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Published on  02.05.2021

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Published on  02.05.2021

The Greenhouse Defect - The most disruptive site on climate science

What is the surface emissivity of Earth?

A straight forward answer would be, we do not really know. There are a couple of issues which make it hard to simply measure emissivity, and that is not just true for Earth, but also for the Moon for instance. Just like within the visible light, the optical properties of the surface within the LWIR range can be very diverse. Just think of the different colors you see when you look outside. What we perceive as colors are effectively different wavelengths of electromagnetic radiation within the relatively tight spectrum of visible light. The fact that surfaces have all these different colors means their optical properties vary substantially and individually over that narrow spectrum. But also the perspective may play a distinct role.

The total LWIR emission spectrum is much wider by comparison, complicating things even further. With the technology we have today it might yet seem easy to solve the riddle. Let us simply shoot a satellite into space which is going to scan the globe and we should have the answer. And indeed that is exactly being done. However the results are underwhelming. The problem has largely to do with GHGs.

For the largest part of the LWIR spectrum the atmosphere is opaque, even with clear skies. You just can not see through it with IR sensors. Also your best view is straight down onto the surface, because it is the shortest path through the atmosphere. The more you look sideways the stronger the atmosphere will impair your vision. So effectively satellites are restricted both in terms of perspective and spectrum. They can only give a partial answer which is a) relatively useless and b) can even be badly misleading.

We can discuss the problems based on a map provided by NASA1, showing emissivity at the wavelength of 8.3µm. 8.3µm btw. is still within the atmospheric window, so there is a good view onto the surface.

What we see are substantial deviations from 1, pointing out that the surface is not a perfect emitter. Some strongly vegetated areas at least come close 1, while arid regions show up to 40% less emissivity. Sand is indeed a very poor emitter at this specific wavelength, but it is not quite as bad for the whole emission spectrum, which roughly ranges from 3µm to 100µm. Equally we can not generalize the overall map.

The more interesting part is actually what the map does not show. There is no information on water nor ice (actually rather snow). For both the angle of emergence is a critical factor and there is not much use in measuring emissivity with a satellite looking more or less only straight down. In a nutshell satellites can do some auxiliary work, but the main job remains undone. It is something to bear in mind when "satellite data" on emissivity, or even emissions respectively, are presented as top notch state of the art "facts". Circumstantially these are rather low quality data products which need to be taken with a lot of salt and must not be generalized. There is one specific instance where this flaw has been exploited to exaggerate the significance of GHGs and everyone fell for it. I am going to feature this problem in an upcoming article.

For now we are still left with the basic question what is surface emissivity, and how we could find out. The best possible approach is water!? Water covers ~71% of the surface, though if we allow for ice shields it will a bit less, and thus is the monolithic dominant surface type. If we know the emissivity of water we can make some educated guesses on the remainder and we should have a good enough approximation.

The optical properties of water are well known physics, even in the LWIR or far-infrared range respectively. It is all there, we only need to look it up. And again, I can not express my disregard for "climate scientists" enough, who are simply too lazy for doing this, or whatever reason one might name. There is no excuse.

At this point I would also like to thank scienceofdoom for their article "Emissivity of the Ocean"2 which I found quite helpful some years ago. They discuss a lot of things we are going to deal with here, but for some reason they did not really go all the way. Also it seems the content of this article does not quite resonate elsewhere on the site. And definitely the article derails in the "Conclusion".

The “unofficial” result, calculating the average emissivity from the ratio: ε = 0.96.
This result is valid for 0-30°C. But I suspect the actual value will be modified slightly by the solid angle calculations. That is, the total flux from the surface (the Stefan-Boltzmann equation) is the spectral intensity integrated over all wavelengths, and integrated over all solid angles. So the reduced emissivity closer to the horizon will affect this measurement.

Right, emissivity to the surface normal is about 0.96 AND the reduced emissivity towards the horizon WILL modify the result slightly. Actually a lot more than just slightly. So let us pull up sleeves and get started.

With Fresnel equations we can calculate (among others..) the reflectivity of water, that is how much light is reflected on the surface of water relative to the angle of incidence. If the water is deep enough (like oceans, but unlike a swimming pool) we can assume the light not reflected is getting absorbed. So absorption = 1 - reflection. And because of Kirchhoff's law absorptivity = emissivity for identical wavelengths. If we consider absorptivity and emissivity to differ, than it is because both are relevant with different wavelengths. Either way, what we need to know is reflectivity and the rest is a no-brainer.

This is what the Fresnel equation3 looks like. What we will need is the refractive index of air (n1 = 1) and that of water, which in the range of visible light is about n2 = 1.33 on average. With these inputs we can calculate a curve for s-polarized and p-polarized light and the average of the two is a valid approximation for the total.

The black line in the chart gives us the share of sun light reflected by water relative to the angle of incidence. It is about 2% only if the sun is high up, but with the sun close to the horizon most of it is getting reflected. If you imagine a sun set, despite the sun is being dimmed a lot by the atmosphere, there will be an intense reflection in the water. This is the physical description of the phenomenon.

