A good friend whose technical credentials i respect recently suggested to me that the reason all the climate scientists agree about global warming is that they are all running the same model that has never been independently developed more than once. I mentioned this to another technically trained friend who said "Wait, you believe in global warming?" Until this point, i have kind of figured that the greenhouse effect was probably real because i (mostly) trust the peer review process and it seems like most of the people with the expertise to do so are saying that it exists. That's not enough anymore. So here is my attempt to model the atmosphere using the good old physics standbys of rough simplification and convenient assumptions.
(This really is a log of my thoughts. Its not the shortest path to the answer. In fact it is much longer than i thought it would be. Calculation 1 is...not my best work. I tried to take a shortcut that cost me a couple of stupid assumptions and ended up not being any shorter. But i found a lot of interesting things along the way so i left it intact.)
First some rules: As of this writing, i have absolutely no training in climate science or geophysics and i have not consulted with anyone in those fields. I will build my model using whatever physics or chemistry seems best to me as i go along. Since the goal is to get a non-expert opinion, i will not reference any text making any claim about climate change. I will source all physical constants from WolframAlpha; if i need another source i will stick to standard reference texts and (inter)national standards offices and i will cite them.
Okay, here we go. The first thing we need is a model for how the Earth heats up and cools down every day. Heat comes from the Sun. I've heard various solar luminosities quoted, but typing (Solar Luminosity)/(4*pi*(1 A.U.)^2) gives me 1368 W/m^2. Radiation that gets blocked by the atmosphere contributes to the planet's heat load but not to luminosities quoted by ground-dwelling solar enthusiasts (who usually estimate 1 kW/m^2), so i'm going to use my number. Since the Sun only illuminates a cross-section of the Earth, i type (1368 W/m^2)*pi*(Earth Radius)^2 and get 1.748x10^17 W or 174.8 petawatts as the incident solar energy. Probably this is high since the Earth is partly reflective, but it shouldn't be radically off.
On the cooling side, i assume the Earth is a black-body radiator with a constant surface temperature. Obviously the poles are colder, but hopefully this cancels out my assumption that they were black-body absorbers. The power radiated per unit area for a black body is the Stefan-Boltzmann constant times the temperature to the fourth power. The area in question is now the surface area of the Earth, 4*pi*R^2. Typing ((174.833 PW)/((Stefan-Boltzmann Constant)*4*pi*(Earth Radius)^2))^(1/4) gives me 278.68 K, which is 42 °F, a bit chilly but certainly a common surface temperature. Looking good so far.
Before going on, i need to know some wavelengths so i can get into a little chemistry. The energy of a photon can be expressed as the Boltzmann constant (k) times its temperature (T) or as Planck's constant (h) times its frequency (f). In addition, the frequency of any wave is its speed (c) over its wavelength (l).
E = kT = hf = hc/l rearranges to l = hc/kT.
A little math on the black-body intensity curve (formalized as Wien's Displacement Law) tells us that a black body radiates most of its power at 3-10 times its thermodynamic temperature with a peak at 5*T. For a 280K planet, this makes the wavelengths of interest 5-17 microns with the peak at 10 microns. This is indeed a good chunk of the middle-to-deep infrared zone. Out of interest, i do the same thing with the surface temperature of the Sun (5780 K) and find that most of the power is in a band around 250-800 nm with the peak at 500 nm. That's the entire visible spectrum plus a bit of the ultraviolet range. Thus, sunsets are red, the daytime sky is blue and we have to wear sunscreen to block UV but we don't worry too much about solar x-rays. That's a good cross-check.
With wavelengths in hand, we now need to see how carbon dioxide changes the picture. This turns out to be tricky. A spectrum taken by Dow Chemical Company in the 1960s (before anyone was thinking about global warming) and kindly digitized by NIST (National Institute of Standards and Technology) shows that CO2 has massive absorption peaks around 15 um and 4.3 um. A 10 cm path through 200 mmHg of CO2 has an absorbance of 0.01 for the infrared light above 17 microns. Between 5 and 13 microns the absorbance is more like 0.005 and right around the peak intensity of 10 microns, the absorbance appears to be zero. This doesn't look good for global warming. CO2 has some absorbance, but it gets progressively smaller the closer you get to the Earth's peak radiation.
Absorbance is kind of a weird unit, so it needs explanation. For a monochromatic beam of light of intensity I_in traveling through a sample, Absorbance = LN(I_in/I_out). The advantage of this arrangement is that if you double the thickness of the sample, the absorbance is doubled rather than having to square a transmission ratio. Absorbance is always positive and higher values mean more blocked light in the way you would intuitively expect.
