What If There Was No Greenhouse Effect?
by Roy W. Spencer, Ph. D.
The climate of the Earth is profoundly affected by two competing processes: the greenhouse effect, which acts to warm the lower atmosphere and cool the upper atmosphere, and atmospheric convection (thermals, clouds, precipitation) which does just the opposite: cools the lower atmosphere and warms the upper atmosphere.
To better understand why this happens, it is an instructive thought experiment to ask the question: What if there was no greenhouse effect? In other words, what if there were no infrared absorbers such as water vapor and carbon dioxide in the atmosphere?
While we usually only discuss the greenhouse effect in the context of global warming (that is, the theory that adding more carbon dioxide to the atmosphere will lead to higher temperatures in the lower atmosphere), it turns out that the greenhouse effect has a more fundamental role: there would be no weather on Earth without the greenhouse effect.
First, the big picture: The Earth surface is warmed by sunlight, and the surface and atmosphere together cool by infrared radiation back to outer space. And just as a pot of water warming on the stove will stop warming when the rate of energy gained by the pot from the stove equals the rate of energy loss by the pot to its surroundings, an initially cold Earth would stop warming when the rate at which solar energy is absorbed equals the rate at which infrared energy is lost by the whole Earth-atmosphere system to space.
So, let’s imagine an extremely cold Earth and atmosphere, without any water vapor, carbon dioxide, methane or any other greenhouse gases – and with no surface water to evaporate and create atmospheric water vapor, either. Next, imagine the sun starts to warm the surface of the Earth. As the surface temperature rises, it begins to give off more infrared energy to outer space in response.
That’s the Earth’s surface. But what would happen to the atmosphere at the same time? The cold air in contact with the warming ground would also begin to warm by thermal conduction. Convective air currents would transport this heat upward, gradually warming the atmosphere from the bottom up. Importantly, this ‘dry convection’ will result in a vertical temperature profile that falls off by 9.8 deg. C for every kilometer rise in altitude, which is the so-called ‘adiabatic lapse rate’. This is because rising warm air parcels cool as they expand at the lower air pressures aloft, and the air that sinks in response to all of that rising air must warm at the same rate by compression.
Eventually, the surface and lower atmosphere would warm until the rate at which infrared energy is lost by the Earth’s surface to space would equal the rate at which sunlight is absorbed by the surface, and the whole system would settle into a fairly repeatable day-night cycle of the surface heating (and lower atmosphere convecting) during the day, and the surface cooling (and a shallow layer of air in contact with it) during the night.
The global-average temperature at which this occurs would depend a lot on how reflective the Earth’s surface is to sunlight in our thought experiment. ..it could be anywhere from well below 0 deg F for a partially reflective Earth to about 45 deg. F for a totally black Earth.
So, how is this different from what happens in the real world? Well, notice that what we are left with in this thought experiment is an atmosphere that is heated from below by the ground absorbing sunlight, but the atmosphere has no way of cooling…except in a very shallow layer right next to the ground where it can cool by conduction at night.
Why is this lack of an atmospheric cooling mechanism important? Because in our thought experiment we now have an atmosphere whose upper layers are colder than the surface and lower atmosphere. And what happens when there is a temperature difference in a material? Heat flows by thermal conduction, which would then gradually warm the upper atmosphere to reduce that temperature difference. The process would be slow, because the thermal conductivity of air is quite low. But eventually, the entire atmosphere would reach a constant temperature with height.
Only the surface and a shallow layer of air next to the surface would go through a day-night cycle of heating and cooling. The rest of the atmosphere would be at approximately the same temperature as the average surface temperature. And without a falloff of temperature with height in the atmosphere of at least 10 deg. C per kilometer, all atmospheric convection would stop.
Since it is the convective overturning of the atmosphere that causes most of what we recognize as ‘weather’, most weather activity on Earth would stop, too. Atmospheric convective overturning is what causes clouds and rainfall. In the tropics, it occurs in relatively small and strongly overturning thunderstorm-type weather systems.
At higher latitudes, that convection occurs in much larger but more weakly overturning cloud and precipitation systems associated with low pressure areas.
There would probably still be some horizontal wind flows associated with the fact that the poles would still be cooler than the tropics, and the day-night heating cycle that moves around the Earth each day. But for the most part, most of what we call ‘weather’ would not occur. The same is true even if there was surface water and water vapor…but if we were able to somehow ‘turn off’ the greenhouse effect of water vapor. Eventually, the atmosphere would still become ‘isothermal’, with a roughly constant temperature with height.
