Illustrating Olbers' paradox

The assumptions are that the universe is infinite and more-or-less static, has a uniform random distribution of similar-type stars per unit volume, and has existed forever (this last assumption can actually be relaxed a little).

I'll add one, for simplicity: all the stars emit light with the same spectrum and intensity as the Sun. This means, among other things, that all the stars emit light at 500 nm at the same rate in all directions. (This can be relaxed too, but it keeps things simple for a first pass.)

Imagine you are equipped with a 500 nm photon detector which only lets photons in that are travelling almost exactly along the axis of the detector. No matter which direction you point it, if the above assumptions are true its line of sight must eventually end on a star's surface, and since the universe is infinitely old there must have been enough time for light to reach the detector from that star. So, the detector detects photons at 500 nm, at whatever rate the star emits them along that precise direction (which, obviously, is the about same rate that it emits photons in any direction).

So, whichever direction you point the detector, it detects photons at the same rate (the rate each star emits 500 nm photons in a particular direction). Therefore, the sky is uniformly illuminated at 500 nm, at whatever intensity the typical star's spectrum has at that wavelength. This holds for every other wavelength, too; therefore the whole sky is uniformly illuminated with the same spectrum and intensity as the Sun's surface.

(You can argue that not all stars are the same; that just gives the sky a sort of average-star spectrum in each direction, and so does not really make things any better.)

 

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The stars never have a zero angular diameter. At a distance r, the angular diameter is reduced by a finite factor that goes as 1/r, and so the solid angle subtended by a star at the Earth goes as 1/r[sup]2[/sup]. Similarly, the solid angle subtended by the Earth at the star goes as 1/r[sup]2[/sup], and in consequence the number of photons reaching Earth per unit time from that star goes as 1/r[sup]2[/sup] too (the familiar inverse square law).

Now suppose for a moment there were only one shell, consisting of stars at distances ranging from r to r + a from Earth (with a very small compared to r). The volume of the shell is 4πr[sup]2[/sup]a, to first order in a/r. If the number of stars per unit volume is some constant N, it follows that the number of stars in the shell is 4πr[sup]2[/sup]Na. By the inverse square law the number of photons per unit time reaching Earth from each star is given by I[sub]0[/sub]/r[sup]2[/sup] for some positive constant I[sub]0[/sub]. Therefore, the total number of photons per unit time reaching Earth from the shell is 4πNI[sub]0[/sub]a, which you can see is independent of r.

Thus, in an infinite large, infinitely old, static universe filled uniformly with stars, every shell sends the same total number of photons towards Earth per unit time. Because there are infinitely many shells, if it were not for the fact that stars can intercept photons from other stars, the assumptions would actually lead to the conclusion that the night sky is infinitely bright. As it is, we just get a finite brightness equal to the surface brightness of a typical star.

 

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Right on the money. Point light source and angular diameter, what these really are and how they impact our conclusion is what this argument actually boils down to.

 

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No, it isn't. Please demonstrate how you came to this (wrong) conclusion.

In the above case you appear to be making a qualitative judgement instead of calculating the actual brightness.

 

Do you believe that is consistent with inverse-square law? I'm afraid not. If that was true then the stars in Hubble Deep Field would be as bright as stars in our galaxy. But they are not, they are up to one ten-billionth the brightness of what the human eye can see, which is why the Hubble telescope had to accumulate their light for almost a month.

There are stars so far away we get no more than a single photon from only every few minutes or hours, even though many of them emit much more photons than our sun. For those stars you can no longer say their brightness is relative to their apparent size (angular diameter), but instead it becomes solely a function of exposure time. This defines the limit where a light source becomes a true point light source.

 

How prescient of you....that is indeed where the OP intends to go.

There's two problems with your logic (and his):

1. Stars only appear to be point sources. They aren't actually point sources.
2. That doesn't resolve the paradox. In fact, it makes the paradox worse: If the stars have no size, then they can't obscure each other and thus instead of the sky being as bright as the surface of the sun, it becomes infinitely bright.

 

That isn't what is being claimed. You need to slow down or you are going to go running off into another mess of false assumptions and interpretations. Here you have misunderstandings of both how the inverse square law works and how the HDF works.

Note for the HDF:
It isn't taking pictures of individual stars, it is taking pictures of galaxies. So each pixel is collecting light from thousands or even millions of stars. An individual star in deep space would indeed be too dim to be visible on the HDF, but a bunch of stars that are close enough together to show up on the same pixel are plenty bright enough.

