The scope of this analysis is to illustrate how one determines the magnitude of the greenhouse increment of ΔT≅33 C often cited in discussions of the Greenhouse Effect (GHE). This is the difference between the average surface of the earth and an effective radiating temperature predicted by a very simple radiation balance.
The analysis will not attempt to explain the physical processes that cause of the greenhouse increment, and will even consider the possibility it is not caused by a greenhouse like process. In later posts, we’ll explore various speculations about the cause that often appear on climate blogs.
This discussion will now proceed as follows:
- Illustrate a baseline model for the earth, and derive a formula.
- Compute the effective radiating temperature of the earth,
- Compare to the measured temperature of near the solid/ liquid surface of the earth, which I will call the surface frm now on. and
- List suggested explanations for the difference found in comments at various climate blogs.
The baseline model: Earth as an isothermal sphere.
The baseline model used to predict the earth’s temperature is a rather simple model that shares some rough similarities with the real earth, but also differs in a variety of ways. I won’t justify the assumptions here or explain why they are somehow better than other oversimplifications of the earth’s climate. I’ll simply list them.
The assumptions used in the base model are:
- The earth is assumed be an isothermal sphere with some sort of surface with a effective radiating temperature Te. (Note, this surface is not necessarily the crust we stand on.)
- The only heat transfer between the universe and the earth is a) the earth receives energy from the sun by means of radiation and b) the earth loses energy by radiation to the universe.
- The earth’s surface reflects some of the sun’s energy; the fraction reflected is called the albedo, α
- The earth surface emits as a black body.
- Nothing, including the temperature of the earth, varies with time; the system has reached equilibrium
The temperature of this model earth can be predicted rather easily, we will call that Te, or the effective radiating temperature of the earth.
Derivation of formula to calculate the effective radiating temperature, Te
To calculate the temperature of the earth, we first draw a cartoon to illustrate our model.
If the earth’s temperature is assumed at steady state, then the net rate of energy from the sun, accounting for reflection, must balance the energy emitted from the earth. In terms of variables show in the figure to the left, conservation of energy requires:
Ps – α Ps = Pe or
Where Ps is the rate of energy intercepted by the earth, α is the fraction of the energy reflected by clouds, ice or anything reflective on the earth’s surface and α Ps is the amount of energy emitted by the earth.
The solar constant, S, describe the energy flux from the sun at the earth’s distance from the sun, and from the perspective of the sun, the earth is seen as as a disk with radius R. So, the rate at which the sun’s energy intercepted by the earth is:
Ps= π R2 S .
Assuming the earth is a black body, the full spherical shell of the earth emits as a black body in all directions and obeys the Stefan-Boltzmann law:
Pe= σ (4 π R2) Te4
where σ is the Stefan-Boltzmann constant and is equal to 5.67 x 10-8 Watts m-2 K-4, and temperature Te must be expressed in Kelvins.
Rearranging and canceling, we obtain
Te4 = (1-α) S / (4 σ)
Calculation of Te
To determine the effective radiating temperature, Teneed values for everything on the right hand side of (4). We make the following selections:
- The Stefan-Boltzman constant, σ= 5.67 x 10-8 Watts m-2 K-4.
- The albedo of the earth is estimated to be α=0.30. This is the commonly used value for the earth; it is a function of the amount of cloud cover, and ice cover. So, it could change if the earth warms or cools.
- The yearly average measured value of the solar constant is S= 1368 W m-2 (See NASA.)
Substituting into (4) results in and effective radiating temperature of Te = 255 K = -18 C = -1 F.
This is the temperature predicted by this very simple baseline model. Since the surface we stand on is not an ice covered snowball, it is obvious that this is not the average temperature at the surface where we all live and breath.