Why would this occur? Infrared absorbers like water vapor and carbon dioxide provide an additional heating mechanism for the atmosphere. But at least as important is the fact that, since infrared absorbers are also infrared emitters, the presence of greenhouse gases allow the atmosphere — not just the surface — to cool to outer space.
When you pile all of the layers of greenhouse gases in the atmosphere on top of one another, they form a sort of radiative blanket, heating the lower layers and cooling the upper layers. (For those of you who have heard claims that the greenhouse effect is physically impossible, see my article here. There is a common misconception that the rate at which a layer absorbs IR energy must equal the rate at which it loses IR energy, which in general is not true.)
Without the convective air currents to transport excess heat from the lower atmosphere to the upper atmosphere, the greenhouse effect by itself would make the surface of the Earth unbearably hot, and the upper atmosphere (at altitudes where where jets fly) very much colder than it really is.
Thus, it is the greenhouse effect that continuously de-stabilizes the atmosphere, ‘trying’ to create a temperature profile that the atmosphere cannot sustain, which then causes all different kinds of weather as the atmosphere convectively overturns. Thus, the greenhouse effect is actually required to explain why weather occurs.
This is what makes water such an amazing substance. It cools the Earth’s surface when it evaporates, it warms the upper atmosphere when it re-condenses to form precipitation, it warms the lower atmosphere through the greenhouse effect, and it cools the upper atmosphere by emitting infrared radiation to outer space (also part of the greenhouse effect process). These heating and cooling processes are continuously interacting, with each limiting the influence of the other.
As Dick Lindzen alluded to back in 1990, while everyone seems to understand that the greenhouse effect warms the Earth’s surface, few people are aware of the fact that weather processes greatly limit that warming. And one very real possibility is that the 1 deg. C direct warming effect of doubling our atmospheric CO2 concentration by late in this century will be mitigated by the cooling effects of weather to a value closer to 0.5 deg. C or so (about 1 deg. F.) This is much less than is being predicted by the UN’s Intergovernmental Panel on Climate Change or by NASA’s James Hansen, who believe that weather changes will amplify, rather than reduce, that warming.
I haven’t read all the comments, so apologies if this question has been asked (and answered):
All atmospheric text books teach about the vertical temperature gradient – the adiabatic lapse rate – which is a result basically of reducing pressure with height. There is nothing I have read which implicates a greenhouse mechanism driving this. Some planetary atmospheres may not even contain green house gases but still have the same temperature profile.
The problem with Dr Spencer’s article is that he simply makes an assertion that greenhouse gases are responsible for the temperature gradient, without giving the physics to explain it. So, I find myself in the position of not being able to accept a statement on trust, going against everything I believe, without a credible explanation.
wayne (01:08:21) :
anna v (00:14:54) :
You sound more versed in thermodynamics than myself. Honestly. Please help clarify my view if you know.
Real but somewhat a thought experiment. Take a single molecule in empty space at rest, no momentum. A photon at correct frequency is absorbed by that molecule, knocking an electron a shell higher.
Right here we are not talking of thermodynamics, the system you are describing is a quantum mechanical system, where there are potential wells and energy levels. These are nonexistent in thermodynamics.
So lets see the quantum mechanical problem.
I assume that until it drops back the molecule now has momentum.
Energy and momentum have to be conserved: the molecule will acquire some momentum and the photon will kick the molecule up to a higher energy level. A pregnant molecule.
When the electron drops back down, here my big question, seems it must be ejected exactly in the same direction that it was traveling before it was absorbed cancelling the momentum.
No, this is not correct. The decay will have a characteristic time of electromagnetic decays, and the direction will be probabilistic, in a type of antenna like pattern, depending on which state has absorbed the photon. Most often two or three or more, much softer photons come out depending on the energy levels available for the molecule as a whole, contributing to thermalization by the kick back of conserving momentum for each decay.
Again this is the quantum mechanical picture.
To this statistical quantum mechanics has to be applied, behavior of i.e. large numbers, and then one can start comparing the view derived this way to the thermodynamic view.
