# A Note: Global Warming Threads

Discussion in 'Earth Science' started by Tristan, Aug 27, 2004.

1. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

Messages:
23,198
There is a difference between "intercept" and "absorb." A small part of the solar radiation incident on the Earth is intercepted by the atmosphere and refracted by it but not absorbed. That is why some sun light falls on the moon during a total eclipse.

I hope you will not try to say that light passing thru a prism was not intercepted by the prism. Perhaps you also tell me that the Brits did not intercept German messages in WWII after they had a working Enigma code device as they did not block it, but let it reach the Germans?

Not that it matters, but I will note that relatively little of the blue light passing thru the Earth's atmosphere falls on the moon during a total eclipse of the moon as it is much more strongly scattered (inverse fourth power of the wave length). That is why a time exposure for the moon with color film, shows it quite reddish moon, not a white like the full moon. Some, especially the Vikings, who don't understand this, think the reddish color is due to the lower reflection by the moon's surface for shorter wave lengths* but then are nearly** at a loss to explain why the full moon is not reddish too after the eclipse is over.

That reddish color gave support to the old Viking explanation for the lunar eclipse. I.e. the moon was being eaten by a great wolf. They would beat their swords against their shields to scare the wolf away and that always worked. The red color was due to a little blood. If the wolf ate the moon, then night time raids to pillage England would be much more difficult. So it was every Viking's duty to make a much noise as he could. Thor might even electrocute those who did not do their duty.

* Admittedly they did not put it in these modern terms, but idea was the same.

** Their myths don't go into this. Perhaps the hungry wolf quickly lapped up the blood before leaving. Some the wolfs did get a good meal as all the Vikings were sleeping - You can see the missing pieces we moderns call "craters." It all makes sense, even predictions and was confirmed in many "noise to save the moon" experiments. I don't know if any wise Vikings concluded the wolf had a quite circular mouth or not from the crescent shape of the partial eclipse and the generally round shape of the craters. They did not publish many scientific observations. Another they could have made was the wolf was so frighten by their noise that it was almost immediately sick and vomited back the bite it had taken during the partial eclipse.

Sing with me: Give me that old time religion, "gibe me" that old time religion! It was good enough for Lars and Olsen so it good enough for me.

Last edited by a moderator: Nov 13, 2013

3. ### AndreRegistered Senior Member

Messages:
889
For all practical purposes, yes looking at the caveats that Billy introduced. But I also showed why it is spurious and misleading to use it in the logic of greenhouse effect. Now would you agree that the radiative equilibrium temperature directly under the sun (1370 w/m2), as if you were standing on the dish, at low albedos is close to the boiling temp of water?

5. ### TrippyALEA IACTA ESTStaff Member

Messages:
10,890
In general, yes, for the purposes of answering that specific question, not really.

Where, in that quote, in the question you were answering, does it say anything about the size of the sphere, or the sphere having an atmospher? Answer? It doesn't. I am referring to a solid sphere and a solid disc of the same size and the same properties.

The question you replied to (deliberately) didn't say anything about the earth, so at the point of answering that question consideration of the atmosphere is irrelevant. Even if it had, we would have been considering a disc with identical properties, which has, well, you know, identical properties.

Apologies for editing the post so late after it was made.

I would like to note, however, that BillyT's attempt at using atmospheric diffraction to justify his answer are at odds with the answer he initially gave:
Which was to say that the disk without an atmosphere will absorb more total energy than a sphere without an atmosphere because of non-normal absorption. I also neither challenged nor commented on his commentary around atmospheric effects, it was only his assertion that the disc without an atmosphere would absorb more total energy than the sphere without an atmosphere because of non-normal absorption that I challenged.

Last edited: Nov 13, 2013

7. ### TrippyALEA IACTA ESTStaff Member

Messages:
10,890
Have you listened to anything i've said? I'm not outright agreeing to anything at this point, and I have already said I intend to address your posts so...

8. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

Messages:
23,198
I agree that a disk or a sphere could have (or could not have) a part that is a gas AND that the question did not indicate one way or the other.

However the last more dozen posts have been concerned with radiative transfer in the atmosphere, I did take that to be the alterative you were asking about. If you wanted the other alternative to be understood, then you could have said "Lets take a detour, that has nothing to do with radiative transfer in the atmosphere and just as a first consider if there is a difference in the sun light intercepted by a thin solid disk or a solid sphere of the same diameters." (or words to that effect.)

Then, being careful and trying to be precise, I might have said: Yes, there is more energy absorb by the sphere than by the disk.

