Talk:Olbers's paradox

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Cosmic microwave background

Main article: Cosmic microwave background radiation

The cosmic microwave background (CMB) helps explain Olbers' paradox. The cosmic microwave background is radiation almost uniformly distributed throughout the sky, like one of the assumptions of the paradox. In addition, the cosmic microwave background, rather than visible light, dominates the radiation energy of the Universe.

Olbers' paradox is concerned with the amount of light in the sky. A way of describing the amount of light in the sky is to use the light density, or amount of light per unit volume. Light consists of particles called photons, so light density can be expressed as photon density, or the number of photons per unit volume. Since each photon has a radiation energy that varies with its wavelength via the Planck constant, light density can also be expressed as radiation energy density. Applying this to the observable Universe, the amount of light in the sky can be described using the average radiation energy density of the Universe.

Olbers' paradox is concerned with visible light from stars. But visible light is radiation energy in just one small range of wavelengths in the electromagnetic spectrum. The extragalactic background light includes radiation energy from a wide range of wavelengths. The bulk of the radiation energy in the Universe is not visible light, but radiation energy in the wavelength range of the cosmic microwave background. The Big Bang theory suggests that the cosmic microwave background fills all of observable space, and that most of the radiation energy in the Universe is in the cosmic microwave background.. Since the cosmic microwave background dominates the radiation energy of the Universe, using the cosmic microwave background helps to explain Olbers' paradox.

A way of describing temperature is to use the relationship between temperature and the wavelengths of radiation energy as given by Wien's displacement law. The wavelengths of the cosmic microwave background suggest that the average temperature of the Universe is about 2.7 Kelvin, which is an extremely cold temperature near absolute zero.

Another way of describing temperature is to use the relationship between temperature and the total radiation energy density as given by Planck's law. One way of using this relationship is to note that if the temperature is limited, then the radiation energy density is also limited. Applying this to the observable Universe, if the average temperature of the Universe is limited, then the average radiation energy density of the Universe is also limited. Since the cosmic microwave background suggests an average temperature of 2.7 Kelvin, the average radiation energy density of the observable Universe is limited.

Finally, since the average radiation energy density of the Universe can be used to express the amount of light in the sky, and the average radiation energy density of the Universe is limited, this means that the amount of light in the sky is limited, so Olbers' paradox is explained.

Mathematically, the total electromagnetic energy density (radiation energy density) from Planck's law is

${U \over V}={\frac {8\pi ^{5}(kT)^{4}}{15(hc)^{3}}},$

e.g. for temperature 2.7K it is 40 fJ/m³ ... 4.5×10⁻³¹ kg/m³ and for visible 6000K we get 1 J/m³ ... 1.1×10⁻¹⁷ kg/m³. But the radiation temperature is as bright as the stars themselves only if these stars are side-by-side. If the mass density is lower, the radiation density and temperature is lower as well (analogous to thermodynamic equilibrium for the Earth (300K) with the Sun (6000K at its surface and 15000000K in its core) at distance 0.000016 ly due to the inverse-square law). For a critical density of about 10⁻²⁶ kg/m³, i.e. one solar mass in a 600 ly cube, the radiated energy can not exceed the binding energy and with respect to the abundance of the chemical elements it results in the corresponding maximal radiation energy density of 2×10⁻²⁹ kg/m³, i.e. temperature 7K. For the density of the observable universe of about 4.6×10⁻²⁸ kg/m³, this is a temperature limit of 3.2K. After subtraction of the cosmic neutrino background energy density, the limit becomes 2.8K (i.e. almost all energy from nuclear fusion is converted into this cold radiation of extragalactic background light). The corresponding photon density is about 3×10⁸ photons/m³, and compared to the baryon density 0.3/m³ the baryon abundance parameter is 10⁻⁹, which agrees with the parameter given by primordial nucleosynthesis and explaining isotopic abundances.