Is this kind of thought of experiment not a perfect use for a GCM? I’d be surprised that this kind of experiment has not been run in a model already. As more complicated processes such as moisture have been omitted it could even be spun up to equilibrium in a fairly basic model giving a good insight to what kind of weather we would have in this situation and also look at the alternatives such as including water vapour but being IR transparent and other variations. Breaking down the weather system into more basic components and running experiments in models, which could never be seen in real life, does seem a good way to provide insight into the importance of the individual processes and interesting thought experiments at the same time. Obviously the interaction of the individual processes produces many other complexities in reality.
Moderator please bear a long one.
anna v (02:34:06) :
OK! That’s the key. I know if there are many emission lines, the momentum will be split between the photon(s) of lower wavelength and some residual momentum may reside in the molecule. But conservation of momentum must always be preserved. One place or the other(s).
Please follow. Think only of the original parcel of momentum and keep track of it.
I have read literally a hundreds of papers concerning this broad subject. But I kept having a reoccurring catch in all of the pure logic used. Let’s take this as Einstein would, one infinitesimal thing at a time.
Take a special IR photon, properties set by me, carrying energy thus momentum, leaving a spot on the equator and heading directly zenith to keep this pure.
The momentum carried by this photon, one way or the other must be preserved. The photon may be absorbed by a molecule in proximity if wavelength allows, that molecule hits other molecules so momentum of the molecule changes and is distributed, this can occur many times. But mother nature by her physics is so frugal with her accounting of her energy and her momentum that she will not miss the tiniest portion of her precious energy and momentum.
That momentum from that zenith heading photon may be divided and scattered over many molecules or subdivided into components of other photons in various directions but the SUM of the net effects of that lonely photon’s momentum will end up in portions of two states. One component’s portion (can be zero) escapes into space carried by the original OR divided between some other photon(s) in some direction(s) possibly at different wavelength(s). The other possible state is to cause an infinitesimal but real and exact expansion of Earth’s atmosphere by increased momentum of a molecule(s). This will initially manifest as an increased pressure but eventually will be expansion. The sum of the momentum of these two states (speaking only of the original photons momentum), seems to me MUST be equal to the original momentum of that lone photon to meet conservation laws.
If the above is correct, from here on only speaking of that original parcel of momentum. If the photon didn’t encounter any interaction on its way up, it merely escapes into space. End of story, momentum preserved. However, if it had any interaction and the portion of momentum escaping into space (can be zero) is less than the original, the only other possible state is an expansion of the atmosphere by the remaining portion of the original momentum.
If I’m still right to this point, you can now view that photon’s momentum generically as radiative pressure. And any layer of the atmosphere as a near spherical membrane that this momentum must cross, one way or the other to satisfy conservation of the momentum.
The gist of the above is to say that energy, thus momentum, of outgoing photons cannot be absorbed and re-radiated IN A NADIR direction without also expanding the atmosphere, thus, by thermodynamic laws cooling the atmosphere in an equal amount of the heat trapped.
I’m stuck on this point. Reoccurs in my mind every paper I read and until I can get a understandable answer, I don’t feel an added portion of CO2 that traps additional heat can help but to expand the atmosphere thus cancelling any effect. Some papers support this view but from other directions. I tried to make mine simple.
Any help, corrections or comments?
RE: Vincent: 02:17:21) :
I believe the only thing that can force the Earth’s atmosphere to follow the adiabatic lapse rate is an active convection process. Note that this lapse rate stops at the tropopause.
Convection can only occur if the upper air is cooler than the wet-air lapse rate. If the atmosphere were perfectly transparent — no clouds or dust, it could not cool by thermal radiation. In that case, all such radiation must come from the earth’s opaque surface. The condition that Dr. Spencer describes is perhaps a ‘perfect inversion’ condition. The stratosphere would extend down to the surface.
The only mechanism that I can imagine that might force convection in that case would be the movement of modified arctic air toward the tropics, but it seems to me that this process would not make the earth’s surface warmer except in the arctic regions. Just remember that the lapse rate is an upper limit, not a lower limit to the rate that temperature decreases with altitude in a stable atmosphere. We usually think of convection as being forced by heating at the surface, but cooling from above is required to keep the process going.
The last line of my previous post should have read:
We usually think of convection as being forced by heating at the surface, but cooling above is required to keep the process going.
Venus. It has a lot more atmosphere than Earth, wrapped around a planet roughly the same size as Earth.
