Inverse square law resolves Olbers' paradox

[tris_d]

Treatment originally used to discard inverse square law as solution to Olbers' paradox was not set up correctly. If we include sensor (camera) in the treatment and model light as photons the result describes what we actually see.

 

[Barakn]

Not really.

 

[tris_d]

Not really.

You see I got the red card for posting this, we are not allowed to discuss it here.

Ah well, I guess this is the point where I get banned.


http://www.asterism.org/tutorials/tut09-1.htm

Since the area of a sphere of radius r is

A = 4p r2 (1)

the volume of such a shell is

V = 4p r2t (2)

If the density of each of the luminous objects within the shell is "n", then the total number of these objects in the shell must be

N = 4p r2nt (3)

Now let us ask just what amount of energy such a shell will send to the Earth. Since the shell's thickness is small, it is reasonable to assume that the entire shell is at a distance "r" from the earth. The energy, E, emitted by any source at distance r, produces an intensity, "I", over a given area, A, on the Earth of (inverse square law)

I = E/4p r2 (4)

The total intensity received on the Earth from all the sources in the shell r units away must then be the intensity produced by each source times the total number of sources or

T = IN (5)

Substituting the value of N previously calculated into the above, we find that

T = tnE (6)

We notice at once that the total energy received from any chosen shell does not depend upon its distance from us (no r in the above equation). The total energy received from all the shells is the sum of the contributions of each shell. If there are M shells this total is

S = tnEM (7)

But there is an infinite number of shells and so the total intensity on the Earth must be infinite. Therefore, the nighttime sky should be blindingly bright!


--//--

They completely ignored sensor surface area, that is some 2-dimensional image receiving this light, like a photo or human eyes, and by ignoring that they get result as if the image has only one pixel. So instead of to "see" many dots, some bright some less bright, they practically sum all the received intensity in only one pixel and thus result wrongly indicates the sky is bright.

They also ignored exposure time. The rate of incoming photons is proportional to distance, due to inverse square law, which is known and accepted fact, that's why very distant stars do not produce any dots on a photo-plate unless we wait long enough. Just by looking at this fact makes it clear to me inverse square law explains it all.

Let me explain with an example. Two stars at distance r would impact photo-plate with intensity I, and four stars at double the distance will also impact photo-plate with the same intensity I. That's what they are saying, and that's true. However, what they are not considering is that two closer stars will produce two dots each with brightens I/2, but four further stars will produce four dots each with brightness I/4.

There is difference between two bright dots and four less bright dots of course, and there is difference between two dots on 10x10 resolution image and 1x1 resolution image. So when they ignore this sensor surface area they practically work with 1x1 resolution image where all the intensity gets summed up at one pixel, and of course all they see is "bright sky". To summarize I draw this conclusion: at infinite distance there will be infinite number of stars and if we had infinite resolution they would produce infinite number of dots, but the brightness of each dot would be I/infinity, which is pretty much nothing but black.

 

[Bandersnatch]

However, what they are not considering is that two closer stars will produce two dots each with brightens I/2, but four further stars will produce four dots each with brightness I/4.

There is difference between two bright dots and four less bright dots of course, and there is difference between two dots on 10x10 resolution image and 1x1 resolution image.

Is there a difference between two halves of the Sun and four quarters of the Sun? They all add up to one Sun.

 

[tris_d]

Is there a difference between two halves of the Sun and four quarters of the Sun? They all add up to one Sun.

Two or more stars do not add up to one star, there is spatial separation between them.

 

[Drakkith]

The Inverse Square Law does not resolve Olbers' Paradox. Instead it is the state of the universe that does so. Consider the following. IF the universe was not expanding, and IF it was infinitely old we would be swamped with visible radiation, since an infinite number of stars lie at every line drawn away from the Earth, leading to an infinite number of photons coming in. However this isn't the case because both of those things are not true. It is the combination of an expanding universe that has a finite age that resolves the paradox.

But even if the inverse square law did happen to "solve" the paradox, it still wouldn't mean anything. Olbers' Paradox is about an eternal static universe. In such a universe the known laws of stellar evolution wouldn't apply, as there would need to be some sort of "recycling machine" to produce new hydrogen from old, heavier matter that stars produce. Otherwise you wouldn't get an infinite number of stars for an infinite amount of time. The fact is that a great many things resolve Olbers' Paradox, from the expansion of the universe, to the basics of nucleosynthesis and more.

 

[Barakn]

Two stars at distance r would impact photo-plate with intensity I, and four stars at double the distance will also impact photo-plate with the same intensity I. That's what they are saying, and that's true. However, what they are not considering is that two closer stars will produce two dots each with brightens I/2, but four further stars will produce four dots each with brightness I/4.

Huh? Maybe you better explain to us what the difference between brightness and intensity is. Most of us view them as being the same, so you've just produced two mutually exclusive statements.

To summarize I draw this conclusion: at infinite distance there will be infinite number of stars and if we had infinite resolution they would produce infinite number of dots, but the brightness of each dot would be I/infinity, which is pretty much nothing but black.

Now you've definitely gone off the deep end. You've just assumed that all stars are infinitely distant. The ones that count are much, much closer. But even at that infinite distance, you've made a bad assumption. The "outermost" shell (in quotes because if there's a shell at infinity, there's another one at infinity+1) has an infinite surface area and therefore has an infinite volume and infinite number of stars. You've essentially divided that by r² where r = ∞ to estimate the intensity here, and you've arbitrarily assumed that the result of dividing one infinite number by another is zero. It doesn't have to be - it could be zero, infinity, or any number in between. The proper approach would involve using limits as r → ∞. Until you show some math, your argument remains just so much handwaving.

 

[tris_d]

Consider the following. IF the universe was not expanding, and IF it was infinitely old we would be swamped with visible radiation, since an infinite number of stars lie at every line drawn away from the Earth, leading to an infinite number of photons coming in.

The rate of incoming photons from very large or infinite distance would proportionally go down, so the chance for even a single photon to hit us from there would be very or infinitely small.

However this isn't the case because both of those things are not true. It is the combination of an expanding universe that has a finite age that resolves the paradox.

I've been e-mailing this to hundreds of professors and researchers at famous universities around the world, and I just got my first reply from one Oxford professor. He said:

- You seem to ignore the key aspect, which is that the further away a star is, the faster it is moving away from us. This motion dims its light in two ways: it shifts the photons to lower frequencies where they have less energy, and it leads to longer gaps between the arrival times of successive photons than there were between the emission of the photons. So the brightness of a star decreases with distance faster than 1/r^2, while the # of stars per unit angle of the beam increases only as r^2.


To which I replied:
- "Thank you for your time. I don't see what you said contradicts what I said, rather just adds up to the effect of making light even dimmer than what would manifest only due to inverse square law.

Can you tell me are we actually able to measure any difference in the arrival time of successive photons of certain galaxies if we compare measurements of today with measurements from say several years ago?"