Measurement indicate the average temperature near the surface of the earth’s crust and over the oceans is closer to Ts≅ 288K (15C, 59F)
So, the difference between the effective radiating temperature and the temperature near the earth surface is
ΔT≅33 C
Any difference between the real temperature of the earth and this Te=255K are to be explained by something not captured in this model. The accepted, (and quite plausible) explanation is the action of greenhouse gases, which will not be explained here. However, in comments at blogs it has been suggested the earths surface is warmer than predicted because:
- the surface is not isothermal,
- the earth is not a black body,
- convection occurs in the atmosphere
- geothermal heat warms the surface,
- the earth is not at steady state
- other
In later posts, I’ll explore various suggestions as I figure out how, and, hopefully, with tips from visitors in blog comments, eventually provide a discuss the effect greenhouse gases which addresses the various sorts of questions those who are skeptical, curious or merely confused actually ask.
Hi, lucia, a useful addition to the local collection of climate facts.
Here’s a suggestion for you to demolish:
The 255K doesn’t refer to the earth’s surface at all, because the vast majority of the IR photons emitted from the surface are absorbed and re-thermalized at higher altitudes before they can reach the tropopause. It refers to an ‘average’ of the temperatures at those altitudes from which IR photons make it out to the stratosphere without being absorbed or thermalized.
You need wv but you don’t need CO2.
#1, Pat Keating:
Just a refinement:
– For the naked-earth model that is on display here, there is no gas, so the 255 properly attaches to ground-level.
– If you now add gases, the issue is the flux coming from the photosphere of the planet at each individual frequency: these need to be added up.
btw, it’s the “Stefan-Boltzmann law”:
Jozef Stefan and Ludwig Boltzmann were two different people.
lucia:
As you indicate, the purpose of this model is to show what would happen in the absence of the GHE, or to pose the question, What really is going on?
In this connection, another useful display would be the “Chinese metal shells” model that Willis was expounding, that gives an attempted explaining in terms of heat shields. It’s kind of cute; regretfully, its failure to provide a real explanation should be made manifest.
@Neil– Thanks for catching my typo.
I know they were two different people. 🙂
With respect to the model, I think it’s difficult to say what the temperature attaches too. The model obviously doesn’t distinguishe atmosphere from crust, and doesn’t necessarily imply we’ve stripped everything off.
Anyway, 255K is the temperature of the outer surface, but what that surface corresponds to is ambigious in this model.
When we get to models with atmospheres, with layers, where each temperature is starts to become clearer. (Though, even in the over simplified RC model in the little book I have, it’s still not clear!)
It just seems to me all the huge over-simplifications need to be shown first. That way, those who don’t know them can understand where the more complicated discussions fit.
Willis’s model is good. I might need to rig it up to have the heat added in an outer shell though.
595 Neal
I’m not sure of the purpose of your post.
First, I’m a physicist and well aware that Boltzmann’s first name was not Stefan!
My post was in response to lucia’s remark at the end:
Any difference between the real temperature of the earth and this Te=255K are to be explained by something not captured in this model.
In other words, the bare model had already been left behind.
Pat– Neil’s bit about the first names wasn’t aimed at you.
I had a typo and wrote “Stefan-Boltzmann’s Law” instead of “The Stefan-Boltzmann Law”. (I didn’t do the same thing for “Stefan-Blotzmann constant”.
It’s non-conventional to write something like “Prandlt-Meyers” function or “Tollmien-Schlichting’s waves”. I’m not quite sure why Neil would jump to the conclusion that one thought that Meyer or Schlichting first name’s were Prandlt or Tollmien because of this typo, but if Neil did, all the more reason to fix the typo.
Typo’s happen. I fixed it. 🙂
lucia,
So we learn from this model that, roughly speaking, the temperature of the Earth should have temperature in the neighborhood of 255 K. The fact that it does not implies that other mechanisms are taking place.
Thus, the Earth will be emitting most of its thermal radiation in the wavelengths in the general neighborhood of kT/hc, which is 1.38-23 * 255/(6.626e-34 * 3e8) = 56.6 microns. Therefore, any absorption band within spitting distance of 57 microns should be taken into account with respect to radiative-transfer effects on the energy-transport question; anything very far away need not be. So we are talking about the IR region.