What nobody has bothered to mention here is one Venusian day is 243 Earth days.
The surface speed at the equator is only about 6.5 kilometers per hour, some humans could actually run fast enough to stay in one spot, assuming said human was wearing an environment suit capable of withstanding molten lead temperatures and 92 atmospheres pressure… 😉
If it had a rotational period quite a lot faster, it’d likely be much cooler. Venus also rotates in the opposite direction from the other planets. There are many theories as to why, mostly involving some sort of large impact, but there’s no evidence of such on Venus’ current surface, which appears to have been all jumbled up and overturned at some point in its history. (Anyone ever bother to think its present condition is because of the impact that knocked it spinning the opposite direction and the hit was so massive the surface disruption completely erased the impact site and the surface hasn’t changed a bit since it solidified post-impact?)
The slow rotation might be a factor in why it’s kept so much atmosphere. Other theories says it’s because Venus doesn’t have a large moon (or any moons at all) to cause strong tides that would’ve long ago stripped away the excess gasses.
The magnetic field of Venus is much weaker than Earth’s. As has been observed on Earth, the field has an effect on the amount of cosmic rays and various types of Solar radiation reaching Earth’s atmosphere and surface. With the weaker magnetic field and being closer to the Sun, plus the slow rotation, no moon… there’s virtually nothing to compare Venus to Earth with respect to greenhouse effect or anything else to do with the atmosphere.
”
Mike Borgelt (23:36:40) :
D. King (21:19:05) :
” Paul (19:34:32) :
wayne (18:51:41) :
Good question about the CO2 and the answer is not so straight forward. Although there are about the same number of molecules of CO2 above any sq meter on Mars as on Earth”
I get about 30 x as many over any square meter on Mars. Would you care to justify your number?
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wayne (23:59:15) :
Mike Borgelt (23:36:40):
Exactly. Paul was saying because the low pressure on Mars, I’m assuming partial pressure, that CO2 on Mars has no effect on temperature or weather at 90% component. I’m questioning why a much lower partial pressure for CO2 on Earth would have any effect either if his statement is correct about
”
Mars has a different diameter and a different mass, hence the gravitational acceleration g is different. It’s atmsophere at the surface is only about 1/100 that of Earth but has 90 to 95% co2. The optical path depends upon the total number of absorbing moleculars of a type in that path. The math will tell you that there is 30 to 40 times the number of molecules of co2 present in the atmosphere of Mars than of Earth. That’s 5 doublings of concentration above and beyond that of Earth. On Earth, a doubling of the molecular count is going to produce an incremental increase of power absorption (in clear skies looking down at an altitude for a straight radiative portion of the problem).
That said, if one looks at the average Martian orbit and its albedo and does a simple incoming radiative calculation for a blackbody and compares it to measured temperatures, one discovers that there is no warming present due to the co2 that is obvious. Whatever exists is basically 0.
The difference being, as mentioned, is that the pressures of both the co2 and other constituents of the air are much lower than lower in Earth’s atmosphere. This is the pressure broadening which smears out the line emissions and absorption likelihoods so that more spectral area is affected. Note that having a high peak and narrow width has little effect compared to a lower peak with greater width when both are pretty much total absorption at the peak after a very small path length.
When one looks at a simple Earth model and accepts h2o in vapor and cloud form being 90% of the ghg effect of 33 degree C rise over a blackbody of similar albedo (0.31), that’s almost 30 degrees C worth and leaves co2 and all the other ghgs a whopping 3 degree C contribution with roughly equal amounts for each time there is a doubling or halving of the total number of a particular molecule’s concentration – as long as that halving doesn’t change so the optical path to extinction becomes large compared to the size of the atmosphere or layers of the atmosphere.
If one looks at clear skies with no clouds, that 90% fraction due to h2o drops significantly and co2 approaches closer to 20% and water closer to 70%, leaving some percentage left for all the other ghg molecules.
Concerning Anna V’s comments, I don’t think you can separate out the radiative thermo from the advent of the quantum mechanics as this is where the classical broke down. Planck’s radiation law introduces that insidious little constant, h, although it is still thought of as being in the classical realm. The blackbody continuum exists for all matter in liquid and solid states (at least sorta). There one has all sorts of interactions between molecules that allow for quantum states ad naseum. Clouds and particulates provide tiny bits of liquids and solids that are no longer limited to only atomic or molecular spectrums but as they get smaller in size, one may not maintain a true BB spectrum – which actually describes the Boltzman distribution or likelihood for atoms and molecules being in a given energy state at a given temperature. If one applies this distribution to the absorption spectra of a molecule, one gets the emission spectra.