Ben Standeven (talk) 19:39, 13 July 2010 (UTC)

References

• The math here seems dodgy; granting the claim that "the radiation temperature is as bright as the stars themselves only if these stars are side-by-side" (which misses the point of Olber's Paradox), we would seem to get an energy density that scales as the two-thirds power of the mass density; if the stars were side-by-side the mass density would be 1.4×10³ kg/m³, the density of the Sun, so the energy density associated with the critical mass density would be 1 J/m³ / (1.4×10²⁹)^2/3 = 3.7×10⁻²⁰ J/m³ = 4.1×10⁻³⁷ kg/m³. This is far colder than the claimed value of 4.6×10⁻²⁸ kg/m³, and in fact represents a temperature of only a few centiKelvins.

• • Your calculation 1 J/m³ / (1.4×10²⁹)^2/3 = 3.7×10⁻²⁰ J/m³ = 4.1×10⁻³⁷ kg/m³ is wrong. How can you mix 1 J/m³ (radiation density at the Sun surface) and some nubmer to obtain density for the Universe? For (observed) visible mass density 4.6×10⁻²⁸ kg/m³ - the total radiation density can not be lower by ten orders (to obtain yours 4.1×10⁻³⁷ kg/m³) but only by about 3 orders (few MeV of the total binding energy per mass corresponding to about 1 GeV). Even if the Universe temperature is lower ("a few centiKelvins") then we can not observe a visible bright sky (but the dark sky in visible spectrum and the CMB).

• • • My bad, the claimed radiation density of the Universe is 4.5×10⁻³¹ kg/m³ not 4.6×10⁻²⁸ kg/m³. It is still six orders of magnitude higher than the value I calculated. But it occurs to me that I may be misunderstanding the logic behind these calculations. What does "the radiated energy can not exceed the binding energy" mean? The binding energy of what? And what volume is the radiated energy being measured over? Ben Standeven (talk) 05:14, 14 August 2010 (UTC)

• • • • The nuclear binding energy (mostly) is limit for radiation energy (mass defect) per nucleon (rest mass/corresponding energy) that is emitted ("created") in fusion from "elementary" particles (electron and protons). The (average) density of radiation and mass is independent on selected volume. —Preceding unsigned comment added by 194.228.230.250 (talk) 11:35, 25 August 2010 (UTC)

• • • • • OK, I get it now. I've changed the text to make it easier to follow, and took out the off-point early paragraphs. Ben Standeven (talk) 18:04, 31 August 2010 (UTC)

• "Thermalisation" can explain this shift. Like LED diodes (with brightness temp. 10000K) that increase temperature ("background" 300K) in a box (this is not correct with "perfect" boundary - it is "cyclic" boundary in the Steady State theory - see Boundary value problem). Like equilibrium of Earth (300K) with Sun (6000K on surface). Thus 2.7K background can be in "planetary" thermodynamic equilibrium with stars. —Preceding unsigned comment added by 194.228.230.250 (talk) 13:41, 9 September 2010 (UTC)

Integrated starlight cannot explain the CMB because the CMB is uniform to one part in 10^5 while even the most uniformitariaun assumptions possible for integrated starlight would result in a uniformity of no more than one part in 10. Thus the integrated starlight model for the CMB is ruled out by almost 4-sigma. Not relevant for this page, but relevant for any attempt to claim that the steady state or static universe can account for the CMB. ScienceApologist (talk) 21:38, 9 September 2010 (UTC)

• See previous comment. The CMB is not direct starlight (like: a temperature increase in this box is thermal/black body (depends on density of LEDs) and it has not brightness temperature of LEDs)

• • There is no mechanism in the vacuum of space by which to thermalize integrated starlight due to the finite lifetimes of stars. If stars stayed turned on forever, you could thermalize to an arbitrary level, but they die. The level to which you can thermalize is approximately 90%. ScienceApologist (talk) 11:57, 10 September 2010 (UTC)

• • • There is not the vacuum - only. The radiation can thermalize (after passing a few Gly) in H I region, galactic halo, cosmic dust, unbounded protons/electrons (not "visible"), etc. (and in many unknown regions like dark matter). This is outline of possibility and it is not proved results/the final theory (There are also some loop-holes in the Big Bang theory and it is not reason to be removed. Some things are also speculative: "On the other hand, inflation and baryogenesis remain somewhat more speculative features of current Big Bang models" - see Big Bang). —Preceding unsigned comment added by 194.228.230.250 (talk) 12:53, 10 September 2010 (UTC)

