TheBrownMamba
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In general terms, but not too general.
The discussion revolves around Olbers' paradox, which questions why the night sky is dark if the universe is infinite in age and size, suggesting that every line of sight should end on a star. Participants explore theoretical implications, alternative models, and the paradox's relation to the universe's expansion and star formation.
Participants express multiple competing views regarding the implications of Olbers' paradox, particularly concerning the universe's expansion, star distribution, and the age of stars. The discussion remains unresolved, with no consensus reached on the interpretations of the paradox.
Participants acknowledge various assumptions in their arguments, such as the distribution of stars and the effects of cosmic expansion, which may influence the conclusions drawn about Olbers' paradox.
russ_watters said:If the universe were of infinite age and size, every line of sight would end on the surface of a star and it would be very bright:
http://en.wikipedia.org/wiki/Olbers'_paradox
skeptic2 said:Notwithstanding the post from Eachus with which I agree, I have always questioned the this explanation. It seems to me that although at twice the distance there would be 4 times as many stars, each star would have 1/4 the angular area. Also there would always be some stars hidden by stars in front of them and the farther they are, the more often that would occur. It seems more likely then that with an infinitely large and old universe the sky would not necessarily be very bright but have an intermediate brightness determined by how sparse the stars are.
Olber's paradox also doesn't address the issue of the age of the stars which is not infinite despite the infinite age of the universe. At some point all the hydrogen in a region would be depleted and no new stars could form. In fact if the universe were infinitely old, would that not mean that there could be no shining stars at all - a truly dark sky.
LeonhardEuler said:Suppose the density of stars is roughly constant in the universe and they are randomly distributed. If you take a ray extending from the earth, then over a sufficient distance, the probability that the ray touches the surface of a star over a constant distance, call it "D", is constant.
So the probability of not hitting a star in the first D light years could be (1-p). Not hitting a star in a distance of 2D would have probability (1-p)2. Not hitting a star ever would have probability
[tex]\lim_{n\to\infty}(1-p)^{n} = 0 \ \forall p>0[/tex]
This assumes the universe has some regular structure for a big enough scale of distance. There are big fluctuations in density over even the scale of light years, because of galaxies compared to intergalactic space, but if there is ever a scale over which the distribution of stars is well approximated as homogeneous and isotropic, then this logic would hold.
Of course, there are some assumptions in this argument. If the stars preferentially aligned themselves behind one another from the Earth's perspective, this would not work. But that would make the Earth a very special place in the universe. The universe could be infinite, but with the density of stars diminishing towards 0 quickly as you head outward. This is why you need the assumption of homogeneity.
Well, in order for there to be enough stars visible at once for a significant number to block some behind them, the sky would already need to be extremely bright.skeptic2 said:The discrepancy is that p is not constant with regards to D because some stars will be blocked by other stars in front of them.
True.russ_watters said:Well, in order for there to be enough stars visible at once for a significant number to block some behind them, the sky would already need to be extremely bright.
russ_watters said:In any case, you're still not thinking about it properly. This is isn't like a lottery where once a number (printed on a ball) is used, it can't be used again. Any individual vector has the same finite odds of hitting a star by a particular distance and the fact that some vectors hit more than one star (and some certainly will!) doesn't change that.
No one is suggesting that each vector would hit only one star!