 

I explained in my following post how the inverse square law fits into this.

 

I assumed my logic (such as it was) was flawed which is why I asked the question instead of making a statement. btr and you have pointed out the obvious flaw - the star is not a point soruce it is just ittsy bittly teeny tiny.

I had not thought of that! That would would do wonders for my tan, prior to my autoignition.

 

Do you claim the number of photons per unit time received from two identical stars, where one is at 5 billion and the other at 10 billion light years away, will be the same?

 

It might not matter. The retina cannot distingush things if they are close enough together. If the image density of point light sources was high enough it would look white.

 

See my earlier post: http://www.sciforums.com/showthread...bers-paradox&p=3195356&viewfull=1#post3195356

 

post #21
So, whichever direction you point the detector, it detects photons at the same rate

Do you say here the amount of photons received from each star is the same?


post #22
By the inverse square law the number of photons per unit time reaching Earth from each star is given by I0/r2

Do you say here the amount of photons received from each star is not the same?

 

You forgot to explain your remark.

You are again talking about ME, for some strange reason. [rolls eyes]

I observed colors and concluded grey is less bright than white. How did you arrive to your conclusion?

Brightness is non-quantitative or subjective property. You appear to be confusing brightness with luminosity or intensity. Brightness is a function of received intensity per unit area, and it is also a function of sensor resolution, sensitivity and exposure time, which vary from person to person. Brightness is in the eye of the beholder.

 

No one has ever claimed that: that's your misunderstanding of how the inverse square law is being explained to you.

No one is claiming that stars have the same brightnesses at different distances: there isn't anything to explain there because you are simply saying pulling that out of thin air.

More rigor would help: You should quote exactly where you think you saw these things and explained why you interpret them the way you do. that's what I mean by slowing down.

By understanding the math. By understanding that 1/4 * 4 = 1.

Qualitatively seeing that grey is less bright than white tells you that each individual star appears dimmer: it doesn't tell you anything about if the whole image is dimmer.

It sounds like you are trying to make this inherently unsolvable and qualitative to avoid the math. The two pictures may as well be photos taken with the same camera at the same exposure time -- you are just adding confusion by adding variables that don't exist. The brightness of those two pictures is measurable. Heck, you're even playing word games by substituting-in the word brightness and then saying it isn't quantitative even as you are claiming to be using the inverse square law. If "brightness" is the wrong word, that's your fault, not mine.

 

I said "whichever direction you point the detector, it detects photons at the same rate." (Emphasis added.)

Every line of sight ends on a star, and I was dealing with an idealised detector with an extremely narrow (strictly, infinitesimal) field of view. Therefore, the photons it receives comes from an infinitesimal point on some star's surface, and the (infinitesimal) photon reception rate is equal to the (infinitesimal) number emitted per unit time by the star from exactly that point in exactly the Earth's direction. This does not depend on the distance between the detector and the star, it just depends on the total number of photons emitted per unit time per unit surface area by the star.

I can improve it so that it doesn't depend on infinitesimals; perhaps that will help. I'll do that at the end of this post.

Yes, because here I was talking about the number received per second from the entire star's surface, rather than from an infinitesimal point on it.

But I was also more specific: "the number of photons per unit time reaching Earth from each star is given by I[sub]0[/sub]/r[sup]2[/sup]." In that same post, I also pointed out that although the number of photons received per second per star varies as the inverse square of the shell radius, the number of stars in a shell varies as the square of the radius, and the two effects exactly cancel. In other words, I showed that the number of photons received per second from the entire shell does not depend on the shell's radius.

(Edited to add: that's not quite right. I actually showed that the number of photons emitted in the direction of Earth per second from the entire shell does not depend on the shell's radius. The distinction does not matter in the scenario with just one shell - or when shells are transparent - but does when multiple, opaque shells are considered.)

OK, now for the improved version of the first argument which I promised above. Let's suppose our grant proposal was accepted and we now have a more expensive detector with an adjustable field of view. We can set the field of view to any solid angle we like, however small (but not zero).

So, we turn the field of view up to maximum (for argument's sake, let's say it can cover half of the sky), pick some completely random direction in space, and switch the detector on. If it so happens that the field of view contains multiple stars, we will reduce the field of view (without changing the detector direction) until it sees part of the surface of just one star. This is always possible, because every line of sight eventually terminates on some star's surface.

Let the star be a distance r away. If I represents the average amount of photons per unit time we'd get from that fraction of a star's surface if it were a unit distance away, then I/r[sup]2[/sup] (here we see the inverse square law) tells us the average number of photons per unit time we'll receive from it in reality.

