I want help (or at least a sanity check) on how to evaluate the reduction in IR escape and cloud reflectivity as function of soot fraction of the clean water droplets cloud. Below is my approximation approach to this, I think, "Ph. D. level problem," but first a few facts graphically presented. Reflectivity Range of Clouds, etc.: Please Register or Log in to view the hidden image!Please Register or Log in to view the hidden image! Note that where warm air is rising even this diagram shows high altitude clouds form. If the since 2005 drought in the Amazon continues, for a few years more, not only will the Amazon be a net source (not sink) for CO2, and thus accelerate the surface temperature increase, but it could burn - making a huge "burp" of CO2 but even worse: high clouds with fine soot – I. e. a great increase in solar heating - throwing more than half of the Earth into the start of at least a local thermal run-a-way process in about 10 days! There are stable states with part of Earth in thermal run-a-way. I.e. with the solar energy absorbed there greater than the IR radiation from that region can send to space. (The excess can be convected to other regions of Earth not in thermal run-a-way conditions.) I believe a large one already exists in the SW Pacific Ocean. Read more about it here: http://www.nasa.gov/centers/ames/news/releases/2002/02_60AR.html I don't think anyone knows for sure, how much solar absorption is required (or IR blockage) before most of the Earth is in "local thermal run-a-away" conditions like that part of the SW Pacific is now. I note that clouds reflect by many (150 or more on average) very small-angle, mainly forward scatterings (not once like a metal mirror) until the photon "random walks" back out of the cloud. So the sunlight may be more than a dozen meters deep inside the cloud before it has net random walked thru angle space by more than 90 degrees to be again headed on its way back out of the cloud. Even after that, it can turn 90+ degrees again and go deeper still into the cloud. Thus, I think /guess, only 1 soot particle for every 500 or so water droplets in the outer dozens or so meters of the cloud's "skin", makes the conversion from high reflectivity to nearly complete absorption. I want to know if that POV, guess is about correct. I think that large forest fires may be the fastest way to serious undeniable global warming problems. Not only would burning much of the Amazon, which is dying as in drought conditions since 2005 and then was actually a source, not sink, for CO2, but ALSO the Hadley (and two other) cells can pump soot up to the high clouds than are the cleanest now with reflectivity of ~2/3. I believe that a very tiny soot fraction can cut the reflectivity in half - down to 1/3 directly. Furthermore in day time, the soot is hotter than the cloud's droplets, so they evaporate (or become smaller) during collision with soot particle. That also lowers the reflectivity and increases the local water vapor concentration - H2O vapor is by far the worst GHG (compared to CH4 or CO2). An approximation solution of this cloud reflectivity vs. soot fraction problem may be feasible without too much work or computer time cost with the following six not completely unreasonable approximations, I think: (1) All drops, the soot included, are spheres of same size, large enough that Mie scattering can be neglected. (2) The distance between successive scatterings is constant (could be the mean free path, but think that if constant the actual value drops outs of the equations.) (3) A one time calculated table of scattering angle probabilities is OK, say with 360 entries for the angle by integer number of degrees. (4) The cloud "edge" is a flat surface, that we call the x, y plain of Cartesian coordinate system with photon approaching it along the +z axis.) (5) The droplet density is uniform insides the cloud and it is so thick that no photon random walks out the "bottom side." (6) We can neglect fact that scattering is sometimes is "up", increases y, (if you take x & z axes to be horizontal) and sometimes it is "down" (decreases y) from the x, z plain. Only change in distance from entry x,y plain (standard step size times cosine of travel angle wrt z axis), of photon from cloud's flat edge surface is important. - I.e. how many scattering on average* without soot would restore z > 0 again? I.e. program just draws random number to add next step away (or back towards) the z=0 entry plain (the x, y plain) in Monte Carlo analysis of many photons to learn the average number of scattering required for escape back to space. (z > 0 again.) If, for example, that turns out to be 1000, then one soot sphere per 1000 water drops, cuts the expected reflectivity in half, I think. One per 500 drops cuts the expected reflectivity by factor of 4, etc. I think, but need to think some more about this. May need to have another random number draw for each scattering to see if absorption, instead of scatter, took place. * Why this is a Monte Carlo study problem.
