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madus
2012-Nov-03, 06:13 AM
So we take our eyes and look towards some distant galaxy, and we see black. Then we take telescope and look in the same direction and we see the galaxy. Is that right? At what point of magnification something becomes visible relative to its distance and light intensity, and what exactly magnification has to do with it?

ShinAce
2012-Nov-03, 09:00 AM
Magnification does you no good. It's all about gathering lots of light, so either long exposures or large mirrors. Magnification is just zooming in on pixels. Digitally, this is child's play.

Although this isn't Olber's paradox.

Jeff Root
2012-Nov-03, 09:09 AM
Magnification has almost nothing to do with it. We can
see galaxies with telescopes because the telescopes have
cameras attached to them, and cameras can accumulate
light over periods of seconds, minutes, hours, or days.
The light gradually builds up an image on a photographic
plate or CCD. Our eyes can only see the light that is
hitting them at that instant.

The greater the magnification, the more detail you can
see, but the larger the diameter of the telescope and
the longer the exposure both need to be to have enough
light to show that detail.

-- Jeff, in Minneapolis

madus
2012-Nov-03, 10:07 AM
That makes sense then. So, when we take the most powerful telescope is it easy to find "black patches", or is it when we make exposure long enough there will always be something wherever we look at?

Light intensity has nothing to do with wavelength (energy), but is about amount of photons in some volume of space. Correct? If so, and if light intensity is proportional to distance, does that not suggest the light dissipates (disperse) on its way?

Shaula
2012-Nov-03, 10:18 AM
Correct? If so, and if light intensity is proportional to distance, does that not suggest the light dissipates (disperse) on its way?
Well, inverse square law for a point and all. Or are you talking about dispersion due to something like vacuum polarisation?

chornedsnorkack
2012-Nov-03, 10:25 AM
Surface luminosity cannot be increased - only decreased.

Eye does have some ability to integrate light over large areas. Take Small Magellanic Cloud. The first sailors to South plainly saw both clouds, not just the Large.

Its magnitude is +2,7. So decrease it by 25 times, 3,5 magnitudes - what you get is +6,2, too dim to see. Since the Cloud is 3x5 degrees, 1/25 of it would be 36x60 minutes. A chunk of Small Magellanic Cloud twice the size of full Moon would be definitely invisible - yet the whole Cloud is plainly visible.

But this ability is limited - the integrated light of gegenschein is quite bright, at -0,3 to +0,5. But very hard to see.

Or take Fornax Dwarf. +9,3 magnitude, and 13x17 minutes

If you magnify it 25 times, to +2,3, it is brighter than Small Magellanic Cloud. But it would span 5x7 degrees... and NO one saw it till 1938! Even though they knew where to look... globular cluster NGC1049 is +12,9 and would be +5,9 under the same manification, but is 24 sec across... at, say, 40x magnification, NGC1049 would be +4,9 and 16 minutes across, and that is fairly visible. And was discovered as early as before 1838.

madus
2012-Nov-03, 10:41 AM
Well, inverse square law for a point and all. Or are you talking about dispersion due to something like vacuum polarisation?

Yes, I mean if you look at the light as photons (particles) then the amount depends on the distance according to inverse square law. Wouldn't that explain the paradox?

http://upload.wikimedia.org/wikipedia/commons/thumb/2/28/Inverse_square_law.svg/420px-Inverse_square_law.svg.png


And now I just learned that is how intensity is actually defined:
- In physics, an inverse-square law is any physical law stating that a specified physical quantity or intensity is inversely proportional to the square of the distance from the source of that physical quantity. In equation form:

http://upload.wikimedia.org/math/d/e/6/de6ee5f95f4bd85100de8d126060c87c.png

Shaula
2012-Nov-03, 11:00 AM
The paradox is that the night sky is not uniformly as bright as the surface of a star. It comes about because as you expand the area over which you look you see more sources. If you assume a uniform distribution of emitting materials (roughly true over huge enough distances) and do the maths you find that the increase in sources in your field of view cancels out the inverse square law dimming. At first the solution was dust - but someone pointed out that eventually all the dust would heat up thanks to the radiation and the entire universe would look hot. Paradox remained. So the next solution was a finite age for the universe - light has not had time to travel far enough for the dust effect to be seen. Expansion neatly fixes it ever further by saying some light will never be able to reach us, imposing an effective horizon to each point in space.

madus
2012-Nov-03, 11:05 AM
Surface luminosity cannot be increased - only decreased.

Eye does have some ability to integrate light over large areas. Take Small Magellanic Cloud. The first sailors to South plainly saw both clouds, not just the Large.

Its magnitude is +2,7. So decrease it by 25 times, 3,5 magnitudes - what you get is +6,2, too dim to see. Since the Cloud is 3x5 degrees, 1/25 of it would be 36x60 minutes. A chunk of Small Magellanic Cloud twice the size of full Moon would be definitely invisible - yet the whole Cloud is plainly visible.

But this ability is limited - the integrated light of gegenschein is quite bright, at -0,3 to +0,5. But very hard to see.

Or take Fornax Dwarf. +9,3 magnitude, and 13x17 minutes

If you magnify it 25 times, to +2,3, it is brighter than Small Magellanic Cloud. But it would span 5x7 degrees... and NO one saw it till 1938! Even though they knew where to look... globular cluster NGC1049 is +12,9 and would be +5,9 under the same manification, but is 24 sec across... at, say, 40x magnification, NGC1049 would be +4,9 and 16 minutes across, and that is fairly visible. And was discovered as early as before 1838.

Is that related to magnification or light intensity dissipation? What's "magnitude" defined with?

madus
2012-Nov-03, 11:58 AM
The paradox is that the night sky is not uniformly as bright as the surface of a star. It comes about because as you expand the area over which you look you see more sources. If you assume a uniform distribution of emitting materials (roughly true over huge enough distances) and do the maths you find that the increase in sources in your field of view cancels out the inverse square law dimming.

Let's make it clear first what is it our measurements say: when we take the most powerful telescope is it easy to find "black patches", or is it when we make exposure long enough there is always something wherever we look at?

I googled to find about this mathematics that disproves "inverse square law" explanation for Olbers' paradox, but I could find very little, just people talking about it on forums. There are cases where people used math to both prove it and disprove it. Wikipedia does not mention it. I'm suspicions as to how they modeled light, do you have some source where I can look at it?

I also googled to confirm inverse square law actually applies to light in vacuum so I can see if our text-books have anything more to say about it, but strangely I could not. Huh?




At first the solution was dust - but someone pointed out that eventually all the dust would heat up thanks to the radiation and the entire universe would look hot.

Is it not mainstream opinion that there is actually dust in space? And why would space dust get any hotter than bright side of the Moon for example?

antoniseb
2012-Nov-03, 01:11 PM
So we take our eyes and look towards some distant galaxy, and we see black. Then we take telescope and look in the same direction and we see the galaxy. Is that right? At what point of magnification something becomes visible relative to its distance and light intensity, and what exactly magnification has to do with it?

Your title is asking about Olbers Paradox. Your question is not about Olbers paradox. Olbers paradox is only hard to explain if we are in an infinite non-expanding (no cosmological red-shift) universe that has existed more or less as our local universe looks now, forever. Of course these assumptions are all wrong, and so it is no surprise that the night sky is not all the brightness of the surface of a star.

Your question is asking about light magnification... which I think has been well explained by the follow-up posts. Just to reword the answer, galaxies do not have bright surfaces. Nice detailed pictures of galaxies require getting lots of photons on to the image sensor (or retina), which takes telescopes with large collecting areas. As has been noted above, geometric magnification is the enemy of light-gathering power. Collecting time and surface area are the allies.

madus
2012-Nov-03, 01:29 PM
Your title is asking about Olbers Paradox. Your question is not about Olbers paradox. Olbers paradox is only hard to explain if we are in an infinite non-expanding (no cosmological red-shift) universe that has existed more or less as our local universe looks now, forever. Of course these assumptions are all wrong, and so it is no surprise that the night sky is not all the brightness of the surface of a star.

Your question is asking about light magnification... which I think has been well explained by the follow-up posts. Just to reword the answer, galaxies do not have bright surfaces. Nice detailed pictures of galaxies require getting lots of photons on to the image sensor (or retina), which takes telescopes with large collecting areas. As has been noted above, geometric magnification is the enemy of light-gathering power. Collecting time and surface area are the allies.

The thread is about Olbers Paradox, opening question is just one of many questions I have. I wanted to understand the measurements first and magnification was explained, now the rest of the thread can be more specifically about the paradox itself. -- So, expanding universe explains the paradox well? And thus you think this belongs to ATM? It's all the same to me.

Aster
2012-Nov-03, 02:29 PM
Yes, I mean if you look at the light as photons (particles) then the amount depends on the distance according to inverse square law. Wouldn't that explain the paradox?

http://upload.wikimedia.org/wikipedia/commons/thumb/2/28/Inverse_square_law.svg/420px-Inverse_square_law.svg.png


And now I just learned that is how intensity is actually defined:
- In physics, an inverse-square law is any physical law stating that a specified physical quantity or intensity is inversely proportional to the square of the distance from the source of that physical quantity. In equation form:

http://upload.wikimedia.org/math/d/e/6/de6ee5f95f4bd85100de8d126060c87c.png

Not exactly so, because some stars are hidden behind others stars and dust.

And final formula is little different, something like this:
Intensity ~ exp(-kr)/r^2;
k - some constant, depends on mas distribution in space (mainly density), and opaqueness - surface of stars, or rather galaxies, ect.

antoniseb
2012-Nov-03, 02:32 PM
... now the rest of the thread can be more specifically about the paradox itself. -- So, expanding universe explains the paradox well? And thus you think this belongs to ATM? It's all the same to me.
I didn't say it belongs in ATM. There have been a number of Olbers Paradox threads. You might do a search and see if you are asking something new and different here. Because it has been heavily discussed the answers in THIS thread may tend to be short and seem to you to be less thorough. If you ask something very specific, you will probably get specific thorough answers.

Buttercup
2012-Nov-03, 02:34 PM
Absence of (lots of) light?

Shaula
2012-Nov-03, 03:10 PM
Let's make it clear first what is it our measurements say: when we take the most powerful telescope is it easy to find "black patches", or is it when we make exposure long enough there is always something wherever we look at?
There is always something in your field of view, but the objects resolved do not fill the field of view.

Try here for some of the maths: http://articles.adsabs.harvard.edu//full/1990IAUS..139....3H/0000003.000.html


I also googled to confirm inverse square law actually applies to light in vacuum so I can see if our text-books have anything more to say about it, but strangely I could not. Huh?
It does. Not sure why you would think it didn't? It is an elementary geometric thing, and if it were not the case then you would be into tired light realms and see blurring and other effects.


Is it not mainstream opinion that there is actually dust in space? And why would space dust get any hotter than bright side of the Moon for example?
Of course it is Mainstream to say there is dust in space. We can see it. The argument was that dust was absorbing and not really re-radiating all of the wavelengths of the energy which would mean it heats up. The Moon re-radiates a lot of what falls on it (50-60% IIRC).

madus
2012-Nov-03, 03:24 PM
I didn't say it belongs in ATM. There have been a number of Olbers Paradox threads. You might do a search and see if you are asking something new and different here. Because it has been heavily discussed the answers in THIS thread may tend to be short and seem to you to be less thorough. If you ask something very specific, you will probably get specific thorough answers.

Yeah, I saw other threads. They did not discuss inverse square law explanation much, and I'd like if we could either prove it or disprove it here.

noncryptic
2012-Nov-03, 03:36 PM
Olbers' paradox stopped being a paradox once it was figured out that the universe is not static, nor eternal, nor infinite wrt observation (see observable universe, CMBR, reshift).
http://en.wikipedia.org/wiki/Olbers%27_paradox

cjameshuff
2012-Nov-03, 03:45 PM
Yes, I mean if you look at the light as photons (particles) then the amount depends on the distance according to inverse square law. Wouldn't that explain the paradox?

The geometry behind the inverse square law works both ways for extended objects...the surface area covered in a given patch of a view also increases with the square of distance, exactly countering the inverse square falloff from any one point of that surface. The total amount of light from a star decreases with distance, but the brightness of its surface does not. In an infinitely old, infinitely large, static universe, every line of sight would end on a star, and the sky would be uniformly as bright as a star's surface.



I also googled to confirm inverse square law actually applies to light in vacuum so I can see if our text-books have anything more to say about it, but strangely I could not. Huh?

It's simple geometry, of course it applies in vacuum.



Is it not mainstream opinion that there is actually dust in space? And why would space dust get any hotter than bright side of the Moon for example?

The sun only covers a section of the sky 0.5 degrees across. Make all the black parts of the sky just as bright, and the moon would reach equilibrium at an equal temperature, and glow just as bright itself. (More realistically, it'd never have formed in the first place.)

madus
2012-Nov-03, 03:46 PM
There is always something in your field of view, but the objects resolved do not fill the field of view.


I'll take that means there are some black patches from where no light comes to us.




Try here for some of the maths: http://articles.adsabs.harvard.edu//full/1990IAUS..139....3H/0000003.000.html


Everything except for inverse square law explanation. Paper mentions Edward Halley and his "rings", it has to do with inverse square law, but the paper doesn't show any math, it just says Halley was wrong. I'm actually not even sure if Halley thought he proved or disproved inverse square law explanation.

Tensor
2012-Nov-03, 04:04 PM
Everything except for inverse square law explanation. Paper mentions Edward Halley and his "rings", it has to do with inverse square law, but the paper doesn't show any math, it just says Halley was wrong. I'm actually not even sure if Halley thought he proved or disproved inverse square law explanation.

Try here (http://www.asterism.org/tutorials/tut09-1.htm).

Aster
2012-Nov-03, 04:09 PM
Olbers' paradox stopped being a paradox once it was figured out that the universe is not static, nor eternal, nor infinite wrt observation (see observable universe, CMBR, reshift).
http://en.wikipedia.org/wiki/Olbers%27_paradox

Very bad reasoning.
Absorption heats body to a temperature of source only under condition, which assumes thesis: whole sky is equally bright.

Laws of thermodynamics, and observations, show that illuminated bodies have lower temperature than the source.

madus
2012-Nov-03, 04:11 PM
The geometry behind the inverse square law works both ways for extended objects...the surface area covered in a given patch of a view also increases with the square of distance, exactly countering the inverse square falloff from any one point of that surface. The total amount of light from a star decreases with distance, but the brightness of its surface does not. In an infinitely old, infinitely large, static universe, every line of sight would end on a star, and the sky would be uniformly as bright as a star's surface.


You seem to be modeling photons as if they are continuous constant rays, or as if there is always a photon right after another coming out from the source all at once. How about if photons gets dispersed and more sparse with the distance like a sprinkler would sprinkle water around?




The sun only covers a section of the sky 0.5 degrees across. Make all the black parts of the sky just as bright, and the moon would reach equilibrium at an equal temperature, and glow just as bright itself. (More realistically, it'd never have formed in the first place.)

Brightness does not equal intensity. How do you define "brightness" by the way? -- Dust particle according to my view would not get bombarded by photons constantly from all the directions all the time, as photons coming from distant sources would miss it more often the further the sources are.

madus
2012-Nov-03, 04:24 PM
Try here (http://www.asterism.org/tutorials/tut09-1.htm).

I see. They don't take time intervals between arriving photons into account. According to me, the further the source is more likely it becomes for its photons to completely miss us. How about that?

Tensor
2012-Nov-03, 04:37 PM
I see. They don't take time intervals between arriving photons into account. According to me, the further the source is more likely it becomes for its photons to completely miss us. How about that?

Show the math.

Shaula
2012-Nov-03, 05:06 PM
Laws of thermodynamics, and observations, show that illuminated bodies have lower temperature than the source.
Very bad reasoning. Bodies do not have a lower temperature than a heat bath which is a closer analogy to what was being supposed to happen.


I see. They don't take time intervals between arriving photons into account. According to me, the further the source is more likely it becomes for its photons to completely miss us. How about that?
Covered by the inverse square law. Photons do not act like little bullets. The intensity is can be expressed as a probability of being hit, so photon count and density is already covered within it.

madus
2012-Nov-03, 05:20 PM
Show the math.

It just occurred to me after I started this thread. I have no idea how to treat that mathematically. But can you see in your mind's eye that would explain it? It fits well in the picture where you have to make longer exposures the further away source is because photons would be incoming less often, right? And so you would agree there is some 'time interval' factor involved, one way or another, wouldn't you?

madus
2012-Nov-03, 05:27 PM
Covered by the inverse square law. Photons do not act like little bullets. The intensity is can be expressed as a probability of being hit, so photon count and density is already covered within it.

I don't see that's how they did it in the link posted by Tensor.

Strange
2012-Nov-03, 05:59 PM
According to me, the further the source is more likely it becomes for its photons to completely miss us. How about that?

Based on that diagram above, the number of photons per unit area would decrease as the area they were spread out over decreased. This area increases with the square of the distance. There you go: the probability of it hitting us falls with the inverse square law.

cjameshuff
2012-Nov-03, 06:28 PM
It just occurred to me after I started this thread. I have no idea how to treat that mathematically. But can you see in your mind's eye that would explain it? It fits well in the picture where you have to make longer exposures the further away source is because photons would be incoming less often, right? And so you would agree there is some 'time interval' factor involved, one way or another, wouldn't you?

That doesn't change anything. You receive photons at a lower rate from a given point on the surface of the star. However, the surface area of star that corresponds to a given angular area increases with the square of distance, so you receive photons at the same rate from the area of sky covered by the star...the surface brightness is independent of distance, regardless of how you treat light.

And yes, the overall rate of photons from a single star decreases with the inverse of distance squared...but with the given assumptions, the number of stars added with each increment of distance increases with the square of distance, again canceling out the inverse square law. You can ignore the spread of light and just consider columns of space along individual sight lines, and with the given assumptions, each either ends on a star or on an object in thermal equilibrium with a sky full of stars and other such objects...said equilibrium being reached when those objects are radiating just as much per unit of surface area as the stars.