The same thing works for calculating emissivity, though there are a couple of things to consider. First of all I did not allow for the extinction coefficient, which is negligible with SW radiation but becomes significant in the LW range. This is complicated since the refractive index will turn into a complex number with the extinction coefficient as imaginary part. I openly admit I have not mastered that calculation yet. However we have the great site refractiveindex.info4 where we can retrieve these data anyhow. I will also admit I first assumed I could ignore the extinction coefficient altogether, since the results perfectly matched in the SW range. That failing by assumption thing affects me as well.

Then both, refractive index and extinction coefficient differ with wavelength. If it was only the refractive index we could try to calculate a weighted average an go from there. With both variables in play that is not an option. Rather I retrieved the results for individual wavelengths (scroll down to "Reflection Calculator", click the 3 lines right under the chart) and do that so often to get a reasonable resolution over the whole emission spectrum from 3-100µm.

Now we can do something pretty cool. We can add up all the information and draw a 3D chart to show the reflectivity of water relative to solid angle and wavelength. You probably have not seen that before. I had to make a compromise with scaling the wavelength to fit it all into one chart. The interval is 1 for 4-20, 2 for 20-30, 5 for 30-50 and 10 for 50-100µm.

To calculate hemispheric reflectivity (or effectively emissivity) we need to allow for the geometry. The density function of the surface of a hemisphere can be written as 1- cos(x). But of course we have to include Lambert's cosine law5 which turns our density function into sin(x)^2. Note: it is "incidentally" the same geometry as you have with a single source of light shining onto a hemisphere! Doing so will give us the hemispheric reflectivity for every data point on our wavelength curve.

You might want to compare the result with Huang et al (2016), Fig. 3.6 If you invert the curve, as 1- reflectivity = emissivity, and consider the logarithmic wavenumber scale they used, you will find the curves are essentially identical. So far we look good.

The only thing left to do now is to weight this curve according to Planck's law and for simplicity I will just assume a surface temperature of 288K. We would get a bit lower reflectivity with higher temperatures, as the emission spectrum gradually moves towards shorter wavelengths, and vice verse the opposite for lower temperatures. It cancels out, so 288K straight is perfectly fine.

The result is 0.0907. For emissivity we get 1 - 0.0907 = 0.909. The hemispheric surface emissivity of water is 0.909, which is pretty consistent with the 0.91 Baehr, Stefan7 name, referenced on german wikipedia8.

Ok, it was a bit technical but hopefully worth it. With water having an emissivity of only about 0.91 we have a substantial deviation from 1. At a temperature of 288K water does not emit 390W/m2, but only 355W/m2. Something very similar goes for the surface of Earth as a whole. As I have said before there are quite some uncertainties as we just do not have all the data, especially in the far-IR range. At least we know land surface emissivity is deviating substantially from 1 as well, so that global surface emissivity can not deviate a lot from 0.91.

For the GHE that insight is a serious blow. Again, we know clear sky emissions are about 270W/m2 while surface emissions are only about 355W/m2. The difference between the two is 85W/m2. So the GHE attributable to GHGs has shrunk from 150W/m2, which is about the usual erroneous claim, to only 85W/m2. Temperature wise that corresponds to roughly 19K, again not 33K! And of course we will have adept our super sophisticated emission graph anew..

Finally it brings up the question what to think about the "Earth Energy Budget" as presented by NASA9, to name just one instance. What you think might be appropriate to call the "scientists" responsible for it? And what would you call the "critical scientists" who say there is nothing to see here, and who disagrees is stupid? Regrettably I can not give you the answers on these questions. ;)

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Comments (12)

  • Alexy Scherbakoff
    Alexy Scherbakoff
    at 07.04.2023
    Critique and not a criticism.
    Check out integrating spheres online. A lot of information available on the subject.
    You ese the terms: emittance , absorptance, transmittance, but you don't use the term reflectance. You use the term reflectivity. To my thinking, you are incorporating specular reflection with reflectance. Different things. I know, fresnel equation. I separate the specular reflection before applying the 'on the normal' fresnel equation. It's near perfect match for the Johns Hopkins reflectance data on the NASA website.
    https://speclib.jpl.nasa.gov/library.
    I hope that's helpful.
    • GHD
      GHD
      at 07.04.2023
      Sorry, I never bothered much about the difference between reflectance and reflectivity. Its a language thing.

      As with the John Hopkins data, they look identical to what I have for reflectivity to surface normal. However these data only reach up to 14µm and spare out the far-IR. Also this confuses me ab bit..

      "Measurement: Directional (10 Degree) Hemispherical Reflectance"

      I mean either it is directional at 10 degrees (which it actually is), or hemispheric..?!
    • Alexy Scherbakoff
      Alexy Scherbakoff
      at 08.04.2023
      Awkward language.
      The monochromator is set 10 degrees off the normal to surface. If it was on the normal, then any specular reflection would come back into the optics of the monochromator and potentially mess with with things. At upto 12 degrees they say it's still ok.
      The reflectance is measured off the wall of the sphere.
      Therefore:"Measurement: Directional (10 Degree) Hemispherical Reflectance"
  • PeterMitchel
    PeterMitchel
    at 20.04.2023
    Hi,

    excelent article.

