# Inverse square law resolves Olbers' paradox

by tris_d
Tags: inverse, olbers, paradox, resolves, square
 P: 740 Good sir/madam, isn't that a little bit too harsh?
P: 162
 Quote by Barakn 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 any one of them?
P: 173
 Quote by tris_d 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.
 Quote by tris_d 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_%2B..._%C2%B7_%C2%B7 ).

 Quote by tris_d 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.

 Quote by tris_d 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.

 Quote by tris_d 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.
P: 162
 Quote by H2Bro 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_%2B..._%C2%B7_%C2%B7 ).
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?
 P: 740 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.
P: 173
 Quote by tris_d Why would you add up intensities when each stars projects onto its own spatial location?
Why did I mention this? Because you said:

 Quote by tris_d 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.

 Quote by tris_d 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.
 Quote by tris_d 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). Lets 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. 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 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.
 Mentor P: 12,006 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.
 Sci Advisor Thanks PF Gold P: 12,261 "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.
P: 162
 Quote by 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 ∞.
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.
Mentor
P: 12,006
 Quote by tris_d 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.
 Sci Advisor Thanks PF Gold P: 12,261 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)
P: 162
 Quote by Drakkith 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?
P: 162
 Quote by H2Bro 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.
Mentor
P: 12,006
 Quote by tris_d 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.
 Sci Advisor Thanks PF Gold P: 12,261 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.
PF Gold
P: 6,505
 Quote by 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.
Boy, howdy, have you ever go THAT right !
Mentor
P: 12,006
 Quote by sophiecentaur 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.
P: 162
 Quote by Drakkith 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.

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