Revisiting the dust issue...light/heat exiting a given region of space is exactly matched by light/heat entering from neighboring regions. There's no sinks for energy to be removed, light absorbed either heats the object or gets re-emitted. Given an infinite amount of time, in an infinite, static universe, any non-stellar objects will have reached the same temperature as the stars, the temperature of the universe being uniform...if anything is visibly glowing, everything should be glowing.

ShinAce
2012-Nov-03, 06:31 PM
Madus, I applaud your patience. It is entirely too normal to take things for granted and not realize what's going on. If it were the other way around, the Greeks would have invented prisms and discovered evolution. But they simply missed it.

The maths isn't too bad. You start with the idea that whatever you're examining is at a point and then see what happens as you get further. That diagram you posted is basically the visual proof of it. At r, there is only a single tile with 9 lines passing through it. We call the lines flux. Then at 2r, there are 4 tiles for 9 lines, or 2.25 lines per tile. Then at a distance of 3r, we have 9 lines and 9 tiles, so one line per tile.

You're entirely correct to say the further away the more the chances of the photon missing you completely. You cannot coax a photon into a telescope. If we have a baseline, we can simply use the inverse square idea to solve. For example, if the earth receives 1 billion photons per second, and Mars is twice as far away from the sun, then Mars will receive =(1 billion photons per second) / 2^2 = 250 million photons per second. Assuming Mars was the same size as the Earth!

In a physics program, we go through the official idea of flux, and surface integrals, and line integrals, and Maxwell's equations. First a student would look at charge. If an electron is 1 meter away and is giving you a force of X, then at 5 meters away it will be X/5^2 = X/25. The inverse square laws applies to electrostatics. It also applies to gravity. As you are seeing, it applies to energy that spreads out evenly in all directions. So in that case, a laser does not follow the inverse square law, but a star does.

When I see inverse square, the first thing I think of is "Is this thing a conservative force?". Imagine going to see the Mona Lisa and someone is standing off to the side of the painting watching intently. Kind of a bad angle to appreciate the work, "don't you think"? We like to do that in physics.

Aster
2012-Nov-03, 06:37 PM
Very bad reasoning. Bodies do not have a lower temperature than a heat bath which is a closer analogy to what was being supposed to happen.

You are using the very thesis as argument in reasoning, i.e.:
p => p is true, because p is true.

That is Your formal proof, that p is true.

Shaula
2012-Nov-03, 06:51 PM
That is Your formal proof, that p is true.
My comment was more about your reasoning which was using a flawed analogy as its starting place.

madus
2012-Nov-03, 07:04 PM
That doesn't change anything. You receive photons at a lower rate from a given point on the surface of the star. However, the surface area of star that corresponds to a given angular area increases with the square of distance, so you receive photons at the same rate from the area of sky covered by the star...the surface brightness is independent of distance, regardless of how you treat light.

And yes, the overall rate of photons from a single star decreases with the inverse of distance squared...but with the given assumptions, the number of stars added with each increment of distance increases with the square of distance, again canceling out the inverse square law. You can ignore the spread of light and just consider columns of space along individual sight lines, and with the given assumptions, each either ends on a star or on an object in thermal equilibrium with a sky full of stars and other such objects...said equilibrium being reached when those objects are radiating just as much per unit of surface area as the stars.

Revisiting the dust issue...light/heat exiting a given region of space is exactly matched by light/heat entering from neighboring regions. There's no sinks for energy to be removed, light absorbed either heats the object or gets re-emitted. Given an infinite amount of time, in an infinite, static universe, any non-stellar objects will have reached the same temperature as the stars, the temperature of the universe being uniform...if anything is visibly glowing, everything should be glowing.

- the surface area of star that corresponds to a given angular area increases with the square of distance

Surface area of star increases?


- number of stars added with each increment of distance increases with the square of distance

Number of stars increases?


You lost me, completely. Is there some article about it like the one posted by Tensor?

madus
2012-Nov-03, 07:16 PM
Madus, I applaud your patience. It is entirely too normal to take things for granted and not realize what's going on. If it were the other way around, the Greeks would have invented prisms and discovered evolution. But they simply missed it.

The maths isn't too bad. You start with the idea that whatever you're examining is at a point and then see what happens as you get further. That diagram you posted is basically the visual proof of it. At r, there is only a single tile with 9 lines passing through it. We call the lines flux. Then at 2r, there are 4 tiles for 9 lines, or 2.25 lines per tile. Then at a distance of 3r, we have 9 lines and 9 tiles, so one line per tile.

You're entirely correct to say the further away the more the chances of the photon missing you completely. You cannot coax a photon into a telescope. If we have a baseline, we can simply use the inverse square idea to solve. For example, if the earth receives 1 billion photons per second, and Mars is twice as far away from the sun, then Mars will receive =(1 billion photons per second) / 2^2 = 250 million photons per second. Assuming Mars was the same size as the Earth!

In a physics program, we go through the official idea of flux, and surface integrals, and line integrals, and Maxwell's equations. First a student would look at charge. If an electron is 1 meter away and is giving you a force of X, then at 5 meters away it will be X/5^2 = X/25. The inverse square laws applies to electrostatics. It also applies to gravity. As you are seeing, it applies to energy that spreads out evenly in all directions. So in that case, a laser does not follow the inverse square law, but a star does.

When I see inverse square, the first thing I think of is "Is this thing a conservative force?". Imagine going to see the Mona Lisa and someone is standing off to the side of the painting watching intently. Kind of a bad angle to appreciate the work, "don't you think"? We like to do that in physics.

I see some agreement, I don't see any disagreement, so do you too think inverse square law actually solves the paradox? -- Do you understand what cjameshuff said?

cjameshuff
2012-Nov-03, 07:19 PM
- the surface area of star that corresponds to a given angular area increases with the square of distance

Surface area of star increases?


- number of stars added with each increment of distance increases with the square of distance

Number of stars increases?

It helps if you read the whole sentences, rather than picking out a few of the words.



You lost me, completely. Is there some article about it like the one posted by Tensor?

Tensor's link is in fact an explanation of it.

madus
2012-Nov-03, 07:28 PM
Based on that diagram above, the number of photons per unit area would decrease as the area they were spread out over decreased. This area increases with the square of the distance. There you go: the probability of it hitting us falls with the inverse square law.

Yeah, but I also think there is a time variable involved. I don't think photons can be emitted from exactly same spot on star surface at the rate where they would be aligned one right after another.

Aster
2012-Nov-03, 07:32 PM
My comment was more about your reasoning which was using a flawed analogy as its starting place.

I didn't use any analogy.

I said a body heats up to the temperature of a source, only under condition: the source covers entire sky - 4pi sr.

So, we can't argue dust warms up to the temperature of stars, because this is typical circular reasoning,
we begin with what we are trying to end up with.
Nothing is proved.

madus
2012-Nov-03, 07:36 PM
It helps if you read the whole sentences, rather than picking out a few of the words.


I did of course, at least five times. Pulled that out so you know exactly what I was referring to.




Tensor's link is in fact an explanation of it.

Ok, I shall study it more carefully, and then I'll be back.

Shaula
2012-Nov-03, 07:45 PM
I said a body heats up to the temperature of a source, only under condition: the source covers entire sky - 4pi sr.
OK and the point is that for an infinite universe, infinitely old, then this is the case. So we can show that the universe is not infinitely old and infinite in extent. What was your point?

antoniseb
2012-Nov-03, 07:53 PM
OK and the point is that for an infinite universe, infinitely old, then this is the case. So we can show that the universe is not infinitely old and infinite in extent. What was your point?
... and not expanding, or otherwise having cosmological red shift.

Strange
2012-Nov-03, 07:58 PM
- the surface area of star that corresponds to a given angular area increases with the square of distance

Surface area of star increases?

Isn't this back to the "3 sides of a triangle" discussion in the other thread?

As the star gets further away, the two sides of the triangle subtending the given angle get longer, therefore the third side (which also forms one side of the surface area of the star) increases proportionally. Therefore the area corresponding to that angle increases as the square of the distance.

Maybe that needs a diagram... but it seems kinda obvious.

Ditto number of stars within that angular area.

ShinAce
2012-Nov-03, 08:04 PM
I see Olber's paradox as a combination of many factors with a simple solution. The universe has an age, and that limited time means you have limited light. As mentioned, it's only a paradox if the universe has an infinite age and/or infinite extent without expansion.

As we look at further and further galaxies, each seems dimmer. You could solve this numerically with the sum of a decreasing monotonic inifinite series. A fancy word for 1/2 + 1/4 + 1/8 + 1/16.....=1 That actually adds up to a finite value at the end. I think that one would be the 'geometric series' and adds up to 1. If it were that simple, there would be a fixed brightness to the night sky. However, as you look further away, you're considering larger amounts of space which contain more galaxies. That's the part that messes up the series and makes it grow to infinity when you try to add it up. 1+1+1+1+1....=infinity.

That's where we look to the age to solve the paradox. As we look further and further back, eventually there stops being galaxies. You can't get infinity if you can only add up a finite number of finite elements.

ShinAce
2012-Nov-03, 08:13 PM
Isn't this back to the "3 sides of a triangle" discussion in the other thread?

As the star gets further away, the two sides of the triangle subtending the given angle get longer, therefore the third side (which also forms one side of the surface area of the star) increases proportionally. Therefore the area corresponding to that angle increases as the square of the distance.

Maybe that needs a diagram... but it seems kinda obvious.

Ditto number of stars within that angular area.

Are we saying that stars get bigger as you look further back? No. That's the problem. The idea is that if a star is close and takes up 1 degree of sky, then a star which is far and takes up 1 degree is larger. But a star in our Milky Way and a star in Andromeda don't both take up 1 degree. The stars in Andromeda take up much less 'sky area' than our local stars. I think that's what is messing up madus. It's an idea, not an assertion. An if-then, so to speak.

madus
2012-Nov-03, 08:32 PM
Are we saying that stars get bigger as you look further back? No. That's the problem. The idea is that if a star is close and takes up 1 degree of sky, then a star which is far and takes up 1 degree is larger. But a star in our Milky Way and a star in Andromeda don't both take up 1 degree. The stars in Andromeda take up much less 'sky area' than our local stars. I think that's what is messing up madus. It's an idea, not an assertion. An if-then, so to speak.

That I understand. But I don't see why would they look at it backwards like that. So are you saying cjameshuff made false premise, or his point still stands?

Strange
2012-Nov-03, 08:48 PM
Are we saying that stars get bigger as you look further back?

That isn't what I meant. To put it another way, the area subtended by a given angle increases with the square of distance. (Hmmm... maybe madus could look up the definition of parsec, see if that helps? Or maybe I'm just confusing things further... :))

But I think your explanation is clearer.

ShinAce
2012-Nov-03, 08:59 PM
That I understand. But I don't see why would they look at it backwards like that. So are you saying cjameshuff made false premise, or his point still stands?

A bit of both, really. While it doesn't apply to comparing the angles subtended by our own sun versus a far away supernova, there is another side to the story. As we look at galaxies that are more redshifted, we can infer that the universe is expanding at a certain rate and there has been a certain time for that galaxies light to get here today. As we think about it a bit more, doesn't that mean that the galaxy was much closer when that light started its trip? You bet. If it was closer, it took up more of the sky. It looks bigger. I believe Ned Wright(spelling?) has a good page about this stuff.

So on the one hand, far away galaxies do actually look bigger than you'd expect. If inverse square was all there was, galaxies would get smaller and smaller pretty quickly as you looked further back. This is not correct. There's that compensation from the fact that they used to be much closer. We're not looking at galaxies that are 13.7 billion years old and 10 billion years away. They are 10 billion years away, but only 3.7 billion years old. It's very important to understand the difference between an image and what you infer is really there. I can watch a Muhammed Ali fight that says live at the bottom right, but the truth is, most of the people I'm watching are now dead. Just cause it says live doesn't mean it is.

But then again, the redshift lowers the energy received. So even for the same apparent size(the angular size, actually), the closer galaxy is brighter. That alone would be a way out of the paradox.

madus
2012-Nov-03, 11:51 PM
Tensor's link is in fact an explanation of it.

I see no relation there with what you said. I also found a mistake.

http://www.asterism.org/tutorials/tut09-1.htm
https://encrypted-tbn1.gstatic.com/images?q=tbn:ANd9GcQenI9HCGhz1vQHukvbYz_pC_nSGLTc6 xuQKf60HyPdIcbjHv2bOg

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) <---


***

N is just a number, it's arbitrary number, depends on nothing but what we choose it to be. We don't need volume and density to come up with some number with no real relation to geometry what so ever. The whole "4p r2t" term serves no actual purpose, so when we get rid of it we are left with inverse square law. And problem solved.

ShinAce
2012-Nov-04, 12:42 AM
It's not inverse, though. The amount of stars is proportional to r2. The amount of power from each star is inversely proportional to r2. Only the second one is inverse.

n is the density, say, one star per cubic light year. Over the whole universe, it's a constant. t, the thickness, needs to remain a constant if you want to be consistent with your results. That's fine.

So we look at N as a function of r.

N=4*Pi*r2*n*t

let k= 4*Pi*n*t = constant

we get N = constant*r2

So when you let r double, N quadruples. Twice the radius, four times as many stars.

Originally we said that if r doubles, then the amount of power received from a star goes down by a factor of four. This is assuming the size of the target is constant(no cheating). But there are now 4 times as many stars. What is 1/4 of the power * 4? It's 1.

In english, that says that the amount of energy you receive from any given shell in the universe, centered on the target, of thickness t, is the same. That's the opposite of problem solved. That's problem in hand. As you look at slices that are further and further away, they stay just as bright!

I talked earlier about series. What is the sum of 1+1+1+1....ad infinitum? Yep, it's infinity.

You're gonna need a different trick to solve the paradox.

cjameshuff
2012-Nov-04, 12:56 AM
N is just a number, it's arbitrary number, depends on nothing but what we choose it to be. We don't need volume and density to come up with some number with no real relation to geometry what so ever. The whole "4p r2t" term serves no actual purpose, so when we get rid of it we are left with inverse square law. And problem solved.

It is not an arbitrary number. It's the number of luminous objects in the shell, which increases with the square of the radius of that shell, thus canceling out the inverse square falloff from the individual stars in that shell...exactly what I've been talking about.

madus
2012-Nov-04, 02:08 AM
So we look at N as a function of r.

N=4*Pi*r2*n*t

let k= 4*Pi*n*t = constant

we get N = constant*r2

So when you let r double, N quadruples. Twice the radius, four times as many stars.


Yes, but we are not calculating number of stars, we are calculating intensity. The number of stars is given, arbitrarily, by proposing there should be some density 'n'.




Originally we said that if r doubles, then the amount of power received from a star goes down by a factor of four. This is assuming the size of the target is constant(no cheating). But there are now 4 times as many stars. What is 1/4 of the power * 4? It's 1.


You are integrating additional shells and calculating number of stars instead of to just calculate intensity for the first shell with whatever number of stars we chose. Additional shells further away will produce their own intensity independently of what is going on with the first shell.




In english, that says that the amount of energy you receive from any given shell in the universe, centered on the target, of thickness t, is the same.


Sphere geometry is completely ignored for intensity calculation. It's like having all the stars at one single point.

ShinAce
2012-Nov-04, 02:17 AM
You don't get to change density, n. Density is defined as a quantity per cubic volume. So, one star every cubic light year. The surface of a sphere has area, which is multiplied by the thickness to get a volume.

The volume will go up with each shell as a function of r2. The number of stars is the density times the volume. So the number of stars goes up as r2, while the intensity of each star goes down as 1/r2. Total intensity is the intensity of each star times the number of stars.

We cannot observe a single shell. When we look at the sky, all shells within the Hubble sphere contribute.

At this point I'm just repeating my previous post. Just know that 'n' is a constant, and the math is consistent. You cannot change n halfway through. Nor can you choose that a single shell is all there is to talk about. The answer to the question is the sum influence of all shells.

madus
2012-Nov-04, 02:18 AM
It is not an arbitrary number. It's the number of luminous objects in the shell, which increases with the square of the radius of that shell, thus canceling out the inverse square falloff from the individual stars in that shell...exactly what I've been talking about.

For some reason you're more concerned with the number of stars than intensity. You can not calculate it all at once. First calculate intensity for the first shell with given number of stars N. Only then calculate number of stars for the second shell and its intensity accordingly. And so on.

ShinAce
2012-Nov-04, 02:22 AM
Sphere geometry is completely ignored for intensity calculation. It's like having all the stars at one single point.

I haven't said any different. All I said with what you quoted as that the intensity from a given shell is a scalar quantity. A simple number. That seems to me like ignoring geometry. But the universe is all shells starting from the earth's surface out to the Hubble sphere. All you do is add up all those scalars from each shell. But like I said, each shell has the same exact intensity. That's the special part of the result. A shell a million light years away should have the same intensity as a shell a billion light years away. I say should because of redshift.

edit: regarding the post above. Calculate the intensity from the first shell. Now calculate the intensity from the second shell. Then show me, don't tell me, why they would have different intensities. If you try it, you'll see that you've jumped the gun. What I did was calculate the number of stars for any shell. Then for that same shell, find the intensity of each star in the shell. Combine the two and I got the intensity for that shell. The combined answer showed that the intensity for a shell is NOT a function of r. It's a constant. Try it!

ShinAce
2012-Nov-04, 02:35 AM
Be very careful what you mean by number of stars. I've said it explicitly that I'm talking about the number of stars per shell. Each shell has a very small thickness, but the thickness stays the same. I used that in my calculations. I calculated one shell. Then to get the intensity of the universe, we need to add up the effect of all shells.

The number of stars, N, is not all the stars from earth out to a certain radius. That one goes as the volume of a sphere times the density. The volume of a sphere is (4/3)*Pi*r3. Nobody here has used r3. If you were able to do that, you wouldn't have to add up the intensity from each shell now would you?

madus
2012-Nov-04, 02:50 AM
You don't get to change density, n. Density is defined as a quantity per cubic volume. So, one star every cubic light year. The surface of a sphere has area, which is multiplied by the thickness to get a volume.

