dI = emission from ds – absorption by ds

dI = n*o*B(lambda,T)*ds – n*o*I*ds = n*o*{B(lambda,T) – I}*ds

Translated into words, this equation says that the net result of absorption and emission of radiation changes the spectral intensity of radiation traveling through an atmosphere toward “blackbody intensity”, B(lambda,T) for the local temperature. The rate (with distance traveled) at which it approaches blackbody intensity depends on the density of GHGs (n) and their absorption cross-section (o). Unfortunately, this explanation has always struck me as not being very tangible.

I stumbled across a way to make this explanation more tangible using Modtran. At altitudes where the spectral intensity of DLR and OLR are equal, absorption and emission are in equilibrium and both have “blackbody spectral intensity”. For example, if you print out and overlay the spectrum of OLR looking down from 9 km and DLR looking up from 9 km, the spectral intensity of both is about 10 W/m2/um. (Tropical atmosphere, standard GHG concentration, no clouds, wavelength, not wavenumber.) The temperature at 9 km is 243.6 K and the center of the CO2 band is just above the blackbody curve for 240 K.

This equality is harder to observe in the bands due to water vapor, where intensity varies quickly with wavelength. However, you can simply look and see how the intensity of DLR at any altitude matches the Blackbody spectra for 220, 240, 260, and 280 K. These are the temperatures at 3.57, 6.57, 9.57, 12.57 and 15.57 km above the tropical surface. Looking up from 3.57 km, DLR is almost perfectly superimposed on the 280K blackbody spectrum at and below 5 um and above 25 um and at peaks between 22 and 25 um (all water vapor) and at 14 to 16 um (CO2). If you want, you can look at a single GHG at a time to assign bands to a particular GHG. Then you can look up from 3.57 km and see where OLR has blackbody spectral intensity for 280 K. At some wavelengths, it is higher than this intensity because the atmosphere is transparent to thermal IR emitted by the surface at these wavelengths.

If you check out 3.57, 6.57, 9.57, 12.57 and 15.57 km, you’ll find that water bands have BB intensity through 6.57 km, but most have dropped below that by 9.57 km. The CO2 band has BB intensity up through the tropopause (17 km), but narrows somewhat. By 24.5 km (in the stratosphere), the temperature has risen to 220 K and the strongest absorption (now a narrow line) is nearly touching the 220K BB curve. By 34 km (240K), this line is below the 220K curve. The density of CO2 at this altitude doesn’t produce enough absorption and emission to maintain equilibrium at this (or any) wavelength.

In conclusion, you can use DLR from Modtran to see what altitudes and wavelengths do and don’t have radiation with BB intensity appropriate for the temperature at that altitude.

]]>Pierre: In the case of radiative transfer of energy, the key quantity is power, not velocity. Power is measured in Watts or more commonly W/m2. Each photon of frequency v delivers hv in energy. n photons per second of the same frequency delivers nhv in power. To get total power, you need to integrate over all frequencies.

There is no need to worry about day and night. The surface of the ocean (70%) varies only about 1 K between and the upper atmosphere doesn’t change that much either. Near land surface, the temperature varies about 10 K between day and night, only 3% in degK. Referring to an average surface temperature is an approximation suitable for most purposes.

All gas molecules are colliding and constantly exchanging kinetic energy. Temperature is proportional to the mean kinetic energy. Occasionally a collision will excite a CO2 or H2O molecule into a higher vibrational state and that state can emit a photon, leaving less kinetic energy to be shared by all molecules, not just the CO2 and H2O. In most of the atmosphere, the fraction of GHGs in an excited state varies with temperature according to the Boltzmann distribution. Most of the time, excited states are created and relaxed by collisions many thousands of times faster than emission or absorption of photons. So the rate of emission of photons (radiative cooling) varies with the temperature of the atmosphere including the nitrogen and oxygen molecules, but not the number photons being absorbed.

