- Despite decades of the prominence of the greenhouse effect, people interested in this discipline haven't discussed the mechanism much
- The fact above creates a large amount of disagreement and misunderstanding
- People don't want to listen to others or think about the matters differently
- Most people brutally underestimate the role of the black bodies in the discussions about the radiation and energy flows

When you heat up an object, it may start to glow. It becomes red and emits a lot of thermal radiation. You may feel that the object is warm. It's not only because the air is getting warmer around the object, thus heating your body. It's also because of direct electromagnetic radiation - mostly infrared radiation - that is directly transmitting heat from the warm object to your body.

**Rayleigh, Jeans, Planck, Einstein**

People have known about these things for centuries. In the 19th century, they began to study the "color" and "intensity" of the heat emitted by warm objects systematically. They found out that black bodies emit radiation that has a characteristic dependence on the wavelength of the radiation:

Recall that the wavelength is inversely proportional to frequency which is directly proportional to the energy of one photon - which will be discussed later. The temperature is also directly proportional to the frequency - the frequency for which the emission becomes maximum (maxima of the curves above): this temperature-frequency relationship is called Wien's displacement law.

Moreover, the classical (non-quantum) physicists were able to correctly derive the radiation emitted by a warm object at very long wavelengths (right portion of the graphs above) - it was the Rayleigh-Jeans law. However, this law implied that the warm objects should emit infinite total energy at high frequencies - a conclusion that is physically impossible (and obviously disagrees with the observations, too).

By looking at the shape of the measured graphs, people could also write a limiting form of the curves for the high frequencies (left portion of the graph above). The required formulae already included Planck's constant. But only in 1900, Planck actually made a great guess how to interpolate between the short-wave and long-wave approximations.

He wrote the full black body formula that turned out to be correct. And amazingly enough, he has also derived it from the assumption that the energy carried by the radiation is not quite continuous: at frequency f, it is divided to quanta E=hf, the minimal amount of energy that a wave with frequency f can carry. This quantization is very important for high frequencies. For low frequencies i.e. long wavelengths, the quantum is very small and the fact that the energy isn't continuous is becoming immaterial.

Planck didn't believe that the quantization of the energy was "serious": he thought it would be just a trick to hide the decline of the radiation for high frequencies. ;-) However, Einstein realized very well that the decline was real and so were the quanta of the electromagnetic radiation, later called photons. In 1905, Einstein used the quanta to explain the photoelectric effect which is what he got his Nobel prize for.

No sane contemporary physicist doubts that the black body radiation is correctly derived and very important. Objects that are black - that don't reflect incoming radiation - also emit the maximum radiation they can emit. This relation is called Kirchhoff's law of thermal radiation. It follows from the equilibrium and the time-symmetry of the microscopic laws of physics. Note that the time reversal of spontaneous emission is absorption of one photon. Because the microscopic laws must give you the same transition probabilities in both directions, you can derive how much objects spontaneously radiate, and the correct answer is the black body curve.

Real bodies are not quite black: they reflect some radiation and only absorb a part of the incoming radiation. By Kirchhoff's argument, that also means that they emit less than the black bodies would emit at the same temperature.

In physics, the cosmic microwave background is the most accurate known natural example of a black body. Its temperature is 2.7 K and even though the Planck curve is very accurate, we can measure the deviations from the perfect curve. They can be mostly parameterized as fluctuations of the temperature - by something like 10^{-5} degrees. These fluctuations, measured by satellites such as WMAP, tell us a lot about the history of the Universe and provide us with indirect evidence for cosmic inflation.

In 1974, Stephen Hawking, building upon visions of Jacob Bekenstein and others, realized that black holes are black bodies, too. This is a conclusion that a linguist could "derive", too: a hole is a body, which is why a black hole must be a black body: they share the adjectives. While I don't want to dispute that linguists are poor scientists in average, this particular argument would actually be morally correct. All objects must radiate. The only question is what is the temperature. Hawking could calculate it and it was proportional to the gravitational acceleration at the surface of the black hole (if you allow me to use Newtonian jargon), with a natural coefficient.

