A Big Picture Look At "Earth's Temperature"

http://climate-change-theory.com says:
January 2, 2012 at 4:36 am
…”If instead we calculate the rate of increase in a more logical, albeit simplistic fashion, from the first maximum to the present maximum we get 0.43 degrees from 1942 to 2011, namely 0.62 degrees per century. I suggest we can’t extrapolate that far anyway,…”
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@ur momisugly climate-change-theory: Good post, but a question. When you do the data for the 1942 peak are you using the data where the peak was adjusted down? I have never seen a logical explanation for the downward adjustments of that trend, or the more recent upward adjustments. I agree completely that 60 years is the minimum time frame one can use to begin to establish a trend.

The snow cover changes are interesting. I am leaning to the idea that the impact of rising co2 levels may not be on temperatures at all. Temperatures are likely to be pegged by the lapse rate to whatever is happening in the upper troposphere/stratosphere. If temperatures were pegged, then extra back radiation must be causing increased evaporation, leading to gradually increasing precipitation.
So there are about 80 W/m**2 of latent heat of evaporation and 320 W/m**2 of back radiation. If one causes the other, that would be a gain of 4, which is completely unstable of course. So each unit of latent heat would cause somewhat less than one unit of back radiation, leading to the positive water vapour feedback.
So increases in precipitation could indicate or at least suggest the strength of this feedback. Not, however, the increment in temperature required to drive it.

peter_ga says:
January 2, 2012 at 11:52 pm
The snow cover changes are interesting. I am leaning to the idea that the impact of rising co2 levels may not be on temperatures at all. Temperatures are likely to be pegged by the lapse rate to whatever is happening in the upper troposphere/stratosphere. If temperatures were pegged, then extra back radiation must be causing increased evaporation, leading to gradually increasing precipitation.
So there are about 80 W/m**2 of latent heat of evaporation and 320 W/m**2 of back radiation. If one causes the other, that would be a gain of 4, which is completely unstable of course. So each unit of latent heat would cause somewhat less than one unit of back radiation, leading to the positive water vapour feedback.

Except, I believe, that the odds of heat being lost by a molecule of CO2 thru thermal contact are higher than of losing it by radiation by approximately 10 to the umpteenth power.

Can anyone answer this question:
Look at the NOAA global temperature record above. The maxima and minima temperatures are clearly displayed. They are also shown on the HADCRU output below this graph. Both show the monthly max/min range about 0.5C before about 1950 and only 0.3C after 1950. This introduces a significant warming bias to the anomaly – of the order of 0.15C.
As minimum temperatures generally occur in the hours of darkness, (i.e. no solar insolation) then the rate of cooling post 1950 could have decreased (cloud density?). Of course it could be down to manipulation of the raw data (homogenisation)..

Peter, another factor to consider is that the ratio, 4 to 1 of LWIR back radiation to evaporation may not be linear at all. (The last ten W/m**2 of back radiation may produce far more evaporation then the first ten.) Also convection may increase the time factor of evaporation to condensation and latent heat release higher in the atmosphere; a quickening of any energy cycle takes work.

“”””” peter_ga says:
January 3, 2012 at 1:17 am
Brian_H says:
“Except, I believe, that the odds of heat being lost by a molecule of CO2 thru thermal contact are higher than of losing it by radiation by approximately 10 to the umpteenth power.”
For sure. That molecule of CO2 will hit a molecule of N2 or O2 most likely. However as the N2 or O2 will not absorb any long wave, neither will it emit any long wave, so that N2 or O2 stays heated until it hits another CO2 (or H20 or other GHG molecule) heating it up and that molecule may very well emit a long-wave photon to cool down. “””””
Why do you assume that a collision between an N2 or O2 molecule and a CO2 molecule, will heat the CO2 molecule. All of the molecules of any species will exhibit some Maxwell-Boltzmann distribution of Kinetic energies, and in any random collision, no-one can say which of the two colliding molecules will gain energy, and which will lose energy.
As for your statement that the N2 and O2 molecules can neither absorb or emit long wave radiation; perhaps you didn’t read Bill Illis’s statement, that the spectrum of the downward radiation from the atmosphere (I presume he has observed it) is in fact a Planckian Black body like continuous spectrum, which is characterized by the temperature of the atmosphere immediately above the measuring instrument. So what is YOUR physical mechanism for the emission of a continuous thermal spectrum from a CO2 molecule, when it is widely asserted that gases CANNOT emit or absorb thermal continuum radiation; but that only certain gas molecular species may emit or absorb molecular resonance type spectral lines, that form relatively narrow absorption bands compared with black body spectra, which go from >0 to <infinity frequency or wavelength.
HOW does CO2 emit a blck body like continuous spectrum.
I would suggest a visit to California's Hiway 280 near Sand Hill road, to look at a two mile long building, where some pretty savvy researchers are quite convinced that N2 and O2, and anything else can emit and absorb long wave radiation.

