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I'm not quite grasping the reasoning behind Olbers' Paradox, or why an eternal, non-expanding and spatially infinite universe would be incompatible with a dark sky. For simplicity, let's suppose that all the stellar matter in the universe is transparent, such that to look at any direction in the night sky is equivalent to looking at an infinite number of stars or light sources. Yet I still don't see why it must follow that every point in the night sky will be infinitely bright (or at least sufficiently bright), provided that the sum of the apparent luminosity for all the stars at a given point converges to some finite value. For example, assume we drew a line to the infinite horizon and the first star intersecting that line had an apparent luminosity of 1 lumen, the second star .1 lumen, the third star .01 lumen and so on and so forth...

Furthermore, if this finite convergent value should happen to be lower than the detectable optical threshold of the human eye, that point of the night sky would still appear to be dark. Of course, this assumes that all the stars are transparent, and that they do not occlude any light. In the real world this effect will be lessened due to stellar opacity, but even this hypothetical scenario involving complete transparency doesn't seem to necessarily invoke a paradox, provided that the stellar density is such that the apparent brightness falls off at a level fast enough.

To put it another way, in an eternal, non-expanding, and spatially infinite universe, every point in the night sky will indeed intersect with some distant star, but there is no reason to think that the received luminosity of that star will be bright enough to be noticed. So, what I am I missing? Why is it commonly assumed that Olbers' paradox precludes an infinite static universe?

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  • $\begingroup$ My first thought was that no matter how far away, every point you look at will receive light directly from a star, but when you mention lumens, I realize that photons are discreet, and since lumens correlates roughly with photons/second, that you're right, there would be points in the sky where the photons/second due to distance would be less than the human eye could perceive, even if it is directly at a star. $\endgroup$ Dec 22, 2022 at 18:23

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Olber's paradox is - as you state - the phenomenon, that the night sky is dark, but would have to be bright as the sun if the universe was infinite and infinitely old. The unspoken assumption here is also that the universe is at least on larger scales isotropic, that is about the same everywhere.

So if the universe is the same everywhere (and with everywhere we can be generous and consider distances between our galaxy and its neighbours still small), then it means that we will see some star or galaxy whereever we look. Galaxies are made of stars, so we will see stars everywhere - and every star is approximately as bright as the Sun - thus we would have a sky as bright everywhere as the disk of our sun. Maybe put the argument differently: consider the sky covered by suns at different distances. But in every direction you see a sun. So in essence it means that every direction you look at a Sun's surface. The contribution of every single Sun may be tiny (depending on how far it is) - but there are infinitely many Suns and thus the surface brightness is the same everywhere. The stars are only so faint because they are so far. Mathematically: you double the distance, you get a quarter of the intensity from every single star (inverse square law). But doubling the distance also means we have four times as many stars (the volume of a spherical shell is $\propto d^2$. Thus both effects taken together, effectively every distance doubled adds the same brightness as the previous one as the reduction in the brightness of each single source is exactly countered by the amount of sources we look at. Now, in an infinite, static universe there are infinitely many such shells, and we add infinitely many constant brightness values, leading to an infinite brightness (or at least the same brightness if we (wrongly) assume that the foreground stars hide the stars behind them.

Now, the argument may be "there is matter between the distant stars and us which absorbs the light". Yes. But we are talking of a static, infinitely old universe. The matter will be heated by the incoming light - and eventually it will become similarily hot as the source(s) which are heating it, thus also reach the temperature of stellar surfaces and emit light at the same intensity. Stellar opacity is not an argument here either - it also just is matter and energy does not get lost and will simply heat up all the same like all other non-stellar matter up to the same temperature as it is heated by: the radiation from the stellar surfaces.

