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## Simple Atmospheric Models – Part Two

In Part One we saw how a very simple energy balance model with some very basic assumptions provided some insight into how the surface and atmospheric temperatures are determined.

We assumed that the atmosphere was transparent to solar radiation (radiation less than 4 μm), that the atmosphere was totally opaque to terrestrial radiation (greater than 4 μm), and that the atmosphere was isothermal (all at one temperature).

All of these assumptions are incorrect.

We derived an average surface temperature of 303K – instead of the realistic 288K – from this simple energy balance model.

In this article we will look at a very basic model to estimate the temperature of the stratosphere. If you aren’t clear about the troposphere/stratosphere, take a look at Tropospheric Basics. For caveats and explanations about simple models, and about radiation and emissivity, please check Part One.

In the troposphere, heat transfer is dominated by convection. The atmosphere transitions to what we call the stratosphere when convection ceases because radiation becomes more effective for moving energy. The atmosphere progressively thins out the higher we go – and as a necessary consequence it becomes optically thinner (note 1). This also implies that the temperature in the stratosphere will not vary significantly with height because radiation can transfer energy across large distances.

In practice, absorption of solar radiation by ozone means that the stratospheric temperature increases with height. However, in this simple model we want to make huge assumptions just to see what results we get, and in this model we continue to assume (incorrectly) that the atmosphere is totally transparent to solar radiation.

Here is an extract from Elementary Climate Physics by Prof. F.W. Taylor.

It is another very simple model: Figure 1

Note there there is a correction (in red) to the diagram (his fig 3.5). Even Professors of Physics make mistakes (or their editors do).

If you followed the method of calculation in Part One then this won’t be too difficult.

We know that the earth/troposphere emits to space at an “effective radiating temperature” of 255K. This term effective radiating temperature causes a lot of confusion. It is convenient shorthand for the temperature at which a blackbody would be to radiate that flux. It doesn’t mean that blackbodies exist – perish the thought. And it doesn’t mean that anyone is assuming that the atmosphere is radiating as a blackbody.

What it means is that the earth/atmosphere emits 239 W/m² to space. We call that an “effective radiating temperature” of 255K. It’s not meant to upset anyone. If we wanted we could just call it 239 W/m². We will do that later just to see the effect.

And following Part One, with the assumption of an optically thick surface/troposphere (read about “optically thick” in that article), we have the emission of surface radiation:

E = σTE4

where TE is the average emitting temperature of the surface/troposphere (as if it was a blackbody)

And if we make some assumptions about the optical properties of the stratosphere we might find some approximate answers that are interesting.

We assume (incorrectly) that the stratosphere is an isothermal layer at temperature, Ts. We assume that the stratosphere is optically thin so that the emission from the surface/troposphere is approximately equal to the emission to space.

The stratosphere has an emissivity, ε, which is very small.

From Kirchhoff’s law, emissivity = absorptivity for the same wavelength ranges. We will come back to review this assumption a little later, but for now note that if the surface/troposphere and the stratosphere are at similar temperatures then this assumption:

absorptivity of stratosphere for surface/troposphere radiation = emissivity of stratosphere

– is approximately correct.

Therefore, from energy balance considerations, the energy absorbed by the stratosphere must be the energy emitted by the stratosphere. So, from figure 1:

εσTE4 = 2εσTs4

therefore:

Ts = TE / 21/4 = 215 K

Now, surprisingly enough for such simple assumptions, this is a reasonable value for the temperature at the bottom of the stratosphere.

Let’s redo the calculations – this time using the actual flux to space from the surface/troposphere instead of the “effective radiating temperature”.

The surface/troposphere radiates (globally annually averaged) 239 W/m². The stratosphere absorbs a small fraction of this, determined by absorptivity = emissivity = ε:

ε x 239 = 2εσTs4

therefore:

Ts = (239/σ)1/4 / 21/4 = 215 K

The value is the same as previously calculated, no surprise to anyone who has got to grips with this subject.

### Using Kirchhoff

Earlier we used the fact that absorptivity = emissivity for the stratosphere, because of Kirchhoff’s law.

