Greenhouse effect and Earth's surface temperature

[Graeme M]

I posed this question in the Earth section of the forums, but I think I wasn't clear enough in how I posed my question, plus I think the question is more physics than Earth science.

I have decided to try posing my question as a series of questions building to the main point. This is because I think I must have some basic misunderstanding and perhaps we can identify that early on.

I loosely follow several blogs that are sceptical of 'global warming'. However, there is a fundamental point that I keep running into that I just don't 'get'. It is the matter of the greenhouse effect causing the Earth to be warmer than it might otherwise be without an atmosphere.

Please note I understand that climate change is not a topic for discussion here. I am not trying to question the greenhouse effect or suggest anything at all about climate change, the science, or societal views about the question. I am simply concerned with the mechanics of the greenhouse effect in terms of measured temperatures.

My understanding is that the theoretical temperature for a body with the same albedo as Earth located in an orbit at the same distance from the sun as Earth and in thermal equilibrium with the incoming radiation can be calculated to be about 255K. The actual measured temperature of Earth is some 287K. That warmer temperature is caused by the 'greenhouse effect' of the atmosphere.

I think this means that an Earth like body without an atmosphere would radiate at a temperature of 255K. I think it follows that the measured 'surface' temperature of that body must therefore be 255K.

Am I correct so far?

 

[jfizzix]

Yes, if the Earth without an atmosphere radiated like a blackbody at a temperature of 255 K, it is reasonable to infer that its equilibrium surface temperature (without an atmosphere) is also 255 K.

 

[russ_watters]

By the way, the ban on climate change discussion has been lifted.

 

[Graeme M]

Thanks for the note about the change in forum policy. However, I am still only wanting to clarify understanding of some of the mechanics.

So far it seems I have it right. The sun heats our theoretical the body which warms in response and at equilibrium its surface is (on average) 255K. That would mean that if we sampled the surface at enough points often enough, an average of that dataset would be the 255K value.

The Earth itself though is a different matter. It’s surface is rather more diverse (ie desert, snow, ocean, grass, jungle) and the atmosphere is an attenuating layer.

In the case of the earth, I understand that the atmosphere is largely transparent to incoming solar radiation. This radiation heats the earth’s surface which then emits long wave infrared radiation. That radiation is absorbed by greenhouse gasses and re-emitted in all directions, including back down.

The end result is that the surface temperature is on average 287K. Hence the ‘greenhouse effect’. But… what is the ‘surface’ temperature that is measured at 287K?

I had understood this to be an average of the various temperature indices, which are measurements of the lowest layer of the atmosphere via thermometers. However, I also understand that satellites can measure the actual surface itself.

Is the 287K measured average derived from thermometer data (atmosphere) or satellite measurements (actual surface)?

 

[russ_watters]

Since the Earth warms the ground just above it, there isn't much if any difference between the actual surface and the air just above, on average. So the answer to your question is: both.

By the way, did you try googling this question? There are lots of links out there:
http://en.wikipedia.org/wiki/Instrumental_temperature_record#Calculating_the_global_temperature

 

[Graeme M]

I have done some digging but generally couldn't find the direct answer. You say temps from just above ground must be similar on average but I've never seen that said anywhere. As far as I know, the major indices like HADCRUT or GISS etc are derived from thermometer data. As your link shows. RSS/UAH are sampling radiance at various altitudes and according to your wiki link, inferring temperature from these. So 'surface' temperature, at least as far as these data go, are actually atmospheric temps.

My question really is whether or not sampling air temperature is the same as directly sampling surface temperature. In the case of the theoretical body, the temperature is derived from a calculation of radiance at equilibrium directly from the surface which should be the same as directly sampling the surface temperature.

With the earth, air temps can differ markedly from actual ground temperature, depending on surface. For example, sea surface temps rarely exceed about 30C as far as I know, but air temps at say 2 metres could be substantially different. Similarly on land, the ground can be quite cool to touch while a warm wind is blowing. A cool change can cause an air temp drop of many degrees, but I doubt the land responds as quickly. Equally at say the poles, what is the radiating temperature of the ice itself? The air temps can get down to what, -30 or less C.

But really, I don't know. It just seemed to me that saying you are measuring surface temperature when you are really measuring air temperatures is not an apples and apples thing.

I would be interested to see if there is any example of data plotted for actual surface temps vs air temps. I'm just having trouble believing at face value that air temps would on average be directly similar to actual surface temps.

 

[russ_watters]

I have done some digging but generally couldn't find the direct answer. You say temps from just above ground must be similar on average but I've never seen that said anywhere.

You can see it in the graph, but think about it logically: if the average surface and air just above the surface temperatures were didfferent, heat would flow from one to the other because they are touching each other.

As your link shows. RSS/UAH are sampling radiance at various altitudes and according to your wiki link, inferring temperature from these. So 'surface' temperature, at least as far as these data go, are actually atmospheric temps.

Where exactly are you seeing that? There was a click-through to more specific info on the satellites that says this:

Satellites may also be used to retrieve surface temperatures in cloud-free conditions, generally via measurement of thermal infrared from AVHRR.

And you can click through a page deeper to get the precise details about how AVHRR works/what it does.

But again, as you can see from the graph, the plots lie pretty much right on top of each other, so I don't see a hair here to split.

My question really is whether or not sampling air temperature is the same as directly sampling surface temperature.

The same how? It is done by a different method, but returns nearly the same results.

I would be interested to see if there is any example of data plotted for actual surface temps vs air temps.

Is there something wrong with the graph I posted?

