03 February 2010

A heuristic for stratospheric cooling

I mentioned in the climate fingerprinting post that if you have more greenhouse gases in the atmosphere, we expect the stratosphere -- the upper atmosphere -- to get colder.  That, naturally, brought on the question 'why'.

I'm far from the first person to make the comment, or to attempt to write up a description of how it works on a blog.  Recently the Stoat took a swipe, or rather referenced a prior attempt and one by Realclimate, and the Rabett has also had a go.  Plus, I'm sure, there are a raft of other efforts in existence.  Yet the questioner is still asking.  That being the case, and having seen prior efforts make the attempt to describe the full situation that you have, I'll aim for a simpler version.

This will be a heuristic description.  It will be capable of being made rigorous, in the sense that you can take the heuristic and put solid math behind it.  But it will be incorrect in many of its details.  The merit of such heuristics is that even though they are incorrect in details, they lead your intuition in the right directions, such that you can then work with and understand the version of the argument that is completely correct in its details.

The simplest heuristic for surface warming in the face of an increase in greenhouse gases is our starting point.  For this, start with the surface at some temperature in balance with the atmosphere at its temperature, and with the incoming solar energy.  Now wave your hand and magically add some greenhouse gas to the atmosphere.  We will say, heuristically, that if more photons come back to the surface than used to, that the surface warms.  The surface emits some photons, same as usual.  But, thanks to the extra greenhouse gas in the atmosphere, some more get absorbed by the atmosphere.  Some of them get tossed out of the atmosphere (where they'd have gone before), but some get thrown back to the surface -- which then warms up.  The real analysis is much more involved than this, but this does give you a correct starting intuition.  It also points to the importance, yet again, of the first law of thermodynamics -- conservation of energy.

For looking at what happens in the atmosphere, let's again track photons.  (Remember those are the packets of energy carried by light).  Again, we'll start with an atmosphere and surface which are in energy balance with solar input.  Add a little bit of greenhouse gas in every layer of the atmosphere.  These cause more photons to be emitted from that layer -- half go towards space, and half towards the ground.  We'll take the heuristic approach that if most of the photons wind up in space, the layer cools.  If most wind up at the surface, the layer warms.  The heuristic also suggests that if almost all photons go to space (eventually), then the layer cools 'a lot' (it doesn't tell us how much, just a relative sense), and if half go to space and half to the surface, then it stays the same temperature.

We need to think a bit about those 'layers'.  They're being caused by energy emission from greenhouse gases.  But greenhouse gases don't emit all photon energies (or wavelengths -- short wavelengths, like blue, are higher energy than longer wavelengths, like red) equally well.  In a band (a small range of energies or wavelengths) that the gas is a strong emitter, it can easily be 1000 times better an emitter than at a wavelength a little bit different.  There is a converse to this.  By Kirchoff's law, any wavelength that the gas emits well, it also absorbs well.  So consider a strong band (15 microns, for instance, for CO2).  That strength means that a photon emitted here will very likely be absorbed before travelling very far.  That 'not very far' means that through the depth of the atmosphere, the photon will be absorbed many times.  So we can turn around and let the number of layers represent how strongly the wavelength is absorbed.  If the gas is a very strong absorber at a wavelength, then we have, say, 1000 layers.  If it's a poor absorber, it might only be one or two.

There's another feature we need in our layers, for the heuristic explanation.  Consider layer 28 (out of, say, 100).  Like all our layers, half the photons it emits go up (towards space) and half go down (towards the ground).  But ... the layers are so thick that photons will be absorbed in the next layer.  So the photons from layer 28 get absorbed in 27 or 29, rather than going to the surface or space immediately.  Now that some are in 27, we can again ask where they go -- and the answer is half go to 26 and half to 28.

I certainly wouldn't try this by hand for 1000 layers, but give yourself 3 layers.  Start with 1024 photons in the top layer (which I'm calling layer 3), send half upwards (space) and half down (layer 2, now has 512 photons).  Send half the photons up and half down, again.  Keep repeating this until all the photons are either in the bucket for space, or for the surface.  Then count up the totals in each bucket.  Repeat the process, starting with photons in layer 2, and then starting with layer 1, and compare your tallies.  You could carry this out with a stack of chips, or coins, whatever.  Or just do it on paper and copy the numbers across (it takes very good writing to carry this out, I discovered).  Or, of course, my solution of writing a short program.