    I am just wondering for the discrepancies in Xianglei Huang et al (https://journals.ametsoc.org/view/journals/atsc/73/9/jas-d-15-0355.1.xml)

    1. in Fig.3 on that paper, and you referenced into your article, its a _simulated_ hemispheric-mean spectral emissivity of calm water. From 850cm-1 to 2000cm-1, more or less, emissivity doesn't go up more than 0.95. But later, when you go to Fig. 1, The measured and simulated spectral emissivity of water surface, after 800cm-1 it goes very near 1. I assume the difference is about methodology, but the author doesn't give information about which one is better or more near to reality.

    2. As far as i understand, Xianglei Huang et al doesn't calculate global emissivity including oceans for section 3, and impact on the LW TOA radiation budget including oceans, right? do you have any idea why?

    thanks in advance.
    • GHD
      GHD
      at 21.04.2023
      1. Water, as ice, is a non-lamberation radiator. Reflections/emissions depend on the solid angle, which is why we need Fresnel equations. The article goes a long way to explain all this. In Fig. 1 Huang et al states "the view zenith angle is 10°", meaning essentially to surface normal, comparable to the restricted perspective of a satellite. Fig. 3 gives hemispheric emissivity. Since we live in a 3-dimensional world, this fits reality.

      2. Not the scope of their paper. It IS dissenting the orthodoxy anyway, but calling them out would not get it past "peer review".
    • PeterMitchel
      PeterMitchel
      at 21.04.2023
      Thanks for the fast answer.

      1. "Fig. 3 gives hemispheric emissivity". But does he use Fig.1 or Fig. 3 for his OLR calculations? thanks by the way to clarify the difference.
      2. I agree 100% with your answer. It would have been a huge punch in the liver. Nowadays, because of the radicalization of science, he only provides a sneak peak with Fig 3. wondering if in the future, other scientists will have the balls to go further. Lots of good paid positions in risk. Its a pitty.

      Thanks a lot for your blog. Its refreshing.
  • PeterMitchel
    PeterMitchel
    at 05.05.2023
    Hi,

    i think you will appreciate this paper. Look at Figure 3.

    https://www.pnas.org/doi/10.1073/pnas.1413640111

    Cheers
  • earthAlbedo
    earthAlbedo
    at 24.07.2023
    Hi again,

    according to IPCC 2021 WG1 report (chapter 7, page 934), the surface longwave radiation flux is 398 W m–2 (395-400 confidence interval). As far as we know, its a measured flux observation from satellite and calibrated with ground stations. If we have to correct something is the T=288, that is not measured, and probably should be higher that these famous value. In fact, it looks like those 288 is an old value that doesn't appear even in the last IPCC report.
    • GHD
      GHD
      at 24.07.2023
      Again, you can not measure surface emissions with a satellite, as only a small fraction of which will be transmitted through the atmosphere. Surface temperature on the other side can be measured with reasonable accuracy, and it is (still) lower than 288K. You may have read about global average temperature exceeding 17°C this July. While this is questionable, the point here is that global average temperature oscillates by over 4K in the annual cycle, hitting highs in July and lows in January.
    • PeterMitchel
      PeterMitchel
      at 26.07.2023
      Something is odd.

      According to Gerrit Lohmann et al 2019 (https://esd.copernicus.org/articles/11/1195/2020/), from eq (3):

      "we gain an average Earth temperature of 288 K (15 ∘C), very close to the global temperature observations and reconstructions (Hansen et al., 2010) at 14 ∘C for 1951–1980. The implicit assumption in Eqs. (2) and (3) is that we have a single temperature on the Earth, although we know that the Equator-to-pole surface temperature gradient is on the order of 50 K and that the incoming solar radiation at the Equator is about twice that at the poles (Peixoto and Oort, 1992)."

      So we know that those 288K is an average measured value with a certain confidence, although as you say, its not constant among the year. But what strikes me is this paragraph above eq (3):

      "and ϵ is the effective emissivity of Earth (about 0.612) (e.g., Archer, 2009)"

      The author goes all the way around. To achieve 288K, he uses an emissivity of 0.6 :-)
    • PeterMitchel
      PeterMitchel
      at 27.07.2023
      Hi GHD,

      those 390W/m2 are emissions from surface temperature (around 2 meter above ground), or surface skin temperature? is it relevant for all those calculations? as far as i understand, 288K (or more recently 289K) are calculations made for near surface, not the last. Is it relevant? does it affect to all this issues? Because maybe, just maybe, skin temperature is some degrees bigger, and then it is compatible with 390 W/m2.

      Cheers.
    • GHD
      GHD
      at 28.07.2023
      Comparing TOA emissions (~240) to assumed surface emissions (~398), then you could assume an emissivity of 240/398 = 60.3% or so for the "surface-atmosphere system", if you will. A meaningless number.
      I think the assumption is that meteorologic and surface temperature are about the same on average. Of course surface temperature have a much larger DTR (daily temp. range). But there could be some difference, I don't know.

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