Many blogs write about over-simplifications of the radiative effects in climate. Many of these blog articles review simple explanations of how it is possible for atmospheric radiative effects to increase the surface temperature – e.g. the “blackbody shell” model.
As a result many people are confused and imagine that climate science hasn’t got past “first base” with how radiation interacts with atmospheric gases.
In any field the “over-simplified analysis” is designed to help the beginner to gain conceptual understanding of the field. Not to present the complete field of scientific endeavor.
This article will try to “bridge the gap” between the over-simplified models and the very detailed theory.
Note – it isn’t possible to cover the whole subject in one blog article and a decent treatment of radiative transfer consumes many chapters of a textbook.
There will be some maths. But I will also try to provide a non-mathematical explanation of “the maths” – or “the process”.
If you find maths daunting or incomprehensible that is understandable, but there is a lot that can be learned by trying to grasp some of the basic concepts.
This means we need to treat each wavelength separately. Why? Because absorption and emission is a wavelength dependent process.
For example, here is the absorption spectrum of one part of NO2:
So when we consider radiation “zooming” through the atmosphere we have to take it “one wavelength” at a time.
There isn’t really any such thing as monochromatic radiation or being able to take “one wavelength at a time” – but that doesn’t stop us analyzing the problem..
A Digression on “Calculus”
How does the world of science and engineering deal with continuous change?
If a force, or a ray of radiation, or a movement of the atmosphere is a continuously changing value, how do we define it? How do we deal with it?
Calculus is the answer. This branch of mathematics allows us to deal with small changes and continuous changes and provide theorems, answers and equations.
For example, if we know something about the variable distance, s, with respect to time, t, then an equation defines the relationship between velocity, v, and these other variables:
v = ds/dt
where the “d”s at the beginning means: “the rate of change of”, so the formula means – in English:
Velocity = the rate of change of Distance with respect to Time
Generally when you see something like “da/db” it means “the rate of change of variable a with respect to variable b“.
It is also common to see Δx and δx – meaning “a small change in x”. This is different from “the continuous change of x”, but the specific rationale behind when we use “dx” and “Δx” isn’t so important for this article.
The other important area of calculus is “summing” results when again there is continuous change. If someone travels at 10 km/hr for 1 hour and then at 20km /hr for 1 hr they will have traveled 10km + 20km = 30km. That’s an easy calculation. But if velocity has continuously changed with time – how do we calculate the total distance traveled?
This means, in (harder to understand) English:
Distance = the integral of Velocity with respect to Time, between the limits of time = t1 and time = t2.
The integral is like the summation of each of the tiny distances covered in each very small time period (between t1 and t2). The integral is also often referred to as “the area under the curve”.
..end of digression
Absorption of Radiation
Let’s define a monochromatic beam of radiation, Iλ, travelling through the atmosphere:
We have some information we can use:
The rate of absorption of the beam of radiation as it travels through the atmosphere is proportional to the amount of absorbers at that wavelength and the ability of that absorber to absorb radiation of that wavelength
This is known as the Beer-Lambert law, and is written like this (note 1):
dIλ = -nσIλ .ds 
which means the same thing in mathematical terms, with n=number of absorbing molecules per unit volume (corrected), and σ=capture cross-section (or effectiveness at absorbing at that wavelength), and the subscript λ indicates that this equation is only true for the radiation at this wavelength
The value σ is a material property and so constant for one gas at one wavelength at one temperature and one pressure, but varies with the temperature and pressure of the gas (see comment). The value n will depend on location in the atmosphere. If we solve this equation between two arbitrary points, s1 and s2, we get:
Iλ(s2) = Iλ(s1). exp [ -∫σn(s).ds ] 
where the integral is between the limits s1 and s2
What it means in English:
The intensity of radiation at wavelength λ is reduced as it travels through the atmosphere according to the total amount of the absorber along the path. “exp” is e, or 2.718, to the power of the value in the square brackets.
