A long time ago I was told that the universe is finite. The provided "proof" (or reasoning), known as Olbers' paradox, was that on infinite universe there would be an infinite number of stars, and that an infinite number of stars in night sky would make the sky bright.

I guess this sounds reasonable, but I have some doubts. If we assume that universe really is infinite and that it has infinite number of stars spread somewhat uniformly on the space, how would one prove (or disprove!) that the night sky really would be bright? What assumptions one would need to make on the way constructing the proof?

I am not looking for a "correct" answer but how to approach the claim mathematically.


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    $\begingroup$ en.wikipedia.org/wiki/Olbers'_paradox $\endgroup$ – Tom Cooney Oct 9 '13 at 10:21
  • $\begingroup$ Title question (v4) has possible duplicates: physics.stackexchange.com/q/24017/2451 and links therein. Question (v4) in main body is possible duplicate of Olbers' paradox, see physics.stackexchange.com/q/11014/2451 , physics.stackexchange.com/q/46353/2451 and links therein. $\endgroup$ – Qmechanic Oct 9 '13 at 11:57
  • $\begingroup$ see physics.stackexchange.com/q/25076 (but note that most of the answers are wrong) $\endgroup$ – Ben Crowell Oct 9 '13 at 14:41
  • $\begingroup$ Comments: (i) Note that the SE software only allows to insert one possible duplicate per close vote. (ii) When asking a question, the phrasing of the title is very important, and it should reflect the question in the main text as much as possible. (iii) Suggestion to new title (v4): What mathematical assumptions about a (possibly idealized) universe go into Olbers' proof that the night sky is bright? $\endgroup$ – Qmechanic Oct 9 '13 at 19:18
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    $\begingroup$ @Qmechanic The title was written to math.se, not to physics.se. It could have been better but I would really expect the question matter more. Also, the "proof" in Wikipedia (or other sources I found) is just hand-waving. The answer of Donkey_2009 is so far the only answer that answers the question. (I think I see your point - problem is that I'm not interested in physics but mathematics and this probably causes a communication barrier) $\endgroup$ – user30793 Oct 9 '13 at 19:31

I have nothing to say about the (possible) infinity of the universe; however, it is the case that infinitely many stars distributed uniformly will create infinite brightness at any point.

Let's throw away all the real physical facts about stars, and make the (incorrect, but good enough for now) assumption that the universe is an infinite volume with some 'star' distributed uniformly throughout it. So there is a uniform 'brightness field' which I'll denote by $\phi$.

Now suppose we are standing at the point $\mathbf 0$. How much brightness are we getting from a small volume $dV$ at a distance $r$ from us? Well, the small volume is emitting light rays at an intensity of $\phi dV$, but as you get further away (say, at a distance $s$) from the volume, the light rays are spread over the surface of a sphere of radius $s$. The surface area of a sphere of radius $s$ is proportional to $s^2$, so the light intensity due to the volume at a point a distance $r$ away from the volume is proportional to $\frac{1}{r^2}$.

(source: ohio-state.edu)

So the volume $dV$ contributes a brightness of $\frac{C}{r^2}dV$ to the total brightness, where $C$ is some constant, and $r$ is the distance of the small volume from $\mathbf 0$. We are now in a position to integrate over a large sphere of radius $R$ to get the brightness due to all the 'stars' at a distance less than $R$ from us:

\begin{align} \textrm{Total brightness at distance less than $R$} &= C\int_{|\mathbf x|<R}\frac1{|\mathbf x|^2}dV\\ &=C\int_0^{2\pi}d\phi\int_0^\pi \sin(\theta)d\theta\int_0^R\frac1{r^2}\times r^2dr\\ &=C\times4\pi\int_0^R dr=4\pi CR \end{align}

So the brightness due to the stars in a sphere of radius $R$ is proportional to $R$. Clearly, if we make $R$ infinitely large, the brightness will become infinitely large as well.

Of course, this ignores several physical realities, such as the expansion of the universe, the geometry of space-time, etc. But that is, at least, the mathematical justification for your friend's claim.

