0
$\begingroup$

In a first approximation the radiative forcing $RF$ is related to the temperature change by the linear relation $$ \Delta T = \lambda \, RF $$ where $$ \lambda\approx 0.8 \, \frac{{\rm K}}{{\rm W}/{\rm m^2}} $$ is the sensitivity parameter which is stated at https://en.wikipedia.org/wiki/Radiative_forcing .

Is it possible to derive this relation somehow? As far as I'm aware the radiative forcing is the effect evoked by additional greenhouse gases (compared to preindustrial times, I think round about 1750) expressed in terms of the same effect which would have been obtained if the solar irradiance would be increased by $RF$.

The basis is the balance of incoming and outgoing flux $$ \frac{1367 \, \frac{\rm W}{\rm m^2} \cdot 0.7}{4} + RF \approx 240 \, \frac{\rm W}{\rm m^2} + RF = \epsilon \,\sigma (T+\Delta T)^4 \approx \epsilon \, \sigma \left\{T^4 + 4T^3 \Delta T + {\cal O}\left(\Delta T^2\right) \right\} $$ where $1367 \, \frac{\rm W}{\rm m^2}$ is the solar constant, the factor $0.7$ on the left accounts for the $30\%$ albedo of the earth and the factor $\frac{1}{4}$ is due to the fact that the incoming total flux is calculated with the effective surface area $\pi R^2$ while it is averaged over the entire sphere with total surface area $4\pi R^2$.

The RHS is the outgoing radiation due to Stefan-Boltzmann. Here $\epsilon$ represents the efficiency of outgoing radiation actually leaving the surface, due to possible absorbers (greenhouse gases) and subsequent diffuse emission. For this value I will assume something like $\epsilon \approx 0.6$ (but maybe somebody knows something better).

This is also how at least I understand the definition of radiative forcing $RF$ i.e. pre-industrially we have $RF=0$ and $\Delta T =0$, giving rise to $$ T=\left({\frac{240 \, \frac{\rm W}{\rm m^2}}{\epsilon \, \sigma}}\right)^{\frac{1}{4}} \approx 290 \, {\rm K} \, . $$ Plugging in to first order we have $$ RF = 3.32 \, \frac{\rm W}{\rm m^2 \, K}\, \Delta T \qquad \Longleftrightarrow \qquad \Delta T = 0.3 \, \frac{\rm K}{\rm W/m^2} \, RF $$ implying $\lambda \approx 0.3 \, \frac{\rm K}{\rm W/m^2}$.

So my question is: Where does the $0.8$ value come from? Is this picture of $RF$ correct?

Improve this question
$\endgroup$
6
  • 1
    $\begingroup$ The answer is no, not usefully. I think your understanding is correct, but between a simplistic black-body model like this and a realistic calculation there is just an enormous gulf. To give two examples: you've assumed a value for $\epsilon$: this is something that is computed in a realistic model; you've assumed a value for the abledo and this is also something that is computed in (still more) realistic models. Finally the value of $\lambda$ isn't actually known very well and different models compute different values for it. $\endgroup$
    – user107153
    Commented Jul 25, 2018 at 19:00
  • $\begingroup$ Do you know that range of variation for $\lambda$ different models predict? $\endgroup$
    – Diger
    Commented Jul 25, 2018 at 19:06
  • $\begingroup$ No. Every time I try to understand this I get confused because what people actually talk about is $\Delta T$ arising from a doubling of CO2, and this seems to me to involve at least two factors: what RF does doubling CO2 correspond to (and from what base value, I presume 1700 or something?), and then what is $\lambda$. My naive physicist's assumption is that those two parameters are not the same thing in models, but people then talk as if they are and I just get depressed about it and give up. $\endgroup$
    – user107153
    Commented Jul 25, 2018 at 19:15
  • $\begingroup$ The base value is CO2 pre-industrially, or? Wasn't it something like $280 \, {\rm ppm}$ and is now round about $400$ ?? What you are talking about may be methane? Also the doubling in CO2 affects RF logarithmically, or did you mean something else? See also: physics.stackexchange.com/questions/13829/… $\endgroup$
    – Diger
    Commented Jul 25, 2018 at 19:28
  • $\begingroup$ sorry by '1700' I meant 'value in 1700', ie the usual preindustrial (but recent) value. $\endgroup$
    – user107153
    Commented Jul 25, 2018 at 19:30

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.