But even if the inverse square law did happen to "solve" the paradox, it still wouldn't mean anything. Olbers' Paradox is about an eternal static universe. In such a universe the known laws of stellar evolution wouldn't apply, as there would need to be some sort of "recycling machine" to produce new hydrogen from old, heavier matter that stars produce. Otherwise you wouldn't get an infinite number of stars for an infinite amount of time. The fact is that a great many things resolve Olbers' Paradox, from the expansion of the universe, to the basics of nucleosynthesis and more.

I see people are interpreting the paradox in different ways. For me it's just about answering the question: why is the night sky dark? And as you say there could be many factors, one of them certainly 'expanding universe', as Oxford professor explained, but what I'm saying is that inverse square law by itself would be enough to account for the night sky not being completely bright as the original treatment concluded.

 

[tris_d]

Huh? Maybe you better explain to us what the difference between brightness and intensity is. Most of us view them as being the same, so you've just produced two mutually exclusive statements.

Good point.

http://en.wikipedia.org/wiki/Brightness
- "Brightness" was formerly used as a synonym for the photometric term luminance and (incorrectly) for the radiometric term radiance. 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


http://www.its.bldrdoc.gov/fs-1037/dir-005/_0719.htm
- brightness: An attribute of visual perception in which a source appears to emit a given amount of light.

Note 1: "Brightness" should be used only for nonquantitative references to physiological sensations and perceptions of light.

Note 2: "Brightness" was formerly used as a synonym for the photometric term "luminance" and (incorrectly) for the radiometric term "radiance."http://en.wikipedia.org/wiki/Apparent_magnitude
- The apparent magnitude (m) of a celestial body is a measure of its brightness as seen by an observer on Earth

Now you've definitely gone off the deep end. You've just assumed that all stars are infinitely distant. The ones that count are much, much closer. But even at that infinite distance, you've made a bad assumption. The "outermost" shell (in quotes because if there's a shell at infinity, there's another one at infinity+1) has an infinite surface area and therefore has an infinite volume and infinite number of stars. You've essentially divided that by r² where r = ∞ to estimate the intensity here, and you've arbitrarily assumed that the result of dividing one infinite number by another is zero. It doesn't have to be - it could be zero, infinity, or any number in between. The proper approach would involve using limits as r → ∞. Until you show some math, your argument remains just so much handwaving.

It's scales proportionally. The further the star the longer is the time interval between successive photons to reach us from there. Basically what I said is that you would need very long or infinitely long exposure time to see any of those stars.

I'm not sure what more math there is. It's pretty simple. The original treatment does not take image resolution into account, so it can not differentiate between two stars with brightness I/2 and four starts with brightness I/4, thus it practically sums up all the intensity from any shell into an image with only one pixel resolution, ignoring all the spatial separation, and therefore the result is not complete. They get correct intensity, but that's not what we see or what camera captures, what we see is 2-dimensional image where each star has its own spatial location, so we need to divide this intensity across all the "dots" to get result indicating what we actually see.

 

[Drakkith]

I see people are interpreting the paradox in different ways. For me it's just about answering the question: why is the night sky dark? And as you say there could be many factors, one of them certainly 'expanding universe', as Oxford professor explained, but what I'm saying is that inverse square law by itself would be enough to account for the night sky not being completely bright as the original treatment concluded.

Then I recommend you stop using Olbers' paradox as your basis for this question. Olbers' paradox is about a static and eternal universe. When I searched for Olbers' Paradox I immediately found this:

In astrophysics and physical cosmology, Olbers' paradox, named after the German astronomer Heinrich Wilhelm Olbers and also called the "dark night sky paradox", is the argument that the darkness of the night sky conflicts with the assumption of an infinite and eternal static universe.

If you are not specifically talking about Olbers' Paradox in an infinite and eternal universe, then simply ask why the night sky is dark.

The rate of incoming photons from very large or infinite distance would proportionally go down, so the chance for even a single photon to hit us from there would be very or infinitely small.

Of course, as there hasn't been time for much light to reach us from distance sources. This of course may not be an issue in an eternal and static universe. It really depends on a great many factors.

Can you tell me are we actually able to measure any difference in the arrival time of successive photons of certain galaxies if we compare measurements of today with measurements from say several years ago?"

Not really. The change in velocity is far too small over such a short period of time. Plus, the arrival of photons at a detector is subject to Poisson Noise, and arrive in a semi-random fashion, which would make it even more difficult to find a change, especially for extremely dim galaxies.

 

[tris_d]

Then I recommend you stop using Olbers' paradox as your basis for this question. Olbers' paradox is about a static and eternal universe. When I searched for Olbers' Paradox I immediately found this:

In astrophysics and physical cosmology, Olbers' paradox, named after the German astronomer Heinrich Wilhelm Olbers and also called the "dark night sky paradox", is the argument that the darkness of the night sky conflicts with the assumption of an infinite and eternal static universe.

If you are not specifically talking about Olbers' Paradox in an infinite and eternal universe, then simply ask why the night sky is dark.

I see what you mean, and finally now I understand why some people thought what I am saying does not fit in mainstream theory. They thought I was trying to argue the universe is infinite, eternal and static, but what I am really saying is that inverse square law would work just the same regardless.

I think it's when they found inverse square law does not answer the question when they started calling it a paradox, and since then it seem like inverse square law was completely ignored and not considered to be even a part of the solution due to conclusions of the original treatment, which I think is a mistake and should be reconsidered.

Not really. The change in velocity is far too small over such a short period of time. Plus, the arrival of photons at a detector is subject to Poisson Noise, and arrive in a semi-random fashion, which would make it even more difficult to find a change, especially for extremely dim galaxies.

Do you know where I could find some actual numbers? I'd like to see how many photons per second we get from two similar stars that are at different distance, and also from two different stars that are about the same distance away from us.

 

[Drakkith]

I think it's when they found inverse square law does not answer the question when they started calling it a paradox, and since then it seem like inverse square law was completely ignored and not considered to be even a part of the solution due to conclusions of the original treatment, which I think is a mistake and should be reconsidered.

It's called a paradox because for most of history the Universe has been thought of as static and unchanging. It wasn't until the early-mid 1900's that we began to even realize the actual size of the universe and that it was expanding. During this time arguments were thrown back and forth for various models, including one that claimed the universe was both eternal and static. For such a model the darkness of the night sky introduces a "paradox" in that IF the universe is static and eternal, the night sky should not be dark. Or at least not as dark as it is.

Do you know where I could find some actual numbers? I'd like to see how many photons per second we get from two similar stars that are at different distance, and also from two different stars that are about the same distance away from us.

I'm not actually sure how to work it all out. Try this article and see if you can figure it out from there: http://en.wikipedia.org/wiki/Apparent_magnitude#Calculations

 

[Chronos]

Actually, the night sky should be about as bright as the surface of an average star under Olber's reasoning - since every possible line of sight falls upon the disc of a star. We are forced to conclude one or more of Olber's premises - the universe is static, spatially and temporally infinite, and infinitely populated with stars - are invalid. Modern observational evidence strongly suggest all these premises are invalid.

