Talk:Olbers's paradox/Archive 1

This is an archive of past discussions about Olbers's paradox. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page.

Archive 1

Archive 2

Myths and alternative explanations

Latest comment: 18 years ago1 comment1 person in discussion

I've added a line about the Charlier Cosmology. Does anyone have a reference to Mandelbrot's paper? Jonathan Silverlight 21:50, 19 October 2006 (UTC)

Wave Structure of Matter

Latest comment: 14 years ago2 comments2 people in discussion

The Wave Structure of Matter theory (see: [Wave Structure of Matter]) proclaims a different solution to this paradox. The distancescale of interaction between distantiated material objects is finite, the size of the Hubble distance. Forces drop down (i.e. drop down faster then expected from 1/R^2) for objects in space which are very far distantiated, because they have less common universe (the Hubble sphere). This also explains that far distantiated star light is red-shifted! (WSM theory sees all matter as caused by standing waves in space). Wouldn't that viewpoint need to be mentioned as a solution? Heusdens 23:16, 22 November 2005 (UTC)

Changed name to "Olbers's Paradox" (from "Olbers'"): Technically that's the grammatically correct spelling of a singular posessive noun :) - qartis

Google gives 542 votes for "Olbers's Paradox" and 6,128 votes for "Olbers' Paradox". It should be changed back, IMO. Kaldari 06:17, 11 Jan 2005 (UTC)

Is this a paradox ? - anon

I know nothing about this, but might not another explanation be that there is so much dust and gas in the universe that light from very different stars is so dim as to be imperceptible, because it is absorbed along the way? Even in a non-expanding universe, you wouldn't expect very very distant stars to be visible anyway... -- Simon J Kissane

That was Olbers' original explanation. It doesn't work because of thermodynamics. Energy isn't destroyed: if light isn't getting to us because it got absorbed by dust or gas, then that dust and gas will warm up, eventually becoming incandescent and glowing as brightly as the stars whose light it is blocking. Shimmin 11:45, Feb 7, 2005 (UTC)

I'm not quite clear about the meaning and correctness of the article's last paragraph. Wouldn't thermodynamics forbid us to recycle radiation into matter? --AxelBoldt

Also, I take it that electromagnetic radiation is converted to kinetic energy (heat) all the time. Why should we postulate a hypothetical method for transferring electromagnetic radiation into matter, when there's another observable explanation for how electromagnetic radiation can be converted into another kind of energy? Beyond that if the universe is not bounded, or if it is expanding there's no reason to believe that anything has to happen to the light -- it can just continue to disperse. (But I am certianly no expert in this field...) MRC

With regard to Axel's question: radiation is routinely converted into matter in particle accelerators: most commonly into electron/positron pairs. This is called pair production. The only problem with such recycling is that known methods of matter production result in equal quantities of matter and antimatter, which is not what is observed in nature. Thermodynamics does not forbid recycling. However, it does suggest that a state of equilibrium will eventually be reached, and maintained thereafter. If there is no conversion of energy into matter then the equilibrium state will probably be one in which all except a small remnant of matter has been converted into energy in the form of radiation.

With regard to MRC's points: Yes, EM radiation is converted into kinetic energy. If this were to take place e.g. in a hydrogen gas cloud, some of the kinetic energy would be converted back into low temperature radiation (radio waves). Somebody has already added a paragraph to the main article regarding this possibility, saying that "it would result in strong radiation which is not observed". This also seems to address Simon Kissane's point. However, *some* radiation from gas clouds most certainly is observed.

With regard to light becoming increasingly dispersed in an expanding universe - this is partly covered by the statement about light becoming increasingly redshifted and diminished in brightness in such a universe. However, increasing separation between photons as a possible cause of diminished brightness should perhaps have an explicit mention.

I'll refrain from modifying the main article any further, because I personally favour the idea that energy is recycled into matter, and I find it difficult to evaluate other possibilities objectively.

--Martin Gradwell.

The universe might be infinite, but that doesn't mean the amount of matter/energy in it needs to be infinite. This would easily resolve the paradox.

True, but then some region of the universe would have more matter/energy than another; the universe would have a "center", which people typically don't like ("why is the center here and not over there?") 207.171.93.45

Are you sure that's right? How would we define the "extent" (not the right term I'm sure) of the universe except in terms of where there is matter/energy? That is, can there actually be parts of the universe that have no matter and no energy? Mswake 04:36 Jul 24, 2002 (PDT)

Actually, Olbers did not propose the paradox to show that the universe is finite, but that the universe is not transparent, being filled with dust that blocks the light of distant stars. Since the time of Newton, it had been appreciated that if the universe was static (not expanding) as was widely assumed, then it must be infinite, or else the combined gravity of all the objects in the universe would cause it to collapse toward its center of mass.

Olbers (believing the universe to be static and infinite) proposed that the darkness of the night sky showed that the universe was not transparent. However, he did not appreciate the consequences of the first law of thermodynamics (which can be forgiven at his time in history), that if interstellar dust blocked the light of stars, then it would heat up until it shone as brightly as the stars.

I am about to change the main page to reflect this.

According to Encyclopedia Britannica, Kepler saw it as an argument against an infinitude of stars. Also, the dust wouldn't necessarily shine "as brightly as the stars": it would be invisible microwave radiation, not visible light. AxelBoldt 01:29 Jan 25, 2003 (UTC)

I stand by "as brightly as the stars." It's actually a bit of an understatement. Let me give you a back-of-the-envelope justification.

Consider an infitnite, static universe filled with a uniform scattering of stars. If the distance to any given star is R, then the light recieved from it falls off as 1/R^2. However, if the universe is filled with a uniform scattering of stars, then the number of stars at distance R increases as R^2 (for the same reasons that the surface area of a sphere is 4*pi R^2). So, the light received from stars at distance R is (R^2)*(1/R^2) = a constant, independent of R. This implies that if light from infinitely far away could reach the observer, then all points in this universe would be bathed in infinite luminosity. To radiate this away, they would have to acheive infinite temperature and shine with infinite intensity at all wavelengths.

Or put another way, if a nonexpanding universe is infinitely old, and has contained an infinite number of luminous objects throughout that time, then at the present it must be infinitely luminous at all points.

Or put another way, suppose a dust grain in an infinitely old universe was at one time cold. It absorbed a visible photon, heated up a bit, and radiated away the energy as several microwave photons. However, neighboring dust grains did the same, and in the meantime, it has absorbed several of these microwave photons, in addition to another visible photon, and is a bit hotter. It will radiate this energy away in the infrared, but in the meantime, its neighbors are doing the same... The problem is energy can't be destroyed, and if an infinitely old universe has contained luminous objects during its entire lifetime, then an infinite amount of energy has been released into it, and that energy has to be somewhere.

[comment and query by Scott Mayers]: I understand that F = L/4πr² , where F = flux, defined as the amount of light that travels through a unit area per unit time, and L the luminosity. If this is more accurate than your value, we should at least say that energy applied to this area at this distance due to light is 1/4πr². Taking in the fact that we would be viewing individual rays coming from us from the source only in direct line-of-sight to the star in question, not the whole spheres of areas covered by all rays radiating from that star in such an area, you can only count the energy expended on particular particles as they interfere in the path of potential sight. This particle would receive much less than the energy of 1/4πr² (particles do not increase as distance from the star as does the sphere of rays of light). I might even increase your uniform rate of scattering to r³ but the final result should still be a tiny number close to zero.
I also question whether the second law of thermal dynamics even applies here. The law makes certain sense with matter. Energetic matter is going to continuously bump into itself and will fill up empty spaces until an average even energy of motion and separation of matter is uniform. But tiny amounts of light from stars hitting particles in space, though may temporarily heat them up, would just as quickly lose it in reflection, refraction, or to the environment after its initial absorption. The quantity of energy in the form of light isn't sufficient to alter most matter without quantity and intensity. I should certainly get more energy from a light bulb on a point of my skin than most specks of dust from a star in the vastness of space.
By the way, if the Steady State Theory cannot explain the detailed behavior of starlight and the background radiation because it has a gap in its explanatory power to connect red-shifted, distant stars to the potential of stars of a microwave spectrum, first, what is the Big Bang's explanation for this gap? And second, as akin to religious-like arguments, why does the lack of an explanation qualify those as one with a higher authority? --70.64.154.45 (talk) 06:19, 27 April 2010 (UTC)Scott.