It's a good idea to draw yourself a diagram here, or [Update] take a look at jg's graphic.

The outcome is, when you have several layers, the photons from the top layer mostly wind up in space.  Photons from the bottom layer mostly wind up in the ground.  The more layers (try the program, or write your own), the more this is the case.  With 100 layers, over 99% of the photons go to space from the top layer (or to the ground from the bottom layer).

So there's our heuristic answer -- the top of the atmosphere cools with an increase in greenhouse gas levels because most photons from the upper atmosphere go to space.  At the same time, the lower atmosphere warms as most photons from the lower atmosphere get caught by the surface.

That said, here's one of the limits to our heuristic description: Most of the photons don't get absorbed and re-emitted immediately, even though that's what our heuristic model says.  Most of the time, a CO2 molecule (or any other greenhouse gas) collides with an oxygen or nitrogen molecule, and hands off the energy to those molecules instead of radiating it away.  This is why nitrogen and oxygen have the same temperature as CO2.  But it also means our heuristic model is incomplete.  In addition to radiation, we should be considering what happens to the temperatures of the layers.  What saves the heuristic model is that after we let the energy slosh around, the now-warm oxygen and nitrogen molecules sometimes crash in to a CO2 molecule.  And sometimes, even if it's rare it does happen, that CO2 molecule emits a photon before it collides with another oxygen or nitrogen molecule.

There are a ton of elaborations that can be made to this heuristic.  One direction of change is to start doing Monte Carlo modeling of radiation.  This is particularly common in dealing with earth radiation in clouds.  The second is the more obvious one of tracking the full conservation of energy.  This, as you get more rigorous, becomes Radiative-Convective Modeling.

carrot eater said...

Let me try the same question I put to Stoat. I think I agree with him, but am unsure.

What happens if you could add greenhouse gas to the troposphere, but somehow left the stratosphere's composition unchanged? Is your cartoon here good enough to provide insight on this question, or do its shortcomings get in the way?

A second question. (I do some work with fluorescence, which is different in many ways, so my intuition can go awry if I'm not thinking carefully.) Let's say a photon is emitted by a CO2 molecule in layer 8 of the atmosphere. It's then absorbed by another CO2 in another layer, say layer 7 or 9. The excitement is relaxed through collision, but eventually collisions lead to excitement and a second emission.

My question: is there a difference in wavelength between the absorbed photon, and the eventually emitted photon? If so, how different? Does this answer change as the temperature difference between the original emitting layer and the absorbing/re-emitting layer increases?

I'm always a bit uncomfortable with heuristics where the lapse rate is set by radiation alone.

Penguindreams said...

carrot:
1) what happens if we add the greenhouse gas only below the atmosphere's midpoint?
The heuristic would say that this leads only to warming in the layer with new greenhouse gas. I'm not sure what the full answer is. But the photons would mostly find their way to the surface, so the heuristic will say warming. More in a moment.

2) Fluorescence?
Note for others: Fluorescence is when one wavelength is absorbed, but a different wavelength is emitted. Fluorescent bulbs use this -- the gas inside is mostly emitting ultraviolet, but the material lining the bulb absorbs that and emits visible light.

For my heuristic, consider that the atmosphere is isothermal. In that case, there's no difference on the average.

For the real atmosphere, there are temperature differences between layers, and that leads to some differences between mean absorbed and emitted energy. The surface layer will emit more high energy photons (since it is hotter) than the mid-atmosphere. The mid-atmosphere greenhouse gas is perfectly willing to absorb those photons. But when it comes time to emit, it emits according to its own -- lower -- temperature. So, on average, emits fewer of the higher energy photons.

For this heuristic the lapse rate (amount of cooling as you go up in the atmosphere) is either 0 (more rigorously true) or is taken as given (not something determined by the heuristic). To determine the lapse rate itself, you want a full radiative-convective model.

Here's a simpler analogy that might still work. Most people have double pane glazed windows. If you take cheap double pane windows, with a small gap, then measure the temperature of the outside pane of glass vs. the inside pane of glass you will see a difference in temperature due to the insulation. If you upgrade to high quality windows, they have a bigger air gap and other changes to make them insulate better. You would then expect that the inner pane of glass is warmer than before, and the outer pane of glass is colder than before. Hence a stronger "greenhouse effect" would cool the outer layer of insulation, as we also see in the atmosphere.

jg said...