 

[H2Bro]

Let me explain with an example. ...

There is difference between two bright dots and four less bright dots of course, and there is difference between two dots on 10x10 resolution image and 1x1 resolution image. So when they ignore this sensor surface area they practically work with 1x1 resolution image where all the intensity gets summed up at one pixel, and of course all they see is "bright sky".

There actually isn't a difference between two dots each I/2 bright, and 4 dots each I/4 bright, so long as each group is seen as a pointsource by the observer.

You keep mentioning Olbers paradox, but you explicitly say your not arguing under the premise that the universe is eternal and static. I think what your getting at is that even IF the premises of Olbers paradox - static and eternal universe - held, the night sky would be black because the inverse square law.

If that's the case then picture this. Imagine concentric shells around an observer, but not filled with stars, instead the surface area of the shell is covered with 50,000K plasma. At distance D from the observer, observer gets a total amount of energy S. At distance 2D you get still get sunlight S, even though the source is twice as far the surface area shining light at you has also doubled, so you get the same.

Now picture half of the shell is at 2D and the other half at D. Still equals S light. Now imagine the shell fractured into billions of points each at different distances, but still forming a continuous shell from the point of the observer. You still get S, and this is why Olbers paradox is a paradox. Also this is independent of what size of sensor your using. So long as your pointing the sensor up at the sphere, it will receive S/(focal arc of the sensor) amount of energy.

If your not arguing within the hypothetical situation of of Olber's premises, then your just saying that the night sky is black because really distant stars aren't as bright as closer stars. Which isn't much of a statement at all, akin to saying you've discovered circles are round.

I've been e-mailing this to hundreds of professors and researchers at famous universities around the world, and I just got my first reply from one Oxford professor. He said:

Have you actually done this?

 

[tris_d]

Actually, the night sky should be about as bright as the surface of an average star under Olber's reasoning - since every possible line of sight falls upon the disc of a star. We are forced to conclude one or more of Olber's premises - the universe is static, spatially and temporally infinite, and infinitely populated with stars - are invalid. Modern observational evidence strongly suggest all these premises are invalid.

It does not matter if every line of site ends up at some star. Light is not some continuous instantaneous rays, as they obviously thought it is at the time they made that conclusion, but there are gaps between incoming photons that increases as the distance increases, so we would not see far away stars without making long exposure time regardless of how many stars there are and regardless of the size of the universe.

http://en.wikipedia.org/wiki/Apparent_magnitude
- Note that brightness varies with distance; an extremely bright object may appear quite dim, if it is far away. Brightness varies inversely with the square of the distance.


You going to get me banned. I don't want to argue the universe is static or infinite, I just want to say the original treatment is incomplete, and even with the result I'm suggesting it does not lead to either conclusion.

 

[tris_d]

There actually isn't a difference between two dots each I/2 bright, and 4 dots each I/4 bright, so long as each group is seen as a pointsource by the observer.

I'm not sure what do you mean. If you see four stars as a single point source than that would be just one star for all it matters.

You keep mentioning Olbers paradox, but you explicitly say your not arguing under the premise that the universe is eternal and static. I think what your getting at is that even IF the premises of Olbers paradox - static and eternal universe - held, the night sky would be black because the inverse square law.

Yes, that's what I'm saying. Inverse square law would make it dark in either case, due to rate of incoming photons that drops with the distance. Expanding universe just makes it even darker.

If that's the case then picture this. Imagine concentric shells around an observer, but not filled with stars, instead the surface area of the shell is covered with 50,000K plasma. At distance D from the observer, observer gets a total amount of energy S. At distance 2D you get still get sunlight S, even though the source is twice as far the surface area shining light at you has also doubled, so you get the same.

That's interesting, but I don't think it compares, as in that case it seems even just the 1st shell would be sufficient to make the night sky bright, and would occlude further shells so they wouldn't even matter.

Now picture half of the shell is at 2D and the other half at D. Still equals S light. Now imagine the shell fractured into billions of points each at different distances, but still forming a continuous shell from the point of the observer. You still get S, and this is why Olbers paradox is a paradox. Also this is independent of what size of sensor your using. So long as your pointing the sensor up at the sphere, it will receive S/(focal arc of the sensor) amount of energy.

Well, that's certainly nice example, it's getting me confused. Let me try the same argument as before: emitted amount of light would be the same, but received amount of light, per unit time, would not be the same as the rate of incoming photons would be slower from more distant shell.

If your not arguing within the hypothetical situation of of Olber's premises, then your just saying that the night sky is black because really distant stars aren't as bright as closer stars. Which isn't much of a statement at all, akin to saying you've discovered circles are round.

Then you should be able to point which one of my sentences is false or does not follow. Like when you said two dots with brightness of I/2 is the same thing as four dots with brightness of I/4. If you can prove/explain that, then I will have to agree with you about everything else.

Have you actually done this?

Yeah. Haha! And it's not the first time I did that either. Every university has very accessible list with all the professors, researchers, alumni, emeritus and all other kinds of associates, friends and visitors. They don't really respond to random e-mails like that so you have to send hundreds of e-mails to get some response. That's my experience anyway. Funny!

 

[Drakkith]

I'm not sure what do you mean. If you see four stars as a single point source than that would be just one star for all it matters.

I think that is exactly what he's saying. If you can't resolve the individual stars, then they can be considered to be a single point source and not multiple ones.

 

[Barakn]

Tris_d's original post includes someone else's mathematical breakdown of the Olber paradox, including:

Since the shell's thickness is small, it is reasonable to assume that the entire shell is at a distance "r" from the earth. The energy, E, emitted by any source at distance r, produces an intensity, "I", over a given area, A, on the Earth of (inverse square law)
I = E/4p r2 (4)

Quite clearly, it has accounted for the 1/r² decrease in intensity (inverse square law), which when multiplied by the area of the shell containing an r² term, cancels out the r's (r⁰ = 1) leading to shells whose brightness doesn't decrease with distance. But tris_d has decided to ignore this and claim that distance matters. It also quite clearly mentions an area A on Earth receiving this radiation, rendering tris_d's claim that they "ignored sensor surface area" false.

I gave tris_d a subtle hint about the dangers of dividing infinity by infinity and the wisdom of using limits instead. But tris_d then wrote

The rate of incoming photons from very large or infinite distance would proportionally go down, so the chance for even a single photon to hit us from there would be very or infinitely small.

Which means he or she is banging his or her head on the same problem in slightly different clothes. In this case the number of photons reaching us from an infinitely far star would be pretty much zero, but there are an infinite number of stars in that shell. Tris_d would have us believe that infinity x 0 = 0, but in reality that is undefined. Once again, the proper approach is to use limits or infinitesimals. The response to my hint suggests that tris_d's grasp of mathematics is weak - probably has never been exposed to calculus and shows only subtle hints of trig.