 

[CWatters]

If it's any help you can find data on soil temperatures on the web. For example...

http://www.builditsolar.com/Projects/Cooling/EarthTemperatures.htm

Picking Florida at random.. it says the mean annual soil temperature is between 72 and 77.

Google suggests the mean air temperature is around 75.

http://www.usclimatedata.com/climate.php?location=usfl0149

Seems reasonably consistent.

 

[Graeme M]

CWatters, that's a good find. Would it be the same for say the ocean? Polar ice? I will read your links when I get a chance, bedtime calls.

russ_watters, ditto re reading deeper at your links, I'll tackle that tomorrow. Thanks for the tip. Logically, heat may indeed flow, but surely for both surface and air to be close in temperature we'd need a fairly static situation? Air and ocean currents, weather systems, local effects must all have some effect? Again, I have no idea so you may well be right but I am not too keen on just taking your word for it. The air temp in the desert can get to something like 55-60C maximum. Yet the moon's surface can get to 123C. How hot could desert sand be? I have read it can be 5-10C hotter than air temps.

I am also not convinced by your suggestion that satellite measurements in your graph are taken directly from the surface. Satellites may be able to do that, but I don't think UAH and RSS represent that. Wikipedia says:

Satellites do not measure temperature directly. They measure radiances in various wavelength bands, from which temperature may be inferred.[1][2] The resulting temperature profiles depend on details of the methods that are used to obtain temperatures from radiances. As a result, different groups that have analyzed the satellite data have obtained different temperature data.

And

UAH provide data on three broad levels of the atmosphere.

• The Lower troposphere - TLT (originally called T2LT).
• The mid troposphere - TMT
• The lower stratosphere - TLS[3]

http://en.wikipedia.org/wiki/UAH_satellite_temperature_dataset

Do you have a source that certifies ground temperatures to be on average very similar to UAH/RSS?

 

[CWatters]

I'm not quite sure what the question is. Perhaps see..

"The Elusive Absolute Surface Air Temperature (SAT)"
http://data.giss.nasa.gov/gistemp/abs_temp.html

 

[Graeme M]

My question is simple, and as you and russ have noted it may be completely beside the point. Without any deep understanding of the physics, I wondered whether the surface temperature is the same as the surface air temperature. If the surface warms the atmosphere, which it does as I understand it, and the air is always losing heat up to space, it seems to me that the average surface air temperature if it's taken at a point several metres above the ground may not be the same as the ground. For all the reasons I mentioned.

For example, the average sea surface temperature is 17C. Right there is a substantial planetary surface warmer by 3C than the theoretical average surface air temperature. And oceans cannot warm over about 30 or cool below 0.

The argument is that the surface is 33C warmer than it should be due to the atmosphere, but strictly speaking the claim is that the near surface air temperature is 33C warmer than the surface temperature of a theoretical blackbody. To me that's a different argument.

So, is the Earth's actual average surface temperature the same as its average near surface air temperature. I have never seen this stated or discussed anywhere.

But then, I've hardly done an exhaustive search...

 

[Bystander]

ground temperatures

Variety of definitions from 10 cm depth, to frostline, to "isothermal."

average surface temperature

Your original question.

average sea surface temperature

Plus these other somewhat nebulous definitions

average near surface air temperature

have by now given you a "feel" for what is and is not defined?

 

[Graeme M]

Bystander, what?

It's a simple question tho perhaps poorly explained. And maybe it's an entirely meaningless question. The average temperature of an earth, otherwise exactly the same as our earth, but minus atmosphere, would be whatever its radiating surface could be measured as. According to explanations of the greenhouse effect, that is calculated to be -18C. The average of the real Earth's near surface atmospheric temperatures is +14C. OK, but what is the average of the real Earth's actual surface radiating temperature?

Is it the same as, more than, or less than, the near surface air temperature? Is there any data anywhere to illustrate the answer? I am after actual data, not people's opinions. russ_waters graphs above I *think* are of tropospheric temps, NOT actual surface.

 

[Bystander]

same as, more than, or less than, the near surface air temperature?

actual data

Clear day, summer, "more than" resulting in thermal convection. Clear night, summer, "less than" resulting in "temperature inversions" (actually a normal density gradient), or cool air piling up against colder ground. Average/mean? Ground temperature is increasing from vernal equinox to solstice along the front range. Daytime onshore breezes on tropical and temperate zone coastlines are another example of land surfaces warmer than air temperatures driving daytime convection; none of the coasts I've visited ever exhibited the complementary nighttime offshore breeze (sea surface temperature driving convection the opposite direction).

Heat capacities, thermal conductivities, thermal expansion coefficients, and phase behaviors of air, ground, and water affect heat transfer processes more than is suggested or implied by "average temperature" data. Definitions of terms with which to ask and answer questions still leave a lot to be desired.

 

[rootone]

Sure, but that is what science is about.
Examining stuff and trying to figure out/imagine what is going on.
as best as possible, (and without having prior conclusions)

 

[CWatters]

The average of the real Earth's near surface atmospheric temperatures is +14C. OK, but what is the average of the real Earth's actual surface radiating temperature?

I don't know but the data appears to exist because the surface temperature is used as a proxy for the air temperature...

http://data.giss.nasa.gov/gistemp/FAQ.html

Weather stations reporting surface air temperatures (SATs) are positioned on land which covers only one third of the planet; the rest is covered by oceans where SAT reports are rare. However, water temperatures (SSTs, sea surface temperatures) are available from ship and buoy reports and more recently there are also SST estimates derived from satellite data. Whereas SATs and SSTs may be very different (since air warms and cools much faster than water), their anomalies are very similar (if the water temperature is 5 degrees above normal, the air right above the water is also likely to be about 5 degrees warmer than normal). This is not true in the presence of sea ice, since in that case water temperature will stay at the freezing level. This allows us to use SST anomalies as proxies for SAT anomalies in regions without sea ice. L-OTI maps show SAT anomalies over land and sea ice, and and show SST anomalies over (ice-free) water.

