Inverse square law resolves Olbers' paradox

[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.

 

[tris_d]

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?

I'm just a computer programmer. I really do not know. I know about gamma though, it's kind of like brightness that screws with the contrast. Anyway, I hear what you are saying and I can not find any flaw there, but if I look at it the other way I see the result doesn't match. I don't know what and how, I guess there must be some assumption in either your view or my view that we are simply not paying attention to, and I do not know what it is.

You help me resolve this and if it turns out I'm right I'll share Nobel prize with you, we'll write a paper together, ok? Heh. Look, all I know is that we know for a fact that inverse square law makes distant star appear dimmer.

http://en.wikipedia.org/wiki/Apparent_brightness
- 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.

That's the only fact I relay on, and according to that those pictures look just right to me, but all in all I'm quite puzzled myself.

 

[tris_d]

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.

Reduction of brightness is due to decreased amount of photons per second. We already established that, I believe that's well documented fact. How apparent size of the disk come into equation, that I do not know.

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.

It would take more exposure time to make it as bright as seen from the Earth, wouldn't it? Anyway, can you make a picture that would look more correct then?

 

[sophiecentaur]

The reason that your simulation goes wrong is, as Bandersnatch has just pointed out, the fact that the inverse square law applies to Point Sources. The distant discs will look the same brightness - just smaller. There will be more, dimmer 'point sources' per m^2

One day you will get to accept that, when you have found that 'Science got it wrong', it may just not have and it is 'probably you'. I find it all the time but, unlike you, I go in expecting to be wrong. I am usually correct in that assumption.

BTW, Gamma is very relevant if you want to get a linear brightness scale - which you did in this case.

 

[sophiecentaur]

Stop bringing photons into this. They are totally irrelevant. Flux is flux, whether it's photons, Watts or golf balls.
That picture you drew (post 52) shows exactly what bandersnatch and I are saying. Your disc has finite width so the inverse square with distance relates to distance from the apex of that black triangle and not from the surface of the disc. Your simulation is ignoring the geometry and that is another reason why it looks wrong. The angle subtended by the discs in the earlier 'demonstration' pictures is massive compared with the angle subtended by stars (approaching zero degrees).

 

[tris_d]

The reason that your simulation goes wrong is, as Bandersnatch has just pointed out, the fact that the inverse square law applies to Point Sources. The distant discs will look the same brightness - just smaller. There will be more, dimmer 'point sources' per m^2

http://en.wikipedia.org/wiki/Apparent_brightness

That's not what Wikipedia says. They will actually look "grayer", and they look black unless you wait long enough, which is why Hubble telescope uses exposure times of several months to get one single pixel turn from black to some brightness. See the picture below, most of the stars are just few pixels in size, and they apparent magnitude goes from black through shades of gray, that's what brightness is.

 

[tris_d]

Stop bringing photons into this. They are totally irrelevant. Flux is flux, whether it's photons, Watts or golf balls.
That picture you drew (post 52) shows exactly what bandersnatch and I are saying. Your disc has finite width so the inverse square with distance relates to distance from the apex of that black triangle and not from the surface of the disc. Your simulation is ignoring the geometry and that is another reason why it looks wrong. The angle subtended by the discs in the earlier 'demonstration' pictures is massive compared with the angle subtended by stars (approaching zero degrees).

It's easy to say something is wrong, but what do you base your conclusions on if you've never done it yourself, so can you show us then how would correct picture look like?

 

[sophiecentaur]

Of course Hubble (and everyone else) gets different results. It's the real Universe out there and not Olber's 'paradoxical' one. Even for Hubble, the stars subtend zero angle and there are no nearby discs out there with equal illumination levels. Also, the mere fact that real pictures in space tend to be faint, makes no difference to the relationships. You are bringing in red herrings again.

I have told you what is wrong and why and if you do the demo strictly correctly then you will see what I mean. You need to get the geometry right and not just 'make up' figures based on inverse square law. Draw a diagram of the situation with two different disc diameters and see how the two 'cones' have different apex distances.
Aaamof, you could do the experiment yourself, easily. Put your camera on manual exposure and take pictures of a round light bulb at different distances. See if the RGB values follow the inverse square law for an object of that finite size. Start with it quite close and you sill see the graph for the value of the central portion hardly changes at all, initially, with distance. That would be the best demo that your 'simulation' is not right. The reason is simply Geometry.

 

[Drakkith]

Tris, this is the last thing I'll say, as I'm done with this thread after this. Re-read the thread and you will find everything you need to know to resolve this issue. It's only that you can't seem to "visualize" it. Well, sorry, but you can't always visualize things. At some point you are going to have to accept that the math is right and leave it at that if you can't visualize it.

 

[russ_watters]

http://en.wikipedia.org/wiki/Apparent_brightness

That's not what Wikipedia says. They will actually look "grayer", and they look black unless you wait long enough, which is why Hubble telescope uses exposure times of several months to get one single pixel turn from black to some brightness. See the picture below, most of the stars are just few pixels in size, and they apparent magnitude goes from black through shades of gray, that's what brightness is.

It was pointed out to you already (by Drakkith, I think) that stars in astrophotos are not good representations of reality. They may appear to be several pixels across, but in reality, they are much, much less than one pixel across.

If you don't listen, you won't learn.