While perhaps not a classic troll, tris_d shows the classic signs of being a crank or crackpot. Tris_d's flawed mental picture reminds him or herself of Einstein's thought experiments and is completely unaware of the sea of math that Einstein backed his ideas up with. This person has already invested a great amount of time trying to promote this flawed notion and isn't seeing the light of day no matter how many different ways the paradox is explained. This individual will probably go to the grave thinking he or she was correct, and everybody else is wrong. It's not worth wasting any more time in this thread.

 

[Bandersnatch]

Good sir/madam, isn't that a little bit too harsh?

 

[tris_d]

Tris_d's original post includes someone else's mathematical breakdown of the Olber paradox, including:
Quite clearly, it has accounted for the 1/r² decrease in intensity (inverse square law), which when multiplied by the area of the shell containing an r² term, cancels out the r's (r⁰ = 1) leading to shells whose brightness doesn't decrease with distance. But tris_d has decided to ignore this and claim that distance matters. It also quite clearly mentions an area A on Earth receiving this radiation, rendering tris_d's claim that they "ignored sensor surface area" false.

You are not even addressing what I said, it's about image resolution.

The original treatment does not take image resolution into account, so it can not differentiate between two stars with brightness I/2 and four starts with brightness I/4. They get correct intensity, but that's not what we see or what camera captures. What we see is 2-dimensional image where each star has its own spatial location, so we need to divide this intensity across all the "dots" to get result indicating what we actually see.

In this case the number of photons reaching us from an infinitely far star would be pretty much zero, but there are an infinite number of stars in that shell.

And when you look at those stars or take a photo of them, what do you see? Bright sky? How long would exposure time need to be for you to see anyone of them?

 

[H2Bro]

I'm not sure what do you mean. If you see four stars as a single point source than that would be just one star for all it matters.

Then you should be able to point which one of my sentences is false or does not follow. Like when you said two dots with brightness of I/2 is the same thing as four dots with brightness of I/4. If you can prove/explain that, then I will have to agree with you about everything else.

Here is such a proof. The easy proof is to say 4 x 1/4 = 1, or in fact any N x 1/N = 1. But stars don't clump together at the same distance, otherwise they would smash together or at least be resolvable into different points. I think you mentioned this resolution problem earlier. So, how to account for having stars at different distances, say 2 at I/4 and 8 at I/16?

Lets take the extreme example of this and imagine what it would look like if there was one star at each level of distance, so one star at I/2, one at I/4, one at I/8, and so on for infinity. To demonstrate what I said about the addition of pointsources, think of each of these stars as being on a single line of perspective from the observer, i.e a single vector from the origin or a single photoreceptor with infintesimal arc resolution that simply sums intensity of light.

To find out the measured value of light for the photoreceptor we simply add up the apparent luminosities (or "brightness") of each star at each distance. Which means I/2 + I/4 + I/8 + I/16 ... + I/2^N as N goes to infinity. Now, finding out whether this summation actually equals one is coincidentally the same problem as resolving whether Zeno's arrow ever hits the target.

As it turns out, the summation of this series gives you I = 1. ( http://en.wikipedia.org/wiki/1/2_+_1/4_+_1/8_+_1/16_+_·_·_· ).

Well, that's certainly nice example, it's getting me confused. Let me try the same argument as before...

Lets be clear and clarify the premises and conclusion, and go from there. Correct me if I am wrong on what your setup is.

Your premises in the thought experiment are
1. The universe is eternal and static
2. There is a constant distribution and number of stars
3. The received light of a star at distance d is proportional to 1/(d^2)
4. The number of stars at distance d is proportional to (d^2)/1
5. ( insert your additional premise on photons? or sensors? unclear to me what it was)
____________
conclusion: the night sky is black in a eternal, static universe

You'll have to fill in premise 5 for me to get where your coming from.

Then you should be able to point which one of my sentences is false or does not follow.

Let's move away from thinking about this as "who's argument is correct" and go towards something like "which description most accurately captures the thought experiment, and why". I don't think its too wise to attach personal preference to one interpretation because I am or you are the one who thought of it. All that does is make you less susceptible to understanding new things.

Yeah. Haha! And it's not the first time I did that either...

Let this be a warning to anyone who would put up their email address on the internet. Also, I do not think you will be banned because so far I think you've demonstrated sincere attempts to understand the phenomenon. But, if you start disregarding math and saying "I'm still right" for reasons you start making up, then its clear your not looking for discussion but just to expound on a conclusion you've already put faith into.

 

[tris_d]

Here is such a proof. The easy proof is to say 4 x 1/4 = 1, or in fact any N x 1/N = 1. But stars don't clump together at the same distance, otherwise they would smash together or at least be resolvable into different points. I think you mentioned this resolution problem earlier. So, how to account for having stars at different distances, say 2 at I/4 and 8 at I/16?

Sum of intensities is the same, but that intensity gets divided by the number of those stars when you project their location onto 2-dimensional image with sufficient resolution where each one of them has its own spatial location.

Lets take the extreme example of this and imagine what it would look like if there was one star at each level of distance, so one star at I/2, one at I/4, one at I/8, and so on for infinity. To demonstrate what I said about the addition of pointsources, think of each of these stars as being on a single line of perspective from the observer, i.e a single vector from the origin or a single photoreceptor with infintesimal arc resolution that simply sums intensity of light.

Yes, single photoreceptor would focus all the light from its field of view and "see" sum of intensities, but human eyes and photographs divide that intensity spatially over projected location of each light source. -- I can not think of many stars lying on the same line because closest star would occlude all the other stars behind it rendering them irrelevant.

To find out the measured value of light for the photoreceptor we simply add up the apparent luminosities (or "brightness") of each star at each distance. Which means I/2 + I/4 + I/8 + I/16 ... + I/2^N as N goes to infinity. Now, finding out whether this summation actually equals one is coincidentally the same problem as resolving whether Zeno's arrow ever hits the target.

As it turns out, the summation of this series gives you I = 1. ( http://en.wikipedia.org/wiki/1/2_+_1/4_+_1/8_+_1/16_+_·_·_· ).

Why would you add up intensities when each stars projects onto its own spatial location? You are measuring light intensity with a single photoreceptor, but photographs and human eyes are not a single receptor, they are 2-dimensional arrays of many receptors where each star projects onto each own spatial image location. You need to divide total intensity by the number of stars to get perceived brightness of each star.

Lets be clear and clarify the premises and conclusion, and go from there. Correct me if I am wrong on what your setup is.

Your premises in the thought experiment are
1. The universe is eternal and static
2. There is a constant distribution and number of stars
3. The received light of a star at distance d is proportional to 1/(d^2)
4. The number of stars at distance d is proportional to (d^2)/1
5. ( insert your additional premise on photons? or sensors? unclear to me what it was)
____________
conclusion: the night sky is black in a eternal, static universe

You'll have to fill in premise 5 for me to get where your coming from.

5. The number of photons received stays the same regardless of distance, but when captured by an image with sufficient resolution those photons spread out over 2-dimensional area where brightness of each star is equal to total intensity divided by the number of stars.