If the Dow Corning sample were made 1 meter square (but still 10 cm thick), it would contain (1 mole / 22.4 liters) * (200 mmHg / 1 atm) * (0.1 cubic meters) = 1.17 moles of CO2. I want to find out how much CO2 would be needed to transmit 1/e of the light so i can calculate the optical depth of the atmosphere. I_out/I_in = 1/e implies an absorbance value of 1. So, for example, if 1.17 moles/m^2 has an absorbance of 0.01, the optical depth at that wavelength is 1.17/0.01 = 117 moles/m^2. If the sample absorbance is 0.005, the optical depth is 1.17/0.005 = 234 moles/m^2.
So how much atmosphere is there and how much CO2 is in it? The pressure at sea level is 14.7 psi. Most of that is N2 gas at 28 g/mol. Typing (1 atm) / (28 grams/mole) / (1 gee) gives me 369,000 moles/m^2. (Aside: I love that we have standardized on 'gee' as the name for Earth gravity to distinguish it from grams in our shorthand.) If the CO2 concentration were 1 part per million (ppm), we would have 0.37 moles/m^2 of CO2 and we wouldn't be worried. The National Oceanic and Atmospheric Administration (NOAA) says that the concentration in recent years is around 390 ppm with some seasonal variation. (The earliest record is 315 ppm in 1960). This gives us about 144 moles/m^2 of CO2 over our heads (or 116.5 moles/m^2 in 1960), which is about one optical depth for deep IR radiation. Interesting...
*************************************(Calculation 1: Playing Around)*************************************
I need to stop and make a model of the atmosphere as a whole. Oh wait, no i don't! The International Organization for Standardization (ISO) publishes an
International Standard Atmosphere (it isn't free, but
many free calculators are available online). To first order, i will assume that blocking more infrared radiation warms up the entire atmosphere evenly without changing the gradients between layers. This obviously isn't true for large temperature changes, but if this calculation ends in even a 10% change we have much bigger problems.
Since we have some numbers handy, let's test the global warming hypothesis by comparing 1960 with now (~2010). On average, radiation emitted half an optical depth or less away from space escapes. Anything emitted from deeper in the atmosphere is blocked. Adding more CO2 moves the emission layer upward (although not uniformly for all wavelengths). The density profile of the atmosphere ensures that the ground is always much warmer than the stratosphere. As radiation is effectively emitted from higher up, the ground will get warmer. Or at least that's a theory. Let's try that on with some numbers.
I'll work my way up the energy scale, starting from the deep infrared where the optical depth is 117 moles/m^2. Half of 117 is 58.5 moles /m^2. In 1960, deep infrared radiation could penetrate 58.5/116.5 or 50% of our atmosphere. Since atmospheric pressure is determined by the weight of the air overhead, i will look for a height where the pressure is 0.50 atm. This turns out to be 5500 meters above sea level. The calculator tells me that the temperature at this altitude is 252.4 K. In 2010, the radiation point was at 58.5/144 = 0.406 atm. This occurs at 7000 meters where the temperature is 242.65 K. If this were the absorbance across the entire spectrum, each of these levels would get fixed at 278.7 K in their respective years with everything else warming up to match. In this case the Earth's surface would have warmed by 9.8 °C in 50 years. The temperature in 1960 would be 312.4 K = 102.6 °F, so its a good thing this isn't the case.
Oddly, when i consider the region of lower absorbance (depth = 234 moles/m^2), i get roughly the same result. Now the 1960 radiation surface is at sea level (288.15 K on the atmosphere calculator) and the 2010 radiation surface is at 0.812 atm -> 1720 meters -> 277.0 K for a temperature increase of 11.1 °C. So let me deal with the absorption peaks. Assuming a sample absorptivity of 0.2, the optical depth is 1.17/0.2 = 5.85 moles/m^2. This means i'm looking for the temperature at pressures of 0.05 atm in 1960 (20500 m -> 217.15 K) and .0406 atm in 2010 (21810 m -> 218.45 K) for a difference of -1.3 °C. Bizarrely, above 18000 meters (216 K), the temperature starts to rise again. The peaks are actually much more absorptive than this, and the scaling down to a surface temperature doesn't work with an inversion. These regions are going to have a small and unclear contribution anyway, so i'm ignoring them for the rest of this calculation.