It should be noted that the calculations given above assume transparent stars. The absorption of stars has never been directly measured because they tend to be brighter than anything behind them, but they display dark line spectra and other signs of internal absorption, so it seems likely they are opaque. In this case, we must add a factor of (1-f)^R where f is the frequency of stars. Since this is exponential, it dominates all existing terms in the limit, and gives us finite luminosity for the sky.

What about the effect on the stars of absorbing all this starlight? Well, they shine a little brighter than they would otherwise, but by a finite amount. Therefore the energy output of a given star is the energy production plus the input, which is a fraction of the output of an average star. So long as the fraction is less than one, this is stable. Calculating the fraction requires arithmatic, but observation shows it to be very small.

Incidentally, none of this changes the fact that the universe must be of finite age or something similar. If every large but finite region of space had been engaging in non-zero amounts of fusion for infinite time, it would contain infinite non-hydrogen. It doesn't, so it hasn't.

I think the title of this article is incorrect. His name was "Olbers" so the title should either be "Olbers' paradox" or "Olbers's paradox". A quick consult to my astronomy textbook published in 2001 prefers the latter and notes that the former is acceptable as well. - 66.81.223.216

Hang on, why has the article been moved to "Olbers' paradox"? Doesn't the above note recommend "Olbers's paradox"? I would certainly prefer it to be there. The omission of the final "s" in the possessive form of a singular noun or proper noun is a rather idiosyncratic rule that a minority of English-speakers follow, and which rather grates on the ears of most of us. -- Oliver P. 14:01 Jan 31, 2003 (UTC)

A quick Googling found 142 instances of Olbers's paradox, while Olbers' paradox rang up more than 2200 (although these may include some results from the former search). Whatever the author of that particular astronomy book's preferences may have been, Olbers' is certainly the form that appears more in print. -- User:Shimmin

Penguin Dictionary of Science has "Olbers' paradox" -- Tarquin 14:07 Feb 5, 2003 (UTC)

The Encyclopaedia Britannica seems to have it your way, as well. I suppose I lose, then. I still don't like it, though... *grumble, grumble* -- Oliver P. 18:53 Feb 5, 2003 (UTC)

A note on English grammar: which one is correct depends more on morphological processes than syntax. If Olbers is a singular noun possessive noun, its structure is: ROOT(Olbers) + POSS('s). If it was plural of Olber, it would be: ROOT(Olber) + PLURAL(s) + POSS('s). Now some people (the Olbers' people) are following this morphological rule: *[ending in s] + POSS('s) => *', thus ROOT(Olbers) + POSS('s) => ROOT(Olbers)+' => Olbers'. The Olbers's people are following this rule: ROOT[singular] + POSS('s) => ROOT's, ROOT[singular] + PLURAL(s) + POSS('s) => ROOT+PLURAL(s)+POSS('). Whose is right? The Olbers' people consider that +'s v. +' is based on morphological considerations, whereas the Olbers's people base it on more grammatical rules.

I reverted to Pakaran's changes, sorry. (MJA, I would notify you on your talk page if you had a talk page.) A paragraph was added which I believe is simply incorrect. In part:

[...] For example, if we were to send a radio signal to the most distantly observed galaxies, the signal would never arrive there, because the rate at which the distant galaxy and ours are receding from each other excedes the rate at which the radio waves travel through space.

This paragraph claims that the two galaxies are receding from each other at faster than c, which is not possible, to my understanding. I saw no support for this concept in any of the external links. I'm far from an expert in the field so feel free to reintroduce the concept if it's true. Tempshill 21:49, 20 Jan 2004 (UTC)

The expansion is faster than light because the space itself is expanding (check out the Universe article, for example). So I guess this text should be put back (it's not immediately apparent where it was, so I can't do it myself and I don't have time at this very moment to sift through the history). Paranoid 16:10, 4 Jun 2004 (UTC)

On an unrelated note, I just estimated how bright should the sky be, knowing that the Universe it's 13-15 billion l.y. big and there are ~10^22 stars. The answer is it should be about ten times less bright as the full moon, give or take a few orders of magnitude. In order for the whole sky to be as bright as the surface of a star the Universe obviously needs to be a million times larger.

I think that we need to add some calculations like this (more rigorous) to give the article reader a sense of what levels of brightness are we talking about. As it is, only about 1 millionth of the sky is the surface of a star (excluding the Sun and our Galaxy, of course, which are anomalously close, by Universal standards).

I hate to be a spoilsport, but is the blurb at the end about the band suitable? Perhaps it could use some NPOV, but then again, I kind of like it. :) --Golbez 02:58, 17 Jun 2004 (UTC)

Maybe I'm wrong, but shouldn't

"There is no known process that can return heavier elements to Helium in the necessary quantities"

be

"There is no known process that can return heavier elements to Hydrogen in the necessary quantities".

Right now it makes no sense

Surely the skies would be dark not light?

My grasp of physics is fairly elementary, but surely the wave nature of light means that this paradox should state that the sky should be dark? Light behaves as a wave. For the uniitiated, Putting a source of light through two slits in a card shows this effect with bands of darkness "rippling" outward. This occurs with any waveform when two waves beocme perfectly inversely corrleated with each other (that is to say the pattern of peaks and troughs of one respectively match the torughs and peaks of the other) and they cancel each other out. If there were an inifinite number of light sources, there would be an infinite area of peaks and troughs in every direction resulting there being no visible light. Would a nice science person be kind enough to comment? Dainamo 12:02, 16 Oct 2004 (UTC)

At any point, the total peaks and troughs would both be infinite; the problem is that you're assuming "infinity minus infinity equals zero", which it doesn't.

It might help to think about tossing coins (heads = 'peak', tails = 'trough') and looking at what's left after you cancel out heads with tails - this is effectively the 'amplitude'. If you toss two coins, you'll average one head & one tail, but the average *difference* will be 1/2. If you toss a hundred, you'll average 50 heads & 50 tails... but on any one toss you probably won't get the same number of each. The average difference between heads & tails will be around 10-20. If you toss ten thousand, you'll average 5000 of each, but the average difference will be ~100-200, and so on.

So, as the number of coins rises to infinity, even though the *average* number of heads still matches the average for tails, the average difference between them doesn't tend to zero. Instead, it also rises to infinity, though rather more slowly (roughly proportional to the square root of the number of coins). It works the same way with light: while much of it cancels out, not all of it does, and the amount of uncancelled light rises as the number of sources rise. --Calair 00:44, 18 Oct 2004 (UTC)

Thank you Calair for a fascinating and enlightening answer. Dainamo 20 Oct 2004 (UTC)

Why lower-case?