The smaller a black hole is, the more intense light it emits. On the other hand, the Hawking (thermal) radiation coming from very large black holes is negligible and you often need something like 10^{100} years before such a huge black hole evaporates.

The total energy radiated by a unit area of a hot surface is proportional to the 4th power of the temperature in Kelvin degrees. This is known as the Stefan-Boltzmann law. The coefficient is fun to calculate - the most complicated piece of mathematics you need to calculate to reach the final result is zeta(4) = pi^4/90. If the spacetime dimension were not 4 but e.g. 10, the 4th power would become 10th power. This fact can be determined by simple dimensional analysis.

**Sun, Earth, thermal radiation**

The black bodies are essential for the energy budget of the Earth, too. First of all, the life on Earth wouldn't exist without the energy coming from the Sun. It arrives in the form of electromagnetic radiation. The temperature at the center of the Sun is 15 million °C. However, the radiation created in the middle of the Sun doesn't get out. It mostly gets absorbed by the outer shells.

What mostly determines the radiation outside the Sun is therefore the temperature of the surface of the Sun. And it is something like 5400 °C or 5700 K. This temperature can be calculated from the energy balance at the surface: the outgoing energy from the Sun - determined by the temperature of the surface - must match the energy that is being produced during the same time inside the Sun via the thermonuclear reactions.

This is how the radiation from the Sun looks like when we decompose it to different colors - or wavelengths:

Look at the yellow area: for very long wavelengths, it beautifully matches the radiation from a 5250 °C or 5500 K black body. Around 1000 nm, the radiation from the Sun is lower than that but in the visible interval - the colors we can see by our eyes - the Sun's radiation is bigger than that of 5250 °C black body.

The excessive radiation may be viewed as radiation coming from somewhat deeper - and hotter - shells of the Sun that nevertheless manages to escape. Because 5250 °C is not the highest temperature of a place inside the Sun that can send us photons, the 5250 °C black body curve is not quite the upper limit for the radiation. On the other hand, the reduced radiation may be interpreted by the Sun's not being quite black: because it wouldn't absorb all the light at the frequencies, it also doesn't emit the maximum it can emit.

However, you should still appreciate how close the solar radiation is to the radiation of a 5250 °C black body. The agreement becomes arbitrarily good for very long wavelengths but it is never too bad, either. Only when the absorption is really strong - like the absorption for extremely high frequencies (see the left margin of the picture, near the "O3" label), the solar radiation is much less than that.

Now, look at the dark orange or red region. This is the radiation measured at the sea level. You see that it is lower than the yellow curve. It's because a part of the solar radiation is absorbed by the atmosphere. How big a part of it is absorbed depends on the wavelength (color).

In the left part of the picture, you see that there is a label "O3" - ozone - and the dark orange function is really low, much lower than the yellow or black graphs. That's because ozone absorbs most of the ultraviolet radiation. The ultraviolet radiation is the radiation whose wavelength is shorter than the wavelength of the visible light. That's why the ozone is important to protect us against the ultraviolet radiation.

The maximum of the solar curve appears in the "visible" band. You see that the atmosphere reduces the amount of light - the dark orange graph is lower than the yellow one - but no colors in this interval are "completely" suppressed. That's why we can see objects of any color in the visible band as long as the object reflects the sunlight - which still contains photons of all visible colors.

For infrared radiation, you see that there are bands that are almost untouched by the atmosphere. And there are bands that are mostly absorbed by the atmosphere, especially by H2O (and, for one sharp line, by oxygen O2, and for one very small band, by CO2).

**Outgoing radiation**

This was the energy that is coming from the Sun. It is being transformed to many other forms on the Earth - but eventually, all these forms become nothing else than thermal energy, general havoc at the atomic level. It's clear that if the Earth were only receiving the energy, it would have to be getting warmer. Imagine that you warm an object by 235 or 342 Watts per squared meter and you try to do it for 4.7 billion years, without allowing the object to cool down. Clearly, it can't work like that.