Thank you for the suggestion. I'm working on an illustration. to draw this. It's very instructive to see it this way (and messy in the first draft--and that's with 3 layers!). I'll share my result when I've had a chance to check my work and clean it up.
jg

Anna Haynes said...

I have a simpler heuristic (though maybe others have it too, and maybe it is excessively simple) -

When you're thinking of weatherizing your house, you have the guy come around in winter with his infrared camera to stand outside and take your house's picture. The more insulation the house has, the cooler it'll appear to the camera.

Now put your house out in space, replace the fiberglass with greenhouse gas, and be sure you warn the camera guy to put on a spacesuit...

Anonymous said...

I don't know that I like the way this is phrased: "We'll take the heuristic approach that if most of the photons wind up in space, the layer cools."

I'd prefer to say (but realize it risks tautology) that "if the CO2 emits more photons than it absorbs, the layer cools".

This, then, addresses both the strat. ozone warming/cooling and the strat. CO2 cooling/warming. Ozone absorbs photons from the sun, more than it releases, because it occasionally transfers the heat by collision with nearby molecules. In contrast, CO2 at high altitudes doesn't absorb much from below (because lower CO2 has already absorbed it) but it does acquire heat from nearby molecules (eg ozone, directly or indirectly) and can lose that heat through emission.

A thought experiment: add a new GHG that is present only in tiny amounts, and absorbs and emits surface IR in a wavelength that no other GHG absorbs: I would argue that addition of this GHG to the atmosphere would warm it* even though above the halfway point in the atmosphere, more than 50% of the photons would make it to space.

*As long as the perceived temperature from the direction of the surface was higher than the ambient gas temperature, which should be true at any altitude if the particular wavelength is unblocked.

-Marcus

Also see: http://www.atmosphere.mpg.de/enid/20c.html

carrot eater said...

Thanks for the clarification. I've been playing with David Archer's Modtran toy, http://geoflop.uchicago.edu/forecast/docs/Projects/modtran.orig.html

I see that at any temperature you might expect to find in the atmosphere, the Planck blackbody emission curve spans 15 microns.

So I guess that no matter where in the atmosphere the CO2 is, it is possible for it to emit in its 15 micron band. The number of photons per time and area will be related to the local temperature.

If I'm on the right track, that is.

Dave W said...

Here's the key sticking point for me:
"We'll take the heuristic approach that if most of the photons wind up in space, the layer cools. If most wind up at the surface, the layer warms."
I don't see why the eventual destination of a photon should affect the temperature of the layer that emitted it.

Adrian - your analogy is appealing, but eventually inapplicable (I think.) Better windows have higher thermal resistance which allows you to maintain the same indoor temperature with less heat loss - and a smaller heat flow through the effective thermal resistance from outer window surface to outdoor ambient results in a smaller temperature difference between outer surface and ambient. (This would apply to Anna's variant, too.)

What we're doing to the atmosphere is more like keeping the heat loss rate the same - in that case, more insulation would raise the equilibrium indoor temperature but leave the outer window surface temperature the same.

Penguindreams said...

jg: You see why I wanted someone much better than myself to do the illustration.

carrot: You are indeed on the right track. 15 microns will always be an active emitter. How much will depend on local temperatures. The absorption is pretty much independent of local temperature. (At atmospheric temperatures, the CO2 is almost always in a ground state.)

marcus:
You touch on a point I didn't make clearly. It's implicit in the 'number of layers' discussion. Your new gas is in 1 layer. What matters to number of layers, for the heuristic, is how much absorption there is at the wavelength you're thinking about. Your gas is the only one absorbing that wavelength, so it is only that gas that matters. If it's a very small amount of gas, then there's only one layer (and the heuristic doesn't work well, it'll say 'no effect' since half the photons from the gas go to space and half to ground).

Dave W:
Also Marcus' initial comment. For the rigorous argument we'll need more serious math. The outcome of it, though, is that if the photons mostly make it to space, the layer cools, and if it's the ground, it warms. Now, if we followed all photons through eternity, we'll find that they all go to space. Energy absorbed by the ground gets emitted back towards space. Eventually that energy gets out of the climate system -- that's how balance is maintained. The heuristic here just allows you do draw a more or less correct conclusion without requiring you to be comfortable with partial differential equations.

Or, that is, playing with MODTRAN, which David Archer has online for folks (a way to get the equations solved for you).

carrot eater said...

OK, now let's see where I'm going off into the ditch. Everybody feel free to ignore my musing, if it's uneducational.