I moved this to Olbers's paradox with a lower-case p because that is the usage followed in the many hundreds, maybe thousands, of pages titled "Smith's theorem", "Smith's law", "Smith's principle", "Smith's hypothesis", etc., etc. See list of eponymous laws (or list of mathematical topics, for that matter). Michael Hardy 02:22, 28 Oct 2004 (UTC)

PS: I've fixed the double redirects. It will take longer to fix all redirects; could others help? Thanks. Michael Hardy 02:23, 28 Oct 2004 (UTC)

apostrophe

Currently the page is called Olbers's paradox, and the first line begins Olbers' paradox. Both are in some sense okay ways to spell, but we should be consistent. Fowler's Modern English Usage (a standard for British English) favours the latter if the post-apostrphal 's' is unvoiced. I like this way of doing it, too, but don't want to cause a nasty grammar spat by just changing it without discussion. What do you think?

Olbers's paradox → Olbers' paradox

Google gives 542 votes for "Olbers's Paradox" and 6,128 votes for "Olbers' Paradox". Kaldari 06:25, 11 Jan 2005 (UTC)

Olbers' Paradox doesn't exist, so it doesn't belong here. See top of the page: Sometimes you want to move a page, but cannot do so because a page of that name already exists. This page allows you to request action by a admin to perform such moves. Correct? Cburnett 06:49, 11 Jan 2005 (UTC)
- Sorry I meant to put Olbers's paradox → Olbers' paradox (second word in lowercase). I've changed the listing above to reflect this. The lowercase version already exists as a redirect page. Kaldari 07:06, 11 Jan 2005 (UTC)

Penguin Dictionary of Science and Britannica both list it as "Olbers' paradox" and this is how it is spelled throughout the article and related articles on Wikipedia. Fowler's Modern English Usage favors the later spelling as well. Kaldari 18:25, 11 Jan 2005 (UTC)

Support: This is primarily an issue of style (possessive proper name ending in 's') and personal preference. "Olbers' paradox" seems to be the more accepted one (8160 vs. 640 google hits and 2700 vs. 560 teoma hits). Looking at the history shows Qartis moved "Olbers' paradox" to "Olbers's Paradox" in Sep 2004 and Michael Hardy moved "Olbers's Paradox" to "Olbers's paradox". Going farther back, the article was originally at "Olbers's paradox" (see Talk:Olbers's paradox just over half-way down). According to Oliver P., the Encyclopaedia Britannica has "Olbers' paradox". This page has a long history of being moved and shuffled around. Google had more hits for "Olbers' paradox" 2 years ago and still does. It stands to be the accepted style is "Olbers' paradox" and not "Olbers's paradox". Cburnett 21:02, 11 Jan 2005 (UTC)
Oppose. Proper grammar. Neutrality^talk 21:15, Jan 12, 2005 (UTC)
- Most grammar sources I have looked at say that it is acceptable to use only an apostrophe at the end if the word already ends in an 's' that is pronounced as /z/. Kaldari 22:27, 12 Jan 2005 (UTC)
Comment. It looks like this is primarily a question of style and grammar. Personally I would have gone with Olbers' paradox, but it looks like the 's on possessive proper nouns ending in s, is an area of grammar which is currently changing. There is a lot of contradictory advice around. I found quite a good summary of current usage with references at google answers. This suggests that although both options are correct, modern usage is moving towards Olbers's paradox. -- Solipsist 10:50, 13 Jan 2005 (UTC)
Support for reasons given by Cburnett and Kaldari. older≠wiser 14:09, Jan 13, 2005 (UTC)
Support I'm too used to "Olbers' Paradox". — RJH 20:40, 14 Jan 2005 (UTC)
Oppose, but suggest that any resolution of this particular question be put on hold, pending the outcome of the general discussion of the apostrophe-s issue at Wikipedia talk:Manual of Style#Possessives of words ending in 's'. JamesMLane 22:03, 15 Jan 2005 (UTC)
Support - primarily for reasons of usage. Also, my education and the style guides I've read suggest no extra s is required, a perhaps rare case in which grammar and pronunciation agree. :) -- Guybrush 09:46, 18 Jan 2005 (UTC)

Brightness, distance

Latest comment: 18 years ago1 comment1 person in discussion

I'm pretty sure I don't understand this sentence:

"The brightness of a surface is independent of its distance, so every point in the sky should be as bright as the surface of a star."

What does it mean for brightness to be independent of distance? A distant light source certainly *seems* less bright than a near one--perhaps there's a technical meaning of "brightness" that is different from the commonplace meaning? But if the technical definition of brightness has nothing to do with how the object is perceived, then what's the point of the paradox? If the sky can be "bright" without looking bright, I mean. Nareek 22:45, 3 March 2006 (UTC)

Isotropic microwave background radiation in Olbers' context

Latest comment: 18 years ago1 comment1 person in discussion

The main article on Olbers' paradox says:

"One explanation attempt is that the universe is not transparent, and the light from distant stars is blocked by intermediate dark stars or absorbed by dust or gas, so that there is a bound on the distance from which light can reach the observer. However, this reasoning does not resolve the paradox. According to the first law of thermodynamics, energy must be conserved, so the intermediate matter would heat up and soon reradiate the energy (possibly at different wavelengths). This would again result in uniform radiation from all directions, which is not observed."

What do you mean, not observed? How about that uniform radiation from all directions that we do observe, the ubiquitous background microwave radiation (which is exactly in the range of interstellar gas secondary emission)? 66.82.53.56 18:41, 22 March 2006 (UTC) Alex Feht

Accepted explanation false

Latest comment: 17 years ago5 comments2 people in discussion

The darkness of the night sky is not due to the universe's finite size; it is due to it's expansion. The night sky in steady state models (which are spatially unbounded) is also dark. --Michael C. Price ^talk 04:44, 10 January 2007 (UTC)

I've added the correct explanation. --Michael C. Price ^talk 17:11, 26 January 2007 (UTC)

But if the universe were infinite in duration, then Olber's Paradox would still hold. The universe could expand all it wanted. Jhobson1 00:42, 17 August 2007 (UTC)

Says who? --Michael C. Price ^talk 02:46, 17 August 2007 (UTC)

Deleted section

==Obsolete argument: what paradox?==

The above discussion^{[clarification needed]} was correct in Olber's time, when the only radiating objects known were stars, or shone by reflected starlight. After blackbody radiation was discovered, one then has to replace the term "star" with the more general term "N degree Kelvin blackbody radiator". The paradox then disappears (N=3).

I've deleted the above text because I think it's not correct. It was also unsourced and rather lacking in explaining why/how the cosmic microwave background resolves the paradox (which it probably doesn't). You might argue that the expansion of the universe implies a CMB (sort of half true) and that the expansion resolves the paradox, but that is a different argument and has already been presented. --Michael C. Price ^talk 17:11, 26 January 2007 (UTC)

Finite age is the dominant effect

Latest comment: 16 years ago4 comments3 people in discussion

In the "Accepted Explanations" section it was stated:

"Two effects contribute to the resolution of Olbers' paradox: the finite age of the universe and the redshift. The latter effect is the dominant effect."

In fact, it should be "... the former effect is the dominant effect." I've made the necessary edit. Redshift only contributes a few factors of dimming. The finite age of the universe (actually, the finite lifetimes of stars, but I won't quibble further) is responsible for the orders of magnitude difference between a "bright" night sky and what we observe (i.e. a "dark" night sky). To find out why, see the papers and books that are referenced in this very entry, especially Wesson's paper and Harrison's book (also the latter's "Cosmology: The Science of the Universe", another fine book). Cragwolf 05:00, 19 February 2007 (UTC)

Finite energy density of the universe please

Can someone explain to me how the finite age of stars in wouldn't explain the paradox? Since the amount of energy contained in a star is limited, the energy density of the universe has a similar upper limit. Realizing that is trivial.

The problem with Olbers' paradox is that it assumes an infinite supply of fuel for the stars, i.e. an infinite energy density of the universe. Of course this yields a result with infinite flux (for point-like stars), one would have to be stupid to suggest anything else.