In the long run, the Earth has to emit the same energy it has received from the Sun. It is emitting the energy back to space in the form of its own thermal radiation. Both the surface and the atmosphere - especially the troposphere, its lower part - are comparably important sources of the thermal radiation.

Now, to make the first key order-of-magnitude estimate of the Earth's temperature, imagine that the Earth is just a black body. The radius of the Sun is 695,000 kilometers which is about 2.3 light seconds. The distance between the Earth and the Sun is 500 light seconds, about 8 light minutes.

The ratio is approximately 220. So the area of the Sun's surface is (220)^2 smaller than the area of a sphere whose radius equals the Sun-Earth distance. However, the energy coming from the Sun is not divided just to the cross section of the Earth, which is the disk of area pi.R_{Earth}^2, but to the whole surface of the Earth, 4.pi.R_{Earth}^2 - which is 4 times larger. So the energy from the Sun is diluted by the factor of 1/(4.220^2), when it's distributed over the Earth.

The radiation goes like temperature^4 - recall the Stefan-Boltzmann law. What is the ratio of the average Sun's surface temperature and the average Earth's surface temperature? It must be such that its fourth power gives us 4.220^2, to compensate for the lower energy density per area. So the temperature ratio is

TThe absolute temperature of the Earth should be 21 times lower than the Sun. If you use the temperature of the Sun at 6000 K, you get 6000/21 K = 286 K, which is exactly the global mean temperature on the Earth, up to 2 degrees._{Sun}/ T_{Earth}= (4.220^2)^{1/4}= sqrt(2.220) = sqrt(440) = 21 or so

The calculation actually gave us a much better result than it should: the error was just one percent or so. But we have neglected various things: the temperature of the Sun is somewhat lower than 6000 K; and the Earth reflects a part of the radiation ("albedo" means "reflectivity") which makes it cooler than if it were absorbing everything as we were assuming.

If you take these things into account, you find the temperature of the Earth which is about 33 K lower than the observed one. However, there is an error on the opposite side, too. The greenhouse effect reduces the overall emissivity of the Earth in the infrared: it doesn't allow the infrared thermal radiation to get from the the surface to the outer space. So the Earth's surface temperature has to get correspondingly higher so that despite the censorship of the infrared radiation, it sends enough energy away. The latter "greenhouse" effect, mostly caused by water vapor, returns the mean surface temperature of the Earth to those 288 K or 15 °C as expected.

The Earth's atmosphere has a lot of complicated dynamics, clouds, convection, conduction of heat etc. These phenomena add a lot of complexities but if you look at it properly, these complexities can't be arbitrary. In particular, the atmosphere can only reduce, but not increase, the amount of radiation that gets through in the corresponding direction: regardless of the wind, conduction, convection, re-emission by greenhouse gases, vapor going up, or whatever else, 100% is the ultimate upper bound on the amount of radiation that gets from one place to the other.

Because the calculation above should have shown you that the energy flows are very close to those expected from the black body, it is right to consider all other effects as small perturbations. You shouldn't forget about the fourth power in the Stefan-Boltzmann law. If you change the energy flows by 4%, the absolute temperature only changes by 1% because the fourth root of 1.04 is about 1.01. The fourth power really means that the relative error in the absolute temperature (in Kelvin) is four times smaller than the relative error of the energy flows.

Any large deviation from the black-body relationship between the near-surface air temperature and the amount of infrared radiation - which crosses the tropopause - is likely to create trouble for your theory. The greenhouse effect may be hypothesized to be amplified by various feedbacks - and there's no problem with that at higher temperatures. However, if you assume that these feedbacks also existed at "just slightly lower temperatures", you will quickly run into problems because your curve for the "total thermal emissions" will quickly exceed the allowed black-body bounds for the temperatures that are a few dozens degrees below the current temperature.

In the same way, if you suppose that the temperature is increasing much more slowly with the energy flows than the black body formula suggests, then you will face the very same problems at higher temperatures - temperatures above the current global mean (surface) temperature.

