In Planck, Stefan-Boltzmann, Kirchhoff and LTE one of our commenters asked a question about emissivity. The first part of that article is worth reading as a primer in the basics for this article. I don’t want to repeat all the basics, except to say that if a body is a “black body” it emits radiation according to a simple formula. This is the **maximum** that any body can emit. In practice, a body will emit less.

The ratio between actual and the black body is the emissivity. It has a value between 0 and 1.

The question that this article tries to help readers understand is the origin and use of the emissivity term in the Stefan-Boltzmann equation:

E = ε’σT^{4}

where E = total flux, ε’ = “effective emissivity” (a value between 0 and 1), σ is a constant and T = temperature in Kelvin (i.e., absolute temperature).

The term ε’ in the Stefan-Boltzmann equation is not really a constant. But it is often treated as a constant in articles that related to climate. Is this valid? Not valid? Why is it not a constant?

There is a constant material property called emissivity, but it is a function of wavelength. For example, if we found that the emissivity of a body at 10.15 μm was 0.55 then this would be the same regardless of whether the body was in Antarctica (around 233K = -40ºC), the tropics (around 303K = 30ºC) or at the temperature of the sun’s surface (5800K). How do we know this? From experimental work over more than a century.

Hopefully some graphs will illuminate the difference between emissivity the material property (that doesn’t change), and the “effective emissivity” (that does change) we find in the Stefan-Boltzmann equation. In each graph you can see:

- (top) the blackbody curve
- (middle) the emissivity of this fictional material as a function of wavelength
- (bottom) the actual emitted radiation due to the emissivity – and a calculation of the “effective emissivity”.

The calculation of “effective emissivity” = total actual emitted radiation / total blackbody emitted radiation (note 1).

At 288K – effective emissivity = 0.49:

At 300K – effective emissivity = 0.49:

At 400K – effective emissivity = 0.44:

At 500K – effective emissivity = 0.35:

At 5800K, that is solar surface temperature — effective emissivity = 0.00 (note the scale on the bottom graph is completely different from the scale of the top graph):

Hopefully this helps people trying to understand what emissivity really relates to in the Stefan Boltzmann equation. It is not a constant except in rare cases. But you can see that treating it as a constant over a range of temperatures is a reasonable approximation (depending on the accuracy you want), but change the temperature “too much” and your “effective emissivity” can change massively.

As always with approximations and useful formulas, you need to understand the basis behind them to know when you can and can’t use them.

Any questions, just ask in the comments.

**Note 1** – The flux was calculated for the wavelength range of 0.01 μm to 50μm. If you use the Stefan Boltzmann equation for 288K you will get E = 5.67×10^{-8} x 288^{4} = 390 W/m^{2}. The reason my graph has 376 W/m^{2} is because I don’t include the wavelength range from 50 to infinity. It doesn’t change the practical results you see.

on February 1, 2017 at 3:55 am |Bob Armstrong“emissivity … is a function of wavelength.”

That’s why the generalization of the computation with produces the ubiquitous 255K meme in the box on http://cosy.com/Science/RadiativeBalanceGraphSummary.html is a function of dot products of the relevant spectra .

on February 1, 2017 at 9:53 am |FrankSOD: Thanks for the post. It is an beautiful explanation – of one facet of the complex phenomena associated with emissivity/absorptivity. (These two phenomena are tightly linked.)

For the atmosphere, the 15 um CO2 band is located at the peak BB emission for 200 K and perhaps 25% below the peak for 300 K at 10 um.

According to MODTRAN (looking up from 0 km), the emissivity of a US standard atmosphere with 400 ppm of CO2 and no other GHGs is 85 W/m2 / 390 W/m2 = 22%. If I offset the temperature by -33 K (to 255 K), 50/240 = 21%. For 220K, 25.5/133 = 17%. You’ve illustrated this phenomena perfectly.

Note: Modtran uses a fixed lapse rate anchored to a surface temperature (with offset) for 10 km of troposphere and leaves stratospheric temperature without any offset. Any offset disappears between 10 and 11 km. So your examples are simpler and cleaner.

Now for the fun: What happens if I reduce or increase the CO2 concentration (with no temperature offset)? Increase from 1 to 2 ppm: +4.5 W/m2. 2 to 4 ppm: +5.4 W/m2. 20 to 10 ppm: +6.0 W/m2. 100 to 200 ppm: +6.4 W/m2. 1000 to 2000 ppm: +6.72 W/m2. 0.1 to 0.2 ppm: +1.7 W/m2.