The volume will go up with each shell as a function of r2. The number of stars is the density times the volume. So the number of stars goes up as r2, while the intensity of each star goes down as 1/r2. Total intensity is the intensity of each star times the number of stars.

We cannot observe a single shell. When we look at the sky, all shells within the Hubble sphere contribute.

At this point I'm just repeating my previous post. Just know that 'n' is a constant, and the math is consistent. You cannot change n halfway through. Nor can you choose that a single shell is all there is to talk about. The answer to the question is the sum influence of all shells.

Awww, I see now. Back to blackboard, one more try.

madus
2012-Nov-04, 11:40 AM
Got it!

The problem was two-fold. We used geometry to calculate the number of stars in each shell, but we ignored it when we were calculating intensity, and more importantly perhaps we also ignored the time. What our result told us is total intensity, from all the directions at once, but that is not comparable to what we actually see.

To realize what we actually see imagine we cover the whole surface of the Earth with a photographic plate, this is how we take geometry into account. All we need to do now is to take the time into account as well. So we start with completely black photographic plate and wait. Photons coming from closer stars are incoming at faster rate and photons from more distant stars at slower rate, so in first few seconds we see bright spots on our plate related to the closest stars, and less bright spots reflecting the position of more distant stars. To make our whole photographic plate bright we would need to wait for months, thus at any given moment what we actually see is not total intensity, but intensity related to the line of sight and the time of exposure. And problem solved.

ShinAce
2012-Nov-04, 02:15 PM
Forget about time. If we can see a star in the first place, there has been time for its light to get here.

Intensity is defined as power per unit area. A photographic plate has area. Power is already energy per second. It has time built in. To compare intensities, you have to use the same exposure each time.

We can forget about geometry because of the symmetry involved. Whether the plate is at the north pole or the equator, the night sky's intensity is the same. The only geometrical assumption we have to make is that the photographic plate is flat to the ground. Or better yet, always used with the same angle with respect to the ground. All the quantities become scalars instead of vectors. Once we've done that, we calculate the intensity of a shell and find out it is a constant. Again this allows us to simplify the answer. So now the final answer is simply the sum of an infinite series(edit: where each term is the same). That is Olber's paradox.

Of course it isn't a paradox since the universe is not infinite in size. Plus further shells are in fact dimmer because of cosmological redshift.

Aster
2012-Nov-04, 02:35 PM
OK and the point is that for an infinite universe, infinitely old, then this is the case. So we can show that the universe is not infinitely old and infinite in extent. What was your point?

This is no any point, but the circular reasoning I showed.
Nothing was proven, nor disproven.

I'm think must exist an equilibrium temperature, and it's should be lower than the sources temperature,
because only in universe composed of the sources alone, the temperature would be maximal - equal to the sources' temperature.
So, in universe containing the additional absorbers, the temperature must be lower.

Actualy dust and gas, and other cold matter, dominate in the universe,
thus the equilibrium or mean temperature should be very low - several kelvins.

ShinAce
2012-Nov-04, 02:44 PM
In argument form, is it not?:
1) if p, then q
2) not q
3) therefore, not p

where p is that the universe is infinite and static
q is that the night sky is infinitely bright

Whether sound or valid or deductive, it is an argument.

Unless y'all are talking about a different set of premises and conclusions.

cjameshuff
2012-Nov-04, 03:00 PM
The problem was two-fold. We used geometry to calculate the number of stars in each shell, but we ignored it when we were calculating intensity, and more importantly perhaps we also ignored the time. What our result told us is total intensity, from all the directions at once, but that is not comparable to what we actually see.

As ShinAce said, intensity already incorporates time. And no, we did not ignore the number of stars when calculating intensity, you were the one who did that. And yes, the result is comparable to what we see. The exact same math applies to any limited field of view...the number of stars visible at a given distance is proportional to the square of that distance.



To realize what we actually see imagine we cover the whole surface of the Earth with a photographic plate, this is how we take geometry into account. All we need to do now is to take the time into account as well. So we start with completely black photographic plate and wait. Photons coming from closer stars are incoming at faster rate and photons from more distant stars at slower rate, so in first few seconds we see bright spots on our plate related to the closest stars, and less bright spots reflecting the position of more distant stars. To make our whole photographic plate bright we would need to wait for months, thus at any given moment what we actually see is not total intensity, but intensity related to the line of sight and the time of exposure. And problem solved.

Your argument boils down to "exposure to light exposes photographic plates". This doesn't solve the problem at all. And no, total exposure integrated over time is not intensity.

cjameshuff
2012-Nov-04, 03:05 PM
I'm think must exist an equilibrium temperature, and it's should be lower than the sources temperature,
because only in universe composed of the sources alone, the temperature would be maximal - equal to the sources' temperature.
So, in universe containing the additional absorbers, the temperature must be lower.

There is an equilibrium...when everything is at an equal temperature. There isn't any way to push heat back "uphill" to the stars, there's no way to keep parts of the universe eternally cooler than the rest. You're arguing against basic thermodynamics.

Aster
2012-Nov-04, 03:28 PM
There is an equilibrium...when everything is at an equal temperature. There isn't any way to push heat back "uphill" to the stars, there's no way to keep parts of the universe eternally cooler than the rest. You're arguing against basic thermodynamics.

Conditions of equilibrium are other.

Only for entire sky equally bright would be equal temperature of any body,
and this means we have universe containing only sources, no absorbers.

Your reasoning leads directly to some global perpetuum mobile.
You can't increase energy density in whole universe!

cjameshuff
2012-Nov-04, 03:43 PM
Only for entire sky equally bright would be equal temperature of any body,

Which is exactly the conditions you'd end up with.



and this means we have universe containing only sources, no absorbers.

What are these mysterious "absorbers", and how does the heat they absorb end up back in the "sources"?



Your reasoning leads directly to some global perpetuum mobile.

Quite the opposite. I'm describing a universe at heat death, with no temperature differences left that can be exploited to do work. You are describing a universe that somehow keeps going, removing heat from some objects and adding it to already-hotter objects to maintain a temperature difference that continues to perform steady work for eternity.



You can't increase energy density in whole universe!

Not in a static one. And?

madus
2012-Nov-04, 03:50 PM
Forget about time. If we can see a star in the first place, there has been time for its light to get here.


It's not about the time it takes light to travel the distance, it's about the rate of incoming photons. Can you forget for a second about starting assumptions and just directly point a flaw in my reasoning - which one of my sentences is false?




Intensity is defined as power per unit area. A photographic plate has area. Power is already energy per second. It has time built in. To compare intensities, you have to use the same exposure each time.


That's the problem, it seems as if it's taken into account, but it isn't as you do not use any area or time when calculating intensity. It's like calculating volume of 2-dimensional geometry.




We can forget about geometry because of the symmetry involved. Whether the plate is at the north pole or the equator, the night sky's intensity is the same. The only geometrical assumption we have to make is that the photographic plate is flat to the ground. Or better yet, always used with the same angle with respect to the ground. All the quantities become scalars instead of vectors. Once we've done that, we calculate the intensity of a shell and find out it is a constant. Again this allows us to simplify the answer. So now the final answer is simply the sum of an infinite series(edit: where each term is the same). That is Olber's paradox.


Beam of light is transparent when you look at it from the side, to see it you need to project it onto something. When you simplify it that way you place all the stars at one point where we are looking directly at all of them with infinite exposure. -- So, let's just suppose I might be right, what would convince you? Math?

madus
2012-Nov-04, 04:10 PM
As ShinAce said, intensity already incorporates time. And no, we did not ignore the number of stars when calculating intensity, you were the one who did that. And yes, the result is comparable to what we see. The exact same math applies to any limited field of view...the number of stars visible at a given distance is proportional to the square of that distance.


To "see" light means to project it onto something with certain exposure time. True, false?

In your model, what surface are you projecting light onto, and how long is your exposure?




Your argument boils down to "exposure to light exposes photographic plates". This doesn't solve the problem at all. And no, total exposure integrated over time is not intensity.

Can you point which one of my sentences is false?

ShinAce
2012-Nov-04, 04:23 PM
it's about the rate of incoming photons.

which one of my sentences is false?


There it is. Can you give me a definition of rate that conflicts with what has been said?

Intensity was defined before any of us were born as the amount of power per unit area.

If you expose a plate, you are counting total photons. But the power is number of photons per second. The intensity is the number of photons per second per pixel. Leaving a plate exposed for a long time does not increase intensity. It simply increases the total energy received. To do that, you multiply intensity by the amount of time exposed and you get total energy.

http://en.wikipedia.org/wiki/Intensity_(physics)

If you want to rederive the equations for intensity, go for it. I did it last semester for Maxwell's equations. Sure enough, the speed of the radiation affects the intensity. But the speed of light is always constant. Nice and easy!

There is a difference between working from the ground up and showing that some assumptions are wrong, and telling someone they're not applicable because the assumptions don't include complete proofs.

Aster
2012-Nov-04, 04:47 PM
Quite the opposite. I'm describing a universe at heat death, with no temperature differences left that can be exploited to do work. You are describing a universe that somehow keeps going, removing heat from some objects and adding it to already-hotter objects to maintain a temperature difference that continues to perform steady work for eternity.

At the heat death the final temperature must be also lower than the sources.
You still stay in the circular reasoning.

Initial conditions + equations => solutions.

I think here is correct way of reasoning, which leads to the true solution of the Olbers' paradox:
http://www.mathpages.com/home/kmath141/kmath141.htm


Recognizing that all the mass-energy which is emitted from a star over its lifetime actually was put into the star during its formation leads to an immediate resolution of the paradox, even in an infinitely old universe.

madus
2012-Nov-04, 05:49 PM
There it is. Can you give me a definition of rate that conflicts with what has been said?


Rate is a ratio between two measurements with different units. The conflict is you don't "measure" the time of exposure.




If you expose a plate, you are counting total photons. But the power is number of photons per second. The intensity is the number of photons per second per pixel.


There is no such thing as "total number of photons", it's always number of photons over some time period. So when we expose a plate we are counting the number of photons per second per surface area.




Leaving a plate exposed for a long time does not increase intensity. It simply increases the total energy received. To do that, you multiply intensity by the amount of time exposed and you get total energy.


Total energy? Is that the same one given by the original result? So to get it we need to multiply by the amount of time exposed, and we never do that in the original treatment of the problem. So something is not right, right?




If you want to rederive the equations for intensity, go for it. I did it last semester for Maxwell's equations. Sure enough, the speed of the radiation affects the intensity. But the speed of light is always constant. Nice and easy!


I don't mean to invent any new equations. I just want to include surface area that light is projected onto and the time of exposure. Can you help me? Just to see what result we get, just for the fun of it, ok?




There is a difference between working from the ground up and showing that some assumptions are wrong, and telling someone they're not applicable because the assumptions don't include complete proofs.

We are just talking. If I am wrong, all you need to do is teach me, make me see the light, that's what you're good at, isn't it? I think you are, I can understand you much better than cjameshuff. He's just too eager to disagree, not trying to understand WHY someone thinks differently, so he keeps explaining things in his own terms instead of to try and formulate explanation in terms of the person he is explaining to. He's still my hero though for telling me stuff about metal detectors and dangers of static electricity. But seriously, I do appreciate everyone's patience, I just hope you guys are enjoying this as much as I do.

ShinAce
2012-Nov-04, 07:47 PM
I can't help you because I would just be repeating myself.

You cannot expose plates for different amounts of time and then consider them all equivalent for the sake of determining intensity. They need the same exposure.

At any rate, everyone agrees that Olber's paradox is not a problem. While we may have different solutions, who cares?

madus
2012-Nov-04, 08:32 PM
I can't help you because I would just be repeating myself.


It looks to me it's you who needs help this time. My point is pretty simple really. Here we have Earth and you standing at south pole looking straight "down". Let's erase south half of all the shells so we are left with infinite number of stars but only at northern hemisphere. Do you see any of those stars? Do they make your nigh sky half-bright, or is it black?



***
* *
* E *
|
|
|
|




At any rate, everyone agrees that Olber's paradox is not a problem. While we may have different solutions, who cares?

Every astronomer and cosmologist should care. And I care, obviously. If you solved some problem that no one else could for hundreds of years I'd say "congratulations" to you.

Aster
2012-Nov-04, 08:39 PM
In argument form, is it not?:
1) if p, then q
2) not q
3) therefore, not p

where p is that the universe is infinite and static
q is that the night sky is infinitely bright

Whether sound or valid or deductive, it is an argument.

Unless y'all are talking about a different set of premises and conclusions.

and if not p -> q then what is going?

madus
2012-Nov-04, 08:43 PM
and if not p -> q then what is going?

You seem to be "truth seeker" like me, can you find any objection to my argument above?

antoniseb
2012-Nov-04, 09:23 PM
... While we may have different solutions, who cares?
Just to be clear: first, there aren't different solutions. The one that I and several others have given here and in previous threads is correct. Second, most of us care.

There is a lot of misinformation and bad analysis by some of the newer guys in this thread. That's not a problem, since we don't think it was deliberate... but if you really want to understand Olbers Paradox, you should probably read the Wikipedia article rather than read this thread.

cjameshuff
2012-Nov-04, 10:02 PM
It looks to me it's you who needs help this time. My point is pretty simple really. Here we have Earth and you standing at south pole looking straight "down". Let's erase south half of all the shells so we are left with infinite number of stars but only at northern hemisphere. Do you see any of those stars? Do they make your nigh sky half-bright, or is it black?

Stars out of view are not visible. Great. How is this relevant? The relationship between number of stars (proportional to r^2) and light received from individual stars (proportional to 1/r^2) holds for any fraction of the sky just as it holds for the whole sky. Further stars individually contribute less, but their number increases such that they collectively contribute exactly the same amount as nearby stars. With the given assumptions, a sky of infinite stars would be as bright as the surface of a star.



Every astronomer and cosmologist should care. And I care, obviously. If you solved some problem that no one else could for hundreds of years I'd say "congratulations" to you.

You have not done this. You've used bad math to arrive at a completely wrong "solution" to an already long-solved problem. There is no flaw in Olber's paradox, the conclusion follows from the assumptions. The real solution is that those assumptions, that the universe is infinitely old, infinitely large, and static, were incorrect. Any one of those assumptions being false would be sufficient to resolve the paradox, and at least two are now known to be false...the universe has a finite age and is not static, but is instead expanding.

madus
2012-Nov-04, 11:55 PM
Stars out of view are not visible. Great. How is this relevant? The relationship between number of stars (proportional to r^2) and light received from individual stars (proportional to 1/r^2) holds for any fraction of the sky just as it holds for the whole sky. Further stars individually contribute less, but their number increases such that they collectively contribute exactly the same amount as nearby stars. With the given assumptions, a sky of infinite stars would be as bright as the surface of a star.


Yes, relationship between number of stars is proportional to r^2. And yes, light received from individual stars is proportional to 1/r^2. But, light received from individual stars is also proportional to relative orientation between sensor surface and light rays, and is also proportional to exposure time. Correct?

Example with half-shell is relevant as it shows the original treatment does not take into account direction of light rays relative to sensor surface orientation, it tells you only how much light is EMITTED, it does not tell you how much light is SENSED. To calculate how much light is sensed we need to model sensor surface as well as exposure time.

So regarding exposure time, let me state that distant stars are "out of the view" as well. But if our eyes had some kind of memory to accumulate light and you starred in "black" sky for months, then you would eventually come to conclusion our night sky is not actually dark at all. Like Bowman (2001, C.Clarke), you would say, "Oh my god, it's full of stars!".

Aster
2012-Nov-04, 11:57 PM
You seem to be "truth seeker" like me, can you find any objection to my argument above?

Yes, I can:
not (p -> q) <=> not p or not q.

and I can choose not q is true: sky is not equaly bright in infinite universe or not.

Strange
2012-Nov-05, 12:00 AM
Yes, relationship between number of stars is proportional to r^2. And yes, light received from individual stars is proportional to 1/r^2.

That is all you need in order to see that every part of the sky is (would be) equally bright.


But, light received from individual stars is also proportional to relative orientation between sensor surface and light rays, and is also proportional to exposure time.

Can you explain why you think that is relevant?


Example with half-shell is relevant as it shows the original treatment does not take into account direction of light rays relative to sensor surface orientation, it tells you only how much light is EMITTED, it does not tell you how much light is SENSED.

The "sensed" part is taken care of by the 1/r2. The bit of the sky you are not looking at is irrelevant; this is not about the whole sky, just the section of the sky you are looking at.


So regarding exposure time, let me state that distant stars are "out of the view" as well.

Why?

caveman1917
2012-Nov-05, 12:17 AM
Yes, I can:
not (p -> q) <=> not p or not q.

and I can choose not q is true: sky is not equaly bright in infinite universe or not.

That is incorrect, it is \neg (P \Rightarrow Q) \Leftrightarrow P \wedge \neg Q

Reality Check
2012-Nov-05, 12:45 AM
Any one interested in actually writing about the title of the thread?
Olbers' paradox (http://en.wikipedia.org/wiki/Olbers%27_paradox): Why is the night sky black?
The paradox is fairly simple to state.
Start off with an static eternal universe. That means that any light from any star has reached the Earth.
Add a finite density of stars (which actually means a infinite number of stars!). That means that wherever you look in the night sky your line of sight will intercept a star.
Thus the night sky will be bright.

ShinAce
2012-Nov-05, 12:48 AM
Just to be clear: first, there aren't different solutions.

And also to be clear, there are different solutions.

1) The universe is not infinitely large. We have the CMBR as a wall to light.
2) The universe is not of infinite age. So even if it did have infinite size, the limited travel time for light would solve the paradox.
2) The universe is expanding, so the cosmological redshift alone could solve the paradox.

madus
2012-Nov-05, 12:50 AM
That is all you need in order to see that every part of the sky is (would be) equally bright.


To "see" light means to project it onto something during some exposure time. True, false?

In your model, what surface are you projecting light onto, and how long is your exposure?




Can you explain why you think that is relevant?


