton movies will explain far better than language does.

Thank you very much for your interest in the blogs.

J. Cao

Thank you so much for the time you have taken to respond to my questions. It's all making sense for me.

ReplyDeleteDr Cao, I'm in a position whereby I can provide a much larger audience for your ideas.

May I have your permission to repost your articles, starting with the pdf in full, with a link back to your blog? (but you'll need to be prepared for an influx of commentors, including the inevitable detractors here at your blog).

Best regards

Oguzhan

Dear Oguzhan;

DeleteIt will be highly appreciated if you can help disseminate the messages. Do you need me to send you pdf files, or you can convert the blogs into pdf?

I welcome all sorts of comments/questions/criticism on my blogs. More discussions, clearer the truth will be.

Best wishes Jinan

"The kinetics of heat transfer by radiation is determined by the equation εσT4" ...

ReplyDeleteWhile this may be true for solid matter, the question is, what is the equation for resonant gaseous molecule emission for the particular LWIR bands each GHG molecule e.g. WV and CO2 posses ...