If there were no convection, the surface temperature of our planet would be about 350 K to drive 240 W/m2 of LWR through our IR-opaque atmosphere to balance the 240 W/m2 of incoming SWR that isn’t reflected back to space. However the atmosphere unstable towards buoyancy-driven convention when when the rate of temperature fall per km rise (lapse rate) is too great. Convection removes heat from the surface fast enough to maintain a marginally stable lapse rate (averaging 6.5 K/km), reducing average surface temperature to 288 K. Near the surface about 100 W/m2 of the vertical flux of energy is provided by convection (mostly latent heat of evaporated water molecules that will condense higher in the atmosphere) and about 60 W/m2 is via thermal IR. As that thermal IR travels through the atmosphere, some photons are absorbed and some are emitted, and the change in radiation spectral intensity is given by the Schwarzschild equation discussed above. Reduction of radiative cooling to space from 2XCO2 (radiative forcing) is calculated by numerically integrating from the surface to space over all wavelengths. This is why this physics is important

dI(λ) = nσ*B(λ,T) – nσ*I(λ)*ds

Since convection involves turbulent mixing, the large grid cells in AOGCMs can’t properly describe convective features the size of a summer thunderstorm. Since convection is handled by parameters, this doesn’t make AOGCMs “total fiction”, but these parameters are tuned and not derived from fundamental physics. So AOGCMs are approximations that are “wrong”, but we can’t say how wrong.

]]>Pierre,

So the atmosphere doesn’t absorb and emit radiation. Is that what you are saying?

The maths is correct but irrelevant because O2 and N2 are transparent to IR? You didn’t mention CO2 or water vapor. Do they absorb and emit radiation?

The Sun’s energy comes in at the speed of light and heats the Earth’s surface on one hemisphere at a time, not 2 like the models say.

The speed of light. Wow. That’s fast. Completely irrelevant to the question of absorption and emission.

“One hemisphere at a time..”. Wow. Who knew? Completely irrelevant for the question of absorption and emission of terrestrial radiation (which is emitted from the surface area of a sphere) by the atmosphere.

“..not 2 like the models say” Wow. You are on a roll! Except.. no.. they don’t. Provide a reference for your “model”.

For other readers, here’s theory (the equations in the article) and observation at top of atmosphere – note the two curves have been offset to make it easier to see. Somehow the calculation by wavelength of absorption and emission match experimental results (graph in next comment due to wordpress randomness):

]]>1) Doesn’t the Schwarzschild equation for radiative transfer leave out one energetic pathway?

2) Can someone check my premise and math below? Thanks.

The Schwarzschild equation for radiative transfer properly includes the process by which absorbed radiation excites vibrational and rotational mode quantum states, and that energy then flows to translational mode of other atmospheric molecules via vibrational-translational and rotational-translational collisional processes in accord with 2LoT.

But I don’t see a term for the opposite process… during which the combined translational mode energy of two colliding molecules is sufficient (in accord with 2LoT and the Equipartition Theorem) to excite vibrational mode or rotational mode quantum states via translational-vibrational collisional processes.

Is this included in the Schwarzschild equation in the radiative term?

At sufficiently high combined translational mode energy of two colliding molecules (one CO2, one any other molecule), this would increase the time duration during which CO2 is vibrationally excited, and thus the probability that it would emit.

We know this takes place:

“The absorbed energy in Reaction (33) once again comes from translation. Two reactions of type (33) must occur for every one of the type indicated by Reaction (32) to maintain the CO2 in thermal equilibrium. The removal of energy from the translational modes by Reactions (32) and (33) cools the CO2 molecular system, and, concomitantly, the air.”

Figuring for the lowest vibrational mode quantum state of carbon dioxide, CO2{v21(1)}, 667.4 cm-1, we find that this equates to a quantum energy level of 0.08280438474000003 eV.

Dividing the energy (necessary to excite that CO2 quantum state from the ground state) into two (since we have two molecules colliding, each molecule would carry half the translational mode energy necessary to excite the vibrational mode), which would require 0.041402192370000015 eV of kinetic energy per molecule to excite CO2’s vibrational mode quantum states via t-v (translational-vibrational) processes.

This equates to a monotonic temperature of 320.3037803910884 K.