So what would happen if there were no molecules with lines in the IR band? It would still pick up heat from direct conduction; and it would not be correct to say that there would be NO interaction with radiation: if there are lower-energy bands, they will be used by the gas to absorb photons.
So one way to think of the cooling problem is by dividing the spectrum into many independent bands, which provide parallel cooling capabilities to the Earth. If a band is free of lines, that cooling capacity is unhindered; but if it has important absorption lines, it will be “blocked” => the same sort of photosphere questions we have discussed elsewhere: a high-altitude photosphere means little power can be transported through that band, whereas an unencumbered band has its photosphere at ground-level. The IR region is important because that’s where most of the Earth’s radiation’s energy is.
Neil. I agree– the earth would have a temperature in the neighborhood of 255K. The model doesn’t imply the temperature would be isothermal– that was an assumption.
We would still have temperature gradients from the poles to the equators, the world would still gain heat on the side that faced the sun. Convection would still occur.
But precisely how that would pan out would required quite a bit of analysis. (After all, you lose the portion of vertical temperature gradient driven by the atmosphere absorbing radiation. Convection tends to smear that out. Also, there is no radiation loss at the top of the planet in the “none absorbing atmosphere” model.
I think the world actually starts to look a little more like jae’s explanation of how the world heats up during the day (which he won’t describe at better length! And, which he won’t flesh out to include the dirt and water on the earth’s surface re-radiating.)
Oh— Neil. I’m doing all the stuff that’s easy for me first. In many cases, that’s answering the repeated questions that don’t explain the greenhouse effect.
But, if you’d like to describe simple models for the radiation that involve analyses of the spectral bands, that would be great. I haven’t done work in that area, and I do want to take the time to answer the questions like “Oh… but the world isn’t isothermal. If we account for that, the greenhouse effect goes away.” Well, no. If we consider the non-isothermal nature, we have to explain away a slightly larger increment. (Say 34K instead of 33K.)
For reasons every researcher and engineer in the world understands, No one answers these, and I think the answers need to be posted.
617, lucia:
I guess I’m working up to it. The challenge is to structure it in more-or-less coherent digestible chunks, so people can understand it, and raise questions on specific points, without wandering around all over the map.
613 was intended to explain (to myself as well as to others) why we talk about GHG vs. non-GHG gases: they are the ones that will affect the most important bands for radiative cooling.
Hhhh… the convention for comments order is confusing! (Old post show oldest first. New posts show oldest last. I need to modify that plugin!)
@Neil– I agree that’s the challenge. That’s why I’m starting at the very beginning. If you start your own blog, decide to send me posts or whatever, the advanced posts can link to the more elementary one.
That’s why that simple radiative balance post is required.
Pat is, of course, correct that the majority of the radiation emitted from the planet are not photons emitted from the surface. They are photons emitted at higher levels. Unfortunately, you can’t just start right out saying that because knowing where they come from pre-supposes you know what the atmosphere does.
What the atmosphere does is the greenhouse effect. The question with regards to the enhanced greenhouse effect (said due to CO2) or the normal effect (due to H20), don’t set aside what the fact that whatever is causing the surface on which we live to be warmer than about 255K shown in the baseline must be explained some how
615, lucia:
I just got an insight from your post about the day/night cycle and spherical Earth: These differences cause temperature differences and pressure differences, that generate convection.
Convection explains why the adiabatic lapse rate should apply.
The traditional instability analysis only proves that the lapse rate cannot be steeper than the ALR, but doesn’t prevent it from being shallower (or even from having a temperature inversion). I was having difficulty coming up with a crisp answer to Gary Moran or gunnar about why I wouldn’t accept that the Earth just shift its LR, but this is the answer: Due to the 3-dimensional aspects, there will be ongoing convection that keeps bringing the LR back to the ALR.
An example of how the insight from 3-dimensions and a time-dependent model can be used to inform the 1-dimensional time-averaged model.