However, if A represents the solid angle which that fraction of the star's surface would have occupied in our sky at a unit distance away, then by basic geometry A/r[sup]2[/sup] is the solid angle it takes up in reality. So, the number of photons we receive per unit time per unit solid angle from that small patch of sky is just I/A, and we see that the distance to the star has cancelled out. The number of photons we receive per unit solid angle per unit time from the fraction of star surface does not depend on how far away the star is! In particular, it must be the same average number of photons per unit solid angle per unit time that we'd get if it were as close as the Sun. This does not contradict the inverse square law; in fact, in a sense it underlies it.

But this was obtained for a completely randomly chosen direction in space. If we look in any other direction, the same argument will lead to precisely the same result. Our detector will be centred on a bit of star surface from which we receive the same number photons per unit time per unit solid angle as we would if the star were as close as the Sun. The sky therefore has a uniformly bright appearance, and in fact every patch of sky with the same apparent size as the Sun will deliver as many photons per unit time to us as the Sun does (assuming the other stars are all much like the Sun, on average).

 

EDIT: never mind, I'll talk about this with btr.

- "As defined by the US Federal Glossary of Telecommunication Terms (FS-1037C), "brightness" should now be used only for non-quantitative references to physiological sensations and perceptions of light."

en.wikipedia.org/wiki/Brightness
its.bldrdoc.gov/fs-1037/dir-005/_0719.htm

I'm trying to be specific and differentiate between luminosity, received intensity and brightness. If you confuse them you will arrive to wrong conclusions. The fact that brightness is non-quantitative or subjective property doesn't mean it is not subject to math. It is still proportionally relative to objective or quantitative properties: intensity and unit area, it's just that it is also relative to other factors that vary from eye to eye. These personal differences in perception are irrelevant, we only need to consider the differences related to measurable and objective properties, that is differentiate between intensity and intensity per unit area.

 

I'm happy to drop the original version of the argument (with the overly-idealised detector) in favour of the improved version in my previous post.

 

OK, on re-reading the thread I think I understand where some confusion has crept in.

The illustration in the OP shows two starfields. The first contains a few (call it N) stars at greyscale 100%, while the second shows 4N stars at greyscale 25%. The stars in both cases appear to be 1 pixel in size.

Assuming I've got that description right, and it isn't just my monitor playing tricks on me, then this is an incorrect depiction of what goes on, perhaps influenced by the finite resolution of bitmap images. As I said in my previous post, the number of photons we receive per unit time per unit solid angle is fixed for a given star, no matter how far away it is, but the solid angle subtended at the Earth by the star's disk decreases as 1/r[sup]2[/sup]. In other words, in reality stars at greater distances are not "dimmer" in the sense of having lower greyscale values, but "dimmer" because they are smaller in apparent size.

So, the first picture should show N stars having an area of 36 pixels (say) and greyscale = 100%. The second pixture should then show 4N stars having an area of 9 pixels and greyscale = 100%. The third picture would have 9N stars with an area of 4 pixels and greyscale = 100%, and so on. Eventually we run into the problem of finite resolution and have to be a bit more cunning, but this is the basic idea.

---

Edited to add: OK, I might have overstated the severity of the problem with the image in the OP.

Basically, if you want to write a program to generate the nth shell bitmap, ideally it should draw n[sup]2[/sup] white circles of radius K/n at a random location in the bitmap, where K is some fixed scale factor. However, because pixels have finite size, if you get to the stage where K/n is smaller than a few pixels, you will have to be careful and use techniques like supersampling.

Alternatively, you can go for an solution similar to the one in the OP, where you make your stars 1 pixel in size and reduce the greyscale values instead of the sizes. This is equivalent to supersampling when K/n is less than a pixel, provided that you plot your stars using an additive blending mode (i.e. if you draw multiple stars at the same pixel location, you must add the greyscale values). If you do this, you will get something reasonably representative of Olber's paradox, provided that you render enough shells.

 

The logic is fine, but reality says otherwise. Take any star in the Hubble Deep Field, it is supposed to emit large quantities of photons in our direction, as if there was a continuous line between us, but apparently they don't all arrive. Due to relative translation and rotation this supposed continuous ray doesn't connect to the same point on the star's surface, but this shouldn't matter much. So then beside dispersion and inverse-square law, these photons also seem to deteriorate somehow on the way. Whatever it is, it could actually be the main reason for our dark night sky.