
#19
Nov1012, 04:49 AM

P: 566

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




#20
Nov1012, 05:44 AM

P: 162

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 2dimensional 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. 



#21
Nov1012, 06:48 AM

P: 168

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_%2B..._%C2%B7_%C2%B7 ). 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. 



#22
Nov1012, 08:46 AM

P: 162

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? 



#23
Nov1012, 10:07 AM

P: 566

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 lightsensitive 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. 



#24
Nov1012, 10:28 AM

P: 168

Let's specify what "sufficient resolution" means. Lets say each photoreceptor cell has infinitesimal focal resolution, so it can only receive light from a single star no matter the distance. Lets 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 Bphotons. 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. 



#25
Nov1012, 11:30 AM

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P: 11,058

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 halfdegree diameter circular section of sky and is very very bright. That same halfdegree 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 halfdegree 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/M^{2} 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. 



#26
Nov1012, 11:58 AM

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P: 11,399

"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. 



#27
Nov1012, 01:25 PM

P: 162





#28
Nov1012, 01:45 PM

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#29
Nov1012, 01:53 PM

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P: 11,399

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) 



#30
Nov1012, 02:06 PM

P: 162





#31
Nov1012, 02:43 PM

P: 162





#32
Nov1012, 03:41 PM

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P: 11,058

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. 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. 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. 



#33
Nov1012, 05:41 PM

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P: 11,399

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. 



#35
Nov1012, 06:42 PM

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#36
Nov1012, 07:08 PM

P: 162

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. 


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