During discussions years ago on the blog of an Austrlian senator I recall Dr. Jinan Cao having to point out that emissivity is wavelegnth dependent. He also alludes to it in his 2012 article at (http://joannenova.com.au/2012/09/do-greenhouse-gases-warm-the-planet-by-33c-jinan-cao-checks-the-numbers/.

Roger puts it this way ” .. The extinction coefficient/absorptivity in the Schwarzschild Equation is wavelength dependent, so the value of I = Io.exp(-KCL) + B[1 – exp(-KCL)] should be calculated for each line or small frequency range .. “.

Best regards, Pete

]]>Roger’s response got into the Schwarzschild equation and I did respond to him that I objected to the use of this equation in regard to CO2 radiance because CO2 has very low emissivity (~.2) . The equation uses emissivity of 1 in both source and sink functions. This is close for the atmosphere as a whole, but way off for CO2 specific radiance.

I heard nothing further from Roger, but I will email him right now.

]]>In Oct. 2015 one of your readers commented aboyt your credibility and invited you to disclose your credentials (https://geosciencebigpicture.com/about/#comment-4648). Your response could well be genuine, considering how you responded to retired science teacher Roger Taguchi’s comment here.

I respectfully suggest that you take up Roger’s offer of E-mail to Over the past 10 or more years he has been involved in trying to enlighten some very well qualified individuals who had misinterpreted the impact of increasing levels of atmospheric CO2. In the process he has put together a very detailed set of analyses which should help you properly understand what those MODTRAN plots are telling us.

BTW, Roger, who earned a degree in Chemistry, undertook post-graduate research under Dr. John Polanyi (http://www.nrcresearchpress.com/doi/abs/10.1139/v70-490) who won the 1986 Nobel Prize in Chemistry, for his research in chemical kinetics

Best regards (and Kaisen),

Pete

The Schwarzschild Equation equation does not assume a Planck black body radiation spectrum, so the term emissivity of CO2 (or any other greenhouse gas) ought not to be used. Instead, we have to apply the Schwarzschild Equation to any frequency (line or band) of interest.

The v=0 to v=1 bond bending absorption band of CO2 is so strong that complete absorption of most of its lines occurs within metres or hundreds of metres from the Earth’s surface. However, by Kirchhoff’s law, that complete absorption is followed by complete emission, so the NET absorption is close to zero in the first tens of metres.

Then why can we observe absorption spectra of CO2 that are well-replicated by MODTRAN computer simulations? Because normal lab absorption spectra are obtained using a warm black body source of infrared (IR) radiation; a very good approximation would be a well-insulated oven with a small hole in its wall, out of which the IR photons stream. Thus the initial intensity, Io, is high compared to that of the 288 K Earth. For example, at 576 K, the Stefan-Boltzmann T^4 law says that the flux will be 2^4 = 16 times that at 288 K.

The coefficient for the second term on the right side of the Schwarzschild Equation B, would correspond to the emission at 288 K, a factor of 16 times smaller.

However, for photons emitted by the 288 K solid and liquid (hard deck) surface of the Earth, Io corresponds to 288 K, and so would be approximately the same as B until the temperature drop with altitude becomes significant. Hence the CO2 “emission” at 667 cm^-1 will be very close to that of the 288 K Planck black body emission from the Earth until the altitude becomes significant.

Your experiments with MODTRAN show this. At 10 km, B/Io can be approximated by (220/288)^4 = 0.34 since the CO2 absorption/emission band occurs close to the maximum of the Planck emission curves for these temperatures.

At frequencies outside the CO2 band, the dry air itself (N2, O2, Ar) does NOT absorb OR emit any significant amount of IR radiation (because the molecules are non-polar, so there would be no changing electric dipole moment). Therefore a spectrometer looking upward would see only the CO2 287 K downward emission band, with none of the other frequencies that make up the spectrum emitted upward by the Earth itself. CO2 is a non-polar linear molecule in the ground state, but in bond-bending there is a changing electric dipole moment perpendicular to the molecular axis, so it can absorb and emit IR around 667 cm^-1. It can also absorb/emit during asymmetric stretch, but the symmetric stretch vibration is IR inactive because the net dipole remains at zero.

Looking upward from 40 km, the CO2 downward emission is very weak because the concentration (partial pressure) there is so small that the lines are no longer saturated. Therefore the peak emission no longer fits a Planck black body curve, and you cannot derive a physically meaningful “temperature” of emission. Because of the temperature inversion from 20 to 50 km caused by the absorption of incoming Solar UV and visible radiation by ozone in the stratosphere, the thermodynamic temperature is actually higher than 220 K, but the emission level is nowhere near that of a hotter Planck black body!

Re the P-, Q- and R-branches of the CO2 spectrum: The photon is a boson with spin 1; this means it follows Bose-Einstein statistics (so many photons can be in the same quantum state, as in a laser beam), and it carries one unit of angular momentum. Therefore in heteronuclear diatomic molecules like HCl, the rotation states change by 1 unit of angular momentum on absorption or emission of IR radiation. This explains the lines in the P- and R-branches of the spectrum (for HCl, see http://hyperphysics.phy-astr.gsu.edu/hbase/molecule/vibrot.html ).

The many, many lines in the Q-branch are formed when J, the rotational quantum number does not change. What happens to the angular momentum of the photon? In the v=1 first excited state, the most probable configuration of the molecule is slightly bent, so the there is angular momentum from the rotation of the three atoms around the equilibrium linear axis of symmetry. Note: the tip of the Q-branch is much higher than the tops of the P- and R-branches because it is made up of many, many (20 or so) very close lines (see what happens if you add 20 Gaussian bell-shaped curves separated by 0.02 standard deviations from each other). The tip of the Q-branch CANNOT be used as a “temperature” probe to deduce the “temperature of emission” and therefore an “altitude of emission”, despite what you might read in textbooks on the physics of climate change.

For the Schwarzschild Equation, see http://www.barrettbellamyclimate.com/ . Hope this helps. For more detail and explanation, contact me at rtaguchi@rogers.com .

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