Here is another example. If we have image with 4294967296 pixels and photograph 65536 stars shining total light of intensity I, we get 65536 dots each with brightness of I/65536 *exposure time, and there is lots of black around them. And then there are 4294967296 stars at double the distance, also shinning total light of intensity I, but overall brightness of the photograph would not change much as each pixel would only increases in brightness by I/4294967296 *exposure time. The end result we see is just those 65536 stars on a slightly brighter background than black, but if our sensor is not sensitive enough and our exposure time is not long enough we would never see any of those 4294967296 stars. Makes sense?

 

[Bandersnatch]

Take a look at the equation (7) from the derivation you had provided. It's the total energy, received from every part of the sky. You can see how it should be infinite in accordance with the paradox's setup(i.e.infinite shells).

Now, let's say you want to observe the whole sky with one light-sensitive pixel.
You get (7) intensity recorded by that pixel - which is ∞.
If you use N pixels to observe that same part of the sky, you're just assigning each pixel a fraction of the original area to observe. So each of the pixels will record intensity equal to (7)/N, which equals ∞, no matter how high the resolution you use. You're just dividing infinity by ever smaller numbers.
It doesn't matter what is the pixels' sensitivity and exposure time, as at infinite brightness of the sky they always get triggered.

 

[H2Bro]

Why would you add up intensities when each stars projects onto its own spatial location?

Why did I mention this? Because you said:

Like when you said two dots with brightness of I/2 is the same thing as four dots with brightness of I/4. If you can prove/explain that, then I will have to agree with you about everything else.

I not only gave you an explanation of the specific condition, but a proof of the general case. Now you need to say why my explanation is wrong or incorrect otherwise you cede the falsity of your position. Asking why I explained this is not sufficient to say the explanation is wrong.

5. The number of photons received stays the same regardless of distance, but when captured by an image with sufficient resolution those photons spread out over 2-dimensional area where brightness of each star is equal to total intensity divided by the number of stars.

Here is another example...

Plasma with temperature of 50,000K emits photons of a specific energy, which is Planck-constant(T)^4 (I'm using 50,000K as this is typical surface of a star temperature). Let's call such a photon a Bphoton. Bphotons from a star 500ly away have the same energy as Bphotons from a star 100ly away in a non-expanding universe (which is one of the paradox's premises). If the total number of photons from all sources is constant then the total energy received is constant.

Let's specify what "sufficient resolution" means. Let's say each photoreceptor cell has infinitesimal focal resolution, so it can only receive light from a single star no matter the distance. Let's also imagine the sensor unit is a sphere floating in space, and that individual cells are infinitesimally small. So, it seems like our sensor sphere has a cell pointing at all possible angles, and each cell only covers that one angle.

Now, if the universe is infinite and static, one never runs out of stars the further one goes from the sensor. Which means every sensor cell's line of sight terminates in a star, and that star emits Bphotons. Perhaps some cells point at planets, or gas, but remember this is a static and eternal universe, so those planets and gas have been heated up to 50,000K by all the surrounding stars. As a result, every cell is receiving Bphotons regardless of orientation. If all angles are covered, then there is an infinite number of angles each receiving B-photons. Which means infinite energy received. In practicality, there are finite sensor cells, but each would still terminate in a star/s, so the sky would be 50,000K.

Edit: I think what follows is the source of the confusion.

Our actual receptors, eye or electronic, have a specific focal resolution. Each imaging cell has a cone that extends outwards from it. all objects in this cone that emit light will be detected by the same imaging unit, and the intensities of each light source are added up to derive the reported or "stimulus" light level.

The number of photons hitting each cell depends on the number of sources and their distance. If increases in distance are compensated by increases in sources, then total number of photons impacting each imaging unit will be constant.

 

[Drakkith]

It appears to me that while stars further away will appear dimmer, you can more of them into the same area of sky than you can with closer stars. For example the Sun takes up a half-degree diameter circular section of sky and is very very bright. That same half-degree circle in another direction could have 4 stars at double the distance, 16 stars at quadruple the distance, or any combination of stars and distances. In fact, with an infinite amount of stars and a static and eternal universe, every section of sky would be packed with stars, whose COMBINED light output would be exceedingly bright.

Consider that 4 stars could fit into the same section of sky that the Sun does if they are twice as distant. We only receive 1/4 as much light from each star as we do the Sun, but combined their output equals that of the Sun if they were identical to the Sun. We could pack 100 stars into that same area if they were 10 times as distant, and the combined light from that same half-degree slice of sky is still equal to what the Sun outputs. You could continue the pattern forever and the situation is the same.

Tris is correct in that we would receive fewer photons per second from distant stars than we would closer stars. However when you pack more stars into the same section of sky their combined light output equals that of a closer star.

Consider an optical device with a resolution of 1 arcminute. This is approximately the resolution of the human eye. We receive about 1.925 W/M² of flux from every square arcminute section of the Sun. But guess what Mars receives from each square arcminute? Almost the same amount of light! It's just that the total apparent size of the Sun is smaller, so Mars receives about half the total amount of light as we do.

What this means is that every section of sky in a static and eternal universe would appear to be approximately the same brightness due to there being an infinite number of stars within every section of sky. It doesn't matter that some are more distant, as more of those distant stars can be packed into the same area of sky, leading to the same amount of photons coming in.

 

[sophiecentaur]

"Image Resolution"??
Image resolution is a red herring. If a received image is blurred then some of its received power is effectively, deflected and appears to come from somewhere else but, in the Olber model, there is an object right next to it that will be blurred and will supply just enough power to make up for that loss. You could merely hold up a white card and observe how brightly it's illuminated, totally forgetting anything about optical imaging.

I don't think Barakn is too far wrong with his criticism of the OP.

 

[tris_d]

Take a look at the equation (7) from the derivation you had provided. It's the total energy, received from every part of the sky. You can see how it should be infinite in accordance with the paradox's setup(i.e.infinite shells).

Now, let's say you want to observe the whole sky with one light-sensitive pixel.
You get (7) intensity recorded by that pixel - which is ∞.

Light intensity is the number of photons per surface area per unit time. Two photons can not occupy the same space in the same time, so it is impossible to have infinite intensity impacting certain area to start with. Conclusion (7) also fails to realize that if each line of sight eventually ends up at some star it will occlude other stars behind it, so again there can not be such thing as infinite intensity coming out from certain area in the sky. Also, the closer the star that meets our line of sight the more area in the sky it will occupy, and so proportionally it will hide from us more of the other stars in the further shells behind it.

If you use N pixels to observe that same part of the sky, you're just assigning each pixel a fraction of the original area to observe. So each of the pixels will record intensity equal to (7)/N, which equals ∞, no matter how high the resolution you use. You're just dividing infinity by ever smaller numbers.

