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## The Hoover Incident

You agree that if earth had 100% nitrogen atmosphere (a non greenhouse gas), the “average” temperature of earth would be different (I’m not entirely sure it wouldn’t be lower) from the 255 K blackbody radiation would suggest.

I got the meaning wrong and said “right, it would be 255K” and wasn’t very specific with what I meant, which was a mistake as the article in question had just explained everything wrong with averages..

He responded with an interesting example of a fictional Latvia, where Latvia got all the solar energy (somehow) and the rest of the world none, and showed that the average temperature of the earth was 0.5K or almost -273°C (and Latvia was quite steamy).

And after I had written a comment as long as a post I thought it was probably a subject that would be worth making a post out of..

“Nice example, and I guess as I’ve been explaining everything that’s wrong with averages I should have been more careful in my explanation.

We will consider a real earth with an albedo..”

The absorbed solar radiation has an average of 239 W/m2. This is averaged over the surface of the earth, which is 5.1 x 108 km2 = 5.10 x 1014 m2.

The incoming absorbed energy is therefore 1.2 x 1017 W.

Note: this average of 239 is a measured value of incoming – reflected solar radiation, divided by 4 for spatial geometry reasons, see The Earth’s Energy Budget – Part One. The total of 1.2 x 1017 W also equals the TSI of 1367 W/m2 x “disc area” of the earth x (1-albedo).

This is also currently the measured average outgoing longwave radiation (OLR) at the top of atmosphere, within instrument error.

### The Hoovering

We’ll come to the energy-grabbing Latvia later (sorry Latvian’s wasn’t my idea).

It’s a typical earth in all other respects but a cosmic being has just “hoovered up” the trace gases like CO2, CH4, NO2 and turned water vapor into a non-absorbing gas. (If water vapor was also hoovered up the oceans and lakes would be a ready source of water vapor and within a month or two the atmosphere would have the same water vapor as before).

As a result radiation >4μm just goes right through it. That is the whole point about radiation – if nothing absorbs it, it keeps on going.

The radiation emitted from the earth just after the Hoover incident is 396 W/m2. Or, putting it another way, the total radiation emitted from the earth’s surface is 2.0 x 1017 W.

This total radiation is calculated by adding up the radiation from every square meter on the earth. And the average is simply the total divided by the area.

Note. In fact across a day and a year this value will change. Even year to year. So 2.0 x 1017 W is just the annual average across an appropriate time period. But it is appreciably higher than the absorbed solar 1.2 x 1017 W.

### Energy Loss through Radiation and Climate Response

The net radiation loss from the planet starts at 0.8 x 1017 W = 2.0 x 1017 – 1.2 x 1017

Who knows what kind of climate response that would generate? And the time for the total response would also depend on how well-mixed the oceans were (because of their large heat capacity), but rather than trying to work out how long it will take, we can say that the earth will cool down over a period of time.

And no matter what happens to convection, lapse rates, and rainfall this cooling will continue. That’s because these aspects of the climate only distribute the heat. Nothing can stop the radiation loss from the surface because the atmosphere is no longer absorbing radiation. They might enhance or reduce the cooling by changing the surface temperature in some way – because radiation emitted by the surface is a function of temperature (proportional to T4). But while energy out > energy in, the climate system would be cooling.

### Clouds and Ice Sheets

It’s possible (although unlikely) that all the clouds would disappear and in which case the net incoming – reflected radiation might increase, perhaps to 287 W/m2. (This value is chosen by measuring the current climate’s solar reflection of clouds, see Clouds and Water Vapor).

If that happened, total absorbed energy = 1.5 x 1017 W.

However, as the earth cools the ice sheets will increase and there’s no doubt that the albedo of the earth’s surface will increase so who knows what exactly would happen.

But it seems like the maximum absorbed solar radiation would at most go from 1.2 x 1017 W to 1.5 x 1017 W and more likely it would reduce below 1.2 x 1017 W.

### A New Equilibrium

Eventually the outgoing radiation would approximately match the incoming radiation and temperature would settle around a value – this would take centuries of course, maybe 1000s of years..

And depending on the ice sheet extent and whether any clouds still existed the value of outgoing radiation might be around 1.0 – 1.5 x 1017 W. This upper value would depend on the ice sheets not growing and all the clouds disappearing which seems impossible, but it’s just for illustration.

Remember that nothing in all this time can stop the emitted radiation from the surface making it to space. So the only changes in the energy balance can come from changes to the earth’s albedo (affecting absorbed solar radiation).

And given that when objects emit more energy than they absorb they cool down, the earth will certainly cool. The atmosphere cannot emit any radiation so any atmospheric changes will only change the distribution of energy around the climate system.

What would the temperature of the earth be?

I have no idea.

