In the three part series on DLR (also known as “back radiation”, also known as atmospheric radiation), Part One looked at the network of stations that measured DLR and some of the measurements, Part Two reviewed the spectra of this radiation, and Part Three asked whether this radiation changed the temperature of the surface.
Very recently, on another blog, someone asked whether I thought “back radiation” heated the ocean. I know from a prominent blog that a very popular idea in blog-land is that the atmospheric radiation doesn’t heat the ocean. I have never seen any evidence for the idea. That doesn’t mean there isn’t any..
See note 1 on “heat”.
The Basic Idea
From what I’ve seen people write about this idea, including the link above, the rough argument goes like this:
- solar radiation penetrates tens of meters into the ocean
- atmospheric radiation – much longer wavelengths – penetrates only 1μm into the ocean
Therefore, solar radiation heats the ocean, but atmospheric radiation only heats the top few molecules. So DLR is unable to transfer any heat into the bulk of the ocean, instead the energy goes into evaporating the top layer into water vapor. This water vapor then goes to make clouds which act as a negative feedback. And so, more back-radiation from more CO2 can only have a cooling effect.
There are a few assumptions in there. Perhaps someone has some evidence of the assumptions, but at least, I can see why it is popular.
As regular readers of this blog know, plus anyone else with a passing knowledge of atmospheric physics, solar radiation is centered around a wavelength of 0.5μm. The energy in wavelengths greater than 4μm is less than 1% of the total solar energy and conventionally, we call solar radiation shortwave.
99% of the energy in atmospheric radiation has longer wavelengths than 4μm and along with terrestrial radiation we call this longwave.
Most surfaces, liquids and gases have a strong wavelength dependence for the absorption or reflection of radiation.
Here is the best one I could find for the ocean. It’s from Wikipedia, not necessarily a reliable source, but I checked the graph against a few papers and it matched up. The papers didn’t provide such a nice graph..
Note the logarithmic axes.
The first obvious point is that absorption varies hugely with the wavelength of incident radiation.
I’ll explain a few basics here, but if the maths is confusing, don’t worry, the graphs and explanation will attempt to put it all together. The basic equation of transmission relies on the Beer-Lambert law:
I = I0.exp(-kd)
where I is the radiation transmitted, I0 is the incident radiation at that wavelength, d is the depth, and k is the property of the ocean at this wavelength
It’s not easy to visualize if you haven’t seen this kind of equation before. So imagine 100 units of radiation incident at the surface at one wavelength where the absorption coefficient, k = 1:
So at 1m, 37% of the original radiation is transmitted (and therefore 63% is absorbed).
At 2m, 14% of the radiation is transmitted.
At 3m, 5% is transmitted
At 10m, 0.005% is transmitted, so 99.995% has been absorbed.
(Note for the detail-oriented people, I have used the case where k=1/m).
Hopefully, this concept is reasonably easy to grasp. Now let’s look at the results of the whole picture using the absorption coefficient vs wavelength from earlier.
The top graph shows the amount of radiation making it to various depths, vs wavelength. As you can see, the longer (and UV) wavelengths drop off very quickly. Wavelengths around 500nm make it the furthest into the ocean depths.
The bottom graph shows the total energy making it through to each depth. You can see that even at 1mm (10-3m) around 13% has been absorbed and by 1m more than 50% has been absorbed. By 10m, 80% of solar radiation has been absorbed.
The graph was constructed using an idealized scenario – solar radiation less reflection at the top of atmosphere (average around 30% reflected), no absorption in the atmosphere and the sun directly overhead. The reason for using “no atmospheric absorption” is just to make it possible to construct a simple model, it doesn’t have much effect on any of the main results.
If we considered the sun at say 45° from the zenith, it would make some difference because the sun’s rays would now be coming into the ocean at an angle. So a depth of 1m would correspond to the solar radiation travelling through 1.4m of water (1 / cos(45°)).
For comparison here is more accurate data:
On the left the “surface” line represents the real solar spectrum at the surface – after absorption of the solar radiation by various trace gases (water vapor, CO2, methane, etc). On the right, the amount of energy measured at various depths in one location. Note the log scale on the vertical axis for the right hand graph. (Note as well that the irradiance in these graphs is in W/m².nm, whereas the calculated graphs earlier are in W/m².μm).
And two more locations measured. Note that the Black Sea is much more absorbing – solar absorption varies with sediment as well as other ocean properties.
DLR or “Back radiation”
The radiation from the atmosphere doesn’t look too much like a “Planck curve”. Different heights in the atmosphere are responsible for radiating at different wavelengths – dependent on the concentration of water vapor, CO2, methane, and other trace gases.
Here is a typical DLR spectrum (note that the horizontal axis needs to be mentally reversed to match other graphs):
You can see more of these in The Amazing Case of Back Radiation – Part Two.
But for interest I took the case of an ideal blackbody at 0°C radiating to the surface and used the absorption coefficients from figure 1 to see how much radiation was transmitted through to different depths:
As you can see, most of the “back radiation” is absorbed in the first 10μm, and 20% is absorbed even in the first 1μm.
I could produce a more accurate calculation by using a spectrum like the Pacific spectrum in fig 6 and running that through the same calculations, but it wouldn’t change the results in any significant way.
So we can see that while around half the solar radiation is absorbed in the first meter and 80% in the first 10 meters, 90% of the DLR is absorbed in the first 10μm.
So now we need to ask what kind of result this implies.
Heating Surfaces and Conduction
When you heat the surface of a body that has a colder bulk temperature (or a colder temperature on the “other side” of the body) then heat flows through the body.
Conduction is driven by temperature differences. Once you establish a temperature difference you inevitably get heat transfer by conduction – for example, see Heat Transfer Basics – Part Zero.
The equation for heat transfer by conduction:
q = kA . ΔT/Δx
where k is the material property called conductivity, ΔT is the temperature difference, Δx is the distance between the two temperatures, and q is the heat transferred.
However, conduction is a very inefficient heat transfer mechanism through still water.
For still water, k ≈ 0.6 W/m.K (the ≈ symbol means “is approximately equal to”).
So, as a rough guide, if you had a temperature difference of 20°C across 50m, you would get heat conduction of 0.24 W/m². And with 20°C across 10m of water, you would only get heat conduction of 1.2 W/m².
However, the ocean surface is also turbulent for a variety of reasons, and in Part Two we will look at how that affects heat transfer via some simulations and a few papers. We will also look at the very important first law of thermodynamics and see what that implies for absorption of back radiation.
Light Absorption in Sea Water, Wozniak & Dera, Atmospheric and Oceanographic Sciences Library (2007)
Note 1 - To avoid upsetting the purists, when we say “does back-radiation heat the ocean?” what we mean is, “does back-radiation affect the temperature of the ocean?”
Some people get upset if we use the term heat, and object that heat is the net of the two way process of energy exchange. It’s not too important for most of us. I only mention it to make it clear that if the colder atmosphere transfers energy to the ocean then more energy goes in the reverse direction.
It is a dull point.