When some value is "proportional" to some other, that means the other value is relevant, and thus if you ignore it you will not get correct result. The question is whether you agree or not that light received from individual stars is proportional to relative orientation between sensor surface and light rays, and also proportional to exposure time. Do you agree?




The "sensed" part is taken care of by the 1/r2. The bit of the sky you are not looking at is irrelevant; this is not about the whole sky, just the section of the sky you are looking at.


The bit of the sky you are not looking at is irrelevant, and yet you take all of the stars into account when calculating "total energy".




Why?

I said distant stars are "out of the view" because you don't SEE them, unless you use long exposure.

madus
2012-Nov-05, 12:59 AM
And also to be clear, there are different solutions.

1) The universe is not infinitely large. We have the CMBR as a wall to light.
2) The universe is not of infinite age. So even if it did have infinite size, the limited travel time for light would solve the paradox.
2) The universe is expanding, so the cosmological redshift alone could solve the paradox.

Is there any room left there for my solution, would it be compatible with current theories or would it make some contradictions?

Jens
2012-Nov-05, 01:02 AM
Add a finite density of stars (which actually means a infinite number of stars!). That means that wherever you look in the night sky your line of sight will intercept a star.
Thus the night sky will be bright.

I'm not sure if this clarification is necessary or not, because I'm not quite sure about the meaning of "finite density." If you said "equal density," then I would agree. Because you can have a situation where there are infinite stars and yet Olber's paradox is resolved. That is a situation where the density of galaxies is smaller than that of stars in a galaxy, and the density of galaxy clusters is smaller than that of galaxies, etc., etc. So the density gets lower as the scale gets larger. In this case, you could avoid the paradox even with an infinite number of stars in a fixed infinite universe.

cjameshuff
2012-Nov-05, 01:10 AM
To "see" light means to project it onto something during some exposure time. True, false?

Details of sensor function, exposure time, etc are irrelevant. It does not matter what you're detecting the light with.



When some value is "proportional" to some other, that means the other value is relevant, and thus if you ignore it you will not get correct result. The question is whether you agree or not that light received from individual stars is proportional to relative orientation between sensor surface and light rays, and also proportional to exposure time. Do you agree?

Also not relevant. All stars in a given field of view have the same range of angles to the sensor, regardless of distance.



The bit of the sky you are not looking at is irrelevant, and yet you take all of the stars into account when calculating "total energy".

For total intensity incident on a given point, yes. Again, however, the same math holds for any fraction of the sky.



I said distant stars are "out of the view" because you don't SEE them, unless you use long exposure.

This is just a statement that the prediction of Olber's paradox doesn't match actual observation. No, it doesn't...this would be why it's called a paradox. If the assumptions of Olber's paradox were true, you would see those stars, uniformly covering the sky.

cjameshuff
2012-Nov-05, 01:11 AM
Is there any room left there for my solution, would it be compatible with current theories or would it make some contradictions?

Your "solution" is incorrect, so no, there isn't.

Strange
2012-Nov-05, 01:23 AM
The bit of the sky you are not looking at is irrelevant, and yet you take all of the stars into account when calculating "total energy".

It is not about the total amount of light falling on your sensor/eye/whatever from all of the sky, is about the amount of light in any given area. The number of stars increases with distance at the same rate that the contribution from each individual star falls off with distance.


I said distant stars are "out of the view" because you don't SEE them, unless you use long exposure.

Each individual star provides a smaller amount of light if they are further away. But there are an increasing number of stars at that distance. So the total light from all the stars at any given distance (within any area of the sky) is constant.

As you have had this explained many different ways and (I assume) read things like the Wikipedia page, maybe you just need to spend a bit more time understanding what the paradox proposes. (And then move on to why it doesn't happen.)

Aster
2012-Nov-05, 01:44 AM
That is incorrect, it is \neg (P \Rightarrow Q) \Leftrightarrow P \wedge \neg Q

The better,
Now I have no choose: q must be false and p must be true:

sky is not equaly bright in infinite universe.

ShinAce
2012-Nov-05, 02:03 AM
The better,
Now I have no choose: q must be false and p must be true:

sky is not equaly bright in infinite universe.

I'm sorry, but I can't follow what you're saying. It's simply incomplete.

cjameshuff
2012-Nov-05, 02:17 AM
At the heat death the final temperature must be also lower than the sources.
You still stay in the circular reasoning.

If there's a temperature difference, it's not at heat death. You are proposing a universe which contains perpetual energy sources.



I think here is correct way of reasoning, which leads to the true solution of the Olbers' paradox:
http://www.mathpages.com/home/kmath141/kmath141.htm


This should come as no surprise, since the premise of the paradox is that the stars are absolute sources of energy, and have been in existence for an infinitely long time, so they must have introduced infinite energy into the universe.

I don't see this as a useful way of looking at the paradox. The paradox doesn't assume they are absolute sources of energy, only that they are now glowing. This would only indicate that they had enough initial heat content to still be glowing after coming to equilibrium with everything else.

That article actually seems to end up taking the same approach through a more convoluted line of reasoning, and to consider it to have been resolved when it has actually only avoided the issue of infinite heat buildup, which wasn't the actual paradox in the first place. Just as I did, it predicts an isotropic distribution of radiation, with stars at equilibrium, absorbing exactly as much as they emit. The emission being in the visible spectrum is then just a matter of temperature, and if anything's hot enough to visibly glow, the same goes for the rest of the sky.

The article also strangely misunderstands the modern use of visual rays, which is simply to describe the geometry of light paths through a scene. The claim that visual rays are no longer used in optics is simply false: it is quite often easier to analyze an optical system in the reverse direction, as you are often only interested in those optical paths that lead to a given target.

They are correct that a fractal distribution can tend toward zero density on the largest scales. This is another possible solution, but I see it as a rather contrived one and an unnecessary one, given that two of the assumptions of the paradox are now known to be false.

caveman1917
2012-Nov-05, 03:44 AM
The better,
Now I have no choose: q must be false and p must be true:

sky is not equaly bright in infinite universe.

I'll second what ShinAce said, I have no idea what you're trying to communicate here.

WayneFrancis
2012-Nov-05, 05:46 AM
Forget about time. If we can see a star in the first place, there has been time for its light to get here.

Intensity is defined as power per unit area. A photographic plate has area. Power is already energy per second. It has time built in. To compare intensities, you have to use the same exposure each time.

We can forget about geometry because of the symmetry involved. Whether the plate is at the north pole or the equator, the night sky's intensity is the same. The only geometrical assumption we have to make is that the photographic plate is flat to the ground. Or better yet, always used with the same angle with respect to the ground. All the quantities become scalars instead of vectors. Once we've done that, we calculate the intensity of a shell and find out it is a constant. Again this allows us to simplify the answer. So now the final answer is simply the sum of an infinite series(edit: where each term is the same). That is Olber's paradox.

Of course it isn't a paradox since the universe is not infinite in size. Plus further shells are in fact dimmer because of cosmological redshift.

One qualifier here, sorry to confuse anyone, but the universe may or may not be infinite in size. The size of the observable universe is finite because the age of the universe is finite.

Grey
2012-Nov-05, 06:16 AM
1) The universe is not infinitely large. We have the CMBR as a wall to light.It's amusing to me that there's a certain sense in which Olbers was on the right track. Most lines of sight actually end at the surface of last scattering, which was in fact as hot and bright as a (relatively cool) star when the light was emitted. We do see the sky glowing in every direction we look!


3) The universe is expanding, so the cosmological redshift alone could solve the paradox.But this saves us from being incinerated from a sky that's as bright as a star in every direction. The glow is still there, but due to cosmological expansion the radiation is a thousand times cooler than when it was emitted, so we only see it in the microwave spectrum. Without expansion, the universe would indeed be filled with blinding light, and we wouldn't be here to be looking at it.

madus
2012-Nov-05, 07:19 AM
Details of sensor function, exposure time, etc are irrelevant.


I note you think exposure time is irrelevant for light intensity measurement. So you think in your model, as per link given by Tensor, "total energy" received on Earth refers to energy that could be measured in an instant? And what is the surface area measuring this "total energy" in that model, whole surface of the Earth?




Also not relevant. All stars in a given field of view have the same range of angles to the sensor, regardless of distance.


You can not see light outside of field of view, so how can you say then relative orientation between sensor surface and light rays is irrelevant? There is no any mention of 'field of view' in that link given by Tensor, so what are you talking about, how is that field of view 'given'?




For total intensity incident on a given point, yes. Again, however, the same math holds for any fraction of the sky.


Tensor's link:
http://www.asterism.org/tutorials/tut09-1.htm
- 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


Is total intensity received on Earth from all the sources the same thing as what a single person can see?




This is just a statement that the prediction of Olber's paradox doesn't match actual observation. No, it doesn't...this would be why it's called a paradox. If the assumptions of Olber's paradox were true, you would see those stars, uniformly covering the sky.

You have your model as defined by the link given by Tensor, and I have my model. They are not the same. My model, unlike yours, takes into account sensor surface area and its orientation as well as the time of exposure. So no, in my model a single person does not see stars are uniformly covering the sky, because in my model a person can not see the whole shell and all the stars at once, nor it can see distant stars without making longer exposure, just like in the real world.

madus
2012-Nov-05, 10:37 AM
It is not about the total amount of light falling on your sensor/eye/whatever from all of the sky, is about the amount of light in any given area.


To calculate intensity per any given area we need to model the impact over some certain area first.




Each individual star provides a smaller amount of light if they are further away. But there are an increasing number of stars at that distance. So the total light from all the stars at any given distance (within any area of the sky) is constant.


I understand now where's your problem. And when I admitted I was wrong before, it was a mistake, because I was right all along. Look, the number of stars increases and intensity emitted increases making it up for the longer distance, but what you don't consider is when this intensity hits the sensor surface it will be divided by that same number of stars, so we get fainter impressions from the distant stars. Edmond Halley was right after all.

Strange
2012-Nov-05, 11:02 AM
I understand now where's your problem. And when I admitted I was wrong before, it was a mistake, because I was right all along. Look, the number of stars increases and intensity emitted increases making it up for the longer distance, but what you don't consider is when this intensity hits the sensor surface it will be divided by that same number of stars, so we get fainter impressions from the distant stars. Edmond Halley was right after all.

You seem to have a mental model in your head that makes no sense to me. I don't know what you are trying to work out but it appears to be the wrong thing. Good luck :)

Shaula
2012-Nov-05, 12:35 PM
I understand now where's your problem. And when I admitted I was wrong before, it was a mistake, because I was right all along. Look, the number of stars increases and intensity emitted increases making it up for the longer distance, but what you don't consider is when this intensity hits the sensor surface it will be divided by that same number of stars, so we get fainter impressions from the distant stars. Edmond Halley was right after all.
You seem to be modelling the limit of infinitesimal pixels. If you insist on taking into account physical detector characteristics then redo the calculation with a pixel size. You will find that the intensity per pixel stays the same because more stars end up binned in each pixel as you look further away.

madus
2012-Nov-05, 01:30 PM
You seem to have a mental model in your head that makes no sense to me. I don't know what you are trying to work out but it appears to be the wrong thing. Good luck :)

Let's speak math then. Number of stars is proportional to r^2. Light received from each star is proportional to 1/r^2. Therefore total intensity 'I' received from any shell is the same. You know this. All you have to do is to imagine photographic plate now. So we can see that number of stars on a photo is also proportional to r^2, and so their photographed brightness is proportional to I/r^2 * exposure. -- Human eyes have roughly one-fifteenth of a second exposure time. If it was longer the night sky would look brighter, proportionally.

ShinAce
2012-Nov-05, 01:50 PM
Let's use an analogy, albeit a bad one.

Let intensity be equivalent to the speed of a car. Let total energy be equivalent to the mileage. Here we will get on the highway and set cruise control to 60 mph. Your car very well does use time to measure the speed. However, what it does is see how long it takes for the wheel to spin 360 degrees. Since the computer knows the circumference of the wheel, and the time it took to make a full turn, then it can figure the speed. Speed is the circumference divided by time.

If we reprogram the car to wait until 5 full revolutions have been made, it will simply take 5 times the circumference divided by the time measured. And it will come back with the same speed, 60 mph.

What I'm saying is that the car is going 60 mph no matter how much time you use to measure the speed. What you're saying is that if we wait 2 hours, the car will travel 120 miles. You keep talking about the mileage as if it's the speed and ignoring the speed.

We're not talking about the same thing. The beauty of this analogy is that police use doppler radar to measure speeds. Doppler measures speed based on shift in wavelength. It doesn't even need to measure time at all to get your speed.

madus
2012-Nov-05, 01:58 PM
You seem to be modelling the limit of infinitesimal pixels. If you insist on taking into account physical detector characteristics then redo the calculation with a pixel size. You will find that the intensity per pixel stays the same because more stars end up binned in each pixel as you look further away.

I can see that if we have only one pixel and focus the light from all the stars in the field of view it would become bright very fast, but I don't think insufficient pixel resolution would provide useful info, rather it would create some error artifact. But yeah, I'd like to take resolution into account, I don't see why not. I just don't know how to even begin, it seems I'd have to treat light as photons rather than energy, and not sure whether to use spherical shells or field of view cone. Any suggestions?

ShinAce
2012-Nov-05, 02:00 PM
I can see that if we have only one pixel and focus the light from all the stars in the field of view it would become bright very fast, but I don't think insufficient pixel resolution would provide useful info, rather it would create some error artifact. But yeah, I'd like to take resolution into account, I don't see why not. I just don't know how to even begin, it seems I'd have to treat light as photons rather than energy, and not sure whether to use spherical shells or field of view cone. Any suggestions?

That's cheating because you're using the large exposed area of a lens to focus it down to a point. You're saying the collecting area is the size of the pixel, when really it is the size of the lens. Keep using bigger lenses and you will get more light, because you're using more area for your measurement. However, intensity is unaffected.

cjameshuff
2012-Nov-05, 02:17 PM
I note you think exposure time is irrelevant for light intensity measurement. So you think in your model, as per link given by Tensor, "total energy" received on Earth refers to energy that could be measured in an instant? And what is the surface area measuring this "total energy" in that model, whole surface of the Earth?

Intensity is not total energy, it is the instantaneous rate at which energy is being received over time per unit of area or solid angle. It is not required and generally makes no sense to specify exposure time or area. As has already been explained.



You can not see light outside of field of view, so how can you say then relative orientation between sensor surface and light rays is irrelevant? There is no any mention of 'field of view' in that link given by Tensor, so what are you talking about, how is that field of view 'given'?

How is the angle to the sensor of objects not even in the field of view in any way relevant?



Is total intensity received on Earth from all the sources the same thing as what a single person can see?

From all the sources in their field of view, yes.



You have your model as defined by the link given by Tensor, and I have my model. They are not the same. My model, unlike yours, takes into account sensor surface area and its orientation as well as the time of exposure. So no, in my model a single person does not see stars are uniformly covering the sky, because in my model a person can not see the whole shell and all the stars at once, nor it can see distant stars without making longer exposure, just like in the real world.

And since these factors are irrelevant, your model is physically nonsensical, predicting an incorrect result. The whole sky is uniformly bright, this means that every fractional view of the sky is also uniformly bright.



I understand now where's your problem. And when I admitted I was wrong before, it was a mistake, because I was right all along. Look, the number of stars increases and intensity emitted increases making it up for the longer distance, but what you don't consider is when this intensity hits the sensor surface it will be divided by that same number of stars, so we get fainter impressions from the distant stars. Edmond Halley was right after all.

That is completely wrong. The intensity received is the intensity measured, it does not matter how many sources it took to achieve it.

madus
2012-Nov-05, 02:49 PM
That is completely wrong. The intensity received is the intensity measured, it does not matter how many sources it took to achieve it.

You do not understand what it means to "see" something. It means to create 2-dimensional image, it involves surface area onto which light is projected. By ignoring surface area you make your image resolution only one pixel. And sure if your resolution is just one pixel and you focus all the light onto it you will see a completely bright night sky with even a single faint star in your field of vision.


a.) Number of stars on a photo is also proportional to r^2

b.) Photographed brightness of each star is proportional to I/r^2 * exposure.

True, false?

ShinAce
2012-Nov-05, 02:52 PM
You were doing so well at being patient and trying things out. Now you're passing judgement on others who are trying to steer you in the right direction.

It would be nice if you would specify whether you're still asking about Olber's paradox, or what we see in reality. One is a thought exercise, the other is an experiment. They don't agree.

madus
2012-Nov-05, 02:57 PM
That's cheating because you're using the large exposed area of a lens to focus it down to a point. You're saying the collecting area is the size of the pixel, when really it is the size of the lens. Keep using bigger lenses and you will get more light, because you're using more area for your measurement. However, intensity is unaffected.

Perceived brightness depends on image resolution. Intensity spreads out over sensor surface area proportionally with the number of stars and their distance.

ShinAce
2012-Nov-05, 03:03 PM
If you take a magnifying glass and use that to focus light to a point, yes, the light is more intense on that spot, but less intense elsewhere. You're not creating intensity, you're stealing it from other areas and bringing it into one place.

Let me be clear. The intensity from the night sky is constant anywhere on earth. If you create a spot where it's higher, you've also created spots where it is lower. That's why when we use the collecting area used in the intensity calculation, we must use the area of the lens/mirror, not the area of the sensor. In which case, if you want more intensity, you have to use larger areas. But the calculation will null that result since it compensates for a standardized area.

If you want to get into exposures again, realize that exposures change the total energy received, not the intensity. See the 60 mph car analogy.

cjameshuff
2012-Nov-05, 03:19 PM
You do not understand what it means to "see" something. It means to create 2-dimensional image, it involves surface area onto which light is projected. By ignoring surface area you make your image resolution only one pixel. And sure if your resolution is just one pixel and you focus all the light onto it you will see a completely bright night sky with even a single faint star in your field of vision.

So? Another pixel looking at an adjacent patch of sky isn't going to change what the first one sees. If it's true for one pixel, it's true for two, four, twelve, or 15 million.



a.) Number of stars on a photo is also proportional to r^2

b.) Photographed brightness of each star is proportional to I/r^2 * exposure.