Unlike most solids and liquids, emissivity isn’t a property of CO2 itself; it’s a property of the amount of CO2. The relationship appears to be roughly log-linear down to a few ppm. So emissivity for CO2 is a function of both temperature and concentration.

on February 1, 2017 at 10:28 am |FrankWhat happens when we put all GHGs in the atmosphere at their normal concentrations? Emissivity in the downward direction is 333 W/m2 / 390 W/m2 (288 K) = 85%. Emissivity in the upward direction is 240 W/m2 / 390 W/m2 = 61%???. About 40 W/m2 of the photons escaping upwards come for the surface, not the atmosphere, so one could argue that the upward emissivity of the atmosphere is closer to 50%. Normal materials usually don’t have emissivities that are different in each direction.

If the downward emissivity of the atmosphere is 85%, does that mean that the upward absorptivity of surface emission by the atmosphere of surface emission is also 85%? The correct answer here is 90% or higher, which is quite reasonable, given that DLR is emitted from a little above the surface where it is cooler than 288 K.

If one carefully accounts for absorptivity, emissivity and transmissivity, everything should work out correctly with absorptivity = emissivity. However, in the case of different upward and downward emissivity from the atmosphere, Grant Petty calculates emissivity using the Schwarzschild eqn.

on February 1, 2017 at 11:32 amscienceofdoomFrank,

Global annual averages are useful but can obscure more than they reveal.

Downward longwave radiation of 333 W/m

^{2}implies an “effective emissivity” of about 85%. In broad brush this number is probably overwhelmed by the higher humidity tropics. You get a lower number for high latitudes and the arctic.Downward longwave radiation is nice and simple because it is only sourced from the atmosphere. There is no (significant) longwave radiation from space.

If the atmosphere was only 500m thick then the upward longwave radiation from the atmosphere would be about the same. And the total upward longwave radiation would be greater than 333 W/m

^{2}because some surface radiation would be transmitted through the atmosphere. The total upward radiation would be around 390 W/m^{2}(a little lower, but not much).Why is the upward radiation at top of atmosphere a lot lower than 390? Because of all the reasons explained in various series including Visualizing Atmospheric Radiation.

Nick suggests (see later comment) some more explanation and I might draw up some more graphs, without going back into the content of the whole series already linked.

on February 1, 2017 at 6:29 pmNick Stokes“Downward longwave radiation of 333 W/m2 implies an “effective emissivity” of about 85%.”But this is a typical gas problem. You can’t deduce any sensible emissivity from 333 W/m2 because you can’t say what the temperature is. It is worse with the frequency dependence. Some frequencies have come from a long way up, while some, eg peak CO₂, from very close.

A paradox is that a thick enough atmosphere with multiple GHGs has emissivity 1 (as an apparent surface), despite the strongly banded spectrum of the gas. You can see this easier by thinking about absorptivity. If it’s think enough everything gets absorbed. There is no reflection or anything else that can happen to the incident radiation.

on February 1, 2017 at 10:17 am |Nick StokesSoD,

Thanks for a good discussion of emissivity of surfaces. I think it would be a good idea if you wrote something about how to understand emissivity of gases. The flux emitted is no longer proportional to a surface area, but to a volume, and as Frank says, depends on gas concentration in that volume. And in that volume there is also absorption. You have described emissivity as a fraction of what a perfect emitter (BB) would put out, but for a gas the idea of a perfect emitter is less clear.

on February 1, 2017 at 11:19 am |scienceofdoomNick,

I think the best way to understand it in an atmosphere is as a “layer” – emitting up and down. Of course a volume emits in all directions, but the atmosphere is like a thin piece of paper with the area of a desk, so the “sideways” doesn’t affect the result.

And then emissivity simply depends on the number of molecules, in a non-linear way (1-exp(-n)).

Maybe I will draw some more graphs.

As far as including emission and absorption, I’ll just point people back to the whole series Visualizing Atmospheric Radiation.

on February 1, 2017 at 9:29 pm |FrankNick wrote: “You have described emissivity as a fraction of what a perfect emitter (BB) would put out, but for a gas the idea of a perfect emitter is less clear.”

Planck’s Law tells us what a perfect emitter would put out, but the derivation starts with an assumption that absorption and emission have reached equilibrium in the medium. As soon as you consider semi-transparent media, this assumption is invalid. (The assumption is still valid if emissivity is produced at a surface.) Can we really apply the S-B eqn. to situations where its derivation is invalid? Having a flexible fudge factor – emissivity – seems to work, but allowing it to be both an intensive property at surfaces and and an extensive property of semi-transparent media is kludgy nomenclature.