If the concentration of the gas doesn’t change along the path the equation becomes a simpler version:
Iλ(s2) = Iλ(s1). exp [ -σn.(s2 - s1) ] 
Optical Thickness & Transmittance
These are some important properties to understand.
Optical thickness, usually written as τ, is the property inside the exponential in equation .
τ = ∫σn(s).ds 
where the integral is between the limits s1 and s2
Transmittance, usually written with a weird T symbol not available in WordPress, but with “t” here, is the amount of radiation “getting through” along the path we are interested in.
t(s1,s2) = exp [-τ(s1,s2)] 
also written as:
t(s1,s2) = e-τ(s1,s2)
The optical thickness is “dimensionless” as is the transmittance.
So we can rewrite equation  as:
Iλ(s2) = Iλ(s1).t(s1,s2) 
The transmittance can be a minimum of zero – although it can never actually get to zero – and a maximum of 1. So it is simply the proportion of radiation at that wavelength which emerges through the section of atmosphere in question:
With optical thickness, τ = 1, transmittance, t = 0.37 – which means that 63% of the radiation is absorbed along the path and 37% is transmitted.
With optical thickness, τ = 10, t=4.5×10-5 – that is, 45 ppm will be transmitted through the path.
A note on definitions – optical thickness is usually defined as = 0 at the top of atmosphere, where z is a maximum and a maximum at the surface, where z = 0:
So τ increases while z decreases, and z increases while τ decreases.
In the absence of scattering (note 2), absorptance, a = 1 -t.
That is, whatever doesn’t get transmitted gets absorbed.
Plane Parallel Assumption
If you refer back to Figure 2, you see that the radiation is not travelling vertically upwards, but at an angle θ to the vertical.
Using simple trigonometry, ds = dz / cos θ 
It’s always an advantage if we can simplify a problem and relating everything to only the vertical height through the atmosphere helps to solve the equations.
Atmospheric properties vary much more in the vertical direction than in the horizontal direction. For example, go up 10 km and the pressure drops by a factor of 5 – from 1000 mbar to 200 mbar. But travel 10 km horizontally and the pressure will have changed by less than 1 mb. Temperature typically changes 100 times faster in the vertical direction than in the horizontal direction.
And as air density is determined by pressure and absolute temperature we can make a reasonable assumption that the density at a given height, z directly above is the same as the density at the same height when we look at an angle of 45°.
Of course, by the time we are considering an angle close to 90° – i.e., horizontal – the assumption is likely to be invalid. However, the transmissivity of the atmosphere at angles very close to the horizon is extremely low anyway, as we will see.
Therefore, making the assumption of a plane parallel atmosphere is a good approximation.
Let’s review the earlier equations using a mathematical identity that reduces “equation clutter”:
μ = cos θ 
And rewrite equation :
t(z1,z2) = exp [-τ(z1,z2)/μ] 
Notice that the equations are now rewritten in terms of the optical thickness between two vertical heights and the angle of the radiation.
It might help to see it in graphical form – and note here that the optical thickness, τ, is for the vertical direction (otherwise the graph would make no sense):
This simply demonstrates that as the angle increases the radiation has to travel through more atmosphere.
So suppose the optical thickness vertically through the atmosphere, τ = 1, then for:
- a vertically travelling ray the transmittance = 0.37
- for a ray at 45° the transmittance = 0.24
- for a ray at 70° the transmittance = 0.05
- for a ray at 80° the transmittance = 0.003
Using Kirchhoff’s law, absorptivity of a material = emissivity of a material for the same wavelength and direction. For diffuse surfaces – and for gases – direction does not affect these material properties, so they are only a function of wavelength. (And for all intents and purposes, absorptance is the same term as absorptivity, and transmittance is the same term as transmissivity-see comment).
Emission of radiation at any given wavelength for a blackbody (a perfect emitter) is given by Planck’s law, which is usually annotated as Bλ(T), where T = temperature.