 

[Graeme M]

OK... but using anomalies to evaluate trends in temperatures is rather a different thing is it not? By striking an average value and then looking at anomalies from that average over time, we can see a trend in a temperature series. That FAQ seems to suggest that sea surface and air temperatures share similar anomaly values when tracked over time (though I think there is a logical inconsistency in what they write there), but that tells us little about the actual values. In other words, SST and SATs may tend to rise and fall in a synchronous pattern, but that does not inform us whether one is typically more than, less than or equal to the other.

They also note that they can "use SST anomalies as proxies for SAT anomalies in regions without sea ice", but I think they mean only insofar as using anomalies to track the series trend.

My interest is the extent to which an average surface temperature is higher or lower than an average near surface air temperature. Your quote above observes that "SATs and SSTs may be very different " and "This is not true in the presence of sea ice, since in that case water temperature will stay at the freezing level". Which sort of supports my idea that an average air temp may be different from an average surface temperature.

 

[klimatos]

My question is simple, and as you and russ have noted it may be completely beside the point. Without any deep understanding of the physics, I wondered whether the surface temperature is the same as the surface air temperature. If the surface warms the atmosphere, which it does as I understand it, and the air is always losing heat up to space, it seems to me that the average surface air temperature if it's taken at a point several metres above the ground may not be the same as the ground. For all the reasons I mentioned.

Most global temperature collections no longer use the term "surface temperature". Instead, they prefer "Near-surface temperature". This is because these readings are taken from ordinary national meteorological stations and from ships at sea. A standard instrument shelter on the ground will have its thermometers located about 1.5 meters from the actual ground surface. Some shelters are on the roofs of various federal buildings. On ships at sea, the measurements may be taken tens of meters above the sea surface and from a moving vessel. For all of these reasons, the term "near-surface" is preferred and is more descriptive.

It is worth noting that countries that report "ground frost" as well as shelter frost show the former to be far more prevalent. Similarly, desert surface temperatures can be tens of degrees Kelvin above shelter temperatures. In any case, most of the outgoing terrestrial radiation comes from the atmosphere. Of the 375 Wm² of outgoing terrestrial radiation, only about 87 Wm² comes directly from the surface.

 

[klimatos]

My understanding is that the theoretical temperature for a body with the same albedo as Earth located in an orbit at the same distance from the sun as Earth and in thermal equilibrium with the incoming radiation can be calculated to be about 255K. The actual measured temperature of Earth is some 287K. That warmer temperature is caused by the 'greenhouse effect' of the atmosphere.

I think this means that an Earth like body without an atmosphere would radiate at a temperature of 255K. I think it follows that the measured 'surface' temperature of that body must therefore be 255K.

Am I correct so far?

Why is the hypothetical emissivity of the atmosphere-free Earth significant in any way? How can this value be used to solve any significant real-world problems. The Earth's estimated surface emissivity is usually given as 375 Wm^2. This is the emissivity that would be produced by a surface at a mean temperature of 285 Kelvin with a coefficient of emissivity of 0.95. This is very close to the 287K you mention as the estimated mean annual near-surface temperature. This is a good starting point for heat budget studies. The imaginary 255K temperature has no predictive value and is useless in global heat budget studies. It just gets in the way.

 

[BillyT]

The kinetic energy, KE, of particles in a small volume is an averages of the KE of their individual ke. These atom or molecules collide and exchange energy with each other. Typically a very fast moving one (higher than the average energy) will leave the collision with less energy than it had just before the collisons. Very rapidly a velocity distribution will be established that can be described with a single free parameter or variable, T, which is the temperature of that small volume of many thousand of particles. We then say that Local Thermodynamic Equilibrium, LTE, exists. However, photons also carry or posses energy and exchange energy with atoms or molecules, both by scattering and by absorption or emission.

Inside a closed box with all the wall materials at same temperature T, the distribution of photons (how many of each wavelength) is also given by a math expression with a single parameter, Planck's equation for "black body radiation." The value of that parameter is the same T that describes the temperature of the wall, both on the Kelvin scale. If there is a tiny, compared to the box's internal surface area, hole in the box, black body radiation will be coming out of that hole.

The atmosphere surrounding the Earth is not confined inside an isothermal box, so one cannot expect the radiation leaving Earth's surface to be black body radiation, not even very locally, say from a square cm which is all at same temperature T; for two reasons.
(1) LTE does not exist as there is a net loss of energy and (2) Black body radiation is called that as it comes only from a surface that is perfectly black. I. e. absorbs all radiation that falls on it. Earth is not "perfectly black."

Many will tell you that a good absorber is a good emitter too, but forget to add "at the same wavelength." Thus a square cm of Earth's surface that has nearly an absorption coefficient, a, of 1, say 0.9, at one wave length may have an emissivity "constant" ,e, at some other wavelength of only 0.4 but for the wave length at which a = 0.9, e = 0.9 also. I won't go into details but just note that if this were not always true, you could violate conservation of energy. - Made a device that produces net energy.

So now to come more to your question: The surface of the Earth is at many different temperatures and has differently varying (with wavelength) coefficients, a & e. As if that were not enough complexity, you need to understand the concept of "optical depth" to understand what satellites looking down at the Earth see as outbound IR radiation.