Per the Maxwell-Boltzmann Speed Distribution Function, a CO2 molecule at that temperature would have:

Most Probable Speed: 347.7243661547154 m/s

Mean Speed: 392.36493066047285 m/s

RMS Speed: 425.8736341058788 m/s

The Maxwell-Boltzmann Speed Distribution Function at 288 K (the stated average global temperature) for CO2 gives a gas fraction for molecular speeds from 347.7243661547154 m/s to 1650 m/s (an arbitrarily high number to encompass all molecules with sufficient kinetic energy to vibrationally excite CO2 upon molecular collision) of 52.73%.

An N2 molecule at that temperature would have:

Most probable speed: 435.8336709427169 m/s

Mean Speed: 491.7856346105226 m/s

RMS Speed: 533.785053266862 m/s

The Maxwell-Boltzmann Speed Distribution Function at 288 K (the stated average global temperature) for N2 gives a gas fraction for molecular speeds from 435.8336709427169 m/s to 1650 m/s (an arbitrarily high number to encompass all molecules with sufficient kinetic energy to vibrationally excite CO2 upon molecular collision) of 52.73%.

Therefore, at ~288 K, there are more atmospheric molecules with kinetic energy sufficient to vibrationally excite CO2 via translational-vibrational (t-v) collisional processes than not.

This increases the time duration during which CO2 is vibrationally excited and therefore the probability that it will radiatively emit.

The conversion of translational mode energy (which we sense as temperature) to vibrational mode energy is, by definition, a **cooling process**.

The emission of the resultant radiation to space is, by definition, a **cooling process**.

You will sometimes read “CO2 doesn’t have time to emit IR because the radiative de-excitation time is much longer than the mean time between collisions”. In conditions where collisions dominate (ie: below the tropopause), CO2 will indeed often vibrationally de-excite via v-t (vibrational-translational) collisional processes. But by the same token it will also often vibrationally excite via t-v (translational-vibrational) collisional processes at a rate dependent upon the ratio of atmospheric molecules which carry sufficient kinetic energy to excite CO2’s vibrational modes, as we calculated above.

Merely because a vibrationally-excited CO2 molecule undergoes collision with another molecule (in conditions where the translational mode energy of the two colliding molecules is higher than CO2’s vibrational mode energy and therefore energy cannot flow from vibrational to translational mode) doesn’t reset the “clock” on CO2’s radiative de-excitation time. Given that out of the three most abundant molecular constituents of our atmosphere (N2, O2, CO2), only CO2 can radiatively emit and break LTE, the net energy flow is to CO2 via t-v collisional processes above ~288 K.

The energetic pathways detailed above:

X (at ~288K+) + CO2{v20(0)} (at ~288K+) –(t-v)–> X + CO2{v21(1)} –> CO2{v20(0)} + 667.4 cm-1

X (at ~288.1K+) + CO2{v21(1)} (at ~288.1K+) –(t-v)–> X + CO2{v22(2)} –> CO2{v21(1)} + 667.8 cm–1 –> CO2{v20(0)} + 667.4 cm-1

X (at ~288.2K+) + CO2{v22(2)} (at ~288.2K+) –(t-v)–> X + CO2{v23(3)} –> CO2{v22(2)} + 668.10 cm–1 –> CO2{v21(1)} + 667.8 cm–1 –> CO2{v20(0)} + 667.4 cm-1

X denotes any atmospheric molecule.

Now, granted, not all molecular collisions are going to be head-on, and the kinetic energy imparted to vibrational mode quantum states is dependent upon angle of collision, but the data above shows that at and above ~288 K (the stated average global temperature), the majority of the molecular constituents of the atmosphere carry sufficient kinetic energy to begin significantly vibrationally exciting CO2 via t-v collisional processes.

We can use the Boltzmann Factor to determine the vibrationally excited population of CO2{v21(1)} due to translational-vibrational (t-v) collisional processes.

1 cm-1 = 11.9624 J mol-1

667.4 cm-1 = 667.4 * 11.9624 / 1000 = 7.98370576 kJ mol-1

The Boltzmann factor at 288 K has the value 1 / (798.370576 / 288R) = 0.36073473 which means that 36.073473% of CO2 molecules are in the CO2{v21(1)} vibrationally excited state due to translational-vibrational (t-v) processes.