With gases that aren’t hot, one has the pressure broadening and a doppler shift broadening associated with temperature. The Sun’s photosphere, compared to Earth’s surface, a moderately decent vacuum at 6000k produces pretty much a BB continuum for that temperature as mostly hydrogen gas wells up rather hot and dumps its radiative load cooling down and returning below that thin “surface” or few kilometers of radiative area.
Cool hydrogen gas is not going to radiate like that. You’ve got the 21cm line based upon flipping spin states but basically, nothing in the visible or IR. In order for the hydrogen to emit in the visible, it must be raised to higher energy states as the electron dropping to the lowest energy state is going to emit in the uV. That’s why (hydrogen alpha) pink hydrogen emitting nebulae exist only where there are bright uV emitting stars nearby. IR (and microwave) tend to depend upon molecular vibrations and rotation modes, unless the atoms or molecules are heated sufficiently to get above the lower energy states.
Just like the qm and classical, while I like Dr. Roy’s presentation and simplifications as an explation, it does have its problems and I’m not sure just how much one can keep that simplified and come conceptually close to describing the system. As for affecting the weather with no ghgs, it certainly would have serious effects. Not having h2o vapor would have serious impacts on the pressure differentials and hence the circulation as well as about everything else from albedo to convection.
If one were simply to drop all but h2o, it’d hardly be noticed, just a few degrees. Maybe that would be enough to trigger an iceage or snowball Earth and then again, maybe it would not without massive extra forcings.
Erm, the atmosphere of Mars is 95% CO2, or nine times the absolute concentration of CO2 on Earth. Since you apparently feel free to insult (wrongly, as it happens), I feel free to mention that I now know, by virtue of your demonstrated ignorance, just how much credence to give anything else you might say in future.
wayne (03:13:52)
That seems plausible but operates counter to the general wisdom that an expansion of the atmosphere may occur but that such an expanded atmosphere is at a higher temperature due to a new equilibrium having been set by the increased length of time the solar energy remains in the Earth system before it is released back to space.
So, one might have an expanded atmosphere but with a higher total energy content than before so a higher temperature will be recorded. You would only see cooling from expansion if the total energy content remained as before but in fact it doesn’t because the increase in the time taken for energy to negotiate the system means that there is more energy in the system to show a higher temperature notwithstanding the expansion.
I hope that makes sense.
Where do all of you get the idea that only CO2 and Water Vapor can absorb and emit radiation? Nitrogen and Oxygen can also perform this trick.
Change your university. Jupiter has minute traces of rock dust, methane and water. In the terrific mass and pressure of Jupiter’s atmosphere, even a minute trace amounts to a huge amount. You may have noticed that Jupiter’s atmosphere is not transparent. It absorbs radiation and therefore also emits it. The planet’s core is about 20,000C, three times hotter than Earth’s. Without wanting to divert a year or two to study the physical processes in Jupiter’s atmosphere, it is nevertheless clear that you haven’t got a professor-stumping question there.
Wayne, I think your basic analysis of the individual photon’s momentum is correct, but it is already taken account of in saying that the atmosphere heats (which means expansion). It is often the case that there are numerous ways of looking at the same thing in physics, especially a ‘bulk’ way (thermodynamics) and a microscopic way (looking at actions of individual particles). Since thermodynamics arises out of the statistical properties of ensembles of particles, these two ways of analysing a problem are exclusive: you can’t take an argument about one photon or molecule and import it into a thermodynamic argument.
Heat is a thermodynamic property of an ensemble; individual particles have mass and velocity. Heat is uncoordinated (statistically randomised) velocities of a large number of particles. Thus from about your third-last paragraph you go off the rails by confusing the microscopic reasoning with the statistical thermodynamic reasoning. Also, nothing in either argument shows the claim in your third-last that the cooling will be to an equal amount. It won’t. The air will heat, expand, then convect (rise) and thus cool.
wayne (03:13:52) :
Moderator please bear a long one.
anna v (02:34:06) :
OK! That’s the key. I know if there are many emission lines, the momentum will be split between the photon(s) of lower wavelength and some residual momentum may reside in the molecule. But conservation of momentum must always be preserved. One place or the other(s).