Imagine that you draw the "most accurate" curve depicting the total outgoing thermal radiation - measured through the tropopause - as a function of the global mean (near-surface air) temperature. I am confident that after you "smoothen" the curve over 1 °C or so, to get rid of some localized wiggles, the slope will never differ from the rescaled black-body slopes by more than 50%. I am actually confident that it will be less than 10%, but just to be sure, more than 50% seems really implausible.

For physical reasons, I think it's downright crazy to deny that any energyflow/temperature relationships we may study are just "somewhat corrected" relationships that can be obtained for black bodies. This also translates to a corollary - another argument why I don't believe that the climate sensitivity can be outside the (0.5 °C, 1.5 °C) interval.

But more generally, I wanted to stress that the black bodies are crucial players in any discussion about the thermal radiation - in black body physics, CMB discussions, or climate science - while all the other complexities and heat flows just determine how effectively some particular radiation can get from one place to another. If you forget that the role of the atmosphere is just to modify various black-body quantities by factors of order one, you can easily end up with completely unrealistic theories.

And that's the memo.

**Bonus: Cosmoclimatology and faint young Sun paradox**

Christopher Karoff and Henrik Svensmark (yes!) have used a Sun-like star, kappa Ceti, as a template for the Sun when it was young.

In the 1970s, Carl Sagan pointed out that the young Sun's luminosity was about 25% lower than today and some people think that it should have been frozen - the simple black body calculation would produce the melting point of water as the global mean temperature.

Fine, but let's accept that the problem was there: the young Sun would naively leave the young Earth frozen and uninhabitable. Various solutions have been attempted in the literature - including a hypothetical higher CO2 (which seems very unlikely). Svensmark and Karoff propose a new solution and argue that the Sun may have ejected some extra stuff - much like kappa Ceti does - which would have protected the Earth against the cosmic rays, strongly reduced the number of clouds, and warmed the Earth.

I present in

ReplyDeletehttp://www.nada.kth.se/~cgjoh/ambsblack.pdf

an alternative mathematical model for the cut-off of high frequencies in blackbody radiation, based on finite precision computation instead of the statistics of quanta which Planck considered to be only a mathematical trick without physical meaning.

Maybe it can be of some interest if blackbody radiation is now hot stuff in climate science.

Lubos wrote: "the black body is one of the most crucial and most universal concepts in thermodynamics."

ReplyDeleteThat's a stretch, Lubos. It's a nice problem in statistical mechanics, but it's no more "crucial" or "universal" than any other

applicationof statistical mechanics like the classical ideal gas or the quantum harmonic oscillator.The problem of the black body radiation was undeniably important in the development of modern physics, but you are exaggerating its importance for thermodynamics. It deserves a chapter in a stat mech text, but one can safely skip that chapter if pressed for time and still learn thermodynamics. Try doing that with truly foundational topics like entropy or the Gibbs ensemble...

Dear Oleg, well, the quantum harmonic oscillator and the ideal gas are surely paramount, too. And in some counting, probably even more important than a black body.

ReplyDeleteBut they're not about the transmission of heat which is what we're talking about here. In that problem, the black body is much more universal because it doesn't make any assumptions about the spectrum or the character of interactions inside the physical system.

It only says that the system is such that it absorbs incoming energy (radiation). For this reason, it's more universal.

I don't know how many sections or chapters there should be about these things and I didn't say any number. This is my first posting fully dedicated to the black body after 3400 postings. ;-) But the idea of yours that one can learn thermodynamics by skipping the black body is entirely preposterous.

That's how you may have "learned" thermodynamics but you haven't really learned it at the end, did you? Half of applications involving heat also include some radiation, and thermodynamics of radiation without the black body is an oxymoron.