With sufficient resolution each pixel will not collect photons from all the stars in the field of view, but each star will project only onto its own specific area on the image. So we have some certain area on the image that can only be brightened by the photons coming from only that particular star in that particular line of sight. Now, consider line of sight that ends up at some very distant star from where we get incoming photons only about every hour or so, and thus the area on the image where this particular star gets projected on will stay black unless exposure time is long enough. Right?

It doesn't matter what is the pixels' sensitivity and exposure time, as at infinite brightness of the sky they always get triggered.

Light intensity is the number of photons per surface area per unit time. By definition then perceived or captured brightness is directly proportional to exposure time. Animal and human eyes too have exposure time, sensitivity as well, and if you had eyes of an owl you would see the night sky is actually quite bright. Brightness is in the eye of the beholder. The night sky IS bright, you just need to use better eyes or longer exposure time to see it.

 

[Drakkith]

Light intensity is the number of photons per surface area per unit time. Two photons can not occupy the same space in the same time, so it is impossible to have infinite intensity impacting certain area to start with.

This is incorrect. Photons are bosons and as such can occupy the same location at the same time.

Conclusion (7) also fails to realize that if each line of sight eventually ends up at some star it will occlude other stars behind it, so again there can not be such thing as infinite intensity coming out from certain area in the sky. Also, the closer the star that meets our line of sight the more area in the sky it will occupy, and so proportionally it will hide from us more of the other stars in the further shells behind it.

True, which is why the paradox doesn't say that the sky would be infinite in intensity, only that it would be as bright as the Sun or something around there.

With sufficient resolution each pixel will not collect photons from all the stars in the field of view, but each star will project only onto its own specific area on the image. So we have some certain area on the image that can only be brightened by the photons coming from only that particular star in that particular line of sight. Now, consider line of sight that ends up at some very distant star from where we get incoming photons only about every hour or so, and thus the area on the image where this particular star gets projected on will stay black unless exposure time is long enough. Right?

But that pixel would have to be incredibly small in order to ONLY get that one star. So small in fact that the same pixel would only get a few photons per minute if aimed at the surface of the Sun.

 

[sophiecentaur]

We can either treat this thing classically or not. Olber, afaik, was dealing with totally classical ideas and so should we, so photons don't come into it (and neither do we need to consider diffraction. The original argument in this thread was about the inverse square law not delivering the Olber paradox.
It does. (Not surprisingly)

 

[tris_d]

But that pixel would have to be incredibly small in order to ONLY get that one star. So small in fact that the same pixel would only get a few photons per minute if aimed at the surface of the Sun.

I think that was my best put argument so far, and I do not see how what you said makes it invalid, so can you please expand on that or describe it with some example?

 

[tris_d]

I not only gave you an explanation of the specific condition, but a proof of the general case. Now you need to say why my explanation is wrong or incorrect otherwise you cede the falsity of your position. Asking why I explained this is not sufficient to say the explanation is wrong.

Ok. It's wrong to sum up intensities of all the stars in the field of view because each star projects onto its own specific area on the image. It's only due to insufficient resolution that we get photons from other stars spill over to pixels that "belong" to some other stars.

Now, if the universe is infinite and static, one never runs out of stars the further one goes from the sensor. Which means every sensor cell's line of sight terminates in a star, and that star emits Bphotons. Perhaps some cells point at planets, or gas, but remember this is a static and eternal universe, so those planets and gas have been heated up to 50,000K by all the surrounding stars. As a result, every cell is receiving Bphotons regardless of orientation. If all angles are covered, then there is an infinite number of angles each receiving B-photons. Which means infinite energy received. In practicality, there are finite sensor cells, but each would still terminate in a star/s, so the sky would be 50,000K.

We know for a fact that distant stars appear dimmer due to inverse square law, so even if every possible line of sight ends up at some star, how can you say that there will be uniform brightness across all of them?

Edit: I think what follows is the source of the confusion.

Our actual receptors, eye or electronic, have a specific focal resolution. Each imaging cell has a cone that extends outwards from it. all objects in this cone that emit light will be detected by the same imaging unit, and the intensities of each light source are added up to derive the reported or "stimulus" light level.

The number of photons hitting each cell depends on the number of sources and their distance. If increases in distance are compensated by increases in sources, then total number of photons impacting each imaging unit will be constant.

Total intensity gets compensated by the number of sources, but when those sources get projected on 2-dimensional surface then this total intensity gets divided by that same number of sources, and so it gets spatially spread out over each one of them.

 

[Drakkith]

We know for a fact that distant stars appear dimmer due to inverse square law, so even if every possible line of sight ends up at some star, how can you say that there will be uniform brightness across all of them?

Look at my explanation of the Sun. As long as the Sun is larger than 1 square arcminute in apparent size, you will see the same amount of light coming from that square arcminute of sky. When the Sun is smaller than 1 square arcminute we can switch to a smaller unit, such as square arcseconds, and the process is still the same. No matter how close or far the Sun is, the amount of light from each square arcminute is exactly the same. The only difference is the total apparent size of the Sun as a whole. A few stars further away than the Sun could be positioned to take up the same apparent area of sky as the Sun did, and the result would be the same.

Now extrapolate from that to other stars. If every star was the same as the Sun, then every single section of sky that falls on a star would be exactly the same brightness. Closer stars would be larger in apparent size, but that is irrelevant as I just explained. Stars further away would add together and fill up the view between the gaps of closer stars.

Total intensity gets compensated by the number of sources, but when those sources get projected on 2-dimensional surface then this total intensity gets divided by that same number of sources, and so it gets spatially spread out over each one of them.

Of course. But the total intensity is exactly the same in both cases. You could have 100 stars that are 10 times further away than the Sun is, and the TOTAL amount of light we receive would be equal to the Sun. And given the right placement, these stars would still occupy the same area of sky as the Sun does. So in essence there is no difference between the Sun and 100 Sun-like stars that are 10 times further away.

I really think you are getting too caught up in the imaging system. Take it away and calculate the amount of light that falls on a section of the Earth and you will see that it is the same.

I think that was my best put argument so far, and I do not see how what you said makes it invalid, so can you please expand on that or describe it with some example?

You aren't WRONG about the inverse square law, let's get that straight right now. What you are missing is that you can add together light from stars at different distances and different apparent sizes.

As an example, imagine a lamp that has 1 bulb. Turn it on and measure the light output from the bulb. Now move it twice as far away and add 3 more bulbs to it to get 4 bulbs. Measure the light and you will find that not only are you receiving the same amount of light from those 4 bulbs as you were the 1 bulb, the apparent size of the 4 bulbs is EQUAL to the 1 bulb when it was closer. You can move the lamp to any distance as long as you add bulbs to it and you will still see this effect. So if you surrounded yourself by lamps so that every direction fell on a light bulb, then you could calculate the total amount of light falling on you. Then move half of those lamps twice as far back and add 3 bulbs to them. The total light doesn't change and every section of your view is still equal in brightness. Yes, you will receive less light from each bulb that is further away, but this is compensated by adding in more bulbs.

