Krammis

Eli has been mining Gerhard Krimms contribution to arXiv and has come up with yet another Krammi. We had a recent discussion of the surface temperature of the moon here at Rabett Run,

The Moon is a good example to contrast with the Earth. It rotates much more slowly, and therefore has a temperature distribution that approaches what is used by Gerlich and Tscheuschner to derive their "Tbut Gerhard shows a contradictory example_{phys}". Each point on the Moon's surface is tolerably close to radiative balance with the solar input at that point.

The Moon has an albedo of about 0.12. It therefore absorbs more of the incoming solar energy than Earth. Using the solar constant of 1369 W/m^{2}, the absorbed radiation for the surface facing the Sun is about 1205 W/m^{2}. Hence T_{eff}for the Moon is (1205/4/σ)^{0.25}= 270K, or -3C. This is the temperature that would radiate back the solar energy, if evenly distributed over the moon. But directly facing the Sun, the temperature will be more like (1205/σ)^{0.25}= 382K, or 109 C. Albedo is not uniform. In any particularly dark patches, the temperature could even get up to (1369/σ)^0.25 = 394K, or 121C. On the night side, however, temperatures will fall toward absolute zero. Bear in mind that as temperatures fall, so too does the rate of emission of energy. Hence it takes a long time to fall all the way to zero. Say rather that temperatures should fall far enough for the emission of energy to be small.

Now consider data on the Moon from http://www.solarviews.com/eng/moon.htm

Average day temperature is 107 C. Maximum day temperature is 123 C. These are close to theoretical expectation, to within a couple of percent.

The mean night temperature is -153C. This about 120K, and radiates a bit less than 12 W/m^{2}. That's less than 1/100 of the solar constant, so the temperature has indeed fallen close to zero, using radiated energy as the basis for comparison.

There's no average temperature given, but the mid point of mean day and mean night temperatures is in the ballpark. This is -23C. And, just as should be expected, it is somewhere between T_{phys}(-120C) and T_{eff}(-3C). But it is closer to T_{eff}, because it is the cool side of the moon that is most different, in absolute temperature, from the unphysical extreme that is the basis of Gerlich and Tscheuschner's T_{phys}

On Earth, fortunately, we have an atmosphere that has to be heated from the surface. By basic thermodynamics, the Earth's average surface temperature is therefore substantially warmer than our airless moon. where surface radiation escapes directly to space.

Figure 5: Moon's disk temperature at 2.77cm wavelength versus moon phase angle φ during two complete cycles from twice new moon via full moon to new moon again (adopted from Monstein.)with a comment below of

Our Moon nearly satisfies the requirement of a planet without an atmosphere. It is well known that the Moon has no uniform temperature. There is not only a variation of the temperature from the lunar day to the lunar night, but also from the Moon equator to its poles.There is nothing necessarily wrong with the Krammigraph (Eli has not checked, but it compares reasonably to similar ones that he has seen. What's up? Work out your answer and send it with to box tops (we are not particular about the boxes) to Rabett Run Krimmi Kontest @ yahno.con. Extra prizes for those who figure out how to send the boxtops.

Using Eqs. (1.4) would provide T 270 K e ≈ when the albedo, 0.12 M α = , and the emissivity, 1 M ε = (black body), are considered. However, as illustrated in Figure 5, the mean disk temperature of the Moon observed at 2.77cm wavelength by Monstein (2001) is much lower than this equilibrium temperature

ANSWER: Look carefully at the wavelength that the brightness temperature was measured at little bunnies. Hmm, 2.77 cm. That's pretty long wave. It turns out that long wavelength radiation (and 2.77 cm is pretty long wave compared to IR even which is micrometers) penetrates through about 4 to 8 times the wavelength. Thus if you use 2.77 cm, what you are measuring is the temperature of the moon 10 - 20 cm below the surface and that is a lot cooler and has a lot less variation with the diurnal cycle. What about if we measured at the surface. Well Apollo 15 left a temperature measuring instrument and this is what it found

Fig. 2. Lunar surface temperature time series from the landing site of the Apollo 15 mission. The series are part of the historical data archive PSPG-00093 of the US National Space Science Data Center. The color-coding of the temperature series used in this diagram is also used in Figs. 3 and 4.The answer is that the Krammi did not show surface temperature, but below surface temperature.