The only place that the S-B eqn really causes trouble is the difference between upward and downward emissivity in the atmosphere. Here Grant Petty (p211) still uses the terms emissivity and absorptivity, but he calculates them by integrating the Schwarzschild eqn. IMO, we should admit that Planck’s and S-B don’t apply to semi-transparent materials and say that the Schwarzschild eqn must be used with them.

If Einstein coefficients had been discovered earlier, we might consider Planck and S-B to merely be corollaries of a more fundamental equation – the Schwarzschild eqn. It describes the interaction between radiation and matter without assuming that radiation is in equilibrium with the medium (as Planck assumed). LTE is required so that a Boltzmann distribution of states can be assumed. The Planck function B(lambda,T) and a single cross-section for absorption and emission arise In the process of deriving the Schwarzschild eqn from Einstein coefficients. However, that derivation is complex and generally ignored. Planck and S-B dominate our view of how radiation is emitted by matter and Beer’s Law dominates our view of how radiation is absorbed by matter. The Schwarzschild eqn – which encompasses both processes – isn’t usually even mentioned by name; the synonym “radiative transfer calculation” is used instead.

Best of all, there is no need for emissivity when calculating radiation transfer thorough semi-transparent media using the Schwarzschild eqn.

Since no one seems to agree with this perspective, I’m probably exaggerating the difficulties of using emissivity properly.

on February 1, 2017 at 7:53 pm |FrankSOD: In the idealized Figures in this post, the emitter has no emissivity below 8 um or above 20 um. What phenomena produce changes in emissivity?

1) In some cases, the emitter doesn’t have any QM-allowed transitions outside of a particular region. For example, carbon monoxide has only a single mode of vibration. In that case, we are talking about a material that is partially TRANSPARENT at wavelengths relevant to radiative cooling. In the case of CO (5 um), the atmosphere is a little too cold emit much energy through its allowed vibrational transitions. So are the stretching bands of CO2 (2.5 and 4 um), which is why we usually discuss only the 15 um bending of CO2.

2) In other cases, the emitter may have QM-allowed transitions and a structure that doesn’t permit all radiation to escape through its SURFACE. I assume this is the case with conductive metals like copper and aluminum, which have emissivity around 10%. A thin layer of oxide on the surface of these metals, however, raises their emissivity to around 90%. A thin layer of low emissivity material (fluorinated tin oxide) is deposited on the surface of glass to make low emissivity windows. Mike M. mentioned that silicon (a poor conductor) is transparent to longer wavelengths of IR, so some metals may have low emissivity because they are semi-transparent. However, low-E glass wouldn’t work if the coating on the surface had low emissivity because of transparency.

So – unless I am badly confused (please let me know if I am) – we have at least two mechanisms that produce low emissivity: semi-transparency and reflection/scattering of radiation at the surface of a non-transparent material. It is easy to see why absorptivity must equal emissivity when low emissivity is produced at a surface. For semi-transparency, we need to remember transmissivity when thinking about absorptivity = emissivity.

The surface scattering mechanism makes emissivity a property that is independent of quantity or thickness. The semi-transparency mechanism produces emissivity that varies with the amount of emitter – as I demonstrated for CO2 in the atmosphere.

If most readers are like me, they don’t immediately associate low emissivity with transparency – especially when using the S-B eqn: W = eoT^4. The average emissivity of the Earth’s surface is 97%, but not because 3% of thermal infrared is passing through the Earth (partial transparency). The same is true for a few inches of concrete (e = 0.91). On the other hand, the downward emissivity of the atmosphere (including cloudy areas) is 85-90% is due to semi-transparency in the “atmospheric window wavelengths”. If clouds didn’t interfere with semi-transparency in more than half of the sky, the downward emissivity of the atmosphere would be closer to 70% (according to MODTRAN for a US Standard atmosphere).

This post does a great job of explaining why emissivity can change with significant changes in temperature. It still leaves many booby traps for those who crudely apply the S-B eqn to the atmosphere.

on February 2, 2017 at 1:38 am |DeWitt PayneFrank,

Emissivity calculated from total emitted flux and a particular temperature using the S-B equation only has some physical meaning if you know that the emissivity is nearly constant over the wavelength range of interest, as when doing optical pryometry for temperatures greater than about 1000K. An optical pyrometer is also the official interpolation instrument for measuring temperatures above the freezing point of gold, 1337.33K. Otherwise, it’s just a number.