Although symmetric molecules like O2 & N2, don't change their "dipole moment" when they vibrate or rotate molecules like CO2 can. CO2 is actually linear molecule: 0---C---0 with the Os having "stolen" some negative charge, on average, from the C. Its dipole moment is zero when in it ground vibrational state but not when excited in the "a-symmetric" vibrational mode.
I.e. with this configuration O--C----0 and then half a cycle later is O----C--O. When this vibrational state return to the ground state, its loss of energy is given to the IR photon emitted. I'll call the wave length of that photon "L."

When a photon of wave length L is emitted by the Earth's surface and there are no clouds in the way, it still has only about 1/3 probability of escape to space now that the concentration of CO2 is about 400ppm. And to even get that ~33% chance, the photon that does leave Earth is not the same as the one that left the surface originally. That original wave length L photon, was absorbed and its energy re-radiated by some other CO2 molecules at a different altitude. I.e.. the escape to space of wave length L photons is a random walk process. Most that do escape, have come from a high altitude, where the temperature is much lower than that of the Earth's surface below them.

They come crudely speaking, from the top "optical depth" of the atmosphere, which is a smaller layer than for most photons not of wave length L. That high layer's colder temperature limits the intensity of this L wave length IR as no surface or volume can radiate more intensely than a black body at the same temperature can. Thus man is lucky that CO2 effect on Global Warming is limited. The CO2 concentration could increase five fold and the IR loss via wave length L radiation would not even double. We say the CO2 absorption bands are nearly "saturated."

Unfortunately that is not true of CH4, which is a 3-D molecule, with more and stronger absorption bands. During the first year after a Kg of CH4 is released it does 120 times more global warming than a Kg of CO2 does, but as its half line in the air is only 12.6 years* during the first decade after puff was released it does only about 84 times more global warming as same mas of CO2 does as CO2 half life is about 1000 years.

I'll stop here as getting too far from your question, but at least you should now understand it was a quite complex question.

* I need to note that CH4's half life is now increasing by about 0.3 year each year, because it is mainly destroyed by the OH- radical but now, unlike the last 800,000 years, the harsh solar UV that make the OH- is not able to do so at the rate CH4 is being released so OH- concentrations are falling and CH4 is both increasing and remaining in the air more years.

 

[haruspex]

Since the Earth warms the ground just above it, there isn't much if any difference between the actual surface and the air just above,

True, but since the net flow of heat is from ground to air it is likely that the mean ground temperature is slightly higher.

 

[Buzz Bloom]

I believe I understand correctly that the average surface TEMPERATURE is, theoretically, NOT the average of the temperatures that might be measured at every point on the Earth's surface. The important concept is that the total RADIATION generated by the Earth at its various surface points divided by the total surface area is the average Earth's radiation. The radiation at a point is proportional to the 4th power of the temperature at that point. Therefore the "average" temperature of the Earth is the 4th root of the average of the 4th powers of the temperatures at all of the surface points.

 

[Buzz Bloom]

Sorry about accidentally making a double posting.

 

[BillyT]

... (1) The important concept is that the total RADIATION generated by the Earth at its various surface points divided by the total surface area is the average Earth's radiation. (2) The radiation at a point is proportional to the 4th power of the temperature at that point. Therefore the "average" temperature of the Earth is the 4th root of the average of the 4th powers of the temperatures at all of the surface points.

On (1):
That would be correct if none of the radiation from the surface were absorbed before it got to space but that is not the case for many of the green house gas wave lengths. Those that do escape often were radiated by much colder molecules at altitudes well above the Earth's surface, as I discussed with that radiation's "random walk" and "optical depth" concepts.

On (2):
The "T to the fourth" radiation is only for black bodies*. Radiation from any point on Earth is less than given by that law for that temperature as its emissivity, e, is not unity at any wave length. The sum of a + e + r where these are coefficients of absorption, emissivity and reflectivity is always unity. This is why space engineers wrap things sent into space, that will pass thru the Earth's shadow or go far away from the sun with highly reflecting foil. (Often gold foil as its a & e are very close to zero for the IR that the spacecraft might other wise radiate heat away with.)

* "Grey bodies" (wave length constant value of a & e but less than unity) do radiate more as T^4 goes, but less than a black body at that T does. If the body is not even grey, it is complex to tell its temperature from the radiation coming from it.

 

[Graeme M]

Hmmm... I will have to reread several of the most recent comments several times, and even then I think I haven't enough of a general physics background to quite follow.

My original question was more a point about method than anything else. When the greenhouse effect is explained, it is always stated in simple terms for average people like me to understand. And the point is that the surface is 33C or so warmer than it would otherwise be due to greenhouse gasses.

Wikipedia for example says: "If an ideal thermally conductive blackbody were the same distance from the Sun as the Earth is, it would have a temperature of about 5.3 °C. However, since the Earth reflects about 30% of the incoming sunlight, this idealized planet's effective temperature (the temperature of a blackbody that would emit the same amount of radiation) would be about −18 °C. The surface temperature of this hypothetical planet is 33 °C below Earth's actual surface temperature of approximately 14 °C."

I have no problem with that idea. I merely wondered if when we say 'surface' for this theoretical black body that is -18C we mean its actual hard surface is -18C, but when we say the Earth's average surface temperature is 14C we mean its near surface air temperature (SAT). And if that's true, is that an apples v apples comparison. I wondered whether, if we measured the average hard Earth surface, it would be more, less than or equal to 14C. Such an average wouild be struck from measurements of sea surface temps (SSTs), land surface temps and ice surface temps.

I wondered because the -18C is a theoretical calculation, but as far as I know the 14C is an average of temperature indices using SATs. But I might be wrong there, the 14C may be a theoretical calculation as well. And maybe the whole simplified "earth is 33C warmer" thing is just too abstract to be considered literally.

