Inverse square law resolves Olbers' paradox

[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.