As cited by Marcus, another physical explanation for stratospheric cooling is given here:
http://www.atmosphere.mpg.de/enid/20c.html

It is stated differently from the heuristic here, but is consistent if you follow penguin's heuristic through.

It says that CO2 can cool a layer of gas, because it's a better emitter of IR than the other gases. But it also absorbs that IR, so it can warm the gas as well. The balance in the stratosphere is to the former, so it cools.

This returns me to my half-half atmosphere question. Using MODTRAN to sit at the tropopause looking down (I'm using tropics, so at 17 km), I can see that the CO2 in the troposphere below is absorbing the Earth's radiation at 15 microns and emitting it at a lower temperature. Increasing the CO2 level increases this effect.

So by increasing the tropospheric CO2, less radiation at 15 microns reaches the stratosphere. I am curious what would then happen in the stratosphere, if you left the CO2 concentration there unchanged. Would it still cool, because less 15 micron radiation is being received? Or after the earth responds to the radiation imbalance and the surface warms up, will the stratosphere also be warmer?

To a limited extent, water vapor acts sort of like what I want; you can increase it in the troposphere but only get minor changes in the stratosphere.

Dave W said...

Unfortunately, the heuristic seems unmotivated to me - I don't see how it's related to any physical principles I know. Can you suggest other, simpler problems in which applying this idea helps in arriving at the correct answer? Maybe thinking about those would help me make the connection.

As for Archer's MODTRAN tool - how would you use it to illustrate stratospheric cooling? I doubled CO2 from the default of 375 ppm to 750 ppm, found I had to increase ground offset by 0.89 C (aha! global warming!) to bring Iout back to the original value of 287.844 W/m^2 (presumably the value required to balance solar input), and compared the output file with the results of the default case. The computed result is a uniform increase of 0.9 C from surface to 10 km, and zero change in temp above that, so no stratospheric cooling.

Andrew said...

I like the better insulation example of either a window or a house on a cold day. Better insulation simple means warmer inside and cooler outside.

The stratosphere is basically the outside of our planet's atmosphere.

The 1000's of layers heuristic is much too long winded and doesn't seem to get to much of a point.

jg said...

I added an illustration of the first 3 layers here: here.

I hope to add a 4th layer for comparison.

jg

I picked windows for my analogy because when the sun comes out from behind a cloud, the interior does get warmer, but the outer glass temperature is constant. Maybe a double-glazed greenhouse would be a good basis for a cartoon.

In practice, modern windows have a coating on the inside that reflects some of the heat back before it gets into the house. And real greenhouses have thermally operated shutters that open to let the hot air out. Unfortunately the earth doesn't...

Anonymous said...

Hmm. I'm still dubious. I think that the ozone in the stratosphere is key to the cooling property of increasing CO2.

New thought experiment: planet absorbs in wavelength X, emits in wavelength Y. CO2 absorbs and emits Y. Multi-layer atmosphere.

Instantaneously double CO2. In the very short term, the flux of Y out will drop, and I would guess that everything above the very first layer will start to cool, while the first layer starts to warm.

But, a warmer first layer will emit more, and so the 2nd layer will warm, and then the 3rd, and so on, until the top layer.

So, my claim is that in this atmosphere where the only absorber and emitter is CO2 is that addition of more CO2 will, in equilibrium, lead to warming of _every_ layer.

However, if you have another source of heat in a layer (such as ozone absorbing solar radiation) then, and only then, can increasing CO2 in a given layer all-else-constant. Where my guess is that the key is whether a given CO2 molecule is more likely to absorb heat from absorbing radiation or heat from collisions.

Or so I theorize,

-Marcus

Stephen said...

Thanks to jg for posting the diagram and to Robert for posting the code.

Obviously, simplifications leave things out and/or gloss over details. Some things left out ... I'd say that the photons emitted do not all go either up or down (to adjacent layers), but some stay in the same layer. So the model ignores that some photons persist within a layer.

The model also seems to assume layers are uniform at least as to absorbing-and-emitting of photons (at whatever wavelength). I'm understanding the real atmosphere is denser at lower layers, and therefore the lower layers would have also a greater density of greenhouse gasses and even more opportunity to "capture" and retain photons than the upper layers. So instead of 50-50 split going up & down, you might have a percent going up, a percent going down and a percent remaining, with the "remaining" percent declining at higher layers: eg, you could say 47% go up and 47% go down and 6% remain in same layer for layer 1; 48% go up and 48% go down and 4% remain in layer 2; etc...(or whatever). That should be fairly easy to incorporate into the model, if reasonable?