True, false?

A as stated is false. The number is infinite, but the number of stars at a given distance r is proportional to r^2.

B is only true if the disk of the star is an infinitesimal fraction of the pixel...and A (the corrected version) is also true on a per-pixel basis, the area of sky covered by each pixel contains an infinite number of stars with equal contribution at every distance, not a single star as you assume.

If you *can* resolve the disk of every star, with no diffraction limits and an infinite resolution sensor, then B is false. The surface brightness of a star does not depend on distance. The area of sensor covered by a star is proportional to 1/r^2, but the pixels in that area record the full intensity and there are infinitely many stars covering the entire area of the sensor.

Either way, the exact same result is obtained. The sensor is an irrelevant detail. Intensity is intensity, there aren't different "kinds" as your reasoning would require.

madus
2012-Nov-05, 03:23 PM
You were doing so well at being patient and trying things out. Now you're passing judgement on others who are trying to steer you in the right direction.

It would be nice if you would specify whether you're still asking about Olber's paradox, or what we see in reality. One is a thought exercise, the other is an experiment. They don't agree.

I'm now explaining how to correctly treat Olbers' paradox with inverse square law. Are you interested to see what result is going to come out?

cjameshuff
2012-Nov-05, 03:25 PM
Perceived brightness depends on image resolution.

False. Total optical power received by individual image elements depends on resolution, but measured intensity must take the sensor area into account. Image resolution is a red herring.

Aster
2012-Nov-05, 04:19 PM
If there's a temperature difference, it's not at heat death. You are proposing a universe which contains perpetual energy sources.

There is no differences, but average is always lower than the maximum (excluding one case, which assumes our thesis at start).

Perpetual energy sources are assumed in Your reasoning:
mass-energy density can not change in an autonomous system.


I don't see this as a useful way of looking at the paradox. The paradox doesn't assume they are absolute sources of energy, only that they are now glowing. This would only indicate that they had enough initial heat content to still be glowing after coming to equilibrium with everything else.

That article actually seems to end up taking the same approach through a more convoluted line of reasoning, and to consider it to have been resolved when it has actually only avoided the issue of infinite heat buildup, which wasn't the actual paradox in the first place. Just as I did, it predicts an isotropic distribution of radiation, with stars at equilibrium, absorbing exactly as much as they emit. The emission being in the visible spectrum is then just a matter of temperature, and if anything's hot enough to visibly glow, the same goes for the rest of the sky.

You permanently ignore dust, gas, clouds, plasmas, stellar winds, ect. - probably over 90% of the universe, which are not sources at all.

madus
2012-Nov-05, 04:21 PM
If you take a magnifying glass and use that to focus light to a point, yes, the light is more intense on that spot, but less intense elsewhere. You're not creating intensity, you're stealing it from other areas and bringing it into one place.


I'm not talking about magnification, but about field of view and resolution. If two photons enter your field of view and get projected onto 10x10 pixels photo-plate you get mostly black image with two dots having some brightness 'b'. But if your resolution is only 1x1 then both of them get projected onto single pixel, and your image contains no black at all, everything is just brightness with value of 2b.

madus
2012-Nov-05, 04:37 PM
From all the sources in their field of view, yes.


There is no any mention of 'field of view' in the link given by Tensor. So what that means, that field of view is zero, or one, or some kind of panoramic 360 degrees view? Or wait, let me guess, it's irrelevant, like everything else I said, isn't it?

cjameshuff
2012-Nov-05, 04:41 PM
There is no differences, but average is always lower than the maximum (excluding one case, which assumes our thesis at start).

These two statements are mutually exclusive. If there are no differences, the average, minimum, and maximum temperatures are the same.



Perpetual energy sources are assumed in Your reasoning:

No, they aren't. I only assume initial energy content sufficient to still glow after coming to equilibrium with the rest of the universe. This is a finite quantity.



You permanently ignore dust, gas, clouds, plasmas, stellar winds, ect. - probably over 90% of the universe, which are not sources at all.

I don't ignore them, they are at equilibrium with the stars. Equal temperature, and glowing just as bright.

ShinAce
2012-Nov-05, 04:47 PM
I'm now explaining how to correctly treat Olbers' paradox with inverse square law. Are you interested to see what result is going to come out?

Seeing as how I already treated the paradox, I would love to see your result. If it's the same, we get verification and confidence. If they're different, there's a mistake somewhere to be found.

Typically when someone does a new proof, they define the quantities they will use. Please feel free to add your definitions.

Cougar
2012-Nov-05, 04:48 PM
I'm now explaining how to correctly treat Olbers' paradox with inverse square law. Are you interested to see what result is going to come out?

That depends on your qualifications in this area of study. Actually, this so-called paradox is no longer a paradox, since it has been answered and explained by astrophysicists who have been studying their field of astrophysics for decades. We now know why the night sky is dark. This is no longer an unanswered question. So if the "result that you're going to have come out" is the mainstream view, then fine, maybe someone who doesn't know the answer will be interested. But if you have an "alternative" answer, then no. Why would anyone be interested in your alternative answer when we don't know if your education has gone beyond the 6th grade or 9th grade or if you have a high-school diploma or if you have any credits from an accredited university or if you have a university degree or if such a degree is in literature or French or ? or if you have a post-graduate degree. You certainly do not seem to be very well studied in this general area. In other words, if you dropped out of school in the 8th grade, then no, I'm not interested at all in how you claim to "correctly" treat Olber's paradox. Did you start considering this topic two days ago or two weeks ago or two years ago or two decades ago? What makes you think anyone would be interested in any alternative view of yours if you've been thinking about the topic for a grand total of two days?

cjameshuff
2012-Nov-05, 04:56 PM
There is no any mention of 'field of view' in the link given by Tensor. So what that means, that field of view is zero, or one, or some kind of panoramic 360 degrees view? Or wait, let me guess, it's irrelevant, like everything else I said, isn't it?

Yes, it's irrelevant. The argument as phrased on that page deals with the whole sphere, but applies equally well to any fraction. 1/100th of a shell has 1/100th the contribution, spread over 1/100th the area.

Intensity is intensity. The assumptions of Olber's paradox lead to the night sky having intensity equal to the surface of a star. Your attempts to account for the dark sky using characteristics of the sensor apply equally well to the daytime sun, which unfortunately for your argument is quite clearly not dark.

madus
2012-Nov-05, 05:04 PM
That depends on your qualifications in this area of study. Actually, this so-called paradox is no longer a paradox, since it has been answered and explained by astrophysicists who have been studying their field of astrophysics for decades. We now know why the night sky is dark. This is no longer an unanswered question. So if the "result that you're going to have come out" is the mainstream view, then fine, maybe someone who doesn't know the answer will be interested. But if you have an "alternative" answer, then no. Why would anyone be interested in your alternative answer when we don't know if your education has gone beyond the 6th grade or 9th grade or if you have a high-school diploma or if you have any credits from an accredited university or if you have a university degree or if such a degree is in literature or French or ? or if you have a post-graduate degree. You certainly do not seem to be very well studied in this general area. In other words, if you dropped out of school in the 8th grade, then no, I'm not interested at all in how you claim to "correctly" treat Olber's paradox. Did you start considering this topic two days ago or two weeks ago or two years ago or two decades ago? What makes you think anyone would be interested in any alternative view of yours if you've been thinking about the topic for a grand total of two days?

I don't know what answer is going to be. I have no answers, just an opinion, but opinions can be wrong. And if I am wrong that's fine, I just want to know for certain. I am simply proposing to include sensor surface in the treatment, that is to model a camera or an eye, and see what will it see, by using mathematics.

ShinAce
2012-Nov-05, 05:12 PM
The energy received per exposure by a sensor will be = intensity*surface area of sensor*exposure time

Notice that the area of the sensor and the exposure are separate from the intensity. Why should the intensity depend on the other two? Be careful. We don't see intensity, we see energy. Our brain infers intensity.

So not only are you moving away from Olber's paradox, you're not even dealing with intensity as a subject matter. Just because the Keck telescope can take brighter images with the same sensors as everyone else doesn't mean the sky is more intense. Or are you saying it does/should?

madus
2012-Nov-05, 05:14 PM
Yes, it's irrelevant. The argument as phrased on that page deals with the whole sphere, but applies equally well to any fraction. 1/100th of a shell has 1/100th the contribution, spread over 1/100th the area.

Intensity is intensity. The assumptions of Olber's paradox lead to the night sky having intensity equal to the surface of a star. Your attempts to account for the dark sky using characteristics of the sensor apply equally well to the daytime sun, which unfortunately for your argument is quite clearly not dark.

Ha! Made you agree with me. Ok, enough talking, shall we do the math now and get this over with? I propose to treat light as photons and model field of vision in either 2D as triangle or in 3D as a cone, so the image would be projected on either line or circle shaped photo negative. How does that sound?

madus
2012-Nov-05, 05:29 PM
Seeing as how I already treated the paradox, I would love to see your result. If it's the same, we get verification and confidence. If they're different, there's a mistake somewhere to be found.

Typically when someone does a new proof, they define the quantities they will use. Please feel free to add your definitions.

Sure, but you need to help me, we need to agree the setup is proper so we don't end up arguing about validity of the result.

madus
2012-Nov-05, 05:42 PM
So not only are you moving away from Olber's paradox, you're not even dealing with intensity as a subject matter. Just because the Keck telescope can take brighter images with the same sensors as everyone else doesn't mean the sky is more intense. Or are you saying it does/should?

I am not moving away but proposing how to correctly treat Olbers' paradox with inverse square law. You said you want see the result, what's the problem all of a sudden?




The energy received per exposure by a sensor will be = intensity*surface area of sensor*exposure time

Notice that the area of the sensor and the exposure are separate from the intensity. Why should the intensity depend on the other two? Be careful. We don't see intensity, we see energy. Our brain infers intensity.


I think the paradox is about brightness, so I propose that's what we should calculate, where brightness of each pixel would be defined by the number of photons that hits it. Sounds good?

ShinAce
2012-Nov-05, 05:43 PM
I don't even know what you're trying to do. Are you trying to figure out how many photons per second fall onto a typical retina?

According to wiki:
"Brightness is an attribute of visual perception in which a source appears to be radiating or reflecting light.[1] In other words, brightness is the perception elicited by the luminance of a visual target. This is a subjective attribute/property of an object being observed."

It is not a scientifically measurable thing to begin with. Apparent magnitude is just that, apparent. My personal experience has been with sound where I realized that there is no rule for "When you increase by X decibels, the sound is perceived as twice as loud". That actually depends on frequency and amplitude but often people just toss around 10 dB.

We can't begin until we get our ducks in order. We need to understand the same thing for: intensity, power, energy, and so forth.

madus
2012-Nov-05, 06:07 PM
I don't even know what you're trying to do. Are you trying to figure out how many photons per second fall onto a typical retina?

It is not a scientifically measurable thing to begin with. Apparent magnitude is just that, apparent. My personal experience has been with sound where I realized that there is no rule for "When you increase by X decibels, the sound is perceived as twice as loud". That actually depends on frequency and amplitude but often people just toss around 10 dB.


Camera should be better. I mean to model infinite stars that emit photons and camera that senses those photons. Intensity does not have to correspond to reality, so we could approximate stars to be emitting just one photon per second, or something like that. We either going to get uniform brightness or many dots with patches of black.




We can't begin until we get our ducks in order. We need to understand the same thing for: intensity, power, energy, and so forth.

Sure. I think we only need photons which we will count per image pixel.

ShinAce
2012-Nov-05, 06:19 PM
Camera should be better. I mean to model infinite stars that emit photons and camera that senses those photons. Intensity does not have to correspond to reality, so we could approximate stars to be emitting just one photon per second, or something like that. We either going to get uniform brightness or many dots with patches of black.




Sure. I think we only need photons which we will count per image pixel.

If you model infinite stars with infinite time like Olber did, you will get uniform brightness.

PetersCreek
2012-Nov-05, 06:38 PM
The Space/Astronomy Questions and Answers forum is where one gets mainstream answers...and does not argue them. Since the question has progressed to more indepth discussion, this thread needs to be moved. Before I do that, madus, I need to know if your treatment of Olber's Paradox conforms to the mainstream or not.

This thread is temporarily closed, pending your answer, which you may give via report or PM.

ETA: Thread reopened in the Astronomy forum for continued discussion.

cjameshuff
2012-Nov-05, 09:14 PM
Ha! Made you agree with me.

Agree? Are you now admitting that your attempts to resolve Olber's paradox using the inverse squared law were fundamentally flawed?



Ok, enough talking, shall we do the math now and get this over with? I propose to treat light as photons and model field of vision in either 2D as triangle or in 3D as a cone, so the image would be projected on either line or circle shaped photo negative. How does that sound?

Once again, you're overcomplicating what is an extremely simple issue. The brightness is uniform. Half of a given patch of sky will have half the overall power and half the area, for exactly the same intensity. Exposure time, sensor resolution, etc do not matter...the sensor will respond the same way to light of a given intensity, regardless of the distance of its source or sources.

The math has been done, over and over again. The number of stars at a given distance scales with the distance squared, exactly canceling the inverse square falloff. It does not matter one bit what fraction of the sky you are looking at. Your "solution" to Olber's paradox is just bad math and bad physics.

Hornblower
2012-Nov-05, 09:28 PM
Madus, your persistence in overcomplicating this exercise suggests that you have significant gaps in your understanding of some novice-level points of physics. There is no shame in that, as we all have to start somewhere, but physics and mathematics are brutally unforgiving of missing prerequisites and attempts short answers in a forum like this can be frustrating. It would be easier to fill the gaps in a face to face exchange at a desk or blackboard where I could ask some leading questions to identify the gaps and use sketches to clarify things.

madus
2012-Nov-06, 05:08 AM
The math has been done, over and over again.

I know about your opinion, you already told me about it. I also know no one has done it by modeling sensor surface and light as photons, which is the only correct way. You said before we'll get the same result, so what's the matter now? I ask you again, do you wanna know what result is going to come out, or do you refuse to look through my telescope?

madus
2012-Nov-06, 05:22 AM
Madus, your...

Is there some particular reason why would you refuse to know what result is going to come out?

madus
2012-Nov-06, 05:32 AM
If you model infinite stars with infinite time like Olber did, you will get uniform brightness.

I acknowledge your opinion. Now, we were on the way to get "our ducks in order" and define the setup, make sure it's proper and approximations valid. So shall we continue, or do you give up? I ask you yet again, do you want to see what result is actually going to come out, or not?

madus
2012-Nov-06, 06:56 AM
Exposure time, sensor resolution, etc do not matter...


Brightness of the image produced by either camera or human eyes does depend on exposure time, it's well tested and confirmed fact, everyday experience and common knowledge. I also explained how insufficient resolution will produce error artifact manifesting as increased brightness. This error is all you get when you ignore image resolution, so naturally you get result indicating uniform brightness.




...the sensor will respond the same way to light of a given intensity, regardless of the distance of its source or sources.


If your image resolution is insufficient, like 1x1 as in the way you treat it, then yes, but otherwise no. 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 you are saying, and that's true. However, what you 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 a difference between two bright spots and four less bright dots. And there is difference between two dots on 10x10 resolution image and 1x1 resolution image. Very distant stars will not produce any dots unless you wait long enough, don't you know that? It's a fact, look it up. So, at infinite distance there will be infinite number of stars and if you 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.

Shaula
2012-Nov-06, 07:11 AM
So, at infinite distance there will be infinite number of stars and if you 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.
But no one is saying all the stars are at infinite distance. It is an integral to infinity, not I/infinity.

madus
2012-Nov-06, 07:53 AM
But no one is saying all the stars are at infinite distance. It is an integral to infinity, not I/infinity.

I gave example with 2 and 4 stars as well, it extends linearly. Try to add additional shells and you get the same intensity recorded with each shell per unit time, but it spreads out over more dots. -- By the way, since rate of incoming photons depends on distance, has anyone tried to measure intergalactic distances by counting individual photons and measuring the time interval between their arrival? It's interesting that you get better resolution the further away galaxy is since the time interval increases and thus makes it easier to time it.

Strange
2012-Nov-06, 08:50 AM
Brightness of the image produced by either camera or human eyes does depend on exposure time

Olber's paradox says that the brightness should be the same everywhere. Yes, integrate over a longer period with a longer exposure and the recorded brightness will be greater. The same is true of any image.


I also explained how insufficient resolution will produce error artifact manifesting as increased brightness. This error is all you get when you ignore image resolution, so naturally you get result indicating uniform brightness.

You can't "solve" the paradox by relying on the non-ideal behaviour of real sensors. Because that is not what it is about. It is about the amount of light arriving. Not how bright a photo is.


So, at infinite distance there will be infinite number of stars and if you 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.

Have you come across the idea of limits?

madus
2012-Nov-06, 09:29 AM
You can't "solve" the paradox by relying on the non-ideal behaviour of real sensors. Because that is not what it is about. It is about the amount of light arriving. Not how bright a photo is.


Olber's paradox is a nothing more than a question: Why do we SEE night sky is black?

I answer that question by using no assumptions, but only well confirmed and tested facts.

It is not about amount of light arriving, it is about how image from that light is being formed.


Cjameshuff and ShinAce already agreed we should get the same result. -- Eyes work very much the same as a camera, eyes too have exposure time. Perception of brightness is relative to image processing instrument. If your eyes had longer exposure you would see brighter night sky, or if your eyes were sensitive to microwave wavelengths you would see night sky is quite bright. Brightness is in the eye of the beholder.




Have you come across the idea of limits?

For more than ten years I've been developing numerical modeling software, applying physics and mathematics to replicate physical processes we see in the real world. My specialty is time integration of those processes, that is visualization and image processing. So to answer your question, yes I have come across the idea of limits, I work with them on a daily basis in a very practical manner. Why do you ask?

tusenfem
2012-Nov-06, 09:35 AM
Brightness of the image produced by either camera or human eyes does depend on exposure time, it's well tested and confirmed fact, everyday experience and common knowledge.