 

[sophiecentaur]

I don't see why so much energy has been expended in this thread on this particular argument. The Olber Paradox has been accepted as 'correct', in as far as it makes the right prediction for a hypothetical Universe of the type that was assumed to exist at the time. It really is a bit late to try to show that people could have go it wrong to the extent that they used the inverse square law in a flawed way. The clever thing about the 'Paradox' was that it forced people to think about and reject their contemporary model because of the consequence of applying a classical rule correctly, within their paradigm.
Science has moved on.

 

[phinds]

I don't see why so much energy has been expended in this thread on this particular argument. The Olber Paradox has been accepted as 'correct', in as far as it makes the right prediction for a hypothetical Universe of the type that was assumed to exist at the time. It really is a bit late to try to show that people could have go it wrong to the extent that they used the inverse square law in a flawed way. The clever thing about the 'Paradox' was that it forced people to think about and reject their contemporary model because of the consequence of applying a classical rule correctly, within their paradigm.
Science has moved on.

Boy, howdy, have you ever go THAT right !

 

[Drakkith]

I don't see why so much energy has been expended in this thread on this particular argument.

Some of us just can't let someone leave the site without knocking some knowledge into their head with a jackhammer cranked up to 11. And I don't think anyone here is saying that the paradox is or isn't valid, it's about explaining WHY it's accepted. That takes a bit longer for some people than for others. Honestly it got me thinking and after about an hour I finally understood how the paradox works. I think that's something.

 

[tris_d]

Look at my explanation of the Sun. As long as the Sun is larger than 1 square arcminute in apparent size, you will see the same amount of light coming from that square arcminute of sky. When the Sun is smaller than 1 square arcminute we can switch to a smaller unit, such as square arcseconds, and the process is still the same. No matter how close or far the Sun is, the amount of light from each square arcminute is exactly the same. The only difference is the total apparent size of the Sun as a whole. A few stars further away than the Sun could be positioned to take up the same apparent area of sky as the Sun did, and the result would be the same.

Now extrapolate from that to other stars. If every star was the same as the Sun, then every single section of sky that falls on a star would be exactly the same brightness. Closer stars would be larger in apparent size, but that is irrelevant as I just explained. Stars further away would add together and fill up the view between the gaps of closer stars.

  \   --   --   --   --    /   |- 2r
   \                      /    |
    \                    /     |
     \                  /      |
      \                /       |
       \  ----  ----  /        |- 1r
        \            /         |
         \          /          |
          \        /           | 
           \      /            |

Ok, here's what I got. There are two stars at distance 1r and four stars at distance 2r. Both of these two shells emit the the same total intensity I= 8. Now we project closer two stars on a photo and we get two circles each with brightness BRIGHT= I/2 = 4, and there is blackness around and between them BLACK= 6. Then we project four further stars on another photo and we get four smaller circles each with brightness BRIGHT= I/4 = 2, and there is blackness around and between them BLACK= 16.

Four further stars not only leave less bright imprints, but there is also more "black" between and around them, so I conclude: while the total emitted intensity is the same, perceived overall brightness of the second shell is much less.

 

[Drakkith]

Ok, here's what I got. There are two stars at distance 1r and four stars at distance 2r. Both of these two shells emit the the same total intensity I= 8. Now we project closer two stars on a photo and we get two circles each with brightness BRIGHT= I/2 = 4, and there is blackness around and between them BLACK= 6. Then we project four further stars on another photo and we get four smaller circles each with brightness BRIGHT= I/4 = 2, and there is blackness around and between them BLACK= 16.

What blackness? What are you talking about? And you need 8 stars in shell two, not 4. Doubling the distance results in 1/4 the intensity or whatever the term is, not half.

 

[russ_watters]

Four further stars not only leave less bright imprints, but there is also more "black" between and around them, so I conclude: while the total emitted intensity is the same, perceived overall brightness of the second shell is much less.

Do you understand that the cone only points in one direction? The paper will have a uniform light level on it, but the point sources are still just point sources and there is blackness between them.

 

[tris_d]

What blackness? What are you talking about? And you need 8 stars in shell two, not 4. Doubling the distance results in 1/4 the intensity or whatever the term is, not half.

"Blackness" are the pixels on the image that were not hit by any photons. I think 2^2= 4.

 

[tris_d]

Do you understand that the cone only points in one direction? The paper will have a uniform light level on it, but the point sources are still just point sources and there is blackness between them.

Yes, the cone points in one direction, like a camera would. The purpose was to compare perceived brightness of two shells. I am not sure what do you mean to say with the second part, is something wrong with my diagram?

 

[H2Bro]

Numerous people have explained this phenomenon from numerous directions, including myself. From what I've seen, at each conclusive explanation tris_d avoids addressing what has been pointed out wrong with his argument, and instead nitpicks over some less relevant aspect of the situation.

tris_d, its time to lay your cards on the table. Explain your thoughts as concisely, as clearly, as plainly as possible, using mathematics where appropriate, to demonstrate why an infinite eternal static universe would not have a sky as bright as the sun. Include all your relevant supporting information but don't include it twice.

Otherwise, this thread should be locked, as its basically someone expounding a personal belief on a totally defunct and pointless topic.

Personally, the only reason I think this is worth explaining is the cringe-inducing notion that this guy is emailing hundreds of professors with this stuff. We should save them the trouble, they have important research to conduct.

 

[Drakkith]

"Blackness" are the pixels on the image that were not hit by any photons. I think 2^2= 4.

No, you cannot just arbitrarily say that X pixels are black. A real imaging device has many factors to take into account, including the size of the airy disk made by the objective and the size of the pixels themselves. In fact, a single point source could eventually put photons on every pixel of the sensor if the pixels were the right size compared with the airy disk and you had a long enough exposure.

And in actuality, the size of airy disk of the stars is exactly the same regardless of how bright they might be. Only the intensity is different, which makes brighter stars LOOK bigger in photographs. More photons per pixel is displayed as being brighter than fewer photons, which makes the outside edges of the disk fade away into the background once the count gets low enough. However the shape and size are still the same no matter how bright the star is.

Honestly trying work this problem by thinking in terms of a camera is doing nothing but confusing you. I still recommend you just imagine it as light shining down on a surface with no optical device to focus it.

 

[tris_d]

No, you cannot just arbitrarily say that X pixels are black. A real imaging device has many factors to take into account, including the size of the airy disk made by the objective and the size of the pixels themselves. In fact, a single point source could eventually put photons on every pixel of the sensor if the pixels were the right size compared with the airy disk and you had a long enough exposure.

Ok, so what numbers do you get? How big circles would those four stars imprint compared to two closer stars, where all the stars are the same, and how much "black" would there be for each of two shells?

 

[Drakkith]

Ok, so what numbers do you get? How big circles would those four stars imprint compared to two closer stars, where all the stars are the same, and how much "black" would there be for each of two shells?

That is irrelevant for this discussion and I couldn't even begin to calculate that without numbers for the optical system and sensor anyways. And please, you need 8 stars, not 4. Besides, the light output is the same for both shells, meaning they are equally bright.
Honestly your question has been answered a dozen times over, what part do you not understand?