The probability of different scattering angles seems to be given (and even in table) here: http://iopscience.iop.org/0370-1328/74/1/317;jsessionid=4EA6D4BC08A3D38A3A678F5CE3A9BE6E.c2 but paper is old and I can only read the abstract, part of which is: " If the light scattering by transparent spheres is calculated according to geometrical optics a large part of the scattered flux falls within a cone of half-angle |m - 1| radians around the forward direction, where m is the refractive index of the sphere relative to the surrounding medium.* ... If I(θ) is the intensity scattered at angle theta and ΔE(θ) the flux scattered within this angle, the light externally reflected (I1) and that transmitted without internal reflection (I2) normally predominate in the forward-scattered light. For 0.75 ≤ m ≤ 2 graphical methods of obtaining I1 and I2, and ΔE1 and ΔE2, are given: this is made possible by plotting I1, I2, and ΔE1, ΔE2, against functions of single variables which themselves depend on m and θ. ... The accuracy obtainable is within about 1% ... To simplify the computation, from the approximations given, of I and ΔE respectively two tables show the procedures suggested in each case." This was my first applicable hit - I'll search for one with the data I need, but at least this strongly confirms my statement that forward angle scatter dominates so that photon enters cloud and goes deep inside it with many small angle scatter before it has even random walked by 90 degrees, so in no longer going ever deeper into the cloud (unless it again turn back to go still deeper). * Water's refraction is accurately 4/3 with small dispersion so most of the scattered light is insides a cone with half angle of one third radian or only about 19 degrees. I believe that the typical scatter angle is less than 7 degree as the scatter is strongly peaked at zero degrees and falls off continuously from that. My next post, as a very quick answer to the OP's question, will how many random right or left scatters are needed to accumulate a 90 degree turn and thus at least stop going deeper into the cloud.
Although " simple one dimensional random walks" are usually considered to be constant LENGTH steps from a starting point (usually zero on the number line with steps of length one either to the right or the left equally probable, the "step" could be constant 7 degree angles either clockwise or counter clockwise. As is well known, how far you get from the starting point (the expected move from it) is simple the step size times the square root of the number of steps. Thus to be traveling (on average) 90 degrees from its initial direction via N randomly right or left 7 degree steps we require: 90/7 = square root N = ~13 successive scattering before the photon entering the cloud normal to the cloud surface ceases to go ever deeper into the cloud. Other entry angles at times far from noon at the equator need to be considered too, but this condition is when soot in a high troposphere cloud can most greatly change the GW effect. When traveling parallel to the surface of the cloud, then next typically 7 degree step has a 50/50 chance of sending it deeper into the cloud (or closer to escaping the cloud surface back to space) by S cos(7degrees), where S is the length of the step. S is of course inversely proportional to the density of the cloud, but its value is of no import as if S = 10 instead of 5 (inches or meters, that too is of no import) then when photon has made 13 steps it will be twice as deep inside the cloud as if S = 5, but also get out to the twice as far away surface in same number length 10 steps as if the length steps were 5. Thus as it is of no import, we set S = one, and forget about it. Now we have the more common random walk in 1D space problem but need to compute first the starting point. I. e. on average, how deep inside the cloud the photon was when it had turned its direction of travel by 90 degrees. The first step takes photon cos(7) away from the surface. The second (of 13) steps, if bending the photon path the same way as the first takes photon cos(14) more away form the surface, but if bending path the other way it is again a step of cos(7) deeper into the cloud and has the photon again traveling normal to the surface. Thus the expected penetration of the cloud when second scattering done and the third is just starting is: cos(7) + [cos(7) + cos(14)]/2 = 1.5 x 0.992546 + 0.5 x 0.97030 = 1.97397 or 0.98698 per step I have removed all that followed this point as it had errors. For example, after the 3d