At 288 K, this leaves a net 63.926527% of CO2 molecules available to absorb radiation. The rest are already vibrationally excited, and thus cannot absorb radiation unless that radiation is of sufficient energy to excite CO2 to the CO2{v22(2)) or CO2{v23(3)} vibrational mode quantum states.

The ratio of vibrationally excited : ground state CO2 will skew proportional to atmospheric temperature, leaving fewer molecules able to absorb radiation and more molecules able to radiatively de-excite with increasing temperature, and vice versa.

—–

The above doesn’t even take into account the other two energetic pathways by which CO2 can act as a net atmospheric coolant above ~288 K:

X (at ~288K+) + N2{v1(0)} (at ~288K+) –(t-v)–> X + N2{v1(1)} –> N2{v1(1)} + CO2{v20(0)} –(v-v)–> N2{v1(0)} + CO2{v3(1)} –> CO2{v1(1)} + 961.54 cm-1

X (at ~288K+) + N2{v1(0)} (at ~288K+) –(t-v)–> X + N2{v1(1)} –> N2{v1(1)} + CO2{v20(0)} –(v-v)–> N2{v1(0)} + CO2{v3(1)} –> CO2{v20(2)} + 1063.83 cm-1

X denotes any atmospheric molecule.

We can use the Boltzmann Factor to determine the vibrationally excited population of N2 due to translational-vibrational (t-v) collisional processes.

N2{v1(1)} (stretch) mode at 2345 cm-1 (4.26439 µm), correcting for anharmonicity, centrifugal distortion and vibro-rotational interaction

1 cm-1 = 11.9624 J mol-1

2345 cm-1 = 2345 * 11.9624 / 1000 = 28.051828 kJ mol-1

The Boltzmann factor at 288 K has the value 1 / (2805.1828 / 288R) = 0.10266710 which means that 10.26671% of N2 molecules are in the N2{v1(1)} vibrationally excited state due to translational-vibrational (t-v) processes.

Given that CO2 constitutes 0.041% of the atmosphere (410 ppm), and N2 constitutes 78.08% of the atmosphere (780800 ppm), this means that 4.1984 ppm of CO2 is excited to its {v3} mode quantum state via collisional translational-to-vibrational (t-v) processes, whereas 80162.3936 ppm of N2 is excited via the same (t-v) processes. This is a ratio of 1 vibrationally excited CO2 to 19093 vibrationally excited N2. You’ll note this is 10.028 times higher than the total CO2:N2 ratio of 1:1904, and **195 times more vibrationally excited N2 molecules than all CO2 molecules (vibrationally excited or not)**. Thus energy will flow from the higher-energy and higher-concentration vibrationally-excited N2 (said N2 vibrational mode quantum states being meta-stable and relatively long-lived because N2 is a homonuclear diatomic with no net magnetic dipole and thus cannot radiatively emit) to vibrationally ground-state CO2.

The conversion of translational mode energy (which we sense as temperature) to vibrational mode energy of N2 is, by definition, a **cooling process**.

The transfer of that N2 vibrational mode energy to vibrational mode energy of CO2, then that energy being emitted to space as radiation is, by definition, a **cooling process**. The resultant radiation from the last two energetic pathways is in the Infrared Atmospheric Window, thus any upwelling radiation has a nearly unfettered path out to space.

An increased atmospheric CO2 concentration will increase the likelihood of vibrationally-excited N2 colliding with CO2, thereby increasing the likelihood of CO2 radiatively emitting, thereby increasing the radiative cooling effect.

]]>Curious: For students, it might be appropriate to say that no one has proposed a location where a significant fraction of the energy from a radiative imbalance could be hiding below the the TOA, but they will be famous if they find one.

