Agreement with measurement is the true test of a theory, right?

As I've shown, standard radiative theory, properly applied, fully accounts for the dayside lunar temperature at all points. Hence, your (increasingly convoluted) ideas about emission temperatures and such must be incorrect.

Please read the comments of Joel Shore of the Rochester Institute of Technology:

http://davidappell.blogspot.com/2012/04/norfolk-constabulary-made-wrong-charges.html?showComment=1333542409674#c5805847283678492801

In particular, this: "There are many temperature distributions with many different average temperatures that are all compatible with a blackbody emitting 240 W/m^2 (which is what the Earth has to emit as seen from space to be in radiative balance). Which one is actually realized depends on details like the body's thermal inertia, thermal transport, etc. However, none of these temperatures is higher than 255 K, which is what a body with a perfectly uniform temperature distribution has to have to emit 240 W/m^2. Nobody has successfully explained how one can satisfy conservation of energy if a planet is at an average temperature of ~288 K, emitting about 395 W/m^2, while the planet (including atmosphere) is only absorbing ~240 W/m^2 from the sun unless you invoke a radiative greenhouse effect. And, it is a fool's errand to try to do so because it is impossible."

----

I'll go through the derivation of the lunar temperatures one more time, filling it all in.

Here are the Diviner measurements:

http://tallbloke.wordpress.com/2012/01/17/nikolov-and-zeller-reply-to-comments-on-the-utc-part-1/

I'll focus on the equator, since it's simplest, but the same physics applies at all latitudes.

Unlike on Earth, when the Sun shines on the Moon you can't assume equilibrium, so radiative theory must be applied pointwise. So, at the point on the Moon where Sunlight is normal to the surface, the temperature will be

sigma*T^4 = S*(1-alpha)

when lunar solar constant is the same as the Earth's, 1362 W/m2. The lunar albedo is 0.11 (via http://www.asterism.org/tutorials/tut26-1.htm), so this gives

T0 = [S*(1-alpha)/sigma]^1/4 = 382 K

Notice that agrees exactly with the Diviner result (black curve) as per the chart referred to above. (There may be some small astronomical factors I'm not going to worry about.)

Elsewhere on the equator, the intensity of sunlight is reduced by a factor of the cosine of the longitude L. This changes the result for T only by a factor of [cos(L)]^(1/4), which exactly describes the shape of the curve in the Diviner plot.

The average dayside temperature will thus be T0 times the average value [cos(L)]^(1/4) from -90 degrees to 90 degrees. Using the standard calculus method to calculate this average, it equals 2.69991/pi. So the average dayside temperature will be

<dayside lunar equatorial temperature> = (2.69991/pi)*T0 = 328 K

Again, the nightside temperature can't be determined by radiative considerations alone, since it depends on heat conduction through the regolith. (See, for example

"Near-Surface Temperatures on Mercury and the Moon and the Stability of Polar Ice Deposits," A. R. Vasavada et al, Icarus 141 (1999) 179-193

http://www.gps.caltech.edu/classes/ge151/references/vasavada_et_al_1999.pdf ).

I'll just call it Tn, which from Diviner is apparently about 95 K.

Hence the entire average is

<lunar equatorial temperature> = 0.5*(328 K + 95 K) = 212 K.

EXACT AGREEMENT, both for the average and for the entire dayside.

I fail to see what is lacking, especially since your ideas have not accounted for any lunar temperatures at all.

David

[deleted]

On 4/27/2012 8:34 PM, David Appell wrote:To: Christopher Monckton

Cc: Andrew Watts, Roger Tattersall, Ned Nikolov

On 4/25/2012 8:19 AM, Christopher Monckton wrote:NASA gives 270.6 K; and the IPCC's value of 255 K for the characteristic-emission temperature is based on similar considerations. In the light of the Diviner results, this value may be too high, in which event so is the Planck parameter. - M of B

Mr Monckton,

*WHERE* does NASA give a value of 270.6 K? More importantly, why do you think it matters? Scientific truth doesn't depend on who wrote it -- it depends on whether theory agrees with experiment.

As I've shown you twice now, a correct application of standard radiative theory gives not only the correct average for the lunar equatorial temperature as measured by Diviner, but it provides all the values (such as the maximum) and shape (~[cos(longitude)]^(1/4)) of the entire sun-side equatorial temperature, at all sun-side longitudes.

[And again, the nightside temperature cannot be determined by radiative transfer alone, but depends on heat conduction through the lunar regolith. See, for example

"Near-Surface Temperatures on Mercury and the Moon and the Stability of Polar Ice Deposits," A. R. Vasavada et al, Icarus 141 (1999) 179-193

http://www.gps.caltech.edu/classes/ge151/references/vasavada_et_al_1999.pdf ]

Thus, standard theory completely -- and easily -- explains the Diviner measurements. Claims that it somehow disproves the canonical calculation of the Earth's greenhouse effect is simply wrong, due an incorrect application of basic science promulgated by Roger Tattersall, Ned Nikolov, and Karl Zeller. They are wrong.

Will you be correcting your WUWT post?

David

----- Original Message -----From: David AppellSent: 04/25/12 03:16 PMTo: Christopher MoncktonSubject: Re: Diviner result

On 4/24/2012 3:33 AM, Christopher Monckton wrote:Now, the method that NASA used in order to derive the 270 K value for the Moon is the same method that is routinely used in climate science to derive the 255 K mean characteristic-emission temperature for the Earth, raising the possibility that 255 K is also too high.

Such a result, whoever did it (I doubt it was NASA), is an incorrect application of the physics of radiation transfer. Unlike on Earth, on the Moon you can't assume equilibrium, so radiative theory must be applied pointwise.

Here's how standard theory gives not just the correct curve, but the correct average too:

On the sunlight side of the moon, the average temperature will be

=B*integral

where, as usual,

B=[S*(1-alpha)/sigma]^1/4 = 382 K

and the integral results from averaging the angular factor (a cosine to the 1/4th power) from -pi/2 to +pi/2, which must be done numerically. So on the sunlight side of the Moon the average temperature is

(2.7/pi)*B

The temperature of the dark side cannot, as you imply, be considered from radiative theory, since it also depends on heat conductance. From the data it is approximately

=95 K.

Averaging these two numbers gives

=212 K

in exact agreement with the data.

David

----- Original Message -----From: David AppellSent: 04/24/12 05:55 AMTo: Christopher MoncktonSubject: Diviner result and standard theory

To: Christopher Monckton

Cc: Anthony Watts, Roger Tattersall

Re: http://wattsupwiththat.com/2012/04/23/why-there-cannot-be-a-global-warming-consensus/

Mr. Monckton:

Your 4/23 post on WUWT saying that standard theory cannot explain the Diviner measurement of lunar temperature is incorrect.

In fact, as I show here:

http://davidappell.blogspot.com/2012/04/norfolk-constabulary-made-wrong-charges.html

standard theory not only easily gives the exact measured value for the average lunar equatorial temperature, but it explains that temperature for all longitudes.

Sincerely,

David

`[deleted]`

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