Dispensing with this fake paradox takes us instead to the question on steady-state/infinite vs finite universe, where the former needs to have a way to prevent heat-death after infinite time.

That, however, is another matter altogether. Let's not try to mix this false paradox into the discussion and pretend it has any merit as an argument for a finite universe.

--ChristofferL 19:04, 23 October 2007 (UTC)

The finite age of stars does not necessarily explain the paradox, because if you had a large enough number of stars per unit volume then that might be enough to still make the night sky bright. Harrison (an author referenced in this article), in his cosmology textbook, talks about a "lookback limit" (*), with an associated lookback time, T, and compares the latter to the lifetime of a typical star, t. If t << T then the night sky is dark; if t ≥ T then it's as bright as the surface of a star. T is determined by the number density of stars: high number density, small T; low number density, large T. So it's not really a finite energy density, but an insufficient energy density that explains why the night sky is not bright. In other words, two factors are fighting each other — number density of stars vs lifetime of typical star — and the winner determines whether or not the night sky is as bright as the surface of a star.

(*) What is the lookback limit? It can be thought of as the average distance of a line of sight from observer's eye to the surface of a star. Cragwolf 12:24, 10 November 2007 (UTC)

I don't agree here. A finite energy density is all that matters. Even if all of the energy of the stars are released at once, and distributed evenly across space, the average energy density would not change. You *could* create a situation where if all the energy is immediately distributed (taking away all the stars), when looking up you'd get a glaring hot sky.. Of course, you *could* also create a situation where you get a far dimmer sky. It would depend on the actual energy density and distribution distances (a high density would create a bright sky, low density would create a dark sky).

Still, the fact is, the energy is not released all at once. In fact, you'd could make the case for an equilibrium condition. The energy density would have to remain uniform in any infinite & static universe. This means you could effectively shuffle the energy around any way you like, you are still going to get the exact same amount of energy wherever you look. The only thing that would vary is the timing and location. This could very well create a situation where, no matter how much time passes, the universe would constantly look the same (within a variance of course). It would be a fact that we would EVENTUALLY return to our current view, no matter how big the variances were.

Think about it. Picture a slice of our infinite universe (that looks just like ours), with a finite energy density. What can you do to that slice that would really achieve the end solution of Olber's paradox? You'd have to release massive amount of energy from the stars in a way that reaches our planet, and doesn't get absorbed by something else. Eventually it WILL get abosrbed by something else (whether its us or not). The energy will just be shifted, and the average energy density will be maintained throughout the rest of the universe. The laws of thermodynamics would dictate that even if our planet did heat up, this would only mean another part of the universe would be cooled by the same amount. In other words, if this super heating in Olber's paradox occurs somewhere, a super cooling would have to occur somewhere else. It would all even out. If you still disagree, try to think what you'd need to do to the universe as a whole to change this.

The way I see it, Olber's paradox is immediately invalidated if you assume Stars do not last forever and do not violate the laws of thermodynamics. Whether or not our universe is static & infinite cannot be proven/disproven by Olber's paradox. Nicknomo (talk) 14:45, 28 February 2008 (UTC)

Contradiction

Latest comment: 17 years ago1 comment1 person in discussion

From the article:

The current scientific consensus is that effects of general relativity relating to the Big Bang and the finite age of the Universe do indeed give a finite size for the observable universe, but that it is the astronomical redshift relationship which really explains the dark sky at night.

From elsewhere in the article:

Three effects contribute to the resolution of Olbers' paradox: the finite age of the universe, the redshift, and the finite radiation life of stars. The first and third effects combined dominate. (Even in steady state theory models, which supposes the universe is infinitely old and spatially unbounded, the night sky would still be dark.)

The second snippet contains a contradiction, and contradicts the first snippet. Please fix it. Shinobu 19:01, 20 April 2007 (UTC)

Splendid new simulation from kmarina86

Latest comment: 17 years ago3 comments3 people in discussion

Why the sky is not bright in night? Why can this old theorem not critic big bang? Redshift cannot compensate it because there should always be enough for us visible stars! Or not? 84.158.86.156 08:08, 17 May 2007 (UTC)

One must consider that the density of the distrubtion of stars is not the same anywhere. The stars within our galaxy alone are certainly not homogenously distributed. Howver Olber's Paradox was formed during a time when Galaxies were unknown. If every star were 4 light years from each other, 10^22 stars (= 100 billion * 100 billion) could fit within a cube 86 million light years on each side (over 300 times smaller). If we were in the center of such a field, the sky would certainly brighter, about 90,000 times brighter (12 magnitudes brighter). But if our horizon was at the edge of the cube, then certainly the sky would not be infinitely bright, but very, very bright indeed.

http://www.stjarnhimlen.se/comp/radfaq.html#10

                                     Luminance       Magnitudes per square
                                     Nit = cd/m2       arcsec   arcmin
 
 Sun                                   3E+9            -10.7    -19.6
 Venus (max elong)                     15000            +1.9    -7
 Clear daytime sky (at horizon)        10000            +3      -6
 Full Moon                              6000            +3.6    -5.3
 Mars at perihelion                     4000            +3.9    -5.0
 Overcast daytime sky (at horizon)      1000            +5      -4
 Jupiter                                 800            +5.7    -3.2
 Saturn                                  700            +5.9    -3.0
 Heavy daytime overcast (at horiz)       100            +8      -1
 Uranus                                   60            +8.6    -0.3
 Neptune                                  30            +9.3    +0.4
 Sunset at horizon, overcast              10           +10      +1
 Clear sky 15 min after sunset (horiz)     1           +13      +4
 Clear sky 30 min after sunset (horiz)     0.1         +15      +6
 Fairly bright moonlight (at horizon)      0.01        +18      +9
 Moonless, clear night sky (at horiz)      1E-3        +20     +11
 Moonless, overcast night sky (at horiz)   1E-4        +23     +14
 Dark country sky between stars (zenith)   3E-5        +24     +15

20 → -10.7

Even making all stars in the observable universe a distance 4 light years from each other is not enough to counter Olber's Paradox....!

However, the Big Bang theory assumes a homegenous universe just at the at the largest scales that expands with time, with decreasing density. During this process, light from a star weakens with the fourth power of the scale factor (http://www.google.com/search?q=%22fourth+power+of+the+scale+factor%22+big+bang) [note: not distance]. With this, the radiation density drops. But what is significant is not the absence of percievable light, but rather the relative brightness it has with respect to the sun. The brightness is cut off due to expansion from a opaque singularity, so it does not have the chance to sum to what would be the brightness of our sun at all points in the sky. More importantly, the magnitude of stars (in relation to the luminosity distance) in the Big Bang theory drops faster in relation to the angular diameter distance. So a star of similar properties in a galaxy 10 times further away in terms of angular diameter distance would be more than 100 times dimmer. In this way, the Big Bang theory solves Olber's Paradox.Kmarinas86 20:17, 17 May 2007 (UTC)

Except it doesn't. Preliminary work by Harrison and later work by Wesson (see references to this article) have shown that "Big Bang theory", i.e. cosmological expansion, darkens the night sky by a few times (compared to the static case, all other variables equal) — which is nowhere near enough to turn a bright night sky into what we currently see. Cragwolf 13:58, 10 November 2007 (UTC)

Debunk paradox people forget the size of the universe...

Latest comment: 16 years ago2 comments2 people in discussion

People how made up this paradox forgot the size of the universe. And the amount of stars (and the distribution of them). Let's do some mind experiment.

Imagine your on a football field, in the darknes, and someone at a random place holds a burning cigaret. Now close to your football field connect hundred other football fields (thats a square about 1KM²) in size. Ofcourse you will be able to see the most nearest cigarets, as you have hawks eyes.

