# Do you think emissivity of air makes sense?

• I
• Fefetltl
In summary, this scientist is asking if the emissivity of a layer of air exists and says that it does not make sense. He goes on to say that if you consider a layer of air between two different media, it does not make sense because it is an intermediate term. The scientist then says that the emissivity of the Earth surface has a sense and that of the sun has a sense. Finally, the scientist says that the emissivity of a cloud surface does not make sense because a cloud is finite.
Fefetltl
TL;DR Summary
Question about the physical sense of air emissivity instead of well define closed surface object's emissivity
Hello guys :)In the frame of finding a physical model for the temperature of Earth's surface, talking about the very "idealized" two-layers model of atmosphere, I ask you now the question to the other physicists or engineers: does it make sens to associate an emissivity to a layer of air (+ some gases in few quantity) ?Why my answer is clearly noThe emissivity is defined between two medias with a discontinuous variation of refractive index, i.e the wall of a oven and the surrounding air, a hot resistance surface in a boiler and the surrounding water etc.

Mathematically speaking we should write ##\epsilon = 1 - t - r##, with ##r## reflection (scattering + specular), ##t## transmission, at thermal equilibrium, Kirchhoff's law.

When you consider now a layer of air with continuous refractive index ##n(z)## at height ##z##, then the layer just upper this one at height ##z + d z## with ref. index ##n(z + d z)##. The transmission between these infinitely close layer is ##1## and the emissivity is clearly ##0##. This is clearly ##0## because you don't have an abrupt interface in atmosphere: it is a continuously varying media (see exponential decrease of gas density and so the same for optical index squared). But for the two layers model, they put that the "upper layer" of atmosphere ##\epsilon## is equal to ##0.78## ... just to adjust the Earth's surface temperature at ##288## K... practical but does not make any physical sense.For you, does the emissivity of a layer of air exist?I ask you the question, because of course I can be wrong. I asked also in another forum, no one can answer...

I get the impression that you are treating air as a chemically uniform medium. It is far from that. Water vapor and CO2, among other components, contribute significantly to the transmission spectrum of Earth's atmosphere within the infrared spectral domain. See for example:

The following might be of interest:
https://www.acs.org/climatescience/atmosphericwarming/climatsensitivity.html
Climate Sensitivity

A Single-Layer Atmosphere Model

A Multilayer Atmosphere Model

Fefetltl said:
When you consider now a layer of air with continuous refractive index n(z) at height z, then the layer just upper this one at height z+dz with ref. index n(z+dz). The transmission between these infinitely close layer is 1 and the emissivity is clearly 0. This is clearly 0 because you don't have an abrupt interface in atmosphere: it is a continuously varying media (see exponential decrease of gas density and so the same for optical index squared).
This sounds like you've missed the point of integral calculus. Yes the index at an infinitesimally small height increase dz has an infinitesimal change, but that doesn't mean you can't add them up to a show a significant change. Sounds a bit like Zeno's paradox to me.

No i don't miss the point of integral calculus, I did some vector spherical harmonics analysis during my Phd in light scattering so don't worry about that ;) (by the way this is not what I prefer to do to be honest). You can easily do the computation of absorption of infrared spectra of earth emitted at ##10## µm by (according to an exponential decresing law of density):

$$A = 1 - e^{- k \sum_i N_i \text{Im}(\alpha_i) h_i }$$

with ##\alpha_i## polarizability, ##N_i## number of particles per unit volume, ##h_i \simeq 5-10## km (depends on species ##i##) and ##i## stands for CO2, H2O and so on (all of gases except for air). You find just for CO2 ##1## by far.
So you are already above the so called "emissivity of atmosphere" which is equal to 0.78... This emissivity, I repeat has NO physical sense.
- Emissivity of earth surface HAS a sense: 2 differents media, rock-water-vegetation / air
- Emissivity of sun surface HAS a sense: vacuum/closed plasma of H-He
- Emisssivity of a cloud surface HAS a sense: air/liquid droplets, a cloud is finite
- Emissivity of air/air ... does not make sense !

I know already the publication, I already read it see this: "The average temperature of the Earth’s surface is about 288 K. For the single-layer atmosphere model, the graph shows that this temperature would correspond to an atmospheric emissivity of about 0.8."

The emissivity of the single layer, is an intermediate term to able to fix the surface temperature at 288 K. But this is not a true physical value.

Do we have engineers here ? For thermal engineers, you have two surface which radiate at ##T_1## and ##T_2## do you consider the air between these two for the thermal radiation equilibrium ? The answer is no (but of course you can consider it for convection or conduction loss, but not for radiation).