No brightness has exactly NOTHING to do with exposure time etc. "as is commonly known."
In mainstream physics brightness is defined as:

The brightness Bν (more properly known as the specific brightness) is the intensity of a radiating source (i.e., the energy flux per solid angle per unit of frequency), also called the radiance or surface brightness. In SI, it is measured in units of J s-1 m-2 Hz-1 sterad-1

What you are looking at is "total received energy" over a certain amount of time. Naturally in "common speach" you say the image gets brighter, howerver that means that more photons have reached the silveriodide plate. You need to know the units of the things you want to measure, and a simple physics textbook would have helped you a lot.

madus
2012-Nov-06, 12:34 PM
What you are looking at is "total received energy" over a certain amount of time. Naturally in "common speach" you say the image gets brighter, howerver that means that more photons have reached the silveriodide plate.

You said image gets brighter as more photons reach the silveriodide plate. That's what I said, and of course the number of photons that reach the plate depends on exposure time.


http://en.wikipedia.org/wiki/Brightness
- With regard to stars, brightness is quantified as apparent magnitude and absolute magnitude

- "brightness" should now be used only for non-quantitative references to physiological sensations and perceptions of light


http://en.wikipedia.org/wiki/Apparent_magnitude
- The apparent magnitude of a celestial body is a measure of its brightness as seen by an observer on Earth

- Brightness varies inversely with the square of the distance


http://en.wikipedia.org/wiki/Exposure_%28photography%29
- exposure is the amount of light allowed to fall on each area unit of a photographic medium

tusenfem
2012-Nov-06, 01:09 PM
You said image gets brighter as more photons reach the silveriodide plate. That's what I said, and of course the number of photons that reach the plate depends on exposure time.


... and I also said "in common speach" which means that the word "brighter" most likely has nothing to do with the physical term.
And although wiki claims it should be used for "non quantative" purposes I disagree, e.g. surface brightness is something often used in astronomy, and still, brightness has units, which do not depend on "exposure time"

The claim "brightness varies inversely with the square of the distance" is no problem, thinking about the units that I gave for brightness (a.o. per square meter) and a point source radiating equally in all directions.

So, the "common speech" that "the image gets brighter" is only so far correct that when you have a longer exposure time, there will be more reflected light from the photographic surface after it is developed. However, this has nothing to do with the "physics speech" brightness of the source, except for the fact that a less bright source will need a longer time to get an equally "brilliant" image on the photo as compared to a more bright source.

The problem here is that you give "common speech" definitions of terms that have specific "physics speech" definitions and that will mess up many a discussion as we often see here on CQ.

Jeff Root
2012-Nov-06, 01:31 PM
madus,

Which light source has greater brightness and radiant intensity?

A 60-watt light bulb shining onto a 35-mm f/4 camera lens for
1/10th of a second,

or

A 60-watt light bulb shining onto a 35-mm f/4 camera lens that
has the lens cap on?

-- Jeff, in Minneapolis

madus
2012-Nov-06, 01:46 PM
... and I also said "in common speach" which means that the word "brighter" most likely has nothing to do with the physical term.
And although wiki claims it should be used for "non quantative" purposes I disagree, e.g. surface brightness is something often used in astronomy, and still, brightness has units, which do not depend on "exposure time"

The claim "brightness varies inversely with the square of the distance" is no problem, thinking about the units that I gave for brightness (a.o. per square meter) and a point source radiating equally in all directions.

So, the "common speech" that "the image gets brighter" is only so far correct that when you have a longer exposure time, there will be more reflected light from the photographic surface after it is developed. However, this has nothing to do with the "physics speech" brightness of the source, except for the fact that a less bright source will need a longer time to get an equally "brilliant" image on the photo as compared to a more bright source.

The problem here is that you give "common speech" definitions of terms that have specific "physics speech" definitions and that will mess up many a discussion as we often see here on CQ.

I am talking about the same brightness Wikipedia is talking about when they say "brightness varies inversely with the square of the distance" in their article about stars and apparent magnitude, the same brightness Olbers' paradox is about. That's not just "common speech", but correct term to describe exactly that. There is no other or more correct term for it. What other word do you think I should use?

I am talking about image brightness as produced by camera, or perceived brightness of an image formed by human eyes. I am not talking about radiance or luminance, which is what you were talking about and Wikipedia says "brightness" is incorrect term to describe those properties, as defined in Federal Standard 1037C, the Federal Glossary of Telecommunication Terms (1996) - brightness: An attribute of visual perception in which a source appears to emit a given amount of light


So all this semantics, does it change anything in regards to my conclusions about Olbers' paradox?

Strange
2012-Nov-06, 01:48 PM
I am talking about the same brightness Wikipedia is talking about when they say "brightness varies inversely with the square of the distance" in their article about stars and apparent magnitude, the same brightness Olbers' paradox is about. ...

I am talking about image brightness as produced by camera, or perceived brightness of an image formed by human eyes.

But those are two different things. Which one are you talking about? Or are you just mixing them up?

madus
2012-Nov-06, 01:54 PM
madus,

Which light source has greater brightness and radiant intensity?


Federal Standard 1037C, the Federal Glossary of Telecommunication Terms (1996)
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."



Do you have anything to say in regards to Olbers' paradox and my conclusions about it?

madus
2012-Nov-06, 01:57 PM
But those are two different things. Which one are you talking about? Or are you just mixing them up?

They are not different things, you are mixing up. Does your semantic objection change anything in regards to my conclusions about Olbers' paradox?

Swift
2012-Nov-06, 02:31 PM
So shall we continue, or do you give up?

They are not different things, you are mixing up. Does your semantic objection change anything in regards to my conclusions about Olbers' paradox?
madus,

I have had just about enough of you. First, you will stop making demands of what other people may write. This is not your private little conversation where you can dictate what people may write or if they may participate or not - that is up to the moderators of this forum.

Second, this is not semantics. People have spent a lot of time explaining the physical definition of terms. Selecting quoting wikipedia or Federal rules for physics terms is not helpful. If you are going to use terms imprecisely, especially after people have pointed out the correct way to use them, you are going to have problems explaining your point.

Finally, you seem to think this entire discussion is about winning, about convincing everyone that you are correct. If you want to do that, take your discussions to ATM, where you can try to prove your non-mainstream ideas. You keep claiming you are "just asking", yet you argue with every single thing that is explained to you. That is not asking and that is not behaving like you are here to learn and to discuss.

You will lose the attitude and you will start following our rules of behavior or you will be banned.

Strange
2012-Nov-06, 02:37 PM
They are not different things

I'm afraid I just cannot understand the way you are thinking about this. If these are the same things, then the stars would go dark if everyone closes their eyes. Their brightness would vary depending how many people are looking into, or taking photos of, the sky.


Does your semantic objection change anything in regards to my conclusions about Olbers' paradox?

It is still obviously wrong. Or, maybe, "not even wrong".

antoniseb
2012-Nov-06, 02:43 PM
madus, ... That is not asking and that is not behaving like you are here to learn and to discuss.
You will lose the attitude and you will start following our rules of behavior or you will be banned.
It's pretty rare that we get someone who fits in so poorly... but in this case let me praise Swift and the rest of the moderator team for their patience and restraint. We all keep hoping that the light will go on, and (obviously interested and energetic) madus will suddenly shift gears and learn what he/she doesn't know and join in on meaningful discussions about real science. But it doesn't seem to be happening. So, madus, participate in a useful way, or I will ban you myself.

Jeff Root
2012-Nov-06, 02:44 PM
madus,

I gave you an alternative to "brightness". So use it.

Which light source has greater radiant intensity?

A 60-watt light bulb shining onto a 35-mm f/4 camera lens for
1/10th of a second,

or

A 60-watt light bulb shining onto a 35-mm f/4 camera lens that
has the lens cap on?

-- Jeff, in Minneapolis

madus
2012-Nov-06, 03:05 PM
madus,

I gave you an alternative to "brightness". So use it.

Which light source has greater radiant intensity?


This thread is about Olbers' paradox. -- I gave you definition that clearly states brightness is not the same thing as radiant intensity, so it is not alternative it's incorrect usage of the term. Brightness is not about what objects emits, it is about what eye or camera perceives. So to answer your question, radiant energy of the BULB is the same in both cases, while the brightness of the PHOTO in 1st case will be higher than in 2nd case, where the photo would be black.


Do you have anything to say in regards to Olbers' paradox and my conclusions about it?

madus
2012-Nov-06, 03:15 PM
I'm afraid I just cannot understand the way you are thinking about this. If these are the same things, then the stars would go dark if everyone closes their eyes. Their brightness would vary depending how many people are looking into, or taking photos of, the sky.


Federal Standard 1037C, the Federal Glossary of Telecommunication Terms (1996)
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


Luminance and radiance are terms that describe amount of light that object emits. Brightness refers to sensed or "captured" amount of light.




It is still obviously wrong. Or, maybe, "not even wrong".

What is wrong? Why is it wrong?

Jeff Root
2012-Nov-06, 03:19 PM
I didn't ask about radiant energy, though. I asked about
radiant intensity.

But assuming that you meant "radiant intensity" when you
said "radiant energy"...

Does Olbers' paradox concern the radiant intensity of the
sky, the brightness of photos of the sky, or something else?

-- Jeff, in Minneapolis

madus
2012-Nov-06, 03:23 PM
It's pretty rare that we get someone who fits in so poorly... but in this case let me praise Swift and the rest of the moderator team for their patience and restraint. We all keep hoping that the light will go on, and (obviously interested and energetic) madus will suddenly shift gears and learn what he/she doesn't know and join in on meaningful discussions about real science. But it doesn't seem to be happening. So, madus, participate in a useful way, or I will ban you myself.

You seem to be suggesting I am not talking about "real science", what are you referring to? Did I make any mistake in my reasoning, made some incorrect assumptions? Do you think my conclusion about Olbers' paradox is wrong?

Strange
2012-Nov-06, 03:31 PM
Did I make any mistake in my reasoning, made some incorrect assumptions? Do you think my conclusion about Olbers' paradox is wrong?

Your reasoning, assumptions and conclusions are all wrong. As has been explained to you, carefully and in great detail multiple times.

madus
2012-Nov-06, 03:35 PM
I didn't ask about radiant energy, though. I asked about
radiant intensity.

But assuming that you meant "radiant intensity" when you
said "radiant energy"...


I hope that means we understand each other now.




Does Olbers' paradox concern the radiant intensity of the
sky, the brightness of photos of the sky, or something else?


Olbers' paradox is not about radiant intensity. It is about brightness as defined by Federal Standard 1037C, the Federal Glossary of Telecommunication Terms (1996)
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.


In other words it is about "Apparent magnitude", as defined by Wikipedia here:
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


It's about what a person can see or what camera captures, in the same sense and context as what Wikipedia means when it states:
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.

tusenfem
2012-Nov-06, 03:40 PM
There cannot be a discussion of Olber's paradox as long as you keep on using well defined physical terms incorrectly.

Brightness or specific intensity is very well explained in e.g. Rybicki & Lightman (http://www.amazon.com/Radiative-Processes-Astrophysics-George-Rybicki/dp/0471827592/ref=sr_1_1?ie=UTF8&qid=1352215790&sr=8-1&keywords=rybicki+%26+lightman) "radiative processes in astrophysics" chapter 1 page 3, to give equation (1.2)

dE_{\nu} = I_{\nu} dA dt d\Omega d\nu

where I_{\nu} is the specific intensity or brightness.

This is the second definition that they give in the book, after they define how the flux from a isotropic source behaves, leading to the inverse square law of flux:

F = \frac{constant}{r^2}

There is nothing here that has anything to do with exposure time. If you take into account exposure time then you are talking about received energy. Whether or not the picture that you take will be called "brighter" not as much a misnomer but describes that that picture will reflect more light than a picture taken with a shorter exposure time.

Please look up Rybicki & Lightman, it is a wonderful book, there are even pdf copies haunting the web.

madus
2012-Nov-06, 03:52 PM
Your reasoning, assumptions and conclusions are all wrong. As has been explained to you, carefully and in great detail multiple times.


Your reasoning, assumptions and conclusions are all wrong. As has been explained to you, carefully and in great detail multiple times.

- 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. What you 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 a difference between two bright spots and four less bright dots.


That is my point. No one disagreed or even commented on that.

Shaula
2012-Nov-06, 03:56 PM
Basically you are taking an imprecisely worded form of Olbers' paradox and arguing when people point out that the original paradox, as was worked out and solved, was about the intensity of starlight received from an eternal, infinite, static universe. What you are discussing is not Olbers' paradox but something leading on from it. I think you are trying to show that for non-ideal sensors you can try to sidestep the paradox by using a short exposure time, effectively thresholding the image to produce artificial dark areas. However what you are missing is that in an infinite universe there are loads of unresolved sources which will integrate up over infinite distance to give infinite flux again. To get around this you are proposing that a short exposure would mean that statistically there would be too few photons arriving from distant sources in a given time so your sensor would not see their contribution? Or assuming zero pixel size.

Shaula
2012-Nov-06, 03:58 PM
- 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. What you 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 a difference between two bright spots and four less bright dots.

Mainly because it is not really relevant? How do you believe that this solves the paradox? As you go to infinity the angular separation of the dots tends to zero. They all merge and when you integrate up ... Infinite flux again.

Strange
2012-Nov-06, 03:58 PM
That is my point. No one disagreed or even commented on that.

One of your false assumptions is that this is somehow relevant. (As has been repeatedly pointed out.)

madus
2012-Nov-06, 04:06 PM
There cannot be a discussion of Olber's paradox as long as you keep on using well defined physical terms incorrectly.

Brightness or specific intensity is very well explained in e.g. Rybicki & Lightman (http://www.amazon.com/Radiative-Processes-Astrophysics-George-Rybicki/dp/0471827592/ref=sr_1_1?ie=UTF8&qid=1352215790&sr=8-1&keywords=rybicki+%26+lightman) "radiative processes in astrophysics" chapter 1 page 3, to give equation (1.2)

dE_{\nu} = I_{\nu} dA dt d\Omega d\nu

where I_{\nu} is the specific intensity or brightness.

This is the second definition that they give in the book, after they define how the flux from a isotropic source behaves, leading to the inverse square law of flux:

F = \frac{constant}{r^2}

There is nothing here that has anything to do with exposure time. If you take into account exposure time then you are talking about received energy. Whether or not the picture that you take will be called "brighter" not as much a misnomer but describes that that picture will reflect more light than a picture taken with a shorter exposure time.

Please look up Rybicki & Lightman, it is a wonderful book, there are even pdf copies haunting the web.

Of course I am talking about received energy, that's what Olbers' paradox is about.

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.

cjameshuff
2012-Nov-06, 04:10 PM
I think the paradox is about brightness, so I propose that's what we should calculate, where brightness of each pixel would be defined by the number of photons that hits it. Sounds good?

No. We are talking about the rate at which photons arrive per unit of area. It has nothing to do with exposure time or sensor area.



I know about your opinion, you already told me about it. I also know no one has done it by modeling sensor surface and light as photons, which is the only correct way. You said before we'll get the same result, so what's the matter now? I ask you again, do you wanna know what result is going to come out, or do you refuse to look through my telescope?

I know what result is going to come out, because it's quite trivial to show that the sensor is irrelevant, and I've already done so in this thread. You have nothing new.



Brightness of the image produced by either camera or human eyes does depend on exposure time, it's well tested and confirmed fact, everyday experience and common knowledge. I also explained how insufficient resolution will produce error artifact manifesting as increased brightness. This error is all you get when you ignore image resolution, so naturally you get result indicating uniform brightness.

Irrelevant. The exposure time has to be taken into account to obtain a measurement of actual brightness, and once again, the same results are obtained from any source of a given brightness.

Your claimed error due to image resolution is simply not physically real, it's a nonsensical result from an apparent misunderstanding of power, energy, and area. It does not exist. Your "solution" to Olber's paradox requires that you also predict that the sun will appear as black as the night sky. This is testable...it doesn't.



If your image resolution is insufficient, like 1x1 as in the way you treat it

Wrong. I treated the issue both with finite and with infinite resolution.

Your argument is fundamentally flawed. Will you be addressing the reasons I gave for why, or simply repeating the same physically incorrect claims?



Luminance and radiance are terms that describe amount of light that object emits. Brightness refers to sensed or "captured" amount of light.

And people will sometimes use simplified terminology when talking to someone who clearly lacks knowledge and experience in a field, as you do. Brightness is radiant intensity or irradiance depending on context, as far as Olber's paradox is concerned. As has been described before in this thread and detailed in mathematical analysis, so the terminology usage was not ambiguous, quit pretending to misunderstand it.

madus
2012-Nov-06, 04:11 PM
One of your false assumptions is that this is somehow relevant. (As has been repeatedly pointed out.)

It is relevant because that is what human eyes or camera will see. You seem to think Olbers' paradox is about emitted light, not perceived light, is that right?

Strange
2012-Nov-06, 04:14 PM
It is relevant because that is what human eyes or camera will see. You seem to think Olbers' paradox is about emitted light, not perceived light, is that right?

If anything, it is about received light. Perception has nothing to do with it.

But is it worth answering this. You have had exactly the same thing explained multiple times, in multiples wyas, by multiple people. But you just keep repeating the same false argument. It is getting a bit pointless.

madus
2012-Nov-06, 04:20 PM
No. We are talking about the rate at which photons arrive per unit of area. It has nothing to do with exposure time or sensor area.

I know what result is going to come out, because it's quite trivial to show that the sensor is irrelevant, and I've already done so in this thread. You have nothing new.


You missed the most important part that actually address the result:

- 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. 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 a difference between two bright spots and four less bright dots.

madus
2012-Nov-06, 04:24 PM
If anything, it is about received light. Perception has nothing to do with it.


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.

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.

cjameshuff
2012-Nov-06, 04:26 PM
- 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. 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 a difference between two bright spots and four less bright dots.