But let’s pick the 1.2 x 1017 W as our average incoming less reflected solar energy, and so when “equilibrium” is reached the earth’s total surface radiation will be at this value. (Strictly speaking equilibrium is never truly the case, but we are considering the case where measured over a few decades the average outgoing radiation is 1.2×1017 W).

Uniform Temperature

Suppose that all around the world the temperature of the surface was identical. It can’t happen, but just to put a stake in the ground, so to speak.

Average outgoing radiation / surface area = 235 W/m2 – (that’s because I originally wrote down 1.2 x 1017 instead of 1.22 x 1017 and so the rounding error has caused a change from 239, but it doesn’t particularly matter, more a note for those in the “rounding police”).

And with a longwave emissivity close to 1 for most of the earth’s surface, this would be a temperature of 254K.

The Energy-Grabbing Latvians

As I mentioned before Latvians and Latvophiles, this wasn’t my idea. And anyway, it probably wasn’t your fault. I’ll adjust the commenter’s numbers slightly to account for the earth’s albedo.

In his example, mythical-Latvia has a surface area of 10,000 km2 = 1010 m2. And mythical-Latvia absorbs all of the energy with none left for the rest of the earth. The rest of the earth is at a temperature of 0K and Latvia is at a temperature of 5080K which means it radiates at 1.2 x 107 W/m2. Therefore the radiation from the whole earth = 1.2×1017 W and so the earth is in equilibrium (with the solar radiation absorbed), but the average temperature of the whole earth = 5080 * 1010 / 5.1×1014 = 0.1K.

Frosty.

And our commenter has nicely demonstrated the same point as in Lunar Madness and Physics Basics – you can have the same radiation from a surface with totally different average temperatures.

### The Maximum “Average” Temperature

So we have had two approaches to calculating our equilibrium hoovered atmosphere, one with a temperature of 254K (-19°) and one with a temperature of 0.1K (-273°C) – both quite unrealistic situations it should be noted.

What’s the maximum “average” temperature? As already noted, at the wavelengths the earth radiates at (>4μ) it is quite close to a blackbody.

In this case, the maximum “average” temperature can’t be higher than what is known as the “effective blackbody temperature”, Teff.

This value, Teff, is a common convention to denote the effective radiating temperature of a body. This is just the temperature-radiation conversion from the Stefan-Boltzmann equation, Teff = (E/σ)1/4 – the rewritten version of the more familiar, E = σT4. It simply converts the energy radiated into a temperature. So 235 W/m2 = 254K.

It doesn’t mean, as already explained, that Teff is the “average” temperature. The “average” temperature (arithmetic mean) can be quite different from Teff, demonstrating that average temperature is a troublesome value.

I created confusion using the concept of Teff in my comment in the earlier article. By saying the earth would be at 255K without a radiatively absorbing atmosphere I really meant that in that situation the earth’s surface would be radiating (averaged globally annually) 239 W/m2.

### The Earth Without an Absorbing Atmosphere

With conventions out of the way – if the atmosphere didn’t absorb any terrestrial radiation the radiation from the surface would slowly fall from its current annual global average of 396 W/m2 to around 240 W/m2.

The climate would undergo dramatic changes of course and no one can say exactly what the equilibrium “effective blackbody radiating temperature” would be as we don’t know how much solar radiation would be reflected in this new climate. Clouds, ice – and aerosols – all play a part in reflecting the solar radiation from the atmosphere and the surface, and if these change the amount of energy absorbed changes.

But without an atmosphere that absorbs longwave radiation there is no way that the radiation from the surface can be greater than the radiation from the top of the atmosphere. And that means that eventually the emission of radiation from the surface would be approximately equal to the absorbed solar radiation.

Therefore, the value of global annual average radiation might be 290W/m2 (unlikely), or it might be less than 239W/m2 (more likely).

The world would be much colder.

With an annual surface radiation of 239W/m², the “average temperature” would be  -18°C or colder.

### 52 Responses

1. on June 5, 2010 at 8:11 pm | Reply Nullius in Verba

“But without an atmosphere that absorbs longwave radiation there is no way that the radiation from the surface can be greater than the radiation from the top of the atmosphere.”

I have mentioned this thought-experiment before, and I know you have a policy on repetition, but I thought it would be worth going through it again.

We have an imaginary planet of about Earth size but with an atmosphere about 100 times the mass made of pure Nitrogen. The surface pressure is of course about 100 bar.

High up in this alien atmosphere, 50 km above the surface, we have clouds of alien life forms, consisting of floating bags of gas. They are coloured black on top (to absorb the sun’s rays), and reflective silver on the bottom, but angled to reflect upcoming rays out sideways – for camouflage, maybe. They’re aliens – they don’t really need a reason.

So, these gas bags bask in the sun, absorbing 90% of the insolation. The remaining 10% shines through to the surface, where it is all absorbed. With the variation with latitude, and with the slow turn of night and day, the heating is differential, and so we get large-scale convection.