The absorptivity of a gas, aλ = 1-tλ =emissivity of a gas, ελ. (Corrected)
For a very small change in monochromatic radiation due to emission:
dIλ = ελBλ(T) .ds [10a]
dIλ = nσBλ(T) .ds [10b]
So if now combine emission and absorption, equations 1 & 10:
dI/ds = nσ.(Bλ(T) – Iλ) 
If we combine this with our definition of optical thickness, from equation :
dIλ/dτ = Iλ – Bλ(T) 
which is also known as Schwarzschild’s Equation – and is the fundamental description of changes in radiation as it passes through an absorbing (and non-scattering) atmosphere.
It says, in not so easy to understand English:
The change in monochromatic radiation with respect to optical density is equal to the difference between the intensity of the radiation and the Planck (blackbody) function at the atmospheric temperature
Sorry it’s not clearer in English.
In more vernacular and less precise terms:
As radiation travels through the atmosphere, the intensity increases if the Planck blackbody emission is greater than the incoming radiation and reduces if the Planck blackbody emission is less than the incoming radiation
Solving Schwarzschild’s Equation
Notice that this important equation contains the Planck term, which is for blackbody radiation (i.e., radiation from a perfect emitter), yet the atmosphere is not a perfect emitter. We definitely haven’t assumed that the atmosphere is a blackbody and yet the Planck terms appears in the equation. It’s just how the equation “pans out”..
I mention this only because so many people have come to believe that there is some “big blackbody assumption” in climate science and they might be concerned by this term. Nothing to worry about, this has not been assumed.
Let’s find a solution to the equation if we are measuring the TOA (top of atmosphere) radiation. Refer to Figure 4:
- at TOA, z=zm and τ=0
- at the surface, z=0 and τ = τm
Now some maths manipulation – skip to the end people who don’t like maths..
First we note a handy party piece:
d/dτ [Ie-τ] = e-τ. dI/dτ – Ie-τ 
Now we multiply both sides of equation  by e-τ:
e-τ.dIλ/dτ = e-τ.Iλ – e-τ.Bλ(T) 
e-τ.dIλ/dτ – e-τ.Iλ = – e-τ.Bλ(T) [14a]
And substituting “handy party piece”  into [14a]:
d/dτ [Iλe-τ] = – e-τ.Bλ(T) 
Now we integrate  between τ=0 and τ=τm:
Iλ(τm)e-τm – Iλ(0) = – ∫ Bλ(T)e-τ dτ 
where the integral is between the limits of 0 and τm
And re-arranging we get:
..end of maths manipulation
Iλ(0) = Iλ(τm)e-τm + ∫ Bλ(T)e-τ dτ 
Which – for those who haven’t followed the intense maths:
Iλ(0) = Iλ(τm)e-τm + ∫ Bλ(T)e-τ dτ 
The intensity at the top of atmosphere equals..
The surface radiation attenuated by the transmittance of the atmosphere, plus..
The sum of all the contributions of atmospheric radiation – each contribution attenuated by the transmittance from that location to the top of atmosphere
Don’t worry about the maths, but it is definitely worth spending some time thinking about the words in colors – to get a conceptual understanding of how atmospheric radiation “works”.
It’s Not Over Yet – Conversion from Intensity to Flux and the Diffusivity Approximation
Remember the note about the Plane Parallel Assumption ?
Getting equations into WordPress is painful and time-consuming so a quick explanation followed by the result, especially as maths fatigue will have set in among most readers, if any made it this far.
Equation  is for spectral intensity. That is, one direction rather than the complete hemispherical power (flux).
To calculate spectral emissive power (flux per unit wavelength) we need to integrate the equation over one hemisphere of solid angle. We re-write equation  in the form of equation  so that the optical thickness references vertical height, z and μ, which is the cosine of the angle from the vertical. Then we integrate from μ=0 (θ=0°) to μ=1 (θ=90°).