One of the earlier commenters pointed to a webpage discussing soil temperatures, but I am not sure that when I say hard surface temperature, I mean soil temps. I mean the temperature of the radiating hard surface because it is this that warms the atmosphere.

I have broadly the same question about the suggestion that greenhouse gas 'back radiation' warms the surface - is that literally the hard surface, or the SATs? The radiation budget diagrams always suggest far less radiation reaching the surface from the sun than from the atmosphere, a claim that I think is quite true when I consider my own physical experience. But does that really translate into a literal "warming" of the hard surface itself, or are we talking only of a warming of SATs?

Is there any physical data showing a trend in hard surface temps over time? Clearly, in the case of SSTs, we do. How about land temps, is there an index of these?

Lastly, if the hard surface temps have increased over time, could we claim this is purely due to greenhouse gas back radiation, or could it be in part assigned to substantial changes in surface composition (ie urban areas, deforestation, land use changes etc)?

Regardless, the main question was simply, is the measured average hard surface temperature of Earth close to 14C or not? Not someone's opinion, is there data or is there not? I have never seen any but then I've not looked that hard. Or is the question itself simply irrelevant because the idea is overly simplified for public consumption.

 

[Buzz Bloom]

On (1):
That would be correct if none of the radiation from the surface were absorbed before it got to space . . .
On (2):
The "T to the fourth" radiation is only for black bodies*. Radiation from any point on Earth is less than given by that law for that temperature as its emissivity, e, is not unity at any wave length.

Good points Billy.

I'll comment on (2) first, since I agree it is correct. In order to calculate the average temperature from the radiation, one would have to weight each point source by the reciprocal of the emissivity, so what I suggested was incomplete. Still, averaging the temperatures would also not be correct.

Regarding (1) I think there is a misunderstanding about what is meant by "the average Earth temperature". If we are talking about the surface temperature, the effective surface temperature, then the atmosphere is irrelevant. If we are talking about the effective average temperature corresponding to the radiation leaving the Earth's as a whole and going into space, then I think the best way to establish this would be to use the same temperature as for an Earth with no atmosphere.

Wikipedia has a good, although incomplete, article about the "Earth's energy budget". The article includes a (incomplete) diagram. The diagram shows the budget at the Earth's surface, with equality between (a) the solar and greenhouse downward radiation energy. and (b) the Earth's upward radiated and convection energy. It also shows the budget o the downward solar radiation at the upper atmosphere and the upward radiation from the upper atmosphere. What is missing is a correct budget for the atmosphere itself.

See the following URL:
https://en.wikipedia.org/wiki/Earth's_energy_budget .
For a comment about the missing part, see the talk page.

 

[BillyT]

... Wikipedia for example says: "If an ideal thermally conductive blackbody were the same distance from the Sun as the Earth is, it would have a temperature of about 5.3 °C. However, since the Earth reflects about 30% of the incoming sunlight, this idealized planet's effective temperature (the temperature of a blackbody that would emit the same amount of radiation) would be about −18 °C. The surface temperature of this hypothetical planet is 33 °C below Earth's actual surface temperature of approximately 14 °C." ...

There is a lot of averaging in that "approximately" 14C. Averaging over areas and times. For example of both: Northern Hemisphere's winter has lower average than Southern for same months. Then there is the diurnal average to consider also - can have frost on ground here in southern Brazil during winter at sunrise, yet a nice swim in the ocean that afternoon. Even shorter term variations of ground temperature on sunny day when it clouds over. Truth is there is a lot of importance in that "approximately" due to the averaging procedure used.

I don't know how that average was done, but suspect airport temperature records were used. How heavy a plane can be at take off drops as the temperature (air density on the runway really) changes. Once after nearly an hour wait in the line to take off, plane I was on for 10 hour flight to USA, had to go back to the gate area to off load some cargo, probably not fuel, as we were then too heavy for the air conditions. (Wind speed and direction wrt the runway may have been part of the problem too.)

As I noted in prior post, some of the IR radiation to spaces comes from different altitudes, some quite high in the center of green house absorption bands. Backing this out along with all the temporal and spatial variation is not easy or exact. That is why, I guess airport thermometer data, available 24/7/365 for decades is mainly where that 14C comes from. Even that is some what inconsistent with the idea that the IR total surface radiation must be same as the total rate of solar energy absorbed for at least three reasons:

(1) most important: is that about half the solar heating is now being stored - the deeper ocean temperature is warming faster (in Kelvin %) than the ocean surface is as the surface cools by evaporation, and considerable part of the "heat of evaporation" does not fall back into the ocean. This deep ocean heat storage is sort of like, for now, that the sun were weaker than it is if your interest is in land surface temperatures. The "scary part" of this, is that the global warming the green house gases are now doing is not yet fully felt. - To get the full "cost" of what man is doing to the Earth now, one must understand that last part of the bill will be presented in about 50 years from now!

(2) A significant part of the outbound from surface IR does not escape but in its "random walk" away, comes back to the surface like the Earth had a weak IR sun shinning on it too (thru IR transparent air) for more hours each day than the sun shines.

(3) And of small import, Earth's long term (centuries) heat loss to space is slightly greater than the radiant energy it gets from space as the core of Earth is both still cooling from the great gravitational collapse energy if got as Earth formed AND radio active isotope decay (K40 now dominate, I think, but U238 was a few million years ago, as I recall reading.)

For Jupiter, in much weaker solar flux and with much larger gravitational collapse energy, more slowly working it way to the "surface" I seem to recall that it radiates (in the far IR of course) about 5% more energy that it receives !