It is very important to understand how to use this law correctly. An excellent example of how not to use it was done by Martin Herztberg in his paper, Earth’s Radiative Equilibrium in the Solar Irradiance, Energy & Environment (2009).

Let me paint a picture with some graphs.

First, the wavelength dependence of blackbody radiation for three different temperatures. Remember that blackbody radiation is just a “perfect emitter”, or the “gold standard” of radiation. Real bodies cannot radiate at a higher intensity at any wavelength, although many come close.

The solar radiation has been normalized to the value at the earth’s surface (because it’s a long way from the sun to the earth – check out The Sun and Max Planck Agree – Part Two): Figure 2

Now, the transmittance for the atmosphere as a function of wavelength. For 4μm and longer wavelengths, absorptance = 1 – transmittance. For shorter wavelengths, especially below 0.5 μm, scattering becomes significant, which means that absorptance = 1 – transmittance – reflectance.

In any case, what should be clear is that absorptance is a strong function of wavelength. See the red marked graph at the bottom:

Figure 3

At any given wavelength, absorptance = emissivity.

So let’s consider the case of a 255 K atmosphere absorbing solar radiation.

The solar radiation is the blue curve in figure 2 – so to calculate how the atmosphere/surface absorbs this radiation we can use the absorptivity (≈ 1 – transmissivity) between 0-5 μm.

The surface/troposphere radiates according to the green curve. We need to multiply this curve by the emissivity = absorptivity at these wavelengths.

So although absorptivity (at a given wavelength) = emissivity (at a given wavelength), it isn’t much use if the source of the incident radiation is at very different wavelengths from the emission. Which is why Martin Herztberg hasn’t passed the competency test in this field. A “rookie mistake”.

You should be able to see from figure 2 that although the surface/troposphere and stratosphere are at different temperatures, an average absorptivity value for absorbing radiation at 255 K will be quite similar to the emissivity value for emitting radiation at 215K.

If we wanted accurate results we would need to use the absorptivity at the relevant wavelength and the emissivity at the relevant wavelength. We also would not assume that the stratosphere was isothermal, or that it was perfectly transparent to solar radiation.

By the way, here are some temperature measurements of the stratosphere in one location: Figure 4

### Conclusion

Calculating the temperature of the stratosphere is a difficult problem. However, as in many fields of scientific endeavor, we can make a very simple model and see how the results compare with reality (and the results of more complex models).

In the example here we make some very simple assumptions and find a result for the stratospheric temperature which is not too far off the mark. However, the only reason for producing (reproducing) this model was to help newcomers to the field gain a conceptual feel for the basics of energy balance and radiative transfer.

By the way, as explained in Part One, no one in climate science believes that the assumptions for this model are correct. Everyone knows that the atmosphere is not a blackbody (perfectly opaque) for all wavelengths greater than 4 μm, and is not perfectly transparent for all wavelengths less than 4 μm.

Notes

Note 1: Optical thickness is proportional to the number of absorbers (molecules that absorb radiation) in the path. So as the atmosphere thins out the density reduces and, therefore, the optical thickness must also reduce. You can read more about the equations of optical thickness in Understanding Atmospheric Radiation and the “Greenhouse” Effect – Part Six – The Equations

### 18 Responses

1. on April 7, 2011 at 4:13 pm | Reply DeWitt Payne

Convection stops at the tropopause because a temperature inversion starts there. When the air above is warmer (or even at the same temperature) and therefore less dense than the air below, you don’t get convection. Temperature inversions over the Los Angeles Basin were why LA had smog problems long before anyone else. You could easily see the inversion boundary when flying into LA on a bad day.

The structure of the atmosphere would be very different if oxygen weren’t present to absorb UV radiation, form ozone and absorb even more UV radiation causing the stratosphere to warm with altitude.

2. on April 7, 2011 at 7:25 pm | Reply scienceofdoom

Taylor also provides examples from Mars & Venus, with very little ozone: alt=””>

3. on April 8, 2011 at 5:18 pm | Reply DeWitt Payne

SoD,

You’re way past LTE conditions at 40 km in the martian atmosphere and the cloud layer and geostrophic winds are going to be important above 60 km in the cytherean atmosphere so not really comparable examples.