Now just imagine that in a random direction somewhere between 50km and 100km away another cluster of footbal fields exist. This is another starsystem (another milkyway). Also overthere there are some people holding cigarets which you might be able to see using optical lenses.

But then the next footbalfield will be away well lets say 50.000km you will not be able to see them. You can keep on repeating this.

But hey let's imagine you've created the ultimate lens a science breaktrough in optic dynamics. Then when looking to your horizon you see on a horizontal line some orange dots of cigaretes. But between the lightdots there will always be a distance. Yes you can zoom into a region but you will only find again (extreme tiny) dots and (and a bit bigger )distances (compared to the new dots size). It is much like a fractal you can never fill the horizon this way

So filling in a horizon this way with cigarete lights will never end up in an orange horizone line.

This is no paradox

Take a look at the photographs by Hubble STS and other distant galaxy images and you see the sky is not just filled with stars but it is filled with galaxies! http://hubblesite.org/newscenter/archive/releases/galaxy/2004/08/ This is only one of many images of numerous galaxies to be found in good viewing. There is no paradox.BillWilliam

If you had any understanding of Physics or indeed cosmology you find that what you are suggesting is not the case. A distant light source only seems dimmed as the photons are scattered by intervening matter. It is true that teh further away a light source is (on Earth) the dimmer it becomes however there is a general lack of matter in space meaning that your interpretation would only hold true if the entirity of space be completly populated by matter in which case distance would affect luminousity. We are not dealing with cigarettes and footballfields with countless moles of gas inbetween but instead unimportant distances. I would check facts further before trying to disbunk a paradox that is likely to be more than twice your age. Your argument about the size of teh light sources is also flawed. The paradox was created during a time of transistion between accepted thinkings, from that of infinite time and space to finite space-time. The paradox only exists when we deal with infinite numbers. No matter how small the points of light are, in an infinite universe there would be infinite numbers of them still creating a solid balnket of light. Olbers' Paradox (I believe should be capitalised) leads us to the finite universe. —Preceding unsigned comment added by Sedecrem (talk • contribs) 08:43, August 27, 2007 (UTC)

I don't think you should've insulted this poster's understanding of physics. Luminosity of the emitted light weakens as you get further away from a bright object due to the fact that there will be the same amount of light in less volume. To have it any other way would induce a violation of thermodynamics. As you put the same amount of energy in a greater amount of space, energy density drops. I think the problem with Olber's paradox is that it makes an error of infinity. Sure there is infinite energy coming at you from all directions, but this does not necessarily mean that you will be receiving infinite energy. If you go to the discussion above, all that really matters is the average energy density (energy per cubic meter). At that point, it just becomes a shuffling match - energy may move around a lot, but it would be counterintuitive to say everywhere starts receiving massive amounts of light. Nicknomo (talk) 14:45, 28 February 2008 (UTC)

Well, I thought I read that the universe doesn't have edges (what would be on the outside, after all?), so, if it's finite, it must curve around to meet itself. But if we live in such a curved universe, then wouldn't the starlight that doesn't directly make it to our eyes be wrapping around the universe and hitting us from another direction? The only thing I can think of is black holes absorbing light. It could be that there's a perfectly good explanation for why black holes wouldn't resolved the paradox, but I am surprised the term 'black hole' is never mentioned even once in the article, even if only to debunk that idea in the next sentence. --Iritscen 15:46, 24 September 2007 (UTC)

Fractal explanation

Latest comment: 16 years ago2 comments2 people in discussion

I was just curious as to the last part of this description, "Mainstream cosmologists reject this fractal cosmology, on the grounds that studies of large-scale structure in combination with the timeline of the universe have not produced any evidence for it." Is there any refrence avaliable as to its rejection? I find an analogy to a cantor set more than sufficient to explain the paradox, fitting infinite points into a finite space. I would like to know the evidence against it, not simply the lack of corroborating evidence. I would think the fractal nature of the sky is self-evident, if it weren't so the paradox would be unresolved. Nazlfrag (talk) 22:52, 29 December 2007 (UTC)

Re Olbers paradox.

The solution to this puzzle is in basic physics, rather than in the circumstantial nature of the universe. According to the old line of reasoning, if the greater universe were infinitely old and spacious and (thus) contained infinite sources of light/radiation then we should see radiation as bright as the sun or even infinitely bright in every direction. As that is not the case it is deduced that the universe must be finite in space and/or time. But this line of reasoning overlooks the fact that whatever the size or age of the universe (i.e even if it were infinite), we would not be allowed to see very bright or infinite radiation because of the conservation of energy principle. That principle rules that there must be a balance between the radiation received by the 'average position' in the universe and the radiation emitted by same. The radiation emitted by the average position is low because, well most of space is empty of sources of radiation. You cannot therefore have a possibility of every position seeing/receiving infinite radiation. The conservation of energy principle is useful in showing up the fault in the old line of reasoning but it does complete the answer. The answer is provided by Einstein's gravitational shift of light. Apply that equation (that led to the idea of a black hole) to an infinite universe and it should be clear; the infinite collective mass of the surrounding infinite universe should erase Olbers theoretical infinite light. That does not mean that all evidence of an infinite surrounding universe would be erased. I theorized, in 1994, that if the universe were infinite then our cosmos/the visible universe should, due to gravity, be accelerating apart, as has been subsequently observed. No need for 'dark energy'.

Mark Bridger 16 May 2008 —Preceding unsigned comment added by 86.144.170.236 (talk) 11:00, 16 May 2008 (UTC)

Original research

Latest comment: 15 years ago3 comments3 people in discussion

The initial paragraph of section 2 looks very much like original speculation to me (and also does not look like very encyclopedic). I added a WP:NOR tag. --Cyclopia (talk) 16:47, 18 May 2008 (UTC)

In the assumptions section it talks about "a more precise way to look..." Shouldn't the Volume of the "doubled sphere" be the real issue, and the fact that it would be filled with 8 times the stars? Since the distance issue is only "removing" light by a factor of four, and the stars are increasing by a factor of 8... you get what I am getting at. The amount of light would in fact Double for every doubling. Infinite doubling of distance would produce infinite doubling of accumulated light.

Thoughts? Jwhmca (talk) 18:38, 28 April 2009 (UTC)jwhmca

The surface of a sphere increases with the square of the radius. So at r times the radius there should be r^2 times as many stars "AT r". With Calculus it can be shown that this still results in the number of stars within r increase by r^3. At r, there are r^2 times as many stars and each star would be r^2 times dimmer. So each layer would be just as bright as the last. So yes, doubling the distance would double the amount of light received. The article already mentions this fact. There is no issue here.Kmarinas86 ^(6sin8karma) 16:08, 20 June 2009 (UTC)

Finite Photon Flux

Latest comment: 15 years ago2 comments2 people in discussion

Another answer has always seemed obvious to me, but I've never seen it discussed and it doesn't appear on the article. Whether a line of sight falls on the surface of a star is irrelevant to whether you see that star. What matters is whether that surface is emitting enough photons in the direction of your retina for you to detect it. Stars emit a finite number of photons, and those photons diverge as they travel greater distances. At a sufficient distance, you will no longer detect a continuous stream of photons from that star even when your line of sight falls on it. If simply having line of sight was sufficient to appear "as bright as the surface of the sun", the all visible galaxies with mesurable angular extent would be intolerably bright. The radial angle between their individual stars is less than the resolution of our retinas, so each galaxy would visually agglomerate into one patch all of which "intersects the surface of a star". But that's not what we see ... because we are so far from them that we are receiving very few photons from any individual star, even though is not beyond the observable bounds of the universe and is not redshifted out of the visual range. IdahoEv (talk) 18:15, 19 June 2009 (UTC)

The problem is that there is a chance of getting a photon from each star along the line of sight [of course, since the stars are opaque, you can't see photons from more than one of these stars], and there are infinitely many. So you're still guaranteed to get a photon from any given direction. This is more or less the same reason we can see a distant galaxy, even though the individual stars are nearly invisible. Ben Standeven (talk) 22:19, 28 July 2009 (UTC)

Conservation of Energy and Gravitational Redshift

The answer to Olbers's Paradox is in the basic laws of physics. Meaning, whatever the size or age of the universe, however mass is distributed, the age of it's stars or if it is infinite and eternal, we would not be allowed to see very bright or infinite radiation because of the Conservation of energy principle, which rules there must be a balance or equality between the radiation received by the 'average position' in the universe and the radiation emitted by the same 'average position'. The radiation emitted by the average position is low because most of space will be empty of sources of radiation. You cannot therefore have the Olbers deduced possibility of every position seeing/receiving infinite radiation.