Penguindreams said...

(reverse order this time)
Stephen:
You're right -- the layers aren't the same geometric thickness. Near the surface, they'll be 10-1000 times thicker than in the stratosphere. But geometry isn't what matters. They're the same thickness -- in terms of the number of absorbing gas molecules. To back up some, I estimated that a photon has to pass something like 6 trillion CO2 molecules between the surface and space. Even with 1000 layers, there are quite a few CO2 molecules in each layer.

I did take the simplification that all photons went either up or down. This is the 'plane parallel' approximation. You can be more elaborate. Not so much by keeping them in the same layer (if we're concerned only about what enters or leaves the layer -- which is indeed what matters for conservation of energy -- this is what matters), but by doing something more geometrically correct. Find a geometric shape (cubes work though are more like the layers I started with, regular tetrahedron? icosahedron?) that you can pack all of space with. Then track the photons going out each face of your shape. The plus here is that not all photons are really emitted straight up or straight down, and you're starting to allow for this. The slanting paths mean that absorption is more likely in a given layer thickness. Doesn't affect the heuristic here, which cares only about the eventual destination, but it does affect real radiation computations.

Paying attention to the absorption within the layers will be important when we try to compute the temperature of the atmosphere itself. I forget offhand whether it's 99.9% or 99.999% of the photons that will actually be absorbed and converted to heat rather than radiated to the next layer. Either way, it's a huge fraction. (On the other hand, the numbers of photons emitted is gargantuan.)

Marcus:
You're now to situations sufficiently complex, or at least you're interested in enough complexity, that heuristics won't do what you want. If you haven't gotten hold of a book on radiative transfer yet, I'll suggest Grant Petty's A first course in atmospheric radiation. (Cheaper there than Amazon.)

jg:
Thank you for the illustration. I'll update the article in a minute.

Belette said...

I don't believe you, because of http://www.antarctica.ac.uk/met/wmc/grey/

In a grey (in the IR) atmosphere, more CO2 warms all levels.

warmcast said...

Having read your heuristic description of the greenhouse effect and Rasmus description "A simple recipe for GHE" at RealClimate, I was inspired to create a simple greenhouse effect simulator in Flash.

It can be found here:

http://warmcast.blogspot.com/

I'm currently thinking how to implement an energy budget meter.
It will probably be something like a cross between a Geiger counter and an election 'swing-o-meter'!

Penguindreams said...

Warmcast: I did take a look shortly after you posted and it looks like a good illustration of the heuristic. Given belette's comment, it may not be an accurate description of the underlying physics, same as my heuristic wasn't. But we have to start somewhere to develop our intuitions. Do keep up the good work.

warmcast said...

Penguindreams I'm assuming you mean that all layers would warm up linearly but a lapse rate would still be present??

There is a lot of 'educational' material out there that oversimplifies the effect, with diagrams of 'blankets', greenhouses and car windscreens. In one case (the BBC i think) a picture of a huge CO2 molecule in the atmosphere with arrows pointing in different directions!

Such diagrams get a person so far and maybe for many that is all they need, but a lot of people want to understand more. So where do they go next?
A physics book?
Suddenly they have to make a big jump from analogies to IR spectrography and most will fall by the wayside.

I think if someone wants to make a journey to understand, they need steps that explain the processes, whether they are precise or not. If the simulator lights a bulb in someones mind and they read further (like me), then it has done it's job.

On the simulator web site, I do point out that the atmosphere is only greenhouse gases (and strangely fixed in space in a neat matrix!). So I think visitors should realise it won't be quite the same as a real and complex atmosphere.

Penguindreams said...

warmcast:
we're in pretty good agreement. I wasn't being negative about your illustration in saying that it isn't entirely accurate about the details of the physics. as you say, we have to start somewhere to develop our understanding, and I think your illustration does a good job of enabling people to make that start. someone will have to be pretty knowledgeable about radiative transfer before your illustration, or my heuristic, become limits on their understanding.

warmcast said...

Hi Penguindreams

No worries.
I have updated the simulator now and added a simple convection algorithm.
This makes the lapse rate steeper (cooler at the bottom).

Really I think my simulator is closer to the Earths Troposphere, with a tiny Stratosphere at the top!

The next thing I am going to do is create a page explaining the differences between the simulator and the Earths atmosphere.