4. on April 11, 2011 at 9:35 pm | Reply macha

Falsi cation Of The Atmospheric CO2 Greenhouse Eff ects
Within The Frame Of Physics
Version 4.0 (January 6, 2009)

Gerhard Gerlich
Institut fur Mathematische Physik
Technische Universitat Carolo-Wilhelmina zu Braunschweig
Mendelssohnstrae 3
D-38106 Braunschweig
Federal Republic of Germany

• on April 11, 2011 at 9:59 pm | Reply scienceofdoom

5. on April 13, 2011 at 3:31 pm | Reply mkelly

εσTE4 = 2εσTs4

This looks like the radiation shield relationship without the third member.

6. on April 17, 2011 at 12:39 pm | Reply Edim

Hello SoD,

I have one question regarding heat budget at the top of the atmosphere, sorry if it’s OT.

What is the direction of the annually averaged net heat flux at TOA, inwards or outwards?

7. on April 18, 2011 at 7:30 am | Reply scienceofdoom

Edim:

The measurement of net TOA flux is zero within instrument uncertainty.

Taking the global average of incoming absorbed solar radiation over a 5 year period from 2000-2004 gives absorbed = 241 W/m2, and emitted = 240 W/m2.

This means the net flux is incoming.

This data is from Toward Optimal Closure of the Earth’s Top-of-Atmosphere Radiation Budget, Loeb et al, Journal of Climate (2009).

Loeb uses CERES data.

Different studies find slightly different results. The main problem is that long term drift is excellent but absolute known accuracy is not quite as good. And we are taking the difference between two different measurements rather than the difference between the same measurement at different times.

8. on April 18, 2011 at 11:07 am | Reply Edim

Sod,

I think the net flux (averaged) should be outwards (outgoing) and equal to the heat flux from the earth interior.

If we define our system (pre-anthropogenic) with the boundaries:

outside – TOA
inside – earth and oceanic crust, ~10 m under the surface

and neglect any heat generation in the system and assume dU = 0

Then these two heat fluxes should be equal. The heat from the earth interior must be dissipated, otherwise the internal energy (U) must increase.

9. on April 18, 2011 at 12:32 pm | Reply Edim

So, the (physical) system is:

atmosphere, oceans and upper ~10 m of earth crust (including oceanic).

10. on April 18, 2011 at 12:42 pm | Reply scienceofdoom

Global average heat flux from the interior is tiny. Less than 0.1 W/m2.

11. on April 18, 2011 at 2:23 pm | Reply Edim

Tiny is relative.

My point is not about the magnitude of geothermal flux vector but rather about its direction.

Its direction is outwards. Assuming dU of the system, as I defined it, is zero, the heat entering the system at the inner boundary (crust, ~10 m under surface), must leave the system at the outer boundary (TOA).

Therefore, the net flux at TOA is outwards (outgoing) and is equal to the geothermal flux (averaged, dU = 0).

Therefore, the system (atmoshere, oceans and upper 10 m of crust) receives heat from earth interior and dissipates it over TOA.

At TOA, the system is losing energy (thermal radiation), despite the insolation!

So back to tiny – it is only tiny because of insolation, without the sun geothermal flux wouldn’t be as tiny.

But tiny or not, it is positive (for my system), contrary to the net heat flux at TOA, which is negative!

We are being heated from the bottom and are losing heat at the top.

12. […] Originally Posted by Ultan Murphy Thanks Mr Joy. If the average member of the public only knew or learned for themselves that CO2 represents just 3.75% of Greenhouses Gases that would be job done for me. Yes you read correctly, 3.75%. That is nothing new or subversive of the physics of the Greenhouse Effect. It is well known that many substances have effects that are out of proportion to raw quanity. Given your philosophy, I presume that you would have no objection to ingesting trace elements of strychnine or arsenic? 2mg per kg of strychnine will kill you. I presume that you also understand that the effect of CO2 is a function of how it retains infra-red radiation in the atmosphere as a result of its molecular structure. In simple terms, a single molecule can absorb and emit many infra-red photons, and the more molecules the more photons absorbed and emitted. Can you explain why the earth has its current average temperature, and why even a —–ist [moderator’s note, please read the etiquette] scientist detects infra red radiation from the sky at night? There are many simple models about to help understanding (Note I am not saying the theory is based on these models, but thay are useful to help explanation). See the diagram on the right. Simple Atmospheric Models […]

