Bill Illis said:
Thanks sylas, I guess I will have to give up this effort.
I've tried similar kinds of effort myself, and ironing out the bugs can be a great way to learn.
But I still have a problem with the inconsistency in all these numbers.
240 watts/metre^2 of solar energy results in 255.0K or 1.06K/watt/metre^2
That's the effective radiating temperature of the planet. Basically, Earth radiates the same amount of thermal energy as an equally sized blackbody at 255 K. However, you can't give a simple linear relation of 255/240 to mean anything. The relation is Stefan-Boltzman, given by
Q = \sigma T^4
with Q = 240, T = 255, and σ = 5.67*10
-8 as the constant.
3.7 watts/metre^2 of 2XCO2 forcing results in 3.0K or 0.8K/watt/metre^2
That's our best guess, but there are significant uncertainties. This forcing is the difference in energy at the top of the atmosphere. This response to a forcing includes feedbacks.
1.0 watt/metre^2 of extra energy in the SB equation results in 0.3K/watt/metre^2
Right; except that is using the surface temperature, not the effective radiating temperature. You can get a rough approximation by introducing an effective emissivity, so that the radiation to space is related to the surface temperature as follows:
Q = \epsilon \sigma T^4
In this case, T is a mean surface temperature; taking a mean temperature over the Earth so as to give the right energy flux into the atmosphere from the surface. The effective emissivity basically relates the energy radiated from the surface to the energy radiated into space.
The full energy details are more subtle than this, of course. You have energy flowing up from the surface by convenction as well, with the balance made by by a backradiation from the atmosphere. However, the equation with ε gets into the ball park.
With this approximation, we have
\frac{dQ}{dT} = 4 \epsilon \sigma T^3 = \frac{4Q}{T}
Thus the change in energy out the top of the atmosphere per unit temperature is 4Q/T = 4*240/288 = 0.3. This is called the no-feedback response, or Plank response, of the planet.
It is the same temperature response you would have for a grey body with a constant emissivity.
288K at the surface is the equivalent of 390 watts/metre^2 in the SB equation (or 0.74K/watt/metre^2)
adding 1.0 extra watt/metre^2 to 390 watts/metre^2 only results in an extra 0.18K/watt/metre^2
You are just doing a simple division here; which doesn't work. You can use the Stefan-Boltzman law to get about the right magnitudes, as shown above.
I think there has been too much averaging in all these estimates and in the climate models and the incremental differentials are not being used (the equations should be logarithmic).
Well, climate models do the best we possibly can with as much physics as they can handle. Without a climate model, we can use the simple averaging methods described above to get approximate answers, but a climate model looks at the whole distribution of temperature around the planet and all the energy flows. They are not complete by any means, but they do use physically sensible equations for the energy flows that they do represent.
One of the things we haven't looked at is the effect of feedbacks. Climate models can represent various other processes; for example they calculate humidity, using equations like the Clausius–Clapeyron relation and others.
Water, as you may know, is the major greenhouse gas on Earth... but you can't increase humidity by just emitting water into the atmosphere. The lifetime of water in the atmosphere is very short, and so the humidity is determined largely by temperature.
Here's a really simple thought experiment.
Suppose that adding a greenhouse gas gives a forcing, and the non-feedback temperature response is given by 0.3 degrees per unit forcing, as estimated above. But suppose also that any rise in temperature gives additional humidity, with a greenhouse forcing of 2 W/m
2 per degree. This is feedback.
What happens to temperature ΔT when we add a forcing F? The total additional forcing is going to be 2 ΔT + F. The equation becomes
\begin{array}{lll}<br />
& \Delta T & = 0.3(2 \Delta T + F) = 0.6 \Delta T + 0.3 F \\<br />
\Rightarrow & \Delta T & = 0.75 F<br />
\end{array}
The number 2 there is not far off, in fact, as far as we can tell... and there are all sorts of other feedback processes as well which give an additional secondary forcing (either positive or negative) in response to temperature change. If f is the feedback in forcing per degree, and λ
0 is the no-feedback response, then the real response turns out to be
\frac{\lambda_0}{1 - f \lambda_0}
This is the way I look at it and I will delete this post (and probably move on to other issues) if people have a problem with it.
I don't think you need to delete it; but that's up to you.
You can extend this chart all the way out to 63,250,000 watts/metre^2 for the surface of the Sun and the temperature of 5,779K will be correct.
The incremental each extra watt/metre^2 at the surface is only 0.18K. The Sun needs to add 50,000 watts/metre^2 to add 1.0K to its surface temperature.
Actually, the Sun is very close to a blackbody, so it is much simpler. The proper equation from differentiating Stefan-Boltzman gives
\frac{dQ}{dT} = 4 \sigma T^3 = 43,800
Try taking the difference between σT
4 for 5779 and 5780 degrees.
Cheers -- sylas