This leaves me with some wavelengths with minor absorptivity which would generate a 10 °C difference in the surface temperature and some wavelengths with no absorptivity where adding more CO2 has no effect on the apparent surface temperature. How do i weight them? Since we're really worried about total power emitted, i'll weight each region by the integral of the Planck black body intensity spectrum over the frequencies they represent (Wikipedia "Black Body Radiation" for more details, but i'm treating this as common knowledge) A few minutes with Wolfram Alpha* tells me that for a 280 K planet, 40.0% of the power is radiated at >16.5 microns, 11.4% in the first absorption peak at 14-16.5 microns, 17.6% in the 'greenhouse' region at 11-14 microns, 19.0% in the 'transparent' region at 8-11 microns and 11.6% in the 'greenhouse' region at 4.5-8 microns. That adds up to 99.6%, so i ignore the wavelengths below 4.5 microns. (One possible flaw: the Dow Corning spectrum only goes up to 22 microns; 23.5% of the power is radiated above that wavelength.)
*The actual formula to generate a weighting factor for a-b microns is:
(15/pi^4)*Integrate[x^3/(e^x-1),{x,50/b,50/a}]
(50 microns = 280 K, 15/pi^4 makes the 0->inf integral equal to 1).
Even assuming the deep IR region remains a greenhouse region above 22 microns, i'm still going to assign it an additional weighting factor of (252.4/288.2)^3 = 0.672 because the radiation it emits is at a lower temperature. The third power is used because that's the exponent for radiation per unit frequency. So my weighting factors are now 0.4*0.67 = 0.269 for the deep IR region with a 9.8 °C increase, 0.176+0.116 = 0.302 for the near IR greenhouse region with a 11.1 °C increase and 0.19 for the transparent region with 0 °C increase for a combined weighted average of 7.87 °C increase over the past 50 years. (Housekeeping: If we assume that CO2 is completely transparent to wavelengths above 22 microns, the above calculation gives a 7.36 °C increase)
*****************(Calculation 2: What I Should Have Done From The Beginning)******************
Wow! That's a huge effect. In Ohio, 7.5 °C is the difference between a hard freeze that kills next year's pests and a ruined harvest. I'm not really sure i want to believe a result that large that depends so heavily on a published standard atmosphere and hand-wavy assumptions. Given the data i've already amassed, what if i do a straight power balance?
If the atmosphere is x optical depths thick in a certain range of wavelengths, then it transmits e^-x of the light in that range. That means it absorbs 1-(e^-x) of the light. That energy will be re-emitted fairly quickly. If x is fairly small, half the re-emitted light will go upward to space and half will return to the surface. So the additional power loading on the surface is (P_emitted)*(1/2)*(1-(e^-x)). For large x (the absorption peak), i'm going to assume 2/3 of the energy returns to the surface. One reason for this is that if i divide the atmosphere into many opaque layers and model the radiation between them, each layer sends 1/2 * 1/2 = 1/4 of the radiation from the previous layer back to the surface. 1/2+1/8+1/32+1/128+... = 2/3. Alternatively, i could just assume that a very opaque atmosphere radiates from its coldest point, which the ISA tells me is 216 K. Turns out (216/288)^4 = 0.316 ~= 1/3.
I can already see that this method is going to lead to a very warm planet even in 1960. Up to now, i haven't accounted for anything that reflects visible light back into space which would reduce the overall power that needs to be radiated away as IR. The technical term for diffusive reflection is albedo, but i can't find any references to Earth's albedo that don't ultimately link back to journals about climate science. (Astronomers also use albedo to describe planets, but any measurement of Earth's albedo from space quickly gets snapped up by climate journals, which violates the rules of this exercise.) Since it doesn't affect the question of recent warming, i'm going to pick 0.3 as the Earth's average reflectivity (so 70% visible absorption), which is the most common value i see on the web. One could imagine that changes in surface temperature could change the albedo, creating feedback effects. Warmer oceans mean more reflective clouds, but less reflective ice. Since these effects are very hard to model and presuppose the victory of the warming hypothesis, i will ignore them.
This leaves me with the equation:
(Stephan-Boltzmann)*T_surface ^4 = 0.7*P_incident/Area + (SB)*T^4 * Sum[*f*(1/2)*(1 - e^-x)]
where f is the fraction of energy emitted in bands with optical depth x. Using the values found in Calculation 1, i get the sum as 0.260 in 1960 and 0.285 in 2010.
Terms In The Sum:
Deep IR f = 40% x = 117/117 in 1960 x = 144/117 in 2010
Shallow IR f = 29.2% x = 58.5/117 in 1960 x = 72/117 in 2010
'Window' Region f = 19% x = 0 in 1960 x = 0 in 2010
Absorption Peak f = 11.4% Use 2/3 in place of (1/2)*(1 - e^-x)
Plugging these back into the power balance equation (recall P_incident = 1.748x10^17 W and Area = 4*pi*R_Earth^2), i get a surface temperature ((0.7*1.748x10^17 W/(4*pi*(Earth Radius)^2)) / ((Stephan-Boltzmann Constant)*(1-0.260)))^(1/4) of 274.8 K = 35.0 °F in 1960 and 277.2 K = 39.3 °F in 2010 for a net change of 2.4 °C or 4.3 °F in 50 years. As far as i know, this is mostly in line with current claims by climate scientists. My overall temperatures are a little low, but they're pretty close and i'm ignoring the greenhouse effect for water (which NIST has digitized here), methane and various hydrocarbons. I'm also ignoring tidal action, volcanism and probably a whole host of minor heat sources.