13. on April 19, 2011 at 7:58 am | Reply scienceofdoom

Edim:

I don’t understand what you are trying to say.
If the earth is in approximate energy balance then the energy in = energy out.

So solar absorbed radiation + geothermal heat = energy radiated from TOA

241 + 0.1 = 240.1

However, the earth may not be complete energy balance and may be gaining or losing energy. In any given month or any given year it is likely to be gaining or losing energy.

If the earth is heating up – as the figures from Loeb et al indicate that it may have been from 2000-2005:

241 + 0.1 -240 = 1.1 W/m2 gained

And in fact the geothermal energy is negligible.

If there was no sun then the geothermal energy would be making a significant contribution to the few degrees Kelvin experienced on the surface of the earth.

We are not being heated at the bottom and losing heat at the top.

We are being heated at the top (99.95% or greater) and losing heat at the top.

14. on April 19, 2011 at 8:47 am | Reply Edim

Sod,

Maybe I am not being clear, sorry.

The system has only two boundaries (inner and outer) and only two NET fluxes:

Inner boundary – geothermal flux (positive – heat is being added to the system continuously),

Outer boundary – net heat flux at TOA (insolation minus earth thermal radiation). This flux is sofar unknown.

Now if we assume dU = 0 (internal energy of the system is constant), the heat entering at the inner boundary must leave the system at the outer boundary (TOA), otherwise U must increase. Therefore, over timescales with dU = 0, averaged net heat flux at TOA must be equal in magnitude to the geothermal flux (but negative).

dU = Q(geothermal) – Q(TOAnet) = 0

• on April 19, 2011 at 2:07 pm | Reply DeWitt Payne

The problem is that we know for a fact that dU ≠ 0. We can measure ocean heat content. It’s been going up. It’s not going up very fast right now but the trend is still up.

1955-2002 (pre-Argo) data as graphed by Bob Tisdale I doubt that even if you average over millions of years you would find that dU = 0.

Tiny is indeed relative. Teff for 240 W/m&sup2; is 255.0683 K. For 240.1 W/m&sup2; it’s 255.0952 for a difference of 0.0269 K or 0.01%. That’s tiny in my book and far less than the precision of our current data.

• on April 19, 2011 at 3:46 pm Edim

Hello DeWitt Payne,

Thanks for the answer. I think the fact that we know that U is variable (measured) is somewhat beside the point. Let me try to explain.

Actually, the first sentence of the OP:

“…how a very simple energy balance model with some very basic assumptions provided some insight into how the surface and atmospheric temperatures are determined.”

reminded me of the system as I defined it, except that I included oceans and upper ~10 m of the earth crust. The system is, let’s say, pre-AGW or better pre-anthropogenic.

Internal energy of the system can go up or down slightly (or more when we have natural changes like glacials/interglacials), my point is that there is continuous geothermal flux from the earth interior, which must mean that heat is being added to the system (positive heat flux at the inner boundary, however tiny).

Therefore, at the outer boundary (TOA), to keep the internal energy somewhat constant (not changing too much and not increasing/decreasing continuously), the net flux must be negative (averaged) – the added heat must be lost at TOA.

So, the net flux at TOA is very tiny, negative (earth radiation is greater than insolation) and “almost” equal to the geothermal flux – the difference is dU.

dU = Qgt + Qtoanet = Qgt + Qsol – Qearth-rad

• on April 19, 2011 at 8:32 pm | Reply DeWitt Payne

I understand all that. What I don’t understand is the point you’re trying to make. Yes the surface temperature will be very slightly higher than if there weren’t radioactive decay in the core of the planet. But the difference is so small it can be ignored. Other than that, what’s your point?