Now, we know that with the gas bags absorbing 90% of the light, they are going to settle at a temperature pretty close to the effective radiative temperature. They absorb heat, radiate it from their tops, and settle out at about 250 K, say.

Now with a convective atmosphere going on below them, we have an adiabatic lapse rate, which will stabilise at around 10 K/km of altitude. Any gas that descends from 50 km altitude to the surface will increase in temperature by 500 K. So the surface will settle at about 750 K.

The surface will be radiating infra-red like mad, of course. And all this radiation will fly straight out into space. It will either go direct, or reflected off the bottom of one of the gas beasts. There will be no back radiation at all. The Nitrogen atmosphere will emit none. The silvery bottoms of the gas beasts will emit none downwards. There will be no downward longwave flux. The only sources of heat at the surface are the shortwave solar radiation, and convective downflow.

And yet, it would appear, we have a 500 K temperature differential between the intensely hot actual surface and the “expected” effective radiative temperature. This sounds a lot like a greenhouse effect. But we have no backradiation!

Oh, what can it mean? Is it possible that without an atmosphere that absorbs longwave radiation there is a way that the radiation from the surface can be greater than the radiation from the top of the atmosphere? Or does the physics go wrong somewhere?

If so, where?

Thanks.

2. Nullius your example is effectively a planet absorbing 25wm-2 sunlight with an atmosphere incapable of absorbing IR.

The average surface temperature of such a planet cannot be warmer than a 25wm-2 blackbody, which is about 144K (the avg temperature could be cooler if the emission is uneven as this article discusses, but it cannot be greater than 144K)

If the surface was 750K it would be emitting almost 18,000wm-2 and only absorbing 25wm-2 the planet will bleed energy until it cools down to 144K.

3. on June 6, 2010 at 1:05 am | Reply John Phillips

On a planet with no IR absorbing gas in the atmosphere, isn’t at least a small amount of heat transferred to the atmosphere at the surface interface? In other words, doesn’t the atmosphere cool the surface, with some heat being transferred to the atmosphere, thus lowering the amount of IR energy emitted from the surface?

4. “With an annual surface radiation of 239W/m², the “average temperature” would be -18°C or colder.”

I would appreciate it if you would apply the logic in the last post to the derivation of this figure. Why do you ignore the fourth power here?

5. Nullius in Verba:

I have mentioned this thought-experiment before, and I know you have a policy on repetition, but I thought it would be worth going through it again.

Not a policy against interesting thoughtful stuff.

Have you drawn out your thought experiment? I just did and if I’ve understood your description correctly I think I can see where the problem lies.

At TOA we have a layer of absorbing atmosphere which, on average, receives 342W/m^2 (=1367/4), absorbs 308W/m^2, transmits 34W/m^2 and radiates the absorbed 308W/m^2 back upwards.

We don’t know the emissivity of the surface at longwave so we don’t know what the temperature will be. However if it has an emissivity of 1, the temperature will be 271K. If the emissivity was 0.5 then the temperature would be 322K.

Now, this is assuming that there is no heat transfer by conduction/convection through to the atmosphere below.

In this case the surface of the earth will be determined by the average incident solar radiation at the surface. This is 34W/m^2.

Therefore the effective surface temperature (not the “average surface temperature” which will always be equal or less) = 156K. (This assumes an emissivity of 1 at longwave).

Now assuming that there is conductivity through to the atmosphere below.

This is the bit that’s important.

How much heat is transferred from the “alien gas” through to the nitrogen atmosphere?

Nothing via radiation. The atmosphere can’t absorb any radiation. So you need to calculate the heat transmitted through the boundary layer – and of course this will lower the temperature of the “alien gas” depending the W/m^2 transmitted vertically down.

It depends on knowing a few parameters including the speed of movement of the nitrogen atmosphere over the boundary layer.

In any case, this is different in a critical way from the earth’s atmosphere.

You have an radiative absorbing layer at the top of the atmosphere which links via convection to the nitrogen atmosphere.

Or maybe I misunderstood something.

• In fact, if I have understood correctly, then you have simply demonstrated how the earth’s climate works – radiation into space from the top of atmosphere being the energy balancing item for the whole planet (ie it will equal the incoming radiation), and that top of atmosphere linked convectively to the surface.

In a non-absorbing atmosphere the only energy balancing item for the total climate’s energy is the surface radiation matching the absorbed solar radiation.

6. JAE:

Why do you ignore the fourth power here?

E = σT^4 and re-arranging-> Teff = (E/σ)^(1/4)

Therefore, Teff = (239/5.67×10^-8)^1/4 = 255K = -18’C

And “average temperature” will be -18’C or less.

What 4th power have I ignored?

7. John Phillips

On a planet with no IR absorbing gas in the atmosphere, isn’t at least a small amount of heat transferred to the atmosphere at the surface interface? In other words, doesn’t the atmosphere cool the surface, with some heat being transferred to the atmosphere, thus lowering the amount of IR energy emitted from the surface?

