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Space Forum / Astronomy / June 2005Cosmic Microwave Background first used in print?
Anyone know (a) when the term "Cosmic Microwave Background" was first used in print (b) first used in print in connection with the Big Bang (c) first used in print as a prediction? Regards, Ian Tresman Derby, UK >>>>> "i" == it <it@knowledge.co.uk> writes: i> Anyone know (a) when the term "Cosmic Microwave Background" was i> first used in print (b) first used in print in connection with the i> Big Bang (c) first used in print as a prediction? The phrase "cosmic microwave background" appears first according to ADS in the paper "Peculiar Velocity of the Sun and its relation to the Cosmic Microwave Background" by Stewart & Sciama (1967, Nature, vol. 216, p. 748). The discovery paper by Penzias & Wilson speaks only of an excess antenna temperature. The companion paper by Dicke et al. is entitled "Cosmic Black-Body Radiation." The original prediction of a cosmic blackbody radiation is typically ascribed to the paper by Alpher, Bethe, & Gamov (the so-called alpha-beta-gamma paper). I haven't been able to find a copy of it, but I bet they speak of a cosmic blackbody radiation rather than a cosmic microwave background.  SignatureLt. Lazio, HTML police | e-mail: jlazio@patriot.net No means no, stop rape. | http://patriot.net/%7Ejlazio/ sci.astro FAQ at http://sciastro.astronomy.net/sci.astro.html Dear Joseph Lazio: >>>>>> "i" == it <it@knowledge.co.uk> writes: > [quoted text clipped - 22 lines] > cosmic blackbody radiation rather than a > cosmic microwave background. Additionally, I believe radiant heat transfer had the "Universe" at about 5 K to achieve night time rates of cooling, based on measured/assumed absorptivity and emissivity of "air". But I don't expect this predates the detection of the CMBR. It does speak to the "prediction", however. David A. Smith > Additionally, I believe radiant heat transfer had the "Universe" > at about 5 K to achieve night time rates of cooling, based on > measured/assumed absorptivity and emissivity of "air". I'm confused. What does the temperature of the CMBR have to do with the nighttime rate of cooling? I suppose observed nighttime cooling sets an upper limit on the CMBR temperature, but that's probably of order 150 K. SignatureSteve Willner Phone 617-495-7123 swillner@cfa.harvard.edu Cambridge, MA 02138 USA (Please email your reply if you want to be sure I see it; include a valid Reply-To address to receive an acknowledgement. Commercial email may be sent to your ISP.) Dear Steve Willner: > In article <rNYqe.7052$7s.5412@fed1read01>, > "N:dlzc D:aol T:com \(dlzc\)" <N: dlzc1 D:cox T:net@nospam.com> [quoted text clipped - 8 lines] > nighttime cooling sets an upper limit on the CMBR > temperature, but that's probably of order 150 K. Actually, no. The Universe (excluding the Sun, of course) accepts radiant heat from the Earth, that does not get absorbed and re-emitted by the air, at a temperature of about 5K, just as I said above. The CMBR is a major contributor to this 5K background... a "floor" if you will. In *very* dry climates, even in the summer, it is possible to freeze thin sheets of ice by diverting radiant heat from the Earth and exposing the water to open sky. David A. Smith > The Universe (excluding the Sun, of course) > accepts radiant heat from the Earth, that does not get absorbed > and re-emitted by the air, at a temperature of about 5K, just as > I said above. The CMBR is a major contributor to this 5K > background... a "floor" if you will. I think you need to show your calculations. In the first place, what do you mean by "does not get absorbed and re-emitted by the air?" What wavelengths would that be? Or are you talking about an object above the atmosphere? You might want to note that at 150 K, half of a blackbody's radiant energy emerges at wavelengths beyond 27 microns, where the atmosphere is pretty much opaque and quite a bit warmer than 150 K. Even if you are considering the situation above the Earth's atmosphere, radiation from the Zodiacal light is a much larger heat source than the CMBR. > In *very* dry climates, even in the summer, it is possible to > freeze thin sheets of ice by diverting radiant heat from the > Earth and exposing the water to open sky. Indeed; I've seen this myself. As I wrote, the equilibrium temperature could perhaps be as low as 150 K, plenty cold enough to freeze water. (I don't think it is really anywhere near that low, even in Antarctica, but it is lower than 273 K.) SignatureSteve Willner Phone 617-495-7123 swillner@cfa.harvard.edu Cambridge, MA 02138 USA (Please email your reply if you want to be sure I see it; include a valid Reply-To address to receive an acknowledgement. Commercial email may be sent to your ISP.) Dear Steve Willner: > In article <OvIue.586$Qo.230@fed1read01>, > "N:dlzc D:aol T:com \(dlzc\)" <N: dlzc1 D:cox T:net@nospam.com> [quoted text clipped - 7 lines] > > I think you need to show your calculations. It was empirical, based on local measurements. I don't have the figures at hand, so discount this if you choose. These were figures that I vaguely recall from "way back when". I do not have a literature citation. A "stab" was taken at the net absorptivity and emisstivity of the air. > In the first place, what > do you mean by "does not get absorbed and re-emitted > by the air?" What wavelengths would that be? The entire blackbody spectrum represented by 5K includes a *lot* of wavelengths. > Or are you talking about an object > above the atmosphere? No. > You might want to note that at 150 K, half of a > blackbody's radiant energy emerges at wavelengths > beyond 27 microns, where the atmosphere is pretty > much opaque and quite a bit warmer than 150 K. I'm talking about 5 K. > Even if you are considering the situation above the > Earth's atmosphere, radiation from the Zodiacal light > is a much larger heat source than the CMBR. A 5 K source radiates more power than a 3 K source, doesn't it? Nearly double, or some such. And the extra power would come from all the point sources represented by all the visible stars. >> In *very* dry climates, even in the summer, it is possible to >> freeze thin sheets of ice by diverting radiant heat from the [quoted text clipped - 4 lines] > enough to freeze water. (I don't think it is really anywhere > near that low, even in Antarctica, but it is lower than 273 K.) Agreed. David A. Smith > A "stab" was taken at the net > absorptivity and emisstivity of the air. It is still not quite clear what you were saying, but if you claim the radiative equilibrium temperature of an object near the Earth's surface could have anything to do with the microwave background, I claim you are wrong by at least 7 orders of magnitude. > I'm talking about 5 K. Don't you think your purported 5 K object will be absorbing radiation from the atmosphere? (I am ignoring conduction, which will also heat it.) > > Even if you are considering the situation above the > > Earth's atmosphere, radiation from the Zodiacal light > > is a much larger heat source than the CMBR. > > A 5 K source radiates more power than a 3 K source, doesn't it? > Nearly double, or some such. Yes, if the emissivities are equal. Power goes as T^4 as you probably know. But I don't see the relevance. The zodiacal light has two components. Both have higher temperatures than 5 K but very low emissivity, but the net heating from both is far below what comes from the atmosphere. > And the extra power would come from > all the point sources represented by all the visible stars. Yes, that power is comparable to the microwave background power and less than the Zodi. And in any case it represents an additional heat source, not a method of cooling. SignatureSteve Willner Phone 617-495-7123 swillner@cfa.harvard.edu Cambridge, MA 02138 USA (Please email your reply if you want to be sure I see it; include a valid Reply-To address to receive an acknowledgement. Commercial email may be sent to your ISP.) |
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