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In general terms, but not too general.

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- Thread starter TheBrownMamba
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In general terms, but not too general.

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russ_watters

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http://en.wikipedia.org/wiki/Olbers'_paradox

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A long-lived universe that does not expand is equivalent to a set of such boxes, or if you prefer, to just one. In either case, Earth, and humans could not survive in such a universe.

In effect, Olber's paradox says we must live in an expanding universe, since we could not exist in a static or shrinking universe, unless it was extremely young. And very young universes have other problems.

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http://en.wikipedia.org/wiki/Olbers'_paradox

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.

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"Eternity is a long time, especially towards the end." - Woody Allen

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LeonhardEuler

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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.

Think of it this way. 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)

[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.

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I concede your point, however I don't believe your example is quite correct, though, even when corrected, it doesn't change the result.

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.

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. As the probability of not encountering a star, [tex](1-p)^{n}[/tex], decreases, the probability for more distant stars being blocked by nearer stars increases. This greatly increases the distance at which the sky would appear very bright but it still will happen.

It still doesn't address the problem of in a universe of infinite age, all the hydrogen and other fusionable elements would have long disappeared.

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russ_watters

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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.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.

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.

No one is suggesting that each vector would hit only one star!

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True.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.

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.Anyindividual 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!

I am suggesting that each vector would hit only one star. It is like a lottery because once a vector hits a star, none of the light from stars behind it adds to the brightness of the sky. When a star's light is blocked by another star, the eclipsed star is essentially subtracted from the total number of visible stars and this has the same effect as if the stars were slightly more sparse which makes p slightly lower. As the distance increases, the number of eclipsed stars increases reducing p with respect to distance.

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LeonhardEuler

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Chronos

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If there is a decreasing density the further you move from your origin, Olber's seems silly, if there is roughly the same density (thus that term "homogenous" being used) every which way you go, then you will not be able to look any direction without seeing a photon.

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