Why do you keep asking questions and then answering them (often injcorrectly)?
This produces polemic not dialogue.
If you have a question please ask it without the intervening incorrect physics.

hutchphd said:
Why do you keep asking questions and then answering them (often injcorrectly)?
This produces polemic not dialogue.
If you have a question please ask it without the intervening incorrect physics.
The thing is I don't accept graphics, I accept equations and definitions.

My question is: How do you define air emissivity ?

Since we are defining emissivity to differenciate how radiates finite surface bodies at the same temperature but different composition, how can I define the emissivity of a shell of air (+ CO2 + ionized O2 in very small quantity and so on) called atmosphere ? The emissivity should be define with a step a refractive index, as Kirchhoff's law suggest.
Then, to perform the integration of radiative transfer theory, the "emissivity of air" should depend on altitude, because the gas density decrease with altitude. Can you give me the expression of the emissivity of air depending on the altitude ? I tell you this because according to climate people (I don't say physicist I say people) we should consider every layers of air to emit as a grey body with different emissivity... which should depend on $z$.

Thank you guys for your interest in this topic.

What is incorrect in what I said hutchphd ? Please tell me, make me more clever, that's all I ask

Fefetltl said:
The thing is I don't accept graphics, I accept equations and definitions.
Yet you give neither explicit definitions nor equations. Thus, let us begin at the beginning: please provide an explicit definition of emissivity that is directly relevant to a planetary system supported by a reference from literature.
Fefetltl said:
Since we are defining emissivity to differenciate how radiates finite surface bodies at the same temperature but different composition, how can I define the emissivity of a shell of air (+ CO2 + ionized O2 in very small quantity and so on) called atmosphere ? The emissivity should be define with a step a refractive index, as Kirchhoff's law suggest.
Emphasis added. The atmosphere is not in an isothermal state.

Hyperfine said:
Yet you give neither explicit definitions nor equations. Thus, let us begin at the beginning: please provide an explicit definition of emissivity that is directly relevant to a planetary system supported by a reference from literature.

Emphasis added. The atmosphere is not in an isothermal state.
We don't care an atmosphere layer is isothermal in case of average of day/night and convections phenomena.

Emissivity for a planet ? We have the one of its surface, that's it.

Guys, you know that with heat equation and electromagnetic propagation we can STILL compute in principle the earth surface temperature right ? You know that radiative transfer theory is an APPROXIMATION of electromagnetism + scalar thermal radiation (which is just an isotropic radiation with no preferencial direction)?

Ok guys apologize for the aggressive behavior but I don't like condescending and superiority attitude, just saying, I talked to moderator also to explain.

That being said, I found the solution in wikipedia (and thks to another forum with this article but the people did not fully understood the question... but anyway).

For a surface at temperature ##T## and of emissivity ##\epsilon##, the surface density irradiance is defined by Planck law:

$$L = \frac{2 \pi h \nu^3 \epsilon (\nu)}{c^2} \frac{1}{e^{h \nu / (k_B T)}-1}$$

For a gas at temperature ##T## and of refractive index ##n##, the surface density irradiance is defined also by Planck law (##c## becomes ##c/n## because the medium of propagation is the gas itself):

$$L = \frac{2 \pi h \nu^3 n^2 }{c^2} \frac{1}{e^{h \nu / (k_B T)}-1}$$

So the emissivity is for surfaces, but for gas we are dealing with the refractive index, that seems fine for me.
And don't forget that there is a difference between saying random sentences and providing the right equations.

hutchphd
And to go into further details for several species ##i## we have:

$$n^2 = 1 + \sum_i N_i(z) \alpha_i$$
with ##z## altitude, ##\alpha## polarizability, ##N## molecule density, then the surface irradiance become (we can deal also with volumic one):

$$L = \frac{2 \pi h \nu^3}{c^2} \frac{1}{e^{h \nu / (k_B T)}-1} + \sum_i \frac{2 \pi hN_i(z) \alpha_i \nu^3}{c^2} \frac{1}{e^{h \nu / (k_B T)}-1}$$

with the second term take into account greenhouse gases.