Good question.

You are correct.. except we have to consider what then happens and what the later equilibrium result will be.

So suppose we are in our equilibrium and the atmosphere starts to take heat away from the surface and cools it down. Therefore less energy is radiated from the surface.

The same amount of energy is still incoming from the sun. Less is now outgoing, so the climate system – as a whole – must be heating up.

This will continue until in the end the surface is back to its “equilibrium temperature”.

Of course, we can’t say for certain that the surface will reach a steady state settled temperature – it might instead oscillate around that value.

But the point is that the atmosphere can have a huge effect on the surface and yet the surface will always be tending towards the equilibrium value of temperature where radiation out = radiation in.

• The same amount of energy is still incoming from the sun. Less is now outgoing, so the climate system – as a whole – must be heating up.

This will continue until in the end the surface is back to its “equilibrium temperature”.

Of course, we can’t say for certain that the surface will reach a steady state settled temperature – it might instead oscillate around that value.

But the point is that the atmosphere can have a huge effect on the surface and yet the surface will always be tending towards the equilibrium value of temperature where radiation out = radiation in.

That sounds illogical to me.

Surely the only place where incoming and outgoing radiation will balance at equilibrium is at top of the atmosphere.

If some of the heat absorbed by the earth and oceans is conveyed away from the surface by convection then that much less will be radiated, reducing the theoretical impact of the long-wave radiation absorbed by the atmosphere.

Since these are know to be non-trivial, they must be included in any calculation. This seems to be one of the under-pinning fallacies of the AGW conjecture.

Paul

• SoD,

So suppose we are in our equilibrium and the atmosphere starts to take heat away from the surface and cools it down. Therefore less energy is radiated from the surface.

If your hoovered atmosphere takes up some heat from the surface through conduction/convection, then it will radiate some energy too, right?

Woudn´t the energy budget at TOA, then, be the thus cooled surface radiation plus this heated atmosphere radiation?

Frequencies would not be our usual longwave (from the atmosphere), but isn´t my imaginary budget here correct?

8. “What 4th power have I ignored?”

Here’s my issue: Remember your “averaging” lecture in the last post; i.e., the part about “take three temperatures….”

Similarly, does it really make sense to “average” the radiation over the whole “Earth-disk” (239 wm-2) and then use the SB equation to get an “average” temperature (-18C), when we have this fourth power relationship between temperature and radiation? What does this “average” mean, when the solar radiation at the zenith for several hours each day is somewhere around 1366 wm-2; and the radiation at the edge of the disk is, I guess, zero? That 1366 wm-2 translates to somewhere around 121 C (forgetting albedo), using the SB equation, if you don’t have an atmosphere. At the edge of the “disk,” I guess the radiation is nil and the temperature is ?? (heat storage??). Anyway, how do you calculate the “average temperature” for a system wherein the radiation decreases nonlinearly from the center of the “disk” to the edge? And when half the time there is no solar radiation (night) Is it really correct to calculate what the temperature of the Earth “should be” from the average radiation?

Basicly, how can you defend the -18 C number?

9. Take a look at The Earth’s Energy Budget – Part One and see if you still have the same questions. Feel free to ask again if you do.

10. JAE

Actually, even though that might be a helpful post I don’t think it really addresses the question.

In the major part of the article I have used total energy in and out, specifically for this reason.

If the energy radiated out from the surface of the earth is 1.2×10^17W the “average temperature” of the surface (with an emissivity of 1) can, at most, be -18’C.

If we use mythical Latvia the temperature will be a frosty sub 1K.

If we have every temperature on the earth constant and the same as every other point, the temperature will be -18’C.

I think it’s all there in the article.

11. I’m sorry, SOD, but I don’t think I got an answer. Please don’t send me off to other articles, and please be more clear. We were talking about the power of four, which you seemed to have dropped, or something.

• on June 6, 2010 at 5:19 pm | Reply Mike Blackadder

JAE,

I’ve been reading through the ‘lunar’ post and I think I understand what SOD is saying.

We know from the example of the moon that the higher the heat capacity, the more uniform temperatures are on the moon.

In all cases the moon gives off the same radiation that it receives but average surface temperature actually gets higher when the moon has higher heat capacity (I don’t think that there was any argument about that).

Where I think you might be confused is in thinking that this contradicts the calculation of -18C for Earth without greenhouse gases. Look at the numbers. When we say -18C as the average temperature of hte earth, this would be an Earth modelled with uniform surface temperature (which is probably a good approximation). If Earth’s surface was like the moon then the average surface temperature would actually be less than -18C.

12. pgtips91:

Surely the only place where incoming and outgoing radiation will balance at equilibrium is at top of the atmosphere.

On the earth’s that’s true. And that is at heart of the radiative-convective model in “modern” atmospheric physics – for the last 40 years+. See for example CO2 – An Insignificant Trace Gas? Part Five.