Please follow. Think only of the original parcel of momentum and keep track of it.
Conservation of momentum is not the same as conservation of energy, in that momentum is a vector and can add up to 0.
We are in the quantum mechanical framework: once the photon is absorbed, the momentum conservation is taken up by the molecule and it is riding happily away at a constant velocity, until it interacts with another molecule or force. Conservation of energy from the same interaction says that there is potential energy in the molecule, the energy left over from the one that became kinetic to conserve momentum.
Take a special IR photon, properties set by me, carrying energy thus momentum, leaving a spot on the equator and heading directly zenith to keep this pure.
The momentum carried by this photon, one way or the other must be preserved. The photon may be absorbed by a molecule in proximity if wavelength allows, that molecule hits other molecules so momentum of the molecule changes and is distributed, this can occur many times. But mother nature by her physics is so frugal with her accounting of her energy and her momentum that she will not miss the tiniest portion of her precious energy and momentum.
That momentum from that zenith heading photon may be divided and scattered over many molecules or subdivided into components of other photons in various directions but the SUM of the net effects of that lonely photon’s momentum will end up in portions of two states. One component’s portion (can be zero) escapes into space carried by the original OR divided between some other photon(s) in some direction(s) possibly at different wavelength(s). The other possible state is to cause an infinitesimal but real and exact expansion of Earth’s atmosphere by increased momentum of a molecule(s). This will initially manifest as an increased pressure but eventually will be expansion.
See, this is what I mean about mixing up systems. Pressure is a thermodynamic variable, entering into the PV=RT formula. In the quantum mechanical frame the excess momentum of one or n molecules will become excess kinetic energy which after millions of similar interactions will appear in the quantum statistical formula which eventually at the limit will manifest as temperature and pressure in the medium, gas in this case.
The sum of the momentum of these two states (speaking only of the original photons momentum), seems to me MUST be equal to the original momentum of that lone photon to meet conservation laws.
No. Once a photon has been absorbed by raising the energy level, rotational vibrational or whatnot, of the molecule, it disappears. Photons are not conserved, there is no photon conservation number. Lets keep it simple , one photon hits one molecule and is absorbed, and the molecule gets a kick in the direction of motion of the photon. The deexcitation of the molecule, suppose it happens with one photon for simplicity, will introduce a new conservation of momentum. The molecule will go one way and the photon the other, but the direction will have no connection with the original photon’s direction. It will be random in direction within a probability distribution given by the state it occupies, and the molecule will kick in the opposite direction. The original momentum conservation law was satisfied by the increased velocity of the molecule that originally absorbed it. After deexcitation,the molecule will have the vector sum of the original momentum plus the opposite momentum to the exiting photon.
If the above is correct, from here on only speaking of that original parcel of momentum.
Well, it is not.
If the photon didn’t encounter any interaction on its way up, it merely escapes into space. End of story, momentum preserved. However, if it had any interaction and the portion of momentum escaping into space (can be zero) is less than the original, the only other possible state is an expansion of the atmosphere by the remaining portion of the original momentum.
Once again, expansion and atmosphere are macroscopic thermodynamic concepts and variables. In this large thermodynamic collective, the increase in the average velocity of molecules manifests as temperature and is connected to pressure through PV=RT in gases.
If I’m still right to this point, you can now view that photon’s momentum generically as radiative pressure. And any layer of the atmosphere as a near spherical membrane that this momentum must cross, one way or the other to satisfy conservation of the momentum.
You are ignoring the fact that infrared photons coming from the ground up are neither coherent nor unidirectional, particularly as the surface of the earth is fractal, though one could see a radiation front forming, if there were no atmosphere, which could be described as having radiation pressure, I do not see what conservation of momentum has to do with it.
The gist of the above is to say that energy, thus momentum, of outgoing photons cannot be absorbed and re-radiated IN A NADIR direction without also expanding the atmosphere, thus, by thermodynamic laws cooling the atmosphere in an equal amount of the heat trapped.
Well, the last time I solved a thermodynamics problem was in 1960, and though it is true that the temperature of the gas will rise , and from PV=RT the pressure or the volume will grow, I cannot tell off hand how the “cooling” and “equal amount of heat trapped” will figure in this.