• • • • The standard calculation gives a level of 10% for integrated starlight. There are also other issues with it: [1]. Right now you are promoting your own original research which is forbidden by Wikipedia. Give a source for your claims or stop making them in article space. ScienceApologist (talk) 22:45, 10 September 2010 (UTC)

• • • • • Why you mean that steady state model is falsified and not the Big Bang model? : "now falsified steady state cosmological model". Please add citation where is "is now falsified". The Big Bang model is also falsified (and more - experimentally falsified!!!). The stability of proton (conserved number) and electron is more than 20 orders larger than age of the Universe in the Big Bang model. It does not matter that it is an unknown process (baryogenesis). It is in contradiction! Your approach is unbalanced. (The section "The mainstream explanation" does not cite any source.)

Absorption section wrong

Being in thermal equilibrium with the surrounding stars does not mean being at the same temperature as them. Those who claim it does are no doubt taking it as an a priori assumption that the sky is already as bright as a star before considering what effect intervening matter has. This situation would never arise though. The intervening matter does not have to block light from stars a trillion light years away, as such light has already been blocked my matter much closer to that star. The radiation reaching any matter would come from a finite region of space, hence the equilibrium temperature would be much lower. Scowie (talk) 19:56, 21 January 2014 (UTC)

Find a reliable source for your speculations, and then we can discuss what weight to give them in the article. Without a reliable source, there is nothing to be done. But to begin with, how exactly do you think the presence of intervening matter that is in thermal equilibrium with everything else is going to reduce the temperature of the radiation that we see? —David Eppstein (talk) 21:19, 21 January 2014 (UTC)

Im using logic and reason here. I am not making any claims; just pointing out the errors in reasoning of others. As for how the temperature of emitted radiation is reduced, that's the way thermodynamics works: When something is in thermal equilibrium the energy density of the incident radiation matches that of the emitted radiation. The temperature does not have to be the same. Few high energy photon absorptions vs many low energy photon emissions. The energy density of the observable universe would determine the equilibrium temperature. Btw, the absorption section is unsourced, as is. Scowie (talk) 00:03, 22 January 2014 (UTC)

Are you assuming significant deviations from Planck's law relating temperature to energy? Why do you think this is a reasonable assumption? —David Eppstein (talk) 00:44, 22 January 2014 (UTC)

No, I'm not assuming significant deviations from Planck's Law. Not sure where you get that idea. It would be pretty much the same as what happens within our solar system. The outer planets are receiving the same temperature of radiation as the inner planets, just less of it, hence they have lower temperatures. The energy density of the absorbed radiation sets the equilibrium temperature. Of course there are other factors involved too, like material properties (and for planets/moons: atmospheres, latent heat from formation, tidal forces). Radiation energy density is the overriding factor. Scowie (talk) 17:52, 22 January 2014 (UTC)

The outer planets are receiving less energy because a much smaller fraction of their sky is covered by the surface of the sun. That wouldn't be the case in the Olbers paradox setup. —David Eppstein (talk) 18:25, 22 January 2014 (UTC)

Due to intervening matter, only starlight from a finite region of the universe would be reaching any location directly. The re-radiated radiation from intervening matter elsewhere has a low temperature so can't possibly heat the matter at the chosen location to star temperature.

Only if your starting premise is that the sky is already bright, do you end up with some cold cloud of matter getting heated up to star temperature. If you start with an observable universe with the energy distribution we observe, i.e. hot stars, with cold diffuse matter in intergalactic space, and make it infinite, how can the cold diffuse matter possible get heated up? That would require the creation of new energy out of no where! The universe has the energy density that it does, whether it is infinite or not. By saying the sky would be bright, you are cooking the books by dreaming up extra energy! Scowie (talk) 18:03, 23 January 2014 (UTC)