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    $\begingroup$ Thanks, that makes sense. I made the error of assuming that any starlight originating at an infinitesimal (or atomic, for a discrete universe) point in the night sky will continue to experience a drop off in luminosity with distance, but obviously this is incorrect since the photon/light-wave energy remains constant. $\endgroup$ Dec 20, 2022 at 17:35
  • $\begingroup$ Would this answer not have to assume that 1) photons can't decay; 2) photons can't be blocked by other objects; and 3) photons don't get deflected by gravity. Surely after some distance the odds of a photon encountering something opaque; decaying (if they do decay); or being deflected out of the way by gravity would tend toward 100% $\endgroup$ Dec 21, 2022 at 17:00
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    $\begingroup$ @ScottishTapWater: I think 1 is true, we do have to assume no photon decay. Regarding 2, Planetmaker already addressed the reason why intervening objects won’t make a difference. As for 3, it won’t matter that photons in the night sky experience curvature, as long as every point in the night sky is filled with photons. While there will be an infinite number of photons deflected away from earth, there will also be an infinite number deflected towards earth. Curvature just redistributes the paths of the individual photons; it doesn’t change the number reaching earth (assuming isotropy). $\endgroup$ Dec 21, 2022 at 17:23
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    $\begingroup$ @ScottishTapWater if photons decay into nothing, then you have a pretty stark violation of energy conservation (which we should be able to observe). If they decay into something else, that should itself be observable, unless they decay into dark matter/energy/something that is very weakly interacting. $\endgroup$ Dec 21, 2022 at 20:07
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    $\begingroup$ @MooingDuck The photons spread out, but the surface area decreases to compensate. Apparent brightness = Total luminosity/4pi(R^2) But notice that the surface area of a star in a night sky is also proportional to the inverse square of the distance. Therefore, apparent brightness is directly proportional to surface area (B=W/m^2). So, the photons per second received per surface area of the night sky is a fixed constant for a given star. That far away sun taking up .25 arcseconds will be just as bright as the close sun observed through a .25 arcseconds pinhole. $\endgroup$ Dec 23, 2022 at 1:52
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It's an easy integral assuming four things:

  • The universe is infinite in extent,
  • Stars are uniformly distributed in this infinite universe,
  • The universe is infinitely old,
  • Newtonian mechanics applies, implying that light from remote stars is inversely proportional to the distance squared.

Given these assumptions, we not only would not have non-dark skies, we wouldn't even exist because the sky would be infinitely bright. As I mentioned at the top, it's an easy integral with these four assumptions: the brightness is $\int_0^\infty \text{some constant}\, dr$. In other words, infinitely bright. This is Olber's Paradox.

General relativity combined with big bang theory offers a solution: The last two assumptions are false. The universe is not infinitely old, and light from ever more remote stars is diminished not only by distance but also by redshift due to expansion.

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A couple of good answers already, but this might be an easier way of thinking about it.

The light you see from a star comes only from the photons emitted at the tiny specific angle just right enough to enter your eye. That angle is very close to zero, but not zero. As a star moves further away, that angle gets smaller and smaller, so more photons miss your eye, the result is the star appears dimmer because a tiny portion of the sky that was sending photons to your eye is now dark sky. Those photons reaching your eye are not less intense, there are just fewer of them. If, instead of dark sky, you fill that darkness with another star (or stars) of equal brightness. So, your eye will interpret that region just as bright as it was before. Extrapolate that across the whole sky, and there is no region not sending photons directly towards your eye at full intensity.

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Another version of Olber's Paradox: consider a cube with side length, say, 1kLY (1000 light years). This cube will have stars inside of it, and the light from those stars will be exiting the cube through its sides. But we can make a symmetry argument that there should also be light coming into the cube through the sides, and for the average cube, the light coming in will equal the light going out; there's no reason for one quantity to be larger than the other. So while there will be individual photons leaving the cube, there won't be any net light leaving the cube, and so we might as well imagine the cube being surrounded by mirrors, reflecting all the light back into the cube. If the stars are infinitely old, and have been producing light infinitely long, and none of that light has left the cube, then there is an infinite amount of light in the cube.

Thermodynamically, if heat is being produced, but the region producing the heat is at finite temperature, it must be shedding that heat to somewhere. And that somewhere must be cooler than the region. But if the universe is at large scale uniform, then each region is on average the same temperature as everywhere else, so there's nowhere for the heat to go.

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