  • $\begingroup$ just assuming it i don't know much about this topic but can it be possible that the light coming from distance star's are affected by the gravity of possible black hole's in their path and deflect them away from earth $\endgroup$ – Dimensionless Oct 9 '13 at 14:36
  • $\begingroup$ @Akash Maybe - I have no idea. Just answering this as a purely mathematical question. $\endgroup$ – John Gowers Oct 9 '13 at 14:53
  • $\begingroup$ @Donkey_2009 Thank you for your effort, you provided an example of the kind of argument I was hoping for. Unfortunately it seems many have difficulties understanding the question. $\endgroup$ – user30793 Oct 9 '13 at 15:55
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    $\begingroup$ @Akash No, and the reason is more math than physics. If the universe is homogeneous (i.e. Earth isn't special), for every bit of light deflected away from us, an equal amount will be deflected toward us that otherwise would have gone elsewhere. Homogeneity begets a symmetry that allows you to assume every light ray travels on a straight line. $\endgroup$ – user10851 Oct 10 '13 at 2:14
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    $\begingroup$ I think you missed that one star gets in the way of another. The result (of an argument neglecting expansion and finite time) is, I think, not infinite brightness but uniform brightness at a temperature dictated by fusion physics in stellar interiors. The argument is merely that the sky would be bright. $\endgroup$ – Andrew Steane Nov 26 '18 at 11:09

There is one thing that you have failed to consider: the expansion of the universe. Since the expansion of the universe is "faster" the further away it is from us, there will come a point where the expansion is faster than the speed of light. (This actually happens, I'll add sources if I have the time to do so.) So, the light from stars at or greater than that distance would not be able to reach us.

Secondly, since infinite time has not elapsed, not all light have reached us.

Edit: great video to answer your question

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    $\begingroup$ Since the expansion of the universe is "faster" the further away it is from us, there will come a point where the expansion is faster than the speed of light. [...] So, the light from stars at or greater than that distance would not be able to reach us. This is a common misconception. There are galaxies that are and always have been moving away from us at $>c$ and that are observable to us. See arxiv.org/abs/astro-ph/0310808 . $\endgroup$ – Ben Crowell Oct 9 '13 at 14:44

If all photons decay after some arbitrarily large time period, Olbers paradox is no longer an objection to the proposed infinite extent of the universe.

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    $\begingroup$ As stated by the OP please address this mathematically if possible. Also add citations to backup your statements, elaborating more your thoughts is also recommended. $\endgroup$ – ohneVal Nov 26 '18 at 11:09

There are things that block out brightness (relative brightness). The sun is the brightest star during the day, and most stars are way too faint and distant to notice them as bright.

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    $\begingroup$ Doesn't solve the problem of Olber's paradox as posed. The question is why the night is not bright, so the Sun is irrelevant. Uniformly distributed stars increase in number with distance just as fast as the intensity of their light here on Earth diminishes, see Donkey_2009's answer. Objects blocking the light will heat up until they reach thermal equilibrium with the incident radiation and themselves emit the same apparent intensity as the light they were previously blocking. You can find more information on Wikipedia. $\endgroup$ – Mark Mitchison Oct 9 '13 at 11:55
  • $\begingroup$ @Mark: But such objects will still scatter the light. It's plausible, but nonobvious (to me), that this effect washes out. $\endgroup$ – Hurkyl Oct 9 '13 at 22:57
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    $\begingroup$ @Hurkyl I don't see how it's plausible. If you give them enough time (Olber's paradox assumes an infinitely old universe) to reach thermal equilibrium with the incoming light then it's simply nonsense, they will radiate at exactly the same temperature. Even if you assume that these objects somehow scatter light away from us without absorbing it, exactly the same amount of light that would otherwise have missed Earth will be scattered towards us from other parts of the sky. This just follows from homogeneity. $\endgroup$ – Mark Mitchison Oct 9 '13 at 23:04
  • $\begingroup$ @Mark: I meant it's plausible, but nonobvious, that the effect of scattering washes out. $\endgroup$ – Hurkyl Oct 9 '13 at 23:14
  • $\begingroup$ @Hurkyl Sorry, don't understand what you mean by washing out. As I explained, exactly the same amount of light scattered away from our path must also be scattered towards us. $\endgroup$ – Mark Mitchison Oct 9 '13 at 23:43

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