 

[Buzz Bloom]

They come crudely speaking, from the top "optical depth" of the atmosphere, which is a smaller layer than for most photons not of wave length L. That high layer's colder temperature limits the intensity of this L wave length IR as no surface or volume can radiate more intensely than a black body at the same temperature can.

Hi @BillyT:

The above quote seems to be saying that the radiation of an IR photon from a high altitude CO2 molecule is thermal. My understanding is that the absorption and re-emitting of IR photons is not thermal because the re-emission is much faster than the time for the molecule to interact with another molecule and covert the internal excited state energy into kinetic energy. Do you have any definitive reference that says it has to be thermal?

Regards,
Buzz

 

[BillyT]

Hi @BillyT:The above quote seems to be saying that the radiation of an IR photon from a high altitude CO2 molecule is thermal. My understanding is that the absorption and re-emitting of IR photons is not thermal because the re-emission is much faster than the time for the molecule to interact with another molecule and covert the internal excited state energy into kinetic energy. Do you have any definitive reference that says it has to be thermal?
Regards,
Buzz

The molecular radiation is spectral lines. My "thermal" as clearly stated was just noting that the intensity at any wave length can not be greater than black body at the local temperature. You are correct. It can be much less than that upper limit. I.e. the upper states must be populated for the radiation to occur. Collisions are not the only way they can be. Some in-band radiation is coming up from lower levels where the temperature and density are both higher and is absorbed. This is why the "optical depth" is a valid concept. I. e. the IR that escapes Earth comes from a top zone where the chance of escape is greater than the chance of being re-absorbed.

 

[Buzz Bloom]

My "thermal" as clearly stated was just noting that the intensity at any wave length can not be greater than black body at the local temperature.

Hi Billy:

Thank you for your post answering my question.

I had one other idea in mind when I said "non-thermal". The spectrum of the re-emitted radiation form CO2 does not match the Planck black-body distribution. You may want to look at the following paper:
http://www.leif.org/EOS/2012GL051542.pdf
You may also find the following thread of some interest.
https://www.physicsforums.com/threads/why-does-the-heat-in-the-atmosphere-mostly-go-down.841737 .
At the end of this thread is a discussion about the random walk you mentioned in your post #24.

Regards,
Buzz

 

[DrStupid]

My understanding is that the absorption and re-emitting of IR photons is not thermal because the re-emission is much faster than the time for the molecule to interact with another molecule and covert the internal excited state energy into kinetic energy.

I can't follow that argumentation because IR absorptions and intermolecular interactions are independent from each other. I would expect a probability of

1 - \exp \left( { - x} \right) \approx x

for an interaction after IR absorption and before IR emission, where x is the ratio between the mean lifetime of the excited state and the mean time period between two interactions. Low values of x make such interactions less likely but not impossible. This just increases the time required to reach the thermal equilibrium resulting in some kind of low-pass filter for the impact of IR radiation to temperature. In order to exclude impact of thermal re-emission you need to show that the relevant variations of IR radiation are too fast to be followed by temperature.

 

[DrStupid]

The spectrum of the re-emitted radiation form CO2 does not match the Planck black-body distribution.

As CO2 is not a black body this isn't very surprising. Do I miss something?

 

[BillyT]

Hi Billy: ... The spectrum of the re-emitted radiation form CO2 does not match the Planck black-body distribution. ...

No one with even the slightest understanding that spectrial line radiation has the energy of the diference between two discrete energy states would think it should be a continuum distribution that Planck BB radiation is.

At high density gases do show an effect of collisions - the spectral lines are not as "sharp" (have slightly wider spacing of the line profile between it half intensity points) This is called "collisional broading." It is caused by the fact that the energy levels of two atoms (or molecules) that are separated by only "atomic distances" are distrubed / not exactly the same as for the isolated radiator.*

The extreme case of this is a hot radiating solid - there the collisional broading is so great that distinguable lines no longer exist, and the the radiation is a continuum.

* My experimental Ph.D. thesis concerned this collisional broading for radiation from a plasma. Mostly Argon ions and electrons. There it is a little more complex and Hans Griem at Un of Md had just worked out a model, which I tested. The near approaching and not too hot (fast) electron does not pass by the ion on a straight line trajectory as a neutral atom would pass by another neutral. The electron path is bent by the net positive ion. So the collision lasts longer and the collision broading is more extreme. Even the peak of the line can be displaced and the line shape is no longer symetical about the peak. - In one line case the peak was shifted by more than an Angstrom.

 

[klimatos]

Re-radiation and Re-emission: A Cautionary Note:

Throughout this series of posts, I have noted a somewhat careless use of the terms “re-radiation” and “re-emission”. It is almost as if the writers were somehow implying that the photons absorbed by a molecule and the photons later emitted by that molecule were somehow identical. This may occasionally be the case, but I think that it is not common. Unless you can prove that the second photon has exactly the same energy level as the first, it is not logically “re-emission”. For example, an atmospheric water molecule may readily absorb a high-energy photon of solar radiation (this accounts for about 83 Wm-2 of the atmosphere’s heat budget), but that molecule is extremely unlikely to emit such a photon. Instead, it is more likely to emit one or more photons of lower energy levels. I would argue that photons are neither re-radiated or re-emitted by either the Earth’s surface or its atmosphere. Both of these entities emit radiation and both of them absorb it. Since both the Earth’s surface and its atmosphere undergo temperature changes due to both emission and absorption, it is obvious that some conversion of internal electronic energy to and from external kinetic energy takes place. Hence, emitted energy does not equal absorbed energy in most atmospheric situations. I think that the use of the “re-“ prefix is both erroneous and misleading. Instead of saying “the later re-emission of a photon”, why can’t we just say “the later emission of a photon”?