This is due to the fact that the 0.5 degree wide sun (seen from the Earth or any 1AU distant point) is ALWAYS with every ray striking the sphere never more than 0.25 degrees form normal at the point on the sphere where it is "high noon," but that is true for the disk only twice each day unless you add, but you did not, that the rotates with a 24 hour period (or what ever period is required to keep the rays from the center of the sun falling normally on the disk. If the disk does not so rotate then 6 hours (or some other number if not 1 AU from the sun) the rays of the sun strike the surface of the disk at 45+ or - 0.25 degrees and thus are more reflected back into space.

As the question was not well posed, In this post I assumed the sun does appear to move in the sky. In post 979, I assumed it did not. That is why one answer is "yes" and the other is "no."

9. ### AndreRegistered Senior Member

Messages:
889
Yes I did read it twice and it just an attempt to build up the case again, step by step. To see where we deviate.

10. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

Messages:
23,198
Not sure what you apologize for, but no problem. Because I am slightly dyslexic, I am often editing my postss for 5 minutes ( or even the next day) I.e. I read what is not correct as if it were and can do so several times before I read what is actually there. With these frequent re-visits ,quite commonly I realize I have not been considering all I should have, so my posts often grow too again even the next day.

11. ### TrippyALEA IACTA ESTStaff Member

Messages:
10,890
Did it explicitly mention an atmosphere? No!
Did it explicitly mention a size? No!
Does an inert sphere 1m in radius have an atmosphere of its own? No!

You even specifically identified this in the reply I was addressing, and your justification for the sphere receiving more energy was non-normal absorption, not atmospheric absorption.

There's a process used in teaching, it involves, essentialy, taking a problem back to the last point of common agreement and working forwards from there. You made assumptions, and you got those assumptions wrong, there is nothing more to it.

No. You need to stop reading things into posts that aren't there. More to the point, when you replied this:
To the question it is clear that you understo

No, they absorb the same total energy.

The more I read this, the less sense it makes.

12. ### TrippyALEA IACTA ESTStaff Member

Messages:
10,890
Good, that is what I intend to do as well, but for now I have no time. Adieu.

13. ### iceauraValued Senior Member

Messages:
25,863
He listens in the same manner as Eliza . You are not in a discussion, with him - you are providing the platform for the generation of responses such as that one there: dodge the question or matter that discredits the most recent claims, repeat bs for which the discrediting was dodged earlier, pose a question designed to lead the thread into some arena in which others are framed as defensive, conned into chasing around to answer a never ending series of irrelevant questions.

Round and round, each turn offering opportunities for repetition of slander and innuendo and attacks on the integrity, character, expertise, etc, of Exxon's enemies.

There is no scientific issue involved, in Andre's posts here. They have nothing to do with CO2 boost global warming itself, in its physical reality.

14. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

Messages:
23,198
Perhaps you don't understand what I mean by "high noon"? Again note my post 979 answer assumed the sun was always at high noon for the disk as well as the sphere. Then the rays are intercepted equally (if no atmosphere) but never more than 0.25 degrees "off normal" for the disk anywhere on the disk, but almost tangent to the surface near the poles of the disk so disk heats more.

In post 986 I assume the opposite - that like the real sun earth case the sun appears to move around the target. The question was not well posed so I have considered all four possibilities with in the poorly posed question. (Moving sun, yes or no + atmospehere, yes or no)

15. ### TrippyALEA IACTA ESTStaff Member

Messages:
10,890
No, I understand what you mean by highnoon. The context of this aspect of the discussion was the method used to calculate the equilibrium temperature of a planet using the Stefan-Boltzman law. The reason for that discussion was because Andre disagrees with the use of the difference between the equilibrium temperature, which is calculated without accounting for the emissivity of the atmosphere, and the observed average temperature as justification for the existence of a greenhouse effect. Andre's stated justification for disagreeing with this is the fact that the equilibrium temperature treats the earth as a homogenous sphere capable of achieveing thermal equilibrium within itself and raidating heat isotropicly. He disagrees with this because this is not how we observe the earth to behave.

This is the context of the question I addressed to Andre. I addressed the question to Andre because, rightly or wrongly, I began to get the impression that he disagreed with the treatment of the earth as a circle of a particular radius to calculate the energy recieved at the top of the atmosphere (it's not actually the top of the atmosphere, the atmosphere behaves as if it has an emitting surface at a specific altitude, and it's the insolation at that level. I was trying to get Andre to explicitly confirm whether or not he agreed with that treatment (he does, it seems) so I could ascertain a starting point for my post in response to him.

It is not the phrasing of the question that is at fault.

Note that intrinsic in the discussion (implied by its context) is the fact that the disc is oriented so that "The sun is at high noon on the disc".

You have provided three or four conflicting answers, all of them wrong.
The correct answer is that under all circumstances a disc, and a sphere with the identical properties, and at identical distances absorb the same total energy. The differences between them are in the distribution of the energy across their surface.

16. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

Messages:
23,198
After the discussion about warm or cold water freezing faster terminated in post 955 then for the next 23 consecutive posts (but with a few brief comment posts interspersed) concerned with some aspect of the GH effect. Including in posts words like: atmosphere, clear sky, IR gases, & their absorption as function of wave length, etc.) After an initial typos Trippy re-state his question here:
Trippy asserts the solar energy given to each* is the same, but I have disagreed as problem is not just one of how much flux fails to continues its original path into space, but one of the relative absorption vs. reflection (or scattering).

At first I assumed that both had atmospheres, due to the 23 prior posts discussing that, but soon Trippy made it clear he was assuming bare disk and sphere. I continued to assume the real sun, 0.5degrees wide at 1AU and realistic absorption coefficient decreasing greatly as the light ray's angle from normal incidence approached 90 degrees. This obviously means the sphere absorbs less energy than the disk does.

I noted however that if the sun angle from the disk normal is not fixed at zero because the sun appears to rotate around the disk like it does around the Earth, then the opposite is true as the sunlight is always falling normally on some part of the sphere. Trippy then clarified that we were to assume the sun remained normal to some point on both sphere and disk.

I, and I think Trippy too, assume that both are geometrically perfect. I. e. no tiny hills, etc. but are smooth at least on the scale of light wave lengths (and of course are completely opaque.)

I still complain that the word “intercept” is troublesome when speaking of “energy intercepted.”An aluminum sphere “intercepts” (gets in the way of) the same flux as a carbon one of same diameter does but scatters most of it back into space instead of absorbs most of its energy as the carbon sphere does.

Thus with Trippy's clarified assumptions (The 0.5 degree wide real sun, 1AU from Earth diameter disk and sphere always remaining stationary on the axis of the disk) Then the answer to the question is that the disk absorbs much more energy (intercepts more energy in spite of getting in the way of the same amount)* than the sphere does due to the rapid decrease in absorption coefficient with increasing angle of incidence as the spot the sunlight is falling on approaches the polar regions of the sphere.

* The thread's subject is about the solar energy absorbed, not about how effectively an energy flux is blocked from continuing its original path. In post 993, Trippy seems to agree that we are concerned about energy absorbed, not just blocked from continuing its original path:
I agree that "heating" too is troublesome word. It can, as I intended when I once used it, mean adding energy to some object, like when evaporating water on a stove at constant temperature 100C, OR when raising the temperature of some object. The first meaning, the one I intended, is fundament and constant temperate while heating does not necessarily mean solid going to liquid, like ice melting at 0C OR water going to steam etc. Only that a phase change is occurring like when body center solid iron is becoming face centered solid iron (or conversely - I forget which is stable at temperatures above this phase change temperature.)

Last edited by a moderator: Nov 14, 2013
17. ### TrippyALEA IACTA ESTStaff Member

Messages:
10,890
I'm not sure why you're doing this, I've told you the context of the question, the context of the question is the discussion between Andre and myself of the validity of using the stefan boltzman law to derive an equilibrium temperature and what that tells us about the greenhouse effect.

Right. Andre proposed we consider the scenario of an atmosphere that is transparent to IR radiation. This scenario has been doing the rounds in a number of blogs run by people opposed to the idea of anthropogenic global warming, and I would like to get back to it, but, you're single handedly derailing that discussion. That is where the discussion started, but, the discussion has moved on since then, it has evolved, we're now talking about a very specific, very important piece of that scenario. First we need to agree on how much energy is intercepted. Then we need to agree on how much of that intercepted energy is absorbed. Then we can begin to discuss what happens to that energy once it has been absorbed. Do you get it yet?

Yes and no. These are considerations in the Stefan Boltzman equation, but they're not relevant to the discussion because the disc and the sphere have identical properties. That's the point. I was making sure that Andre and I were in agreement on the total flux intercepted by the earth, and the treatment of that, before I moved onto more complicated questions like how much of this is absorbed, scattered, or reflected.

In otherwords, this is because you have failed to keep up with the discussion. I've given you the correct context of the question - Andre and I were discussing the Stefan Boltzman law and the relevance of equilibrium temperature in discussions of global warming. Calculating the equilibrium temperature requires knowledge of how much flux is intercepted by the object as well as knowledge of how much of that energy is absorbed or reflected - IE the albedo of the object. I was making sure we agreed on the very first step before we moved on.

As I have stated repeatedly the context of the discussion is the stefan boltzman equation and the equilibrium temperature.

You've confused yourself, and you're confusing the conversation. Here you're talking about Albedo, which is a consideration in the calculation of the equilibrium temperature, but not directly related to the question asked. The disc and the sphere have the same properties. The first step of calculating the equilibrium temperature is to calculate the total flux absorbed which is dependent on the total flux intercepted and the albedo. The total power striking an objects surface is proportional to its cross sectional area and its distance from the emitting body - this is why a sphere and a disc with identical properties intercept and absorb the same amount of total energy, because they have they same cross-sectional area. The disc is the cross-section of the sphere.