@neil– Yes. One difficulty is the first non-trivial result for the non-radiative atmosphere involves the differential radiant applied at the poles and equator, and the diurnal cycle. Analysis requires us to permit the earth’s surface temperature to vary spatially. The air will then be heated only by convection from the surface. I’m not sure whether the spacial or temporal variations are more important, but it might be possible to do each case one at a time. (The question is whether it’s worth it . The answer to that depends on how many people’s views are based on this issue.)
The problem might be possible to attack as a perturbation problem. If you use a 2D approximation to start (Temperature varies only as a function of latitude),
Step 1: you assuming the surface temperature varies as it would with no atmosphere at all, then you.
Step 2: You then see how much convection this drives, and estimate the rate of heating of the surface, find the new surface temperature. (This will involve some sort of conduction/convection model for the surface of hte earth, with some estimate of heat capacity.)
Step 3: Using new surface temparature, redo 2.
Continue until you have a steady temperature profile as a function of depth into the planet and as a function of elevation off the planet.
This might be do-able. Or, if you wrote down a set of approximations, there might be some other solution possible.
What is certain: Convection will tend to equalize the temperature differences as a result of latitude in this problem. If later one were to do the diurnal problem, convection from the hot side facing the sun toward the cool side will reduce temperature differences. Plus, if the planet rotates relatively quickly, and heat capacity of the earth and will minimize day/night variations causing the solution to approach the problem where latitude only matters.
While interesting as a puzzle, it would be highly unlikely you could publish it. This planet is truly fake. If the problem isn’t a limiting case for something real, where does all the trouble get us?
Another 3-d effect to keep in mind is the Coriolis effect. It isn’t just natural convection that drives the winds.
622 Neal
Now, I’m scratching my head.
The adiabatic lapse rate applies to the situation where there is no heat exchange (that’s what ‘adiabatic’ means). When vertical convection occurs, the correct lapse-rate is the ‘environmental’ lapse rate, which is lower because of vertical convection (in part) moving thermal energy upwards.
#622, Pat Keating:
No the adiabatic lapse rate refers to the case when gas rises upwards or downwards adiabatically = without exchanging heat horizontally: It is a calculated lapse rate. The environmental lapse rate is the actual one that occurs in real life. Since we are doing a model calculation, the appropriate rate is the ALR: You can’t keep changing the model every 5 minutes. See: http://en.wikipedia.org/wiki/Lapse_rate .
#624, Larry R.:
The Coriolis effect needs to be taken into account when considering the generation of air-mass motions. I don’t see that it has a 1st-order effect on global warming – particularly not in a 1-dimensional calculation. In the point I made, the insight that convection will be taking place comes from the 3-dimensional understanding, but it is reflected in the 1-dimensional calculation by setting the temperature profile in accordance with the adiabatic lapse rate.
#623, lucia:
My next step would be to add gases, including both GHGs and non-GHGs, and see what the temperature and density profiles would be. That will determine the radiative behavior.
You can’t take circulation into account in a 1D model, regardless of its cause. But the Coriolis effect does have an influence in that it drives the jet stream, and contributes to tropospheric convection, which will manifest itself in the 1D model as more effective convection. If you are approximating convection as dominating radiation in the troposphere, then I’d agree that there’s no need to consider it specifically. If we had a very good 3-d model, it definitely would decrease the greenhouse effect in some measurable way.
@Larry and Pat– Yes. I think we all agree we can’t account for convection in any detailed way when doing 1-d models of the atmosphere. We also can’t account for coriolis effects if we have a 2-D model only accounting for the effects of latitude and elevation off the surface.
But, obviously, these winds will affect things!
So, in a fully 3D model, we might be able to capture them. (Can we do it right? It might take a good GCM. That gets us back to the whole question of whether or notwe trust GCMs!)
But, I think we can (and others have) learned useful thing about which effects are leading order from 0D, 1D and 2D models. They are admittedly, and obviously imperfect (but in someways, I think that’s good. You can’t hide the warts.)
What I am after is an explanation that is free of the kool aid. In other words, list of things that you just have to accept in order to follow the argument, things that you can’t find clear, convincing, and complete explanations of in texbooks that is.