 

[rootone]

It's a note about semantics rather than the physics here, but yes 'the later emission of a photon' leaves less room for misunderstanding.
're-emission' might lead some to conclude that what is being emitted is the exact same photon that was originally 'captured'/absorbed, but that photon of course no longer exists.

 

[BillyT]

It's a note about semantics rather than the physics here, but yes 'the later emission of a photon' leaves less room for misunderstanding.
're-emission' might lead some to conclude that what is being emitted is the exact same photon that was originally 'captured'/absorbed, but that photon of course no longer exists.

Agreed. The photon which is emitted after one has been absorbed will be traveling 99.9% of the time in a significantly different direction - certainly is not the absorbed photon "re-emitted"

Also if an energetic photon is absorbed and produces a higher than least excited state of the absorber, it is very likely that the decay of that excited state will be via emission of two or more lower energy photons. This is because, crudely speaking the transition probably for a state to a lower state is greater when the two states (upper and lower) are less separated.* I. e. when a high state is excited, the result is usually a cascade down thru several intermediate states, not a single radiative transition down to the ground state.

For example, in deep space most of the hydrogen atoms are just protons, but they do occasionally become neutral atomic hydrogen by capture of a passing electron into a very high bound state. To even do that there needs to be some "third body" to take up the electron's greater than binding energy. Thus it is very improbable that the "third body" can absorb all the energy the free electron had (More than 13.6 ev) so it "falls" directly into the ground state. It might for example be captured into the n = 23 excited state, then "drop down" with IR radiation to the n = 22, 21, or 20 state, from which it will drop down in cascade fashion thru more lower states still. Eventually (after a fraction of a second) it may emit some visible light of the Balmer serie and then a UV photon of the Lyman serie as it become part of a hydrogen atom in the ground state.

* This is because closely spaced quantized energy level wave functions overlap much more than pair with greater difference in energy levels. Again crudely speaking the radiative transition probabily between two state is proportional to the overlap of their wave functions. (Assuming, of course, that such transitions are "permitted" - obey the "selection rules.")

 

[Buzz Bloom]

Hi Billy:

Thanks for your post explaining my misunderstanding.

No one with even the slightest understanding that spectrial line radiation has the energy of the diference between two discrete energy states would think it should be a continuum distribution that Planck BB radiation is.

The extreme case of this is a hot radiating solid - there the collisional broading is so great that distinguable lines no longer exist, and the the radiation is a continuum.

* My experimental Ph.D. thesis concerned this collisional broading for radiation from a plasma.

Not being well educated on the topic, I had the misunderstanding that the "broadening" due to the range of the kinetic energy of molecules caused the shape of radiation from a gas to approximate a Planck distribution, as you say it does for a solid, and I am guessing it does also for a liquid. Can you suggest a reference that discusses the degree of broadening by a gas or plasma as a function of temperature?

Regards,
Buzz

 

[Buzz Bloom]

Hence, emitted energy does not equal absorbed energy in most atmospheric situations.

Hi klimatos:

I understand that there are very many excited states of electrons bound to nuclei, and that the energy absorbed which excites an electron to a higher state than the minimum excited state has more than one way to release energy while returning to a lower energy state. I gather from what you are saying that for the excited molecular states of such molecules as CO2, these states also show the same behavior of as bound electrons' states. This seems a bit odd to me since the physical changes in the molecular shape when it it is in an excited state do not seem to have the quantum mechanical characteristic of electrons bound to a nucleus, in which an electron does not actually have a moving position in an orbit, but rather a probability distribution of its position. Is this also true of the position of the carbon and oxygen nuclei relative to each other in defining the molecule's excites states? Do all excited states of a molecule have multiple transition paths to lower states as do bound electrons?

I have no access to a technical library, and limited access online to more than abstracts of articles. Can you recommend any reference that I might be able to access that describes the excited energy states and transitional possibilities of those states for CO2?

Regards,
Buzz

 

[Buzz Bloom]

Hi @BillyT:

Here is one more thought about themeasured CO2 IR spectrum not being thermal, and what this means.

Even if the spectrum is not continuous, that is it is a collection of spikes at discrete frequencies, I would expect that the spectrum will be different than that of CO2 gas at a controlled fixed temperature, or a combination of such spectrums for several fixed temperatures. My thought is that the CO2 in our atmosphere is mostly concentrated at a few distinct levels, and that the temperature at these levels depends mostly on altitude, although there are variations due to latitude and season and time of day. I would expect the measured distribution to be more like that of CO2 at the Earth's surface temperature (rather than the temperatures atr the altitudes of CO2 concentrations) adjusted for the variations of the absorption coefficient for different frequencies.

AFAIK, this analysis of the data in the http://www.leif.org/EOS/2012GL051542.pdf paper has not been done. What would you expect such an analysis to show?

Regards,
Buzz

 

[Buzz Bloom]

Throughout this series of posts, I have noted a somewhat careless use of the terms “re-radiation” and “re-emission”. It is almost as if the writers were somehow implying that the photons absorbed by a molecule and the photons later emitted by that molecule were somehow identical.

Hi @klimatos:

It was my intention to imply that the IR photon emitted by a CO2 molecule would be have the same energy as the photon it had recently absorbed. I did not intent that it was the same photon. What not-too-wordy vocabulary would you suggest for expressing this idea?

I understand that the recent posts have suggested that this intent of mine is incorrect, but that is a separate issue which I am trying to resolve.

