Introduction to the problem

I computed the absorption by carbon dioxide in earth atmosphere at $14,7$ µm wavelength (absorption band of CO2 where the earth emission is the strongest by far), taking into account the carbon dioxide density decreasing with altitude, the exact solution with hyper geometric series and I found something which is very close to unity, which is completely wrong obviously, so this is a big issue ! That means an absorption of closely $1$ for $400$ ppm of CO2 without taking into account the other atmospheric phenomena ! Let's look into the details, and I think the major issue is the semi-classical expression of polarizability, which is not pressure-dependent.

Maxwell equations

First to compute absorption, we need to consider, to simply one polarization (we need to compute also the other one and do an average for the transmission but this is not a big deal here). For a plane wave who propagates in $z$ direction, we assume an electric field $E_x(z)$ who obeys to:

$$ \left[ \partial_z^2 + k^2 (1 + f(z)) \right] E_x(z) = 0 $$

with $1 + f(z) = \epsilon(z)$, with $\epsilon$ the relative permittivity of CO2 and $k = 2 \pi / \lambda$. It is obvious that $f(z) << 1$. We can rely permittivity to molecular polarizability by Debye formula:

$$ \epsilon - 1 \simeq N(z) \alpha = f(z) $$

with $\alpha$ the polarizability of one molecule of CO2 and $N(z)$ the number of CO2 particles by unit of volume. We assume the form of $N$ with exponential approximation:

$$ N(z) = N_0 exp(-z/H) $$

With $H \simeq 6$ km. The solution is exact in terms of hypergeometric serie, but we can have a more comprehensive approximation by posing into the propagation equation:

$$ E = A e^{i k z + w(z)} $$

with $e^{w(z)}$ non oscillatory at infinity, $E(0) = E_0$,and assuming (easy to verify) $\partial_z << k$:

$$ w'(z) = i f(z) k/2 $$

We find transmission coefficient:

$$T = \frac{ \vert E(+ \infty) \vert^2 }{E_0^2} = e^{- N_0 \text{imag}(\alpha) k H} $$

We need now the expressions of $\alpha$ and $N_0$.

Value of number of particles of CO2 per unit volume at earth surface, $N_0$

To determine $N_0$, we can use what we know as there is a ratio of $\beta = 400$ ppm of CO2 molecules in all the troposphere. So we can write:

$$ N_0 = \beta N_A \mu_{a} / M_a $$

with $N_A$, $\mu_a$ and $M_a$: Avogadro's number, $\mu_a$ mass density of air at surface and $M_a$ air molar mass.

Now we have to determine $\alpha$, the sensitive point.

Value of CO2 gas polarizability

For this paragraph I used this publication. According to a Drude model we have the following expression:

$$ \alpha = \sum_{i=1}^3 \frac{f_i e^2/m}{\omega_i^2 - \omega^2 - i\gamma_i \omega} $$

Since this expression must hold for a single molecule of CO2, I took the value from the curve fitting in the paper. This as been measure with 1 bar of CO2 pressure at 0°C.

Expression of transmission (or absorption) and problem exposed

Using the polarizability coefficients provided in the paper, I found $T \simeq exp(-51) \simeq 10^{-22}$ at $14.7$ µm for the transmission coefficient, which is way to small ! And that gives an absorption, almost equal to unity, by the way superior to the emmissivity of the whole atmosphere, which is just crazy.

Question, discussion

Is the value of the polarizability right ? Or did I make a basic mistake anywhere else ?

Don't you think the relaxation time $1 / \gamma $ or the oscillator strength are strongly dependent on inter-molecular force (and so on the pressure or density)? I don't have enough info on this. Maybe quantum/statistical physicists may help me to find the right expression of polarizability.

Thanks for your interest in the topic.


1 Answer 1


This YouTube video goes into depth about the real explanation of global warming. Indeed, as you point out, infrared radiation from the earth is always absorbed and reemitted in the atmosphere many times. And infrared radiation from the Earth's surface only makes it a few meters before being absorbed. What it does change with increasing CO2 concentration is the height in the atmosphere at which this is no longer true. In other words - the height at which the probability of being reabsorbed on the way to space is no longer very close to 1. Earth's "effective radiation temperature," and by extension how much power the Earth radiates, is determined by the temperature of that layer of the atmosphere.

The video goes into a lot more detail, especially with how this effect varies with wavelength and how the global increase in CO2 concentration has affected this.


  • $\begingroup$ Thanks I will rewatch the video of Sabine later. The absorption and reemission of CO2 is already taking into account inside the polarizability... which take into account all the light/matter process and we should have the effect you mention inside this one. And come on, in 2023, we are talking about black holes, dark matter... but the polarizability of the CO2 is a blocking topic ? I'm wainting for the quantum physicist response $\endgroup$
    – fefetltl
    Commented Apr 3, 2023 at 12:56
  • 1
    $\begingroup$ So in fact my computations are fully right according to the video, no issue in polarizability... This clearly counter the climate model of 2 surfaces. $\endgroup$
    – fefetltl
    Commented Apr 3, 2023 at 13:35
  • $\begingroup$ Yeah maybe I should have summed it up with something like this: your model looks fine to me, and the results look about right. It's not a catastrophe that the absorption of infrared light from the earth is 99.999...% Climate models are well aware of this - it's just your high school physics teacher and maybe Bill Nye who oversimplified this and gave you a totally incorrect model for how global warming works. $\endgroup$
    – AXensen
    Commented Apr 3, 2023 at 14:43
  • $\begingroup$ Thank you I see thé point, i took the basic model of earth 2 face layer in fact, the basic model equating Stefan law and sun radiation. But according to the video, it is in fact that each layer of gas is considered as a grey body emmitter... That being said, i need to dive inside radiative transfer equation. But the quantity of co2 which can heat up the planet surface is not a tricky problem, we can say so... $\endgroup$
    – fefetltl
    Commented Apr 3, 2023 at 16:18
  • 1
    $\begingroup$ I gave you one point $\endgroup$
    – fefetltl
    Commented Apr 3, 2023 at 16:19

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