My goal, and I am pretty sure it will take me a while, and maybe some of you could short cut it, would be to add a dry atmosphere, build the structure first, then start introducing termites like convection, H2O line overlap, etc.
Here is my plan,
– Start with the black body above
– Add a dry atmosphere and numerically integrate the IR absorption at temps and pressures starting at the surface into the stratosphere at some resolution of wavenumber and altitudes low enough that I can manage it with a simple C++ program and high enough to have some physical meaning. Do this for each 10 degrees of lattitude. On the assumption that pressure broadening would be less at the poles, more at the equator.
– Add albedo by lattitude.
– Add solar insolation by lattitude.
– Run the model month by month.
– No real plan after this because I have no idea of how I would add H2O, convection.
If this is completely idiodic, please tell me why. If anybody knows where I can get asorbtion figures for CO2 at various temps and pressures for a fixed lengths of atmospher, or a solid forumla that I could use that takes temp, pressure, and wavelenght into account, that would be great.
– The way the 1-d model takes convection into account is to ASSUME the adiabatic lapse rate.
– I believe it makes sense to apply the ALR (= convection) before working on the radiation problem. The radiative-transport issue depends on the temperature profile of the atmosphere, so why not assume ALR to start with?
Neil– The reason to show the result without convection is to show Dan that, yes, if we permit the atmosphere to persist in a hydrodynamically/ thermodynamically unstable state (as if it were some sort of absorbing solid pane of glass), then yes, accounting for the opacity of the atmosphere does over predict the ground temperature.
The difficulty with the perpetually repeated arguments is that everyone will point to some specific issue and claim that because that’s not discussed, it was overlooked.
It’s easy enough to show both at once.
(Strangely, I got diverted, and I think I just did moderately interesting analysis about climate sensitivity. It may even be moderately original.)
615, lucia:
I would if I knew how. It just intrigues me that the amount of solar energy received in a day in, say, July at 45 deg. N. Lat. is almost exactly what it takes to heat the column of air up to the tropopause by the amount of the diurnal variation. For example, a typical diurnal temperature range for that latitude is 12 C. The amount of air in the air column at STP weighs about 10,000 kg/m^2. 80 percent of that air, or 8,000 kg/m^2 is below the tropopause, which is where 0 heating occurs during the day. Assuming a linear relationship between temperature change and altitude, then the total column would be heated by half the diurnal variation (i.e., the average for the column), or 6 C. One can then calculate the amount of heat input necessary (assume Cp air = 1,000 joules/kg/K): (8,000 kg/m^2)(6 C)(1,000 joules/kg/K)(2.778 e-4 joules/watt-hr) = 13,800 watt-hr/m^2. This is very close to what is actually received at TOA at this latitude in July. In my heat storage “model” I’m just assuming that about the same amount of heat is lost to space during the night by convection and radiation. I’m saying that all gases are greenhouse gases and that the CO2 and HOH simply help thermalize the N2 and O2. Now, folks, please realize that I am just throwing this out for fun and do not consider myself any kind of expert on thermo or greenhouse gases. I may be totally off-base. It’s just an idea that I find fascinating.
#647, lucia:
What I don’t like about unphysical models (such as convection-less gas) is that you are positing something that not only does not exist, but may involve self-contradictory unphysical properties.
When you do that, you can get all kinds of nonsense.
There is a fine art to gedanken experiments: they can be idealized, but not self-contradictory. In my view, convection-less gas is beyond the pale. If you want to talk about panes of glass, let’s talk about panes of glass, not convection-less gas.
More on the heat storage model: On second thought, I don’t need to “flesh out” the amount of heat or radiation coming from the dirt and water. It’s all subsumed in the model, which represents 30-year averages. All the heat from the dirt and water ends up in the air column eventually. (I’m not including oceans here). Also, under static conditions (no flux from the oceans or other air masses) the amount of heat lost at night MUST approximately balance the amount of heat received during the day. Otherwise, it would get hotter and hotter or colder and colder through time. And, of course, that’s what happens when there is a change in temperature due to clouds; less radiation is received, maximum temperature doesn’t get as high, and diurnal variation is less. It balances out. During the transition from summer to fall, less and less energy is received, maximum temperature is lower, and diurnal variation is less.
