Timeline for Approximately what percent of the sky has nothing in it?
Current License: CC BY-SA 4.0
16 events
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| Feb 6, 2023 at 1:13 | comment | added | Loren Pechtel | @MichaelSeifert That -6.5 is looking at the whole visual spectrum rather than one frequency. I don't really think it matters that much overall, though--it's still going to be less than 1 part in a billion. | |
| Feb 5, 2023 at 13:32 | comment | added | Michael Seifert | @LorenPechtel: I'm calculating the spectral power density at 550 nm ≈ 545 THz, which is the nominal center of the visible magnitude band (using the figures here). The power density radiated at a fixed frequency/wavelength doesn't go as the fourth power; it's just proportional to the $1/(e^{h \nu/kT} - 1)$ factor in the Planck law. (It's only when you integrate over all frequencies that you get the Stefan-Boltzmann law.) | |
| Feb 5, 2023 at 4:22 | comment | added | Loren Pechtel | @MichaelSeifert I think something must be wrong with your energy calculation because radiated power goes at the 4th power of the temperature. It should have a bigger effect than you're getting. | |
| Feb 5, 2023 at 1:37 | comment | added | John Doty | Of course, stars aren't the only things that make light. The interstellar medium is very faint in the optical, but it's present on every line of sight. It's brighter at x-ray wavelengths. | |
| Feb 4, 2023 at 14:28 | comment | added | Michael Seifert | @ProfRob: I tweaked my preamble just now to mention "a sufficiently large telescope." :-) | |
| Feb 4, 2023 at 14:27 | history | edited | Michael Seifert | CC BY-SA 4.0 |
deleted 2 characters in body
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| Feb 4, 2023 at 14:27 | comment | added | ProfRob | Well of you have to see it, then a consideration of angular resolution is needed. But your calculation is fine for answering whether an infinitesimally narrow sightline intercepts the surface of a star. | |
| Feb 4, 2023 at 14:19 | history | edited | Michael Seifert | CC BY-SA 4.0 |
Modified calculation for red dwarf & inserted caveats
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| Feb 4, 2023 at 14:15 | comment | added | Michael Seifert | @ProfRob: That's a fair point. My calculation is focused on the things you could literally "see", since that's what the OP seemed to primarily be talking about. I've put a caveat at the start to reflect this. | |
| Feb 4, 2023 at 14:13 | comment | added | Michael Seifert | @LorenPechtel: Fair critique. I've edited the numbers for a 3000-K blackbody instead. It only changes the result by a factor of 2, though. (Even a 2000-K blackbody only changes my original result by a factor of 3.) | |
| Feb 4, 2023 at 14:12 | history | edited | Michael Seifert | CC BY-SA 4.0 |
Modified calculation for red dwarf & inserted caveats
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| Feb 4, 2023 at 8:40 | comment | added | ProfRob | By that, I mean that every sight line does end up on something that had the surface temperature of a star - it's the cosmic microwave background. The argument presented at the moment is only that the sky cannot be (geometrically) full of the discs of unredshifted stars. | |
| Feb 3, 2023 at 23:26 | comment | added | ProfRob | Also seems to ignore redshift. | |
| Feb 3, 2023 at 23:01 | comment | added | Loren Pechtel | I disagree on your math--you are assuming a star as bright as our own and in reality most stars are red dwarfs and you are neglecting the luminosity drop off as you approach the limbs of the star. However, I fully agree with the basic concept, you're at most a few zeroes off and that doesn't meaningfully change the basic concept. The sky must be empty or we cook. | |
| Feb 3, 2023 at 22:19 | comment | added | Michael Seifert | Calculations like the above are not really in my wheelhouse, so please take the above numbers with a grain of salt, and if someone would be so kind as to check my work I'd be grateful. | |
| Feb 3, 2023 at 22:18 | history | answered | Michael Seifert | CC BY-SA 4.0 |