Another way to see this is to realise that if one could see the light from objects an infinite distance away it would mean that each object could be radiating it's light in an infinite number of directions to infinite recipients, which would mean it was giving out infinite radiation. This problem arises in quantum physics and the solution is to put a finite limit on the number of directions/recipients of the emitted radiation, and that would mean light does not in reality travel an infinite distance. But how is a finite limit put on the number of recipients? This sounds a bit arbitrary. Fortunately there is another consideration which completes the answer: Gravitational redshift (the equation relevant to black holes).

So, what happens when you apply Einstein's equation for Gravitational redshift to the light from a surrounding infinite universe? Here is the equation:

                                            F’/ F  =  1 - GM / c2 R

where F’ is the frequency of the shifted light, F is the frequency of the original light, G is the Gravity constant, c2 is the speed of light squared, M is the mass of the object and R is it’s radius.

In the case of an infinite surrounding Universe the mass M would be infinite and so would the radius R, suggesting the result is indeterminable. However, the larger an object, the greater it’s mass to radius ratio tends to be (because volume increases cubically with radius) so the mass M could be considered the predominant infinity. In that case the mass to radius ratio becomes infinite, suggesting the resultant Gravitationally shifted radiation from an infinite surrounding universe would be infinite frequency zero wavelength radiation. That would be invisible and immeasurable - but may explain why the space in our Cosmos can have a field of infinite energy, as suggested by quantum physics/virtual particles/multiverse theory and the emergence of our Cosmos out of, seemingly, a point of space. Mark Bridger 14/12/09

Animation is incorrect

Latest comment: 14 years ago10 comments4 people in discussion

Smaller stars should be in the background, not bigger stars. Paradoctor (talk) 15:33, 7 April 2010 (UTC)

Are you the kind of person who claims that the "ufos" in the Nasa STS-75 Tether scene (link) are behind the tether?Kmarinas86 ^(6sin8karma) 20:12, 7 April 2010 (UTC)

There is probably an entirely rational and valid explanation for your comment. I would really like to hear it. Paradoctor (talk) 21:56, 7 April 2010 (UTC)

I believe Paradoctor is referring to the animation at the top right of this article. It starts with a few large stars, then starts filling in the blank space with progressively smaller stars. These later stars are clearly placed in the gif on top of the larger stars, which would seem to be backwards - new layers should go behind the old ones. - 2/0 (cont.) 03:33, 8 April 2010 (UTC)

I agree that the animation is incorrect. The way to fix it is to disassemble this .gif animation into its frames, and reassemble the frames into a .gif animation using layering, so that each successive layer goes underneath instead of on top. Photoshop or another graphics tool could be used to do this. Obankston (talk) 18:02, 1 June 2010 (UTC)

Ok. Now I get it. I use the "screen" filter for the stars, and you don't want to see the stars behind the others. I'm working on that then.Kmarinas86 ^(6sin8karma) 18:07, 1 June 2010 (UTC)

Good work. I think we can do without the final big star, though. It kind of muddles the message that the distant stars cover up everything, eventually. Paradoctor (talk) 05:26, 2 June 2010 (UTC)

I agree that the animation now correctly demonstrates Olbers Paradox, except the big star at the end is unnecessary. Obankston (talk) 17:12, 26 June 2010 (UTC)

One happy customer here. :) Paradoctor (talk) 08:42, 27 June 2010 (UTC)

Animation looks good now. Obankston (talk) 23:45, 2 July 2010 (UTC)

Cosmic Microwave Background

Latest comment: 14 years ago3 comments1 person in discussion

I wrote the first half of this section, but there are at least two technical flaws in the explanation I provided, so this is in the right place in the article as a non-mainstream explanation. (1) the cosmic microwave background radiation represents radiation leftover from the Big Bang, and does not necessarily represent the radiation representing the current average temperature of the Universe; (2) the cosmic microwave background implies a temperature that limits the amount of light, but the explanation does not calculate how high that limit is, so that perhaps the limit is so high that the nighttime sky would be bright. Obankston (talk) 14:55, 13 May 2010 (UTC)

The intensely technical second part of the section mentions the cosmic microwave background, but only because the implied temperature is consistent with yet another different explanation for Olbers' paradox. Perhaps it should be in a section of its own titled Baryon Abundance or Baryon Abundance Number or Baryon Abundance Parameter. Some material added by User:Ben_Standeven was applicable to this, but the material was subsequently deleted. The material is visible in the older version at Revision as of 06:42, 11 May 2010. Obankston (talk) 16:04, 13 May 2010 (UTC)

I created the new section and added the material from Revision as of 06:42, 11 May 2010. Obankston (talk) 17:08, 20 May 2010 (UTC)

Hydrogen or Helium

Latest comment: 13 years ago4 comments3 people in discussion

I just read this page to get information. I'm not a physicist.

The section on 'Alternative Explanations', 'Steady State' contains the following:

"assuming that essentially all atoms other than hydrogen-1 are non-primordial (in keeping with the original form of the steady-state hypothesis) the corresponding maximal radiation energy density of 9.2×10−31 kg/m3, i.e. temperature 3.2K[4]. This is close to the summed energy density of the cosmic microwave background and the cosmic neutrino background. The Big Bang hypothesis, by contrast, predicts that the CBR should have the same energy density as the binding energy density of the primordial helium,"

Note: 'all atoms other than hydrogen-1 are non-primordial' vs 'the primordial helium'

Should 'helium' here read 'hydrogen'? Or did I miss something?

[Sorry, I'm new to this. Same post appears below with signature.]

OK, the primordiality was not my work. This adjective is irrelevant in the (infinite) 'Steady State explanation'. I will remove it. —Preceding unsigned comment added by 194.228.230.250 (talk) 08:45, 8 February 2011 (UTC)

The Olbers ' paradox is ( or should be ) the proof that the universe is finite ... The galaxies are distant about 500.000 y. l. each other, the galaxies clusters about 10 million y. l. each other: so our observation ' point ( interior to the Milky Way ) is special and in a middle spatial probable point we should see only darkness ( with the sensibility of our eyes which with difficulty can see the giant Andromeda ' galaxy , at 2 millions y. l. ) ... and so this paradox should not rise . Also in our privileged position ( in a galaxy of 100 milliards of stars ) the most powerful thelescopes cannot see the star ' ligth jointed (like supposed by Olbers) because at the distance of 100 y. l. that-ones are almost like-point ( a point has infinitely small dimension ! ) . We remember also that the addiction of infinite small quantities can give a finite value ( 1+1/2+1/4 ...gives 2 ) ; then it is important to consider 1) our optical sensitivity ( which cannot be infinite ) 2) the separation ' power of the eye ( two ligth ' sources can appear jointed if the separation ' power has a limit and however the limit excludes to see a part of smaller sources ; perhaps some insects , having more sensibility and less separation power than us , can see the nigth ' sky luminous like hipotized by Olbers ) 3) the probability of spatial absorption known and unknown ..: the conclusion : good ' mind exercise , but nothing more .. no solutions in sigth . —Preceding unsigned comment added by 188.153.131.141 (talk) 09:10, 14 February 2011 (UTC)

The explanation can be more simple. The light is not directional (Geometrical optics) with infinite energy (zero wavelength approximation) but it is a wave ... —Preceding unsigned comment added by 194.228.230.250 (talk) 10:08, 22 February 2011 (UTC)

- The esplanation can be more simple , but not enought understandable , for meee...93.144.144.205 (talk) 15:39, 2 April 2011 (UTC)oldogf@yahoo.it

Is Olbers' paradox based on a false assumption?

