I am aware that I am losing my audience here, but I'm drafting chapters for a book that will never be published!
--------------------------------------------------------
In science, as in real life, you are supposed to compare like-with-like.
If you run experiments to see how high a ball bearing will bounce, you are only supposed to change one variable, so...
a) You drop a ball bearing, from the same height, onto different surfaces (concrete, rubber, wood); or
b) You drop the same ball bearing, onto the same surface, from different heights; or
c) You drop a different size ball bearing (made of the same material as the small one) from the same height onto the same surface.
The results tell you
a) How bouncy different surfaces are; or
b) How the drop height affects bounce height; or
c) How size of the ball bearing affects bounce height.
There is no point dropping a small ball bearing, made of steel, from 10 metres, onto concrete, and measuring how it bounces. Then dropping a large ball bearing, made of copper, from 20 metres, onto rubber, and measuring how high that one bounces. The latter will bounce a lot higher, we assume.
But what conclusion can you draw? That large ball bearings bounce higher than small ones? That copper is bouncier than steel? That a ball bearing bounces higher if you drop it from higher? That rubber is bouncier than concrete? It's actually not conclusive.
Clear so far? Back to the actual topic...
-------------------------------------------------------
Q1. Is there compelling evidence for a Greenhouse Effect at the hard surface?
A1. Of course there is! Broadly speaking - for a given albedo and distance from the Sun - the thicker the atmosphere, the higher the temperature of the hard surface (or the surface of the oceans, if your home planet is lucky enough to have vast amounts of liquid water on its surface, I don't wish to offend my few remaining readers from other planets). You can calculate it, you can measure, it's there.
Q2. How do you calculate the amount of the Greenhouse Effect at the hard surface?
A2. First you calculate the 'effective temperature', by working out solar radiation in W/m2, adjusting for albedo - the amount of solar radiation reflected straight back to space - and applying the Stefan-Boltzmann Law. Then you compare that with the actual measured temperature at the hard surface. The warmer the hard surface is compared to the 'effective temperature', the bigger the Greenhouse Effect at the hard surface.
Q3. Why is there a Greenhouse Effect at the hard surface?
A3. Basically, the 'effective temperature' is what it is - at the 'effective surface', which in the case of cloudy planets (like Earth or Venus) is high up in the sky. Don't forget that you take into account the albedo of cloud cover when calculating the 'effective temperature', so that 'effective temperature' tells you the likely temperature of the cloud cover at the altitude of that cloud cover.
For every km that the hard surface is lower than the 'effective surface' (the cloud cover), the temperature increases by 6.5 K/km (Earth) or 7.8 K/km (Venus), aka the 'lapse rate' (easy to calculate, tricky to understand, it's acceleration due to gravity divided by specific heat capacity, yes I know that sounds nuts the first time you hear it, but it makes sense if you think about it long enough).
So you can do like I do and explain it with a few constants and formulas, using basic physics (all that matters is the specific heat capacity of the constituent gases, it doesn't matter if they are classed as 'greenhouse gases' or not), acceleration due to gravity on your planet, several other constants etc.
The maths is then easy.
The 'effective temperature' of Venus at its 'effective surface' (its cloud cover) is +/- 230 K. The 'effective surface' is +/- 65 km above the hard surface. So the hard surface is +/- 65 km below the 'effective surface'. The lapse rate (calculated independently and undisputed) is 7.8 K/km. So the expected hard surface temperature is +/- 230 K plus 65 km times 7.8 K/km = +/- 735 K (which is what the actual measured average hard surface temperature is, unsurprisingly).
On Earth, the 'effective temperature' is +/- 255 K. The 'effective surface' (its cloud cover - on average two-thirds of Earth is covered in clouds) is +/- 5 km above the hard surface. So the hard surface is +/- 5 km below the 'effective surface'. The lapse rate (calculated independently and undisputed) is 6.5 K/km. So the hard surface temperature is +/- 255 K plus 5 km x 6.5 K/km = +/- 288 K (which is what the actual measured average hard surface temperature is, unsurprisingly).
Easy peasy, GCSE-level basic physics and a bit of fun maths!
Or you can explain it using the pseudo-science of 'greenhouse gases'...
Q4. What is the most compelling evidence for the influence of 'greenhouse gases' on the temperature of the hard surface?
A4. The poster boy (or girl?) for this is the planet Venus.
A few agreed facts:
- Venus' atmosphere is mainly CO2, there are about one thousand tonnes of CO2 per m2 surface (as opposed to about 6 kg per m2 on Earth).
- the 'effective temperature'. You calculate this by working out solar radiation in W/m2, adjusting for albedo - the amount of solar radiation reflected straight back to space by the high clouds - and applying the Stefan-Boltzmann Law. The answer is +/- 230 K.
- the actual temperature at the hard surface is 500 K hotter than that.