RE: kadaka (23:28:44)
“To me, he seems to basically be describing Venus, which although it has CO2 in abundance it does not really function as a GHG in that planet’s atmosphere as CO2 does in ours, you could replace it with nitrogen for about the same effect.”
I believe you are missing the point. Dr. Spencer is describing a planet with a *perfectly transparent* atmosphere — no greenhouse gasses, no clouds, and no dust in the upper atmosphere. Water vapor is one of the most effective greenhouse gasses. It also acts as a natural refrigerant in the lower atmosphere.
I do not think Venus can be said to have a transparent atmosphere. Even though the atmosphere of Venus is 97% carbon dioxide, I suspect that most of the greenhouse effect there is due to sulfuric acid in the atmosphere which also causes the optical haze preventing our view of the surface. H2SO4 is a relatively complex molecule with many more possible quantum energy states than CO2. (I wonder how much anthropogenic H2SO4 is being added annually to our atmosphere.)
Earth sans Greenhouse=Earth sans Al Gore…Obvious!
I think this exercise shows the difference between engineers and theorists. Engineers have a great deal of trouble letting go of so many real world issues. That is very good, however, I find the difference rather enlightening. 😉
Oh no… someone posted a link above (http://theevolutioncrisis.org.uk_testimony2.php) where one can read Dr Spencer’s less than enlightened views on evolution and the historicity of the Biblical Gospels, among other things. I must admit I was quite shocked by what I read. I can’t avoid being sceptical of anything he writes on climate science from now on, even if it isn’t related to his religious/anti-evolutionist views.
“John A (16:34:15) :
[…]Unfortunately Willis’ “Steel Greenhouse” is unphysical rubbish. Putting hot food in a Dewar flask does not cause the food’s temperature to rise – it simply reduces the rate at which the food cools.”
I didn’t see this comment… but a word of defense is necessary. First, Willis postulated an energy source inside his imaginary planet to explain surface radiation, he used radioactive decay as an example. Your analogy with the hot food storage is invalid. Please read it properly.
Second, i don’t care much about the peculiarities of his entire setup – what was important for me was his explanation of the Stefan-Boltzmann law which defines the relationship between surface temperature and upward radiated flux, and most importantly the 50:50 splitting up of this flux by the steel shell / GHG shell above a greenhouse-affected surface.
His steel shell experiment explains in clear terms that it doesn’t matter of which material your shell is made when it completely absorbs and re-radiates a certain part of the spectrum; it will re-radiate to the inside and to the outside of the shell in equal parts. And here Willis’s Gedankenexperiment meets nicely with the work of Ferenc Miskolczi. And that’s why it’s important.
I don’t want to be brash but i think it’s important to understand this if you want to understand the thermodynamics and the energy balance. Because it deflates the feedback / tipping point voodoo claims made by the AGW proponents.
For all latecomers here’s the link to the Steel Greenhouse writeup again.
http://wattsupwiththat.com/2009/11/17/the-steel-greenhouse/
Re: bob (19:17:07) :
“RE: Alex Harvey (16:39:03) :
Alex, all matter emits blackbody radiation, although you are right that gases are not blackbodies.
Some poster here seem to confuse a substances Infared emmision spectrum with its absorption spectrum, they are two separate and different things.”
Thanks, this has to be answered with some care. It is true that absorption and emission spectra for the same molecule are different things, but that is not to say that a molecule can emit radiation at frequencies that it can not absorb radiation or vice versa.
There is thought experiment that covers this, of which I cannot recall all the details, but basically it would allow for heat to flow from a colder bodies of gas to warmer bodies, if I recall correctly.
Also the spectra come into alignment when radiative equilibrium is attained.
The situation with mixtures of gases is a little more complicated as radiation absorbed by CO2 can be passed thermally to H2O and emitted at frequencies that CO2 simply could not.
But finally, if anyone is saying that a gas like CO2 can emit radiation with a continuous spectrum, such as a black body does, then that is not the case.
It is true that the black body spectrum is important when calculating the emissions from a gas, as the temperature of the gas changes the relative (and absolute) intensities of the radiation at different frequencies. But that is not the same as saying it radiates like a blackbody does. Not in my book anyway.
