I am aware that I am losing my audience here, but I'm drafting chapters for a book that will never be published. I trust you are familiar with the various concepts and calculations by now, it's a bit tedious repeating them all.
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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 high it bounces. Then dropping a large ball bearing, made of copper, from 12 metres, onto rubber, and measuring how high that one bounces. The latter will bounce a bit 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? At least one of those things must be true, but some might not be.
Clear so far? Back to the actual topic...
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Q1. What is the Consensus' most compelling evidence for the influence of 'greenhouse gases' on the temperature of the hard surface?
A1. The poster boy (or girl?) for this is the planet Venus.
A few agreed facts:
- if you look at it through a visible-light-telescope, you can see that the surface is white (clouds made of H2SO4*, although we don't know that yet), with an albedo 0.75. If you are Isaac Newton, you can also work out its distance from the Sun, its size and its mass (and hence acceleration due to gravity).
- the 'effective temperature' can then be calculated as ~ 230 K (this calculation is really tricky, but let's accept the result as correct).
- with infra-red-telescopes and actual space probes, we have since measured the actual temperature at the hard surface at 735 K, and know that the atmosphere is 95% CO2. There are about one thousand tonnes of CO2 per m2 surface (as opposed to about 6 kg per m2 on Earth).
- therefore, the Greenhouse Effect on Venus ≈ 500 K (on Earth, the same calculation suggests ~ 33 K).
So far so good, no problems there. You know the 'effective temperature'; how high the visible clouds are (50 km - 80 km. let's take the mid-point to be the 'effective surface' = 65 km); what the main gas in the thick atmosphere is and its specific heat capacity etc. You can then work out likely lapse rate (8.87 m/s2 ÷ 1,126 J/K/kg** = 7.8 K/km) and estimate the likely surface temperature by simply adding lapse rate x altitude (~ 7.8 K/km x ~ 65 km ≈ 500 K) to temperature of 'effective surface' (~ 230 K) ≈ 730 K. Bingo!
But the Consensus skips the basic physics, logic and maths and jumps to "It was the one thousand tonnes of CO2 wot dunnit!"
Q2. Why is this a Diagonal Comparison?
A2. Because they are not comparing like-with-like. They calculate the 'effective temperature' of the clouds and then compare it with the temperature of the hard surface. Of course the hard surface is a lot hotter - because it's much lower down than the clouds (which form the basis of our calculation of 'effective temperature') and there's a lapse rate (temperatures go up as you descend from the clouds). You might as well calculate how high our ball bearing will bounce if you drop it from 10 metres; then drop it from 5 metres anyway and be surprised that the observation does not match the prediction.
What if it turned out that Venus is actually a pale grey planet with only a thin layer of white clouds? Our calculation of the 'effective temperature' (based on what you see through a visible-light-telescope) is the same, but this doesn't need much adjustment for lapse rate x altitude to estimate the temperature at the hard surface.
Hey presto, Greenhouse Effect nearly vanished - like on Mars. Mars serves as a good counter-example. It has hardly any clouds, so the 'effective surface' for which 'effective temperature' (based on low albedo of soil = 0.25 i.e. dark red-grey) is calculated is pretty much at the hard surface. That is why there little apparent discrepancy between 'effective temperature' and hard surface temperature and using the Consensus approach, the Greenhouse Effect is only 6 K. The approach is deeply flawed anyway - see A4. below.
Q3. "Hah!" shouts the Consensus, "So you admit that if there were less CO2, the [hard] surface temperature would be lower? So we are correct - less CO2 => lower temperatures; more CO2 => higher temperatures. So CO2 must be a Greenhouse Gas!!"
A3. Sure, if we just remove a lot of the CO2, the temperature at the hard surface would be lower. The thicker the atmosphere, the greater the Greenhouse Effect and vice versa. But if we replaced the CO2 with the same amount of N2 or O2, the temperature of the hard surface would go up a bit (I think), because the specific heat capacity of those gases is lower than for CO2 so the lapse rate would be higher.
Also, saying "So you admit..." is pure polemic and of no relevance to a scientific discussion. And, if you want cheap shots, there is twenty-five times as much CO2 per m2 on Mars than there is on Earth but officially barely any Greenhouse Effect.
Q4. Isn't it a lazy and logically flawed short-cut to compare 'effective temperature' with hard surface temperature in order to estimate magnitude of the Greenhouse Effect anyway?
A4. Yes, good question and I'm glad you asked.
Let's imagine Venus had no clouds whatsoever, and let's assume that like on Earth, the surface temperature goes up 'a bit' as a result.
We then re-calculate 'effective temperature' based on an albedo of (say) 0.25 (dark) instead of 0.75 (nearly white), which means about three times as much solar radiation being absorbed and converted to kinetic energy (heat); which means the 'effective temperature' you calculate will be a lot higher. So the apparent discrepancy between 'effective temperature' and hard surface temperature will be approximately halved (I haven't done the exact number yet).
Does that mean that the very real Greenhouse Effect has halved (or whatever the exact number is)? Of course not - lack of clouds means the temperatures go up, so there's now slightly more Greenhouse Effect!
The reverse logic applies to Mars which has hardly any clouds and so no apparent Greenhouse Effect (using the flawed Consensus approach to calculating it). Nonetheless, Mars has a predicted lapse rate of about 5 K/km (3.71 m/s2 ÷ 736 J/K/kg - the measured value is half that, apparently). But if it did have a layer of clouds at 10 km altitude (I'm not sure if that's physically possible, but let me illustrate the point) it would have an albedo of 0.75 (like Venus); Mars would only absorb one-third as much solar radiation; and the 'effective temperature' we calculate would fall from 209 K to 159 K ((0.333 ^ 0.25) x 209 K).
So - using my method - we would estimate the hard surface temperature on cloudy Mars to be 158 K plus 10 km x 5 K/km = 209 K. This is a bit less than the measured hard surface temperature of 215 K, which makes sense as the clouds would cool it down a bit.
So now, even though the hard surface temperature of Mars has gone down 'a bit' (and the actual Greenhouse Effect has gone down 'a bit' as well), the apparent Greenhouse Effect would go up from 6 K to 50 K! Even though in reality, not much has changed, which means all these figures - 500 K for Venus, 33 K for Earth and 6 K for Mars are meaningless.
* "Poor Jones is dead and gone,
his face will be no more.
For what he thought was H2O
was H2SO4"
** Actually, this is a slightly circular calculation. You would start by assuming the highest layer below the clouds is ~ 230 K, and the specific heat capacity of CO2 at that temperature is lower (0.763 J/K/kg) so you would predict a higher lapse rate of 11 K/km; so you would have to work your way down, km by km and use a different lapse rate each time until you're down at the surface; then work back upwards again until it is all in balance.
Insults
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