It’s a completely different number if you use the temperature of the atmosphere at 5km, which is the average effective height of emission or assume that the emissivity is one and calculate the effective temperature. Effective temperature may or may not have physical meaning. In the real world, emissivity is always a function of wavelength and for solids or liquids, a function of viewing angle. The same goes for absorptivity and transmissivity.

Saying that the Earth has an emissivity viewed from space of ~0.6 using the S-B equation and the surface temperature of the Earth is a pointless exercise. The same goes for downward radiation from the atmosphere. It wasn’t all emitted from the same altitude and temperature.

on February 3, 2017 at 8:23 amFrankDeWitt wrote: “Saying that the Earth has an emissivity viewed from space of ~0.6 using the S-B equation and the surface temperature of the Earth is a pointless exercise.”

I pointed out the difference in directional emissivity/absorptivity of the Earth to illustrate the absurdity of applying the S-B equation in some situations. The question I am trying to get agreement about is: What situations give absurd answers and why? When is the S-B eqn suspect or invalid?

1) Like the Earth and Venus, objects without a homogeneous temperature MAY be suspect. Ocean temperature varies with depth, but the surface has a well-defined emissivity in the thermal infrared, because essentially all of the outgoing photons are emitted and incoming photons are absorbed in the top 1 mm, where the temperature is locally homogeneous. So we need temperature homogeneity within a few tau (optical depth) of the emitting surface. The corona of the sun is much hotter than the 5700 K “surface”, but there is negligible optical density in the corona. The temperature of our thermosphere don’t have any impact on TOA OLR for the same reason.

Interestingly, Wikipedia discusses emissivity in terms of “reduced emission”, but the author(s) of the article about the S-B eqn approach the problem from a different direction. “A body that does not absorb all incident radiation (sometimes known as a grey body) emits less total energy than a black body and is characterized by an emissivity less than 1”. I’d add: “… and equal to its absorptivity (assuming LTE)”. So here a GRAY BODY is defined by its ABSORPTIVITY, which is [mostly] independent of temperature. Emissivity isn’t a fudge factor we can select to correct for temperature inhomogeneity or some other weaknesses of a model. (:))

Those who attempt to treat the atmosphere as a black box with an arbitrary temperature and emissivity run into problems with this definition. The absorptivity of their black box must be the same from the top or bottom. So they aren’t allowed to postulate different emissivities from the top and bottom. (They need to break the atmosphere up into a vertical stack of black boxes thin enough that emissivity is the same in both directions.)

2) Objects that are semi-transparent appear to be OK if they have a homogeneous temperature and W = eoT^4 is applied over a modest range of temperature (less than 2X in degK).

3) The S-B eqn is invalid if it disagrees with the Schwarzschild eqn. The latter properly accounts for the rate of emission and absorption of photons; the former is derived assuming an equilibrium that may not exist. (Grant Petty uses the Schwarzschild eqn when discussing the emissivity and absorptivity of the Earth’s atmosphere.)

on February 3, 2017 at 3:39 pmDeWitt PayneFrank,

I think you’ve answered your own question. The key word for emissivity less than one is gray. If the emission spectrum has more or less the same shape as a blackbody but is lower everywhere by very close to the same proportion, then the S-B equation is useful.

The emission spectrum of the Earth’s atmosphere is not even close to gray. Nor is the example given by SoD in this post.

However, the S-B equation

canbe used to determine total radiative flux even if the flux spectrum isn’t gray. That’s how a pyrgeometer works. It works because the detector has constant emissivity so the effective temperature of the source has physical meaning even though the spectrum isn’t gray. The pyrgeometer measures the effective temperature, but the data out is W/m².on February 2, 2017 at 1:41 am |DeWitt PayneTo put it another way: Radiative imbalance is expressed in W/m², not temperature.

on February 4, 2017 at 1:56 am |curiousdpI think there is related confusing issue here. Usually it is o.k. to assume that thermal flux incident on the earth will be absorbed, even if the emissivity is less than one. But not always, apparently. In Feldmanm et. al. PNAS, no. 46, pp.16297 – 16302 he finds that sometimes reflected thermal IR flux will make it out to TOA but only in a few arid regions. In most places, it is absorbed by water vapor. Of course,and fortunately, sunlight has no problem leaving once reflected, but under most circumstances thermal IR can’t escape.

on February 7, 2017 at 9:35 pm |Princeton UniversityIt’s surprising to find on scienceofdoom.com a resource so precious

about equations.

We will note your page as a benchmark for Basics – Emissivity and the

Stefan Boltzmann Equation .

We also invite you to link and other web resources for equations like http://equation-solver.org/ or https://en.wikipedia.org/wiki/Equation.

Thank you ang good luck!