 

[sophiecentaur]

Numerous people have explained this phenomenon from numerous directions, including myself. From what I've seen, at each conclusive explanation tris_d avoids addressing what has been pointed out wrong with his argument, and instead nitpicks over some less relevant aspect of the situation.

tris_d, its time to lay your cards on the table. Explain your thoughts as concisely, as clearly, as plainly as possible, using mathematics where appropriate, to demonstrate why an infinite eternal static universe would not have a sky as bright as the sun. Include all your relevant supporting information but don't include it twice. Otherwise, this thread should be locked, as its basically someone expounding a personal belief on a totally defunct and pointless topic.

Personally, the only reason I think this is worth explaining is the cringe-inducing notion that this guy is emailing hundreds of professors with this stuff. We should save them the trouble, they have important research to conduct.

I think we're in yet another a situation where youth and testosterone are taking on the scientific reasoning of countless very well informed people throughout the time since the Enlightenment. You never get this sort of stuff from 'older' and well qualified (or informed) contributors. Just give it time. Sense will prevail, I'm sure. At least he is keen.

Professors earn plenty - they are paid to handle this sort of thing!

 

[tris_d]

That is irrelevant for this discussion and I couldn't even begin to calculate that without numbers for the optical system and sensor anyways.

That was your own example. You said:
- For example the Sun takes up a half-degree diameter circular section of sky and is very very bright. That same half-degree circle in another direction could have 4 stars at double the distance, 16 stars at quadruple the distance, or any combination of stars and distances.

And then you said:
- Look at my explanation of the Sun... Now extrapolate from that to other stars. If every star was the same as the Sun, then every single section of sky that falls on a star would be exactly the same brightness. Closer stars would be larger in apparent size, but that is irrelevant as I just explained. Stars further away would add together and fill up the view between the gaps of closer stars.So I did what you said, how is it not relevant all of a sudden? We already know the intensity each star in the second shell would have, you only need to calculate apparent size of those stars relative to stars in the first shell, so we can compare.

And please, you need 8 stars, not 4. Besides, the light output is the same for both shells, meaning they are equally bright.
Honestly your question has been answered a dozen times over, what part do you not understand?

Ok, eight stars each with brightness of I/8, the point stays the same. The part I do not understand is why do you think these two pictures bellow have the same apparent brightness. Appearances can be deceiving, and there is no need to make assumptions, so if you mean to claim that I'm wrong then please show the result you get.

http://img803.imageshack.us/img803/1908/starsv.jpg

 

[sophiecentaur]

@tris_d
Forget about photons. They really don't help this at all; Olber's model was Idealised so you have to follow his rules if you want to show its conclusions are wrong!
Just imagine yourself at the centre of a uniformly illuminated sphere (1W/m^2) Measure the illumination on your face. Now inflate the sphere to ten times its radius (100 times the area) and keep its illumination the same, at 1W/m^2. That would, of course, involve 100 times the amount of power supplied. What would the illumination on your face be? More or less? How could it be different? Do the same thought experiment with a Sun-sized disc and a disc, ten times as far away and with the same surface illumination. Same illumination of your face in both cases. Extend this idea to a universe sized sphere, with each m^2 supplying the same energy as each m^2 of the Sun, with the Sun still where it is. Would you see the Sun as any different from the background? The Olber model assumes that there are so many stars out there and that they are stationary, so there is, effectively, a massive sphere with uniform illumination. Its radius is, in fact, irrelevant to the basic argument. All that's necessary is that it should be FULL.

There is also the issue that every one of the 'near' stars, including the Sun, is illuminated from behind by the outer stars (receiving energy) - so, assuming that equilibrium has been reached, they will all be hotter than they would have been (in isolation) and are 'passing on' the received energy from the outer stars and not actually blocking them. In fact All stars will be getting energy from All Other stars! by the same process.

Your picture, above, is not a representation of the Olber model. If it were, then put an integrating light meter in front of either of those pictures and, if you have got it right, the meter would give the same reading. 'Seeing' is not necessarily believing. You would need to callibrate your demo properly in order to prove anything- it would not satisfy Advertising Standards as it stands! You may find it easier to use squares rather than circular discs.

 

[tris_d]

@tris_d
Just imagine yourself at the centre of a uniformly illuminated sphere (1W/m^2) Measure the illumination on your face. Now inflate the sphere to ten times its radius (100 times the area) and keep its illumination the same, at 1W/m^2. That would, of course, involve 100 times the amount of power supplied. What would the illumination on your face be? More or less? How could it be different?

I do not know, you tell me. I think you are not taking exposure time into account and that emitted amount of light does not compare with perceived (captured) amount of light, for some reason. Perhaps you are not taking into account apparent size of the stars decreases with distance, or something. I do not know. I did these pictures in Photoshop where I could set up "brightness" property exactly, so while the size of those circles are bit arbitrary the brightness is given by Photoshop itself. So you tell me, did I make some mistake when making these pictures, is Photoshop wrong, or what?

http://img15.imageshack.us/img15/9423/stars2r.jpg

Your picture, above, is not a representation of the Olber model. If it were, then put an integrating light meter in front of either of those pictures and, if you have got it right, the meter would give the same reading. 'Seeing' is not necessarily believing. You would need to callibrate your demo properly in order to prove anything- it would not satisfy Advertising Standards as it stands! You may find it easier to use squares rather than circular discs.

These pictures are supposed to be representation of inverse square law and its impact on perceived brightens relative to distance. It is calibrated by Photoshop. Star on first picture has brightness 100, four stars at second picture each have brightness 100/4, nine stars on third picture each have brightness 100/9, and sixteen stars on fourth picture each have brightness 100/16. Try it yourself.

 

[sophiecentaur]

Have you heard of display gamma? What counts is how much energy arrives at one place, so sizes need to be correct. I am not in favour of simulations where all variables are not know. I would rather.rely on calculation and the right sums give the correct, established, answer.
What has "exposure time" got to do with it?
How can you not appreciate that four sources with a quarter of the power flux each must be the equivalent flux to one unit. Where else can the power go?

 

[Bandersnatch]

What you did there on your pictures is a case of double counting, i.e. the reduction in brightness from the inverse square law comes from the star disks being smaller by that factor, nothing more. Just by making a disk's diameter two times smaller(because it's two times farther away), you receive four times less light from it. It should not be dimmed additionally.
Their surface brightness per unit area stays the same as with the closer ones, it's just that the areas are proportionally smaller/larger.

That's why when you add four stars, each with 1/4th of the area of a single star, you get the same amount of light coming at you from that direction.

To put it differently, the Sun is not 30^2 times "grayer" as seen from Neptune when compared to how we see it from Earth. It's just 30^2 times smaller, and so Neptune receives 30^2 times less energy. According to your picture example, it would have to be both 900 times smaller, AND have 900 times less emissivity per unit area of its visible disk, making it 810000 dimmer than on Earth.