(Housekeeping: If i assume that CO2 is transparent to IR above 22 microns (f = 16.5% for deep IR), i get Sum values of 0.186 in 1960 and 0.2015 in 2010. This give temperatures of 268.3 K = 23.4 °F in 1960 and 269.6 K = 25.7 °F in 2010 for a change of 1.3 °C or 2.3 °F in 50 years.)
***********************************************************************************
I have a vague memory of seeing somewhere in the news a claim that the average surface temperature of the Earth has raised by 1-2 °C since we started recording it. If you're reading this and you want to know the details, you should go talk to a climate scientist; they've spent decades on this while i've spent a few hours. Thinking about other heat-movers, the biggest thing i've left out is convective cooling. Since the vapor pressure of water changes rapidly with temperature, i could imagine convection currents increasing as the ground/sea warms up, carrying heat above the greenhouse absorbers. That would have a stabilizing influence, but "convection bubble driven by warm moist air" is meteorologist-speak for "giant storm". Right now, i'm in favor of anything that causes more rain to fall on the Midwest. In the long run, i'm not convinced that dumping that much power into hurricanes and thunderstorms is a net gain.
If you think the climate scientists are all toeing a party line, please consider me as an outside adjudicator. I am a politically and socially conservative, scientifically-trained Christian who can't possibly have been indoctrinated into any sort of grand cover-up because (1) i disagree with almost all of the political and moral statements made in the name of global warming and (2) i haven't been paying enough attention. I believe the world will end when God is good and ready to end it and not a moment before. However based on the above calculation, i believe that the science behind the greenhouse effect is sound. We will not end the world, but we are changing it.
(Aside: I'm all for arguing over the implications of climate change. For that, please address your complaint to the relevant politicians and activists. It might be helpful to know how a 1 degree change affects various ecosystems, but i have no idea how to model that. Probably its hard to separate the warming component from other variables which are more obviously human-driven like toxins and GMOs.)
I'm curious now about the 'anthropogenic' question. Is the increase in CO2 man-made or natural in origin? The total recorded increase in the carbon content of the atmosphere is (144-116.5 moles/m^2)*(12g/mole)*4*pi*(Earth Radius)^2 = 169 billion tons. (The oxygen was already in the air so its mass shouldn't be counted.) WolframAlpha helpfully tells me that this is "~1.7 x estimated mass of all oil produced since 1850 (upper limit)" although who knows where that information comes from. Looking around the internet, sites like this one seem to agree that the average global oil consumption in the past 50 years is about 3 billion tons per year for a total of 150 billion tons. (I can't find an authoritative source that isn't buried in government-speak, but the exercise is over so i can bend the rules) A lot of petroleum ends up as plastics in landfills, and even the oil that gets burned sheds a little mass as water (which promptly rains out).
On the other hand, we are burning forests at a pretty spectacular rate, which might make up the difference. Conservationists on the Internet seems to agree on "1.5 acres/second" as the current rate of rainforest destruction. That works out to 4.6e7 = 46 million acres per year. A search for timber yield suggests that lumber and paper companies are getting about 100 green tons per acre when they clear-cut U.S. forests. (Unsurprisingly, no one is publishing how much money they're making by logging rainforest.) As a rough guide, this suggests that 4-5 billion tons of rainforest are destroyed each year. That's more carbon than shows up in the atmosphere, but one could imagine scenarios where most of the mass ends up buried or converted to lumber (which is ultimately land-filled) instead of burned. Anyway, it looks to me like we're emitting carbon at a rate at least comparable to the observed increase.
Given the other unexpected coincidences that happened during this exercise, i'm not going to say that means we are definitely the cause of global warming. Apparently plankton absorb about 10 billion tons of carbon from the air each year, then die and carry it to the bottom of the ocean. Since the CO2 concentration continues to rise, there must be other sources that offset this. So i'll say with some confidence that i think additional atmospheric CO2 is causing a global increase in temperature, and with somewhat less confidence that we are the cause of the additional CO2.
Well, that took longer than expected. In the interest of scientific honesty, i'm going to post this before asking a climate scientist to evaluate it. I think the blog format lets me comment on my own posts, so we'll see whether this model is even close to theirs. If so, then yes the model can be independently derived, even by a bumbling physicist like me.
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