AMEN to that!
I understand what you mean.
I can call them panes of glass. 🙂
That’s actually sort of like Willis’ solid shell model. By making them panes of glass, I can discuss absorption and separate out the direction of the effects: in a 1 d model, we only have temperature gradients because of radiation. Convection lessens these.
(In a 2 D model, we have completely different types of gradients.)
#653, lucia:
So, what do we have? I guess it’s the following:
– In 1-d, a ground with a constant incoming radiation flux (numerically equal to the 3-dimensional flux averaged over the sphere and over time)
– Somewhere over the ground, a pane of glass that has 0 absorption coefficient except in a few frequency bands in the IR
Is that it? Or am I forgetting something important?
Step 2: What do glass panes do?
In Step 1, we took a “naked Earth” model into 1-d and found the temperature at which the total thermal radiation from the Earth would balance the total radiation from the Sun (suitably attenuated by 1/R^2, and taking into account the spatial average over the Earth with the factor of 4). This was done by integrating over all frequencies, so that we could use the Stefan-Boltzmann law and obtain the well-known naked-Earth temperature of 255 K.
Now I want to “add a bit of glass” to indicate the effect of a frequency-dependent alteration. This is still not the same as the real effect of C-O2, but it leads us to thinking in that direction.
Step 2.1: Glass as a 0-% / 100-% filter:
Suppose we can divide up the electromagnetic spectrum into three broad ranges:
– Range A: Frequencies in which the glass passes 100% of radiation.
– Range B: Frequencies in which the glass passes a variable amount of radiant power, depending on the concentration of substance X. Let’s parameterize this simply: Let x = the normalized transmission (x = 1 if X = 0; x = 0 if X gets large enough). What this means is that if you shine a beam of radiation in range B, if x = 0, none of the beam will get through, so the only radiation you will see coming out at those frequencies will be the thermal emission from the material of the glass itself. Whereas if x = 1, the beam will pass through without any attenuation, and there will also be no thermal emission at those frequencies.
– Range C: Frequencies in which the glass passes 0% of radiation.
It should be clear that range A are the set of frequencies that act like range B frequencies, but with x = 1 permanently, and range C are the set of frequencies that act like range B frequencies but with x = 0 permanently.
In the first run at this problem, I will set aside range B: All frequencies are range A or range C. In particular, I will assume that visible light is in range A, and that the lower frequencies are in range C.
So when we stick the glass above the ground, range A is unaffected: all the radiation from the Sun in the visible range passes through unattenuated; however, the Sun’s energy in the lower frequencies is in range C, and is blocked.
How does this affect the energy flux from the Sun into the glass (and hence into the Earth)? The bulk of it will not be affected, because the higher proportion of the Sun’s energy is in the visible. The range-C stuff will be blocked: this means that its power will go into heating up the glass, which will thermally radiate it both upwards and downwards equally, so half of it will head into the Earth and half of it will head out into space.
So, whereas before the glass was installed, the radiant input to the Earth was:
Solar(range-A) + Solar(range-C)
now the amount that makes it past the glass heading downward is:
Solar(range-A) + Solar(range-C)/2
(By the way, I am not claiming that the range-C stuff power remains in the same frequency band: Once that power has been absorbed, it will be radiated out over any frequencies, depending on the blackbody formula for the glass temperature and on the emissivity of the glass.)