Latest comment: 13 years ago2 comments2 people in discussion

One explanation of Olbers' paradox is that it is based on a false assumption, namely that the sky is dark at night. In fact, the night-time sky is bright with microwave radiation, but our eyes cannot see it. To a radio telescope tuned to a frequency of 160.2 GHz (corresponding to 2.73 degrees Kelvin), the night-time sky is bright in every direction. So the paradox depends on who (or what) is doing the observing. — Preceding unsigned comment added by 66.212.78.220 (talk) 21:40, 5 July 2011 (UTC)

"Dark" is in terms of the total luminance (total energy flux per solid angle). These ~3 K are much lower than ~6000 K expected from star surfaces in the infinite stationary universe. — Mikhail Ryazanov (talk) 03:29, 29 July 2011 (UTC)

Absorption section needs work

Latest comment: 12 years ago9 comments8 people in discussion

The section on why absorption cannot be an explanation needs work. It completely ignores ways for photons to be absorbed in a manner that precludes re-radiation of another photon (black holes come to mind). Kurt 01:51, 4 July 2007 (UTC)

Even black holes radiate - but the section does need to cover the case explicitly. I guess the "black holes as absorbers" solution wouldn't be be stable in the sense that they would either explode (if hotter than the stellar environment) or grow without limit (if cooler than the stellar environment) and hence be "seen". --Michael C. Price ^talk 16:49, 24 September 2007 (UTC)

I just noticed Kurt's above post, which is similar in thought to my own post above of a few days ago. I will just second the notion (or third it) that it would be good for someone knowledgeable about black holes to explain why they could not be absorbing light. Perhaps MichaelCPrince is up to the task; he certainly knows more than I do on the subject. But of course there's the very theoretical idea that black holes might have "exits" elsewhere in the universe, which, if true, would certainly explain why black holes are not the answer to Olbers' paradox, if for no other reason. --Iritscen 18:37, 27 September 2007 (UTC)

Olbers; Paradox is usually posed as a problem with universes that are heterogenous, infinite in extent, and infinite in age. Along with the infinities comes the implicit assumption that the universe is heterogenous with respect to space, and also with respect to time. The problem also tends to assume that basic thermodynamic rules are operating.

The difficulty with with introducing GR1915's black holes to the problem is twofold: "traditional" GR black holes don't obviously obey thermodynamic principles (until we retrofit quantum mechanics), and their one-way nature means that they also aren't obviously compatible with the idea of an infinitely old universe in equilibrium. If black holes kept sucking the energy out of space, and the universe was infinitely old, then one might expect that the "equilibrium state" would be an arbitrarily "cold" universe in which there were no visible stars at all! Which isn't what we see. With the advent of quantum mechanics, black holes can re-radiate, but if the universe is to have looked like ours for an infinitely long period of time, we'd need the community of black holes (on average) to be reradiating back into the universe at the same rate that they absorb from it ... or we'd need some other compensating mechanism for reintroducing energy into the universe.

Now, given that the conditions for black hole formation are more demanding than the conditions for luminous matter, and that it tends to be easier for luminous matter to obscure a black hole than for a black hole to obscure lumiunous matter, and that black holes have a habit of accumulating clouds of radiating matter around them, even if the universe //did// contain an infinite number of black holes without surrounding accretion discs or infalling clouds, or surrounding galaxies, the number of lightsources would still be expected to be a much //larger// infinity than the number of black holes. So //almost// every point in the night sky might be expected to appear illuminated. You might get around that my suggesting non-homogenous distributions of holes and matter, but if you go down that path, you're into "Mandelbrot" territory, in which case you don't need to posit the existence of black holes anyway.

I don't think that any of this needs to go into the article. The main //historical// importance of the problem was that it helped us to move towards the idea that the universe wasn't static and spatially and temporally infinite. Now that we have Hubble images appearing to show a statistical variation in the distribution of stars and the structure of galaxies that depends on distance from us (i.e. depends on age), the idea of a static universe doesn't seem as sensible as it once did.

Olbers' paradox has probably done its job. ErkDemon (talk) 00:40, 24 October 2008 (UTC)

Vincent - Maybe Im not quite grasping something here but as I understand it- the proposal that the universe might be infinite and static but the infinite radiation is absorbed by some kind of "dust" is rather well debunked by the thermodynamics argument, but how about if this "dust" was dark matter? The topic of dark matter seems to be shrouded in mystery at present. But dark matter itself must have no (or very very little) interaction with electromagnetic radiation whatsoever - this is why it's invisible to us and not directly detectable... Could it be that the "dust" idea is not as weak as the article would have us believe?? 15:26 9th October 2007 —Preceding unsigned comment added by 137.205.8.2 (talk) 14:26, 9 October 2007 (UTC)

Maybe it's because I nearly flunked my physics final all those years ago, but I don't immediately see why the thermodynamic argument necessarily debunks the absorption proposal. There's no reason that the absorber must reach thermal equilibrium with its surroundings immediately - couldn't it be of the order of the lifetime of the heat source objects? Suppose there's a population of stars within a 5Gy light cone from a planetar: by the time the planetar would have absorbed enough energy to be as hot (and hence as bright) as the stars, the stars are dying. In fact the planetar starts radiating only as the origin stars are stopping, becoming the new absorbers. I must be missing something, but I'd be curious to know what, exactly. Bernd Jendrissek (talk) 14:02, 8 April 2010 (UTC)

You've forgotten that as the old stars die, new ones are formed. Basically the same mistake as in the Finite Lifetime of Stars section. Ben Standeven (talk) 01:55, 9 May 2010 (UTC)

yes this smacks of 'original research'. Can we PLEASE PLEASE PLEASE find a citation somewhere? We are talking about SINGLE PHOTONS traveling very long distances, and we really think that the intermediate absorber, millions of light years away hit by a periodic stream of single photons necessarily reaches THERMAL EQUILIBRIUM, let alone REEMIT in the same frequency?? If the sky was white with stars the intensity of a spotlight BUT the earth was surrounded by black paper between the sky and the earth, the black paper would NEVER reemit white light no matter how long you wait. --24.191.102.86 (talk) 16:09, 14 September 2012 (UTC)

"Gravitational" counterpart

PS, the "black hole" issue (above) also calls up a related paradox that can be thought of as the gravitational counterpart of Olbers' paradox ... the idea that if the universe is reasonably homogenous, and we mark out larger and larger spheres of space, then since the surface area of the spheres depends on radius squared, but the contained mass increases with radius cubed, we get a certain radius at which the Newtonian escape velocity at the sphere's surface exceeds the speed of light - for an infinite universe, in which there exists a star at every position in the night sky, the gravitational flux density might be expected to be infinite.

