Horace Heffner wrote: > > On Jul 27, 2009, at 10:07 AM, Stephen A. Lawrence wrote: > > >> Suppose you took a lump of glass and placed it in an (evacuated) oven. >> Suppose further that the walls of the oven are dead black, absorbing >> (nearly) all radiation which falls on them, and assume that they radiate >> about as you'd expect a blackbody to radiate. >> >> Suppose further that the oven and the lump of glass are at the same >> orange-hot temperature (and let's ignore the fact that the glass has >> melted all over the bottom of the oven because that adds unnecessary >> complexity to the experiment -- maybe we put the whole thing in >> free-fall, or whatever). >> >> Now the walls of the oven are giving off a cheery orange glow. Assume >> the glass is glowing orange, too, and assume further that it's glowing >> just as brightly as the walls of the oven. (This is an assumption; we >> know glass glows *some* but we haven't confirmed that glass glows as >> brightly as something which starts out black.) > > I think there might be a misconception here about the difference in > behavior between heated surfaces and heated black body cavities. > Cavities, at least peep holes into cavities, act as almost perfect black > bodies. See: > > http://en.wikipedia.org/wiki/Black_body > > "In the laboratory, black-body radiation is approximated by the > radiation from a small hole entrance to a large cavity, a hohlraum. > (This technique leads to the alternative term cavity radiation.) Any > light entering the hole would have to reflect off the walls of the > cavity multiple times before it escaped, in which process it is nearly > certain to be absorbed. This occurs regardless of the wavelength of the > radiation entering (as long as it is small compared to the hole). The > hole, then, is a close approximation of a theoretical black body and, if > the cavity is heated, the spectrum of the hole's radiation (i.e., the > amount of light emitted from the hole at each wavelength) will be > continuous, and will not depend on the material in the cavity (compare > with emission spectrum)."
Yes, I understand this. > > I know from personal observation that it goes beyond this. As a cavity > and its contents heat up, everything in the cavity eventually disappears > from view through the peep-hole. I have personally sat and watched > through a gas forge observation port, which I kept open, the cover > lifted, as that gas forge, which was about 1' by 2' by 2', heated up. > Initially, I could clearly see the far walls of the forge and things in > it through the port. When the temperature rose to an orange glow, > suddenly nothing was visible inside the forge. There was a pure orange > glow coming from the observation port that had nothing to do with the > contents of the forge. One moment I could see the other side of the > forge, which had some hot spots and dark spots on it, and the next it > was replaced by flat orange glow. I could see nothing at all inside the > port. It was as if the hole surface itself (which is not a physical > thing) was radiating. Cool! I like your description -- I hadn't realized it would be so dramatic, but I suppose it must, when everything inside is glowing equally. Actually I do understand the distinction between heated ordinary bodies and an ideal blackbody or the peephole in an oven. Furthermore, the reason the peephole in an oven works that way, regardless of the material in the oven, is exactly the fact that a hot real-world object radiates exactly to the extent that it also absorbs. As a result, the sum of radiated, reflected, and transmitted radiation from the object (when it's in an oven of a fixed temperature) is identical to the radiation from an "ideal blackbody" of the same temperature, and since everything in the oven behaves that way, everything in the oven looks identical (when it's all glowing at a uniform temperature). And that's why I was wondering about the opacity of *glowing* glass. > > Best regards, > > Horace Heffner > http://www.mtaonline.net/~hheffner/ > > > >