But the “post-Hoover” world is a different story which is why it is painted here, as an aid to understanding the role of atmospheric absorption.

In the “post-Hoover” world there is no absorption or emission of radiation from the atmosphere.

This means incoming radiation can only be balanced by outgoing radiation from the surface.

• That’s a non-sequitur.

It is well known that most of the heat from the oceans and land is conveyed to the lower atmosphere in other ways than radiation, is then conveyed to the upper atmosphere by convection, and from the upper atmosphere is conveyed to outer space via radiation.

Hence, even in the absence of radiative absorption in the atmosphere the balance of radiation takes place at the upper levels, not at the surface.

Paul

• on June 8, 2010 at 1:45 am Mike Blackadder

This is exactly what I’m trying to wrap my head around.

The atmosphere is basically transparent to light and warming of the surface is transferred to the lower atmosphere because they are in contact with one another and is not transferred through radiation.

Doesn’t the insulating properties of the atmosphere impact the surface temperature (regardless of whether or not it contains greenhouse gases)? Like when you wear a jacket in winter time.

I notice that it is colder at high altitudes than low altitudes. Is this because I’m not getting enough back radiation from greenhouse gases?

13. JAE

Please don’t send me off to other articles, and please be more clear. We were talking about the power of four, which you seemed to have dropped, or something.

– Perhaps it’s not clear because you haven’t taken the time to read it and try and understand it.

– Perhaps it’s not clear because I haven’t explained it very well.

– Perhaps it’s not clear because it is full of errors.

All of these are possible scenarios. But please don’t ask every question and make every statement as if the last one is the case, otherwise it is much more unpleasant to try and deal with your “questions” and claims.

You say:

..What does this “average” mean, when the solar radiation at the zenith for several hours each day is somewhere around 1366 wm-2; and the radiation at the edge of the disk is, I guess, zero?..

We are totalling the solar radiation into the planet. And we are totalling the radiation out of the planet. No averages are needed.

Up to the point where we have calculated the total radiated power leaving the surface of the planet do you understand the approach? And do you think anything is wrong with the approach?

Then we used this total power to calculate two easy scenarios: Uniform Temperature and Mythical Latvia.

In the first case we can easily do averages because every point on the earth is the same. And converting 235W/m^2 into 254K is the Stefan-Boltzmann 4th power relationship as laid out in detail below that section.

Do you have a problem with that?

In the second case we calculate the temperature in mythical-Latvia from the same relationship. This time the equation wasn’t written down.

“and Latvia is at a temperature of 5080K which means it radiates at 1.2 x 10^7 W/m^2”

This uses the Stefan-Boltzmann relationship like in every other case.

Do you have a problem with that?

Then finally a mathematical statement is made, but not proven – “In this case, the maximum “average” temperature can’t be higher than what is known as the “effective blackbody temperature”, Teff.”

Do you have a problem with that?

14. Alexandre:

If your hoovered atmosphere takes up some heat from the surface through conduction/convection, then it will radiate some energy too, right?

Woudn’t the energy budget at TOA, then, be the thus cooled surface radiation plus this heated atmosphere radiation?

Frequencies would not be our usual longwave (from the atmosphere), but isn´t my imaginary budget here correct?

If the atmosphere has had (longwave) absorbing gases removed then the atmosphere cannot radiate (at longwave) either.

The atmosphere from top to bottom is not radiating any energy (in longwave). Therefore any movement of heat within this atmosphere has to be convection or conduction. And these are zero at TOA into space.

• SoD,

Thanks for your response. I´ll try to articulate my question better.

If the atmosphere has had (longwave) absorbing gases removed then the atmosphere cannot radiate (at longwave) either.

Yes, this part was clear on your post. Understood.

If its temperature is not 0ºK, it must radiate, right? Even if it´s not longwave. But it has to comply with Stefan-Boltzmann law.

The hoovered atmosphere is heated by the surface to some degree. Not by radiation – only conduction/convection.

Then, my question: Wouldn´t the TOA outgoing radiation be the sum of this atmospheric radiation (at whatever wavelength – it´s not 0ºK) and surface radiation?

15. The Stefan-Boltzmann law contains the emissivity term, ε, and if that is zero then there is no radiation even if temperature is above 0K.

16. Just as a clarification, that earth would radiate at 396 W/m2 without atmospheric absorption/reflection only if earth would have a temperature of approx. 288 K and be isothermal (I mean the temperature would be 288 K at every point on earth’s surface). For realistic temperatures of earth surface, the value is probably a bit higher (again due to the 4th power relationship between energy emitted and temperature). This assumption is mentioned in the latter part of the post, but it applies to this value as well. (I need to add as a disclaimer, that I don’t know how the 15C value is found).

