In article <1993Apr28.002214.16544@Princeton.EDU>, richmond@spiff.Princeton.EDU (Stupendous Man) writes:  >    If that's the case, let me point out that interstellar dust and  > molecules provide many instances of things that are, well, not-too-far > from being blackbodies.  Many different observations, including IRAS > and COBE, have determined that interstellar dust grain temperatures > can range from 40K to 150K.  Interstellar grains are not at all close to blackbodies.  The "large" grains have sizes of order 0.1 micron and absorb visible light with fair efficiency.  However, at temperatures below 100 K, 90% of the thermal emission will be beyond 22 microns, where radiating efficiency is poor.  (A small antenna cannot easily radiate at long wavelengths.)  Thus the grains must heat up more in order to radiate the energy they have absorbed.  Moreover, the IRAS observations had a maximum wavelength of 100 microns.  Grains colder than 30 K will radiate primarily at longer wavelengths, and IRAS would be relatively insensitive to them.  In the extreme limit, grains as cold as 5 K will be almost undetectable by any conceivable observation.  Worse still, IRAS color temperatures are heavily contaminated by a population of "small" grains.  These grains have only perhaps 50 atoms, and when they are hit by a single photon they heat up to temperatures of several hundred or 1000 K.  Of course they cool quickly and then stay cold for a while, but _when they are radiating_ the characteristic temperature is several hundred K.  Even a small population of these grains can dramatically raise the observed "average" temperature.  A model for local infrared emission consistent with COBE data has three components.  These represent scattered radiation from Zodiacal dust (color temperature 5500 K), thermal emission from Zodiacal dust (Tc = 280 K), and thermal emission from Galactic dust (Tc=25 K).  At the ecliptic poles, the emissivities or dilution factors are respectively 1.9E-13, 4E-8, and 2E-5.  The first two are roughly doubled in the ecliptic plane.  To find the thermal equilibrium temperature, we add up the dilution factor times the fourth power of temperature for all components, then take the fourth root.  In the table below, starlight comes from Allen's number that stellar emission from the whole sky is equivalent to 460 zero mag stars with B-V color of 0.75.  No doubt careful work could do much better.  (The person who suggested starlight had a dilution factor of E-4 must have been remembering wrong.  We would be cooked if that were the case.  In any event, the energy density of starlight comes out about the same as that of the microwave background, and I believe that to be correct.)                           Dilution   Temp.   DT^4 Microwave background         1        2.7     53 Galactic dust              2E-5      25        8 Zodiacal dust (emission)   6E-8     280      369 Zodiacal dust (scattering) 3E-13   5500      275 Starlight                  1E-13   5500       92                                            -----                                              797  The fourth root of 797 is 5.3 K.  Outside the Solar system, the result would be 3.5 K.    I find these results surprising, especially the importance of Zodiacal dust, but I don't see any serious mistakes.