I didn't "miss" it at all, it's exactly I've explained to you over and over. And no, there isn't a difference: (4)*(1/4) = (2)*(1/2)
The same holds for the plate as a whole or for any finite region of it. By your argument, cutting a plate in half would make the exposed image half as bright.

madus
2012-Nov-06, 04:39 PM
I didn't "miss" it at all, it's exactly I've explained to you over and over. And no, there isn't a difference: (4)*(1/4) = (2)*(1/2)
The same holds for the plate as a whole or for any finite region of it. By your argument, cutting a plate in half would make the exposed image half as bright.

What about all the "black" around those dots? Sum of intensities is not what you see or capture with camera, it spreads out over each "dot". What you see is 2-dimensional image and each dot has it's own spatial location.


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.

cjameshuff
2012-Nov-06, 04:54 PM
What about all the "black" around those dots?

In Olber's paradox, there isn't any. Just the stars in the background. Every line of sight ends on the surface of a star.



Sum of intensities is not what you see or capture with camera.

Actually, yes, it is. The only thing that matters is the total light landing on each sensor site. The source of that light is irrelevant.

caveman1917
2012-Nov-06, 05:17 PM
madus i don't understand why you're dragging brightness, apparent magnitude, exposure or anything else into it. As cjameshuff said it is quite simple, every line of sight asymptotically almost surely ends on the surface of a star, ergo the night sky is fully lit.

Are you perhaps considering the stars as point sources? Because in that case you would indeed get the wrong result that the night sky is black.

cjameshuff
2012-Nov-06, 06:21 PM
Are you perhaps considering the stars as point sources? Because in that case you would indeed get the wrong result that the night sky is black.

Actually, considering the stars as point sources gives the result that the sky is infinitely bright, rather than the brightness of a star's surface, as no stars are occluded by nearer stars. This was pointed out in the page Tensor linked way at the beginning.

caveman1917
2012-Nov-06, 07:09 PM
Actually, considering the stars as point sources gives the result that the sky is infinitely bright, rather than the brightness of a star's surface, as no stars are occluded by nearer stars. This was pointed out in the page Tensor linked way at the beginning.

Interesting, i'm getting the result that the brightness of the sky is zero.

Let's take our shell of finite neglegible thickness at radius r containing n point sources. The chance that a line of sight misses those point sources is 1 (the set of point sources in the set of all points on our shell has measure zero). Taking the limit over all consecutive shells doesn't change this, since (even without occlusion) the set of all point sources is countable whereas the set of all lines of sights is uncountable, therefor any light of sight almost never crosses a star. It is true that the energy received is infinite, yet it is contained in only a countable subset of points, and thus taking the integral over the entire sky (and noting that this subset has measure zero and thus doesn't influence this integral), the sky has brightness zero.

cjameshuff
2012-Nov-06, 08:22 PM
It is true that the energy received is infinite, yet it is contained in only a countable subset of points, and thus taking the integral over the entire sky (and noting that this subset has measure zero and thus doesn't influence this integral), the sky has brightness zero.

Okay, considering lines of sight to be zero width rays and considering the sources to be points leads to that result. But any finite fraction of the sky, any finite solid angle, will have an infinite number of such bright points. I think you've just proven that there's limits to how far you can simplify the problem.

Finite view angle and finite sources: sky is uniformly the surface brightness of a source.
Zero view angle and finite sources: sky is uniformly the surface brightness of a source.
Finite view angle and point sources: sky is infinitely bright, as an infinite number of sources contribute to each field of view.
Zero view angle and point sources: sky is dark, as probability of a source contributing is infinitesimal.

madus' reasoning is pretty obscure, but seems rooted more in a fundamental misunderstanding of light, inverse square falloff, basic geometry, etc. He seems to think that if you divide a sensor in half, that because each half receives half the total power, that each half measures half the intensity, not accounting for the fact that each half spreads that received power over half the area. The same basic mistake he made early on when he failed to account for the increasing number of stars with distance.

caveman1917
2012-Nov-06, 11:13 PM
I think you've just proven that there's limits to how far you can simplify the problem.

Yes i did not intend this to be a serious solution, but since madus was complaining about "all the black dots around the bright dots" i thought he might have been using that line of reasoning in considering point sources.

I specifically wouldn't consider it a serious argument since it represents a (removable) discontinuity in the brightness as a function of viewing angle or source size,

\lim_{\theta \to 0} I(\theta) \neq I(0)

where the LHS is infinite (or surface brightness in the case of \theta representing source size) and the RHS is zero. Whenever you come across something like that in a physical model it's a good indicator that you've hit an unphysical mathematical curiosity, and the physical solution is probably in removing the discontinuity by redefining the function value as the limit.

Hornblower
2012-Nov-07, 03:16 PM
Interesting, i'm getting the result that the brightness of the sky is zero.

Let's take our shell of finite neglegible thickness at radius r containing n point sources. The chance that a line of sight misses those point sources is 1 (the set of point sources in the set of all points on our shell has measure zero). Taking the limit over all consecutive shells doesn't change this, since (even without occlusion) the set of all point sources is countable whereas the set of all lines of sights is uncountable, therefor any light of sight almost never crosses a star. It is true that the energy received is infinite, yet it is contained in only a countable subset of points, and thus taking the integral over the entire sky (and noting that this subset has measure zero and thus doesn't influence this integral), the sky has brightness zero.


Okay, considering lines of sight to be zero width rays and considering the sources to be points leads to that result. But any finite fraction of the sky, any finite solid angle, will have an infinite number of such bright points. I think you've just proven that there's limits to how far you can simplify the problem.

Finite view angle and finite sources: sky is uniformly the surface brightness of a source.
Zero view angle and finite sources: sky is uniformly the surface brightness of a source.
Finite view angle and point sources: sky is infinitely bright, as an infinite number of sources contribute to each field of view.
Zero view angle and point sources: sky is dark, as probability of a source contributing is infinitesimal.

madus' reasoning is pretty obscure, but seems rooted more in a fundamental misunderstanding of light, inverse square falloff, basic geometry, etc. He seems to think that if you divide a sensor in half, that because each half receives half the total power, that each half measures half the intensity, not accounting for the fact that each half spreads that received power over half the area. The same basic mistake he made early on when he failed to account for the increasing number of stars with distance.


Yes i did not intend this to be a serious solution, but since madus was complaining about "all the black dots around the bright dots" i thought he might have been using that line of reasoning in considering point sources.

I specifically wouldn't consider it a serious argument since it represents a (removable) discontinuity in the brightness as a function of viewing angle or source size,



where the LHS is infinite (or surface brightness in the case of representing source size) and the RHS is zero. Whenever you come across something like that in a physical model it's a good indicator that you've hit an unphysical mathematical curiosity, and the physical solution is probably in removing the discontinuity by redefining the function value as the limit.
This appears to show some pitfalls that can happen in a curious combination of overcomplicating an exercise and then muffing the math in a subsequent simplification. I have not analyzed the math thoroughly but it may have been somehow getting a result of zero when it should have been indeterminate.

Jeff Root
2012-Nov-07, 04:45 PM
A result of infinite or zero surface brightness pretty obviously
means that you can't call the size of something "zero" when
the size is a factor in the quantity you are trying to model.

I don't see that the result would be indeterminate. The angular
size of distant stars is always greater than zero, so the sky will
be completely covered by a finite number of them.

One problem that didn't come up explicitly is that different
stars have different surface brightnesses. If the majority of
stars have lower surface brightness than the Sun, then the
average surface brightness of the sky would be less than that
of the Sun. But, just like with dust grains, the surfaces of
cooler stars would be heated so that they reached equilibrium
with hotter stars, which would remain brighter spots in a
bright sky. Olbers' paradox was formulated with the naive
assumption that all stars are the same as the Sun.

-- Jeff, in Minneapolis

cjameshuff
2012-Nov-07, 05:12 PM
Olbers' paradox was formulated with the naive
assumption that all stars are the same as the Sun.

I'd rather state this as an observation that they are clearly now bright enough to shine, which for the purposes of the paradox is all that matters.

Hornblower
2012-Nov-07, 07:46 PM
I just did some thought exercise calculations, using a cosmic model consisting of stars like the Sun about 1000 lightyears apart throughout non-expanding space. I used that as an average for what we actually get from galaxies, and assumed that density by having the stars about 100 times as far apart as they are in this part of our galaxy. That is approximately the same ratio as that of intergalactic hydrogen to local interstellar hydrogen. Integrating that out to a radius of 10 billion lightyears gave me less than a millionth of the amount of light we get from the Sun, and it would be spread out over the whole sky. It would roughly resemble the skyglow we get from the full Moon.

I realize this might not be an accurate estimate of the total starlight in the cosmos, but even if the stars were as abundant as in our part of the galaxy, the total light from this non-expanding model would be less than what we get from the Sun, and again it is spread out over the whole sky. The surface brightness would be on the order of magnitude of our daytime sky. Clearly Olbers was envisioning a vastly larger radius to get the predicted surface brightness up to that of the Sun.

glappkaeft
2012-Nov-07, 08:34 PM
Clearly Olbers was envisioning a vastly larger radius to get the predicted surface brightness up to that of the Sun.

He specifically assumed a infinitely old, infinitely large, static universe with an infinite number of stars at least somewhat evenly distributed.

caveman1917
2012-Nov-07, 09:59 PM
This appears to show some pitfalls that can happen in a curious combination of overcomplicating an exercise and then muffing the math in a subsequent simplification. I have not analyzed the math thoroughly but it may have been somehow getting a result of zero when it should have been indeterminate.

The math is fine, the result is zero and not indeterminate. Specifically, when f(x) = 0 almost everywhere in a region R then \int_R f = 0

It's just not a very physical solution, but it is the only one i could think of that would reproduce madus' result of a black sky.

TooMany
2012-Nov-08, 12:34 AM
The Moon re-radiates a lot of what falls on it (50-60% IIRC).

I think you mean "reflects" rather than "re-radiates". The Moon re-radiates all of the radiation that falls on it (otherwise it would heat up).

glappkaeft
2012-Nov-08, 01:23 AM
Also the moon only reflects 12% (the dark gray of worn asphalt) of the light that hits it.

Hornblower
2012-Nov-08, 02:23 AM
He specifically assumed a infinitely old, infinitely large, static universe with an infinite number of stars at least somewhat evenly distributed.

Yes, I know that. My point was to show how vast a finite expanse of star-filled space would be needed to come close to that light level.

Aster
2012-Nov-08, 06:53 PM
I just did some thought exercise calculations, using a cosmic model consisting of stars like the Sun about 1000 lightyears apart throughout non-expanding space. I used that as an average for what we actually get from galaxies, and assumed that density by having the stars about 100 times as far apart as they are in this part of our galaxy. That is approximately the same ratio as that of intergalactic hydrogen to local interstellar hydrogen. Integrating that out to a radius of 10 billion lightyears gave me less than a millionth of the amount of light we get from the Sun, and it would be spread out over the whole sky. It would roughly resemble the skyglow we get from the full Moon.

I realize this might not be an accurate estimate of the total starlight in the cosmos, but even if the stars were as abundant as in our part of the galaxy, the total light from this non-expanding model would be less than what we get from the Sun, and again it is spread out over the whole sky. The surface brightness would be on the order of magnitude of our daytime sky. Clearly Olbers was envisioning a vastly larger radius to get the predicted surface brightness up to that of the Sun.

The brightness does not depends on the radius.

Radiation energy density:
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/raddens.html

u = Energy / Volume = a T^4

Maximal energy of radiation, brightness and temperature,
will be when all the mass will be converted to radiation energy:
u = ro * c^2 = aT^4
ro - mass density of the universe.

The density is probably around: u = 1e-27 kg/m^3

T = (ro*c^2/a)^1/4 = 18.5 K, maximal temperature, regardless of the radius, finiteness, of the Universe.

Hornblower
2012-Nov-08, 07:55 PM
The brightness does not depends on the radius.

Radiation energy density:
http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/raddens.html

u = Energy / Volume = a T^4

Maximal energy of radiation, brightness and temperature,
will be when all the mass will be converted to radiation energy:
u = ro * c^2 = aT^4
ro - mass density of the universe.

The density is probably around: u = 1e-27 kg/m^3

T = (ro*c^2/a)^1/4 = 18.5 K, maximal temperature, regardless of the radius, finiteness, of the Universe.

Your exercise may be accurate for whatever it was intended for, but it does not apply to my thought exercise model, which if extended to infinity is Olbers' model. I stand by my findings, and by the other posts in this thread which in my opinion are correct.

Aster
2012-Nov-08, 08:39 PM
For temperature T = 6000K we get radiation energy density:
u = aT^4 = 1 J/m^3

in such universe mass density is: ro = u/c^2 =~ 10^-17 kg/m^3

1e-17 / 1e-27 = 1e+10 times too big density of the mass-energy.

(10^10)^1/3 = 4640 lower averrage distance between galaxes.

Now the average distance is probably about 1Mpc ~ 3 mln ly:
3 mln / 4640 = 646 ly only!
A stone's throw to Andromeda...

tusenfem
2012-Nov-09, 07:58 AM
Aster your calculations have absolutely nothing to do with the topic and maybe not even with actual physics.
Please refrain from making random calculation posts.

Jeff Root
2012-Nov-12, 03:05 PM
You seem to think Olbers' paradox is about emitted light, not
perceived light, is that right?
No.

We think it is about the radiant intensity of light in space.

Olbers paradox is the question of why, if the Universe is
infinite in extent, the sky is not completely filled with the
surfaces of stars similar in radiant intensity to the surface
of the Sun.

It must be filled that way if the Universe is infinite in extent,
infinitely old, and has always been pretty much the same
everywhere. Every line of sight would end on the surface
of a star, so every part of the sky would be as bright as the
surface of a star.

Dimming with distance cannot explain the darkness of the
sky, because surface brightness does not decrease with
increased distance. If you double your distance from a star,
the rate light reaches you from it is reduced to 1/4 because
the visible area of the star is reduced to 1/4. The star's
surface brightness is unchanged.

In Olbers' paradox, the reduced visible area of the star is
exactly made up for by more distant stars filling in the
background.

To express that another way...

Consider an infinite flat wall as bright as the surface of the Sun.
Nomatter how close you are to the wall, or how far away, the
wall will always appear as bright as the surface of the Sun. The
same amount of light will always be reaching you from the wall,
nomatter what your distance from it. When you are close to
the wall, you will receive a lot of light from a small area. When
you are far away, you will receive less light from each small area,
but the total area that you receive light from will be larger, in
exactly the same proportion. You will always be very hot.

Olbers' paradox is similar to the glowing interior of a sphere
rather than a glowing infinite, flat wall. The intensity of light is
the same everywhere inside the sphere, nomatter how big the
sphere is. Given that the interior surface of the sphere has the
surface brightness of the Sun, it would be very hot everywhere
inside the sphere. As hot as the Sun's surface. Everywhere.

-- Jeff, in Minneapolis

Aster
2012-Nov-12, 03:46 PM
Consider an infinite flat wall as bright as the surface of the Sun.
Nomatter how close you are to the wall, or how far away, the
wall will always appear as bright as the surface of the Sun. The
same amount of light will always be reaching you from the wall,
nomatter what your distance from it.

The wall can't be bright infinite time, because there is no inexhaustible energy sources.

Time: T = E/P
E - energy, P - power

E and P may be infinite or not, but a limit T = E/P is always finite.

cjameshuff
2012-Nov-12, 04:01 PM
The wall can't be bright infinite time, because there is no inexhaustible energy sources.

Your objection is simply not relevant to Olbers' paradox, and is incorrect anyway: there are no infinite energy sinks either. After infinite time, a static universe will be at uniform temperature. It being bright in the visible range is only a matter of that temperature being high enough.

Aster
2012-Nov-12, 04:43 PM
You permanently get lost in these infinities.

The temperature of environment, space, or something other,
is always lower than the temperature of the sources,
without excluding the case of complete exhaustion of any sources!

Furthermore, two bodies with the same temperature, as stars or galaxies, can't heat up each other.

Jeff Root
2012-Nov-12, 05:26 PM
Aster,

If the entire sky has a surface brightness equal to that
of the surfaces of stars, then all dust, asteroids, and
planets will be heated to a temperature equal to the
mean temperature of the star surfaces, which is much,
much cooler than the interiors of the stars, where the
energy originates.

Stars with higher surface temperature will heat the
surfaces of stars that have lower surface temperature,
until the temperatures are at equilibrium. The stars
will not end up with the same surface temperature,
but they will all be hotter than if they were radiating
into empty space.

If stars with the same surface temperature are close
together, they will heat each other's surfaces, because
the energy cannot escape. The interiors of the stars,
which are much hotter than the surfaces, are largely
unaffected.

-- Jeff, in Minneapolis

Aster
2012-Nov-12, 06:13 PM
This reasoning adds nothing new.
You redefined the sources only, locating them under a surface of the stars:
the temperature was increased, and the surface was decreased, the stream of energy remains intact.

You can increase the stream of energy too, but this still nothing changes.

Hornblower
2012-Nov-12, 06:41 PM
Olbers' paradox is part of a thought exercise of two or more centuries ago, before anything was known about how stars generate light. Let's not go overboard in applying modern physics to it.

I stand by my previous findings for a hypothetical static and eternal set of stars distributed evenly through space. If the radius is finite at a few billion lightyears, the resulting massed light reaching us will be rather dim. As we increase the radius of this set without limit, while keeping the density of stars the same throughout, the brightness of the massed light will increase without limit as has been shown mathematically more than once. If anyone here still disagrees, please explain why in appropriate mathematical detail, or refer us to previous posts if the answer is in one. I am not going to waste my time and effort searching this long thread.

Jeff Root
2012-Nov-12, 07:24 PM
Hornblower,

You don't really mean that the brightness of the sky would
increase without limit, do you? You mean that it would
increase until the entire sky was as bright as the stars the
light came from. The notion that it could increase without
limit depends on the stars having zero area on the sky but
finite radiant intensity -- which is obviously wrong.

The reverse notion, raised by caveman, that the sky would
be black, depends on the stars having zero radiant intensity
because of their zero area -- which is also obviously wrong.
Nomatter how far away a star might be, it wouldn't have
zero area on the sky, and that makes the difference.