On a somewhat different topic, can anyone tell me how the albedo value is found. It seems a bit trickier than measuring it for all over the world and finding the average. For example, ice on the equator would reflect significantly more energy than than ice on the poles.

17. Alexandre:

Does that clear things up for you? Gases that don’t absorb don’t radiate..

• SoD,

Yes… I guess.

Let me test my own understanding: if the spectrum defined by the Planck function does not coincide with the band where the gas´emissivity is greater than zero, than there´s no emission.

Does this non-emited energy manifest itself in some other way in this case?

Anyway, thanks SoD and Cthulhu for your helping.

18. on June 6, 2010 at 6:59 pm | Reply Nullius in Verba

SoD,

“In fact, if I have understood correctly, then you have simply demonstrated how the earth’s climate works – radiation into space from the top of atmosphere being the energy balancing item for the whole planet (ie it will equal the incoming radiation), and that top of atmosphere linked convectively to the surface.”

Yes, that was more or less the idea.

19. Nullius in Verba,

Concerning your experiment – I think adiabatic lapse rate doesn’t work that way (downwards). I can’t really think of any force that would cause the heated air to descend down to the surface in your model, meaning it would have warm air at 50 km, then going downwards the temperature would drop, until we get close to surface again.

To be honest I’m not very familiar with the adiabatic lapse rate concept, but from what I’ve understood it wouldn’t work downwards (towards the gravitational center) very well.

20. “sorry Latvian’s wasn’t my idea”

Grammar trivia: the correct term for people of Latvian nationality is “Letts.” Really.

• Actually, that parenthetical phrase has much more serious problems.

There’s no comma after “sorry”, which suggests that it’s modifying the word that follows (i.e., our host is referring to some particular Latvian/Lett who is apparently either apologetic or pathetic): “See that Lett standing over there in the rain? That sorry Lett forgot to bring his umbrella.

Next, of course, the apostrophe followed by an “s” would only be to indicate either a possessive (That sorry Lett’s clothes are getting soaked) or a contraction of “is” (That sorry Lett’s going to catch pneumonia; does anyone have an umbrella we could loan him?).

Next, it’s really not clear what exactly our host is insisting was not his idea. Is there an implied subject? (Sorry, Letts, incinerating your country with solar irradiance wasn’t my idea.) Alternatively, perhaps our host was disclaiming responsibility for the existence of apologetic/pathetic Letts. (It was not my idea to have sorry Letts.

The moral of this story: you should endeavor to make your grammar as clear as your mathematics!

21. S.O.D.:

OK, I’m still confused about the fourth power issue. The “experts” say that the temperature of the Earth, sans greenhouse gases, “should be” -18 C. This is based upon simply calculating the AVERAGE RADIATION received by a disk, defined by the Earth’s diameter, exposed to the solar constant of 1366 wm-2, adjusted for the albedo of about 0.3. This comes out at -18 C.

The calculation of temperature is based on a simple average radiation figure, right? Right?

If so, I have some problems with that calculation. For one thing, it is a hemispheric number and seems to ignore night and day (it’s OK for day, but what about night?) For another–my biggest concern– it is a linear calculation and fails to account for the fact that temperature is related to the fourth power of the radiation.

I understand that the total HEAT can be calculated with the disk equation. But temperature???

Can you help?

The atmosphere is basically transparent to light and warming of the surface is transferred to the lower atmosphere because they are in contact with one another and is not transferred through radiation.

Doesn’t the insulating properties of the atmosphere impact the surface temperature (regardless of whether or not it contains greenhouse gases)? Like when you wear a jacket in winter time.

I notice that it is colder at high altitudes than low altitudes. Is this because I’m not getting enough back radiation from greenhouse gases?

I’m assuming we are talking about normal earth..

Heat is transferred from the surface to the atmosphere via radiation as well as via convection and conduction.

The atmosphere is transparent (mostly) to solar radiation but is not transparent to the terrestrial radiation (4-100um).

The main mover of heat in the lower atmosphere is convection – see Tropospheric Basics and

23. JAE,

-18C is so called effective temperature of earth, which means that it is the temperature of a blackbody which emits the amount of radiation earth is absorbing from the sun. It is more appropriate to say it uses the total value rather than an average. It is not a value that should be compared with the average temperature of the earth. Why it is used so much is beyond me as it is a pretty meaningless value in my opinion. I suppose it is just relatively easy to find and hence good for explaining the concepts without making things too complicated. I don’t like it’s usage myself as it is somewhat misleading in my opinion.

24. on June 9, 2010 at 9:45 pm | Reply DeWitt Payne

Nullus in Verba

There might be convection in your thought experiment, but it would be above the cloud layer, not below. If the surface were at 750 K, it would radiate 17,940 W/m2 compared to an energy input from radiation of 34 W/m2. It takes work to compress a gas from 250 K at whatever pressure to 750 K at a much higher pressure. Where is this work coming from? In reality, what you would have is more like the ocean where the temperature at the bottom would be colder than at the top, where all the sunlight is absorbed, and there would be little vertical convection. The adiabatic lapse rate follows from warm air being less dense and more buoyant than cold air. If the air below is sufficiently warmer than the air above, then it rises and the colder air sinks and is warmed by compression and contact with the surface. But if the air above is warmer than the air below, it stays where it is.