This is going nowhere. Find a reliable source for your misapprehensions and then maybe we can discuss some more what to do with the article. Without a source, there is nothing to do and continuing to discuss it is a waste of time. —David Eppstein (talk) 18:48, 23 January 2014 (UTC)

Temp lock this article

You might want to temporarily lock this article. In a xkcd "What If?", Randall Munroe casually jokes that he'd been tempted to vandalize this article by placing {{citation needed}} every time the article said the sky was dark (I suppose how dark it is depends on where you live?). It was a joke, but I see we've already had one vandalation, and the xkcd page in question just went up. What does everyone else think? Is a temp protect needed? — Gopher65_talk 14:17, 21 August 2014 (UTC)

Two instances, actually, but nothing in several hours. I don't see an issue here yet. Powers ^T 14:21, 21 August 2014 (UTC)

Ok:). Just wanted to let people know what the source was, so they aren't left wondering what's going on if it gets worse. — Gopher65_talk 14:26, 21 August 2014 (UTC)

Yeah, there'll probably be a few more, but it's ultimately a harmless joke and it'll die down. I honestly came here hoping to see a [citation needed] that hadn't been reverted yet. SSSheridan (talk) 15:04, 21 August 2014 (UTC)

The XKCD joke is not immediately obvious: it's buried in a Reference 5 (you have to click on the 5 to see it), in a what-if article. Hopefully that cuts down on the number of pranksters who'd see it. 128.232.254.133 (talk) 18:44, 21 August 2014 (UTC)

You might want to consider that it's a well-known fact among xkcd fans that the comics always have witty alt text, a tradition that was carried along into what-if's "references". It is not unlikely that the majority of readers will open all those references just to have an extra laugh. While the amount of pranksters that actually read what-ifs might be indeed low, I'm seeing [citation needed] marks even in this article's references themselves, and there's been more edits today than in two months. Good luck :) --186.136.111.144 (talk) 20:36, 21 August 2014 (UTC)

Technical note: Randall should really pre-notify Wikipedia if he intends to mention Wikipedia article in his work.

(On the other hand, this can be ended creatively, by actually providing a very scientific source for the fact, that the night sky is dark. Come on, there MUST be some!) — Preceding unsigned comment added by 82.99.189.14 (talk) 22:36, 21 August 2014 (UTC)

The purported need to cite the dark night sky brings to mind User:Jnc/Astronomer vs Amateur. But perhaps Munroe missed the fact that, for nearly a year, one of the references of this article has included the statement that "the night sky is dark" in a direct quote from the reference? —David Eppstein (talk) 03:33, 25 August 2014 (UTC)

I think some of us will just come to see if someone else messed with it. A few of us will come here to see if it's been an issue and to have a laugh. A virtually insignificant number of us will comment here. Even fewer still vandalize your precious post... but I feel it's obligatory at this point to at least make a mockery of the whole thing. -Signed an XKCD fan and wiki lover. 76.113.75.7 (talk) 22:42, 22 August 2014 (UTC)

I'm actually impressed with the small number of vandalism attempts before the article was temp protected for a week against anon edits and edits by newly created accounts (last 3 days I think it is?). Apparently xkcd's (it's all lowercase!) community is more mature than I imagined. — Gopher65_talk 02:05, 23 August 2014 (UTC)

I hate to point out the obvious, but... why is the night sky dark? It is obviously easy to criticise a (very bright) writer of web comics, but none of the comments here consider the point he made! Yes, the article itself addresses why the night sky is not 'shining with the brightness of infinite suns', but there does not appear to be any Wikipedia article addressing the direct question of why the sky is dark. Google's first answer to the question is a link to this article. It also sends me off to NASA and then to Scientific American, both of which focus on Olbers' Paradox. There are plenty of other answers that Google wants me to read, but all of them appear to focus only on this sub-set of the question Mr Munroe raised.

So - we have an answer to "Why is the night sky not covered in light from an infinite number of stars?", but that is not the same as answering "Why is the night sky dark?" They are different questions, and this article - as Mr Munroe quite correctly points out - only answers a sub-set of the latter. Is there any chance that we might one day see a Wikipedia article that addresses the full question?