Therefore, the change in radiant input to the Earth is:
Delta-Inward-C = – Solar(range-C)/2
Using the same analysis for the outward radiant power: The range-A stuff will not be affected (but there really isn’t any, because the Earth is not hot enough to generate significant visible light); the range-C stuff will be absorbed, and half of the power will find it’s way out. So, just as in the case of the Sun, the change in radiant output from the Earth is:
Delta-Outward-C = – Earth(range-C)/2
Now, in the pre-glass situation, we had:
Inward = Outward ; so now we have:
New-Inward = Inward + Delta-Inward-C = Inward – Solar(range-C)/2
New-Outward = Outward + Delta-Outward-C = Outward – Earth(range-C)/2
New-Inward – New-Outward = (Inward – Outward) – (Solar(range-C)- Earth(range-C))/2
= (Earth(range-C) – Solar(range-C))/2
The point to notice is that if range C is in the lower frequencies (as much of it is in real life), it is a much bigger portion of the Earth’s radiation budget than it is of the Sun’s radiation budget, because the Sun’s radiation-energy proportions are set by the blackbody radiation at a much higher temperature. Since the total radiation budgets for the Earth and the Sun (at the distance we are at from the Sun) are the same (see Step 1 of this model), since the Earth’s proportion of range-C energy is higher, the quantity
(Earth(range-C) – Solar(range-C))/2
is positive. This means that there is now more radiant energy coming into the Earth than is going out, and this is the radiative forcing.
The result will be that the temperature of the Earth will have to increase until
Earth(T, range-A) + Earth(T, range-C)/2 = Solar(range-A)+ Solar(range-C)/2
In other words, until the temperature has increased until the Earth’s blackbody radiation integrated over ranges A and C = the Solar radiation flux integrated over the same ranges.
Step 2.2: Glass with variable x:
Now let’s add range B. I am assuming that the same parameter x affects all frequencies in range B in the same way and at the same time. We can relax this point later.
Since a frequency in range B has transmission x, that means that the the inward budget due to range B goes from:
Solar(range-B) (this term was previously left out)
to:
x * Solar(range-B) + ((1-x)/2) * Solar(range-B)
The first term: this is what survived of the inward beam in range B. When x = 0, it disappears, as in range-C.
The second term: the power that did not survive (= (1-x)* Solar(range-B)) is absorbed, thermalized, and re-emitted, and half of it (1/2) heads downward.
So the impact of considering range B is that we have to add a term to the delta equations (remember in the pre-glass situation, x would = 1; so that has to be subtracted out):
Delta-Inward-B = ((x-1)/2) * Solar(range-B)
Delta-Outward-B = ((x-1)/2) * Earth(range-B)
(Note that when x = 0, this is like range C; when x = 1, it is like range-A.
Again, we note that since range B is at low frequencies, the impact on the outward budget is larger than on the inward budget, so this again increases the temperature the Earth must reach before radiation-power steady-state is reached. The new full equation for equilibrium would then be:
Solar(range-A) + Solar(range-C)/2 + ((x+1)/2)*Solar(range-B)
= Earth(range-A) + Earth(T, range-C)/2 + ((x+1)/2)*Earth(T, range-B)
This can be simplified if we consider that the transmission through the glass is a function of frequency x(f) in the range [0,1], with x for range A = 1 and x for range C = 0:
0 = Integral[((x(f)+1)/2)*(Solar(f) – Earth(T,f)]
Here Solar(f) is the blackbody function for the Sun (about 6000 K) attenuated by the distance and factor of 4; and Earth(T,f) is the blackbody function for the Earth temperature.
In principle, this is a perfectly determined calculation, once you determine x(f) for the glass. (But if you wanted to test this in the lab, you would have to further take into account that a real glass still has the atmosphere on top of it, so there would have to be a further factor of transmission through the real atmosphere. Or you could replace the Sun by a spectrally well-measured source of radiant energy, and see how the glass affects it.)
However, all I really wanted to show with this argument is that one does, in fact, expect that adding glass will lead to a higher steady-state temperature for the Earth. In brief, the argument is: The glass impedes outward radiation more than inward radiation, because the bulk of the inward radiation is at frequencies unaffected by the glass, whereas a much larger proportion of the upward radiation is affected by the glass.
As you will know, this is not, of course, how C-O2 affects radiation equilibrium. A key point is that this piece of glass is characterized by having a single temperature, whereas a key point in the C-O2 case is that the atmosphere has a temperature profile. This means that the nature of x(f) has to be carefully unpacked. Doing so takes the argument down an entirely different path.