Alex
While the notion of putting food in a good insulator will reduce the time it takes for the food to cool, the Earth has two sources of heat continually applied. The big one is the Sun, the lesser one is the geothermal heat. If you increase the insulation for the Earth, it will heat up from either of these sources. If somehow you created a perfect insulator – even if the Sun totally went away, the geothermal heat would raise the surface to thousands of degrees and the whole Earth would reach true thermal equilibrium – and it would be above the temperature of molten lava everywhere. Right now, the heat flow at all points above the interior of the Earth is such that there is a large temperature differential from the inside to the outside. If ultimately enough insulation could be used to bring the heat flow down to zero, that temperature differential would decrease to zero and the temperature would average out, to thousands of degrees.
Gregg E – I think the explanation is that the rotation rate of an inner planet was determined by the final few collisions of larger pieces that created the planet. There is a theory that a Mars-sized world collided with early Earth creating the moon.
Venus’ atmosphere contains four times the nitrogen of Earth’s. If both planets had similar origins, then the early Earth collision theory explains the loss of 3/4 of Earth’s primordeal atmosphere. The CO2 on Earth was removed by chemical (and later biological) action in the oceans. Venus was presumably too hot to have oceans so the CO2 stayed in it’s atmosphere. The slow rotation of Venus would have made the sun-facing hemisphere even hotter.
Note also that Venus’ rotation rate is phase-locked with Earth, such that the same face of Venus always faces the Earth when Venus and Earth are closest in their orbits.
Mike Borgelt (00:05:29) :
Paul (19:34:32) :
“Good question about the CO2 and the answer is not so straight forward. Although there are about the same number of molecules of CO2 above any sq meter on Mars as on Earth, they are not nearly as effective as a greenhouse gas because the total atmospheric pressure is much lower. ”
I believe you are flat out wrong about this:
Assuming a Mars surface partial pressure of CO2 of 6hPa on average (this is at the low end of the variable surface pressure on Mars and 95% of the atmosphere is CO2) and Earth at 1013hPA of which the partial pressure of CO2 is 0.58hPa (390ppmv and CO2 has 1.51 times the molecular mass as oxy-nitrogen) this would make the number of molecules of CO2 above each square meter of mars TEN times that over each square meter on Earth, however pressure is WEIGHT and Mars has only 0.38 the surface gravity of Earth so you need 2.63 times as much mass to get the pressure so we get at least 26.3 times as many molecules of CO2 over each square meter of Mars as on Earth. And that was the low end of the martian surface pressure so around 30 x seems reasonable.
And yet the greenhouse effect on mars is “feeble”?
As for this:
“The higher density of gas in Earth’s atmosphere works to spread the absorption of IR by CO2 over more of the wavelength scale. The same amount of CO2 (or H2O) is more efficient on earth because of all the N2 and O2 colliders (this is called pressure broadening. There is a reasonable wiki page http://en.wikipedia.org/wiki/Spectral_line but it could use to be updated to discuss the role of broadening on climate.”
OK great, but isn’t one of Gavin’s arguments at RC that extra CO2 in Earths atmosphere causes extra pressure broadening? Just how much extra pressure broadening do you get by doubling CO2 from a partial pressure of 0.44hPa to 0.88 hPa(corresponding to CO2ppmv going from 290 to 580ppmv) out of a total pressure of over 1000hPa?
I’m really pleased you’ve cleared that up that it is the TOTAL pressure that causes pressure broadening of absorption spectra minor constituent gases of the atmosphere like CO2 and not the partial pressure of CO2.
Here’s a comparison between CO2 spectra at earth and martian conditions, note the extra broadening:
http://i302.photobucket.com/albums/nn107/Sprintstar400/Mars-Earth.gif
Spector,
Thanks for your explanation. I think the idea that a perfectly transparent atmosphere cannot cool by thermal radiation would probably account for the peculiarity of a zero lapse rate. I had assumed that the top of atmosphere must be continually cooling by radiation into space and must be colder than the ground, but obviously not.
“John A (16:34:15) :
[…]Unfortunately Willis’ “Steel Greenhouse” is unphysical rubbish. Putting hot food in a Dewar flask does not cause the food’s temperature to rise – it simply reduces the rate at which the food cools.”
DirkH has already commented on this, but I would just add that to make a correct analogy with Wilis’ steel shell, the flask would need to contain a heating element. In that case it would definately get warmer.