As a side note, are poets considered to be authoritative sources for articles on astronomy? That is, plenty of 'famous people' have made crazy statements about fields in which they have no knowledge, that later turn out to be at least partly true. Does Wikipedia accept such statements as valid sources? Ambiguosity (talk) 08:25, 28 February 2016 (UTC)

Units?

I'm not a physicist or astronomer, so I'm not going to touch the article itself, but the numbers at the end of the "Mainstream Explanation" section seem like they need units or a mention of whatever scale is being used, probably with a link to the relevant article. I was left wondering "5 to 8.6 what?" 184.63.10.160 (talk) 19:17, 21 August 2014 (UTC) Gabe Burns

It's a ratio of wavelengths, so the units cancel out. See Redshift#Measurement, characterization, and interpretation73.186.239.20 (talk) 03:34, 22 August 2014 (UTC)

The only thing that we could really add would be "z=####". That doesn't tell you too much, but it does indicate that the number after the z refers to a redshift. — Gopher65_talk 15:15, 22 August 2014 (UTC)

Absorption Section Removed

I removed the absorption section since it is self refuting and had already been brought up and refuted earlier in the article. The section was redundant and did not improve the article. I may also remove the finite age section since it sounds like OR. Dr. Morbius (talk) 17:02, 11 October 2014 (UTC)

Reference to CMB Dipole Moment

Regarding the last sentence of the article and the discussion about a 'fractal star distribution' I see no reason to cite G. F. Smoot's paper and the dipole moment of the CMB. This paper is now 40 years old but it gives the cause for the observed dipole moment right in the abstract: the motion of the observer with respect to the CMB restframe. This effect is expected and has to be there. Observing a vanishing dipole moment would instead be worrying. Anyway, the cosine anisotropy has nothing to do with the large scale structure but is interpreted best as a completely local phenomenon attributed to the observer alone. In contrary, Smoot constrain the remaining anisotropy to be smaller than 1/3000. Therefore, the last sentence in the article is nonsense should be removed. The only reason I have not done it myself is that, for totally different reasons, really like that paper. --2.246.187.19 (talk) 19:33, 6 April 2016 (UTC)

Really? The brightness gives an argument?

"Suppose that the universe were not expanding, and always had the same stellar density; then the temperature of the universe would continually increase as the stars put out more radiation."

"Suppose that the EARTH were not expanding, and always had the same MATTER density; then the temperature of the EARTH would continually increase as the EARTH gathers more radiation."

(1) Where is the radiation from? => There, where it comes from, there isn't it anymore!

Infinite directions?

Not sure I understand the initial paradox correctly. It assumes that if there were an infinite number of static stars, then a line from the observer in any direction in the night sky should end on the surface of a star, right? But does that mean that that star has been sending out photons along the same line back to the observer? An active star glows, but not in an infinite number of directions, it does not send out an infinite number of photons every second, or am I wrong? Iago2¹² 15:47, 23 November 2016 (UTC)

The brightness per unit of solid angle covered is constant regardless of distance, assuming the standard inverse-square laws. —David Eppstein (talk) 16:43, 23 November 2016 (UTC)

What does that have to do with my question? Iago2¹² 21:37, 23 November 2016 (UTC)

It means that it looks the same brightness no matter how far from it you are; your distance controls how small it looks but not how bright it looks. If you insist on trying to understand it in terms of photons: the photons should not be interpreted as being emitted periodically, but rather randomly, at a given rate. Similarly, it does not emit photons along some fixed finite set of rays, but in all directions, randomly. Given this random process, the number of photons per second that reach your eye from any star (given a fixed value of its surface brightness) is proportional to the area of the sky that it covers from your viewpoint, but does not otherwise depend on the distance to the star. So if you see a star in every direction you look (far enough away in that direction), you would see the whole sky at the surface brightness of a star (the same brightness that we see the surface of the sun). —David Eppstein (talk) 21:47, 23 November 2016 (UTC)

OK, so would not in all directions, randomly man that it has to emit an infinite number of photons to be seen from an infinite number of directions? Iago2¹² 22:17, 23 November 2016 (UTC)