Gorge E Smith says:
“Why do you assume that a collision between an N2 or O2 molecule and a CO2 molecule, will heat the CO2 molecule…”
My understanding is that if a mixture of GHG and non-GHG gases is warming via the mechanism of electromagnetic absorption, then there will be a net transfer of energy in energetic collisions from the GHG to the non-GHG gases. If cooling, then the transfer of energy is in the other direction. If neither cooling nor warming, there will be no net transfer of energy. Occasionally a GHG molecule will absorb a photon, become more energetic, and on average will transfer its energy on the next collision. Occasionally when a GHG molecule hits another molecule, as well as transference of kinetic energy, a photon may be emitted by an atom in the GHG molecule.
The radiation properties of gases is basic physics. For example, to quote my old textbook:

“Elementary gases such as O2, N2, H2, and dry air have a symmetrical molecular structure and neither emit nor absorb radiation unless they are heated to extremely high temperatures at which they become ionized plasmas and at which electronic transformations occur. On the other hand, gases which have polar molecular forms with an electronic moment such as a dipole or quadrupole absorb and emit radiation in limited spectral ranges, called bands. In practice the most important of these gases are H2O, CO2, CO, SO2, NH3 and the hydrocarbons. These gases are asymmetric in one or more of their modes of vibration. During molecular collisions, rotations and vibrations of individual atoms in a molecule can be excited so that atoms which possess free electrical charges can emit electromagnetic waves. Similarly, when radiation of the appropriate wavelength impinges on such a gas, it can be absorbed in the process.”

from “Principles of Heat Transfer” by Frank Kreith, 3ed, 1976, pp273-274.
And a quick google reveals this posting showing bottom-of-the-atmosphere spectra that do seem rather coloured. (Apologies in advance as I am sure to stuff this link up. Cannot find any code guide.)

So when I said that a GHG molecule emits a photon, a single photon will have a fixed frequency, from E = hf, and GHG emissions will not be blackbody, but banded, with emissions at a range of fixed frequencies.