25. on June 11, 2010 at 8:06 pm | Reply Bill Stoltzfus

This is OT, but I was wondering if you were going to do an article on attribution at some point. Mostly I am wondering what avenues have they investigated, and is it really as simple as “we’ve emitted more than enough CO2 to account for it, so we did it”. I have read and understood about the C13/C12 ratio, but I still have this feeling that there is a gap somewhere in there that I just can’t put my finger on.

Also I was wondering if it would be helpful to anyone else (newbies like myself mostly) if you did a post about the best ways to search for scientific articles, sites that have paywalls, better and worse word choices in search strings, that kind of thing. I see people comment everywhere, “Go read the literature”, and I would, but I really don’t know where to start.

Thanks.

26. Mostly I am wondering what avenues have they investigated, and is it really as simple as “we’ve emitted more than enough CO2 to account for it, so we did it”.

The carbon cycle isn’t fully understood (at least in the sense of being fully quantified), but it’s hardly not known.

The wikipedia article seems fine:

http://en.wikipedia.org/wiki/Carbon_cycle

The real question to ask is, “why does half of the CO2 we dump in the atmosphere disappear?”. People who study the carbon cycle will tell you that most of it is disappearing in the ocean, as adding CO2 to the atmosphere increases the partial pressure of CO2, and the ocean has not yet reached equilibrium with the increasing partial pressure.

Measurements show that the PH of the upper levels of the ocean are slowly declining, which is observational evidence in support of what physics tells us to expect.

If you can put your finger on the gap you perceive, let us know.

If our dumping of 2x the amount of CO2 to account for increased CO2 isn’t the cause, then

1. where is it all going? It has to be going *somewhere*, it’s not being “poofed” out of existence.

2. where is the additional CO2 coming from, and why isn’t it simply just disappearing into the same sink that’s devouring our emitted CO2?

27. Bill Stoltzfus

Attribution is definitely a challenging subject. So far Stratospheric Cooling is the only article that I have done on this. More will be forthcoming at some stage.

On papers etc this is a good question, not easy to answer but it is worth posting something about this. Watch out for it.

28. on June 14, 2010 at 3:24 pm | Reply Bill Stoltzfus

dhogaza and scienceofdoom:

Thanks for the replies, every little bit is helpful. I did read through some of the RC posts about attribution, including an older one that William Connoly was moderator on at the end of 2004, I think (can’t find it now, of course). Responding to one of the comments, he wrote that if it isn’t human-induced, then you’d have to postulate a sudden CO2 source and a matching sudden CO2 sink of sufficient magnitude to account for the changes that happened to coincide with the timeframe of our contributions, neither one of which has been accounted for before. That rebuttal makes sense, but I don’t think they ever got into listing how they investigated all of the sources and sinks–maybe that’s the “gap” I’m having trouble with. I will keep thinking about it.

Thanks again.

29. on June 23, 2010 at 5:40 pm | Reply Bill Stoltzfus

dhogaza:

Actually, my questions about the ocean uptake are, “Why is it only about half of what we put into the atmosphere and not a larger percentage? And why has that fraction stayed stable? And if nature is absorbing about half of our emissions now, why was it not absorbing more than half 30 years ago when our total output was less?”

The sizes of the natural stores/sinks in GtC are huge: atmosphere ~750, land surface ~1,500, and upper ocean ~2,000 (with the potential for closer to 40,000 in the deep ocean). With the sizes of the natural sinks being so large and with them having obviously rounded values, my thoughts always tended toward, “how much could the measurements be off, and what does that mean in relation to our rather small contribution of about 8GtC?” But I realized this morning that since they all have to be in equilibrium, the size of the sinks doesn’t matter at all—even the amounts being transferred between them don’t matter—if you’re adding an unbalanced amount into one place the equilibrium will have to shift, regardless of the amount. It makes sense that natural systems can only absorb so much each year in their progress towards a new equilibrium, but is there any data regarding how much extra “storage” per year we think is out there? Will the 50% figure continue to hold, and if so, for how much longer? If we decrease the amount we’re putting into the atmosphere at some future point, will nature “catch up”, or will the uptake amount decrease and the percentage stay around 50?

In relation to my perceived “carbon attribution gap” that I am still having trouble verbalizing, perhaps it’s just that I see the uncertainty in one part (we don’t really know why only half is taken up, and are unsure exactly which sinks are responsible and how, etc.) and project that uncertainty onto the rest of it, whether that is valid or not.