scattering the photon is traveling at either 7 or 21 degrees to the surface normal and 7 is three times more probable than 21. Thus I need better notation, than first used to do even this crude approximation to the approximation (six not too bad assumptions) in the PO. That better notation is to use R for a 7 degree bend to the right and L for a 7 degree bend to the left from the photon's prior to scattering direction of travel. For example, the 3d scatter could be made as: RRR, RRL, RLR, LRR and by the four mirror images LLL, LLR, LRL, RLL, but henceforth to avoid list all, only one set, not its mirror image will be listed. Note there is no way that after 3d scattering photon can be traveling at 14 degrees to the surface normal. Thus just as the 4th scattering is starting, the photon is: cos(7) + [cos(7) + cos(14)]/2 + [0.75xcos(7) + 0.25xcos(21)]/2 from the surface of the cloud, on average. Or with cos(7) = 0.992546 we have expected penetration at end of the 3d scatter of: 0.992546 x (1 +0.5 +0.375) + 0.5 x cos(14) + 0.125 x cos(21) = 1.8610 + 0.97030 /2 + 0.125 x 0.933580 = 2.4628+ or 0.820933... per step This is getting complex, but lets consider the sets of bends the 4th scattering can leave the photon path in for the 5th scattering, but not showing the mirror symmetric cases (i.e. show RRRR, but not LLLL). They are: RRRR(a net 28 bend), {RRRL, RRLR, RLRR, LRRR(a net 14 bend)}, RRLL, RLRL, LRRL(no net bend. Also note LLLR is the not being listed as that is the mirror image of RRRL and that a 7 degree net from 4 bends is impossible.) Hence the expected step away from surface made by 4th scattering alone is: [(1/8)cos(28) + (4/8)cos(14) + (3/8)cos(0) = 3/8] = 0.125 x 0.88294759 + 0.5 x 0.97030 + 0.375 = 0.97052, which added to the expected separation from cloud surface the first three scatterings made is: 3.4333 ... or 0.85833... per step. I. e. slightly more than the average per step of three steps, so I must have error in one (or both?) of these last two calculations. Each added scattering should slightly lower the magnitude of the per step separation from the surface. I don't feel like trying to find error now. Plan was than when I knew how deep inside the ~13 steps took photon, i.e. got it traveling parallel to the cloud surface to see how many random walk steps of sin(7) from there got either back to the surface (on average) or twice as deep into the cloud. I. e. crudely half the time one might expect ~26 steps for photon to "reflect" and half the time expect it to be twice as deep inside the cloud and still not "reflected" back to space. I. e. half the time one soot particle in 25 drops or 4% kills the reflection by soot absorption and half the time far lower concentration kills the reflection. This is some progress but perhaps the Monte Carlo approach is not that much harder than this even less accurate "random wall" analysis, if you have access to Monte Carlo code you can modify for this problem.
I approve of this analysis. But don't higher oxygen levels (as opposed to CO2) promote forest fires also? You can't win this game, no matter how dire your climate / scattering analysis prediction might be. I'm reminded once again of the Continental Drift bumper sticker. It undoubtably happens. It's going to be dangerous to those near where it's happening. Not a blessed thing anyone can do about it, other than to tout it to some imagined political advantage, which benefits few other than fear mongering politicians, a not insignificant source of more hot air. Of greater concern to me is the headline that appeared in the BBC Science and Environment news just a few weeks back that indicated a substantial increase in the release of methane from beneath the Atlantic Ocean all along the East Coast of the US. This is exactly the scenario that was predicted to cascade into an environmental nightmare of biblical proportions, in terms of global warming and other detrimental environmental effects on the ocean ecology. NOT ONE PEEP of this new finding seems to have hit the newspapers in the US. Why is that?
{rest removed} as this is NOT to become yet another GW thread. I'm not after approval, but comment on the OP method's errors or how to do some approximations to even that approximation (as I am attempting in post 3, still a "work in progress."
You have my express permission to remove (instead of editing) my post from this thread, Billy T. I wasn't trying to promote a message countering global warming, which I agree with, independent of this thread quantifying the process. Your new book looks awesome.