“”””” peter_ga says:
January 3, 2012 at 7:13 pm
Gorge E Smith says:
“Why do you assume that a collision between an N2 or O2 molecule and a CO2 molecule, will heat the CO2 molecule…”
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The radiation properties of gases is basic physics. For example, to quote my old textbook:
“Elementary gases such as O2, N2, H2, and dry air have a symmetrical molecular structure and neither emit nor absorb radiation unless they are heated to extremely high temperatures at which they become ionized plasmas and at which electronic transformations occur. On the other hand, gases which have polar molecular forms with an electronic moment such as a dipole or quadrupole absorb and emit radiation in limited spectral ranges, called bands. In practice the most important of these gases are H2O, CO2, CO, SO2, NH3 and the hydrocarbons. These gases are asymmetric in one or more of their modes of vibration. During molecular collisions, rotations and vibrations of individual atoms in a molecule can be excited so that atoms which possess free electrical charges can emit electromagnetic waves. Similarly, when radiation of the appropriate wavelength impinges on such a gas, it can be absorbed in the process.” “””””
Well Peter the CO2 molecule is also quite charge symmetrical, just like those diatomic molecules H2, N2, O2, and not to mention Ar, it is NOT a polar molecule like H2O. So how then does CO2 emit a continuum thermal radiation spectrum, that matches the Planck spectral irradiance spectrum of a black body at the same Temperature ?
Also which of those atoms mentioned in your old text book possess free electrical charges (under STP ordinary atmospheric conditions).
Your old text book also says: “””””These gases are asymmetric in one or more of their modes of vibration. “”””” Too bad it doesn’t go on to say that these modes of vibration are mechanical resonances, and can emit or absorb, only those specific photon energies that correspond to those frequencies; those are discrete line spectra, which in molecules tend to be broader than in the lines of atomic spectra; which also are NOT thermal radiation spectra.
What the molecules need, in order to absorb or emit EM radiation, is not free electric charges; but a non zero electric dipole moment, and CO2 in its degenerate bending mode at 15 microns, or its asymmetrical stretch mode at 4.0 microns, does possess such a non zero electric dipole moment, but it can emit or absorb, only photons at those specific frequencies, so that does not yield a thermal continuum spectrum.
Your old text book does also say: “”” neither emit nor absorb radiation unless they are heated to extremely high temperatures at which they become ionized plasmas and at which electronic transformations occur. “””
Too bad it doesn’t say anything at all about why or how,they can emit or absorb while they are ionised plasmas; just a simple declarative statement; ie appeal to authority.
Well yes they do at least emit a continuum spectrum under those conditions; but sadly, it is not a black body spectrum corresponding to the gas or plasma Temperature, in fact it is simply appended on to the [asymptomatic] end of the “Bohr” atomic spectrum.
The reason for such a spectrum, is because the emission results from the capture of a free electron, by the ion, and eventual transfer to some atomic energy level. Since the free electron can have any value of energy at all, the energy of the emitted photon can also have any frequency at all; so long as it is above the end of the atomic line spectrum.
So back to those diatomic or monoatomic gases, that are charge symmetrical in their ground state, so have zero electric dipole moment.
Now the physical manifestation of the Temperature of such a collection of molecules, is the Maxwell-Boltzmann distribution of kinetic energies of those molecules, which is how the thermodynamic scale of Temperature is defined.
So those mono or diatomic molecules are in constant collisions with each other, and at ordinary atmospheric STP conditions, their velocities are non relativistic, and at that atomic scale, ordinary Newtonian dynamics applies, as does Coulomb’s law for the force between charges.
The positive charge is essentially (at atomic scale) a single point, and coincident with the nuclear mass, which is of the order of 3675 times the mass of the equal and oppositely charged electron cloud. The proton/electron mass ratio is 1837, and 1838 for the neutron/electron mass ratio.
So virtually the entire kinetic energy and momentum , that represents the Temperature of the neutral gas, is resident in the nucleus. Also the nucleus is relatively unaffected by the charge of the electrons, which in the free flight state, is roughly spherical, and the Biot-Savart law applies, for the net electric field inside a charged conductor. In free flight, the center of charge of the electron cloud, and the nucleus are coincident, which is why the electric dipole moment is zero.
But when two such molecules collide, the electron clouds repel each other, and in a head on collision, they will come to a complete stop (in center of mass space). Meanwhile the nuclei, having all the momentum, and no greater Coulomb force than the electrons carry on towards each other until they are much closer, and the inverse square law, eventually takes charge, and the nuclei also come to a stop, and then a reversal acceleration; they better, because the electron clouds are already taking off in the reverse direction.
So during the collision, the electric dipole moment of the atoms takes on a non zero value, and while the charges accelerate, they MUST radiate EM waves according to Maxwell’s equations. The radio-physicist would simply describe this as a varying electric current flowing in a non zero length antenna; and once again the laws of Physics would require that it radiate or absorb EM waves, during the transient asymmetry of charge during the collision. A simple Fourier transform of the collision profile, will lead to the spectrum of the emitted “pulse” of EM radiation, and because of the totally random nature of the collisions as to velocities and trajectories, the total spectrum of emissions or absorptions by a large collection of molecules, will radiate a thermal continuum spectrum, which is characterized by the Thermodynamic Temperature of the gas. That is the only way, that the downward radiation from the atmosphere can have a Planckian black body like spectrum, which Bill Illis says it certainly is.
But since the density of gas molecules is much lower than for liquids, and even lower still than for solids, the amount of volume of gas it would take to be a total absorber, which some solids are close to, and which 3,000 feet of sea water is close to, the earth doesn’t have near enough atmosphere for it to behave like a black body radiator or emitter; but the spectrum does match the appropriate BB spectrum.
Neither Raleigh nor Jeans, or Max Planck addressed the basic physical source of black body radiation; they simply calculated the statistical mechanics of its distribution; incorrectly in the case of Raleigh and Jeans.
It was up to Heinrich Hertz and Maxwell to derive the emission of EM radiation from varying electric currents in antennas; which is exactly what accelerated charges are.
The Stanford 2 mile Linac exists simply because Maxwell insisted that accelerated electric charges MUST radiate EM waves; and must do so, even in Bohr’s planetary atom.
Bohr simply rejected that assertion, and quite arbitrarily, whithout any basis in observation declared that Maxwell’s equations be null and void, when the atomic electrons, were in certain magic orbits, and no matter their huge radial acceleration they simply disobeyed Maxwell’s equations.
Well not so. The quantum mechanical picture of Bohr’s “magic” orbits, became simply a statistical probability of where the electron might be found. So now there is no assumption of any charge motion at all, let alone acceleration; just probability of location; so Maxwell’s laws of EM radiation are re-instated; and actually rule the roost, since they alone determine the velocity of light; which now is assigned an absolute value, with no error in value, as do epsilon nought, and mu nought, which (c) is calculated from.
As Bill Illis has noted, the ordinary atmospheric gases, most certainly are absorbing and radiating thermal continuum spectra that match the BB spectrum for their gas Temperature.
And if you read the last two sentences excerpted from your old Physics book, you will see that is exactly what your book says.

George E. Smith; says:
January 3, 2012 at 8:51 pm
Thank you for your reply.
A partial quote from your reply : “Of course it does, in several spectral bands starting at about 700 nm wavelength, beyond which about 45% of the solar energy exists, and those water bands seem to absorb about half of the total spectral range from there out to around 4.0 microns, beyond which, only 1% of solar energy remains.”
Would your reference to 4.0 microns be the condensate size (droplets) of the H20 ?
Would there be an effective ‘size’ to the H20 at say 50% humidity?