## Monday, February 08, 2010

### Another Try

UPDATE: Eli has started working on the graphics and text as per suggestions. When he is done he will move the original text with comments to Eli Rabett's Memory Hole and repost the modification. 2/10

Periodically Eli tries to explain the greenhouse effect to someone who doesn't want to believe in it, and like explaining atomic or nuclear structure, it really cannot be done without the person with hands over ears knowing a fair amount of stuff. The really short version is in caps at the bottom. The numbers used are based on measurements or averages of measurements. Let's list a few of the preliminaries and put quotes around concepts that anyone interested in learning more can google

1. Eli is going to use the "Kelvin temperature scale" which is the appropriate one for all thermodynamical stuff.

2. In the atmosphere it gets colder the higher you go up to about 12 km. This is called the "adiabatic lapse rate" and is a result of gravity compressing the atmosphere. **

3. A solid body emits thermal radiation in the infrared (IR), this is called "black body radiation" and the amount emitted as a function of frequency (or wavelength) is described by the "Planck radiation formula". The total amount emitted per unit area is proportional to the fourth power of the temperature (ecT^4) according to the "Stefan Boltzmann law". e is a constant called the emissivity, c usually written as the Greek sigma, is the "Stefan Boltzmann constant". e is close to unity for all solids or liquids.

4. The total amount of energy in the sunlight absorbed by the earth per unit time, has to be emitted to space for the Earth to remain at a constant temperature. The emission is all in the thermal IR. There is very little, to no overlap between the "solar spectrum" and the thermal IR emitted by the Earth. If less energy is emitted, the Earth warms, both the surface and the atmosphere until the temperature is high enough (see "Stefan Boltzmann Law") to restore the balance. If more energy is emitted, the Earth cools, both the surface and the atmosphere, again, until the temperature is low enough to restore the balance.

With the preliminaries out of the way we can look at what the emission from the Earth through the atmosphere looks like at 20 km.

Look at the figure. The x-axis is in frequency units, wavenumbers, used by spectroscopists. It is the inverse of the wavelength in cm. The y-axis is the intensity of emission in W/m^2 per unit wavenumber. The red curve is the emission, the others are "Planck function" curves for various temperatures. If you look to the right, you see that there are parts of the curve that roughly follow the Planck curve for ~290 K which is the temperature of the ground. This is radiation that is going through the atmosphere essentially undisturbed. The chaff is absorption by water. The dip at about 1000 wavenumbers is from absorption by ozone, the big dip at 675 wavenumbers is absorption by CO2. There are continuum absorptions at the high and low frequency ends, mostly due to (H2O)2 (two water vapor molecules stuck together) and other complicated things consideration of which we leave for the next class.

At 375 "ppm", the current value, emission in the center of the CO2 band is characteristic of the 220 K Planck curve, meaning that the emission to space comes from an altitude at which the temperature is ~ 210 K. Because the temperature is very low, the amount of energy radiated to space is low, about 30% of the amount of thermal radiation leaving the ground at those frequencies (210^4/290^4 = .28). If we double the amount of CO2 in the atmosphere to 750 ppm the CO2 band widens blocking MORE thermal radiation

To restore the balance the Earth system has to warm.

Let's talk about that. A lot of this discussion, as you have noticed has to do with rates at which energy is transferred. Thermodynamics requires is that no NET energy be transferred from a hotter to a colder body.

The rate at which sunlight is absorbed by the ground is ~170 W/m2. If there were no absorption of the thermal IR in the atmosphere the average temperature of the Earth would be AT MOST ~255 K*

*(Eli is making a simplification here having to do with cloud reflectivity and absorption in the atmosphere, but the bottom line is the same)

What happens when the "greenhouse gases" absorb the thermal IR? The molecules almost immediately and completely transfer that energy by collision to the nitrogen and oxygen in the atmosphere, slightly warming it, BUT, collisions also vibrationally excite the greenhouse gases, including CO2. The net result is that there is an equilibrium amount of vibrationally excited greenhouse gases in the atmosphere, and this equilibrium amount depends on the local temperature.