-- Jeff, in Minneapolis

Aster
2012-Nov-12, 07:26 PM
I stand by my previous findings for a hypothetical static and eternal set of stars distributed evenly through space. If the radius is finite at a few billion lightyears, the resulting massed light reaching us will be rather dim. As we increase the radius of this set without limit, while keeping the density of stars the same throughout, the brightness of the massed light will increase without limit as has been shown mathematically more than once. If anyone here still disagrees, please explain why in appropriate mathematical detail, or refer us to previous posts if the answer is in one. I am not going to waste my time and effort searching this long thread.

Impossible.
Probably in this simple model of radiation propagation in straight lines,
we should treat stars as any other obstacle for the light.

When it is snowing, the visibility decreases - there is a maximum distance,
the horizon of visibility, which depends only on the sizes and density of the snow flakes.

And that means that nothing comes to us from outside of the horizon,
thus increasing the radius to infinity does not change anything here.

Hornblower
2012-Nov-12, 09:43 PM
Hornblower,

You don't really mean that the brightness of the sky would
increase without limit, do you? You mean that it would
increase until the entire sky was as bright as the stars the
light came from. The notion that it could increase without
limit depends on the stars having zero area on the sky but
finite radiant intensity -- which is obviously wrong.

The reverse notion, raised by caveman, that the sky would
be black, depends on the stars having zero radiant intensity
because of their zero area -- which is also obviously wrong.
Nomatter how far away a star might be, it wouldn't have
zero area on the sky, and that makes the difference.

-- Jeff, in Minneapolis

Impossible.
Probably in this simple model of radiation propagation in straight lines,
we should treat stars as any other obstacle for the light.

When it is snowing, the visibility decreases - there is a maximum distance,
the horizon of visibility, which depends only on the sizes and density of the snow flakes.

And that means that nothing comes to us from outside of the horizon,
thus increasing the radius to infinity does not change anything here.

Here is why I think the brightness should increase without limit in this thought exercise. When a foreground star intercepts light from one behind it, the energy in that light does not go away. It should heat the star above and beyond the heating that is caused by the star's internal source, thus making it brighter.

Of course this is a purely academic curiosity, as the actual universe is very different from the hypothetical one that yielded this paradox.

Aster
2012-Nov-13, 02:06 AM
Here is why I think the brightness should increase without limit in this thought exercise. When a foreground star intercepts light from one behind it, the energy in that light does not go away. It should heat the star above and beyond the heating that is caused by the star's internal source, thus making it brighter.

It's incorrect again.
There is no special region in space around us.

The same amount of energy escapes from any region, what enters inside the region.

Jeff Root
2012-Nov-13, 12:53 PM
Here is why I think the brightness should increase without limit
in this thought exercise. When a foreground star intercepts light
from one behind it, the energy in that light does not go away.
It should heat the star above and beyond the heating that is
caused by the star's internal source, thus making it brighter.
Ah! Your're right. I was thrown by the weirdness of the
properties of this hypothetical universe. The stars not only
keep pouring out energy from their interiors at the same rate
forever into the future, but have been doing so forever in the
past, so they must always be at equilibrium, but there can
be no equilibrium because the amount of energy is constantly
increasing -- even though it has always been infinite.

Having more energy isn't always a good thing.

-- Jeff, in Minneapolis

Aster
2012-Nov-13, 03:19 PM
Olbers did not assume the infinite energy in every star.

Hornblower
2012-Nov-13, 03:46 PM
Olbers did not assume the infinite energy in every star.

But I did in my model, or else continually replaced exhausted stars with new ones, somewhat along the lines of Fred Hoyle's continuous creation theory, but without the expansion that is included in his model. The idea is to keep the overall averaged characteristics of a static cosmos unchanged forever, as in the archaic models offered by Olbers and his contemporaries and predecessors. As I said before, this is purely an academic curiosity of an exercise, not intended to be an accurate model of the cosmos as we now envision it in theory.

Madus tried to have us believe that he resolved the paradox with a geometrically correct mathematical analysis. It appears to me and many others here that his concept and execution were faulty, and that he did not know enough about some rather elementary points of physics to understand the responses to his posts.

antoniseb
2012-Nov-13, 04:06 PM
Olbers did not assume the infinite energy in every star.
In Olbers Paradox, the assumptions had nothing to do with Conservation of Energy or sources of energy. It was simply shown that each shell in a series of concentric sphere surfaces of equal thickness should supply the same amount of solid angle as seen by us supplying sun-like brightness, and that if the universe is big enough in age and extent, and things have never changed, then the sum of all the contributions of each shell looking out far enough will equal or exceed the solid-angle of the sky.

Now in this thread a few things have been highlighted that Olbers didn't express. One is that (as Aster pointed out with his nice Snowfall analogy), the universe doesn't have to be infinite for this to work... it just has to be large enough and old enough and not expanding (or have any other kind of cosmological dimming at distance). It is also the case that he didn't consider the possible energy sources that could keep the whole universe brightly lit for that long. They don't exist, but we know that now... not when Olbers first asked the question.

Getting back to Aster's quote above... implicit in Olbers Paradox is that the universe doesn't run out of energy over a long enough time to get to light from every possible direction to us. Not infinite. No assumption that stars last for ever, but somehow each unit of large enough space radiates the same amount of light for all time. Super-long-lived stars? Is it steady influx of new stars to replace dead ones? Olbers doesn't care.

Aster
2012-Nov-13, 07:47 PM
No assumption that stars last for ever, but somehow each unit of large enough space radiates the same amount of light for all time. Super-long-lived stars? Is it steady influx of new stars to replace dead ones? Olbers doesn't care.


Yes. It's probably a key:
if number of stars is constant, then inflow of energy to any star must be equal to the radiated energy by the star.

This leads directly to the Olbers' paradox.

But here is a very nice compliance:

a radius of visibility, influence, is around 10 billions of years x c,
and it's an average star's life span.

I think this explains all.

madus
2012-Nov-21, 10:00 PM
I didn't "miss" it at all, it's exactly I've explained to you over and over. And no, there isn't a difference: (4)*(1/4) = (2)*(1/2)
The same holds for the plate as a whole or for any finite region of it. By your argument, cutting a plate in half would make the exposed image half as bright.

I am sorry I could not reply sooner. There was some problem with the forum and it kept telling me I was banned. Imagine that!

http://img94.imageshack.us/img94/4757/stars4g.jpg

So, anyway... there is your problem, you sum up all the intensity into a single value, and such result will always indicate increased and uniform brightness. If you sum up all the intensity into a single value (1x1 image) then even with these two images above you would get result that indicates they are bright and uniform. So you see now that math is simply wrong and unable to tell you anything about actual brightness and its distribution.

Without modeling resolution you don't have pixels which brightness you need to compare in order to deduce whether their brightness is the same or not across the sensor surface area receiving the light. You should not expect to be able to deduce whether the brightness is uniform or not if you sum all the intensity. Of course the result will be uniform when your result can only hold one value. Single number is uniform just by being a single value. You need an array, a matrix, that can hold multiple values related to spatial distribution of received light if you want to know what you would ACTUALLY SEE.

ShinAce
2012-Nov-21, 11:05 PM
Let's say I take a coin and flip it 10 times. I have no idea what the 4th flip came out as, I can only give a 50/50 guess. However, I can correctly deduce that the most likely total of heads is 5. If I flip the coin a 2 million times, I will get 1 million heads +/- a small difference. The fact that I actually get 999,996 heads and 1,000,004 tails does not mean the coin isn't a perfect 50/50.

What we're saying is that as you add up the effect of more and more shells, it doesn't matter what each shell is doing. Since we're using a ridiculously large number of shells, the answer observed will match extremely closely the statistical prediction.

Why would someone bother with small sample statistics when the problem uses large number statistics? Even if you take the trouble to do it your way, you will still find that the final answer (after summing all shells) is a uniform brightness. If you insist on doing it that way, go ahead. Once you have done so, you will see that it comes out to the same answer.

The only difference is that you will be able to say that the brightness of each pixel is the same plus or minus a very very small intensity. You should be able to derive the standard deviation of the expected pixel brightness. However, if you integrate an infinite number of shells, the standard deviation, by definition, will be zero.

So...your approach will allow you to find the standard deviation of a pixel's brightness, which is predicted to be 0 when integrated to infinity. If you do not integrate to infinity, you are no longer working with Olber's paradox. You might as well call it Madus' non-paradox(or even Madus' bounded universe solution to astronomical light bucket statistics).

madus
2012-Nov-21, 11:50 PM
Let's say I take a coin and flip it 10 times. I have no idea what the 4th flip came out as, I can only give a 50/50 guess. However, I can correctly deduce that the most likely total of heads is 5. If I flip the coin a 2 million times, I will get 1 million heads +/- a small difference. The fact that I actually get 999,996 heads and 1,000,004 tails does not mean the coin isn't a perfect 50/50.

What we're saying is that as you add up the effect of more and more shells, it doesn't matter what each shell is doing. Since we're using a ridiculously large number of shells, the answer observed will match extremely closely the statistical prediction.


It matters because stars in closer shell occlude stars in further away shells, and the stars in further away shell can only get dimmer. Look at these two images below, they represent first two shells. Can any stars behind those two shells make the stars in the second shell look as bright as the stars in the first shell?

http://img94.imageshack.us/img94/4757/stars4g.jpg




Why would someone bother with small sample statistics when the problem uses large number statistics?


Because the calculation behind that statistics is posed wrong. Your eyes do not have one pixel resolution. If human eyes did have single pixel resolution then you would indeed see the night sky is uniformly bright. Anything you would see with such eyes would appear uniformly bright, you would be blind.




Even if you take the trouble to do it your way, you will still find that the final answer (after summing all shells) is a uniform brightness. If you insist on doing it that way, go ahead. Once you have done so, you will see that it comes out to the same answer.


I've done it, here it is:

http://img94.imageshack.us/img94/4757/stars4g.jpg

These are first two shells, you don't see any other stars in further away shells as they are too dim, as bright as black. However, if you increase exposure time, then you would see further away stars as well, but proportionally the closer stars would over-expose the image and they could never appear to have the same brightness, just like in the real world.

ShinAce
2012-Nov-22, 12:00 AM
I'm sorry, but generating two photos to represent two shells is not a solution. You can claim to have worked through it, but you've limited your scope and are simply extending the results without generalizing them.

By your logic, all galaxies should be invisible since stars at that distance are too dim to be seen individually. The fact that a collection of billions of such stars is indeed visible without being able to see single stars seems to go against your 'brightness' argument.

Not to mention that eyes having single pixel resolution makes no sense. Our 'pixel' size is defined as an angle or arc. But the brightness of a shell for a given angle of arc is constant. How does a completely new and alien definition of human sight invalidate statistics?

madus
2012-Nov-22, 12:12 AM
I'm sorry, but generating two photos to represent two shells is not a solution. You can claim to have worked through it, but you've limited your scope and are simply extending the results without generalizing them.


Can any stars behind those two shells make the stars in the second shell look as bright as the stars in the first shell?




By your logic, all galaxies should be invisible since stars at that distance are too dim to be seen individually. The fact that a collection of billions of such stars is indeed visible without being able to see single stars seems to go against your 'brightness' argument.


No, it's just that for simplicity we take all of the stars are the same size and of the same absolute brightness. In reality though further away stars can appear brighter than closer stars, due to their larger size or greater absolute brightness.




Not to mention that eyes having single pixel resolution makes no sense. Our 'pixel' size is defined as an angle or arc. How does a completely new and alien definition of human sight invalidate statistics?

Of course single pixel resolution eyes do not make sense, but that's the logic used in the original treatment of the paradox. That's what single value result of those equations practically means.

Strange
2012-Nov-22, 12:20 AM
Of course single pixel resolution eyes do not make sense, but that's the logic used in the original treatment of the paradox.

No, it is just something bizarre you have invented.

ShinAce
2012-Nov-22, 12:22 AM
Can any stars behind those two shells make the stars in the second shell look as bright as the stars in the first shell?


Nobody said that the intensity of the stars themselves are constant from one shell to the next. The intensity of the shell is constant. They get dimmer, but multiply in numbers at the same time. We covered that. Besides, if you want to go into that level of detail, you might consider include gravitational lensing, which will amplify what we see of the background star.

Secondly, while a nearby star can very well obscure what's behind it, the area of stars versus area of the shell they're in is a ridiculously small fraction. There is just too much space between stars to bother worrying about how much space they take up. It's not significant for this purpose, especially if you integrate to infinity.

madus
2012-Nov-22, 12:25 AM
No, it is just something bizarre you have invented.

Is result of the original calculation a single number? What kind of sensor resolution do you imagine a single value represents?

madus
2012-Nov-22, 12:28 AM
Nobody said that the intensity of the stars themselves are constant from one shell to the next. The intensity of the shell is constant. They get dimmer, but multiply in numbers at the same time. We covered that. Besides, if you want to go into that level of detail, you might consider include gravitational lensing, which will amplify what we see of the background star.

Secondly, while a nearby star can very well obscure what's behind it, the area of stars versus area of the shell they're in is a ridiculously small fraction. There is just too much space between stars to bother worrying about how much space they take up. It's not significant for this purpose, especially if you integrate to infinity.

You did not address the question.

http://img94.imageshack.us/img94/4757/stars4g.jpg

These are first two shells, can any stars in further away shells make the stars in second shell appear any brighter? YES/NO

cjameshuff
2012-Nov-22, 12:29 AM
It matters because stars in closer shell occlude stars in further away shells, and the stars in further away shell can only get dimmer.

The portion of the sky covered by a star is as bright as the surface of a star. It does not matter what they occlude.



Can any stars behind those two shells make the stars in the second shell look as bright as the stars in the first shell?

No. They make the whole sky look as bright as the surface of a star.



Of course single pixel resolution eyes do not make sense, but that's the logic used in the original treatment of the paradox. That's what single value result of those equations practically means.

I've explained the many issues with your "single pixel" argument in detail. It's been quite thoroughly demolished. Are you going to just keep repeating a statement that's been demonstrated to be false, or will you actually address the problems I've pointed out?

cjameshuff
2012-Nov-22, 12:30 AM
These are first two shells, can any stars in further away shells make the stars in second shell appear any brighter? YES/NO

They don't need to be made to appear brighter than a star.

Strange
2012-Nov-22, 12:41 AM
What kind of sensor resolution do you imagine a single value represents?

Sensor resolution has nothing to do with Olber's paradox. As has been explained in great detail many times.

madus
2012-Nov-22, 12:47 AM
No. They make the whole sky look as bright as the surface of a star.


According to what logic, what optics and what physics?




I've explained the many issues with your "single pixel" argument in detail. It's been quite thoroughly demolished. Are you going to just keep repeating a statement that's been demonstrated to be false, or will you actually address the problems I've pointed out?

I've explained the issue with TOTAL INTENSITY and single value result of the original equations. Are you just keep repeating blank statements that's been demonstrated to be false, or are you going to address my point?

http://img94.imageshack.us/img94/4757/stars4g.jpg

HOW and WHY do you imagine any stars behind those two shells can make the stars in the second shell look as bright as the stars in the first shell?

madus
2012-Nov-22, 12:49 AM
Sensor resolution has nothing to do with Olber's paradox. As has been explained in great detail many times.

As I explained to you in great detail resolution has everything to do with why you see what you see the way you see it.

ShinAce
2012-Nov-22, 12:54 AM
You did not address the question.


Actually, I did. In what you just quoted, it's the first sentence of the second paragraph.

madus
2012-Nov-22, 12:58 AM
Actually, I did. In what you just quoted, it's the first sentence of the second paragraph.

- "Nobody said that the intensity of the stars themselves are constant from one shell to the next."

This sentence? Well, then we agree. The original conclusion of the paradox is that the night sky would be UNIFORMLY bright.

ShinAce
2012-Nov-22, 01:03 AM
No, not that sentence. That was the first paragraph.

The second paragraph was:

Secondly, while a nearby star can very well obscure what's behind it, the area of stars versus area of the shell they're in is a ridiculously small fraction. There is just too much space between stars to bother worrying about how much space they take up. It's not significant for this purpose, especially if you integrate to infinity.

Original conclusion? Has the conclusion changed? The sky would be uniformly bright according to Olber is still the same conclusion. It doesn't matter if there are stars hidden or not when 100% of the sky is covered with "star surface" of a given temperature. As soon as a significant portion of stars is being hidden by nearer ones, then you're getting close to completing a numerical integration of Olber's paradox. It does not go against the paradox in any way, nor does it solve the paradox into a black sky.

madus
2012-Nov-22, 01:45 AM
No, not that sentence. That was the first paragraph.

The second paragraph was:


Original conclusion? Has the conclusion changed? The sky would be uniformly bright according to Olber is still the same conclusion. It doesn't matter if there are stars hidden or not when 100% of the sky is covered with "star surface" of a given temperature. As soon as a significant portion of stars is being hidden by nearer ones, then you're getting close to completing a numerical integration of Olber's paradox. It does not go against the paradox in any way, nor does it solve the paradox into a black sky.

That does not address the question.

http://img94.imageshack.us/img94/4757/stars4g.jpg

First few shells would cover more than %50 of the sky, and if not the stars in further shells can only get dimmer. So how and why do you think any stars behind these closer shells can make the stars in, say second shell look as bright as the stars in the first shell?

cjameshuff
2012-Nov-22, 02:04 AM
According to what logic, what optics and what physics?

It's been explained over and over, from the very beginning of this thread.



I've explained the issue with TOTAL INTENSITY and single value result of the original equations.

No, you haven't. You've simply asserted that there is such a problem, and ignored all the flaws and holes people have found in your argument. Your claims have been shown to be completely at odds with basic physics, logic, and mathematics. Mindlessly repeating your claims will not change this. Nor will mockery.



HOW and WHY do you imagine any stars behind those two shells can make the stars in the second shell look as bright as the stars in the first shell?

They are already all of equal surface brightness. As has already been repeatedly explained.

Swift
2012-Nov-22, 02:11 AM
OK, we're done.

madus, this has been explained multiple times. I suspect you are just trying to advocate a non-mainstream idea. If you want to do that, then start a thread in ATM. We're done here.