## 9 comments:

Kramm is the energizer bunny of Gerlich and Tscheuschner defense! But how does he get his coauthors (Dlugi, Zelger in this case) to go along with such illogical silliness?

Anyway, thanks for digging up this one. Ironically, the only real substance in this very long comment on my proof seems to be section 5 which proves, as I did, that infrared absorption raises the temperature of the surface - *completely contrary to the claims of Gerlich and Tscheuschner*. I rest my case :)

Gerhard Krimm? Surely you mean 'Grimm'? Freudian? More fairy tales...

:wq

I don't have a problem with Kramm's graph here. It still serves to give you a mean surface temperature, and show that it is less than Teff.

I'm more than happy to include "surface" as incorporating a couple of centimeters of surface material. In his graph, the mean temperature works out to be about 215 or 220K, which is about -58 to -53C. This rather less than the very top surface, which I estimated previously as about -23C. And it is certainly less than the Teff of -3C.

Of course, in the original G&T nonsense, there was no attempt at damping, and so the very top surface is a better fit to the G&T paper; so I would continue to use the surface that radiates into space as more directly relevant.

The real problem with Kramm's paper and the discussion of this graph remain the same things that he got wrong here consistently. He doesn't understand surface integrals well enough to handle change of variables, or get the units right. He doesn't understand a weighted average.

OT: Eli, I don't recall seeing you or anyone blog on the recent bloodletting (documented here) over at the APS Physics & Society newsletter. Also, I would love to hear any skinny that may be available regarding the APS Council's review of their climate change statement (assuming this was described accurately by S. Fred et al's recent correspondence in Nature).

Steve, OTOH in things like that people wear gloves so it takes at least a few conferences to get the details. S. Fred OTOH is attempting an interesting sidestep back to semi-respectibility as WAH details and other noises.

Steve: what's the question, i.e., what did you expect @ APS?

This (wholesale replacement) was pretty much predictable as of last July/August, given the Monckton mess....

Hmm - on the problem with the graph (and explication) - isn't the "disk temperature" determined radiatively - i.e. it's effectively an average over the Moon's disk at any given time of <T^4>, not the average <T> he claims?

There's also something funny about measuring radiation from a disk when the actual radiation is from the sphere - but I haven't thought about it long enough to decide whether it makes any difference to the average. Didn't see any discussion of the difference in Kramm's paper though.

An aside - I also don't see how he can claim it's "two complete cycles" when it's just one, but that's pretty minor...

Ah, no I was wrong about the <T^4> - of course this is measurement at a single very long wavelength, where Planck's formula gives an intensity pretty much linear in temperature, so it really would give <T>.

However, of course disk-averaging is not quite the same as sphere-averaging - it actually over-weights the equatorial regions and under-weights the poles. You can see this by looking at the area of the portion of a disk or a sphere with |x| <= 1/2 r (y and z unconstrained). For a disk that close-to-equator area is (pi/3 + sqrt(3)/2) r^2 which is about 61% of the total surface area; for a sphere that area is just 2 pi r^2, or half the total.

So, he's looking at that average is still not quite right. Not too far off though.

In other news, it appears that Kramm is back in Deltoid spouting his usual nonsense. Think he chickened out from here after getting the academic pistol-whip from Arthur, DQ, Eli and everyone else here. Tsk Tsk.

Unlike RPSr. who has at least maintained a semblance of decorum in his harrumphing, it's hard to feel pity for GK when he resorts to puerile arguments and mudslinging. What a joke of a (previously) fine mind.

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