Dear Lubos,

ReplyDeleteSince long time I wonder how the Planck-black-body law should be modified when applied to the Earth solid (or a liquid) surface to account for evaporation and to direct heat transfer from the surface to air (and next upward via convection). You correctly write that the Planck law constitutes a limit. An emissivity coefficient (smaller than 1) should be introduced but how this coefficient will depend on frequency an on the kind surface (tropical forest ,ocean)?

How one can derive such an emissivity coefficient?

Dear Lubos,

ReplyDeleteSince long time I wonder how the Planck-black-body law should be modified when applied to the Earth solid (or a liquid) surface to account for evaporation and to direct heat transfer from the surface to air (and next upward via convection). You correctly write that the Planck law constitutes a limit. An emissivity coefficient (smaller than 1) should be introduced but how this coefficient will depend on frequency an on the kind surface (tropical forest ,ocean)?

How one can derive such an emissivity coefficient?

Dear Stefan,

ReplyDeletegood questions.

If you measure the emissivity of the surface locally, the Earth's surface has 0.96-0.99, except for pieces of desert that can drop up to 0.7. For the higher figures, they're so high and close to one that it doesn't matter too much if you ignore the frequency dependence because it's so close to one that it looks "grey", anyway.

Clouds have emissivity 0.5. The average over the Earth is "0.612" - the real accuracy isn't that high.

By Kirchhoff's law, emissivity is the same thing as absorptivity. So while the figures above have been measured, you could also calculate them from the first principles. You prepare the initial state including an incoming photon and you try to calculate the probability that it is absorbed and/or reflected: a complex version of some QM scattering exercises. Of course, it's tough for realistic surfaces because the absorptivity also depends on their precise "fineness" of the material comparable to the wavelength and many other things. Note that the diamond's and graphite's emissivities - and colors - are very different.

If you include the troposphere to the Earth, as I did, this part of the atmosphere will act as a coating that reduces the emissivity of the surface, too: not everything about the surface's IR radiation gets out through this coating. But I don't know of any simpler way to calculate the precise value of the emissivity as reduced by the phenomena in the atmosphere - than to go through the normal climate science. The problem is all of atmospheric or climate science here. However, I still think that the upper bound holds.

Cheers

LM

Hi Lumo:

ReplyDeleteI have written a couple of posts Climate Sensitivity 1-5 on my blog

http://claesjohnson.blogspot.com/search/label/climate%20sensitivity

claiming that basic climate sensitivity is 0.15 C (by Fourier's Law), rather than 1 C (by Stefan-Boltzmann's Law) as IPCC alarmism rests on. Would appreciate if you take a look at the argument. Seems central to me.

Claes

Dear Claes,

ReplyDeleteit's surely a very original approach ;-) but I don't think that the details are right.

For others, he says that CO2 doubling adds 1% to the radiative forcing, which improves the insulation by the troposphere by 1%, which warms up the surface-tropopause difference by 1% of the old difference 15 - 0 °C = 15 °C, so the climate sensitivity is 0.15 °C.

That's a great argument. However, the temperature of the tropopause is about -55 deg C, rather than zero, so it is 70 deg C below the surface temperature.

One percent of it is 0.7 °C which is pretty close to those 1.2 °C, given the approximate nature of your argument.

Cheers

LM

No, the temperature on top of the troposphere is

ReplyDeleteroughly 0 C. See e.g.

http://apollo.lsc.vsc.edu/classes/met130/notes/chapter1/vert_temp_all.html

http://www.windows.ucar.edu/tour/link=/kids_space/temp_profile.html

Right?

Dear Claes, next time, I will always be losing my patience.

ReplyDeleteBoth graphs you show are identical, and both of them show that the top of the troposphere - the tropopause - is near -55 deg C in average.

That shouldn't be shocking for you if you ever - in your life - flew in an airplane and looked at the outside temperature.

The temperature graph is the wiggly one, and you want to look at the lowest wiggle how it goes on the left.

The lapse rate is about -6.5 deg Celsius per kilometer and the tropopause is about 10 km above the surface.

Cheers

LM

No I mean top of the stratosphere! About 0 C.

