palerider said:
So lets see that measured observation of energy moving from a cooler object to a warmer one...you claim that the surface of the earth absorbs more energy from the cooler atmosphere than it does from the sun...it should be no problem to provide a measured observation of it happening.
I already gave you an example of exactly that. That question was answered in post #299 and two later posts. Apparently you didn't understand it, or realize the significance. I will reword it to make it simpler for you.
Suppose we do a simple experiment. You have an insulated box with a large hot pipe wrench inside at 200 degrees F. The box is then filled with pure cold nitrogen gas at 0 degrees F. The box is then sealed. Everyone knows the thermal energy of the hot solid will be transferred to the cold gas. The kinetic motion of the atoms of the cold gas must strike the hot solid. If you think the atoms of the cold gas can never hit the hot solid then you have no way of explaining how the gas gets hotter.
The conclusion is that the kinetic energy of atoms of a cold gas must strike the hot solid. That is an example of a two way energy transfer between a colder object and a warmer one.
palerider said:
From the paper I provided:
Figure 6: “Outside the System” view of the radiative interaction. Earth is in fact, on average, cooler than the solar radiative input temperature. With this single physical reality, the need to postulate a radiative greenhouse effect evaporates.
The author is using totally erroneous physics. On page 35 he replaces the idea of energy flux with a black body "radiative temperature", and then starts arithmetic with
temperature(!) in the third paragraph.
That is totally wrong wrong wrong ! The energy flux and temperature are related by the fourth power of temperature. That is not equivalent to doing arithmetic with energy flux. Energy has a conservation of energy law. Temperature does not.
Here is an analogy that anyone should understand. Suppose you are selling loose fertilizer at $1 per cubic foot. A customer comes in and fills a cubical box 3 feet on a side, and then insists he owes you $3 because it is a 3 foot box. Would you fall for that smoke and mirrors? Apparently you did with the paper you cited. He would really owe you 3 cubed = $27. With temperature it is even worse since it is the fourth power law. The customer is confusing weight of fertilizer with a linear dimension in feet. This analogy about a box of shit is also a metaphor on the author's concept.
In the third paragraph on page 35 he tacitly acknowledges the above problem and says, "
But we cannot and do not treat this [solar radiative temperature] as a physically real average ..." The irony is that he does just that.
That leaves a curious question. He assumes 1370 W/mm solar input and a reflective albedo loss half that. And he understands that half the earth is in the shade and another quarter diminished by the cosine law. So he is assuming that 1/4 of the earth surface gets 644 W/mm at any instant in it's rotation. Let's accept his approximation.
Let's follow his example and divide the earth into four equal area sections as in his Fig. 6. The 1/4 of the earth at the zenith
absorbs 644 W/mm and
radiates 396 W/mm. At the 1/4 periphery on the sunny side the solar flux absorption becomes zero (according to the authors approximation) while that same area
radiates 396 W/mm. On the dark side at the bottom left quartile, the solar flux is zero while that area
radiates 396. Same on the dark quartile at the
right: 396.
If you add up all the radiative energy, you have
644 W/mm short wave solar input and 396 x 4 =
1584 W/mm earth long wave output. The author ignores that obvious and huge discrepancy.
If you divide both sides of the flux by 4 you get the average solar input flux over the entire earth is 161W/mm while at the same instant the entire earth is radiating 396 W/mm. That is the same as Trenberth, so the author doesn’t assume anything different than Trenberth. Rather than a flux in Watts per meter squared, the author's units are basically Watts per 4 meters squared for both, input and output flux.
In Fig. 7 the author addresses the legitimate idea of an uneven temperature from day to night.
Let's calculate what that might be. If you use the standard average of temperature of the earth, 14.5 deg C, the SB equation says you radiate 373 W/mm. That is a simple computation which is only 5.8% lower than 396 in the diagram.
However the maximum difference of temperature from a daytime high to a nighttime low at latitude 45 deg is about 10 degrees C. Let's ignore the fact that the temperature is a smooth transition and do a worse case check and assume a worse case of 14.5 + 5 deg C all day to 14.5 - 5 all night extreme, and then average the two resulting fluxes The result is 374 W/mm. So the worst case is that the author's suggestion increases the average earth flux from 373 to 374. That is a 1 W/mm difference or a very small 0.026% correction that Trenberth already did.
Try to remember that the above analysis has
nothing to do with greenhouse gases. It uses measurements of the solar flux input, and compares that to the earth radiation flux output from the SB equation.
How do you and the author reconcile the fact that the earth is radiating over two and a half times what it is absorbing from the sun? Hint: look at Trenberth's diagram.
You do have a sense of humor! That is the first time I've seen it. My bad. It's post 223. 40 posts ago; and repeated a few after that.