Limiting cases for Schwarzschild equation have trivial – but commonsense – analytical solutions. When temperature is low and B(lambda, T) is much smaller than incoming radiation (I), the equation reduces to Beer’s Law. When radiation has passed far enough through a homogeneous medium that absorption and emission have come into equilibrium, I = B(lambda, T). Finally, radiation not in thermodynamic equilibrium with a homogeneous medium through which it is traveling has an intensity that approaches (via a negative exponential) blackbody intensity at a rate proportional to n*o*s (optical depth). If B(lambda, T) is independent of position, one can replace B-I with a new variable W. dI/ds = dB/ds – dW/ds and dB/ds = 0. However, I don’t have an intuitive

]]>2. I think our disagreement is whether or not this is obvious to the student. After all, if I heat a container of water with a blow torch the temperature will increase but also I may produce motion in the water from convection.

3. I seek anyplace in any text book which discusses the point.

4. By the way, there are no numerically worked out solutions to Schwarzchild’s equation in any text. To slightly alleviate this deficit I published the following in the Bulletin of the 2019 Phoenix annual meeting of the AMS.

https://ams.confex.com/ams/2019Annual/meetingapp.cgi/Paper/349424

]]>Hi Frank

1. Again I agree with you that the amount of forcing going into a non thermal , i.e. mechanical, component in the final internal energy is probably small.

2. As far as I know, this issue is not even discussed in any text book.

3. It is not “obvious” to a student, who might point out that if she heats up a container of water there may be convective circulation of the water as well as a temperature increase. What I am looking for is a text – book or paper which discusses the issue.

5. So there is no disagreement between us, except on whether this is “obvious to the student”

Speaking of “obvious to the student” here is a numerical solution to S.E.I presented at the Phoenix Annual Meeting of the AMS; one cannot find such a numerical solution in any text book.

https://ams.confex.com/ams/2019Annual/meetingapp.cgi/Paper/349424

]]>Curious wrote: “All these texts make the tacit assumption without justification that a forcing due to increased GHG is always reflected in an increase of temperature.”

Climate change is physics. The fundamental principle is conservation of energy. If the power from a positive forcing is not going anywhere else, it must be raising internal energy somewhere beneath the TOA! If you want to be a devil’s advocate, it might worth briefly considering how much power might be consumed by other processes: Power our society uses. Kinetic energy in ocean currents, Chemical reactions/photosynthesis (which ends up as heat when vegetation decays or we eat food). Hurricanes (accumulated cyclone energy, ACE per month during hurricane season). Winds and currents can only consume power if they are accelerating. This might explain why we make your “tacit assumption”, which is justified as best I can tell.

Energy is proportional to a temperature change; power translates to a RATE of temperature change. Power per unit area translates to a rate of uniform warming in a mixed layer (a depth, which times area is a volume) of the ocean and the mass of the atmosphere. A 1 W/m2 radiative imbalance is a massive amount of power: In a closed system, 1 W/m2 would warm a 50 m mixed layer of the ocean and the atmosphere at a rate of 0.2 K/year. Since 1970, radiative forcing has been increasing at 0.4 W/m2/decade. It is difficult for me to imagine where, below the TOA, this much energy could be hiding. IMO, it is obvious that this much power must be going into the deep ocean or out to space (due to higher temperature), because the mixed layer and atmosphere certainly aren’t warming at 0.8 K/decade.

These calculations can be made more tangible for students by discussing seasonal warming in response to a seasonal change in irradiation. 1000 W/m2 of post-albedo SWR comes directly down on the Tropic of Cancer in summer and at an angle of 46 degC in winter. Lambert’s cosine law says the power delivered is reduced by the cosine of 46 deg. That’s a maximum seasonal difference of 300 W/m2 and an average seasonal difference of perhaps 200 W/m2. Thats enough to warm the atmosphere and mixed layer of the ocean at a rate of 40 K/yr or 3 K/month changing from winter to summer. (Pretty decent for a back-of-the-envelop calculation.)

BTW, converting a forcing to an amount of warming create a problem with units (or dimensional analysis). Warming is a change in internal energy. Forcing is measured in terms of power. As the planet approaches a new steady state over time, the radiative IMBALANCE approaches zero. The radiative imbalance integrated over time provides the energy that is required for warming.

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