If we have a finite universe that's larger than the critical diameter, then lightbeams can be wrapped around on themselves by the region's gravitation, and if the distribution of matter and gravitation obeys a certain tidy set of rules, then every part of that finite universe can be equivalent, with no "edges" anywhere. Trouble is, it's not stable, so it can't be homogenous with time unless we invent new things specifically to generate that result (e.g. Einstein's "Cosmological Constant") -- otherwise, the descriptions tend to end up giving an evolving, initially-expanding spatially-closed hypersphere. Which is pretty much what modern cosmology describes by default. ErkDemon (talk) 00:59, 24 October 2008 (UTC)

/* Cosmic microwave background */ moving to talk page.

Latest comment: 14 years ago9 comments3 people in discussion

Cosmic microwave background

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.^[1]. 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.^[2]

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

References

^ Hobson, M.P.; Efstathiou, G.; Lasenby, A.N. (2006). General Relativity: An Introduction for Physicists. Cambridge University Press. p. 388. ISBN 0521829518.
^ Steigman, Gary (15 November 2005). "Primordial Nucleosynthesis: Successes and Challenges" (PDF). International Journal of Modern Physics E. World Scientific Publishing Company.

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.)

How bright would the sky be?

Latest comment: 12 years ago1 comment1 person in discussion

I don't think I fully understand this paragraph, but isn't the idea much simpler? Since we'll see the surface of a star in any direction we look at and assuming the luminosity per surface area of all stars is the same as that of the sun's, the sky would look like as if many sun's cover the entire sky which is the same brightness as if we were standing on the sun and looking into it. That would make it ( 1/2 * 4*pi * (180/pi)^2 / (pi*(1/4)^2) ) about 100000 times brighter than current day light (sun has a radius of about 1/4 degree as viewed from earth and the half sphere has about 20500 square degrees).

Also, for this to happen we don't even need an infinite universe, a sufficiently large but finite universe can also achieve this. I made a quick calculation of the average distance to the next star at any random line of sight assuming uniform spacial distribution of stars which are all of the same size (can anyone confirm or correct this formula?), which is exponentially distributed with

avg distance to next star in line of sight = ${\frac {4}{\pi }}\left({\frac {dist}{diam}}\right)^{2}dist$

where dist means the average distance between two nearest stars (i.e. there is one star on average in a cube with each side dist long) and diam the diameter of a star (this assumes all stars are of the same size).

Assuming the density of stars in the milky way (about 1 star per 1pc cube?) everywhere and stars the size of the sun that would give an average distance of $5\times 10^{14}$ pc to the next star at any line of sight. Using the proper density of stars in the observable universe (about 1 star per 1kpc cube?) we'd get $5\times 10^{23}$ pc. Both values are way beyond the observable universe of radius $15\times 10^{9}$ pc. — Preceding unsigned comment added by 90.199.161.111 (talk) 22:23, 24 April 2012 (UTC)

Inverse square law actually does explain dark night sky

Latest comment: 11 years ago4 comments2 people in discussion

In the original treatment of the problem they completely ignored sensor surface area, that is some 2-dimensional image receiving this light, like a photo or human eyes, and by ignoring that they get result as if the image has only one pixel. So instead of to "see" many dots, some bright some less bright, they practically sum all the received intensity in only one pixel and thus result wrongly indicates the sky is bright.

They also ignored exposure time. The rate of incoming photons is proportional to distance, due to inverse square law, which is known and accepted fact, that's why very distant stars do not produce any dots on a photo-plate unless we wait long enough. Just by looking at this fact makes it clear to me inverse square law explains it all.

Let me explain with an example. Two stars at distance r would impact photo-plate with intensity I, and eight stars at double the distance will also impact photo-plate with the same intensity I. That's what they are saying, and that's fine. However, what they are not considering is that two closer stars will produce two dots each with brightens I/2, but eight further stars will produce eight dots each with brightness I/8.

There is difference between two bright dots and eight less bright dots of course, and there is difference between two dots on 10x10 resolution image and 1x1 resolution image. So when they ignore this sensor surface area they practically work with 1x1 resolution image where all the intensity gets summed up at one pixel, and of course all they see is "bright sky". To summarize I draw this conclusion: at infinite distance there will be infinite number of stars and if we had infinite resolution they would produce infinite number of dots, but the brightness of each dot would be I/infinity, which is pretty much nothing but black. Ze-aksent (talk) 20:39, 11 November 2012 (UTC)

This is not a forum on the article subject

. --Cyclopia ^talk 21:01, 11 November 2012 (UTC)

What I said is either true or false. Which is it?

— Preceding unsigned comment added by Ze-aksent (talk • contribs) 12:58, 12 November 2012 (UTC)

That image above is a bit wrong. Light source needs to be a point source for its apparent brightness to fall of with the square of the distance, otherwise it just shrinks in size until its angular size can not be resolved by sensor resolution anymore, at which point it becomes point light source and from that point on it starts to appears dimmer proportionally to inverse square of the distance.

Olbers' paradox:

a.) there would be four times as many stars in a second shell

b.) each star in it would appear four times dimmer than the first shell

c.) the total light received from the second shell is the same

a.) left image representing first shell contains 10 bright stars

b.) right image representing second shell has 40 stars each 4x less bright

c.) total light received is the same, but does that make them equally bright?

Ze-aksent (talk) 13:54, 24 November 2012 (UTC)

Legible Equations?

Latest comment: 11 years ago1 comment1 person in discussion

Can something be done about the equations which appear as a grey "fuzz" against a dark background? I've noticed this in most Wiki articles that contain equations. If the equations can't be rendered with ASCII characters, how about at least inserting them as legible jpegs? (For example, the equations under "Alternative Explanations: Steady State" are completely illegible, whereas the equation under "Fractal Start Distribution" is perfectly legible.)

— Preceding unsigned comment added by 74.92.174.105 (talk) 20:29, 10 December 2012 (UTC)

BB and helium - fails

Latest comment: 11 years ago1 comment1 person in discussion

"The Big Bang hypothesis, by contrast, predicts that the CBR should have the same energy density as the binding energy density of the primordial helium, which is much greater than the binding energy density of the non-primordial elements; so it gives almost the same result."

This is incorrect. The expansion of space stretches a wavelenght (of photons), and because: E = hf = hc/l, the energy of radiation decreases proportionally: E_now = E_0 / (z+1),

Acording to BB: z = 1100 for CMB, thus this model predicts the temperature of CMB: 1100^1/4 = 5.76 times lower, i.e. about 0.5 K only. — Preceding unsigned comment added by 83.22.180.81 (talk) 02:05, 24 December 2012 (UTC)

Absorption section wrong

Latest comment: 10 years ago8 comments2 people in discussion

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)

Units?

Latest comment: 10 years ago3 comments3 people in discussion

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 interpretation 73.186.239.20 (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

Latest comment: 10 years ago1 comment1 person in discussion

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

Latest comment: 8 years ago1 comment1 person in discussion

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)

Infinite directions?

Latest comment: 7 years ago7 comments2 people in discussion

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? Iago 2¹² 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? Iago 2¹² 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? Iago 2¹² 22:17, 23 November 2016 (UTC)

No. Because your eye is not a mathematical point. If it emits a single photon per second, in a uniformly random direction, that still gives it a nonzero brightness in all directions. —David Eppstein (talk) 00:17, 24 November 2016 (UTC)

How does a random distribution of a finite number of photons give you all directions? Iago 2¹² 09:33, 24 November 2016 (UTC)

[1] Hobson, M.P.; Efstathiou, G.; Lasenby, A.N. (2006). General Relativity: An Introduction for Physicists. Cambridge University Press. p. 388. ISBN 0521829518.

[2] Steigman, Gary (15 November 2005). "Primordial Nucleosynthesis: Successes and Challenges" (PDF). International Journal of Modern Physics E. World Scientific Publishing Company.

[1]

[2]