Transmittance, t(z,0) = 2 ∫ e-τ(z,0)/μ μ.dμ
where the integral is from 0 to 1
This equation doesn’t have an “analytical” solution, meaning we can’t rewrite it in a nice equation form without the integral. But with some clever maths that I haven’t tried to follow – but have checked – we can produce an approximation which is known as the diffusivity approximation:
2 ∫ e-τ(z,0)/μ μ.dμ ≈ e-τ/μ’
Where μ’ is the “effective angle” which produces a close approximation to the actual answer without needing to integrate across all angles (for each wavelength). The best value of μ’ = 0.6.
Here is the calculated integral (left side of the equation) vs the approximation with μ’ = 0.6, as a function of optical thickness, τ, demonstrating the usefulness of the approximation:
There are some other refinements needed, for example, the reflection of atmospheric radiation for a surface emissivity < 1, which is then attenuated by the absorptance of the atmosphere before contributing the TOA measurement. But these factors can all be introduced into the equations.
Full Color Solution
What we have produced so far is a solution for each monochromatic wavelength, λ.
Also, we haven’t explicitly stated the fact that the optical thickness depends on the concentration and “capture cross section” of each absorber for that wavelength. So the optical thickness, and transmittance, for each height requires combining the effects of each active molecule.
The solution for flux, W/m², requires integrating the equations across all wavelengths.
Wow. Conceptually straightforward. But computationally very expensive – check out Figure 1 – the absorption characteristics of each radiatively-active gas vary significantly with wavelength.
The equation for radiative transfer is commonly known, (in differential form) as Schwarzschild’s Equation. It relies on fundamental physics.
To solve the equation requires some maths.
To solve the equation in practical terms the plane parallel assumption is used. This relies on the fact that variations in temperature and pressure (and therefore density) are negligible in the horizontal direction compared with the vertical direction.
The equation could be solved without this plane parallel assumption, but the variations horizontally in pressure and temperature are so slight that the same result would be obtained, unless extremely high quality data on temperature, pressure, density and concentration of absorbers was available.
To solve the equation in practical terms we need to know:
- the temperature (vs height) in the atmosphere
- the concentration of each absorber vs height
- the absorption characteristics of each absorber vs wavelength
In any practical field, the “proof of the pudding is in the eating”, and so take a look at Theory and Experiment – Atmospheric Radiation – where theoretical and practical results are compared.
And lastly, the Stefan-Boltzmann equation, correct and accurate though it is (check out Planck, Stefan-Boltzmann, Kirchhoff and LTE) is not used in the actual equations of radiative transfer in the atmosphere. Nor is any assumption of “unrealistic blackbodies”.
I only note these last points due to the high quantity (but not high quality), of blog articles and comments demonstrating the writers haven’t actually read a textbook on the subject, but still feel qualified to pass judgement on this field of scientific endeavor.
Part One – a bit of a re-introduction to the subject
Part Two – introducing a simple model, with molecules pH2O and pCO2 to demonstrate some basic effects in the atmosphere. This part – absorption only
Part Three – the simple model extended to emission and absorption, showing what a difference an emitting atmosphere makes. Also very easy to see that the “IPCC logarithmic graph” is not at odds with the Beer-Lambert law.
Part Four – the effect of changing lapse rates (atmospheric temperature profile) and of overlapping the pH2O and pCO2 bands. Why surface radiation is not a mirror image of top of atmosphere radiation.
Part Five – a bit of a wrap up so far as well as an explanation of how the stratospheric temperature profile can affect “saturation”
Part Seven – changing the shape of the pCO2 band to see how it affects “saturation” – the wings of the band pick up the slack, in a manner of speaking
And Also -
Theory and Experiment – Atmospheric Radiation – real values of total flux and spectra compared with the theory.
Note 1: There are many formulations of the Beer-Lambert law and even much dispute about who exactly the law should be attributed to.
Other formulations include using the density of the gas and a matching coefficient for the effectiveness of the gas at absorbing.
Note 2: When considering solar radiation (shortwave), scattering is important. When considering terrestrial radiation (longwave), scattering can be neglected. In this article, we will ignore scattering, so the results will be appropriate for longwave but not correct for shortwave.