The vibrationally excited greenhouse gas molecules emit IR in all directions, including back to the surface. With the greenhouse effect the rate at which thermal energy leaves the surface is ~390 W/m2. The rate at which the thermal energy returned to the surface is, ~325 W/m2 (go look at the Trenberth diagrams referenced below). Therefore the NET amount of energy leaving the surface due to radiation is ~65 W/m2. Of course this neglects convection, evaporation of water and a few other things. Add everything up and you get that on NET the rate at which the surface radiates is 170 W/m2 but here is the joker

BECAUSE OF THE GREENHOUSE GASES THE SURFACE HAS TO BE WARM ENOUGH TO RADIATE 390 W/m2, AND THAT MEANS THAT IT IS AT ~290 K RATHER THAN LESS THAN 255 K

** the atmosphere cools because of the adiabatic lapse rate up to the tropopause at ~ 12-15 km. Above that it warms because of absorption of UV light from the sun by ozone, but the greenhouse effect is pretty much confined to below the tropopause. Eli has a post which touches on this

carrot eater said...

either i can't find it, or you didn't give us a promised footnote ** about the lapse rate.

The lapse rate being the missing feature from most simple cartoons of the thing.

EliRabett said...

It was a minor point, and has been added. Something of a distraction for an explanation at this level. Indeed Eli would appreciate pointers to stuff that could be taken out.

Harrywr2 said...

Given that China hit 'Peak Coal' in 2006 and instituted policies based on national self interest to limit net imports of coal, how does one get to 750 ppm CO2?

The Business as Usual Scenario in the Copenhagen non-agreement shows China's CO2 output doubling by 2020.
They already consume 42% of the worlds coal production. How do they double CO2 output again?

Anonymous said...

Harrywr3 - don't know from where you are getting the idea that China hit "peak coal" in 2006, but I would take any claim of Chinese coal reserves with a really big grain of salt. Given the wide range of substances that can be labeled as "coal" it is also a good idea to question the concept of "peak coal" as it is probably poorly defined.

carrot eater said...

Peak coal? That's a new one on me. There are huge amounts of coal left; no need to worry about that.

China is limiting their coal imports? Wouldn't know it by looking at how quickly they're buying up Australian coal.

Anonymous said...

- In #3 you can omit discussion of emissivity and Stefan Boltzmann constants, and leave them out of the parenthetical term. Since you are discussing proportionality to temperature, these are a diversion.
- You introduce the impact of doubling CO2 concentration over current levels and imply you are going to discuss it further, but then in the discussion reference current CO2 vs. no CO2 (I think). This is unnecessarily confusing.
- I am unsure why you don't just include the 210 K Planck curve, and state that the "emission...is characteristic of the 210 K Plank curve..." rather than referencing 220, and then using 210 in your calculation.
- I am unclear on how the "radiative equilibrium temperature of -18 C" as discussed in the Copenhagen Diagnosis Executive Summary (page 12) relates to your discussion and think it would be helpful if you introduced that.

The Wonderer

DeWitt said...

Either don't explain the cause of the lapse rate, probably the better choice for something like this, or do a better job. As I'm sure you know, an atmosphere with a lapse rate less than the adiabatic rate is stable absent radiative heat loss. And the magnitude of the lapse rate is controlled by the heat capacity as well as the gravitational acceleration. Add water vapor and things get really ugly. Anyone who has flown recently in an airplane with a seat back display or climbed a mountain knows it gets colder as you go higher. I don't think leaving out an explanation would confuse anyone. It would be better, IMO, to just put in a link to a more detailed explanation. Compression by gravity is part of one of the nastier myths about why there isn't a greenhouse effect so you definitely want to leave that out.

EliRabett said...

The last two comments are just the sort of thing Eli wanted. Comments on them or further comments are requested.

Unknown said...

Is "Planck radiation formula" the same as "Planck function"? If so, just pick one.

Point 3 describes thermal radiation from solids and liquids - a note about gases could be helpful. E.g., they follow the Planck function at some frequencies but not others?

The use of the term "wavenumber" seems needlessly confusing - you have to define it in terms of the more familiar "wavelength" anyway, so maybe it's worth getting a plot that uses wavelength on the x-axis.

Arthur said...

Eli, black body radiation at a given wavelength is *not* proportional to T^4, the T^4 applies only to the integrated spectrum. At low temperatures, the Planck formula become exp(-h nu/kT), i.e. the intensity drops exponentially to zero for a given wavelength (I would just leave out the T^4 discussion in favor of saying something like "increases rapidly with temperature" if this is meant to be simple...)

carrot eater said...

The wanderer: See the sentence in all caps. 255 K is the -18 C you're looking for. The 210 K blackbody curve is not included because Eli is a lazy bunny (he's told us as much), and he just copied and pasted the figure from another source. But Eli knows Planck's Law, and could add it himself.

Arthur: Where does Eli say that? Up top, he says 'total amount emitted', implying the integration over wavelength and solid angle. Ah, do you mean the part "at those frequencies (210^4/290^4 = .28)"? Yeah, that's problemmatic.