- therefore, the Greenhouse Effect on Venus is 500 K (on Earth, the same calculation suggests 33 K)
So far so good, no problems there. All stacks up, you get the same end answer as I did in my answer to Q3.
Then the Consensus skips the basic physics, logic and maths and says "It was the CO2, all one thousand tonnes of it, wot dunnit!"
Q5. So, Mr Wadsworth, where's the Diagonal Comparison? What should we be comparing with what? How's it not the CO2 wot dunnit?
A5. [clears throat, steps up to the microphone]
"Because the Consensus is doing a Diagonal Comparison. They are comparing 'effective temperature' calculated on the basis of the color and altitude of clouds at the 'effective surface' with the actual temperature of the hard surface which is at a much lower altitude (and so is much warmer).
Allow me to take a step back...
Let's imagine that all we know about Venus is what we can see through a telescope. So we know the color of what appears to be the surface (very white clouds made of H2SO4*, although we don't know that yet, with an albedo 0.75) and its distance from the Sun (so we know how much solar radiation it gets in W/m2).
How would we work out its temperature of that white-ish surface'? Usual rules apply - work out solar radiation in W/m2, adjust for albedo, do the maths and there's your answer, that's the 'effective temperature'. That is how to work it out, that is what they do. It comes out at 230 K. No dispute there.
Clever scientists and rocket scientists have found out a lot more about Venus since the telescope was invented. The composition and thickness of the atmosphere, the temperature-pressure gradient; the temperature of the hard surface; and so on. Great. They have measured what they have measured.
And measurements show that the top layer of white cloud is indeed +/- 230 K. So a good match between predictions and results. That's a fair comparison, that's comparing like-with-like. Compare expected temperature of something with actual temperature of something, and see if you got the logic and calculations right (which they did).
But...
i. Let's imagine that it turned out that Venus is actually a large pale gray ball (not like Mars which is dark reddish-grey by all accounts) with little atmosphere made of XYZ gas and no clouds (just like Mars) and albedo 0.75. If we calculate the 'effective temperature', we would still get 230 K. And that would still be correct.
ii. Or let's assume that Venus is the same large pale gray ball, but there is a really deep hole somewhere, miles below normal surface level. If we calculate the 'effective temperature' of the normal surface we still get 230 K and that's still correct.
iii. Or let's assume it turned out that Venus core is a smaller pale pale gray ball, surrounded with a thick layer of fluffy marshmallow like pale gray material, with the same XYZ gas atmosphere, with a few wispy pale gray clouds above that, with overall albedo 0.75. If we calculate the 'effective temperature' we still get 230 K and that's still correct.
iv. Or let's assume has a tiny solid core and has an very, very thick atmosphere of XYZ gas with white clouds on top (like a mini-Gas Giant, if that's not a contradiction in terms). If we calculate the 'effective temperature', we still get 230 K, and that's still correct.
In cases i. to iv., the XYZ gas that makes up the atmosphere is exactly the same (compare like-with-like).
In any case, whatever what Venus is actually like below the visible surface (visible with a telescope), what does any of this tell us about 'surface temperature'? That depends on how you define 'surface', how far below the visible layer the surface is etc. You have to go there and measure it.
i. If Venus were a large pale gray ball with little atmosphere, the actual hard surface temperature would be barely more than the 'effective temperature' (like on Mars).
ii. If Venus were the same large pale gray ball [etc] but with one really deep hole, the temperature at bottom of the really deep hole would be noticeably higher than the normal surface temperature.
iii. If Venus were a small, pale gray ball, surrounded by fluffy marshmallows [etc], the hard surface of the small pale gray ball will be the 'effect temperature' plus the lapse rate multiplied by the altitude of the top visible layer = a lot higher than the 'effective temperature'.
iv. If Venus had a tiny solid core with an very, very thick atmosphere [etc], we would find, like on Jupiter and Saturn, that the hard surface temperature is incredibly high. That's just the same 'effective temperature' plus the same lapse rate multiplied by a much higher altitude of the top visible layer = a lot, lot higher than the 'effective temperature.
In cases i. to iv., the constituent gas of the atmosphere is the same XYZ gas, but the measured hard surface temperature would differ. Would we conclude that the atmosphere in case i. is 'non-greenhouse gas'; that in case ii. it is 'a bit greenhouse gassy'; in case iii. there is a 'strong greenhouse effect' and in case iv. there is a 'runaway greenhouse effect'?
I hope not!
Because what matters most is how much lower the hard surface (as arbitrarily defined - is the fluffy marshmallow stuff hard surface or atmosphere?) is than the 'effective surface' (or how far above the hard surface the 'effective surface' is). That is 99% of what dictates the measurable Greenhouse Effect at the hard surface."
[Steps back from microphone and waits nervously for questions].
* "Poor Jones is dead and gone,
his face will be no more.
For what he thought was H2O
was H2SO4"
Debate Magazine