Regards,
Buzz

 

[Buzz Bloom]

I can't follow that argumentation because IR absorptions and intermolecular interactions are independent from each other. I would expect a probability of

1−exp(−x) ≈ x​

for an interaction after IR absorption and before IR emission, where x is the ratio between the mean lifetime of the excited state and the mean time period between two interactions. Low values of x make such interactions less likely but not impossible. This just increases the time required to reach the thermal equilibrium resulting in some kind of low-pass filter for the impact of IR radiation to temperature. In order to exclude impact of thermal re-emission you need to show that the relevant variations of IR radiation are too fast to be followed by temperature.

Hi @DrStupid:

Thank you for your discussion about "x". I apologize for being careless in my exposition in ignoring what I believe (based on email correspondence with an expert) is a very small value of x.

Regards,
Buzz

 

[DrStupid]

I apologize for being careless in my exposition in ignoring what I believe (based on email correspondence with an expert) is a very small value of x.

Very small values of x just mean, that the time to reach the thermal equilibrium is very long compared to the life time of the excited state. What you need for your argumentation is a comparison of the time to equilibrium with the rate of IR variations.

 

[klimatos]

I have no access to a technical library, and limited access online to more than abstracts of articles. Can you recommend any reference that I might be able to access that describes the excited energy states and transitional possibilities of those states for CO2?

Sorry, Buzz, my field is the atmosphere. You need a QM guru for that task.

 

[klimatos]

It was my intention to imply that the IR photon emitted by a CO2 molecule would be have the same energy as the photon it had recently absorbed. I did not intent that it was the same photon. What not-too-wordy vocabulary would you suggest for expressing this idea?

Why not just say, "emits a photon of the identical energy level"? As you went on to note, I'm not sure how often this happens except for lowest excitation level.

 

[klimatos]

Some More-or-Less "Hard" Numbers

NASA’s Global Observatory has released some interesting numbers relevant to this thread:
1) 1880 to 2014 global near-surface temperature increase is 0.75°C (1.35°F)
2) October 2015 mean global carbon-dioxide level is 401.58 ppm
3) Rise in world mean sea level:
Satellite measurements: 1993 to 2015: 3.24 mm per year (about a foot per century)
Tide gauge data: 1880-2014: 1.87 mm per year (a bit over seven inches per century)

http://climate.nasa.gov/vital-signs/

 

[Devon Fletcher]

Of course, the 'average' temperature of anything is a rather nebulous concept, and, in the case of the Earth, things like the albedo, relative temperatures, and radiated energy, are constantly changing, and vary widely across the globe.
As the atmosphere absorbs some of the radiation coming from the Sun, that energy is not all being re-radiated back to Space, but some of it hangs around, warming things up down here.
For a really comprehensive look at all this stuff, I can't recommend highly enough 'Experimenting on a small Planet', Wm. W. Hay, Published by Springer.
It's a great read, includes a comprehensive history of Science, a self-indulgent memoir, and lucid, entertaining explanations of all this climate stuff.
This may have been mentioned before. I'm new here, and haven't read all the threads... but, get it if you haven't already.

 

[Buzz Bloom]

Why not just say, "emits a photon of the identical energy level"? As you went on to note, I'm not sure how often this happens except for lowest excitation level.

HI klimatos:

Thank you for your post responding to my questions.

You suggestion "emits a photon of the identical energy level" is excellent, except that it has eight words. I think this would be inconvenient to repeat multiple times in a long discussion about the concept.and its implications. I would prefer to continue using "re-emits" with a one time additional sentence using your suggested language to explain what "re-emits" is intended to mean.

I am also not "sure" about the relative frequency of re-emits compared with the emitting a photon of a different energy than the one absorbed. I have been trying to get more information about this for some time, but it extremely difficult for me with my limited access to a technical library. My belief that CO2 IR re-emitting is quite frequent is a conclusion I came to from a brief email correspondence with a professor of atmospheric physics at MIT. As you can see from the unanswered questions I have asked in this thread about this topic, easy answers don't seem to exist.

Regards,
Buzz

 

[Buzz Bloom]

For a really comprehensive look at all this stuff, I can't recommend highly enough 'Experimenting on a small Planet', Wm. W. Hay, Published by Springer.

Hi Devon:

Thank is for the recommendation. The book is not available in my local library, nor in the inter-library network to which my local library belongs. I will request the help of my library's research librarian to get a copy for me from some other source.

Regards,
Buzz

 

[Buzz Bloom]

What you need for your argumentation is a comparison of the time to equilibrium with the rate of IR variations.

Hi DrStupid:

Thank you for responding to my post. I acquired some understanding about the non-thermal emission of IR photons by atmospheric CO2 molecules from a brief email conversation with an expert. My qualitative understanding that the time between the absorption of a IR photon by a CO2 molecule and the subsequent emission of a similar photon is very much less than the time for the molecule to interact with another molecule. I also came to understand that it is molecule-to-molecule interactions that transforms the absorbed energy from a photon into kinetic energy, and thereby to a rise in temperature of the gas.

Since that conversation I have been trying to acquire some specific quantitative information about this topic, including asking multiple questions on this topic in the Physics Forum, and so far I have been unsuccessful. Can you help me?

Regards,
Buzz

 

[Devon Fletcher]

According to Hay, the CO2 (and other) molecules can have an electron 'promoted' to a higher orbit (and therefore capable of an emission), by a photon of the appropriate ...or slightly higher...energy. The energy excess to the promotion is absorbed as translational (temperature).energy.
All these emissions occur at very specific frequencies, but with multi-atomic molecules, the number of possible appropriate photon energies are many, and, as a result can virtually blanket whole swathes of the spectrum.

Here is a graph of the various greenhouse gases and theirabsorption spectra. It is from the Wm Hay book, and is typical of the many illustrations therein.