Fefetltl said:
And to go into further details for several species ##i## we have:

$$n^2 = 1 + \sum_i N_i(z) \alpha_i$$
with ##z## altitude, ##\alpha## polarizability, ##N## molecule density, then the surface irradiance become (we can deal also with volumic one):

$$L = \frac{2 \pi h \nu^3}{c^2} \frac{1}{e^{h \nu / (k_B T)}-1} + \sum_i \frac{2 \pi hN_i(z) \alpha_i \nu^3}{c^2} \frac{1}{e^{h \nu / (k_B T)}-1}$$

with the second term take into account greenhouse gases.
Thanks for raising the question on emissive atmosphere a few months ago that I'm interested in understanding further.

I have trouble with the concept of gases under Earth's atmospheric conditions being modeled with the T^4 relationship as a blackbody source. I think you've been persuaded differently now from seeing your last couple of posts but correct me if I have misunderstood.

During my consideration of the topic, I came across this physics letter (attached) suggesting that gases are not blackbody emitters - see Blackbody Radiation in Optically Thick Gases which indicates that only condensed matter is capable of blackbody radiation. There are other references that state in general that gases are not blackbody emitters and so it would be my understanding that only clouds and haze in the atmosphere would be sources of blackbody radiation apart from the Earth’s upward radiation from the surface.

If the theory of those saying the atmosphere is emissive like a blackbody is true it brings out several questions for me. Here are a few examples for anyone to chime in on:

1. Wouldn’t all gas molecules release photons with collisions, not just GHGs? If it is just GHGs, and just in their resonant bands, this is not blackbody radiation.

2. If 1. is true, thermal imaging cameras (8-14um) would be drowned in atmospheric emission if objects were many miles away as the signal would be lost in the noise.

3. Although there are 3 modes (rot,vib,trans) of heat energy storage for a molecule, a photon if released would have more energy than the KE of the molecule. For example, a molecule of CO2 if emissive in a blackbody sense at T=288K, its KE = 5.1E-21 Joules and for an arbitrary photon at wavelength 10um, its energy is 1.99E-20 Joules. Even though the photon may derive its energy from all 3 modes, the point is the gas would rapidly be depleted of all of its thermal energy. In comparison, for a condensed material, like a body of water, the surface molecules at 288K release approximately 3600 photons per second per molecule in blackbody radiation.

Also, related to GHGs, I’d like to find a unified understanding where the energy distribution is computationally connected. Here are some of the various GHG explanations one can find in searches (books, papers, and institutional websites) that don’t seem collectively consistent or at least proportionally rationalized:

4. GHGs absorb photons in their respective bands and release the energy sometime later through photons in a random direction.

5. About 50% of the energy that GHGs intercept is absorbed by the Earth and 50% is released to outer space

6. GHGs directly heat the atmosphere (the explanation being a quantum mechanical redistribution of vibrational energy across spin, vib, trans and so temperature will rise)

7. GHGs are emissive in a blackbody sense (the topic you raised)

To explore the proportional rationalization, consider Earth’s emission at 288K and 1 atm, and consider the peak of the 15um band for CO2. I wonder between 4, 6, and 7, what is the distribution of energy and do 50% make it skyward and 50% get returned to Earth in that region or not.

So if gases are not blackbody emitters, then the radiation back to space is coming from emissions from Earth's surface (direct and absorbed and re-emitted by GHGs throughout the atmosphere, perhaps many times), clouds, and haze.

Anyone can chime in if they have some insight to share. I’ve numbered the points to reduce confusion in a response if there is one.

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Fefetltl said:
And don't forget that there is a difference between saying random sentences and providing the right equations.

LightTopics said:
can chime in if they have some insight to share.
I find it interesting that the reference supplied by @LightTopics has not a single equation in its many paragraphs of hand flapping and waving. Of course heated gases have convection currents!! Heated solids have moving phonons and dislocations !! Let's do physics here.
There are many circumstances where an exponential decay is simplified by truncation to the decay length (optical density, for instance). If it leads to a wrong result then show that explicitly, please, without adding to the general noise level by saying it over and over again.

DrClaude
hutchphd said:
I find it interesting that the reference supplied by @LightTopics has not a single equation in its many paragraphs of hand flapping and waving. ...
And the author of the paper is not spoken of very highly over at Rational-Wiki: Pierre-Marie Robitaille

Nor the journal he published the paper in:

...and pseudojournal alternative science journal Progress In Physics.

hutchphd
OmCheeto said:
And the author of the paper is not spoken of very highly over at Rational-Wiki: Pierre-Marie Robitaille

Nor the journal he published the paper in:
I hadn't researched the author, but what you found about the author does make me question the document's quality. Thanks for finding it and pointing it out.

hutchphd

Due to multiple issues (some of which are in deleted posts), this thread will remain closed. Further questions can be posted in a new thread.

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