I agree, if this is meant to be as simple as possible without being wrong, the reference to Planck's formula can be left out. Just add a hyperlink below the figure from the Modtran diagram, for those wanting to know where the blackbody curves came from.

Dave W: I'm with you. I realise the chemists in this field think in wavenumbers, but most people think in wavelengths. The figure is copied and pasted from another source, but Eli could reconstruct the graph in wavelength.

Anonymous said...

MarkeyMouse asks: What is the exact physical mechanism that keeps the atmosphere attached to the planet? Don't just say "gravity". I want to know what force is holding an atom or molecule in position towards the earths surface.

Anonymous said...

NAAP Atmospheric Retention Lab

Danger. Scientific information derived from real scientists (i.e. you can't believe a word they say) will be imparted.

Gravity. Temperature. Escape velocity. (Jeans' theory application might also help later.)

Cymraeg llygoden

Anonymous said...

Having said the above ...

"Thus, a methane (NH3) molecule would have ..."

ROFLMAO. (I've let them know.)

Cymraeg llygoden

EliRabett said...

Hey, it's a hydride with sp3 bonding, that gets a few points.

carrot eater said...

Playing around with Modtran. Does setting 'water vapor scale' to 0 not get rid of all the water?

I can get a pure blackbody curve at zero altitude, but I lose it as I move further up, no matter what I do.

Anonymous said...

I was thinking more in terms of a discussion of how the emission spectrum determines the hight at radiative equilibrium, and then how that in turn relates to the radiation budget diagram.

The Wonderer

Anonymous said...

What does the emission from the Earth look like at 3 Meters? What is the % of atmosphere which is CO2?

DeWitt said...

carrot eater,

No,it doesn't. It gets rid of all water vapor below 10 km, but there's still water vapor at 10 km and above. Select save text for later retrieval and and then after calculating, click on the 'view the whole output file' link below the graphs. The relative humidity data is in the far right colum of the second atmospheric profiles table. Much further down are radiance tables in both per cm-1 and per micron

carrot eater said...

DeWitt,
Thanks. I've actually been working from that text output, and didn't notice the humidity column. Stupid me.

Is there any way to get an atmosphere completely devoid of anything that absorbs or emits in this wavelength window?

EliRabett said...

I don't think so from the U Chicago interface, but you can download MODTRAN and do it in batch mode or write your own graphical interface

David B. Benson said...

Don't quite know how to fit it in, but more on lapse rate would help me at least.

In particular, as the globe warms one supposes the lapse rate changes. So that will affect the height of the tropopause?

carrot eater said...

darn it. carrot eater is a lazy bunny...

maybe someday

Arthur said...

David - lapse rate declines (due to the moist lapse rate being less than the dry one, and lower for higher temperatures), the surface warms, and so from both effects the tropopause has to be higher. Actually, that's slightly mixing up cause and effect - you can look at the lapse rate change as a negative feedback as is usually done (ch. 8 of IPCC AR4 WG1 goes into this, with plenty of references) and the tropopause increase as, to some extent, causative; in any case, once you re-equilibrate all three changes have happened.

David B. Benson said...

Arthur --- Thanks. Your comment is clearer (and shorter) than A. Hartmann's 2000 lectures notes in
571_Water_Feeback_Notes.pdf
which anyway has a mistaken equation regarding feedback.

Anyway, upon reflection, lapse rate and tropopause height and so-on is of interest once the main point of Eli's efforts has been digested. I opine it ought to be a different page.

Anonymous said...

I've not had physics since high school and that was a painfully long time ago, so with some trepidation I'll suggest that the statement "Thermodynamics requires is that no NET energy be transferred from a hotter to a colder body." is both grammatically and thermodynamically flawed.

Rick B.

EliRabett said...

got it backwards. sorry

William M. Connolley said...

The really short version is... "The surface is warmer with an atmosphere because it receives radiation from both the sun and the atmosphere".

For some reason this brilliantly simple and easily understandable explanation pleases no-one :-(

William M. Connolley said...

"** the atmosphere cools because of the adiabatic lapse rate up to the tropopause... Yeeesss: so more carefully, it reduces with height because the atmosphere is heated from below and is equilibriated by convective overturning; that, together with gravity, explains it.

Which makes it clearer why this breaks down in the stratosphere.

William M. Connolley said...

"looks like at 20 km..." What are you showing this? Your ** footnote tells us that "the tropopause at ~ 12-15 km... the greenhouse effect is pretty much confined to below the tropopause".

Consistency, man!

carrot eater said...

I'm impressed when the spammer says something that's fairly relevant to the topic.