And this may sound stupid, but how much carbon is locked up in plastics? They don’t biodegrade, so it’s a permanent sink, and we’re making more all the time. Or does that matter at all since we made them all out of fossil fuels?

Okay, that was too many questions—sorry. As always, thanks in advance for any answers.

30. Mike Blackadder: “Like when you wear a jacket in winter time.”

Do you not think this would be quite a stunning difference?

31. Alexandre “”Therefore, how can it radiate?”

If its temperature is not 0ºK, it must radiate, right? ”

But how can it radiate if it can’t create an IR photon? The process for photon creation is the time-reversed (time symmetric) process for photon absorbtion. So if it can’t absorb an IR photon, how can that when time-reversed, cause an emission of a photon?

32. PS Alexandre, temperature is the amount of kinetic and internal molecular energy of a gas. It doesn’t have anything to do with radiation. Therefore an ideal gas will as long as it is moving about a temperature greater than 0K.

33. Bill: “Why is it only about half of what we put into the atmosphere and not a larger percentage? ”

Me: Why do you think it would be larger?

100% is out because that would increase the CO2 concentration of the ocean beyond equilibrium and that density gradient would make the ocean emit CO2.

It couldn’t be 0% else the atmosphere would be out of equilibrum and the concentration of the ocean would cause CO2 to diffuse from the atmosphere into the ocean.

Since half is what we see, why must it be different?

34. PS Bill: “And if nature is absorbing about half of our emissions now, why was it not absorbing more than half 30 years ago when our total output was less?”

Because that would then have someone asking “why is nature taking half our emissions now when it was taking more than half 30 years ago”.

This half-and-half is merely showing that a CO2 molecule is as likely to go into the ocean from the air as go from the air into the ocean.

All this really says is that our output of CO2 is bringing ocean and air equally out of equilibrium.

35. on July 9, 2010 at 6:48 pm | Reply Bill Stoltzfus

Mark:
Thanks for the replies. I was thinking that the oceans (or whatever the sink is that’s taking half our emissions) should have been able to absorb a greater fraction 30 years ago because 30 years ago the ocean as a whole was colder, and colder water can absorb more CO2 than warmer water. So it doesn’t make entire sense to me that the ocean is the only CO2 sink that we should be talking about.

But I would say that still leaves the puzzle over why the absorbed fraction is relatively constant, while the amount absorbed has increased.

At some point I hope to go from asking questions all of the time to being able to answer them some of the time, but in the meantime, I am grateful for the help I am getting here and elsewhere.

• on July 10, 2010 at 3:40 pm | Reply UncertaintyRunAmok

Bill,

There is no mystery here for someone who understands the underlying processes and what the measurements actually mean. My field is spectroscopy (analytical chemistry to laypeople), and I can explain exacty where this discrepancy begins. It is traditionally described as comparing apples and oranges.

The ratio of the carbon to oxygen in a CO2 molecule is ~3/8 based on atomic weight. When we analyze a sample (normally a very small fraction of the material being tested), we quote the results as percent concentration per volume or mass. The total mass of the material being analyzed is irrelevant.
The ratio of the carbon to oxygen in a CO2 molecule is ~3/8 based on atomic weight. And this is where simply doing calculations without thinking about the underlying processes leads to confusion. Combustion takes O2 nearly 100% from the atmosphere, and combines it, in the case of hydrocarbons, with H and C, primarily releasing H2O and CO2. Now think; if you take 3 tons of C from the ground, and add 8 tons of O from the atmosphere, what is the weight which has actually been added to the atmosphere? Right, only 3 tons of C have been added to the atmosphere. The O2 was already there. When we do the chemical analysis, the atmospheric O2 has also been reduced by the amount of O2 which has been bonded to the C atoms(as well as the amount of O2 bonded to H, but that is another discussion). No, we will not run out of O2. The question always seems to pop up.
So, if you compare the results of the chemical analysis with the calculations based on CO2 emission versus total (calculated) atmospheric mass, you will get two totally different answers, especially since they never seem to adjust the calculated mass of the atmosphere upwards by the equivalent of the amount of CO2 emitted. Go figure.

36. on July 10, 2010 at 8:04 pm | Reply UncertaintyRunAmok

Oops! I obviously clicked on copy when I intended to click on cut.
Anyway, the point is, there is no “missing” CO2.

37. […] The Hoover Incident I explained what would happen if the atmosphere didn’t absorb or emit radiation – i.e., […]

38. […] or emit radiation the surface of the earth would radiate at an average of around 240 W/m² (see The Hoover Incident, CO2 – An Insignificant Trace Gas? and many other articles on this […]

39. […] If the atmosphere had no radiative absorbers (no “greenhouse” effect) then F= σT4, which means G=0. See The Hoover Incident. […]

40. […] the atmosphere the surface radiation would not change on its journey to the top of atmosphere. See The Hoover Incident for more on this and the […]

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