No, if they have identical properties, they intercept and absorb identical amounts of total energy. We're talking about total energy integrated across the whole body, not the energy at the surface at a specific location. As I keep saying to you, you're considering the energy distribution across the surface not the total energy absorbed by the body.

First rule of holes. Stop digging. Andre and I were taking the same approach, taking the discussion back to the last thing we could agree on. As a consequence, we've gone right back to the Stefan-Boltzman law, and I was attempting to make sure that Andre agreed on the calculation of the total power absorbed at the surface. In order to calculate the total energy absorbed by an object that is radiatively heated we first need to know how much total energy is blocked by the object. The energy absorbed is a fraction of the energy blocked, the rest is reflected, however, before we can discuss how much is absorbed or reflected at specific locations on the surface, and then finally how that energy is redistributed accross the entire surface.

The simple fact of the matter is that the equation for calculating the power emitted by the sun is $P_{Semt} = 4\pi R^2_s \sigma T^4_s$ and the power intercepted by the earth is $P_{SE} = P_{S emt}\frac{\pi R^2_e}{4\pi D^2}$ in other words, it's the total power emitted by the sun multiplied by the cross sectional area of the earth as a fraction of the area of a sphere with a radius of 1AU because a disc and a sphere with identical properties absorb the same amount of total power because the total power is proportional to the cross sectional area of the sphere which is the same as the surface area of the disc.

Last edited: Nov 14, 2013
18. ### iceauraValued Senior Member

Messages:
25,863
That's not what Andre is doing. Andre is deflecting the thread into a morass of irrelevancies camouflaged in Physics 101 vocabulary. He's happy there, and will stay as long as you abet the "discussion". When you do corner him in error or incoherency, he will change the subject and ask a new question phrased as a challenge, demanding a response.

Do this often enough, you will notice the new question is from a matter already dealt with in just this way, some time in the past - without the part about having been cornered in error or incoherency, as that is to be reenacted. In fact, if past patterns repeat, this very discussion you are having now will recur, in Andre's posts, starting from the very beginning claims of grey-body vs blackbody and nobody paying attention to the diurnal cycle of energy flow and so forth.

In particular, this is a vain expectation:
You will never get to the part about the actual, ongoing greenhouse effect (or anywhere else of thread relevance). You may get as far as a large muddle concerning a theoretical greenhouse effect, if you can get past that "inert" atmosphere he wants to discuss at length as a "null hypothesis".

19. ### AndreRegistered Senior Member

Messages:
889
Right. So the numerical outcome of that calculation, the solar constant, is measured to be around 1361 W/m2 with a few W/m2 variation and the next step would be to see how this results in temperature variation in the diurnal cycle, using Stefan Boltzman, given a completely transparant atmosphere. Right?

20. ### TrippyALEA IACTA ESTStaff Member

Messages:
10,890
look again. You've got R[sup] 2[/sup] on the top line and D[sup] 2[/sup] on the bottom line, both are measured in meters. The units cancel leaving you only with watts . The numerical answer provided by the equation is the total power intercepted by the earth, which is on the order of Terrawatts.

The solar constant is the power density of the sun at 1AU.

21. ### AndreRegistered Senior Member

Messages:
889
Yes, you're right, one step too fast, So that would be 1.740 * 10[sup]17[/sup] Watts, or 174 Peta watts but I think that in the next step we have to take the flux per m[sup]2[/sup] into account.

22. ### TrippyALEA IACTA ESTStaff Member

Messages:
10,890
No, I don't think so anyway. I don't have reference material handy, but, the stefan boltzman calculations consider total power in, apply the conservation of energy and assume that total power out is equal to it, which is where, as I understand it, your objection comes into it. Your objection of course relating to the mechanics of heat distribution and variations in both outgoing and incoming power density at the surface ad a function of the angle from the surface. It's these details that explain things such as the temperature variations on the moon and why they deviate from the equilibriun temperature of the moon.

23. ### AndreRegistered Senior Member

Messages:
889
I don't know, are we approaching one area of misunderstanding here? It's just how you apply Stefan Boltzman when you don't have a real grey/black body. I mean Earth and moon do have a specific heat capacity where for a black body it's is zero, but more importantly, as said before, a black body is a perfect heat conductor where Earth and moon are near perfect heat insolators. That means that the moon's spherical surface does not emit radiation energy uniformly at one fourth of the energy that the disc surface receives. Instead it's boiling hot where the sun is in zenith and bitterly cold at the night site.

Does that make sense?