ReplyDeleteThe temperature decreases though the troposphere, you are right, but increases through the stratosphere, and thus is represented by the broken line in the model fig in

http://claesjohnson.blogspot.com/2010/04/on-climate-sensitivity.html

Right? I think this is very essential.

Dear Claes, stratopause - the top of stratosphere - is near minus 10 degrees Celsius, close to your zero (freezing point), but I have no idea what stratopause has to do with the surface warming.

ReplyDeleteIn your insulation model, which is kind of a nice intuition, both troposphere and stratosphere may be viewed as insulating layers. But because of the reversed lapse rate, the stratosphere isolates the adjacent layers against the "opposite" flow of heat.

Why don't you go up to e.g. mesopause?

Your initial argument was nice. One can say that CO2 increases the isolation by 1.5% between the surface and the tropopause (3.7 W/m^2 is 1.5% of 245 Watts/m^2, the IR outflow, or so), so it allows the surface-tropopause temperature difference to jump by 1.5% of those 70 °C today - by 1.2 °C.

With realistic numbers, it gives you realistic results. It's about the isolating power of the troposphere. Troposphere combined with stratosphere - which have the opposite signs of the lapse rate - is a much more complicated and much less relevant quantity.

Cheers

LM

Fine, we agree so far. And the model is simple. It remains to be seen if it describes the coupled tropo/stratosphere as window towards outer space.

ReplyDeleteNevertheless, what is clear, to me at least, is that Stefan-Boltzmann should be not used to draw conclusions about climate sensitivity. It is as crazy as using Stefan-Boltzmann to compute the U-factor of a window. This is basic physics and should be understood by many.

Dear Claes,

ReplyDeleteis it "clear" to you that the Stefan-Boltzmann calculation is wrong?

Haven't you just agreed that your method, when corrected, does *confirm* the Stefan-Boltzmann result?

You surely have no arguments against the S-B picture that would still be alive.

Best wishes

Lubos

If you believe that SB applies to global climate, I suggest you compute the temperature in the room your are sitting in, using SB, because this is a similar situation. What result do you get?

ReplyDeleteTo believe that 1% change of albedo will change

surface temperature by 1 C is nothing but black body magic, as far as I can understand.

Dear Claes,

ReplyDeletethe window in this room is open, so the temperature is aligned with the temperature outside.

I calculated the temperature outside - the mean Earth's surface temperature - by matching the Sun's black body radiation to the outgoing Earth's black body radiation in this very article, and the result is 15 deg Celsius.

When I look at the thermometer, it is 20 deg Celsius, because it is a pretty warm day (14 deg C outside) and I kept the building somewhat warmer than the exterior. At any rate, the agreement is impressive.

What else did you want me to calculate?

1% change of albedo obviously has to bring O(1) degree Celsius of temperature change. 100% change of albedo changes the temperature of the Earth by hundreds of degrees - do you have doubts about it? If the albedo were 100%, the Earth's temperature would drop close to absolute zero because the solar radiation would be reflected.

Even if you consider less singular 50% increase in the albedo, it's clear that a substantial portion of the energy would be reflected, and the absolute temperature of the Earth would drop by a substantial part of the current 288 K. I don't understand what's your problem.

Best wishes

Lubos

In your article you assumed a Sun surface temperature of 6000 K, and obtained 15 C as temp on Earth.

ReplyDeleteBut 6000 K seems too high. 5700 K appears to be more correct which gives 0 C on Earth.

From where did you get 6000 K?

Dear Claes,

ReplyDelete6000 kelvins is the standard rounded figure - click the link.

At any rate, it differs by 5700 by 5% only - which can't account for your estimates that differ from the right figures by 900% (of your estimate).

Best wishes

Lubos

Dear Lumo,

ReplyDeleteNo the effective Sun temp is 5778 K according to

http://nssdc.gsfc.nasa.gov/planetary/factsheet/sunfact.html

which gives 5778/21 = 275 K . If you allow yourself to fiddle with the temp, then you can any number between 0 C and 15 C, from SB. But that is not science.

Best, Claes