Illustrating Olbers' paradox

I'm talking about real world images and real world cameras, not your hallucinations. Wake up!

Do you understand the number of visible shells would depend on exposure time and film speed?

 

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Again: up to this point, we have been discussing my extension of your simulation. There is nothing wrong with it (it does match the laws on how reality works) and it, on its own, answers the question of the thread.

I have, in fact, taken pictures similar to your image of the first shell. I could repeat the simulated extension starting with such a picture, but there would be no point because the analysis works the same way and the result is the same.

That isn't true either in the real world or in the simulation. Again: I've done the analysis that shows it. You don't believe it, but you haven't found a flaw.

You are simply staring reality in the face and refusing to accept it.

 

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My simulation is based on real world facts. My image has pixel sensitivity and my camera has exposure time. In my simulation there is time and it flows.

Your simulation is unrealistic nonsense, it has nothing to do with real world physics and logic. You will never be able to know the brightness of Olbers's paradox night sky, you don't know any actual numbers in unit time and per unit area.

Hubble Deep Field is visible to the human eye?

Each star is equally bright regardless of distance?

 

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And my extension of your simulation follows real-world rules. So we should be all good.

Of course. That's true of mine as well.

Yes it does. Again, I laid-out the analysis in detail. If there is a flaw, show it to me.

You could start from scratch to generate your simulated image if you want, but as I have shown, it doesn't matter. Here, try this:
The camera is a Meade DSI III at f7.5, with a 1 second exposure. Take a 250x250 pixel piece of the image.
The star is the sun, duplicated 10 times, and at 10 light years distance.

From that information (you can look up the specs of the camera) you can re-create your simulated image. Then I would take that image, apply the same logic/math and get the same result. If you want to do that, go for it. It is your responsibility, after all.

No and no and I never claimed either of those. From the math/logic, it should be evident that what the picture shows is one star is as bright as four stars at twice the distance, shining light on the same pixel. What this really comes down to, still, is that you don't understand the inverse square law and how it applies here.

You don't understand the logic I used to generate the simulations. You should be working harder to understand the logic. Reread it and ask specific questions where you have specific issues.

 

The flow is undefined exposure time and pixel sensitivity. In the real world you can not measure any brightness without them.

You were never supposed to do anything else but use real world physics and real world instruments.

All you need to say is how much total energy is received by one square meter surface area per one second time interval.

Your simulation?

simulation without time
is like a man without head
not animated
but completely dead

it is nice to have a head
you can smile and such
but when your head is dead
it doesn't matter much


Always wear your head on!

 

You're having trouble getting your arms around the process here, so perhaps if I listed the steps more simply/concisely, it would be a little easier for you to grasp what has happened here:

Step 1: Generate the simulated first shell. <- You did that.
Step 2: Generate the next three "visible" shells. <-You generated shell 2, I generated 3 and 4.
Step 3: Later shells are not visible individually, only collectively, and are each uniformly "bright" after about shell 8 or 10. Calculate average shell "brightness", use it to generate later collections of shells. <-I did that.

You have objected to the results of Steps 1 and 3.

Your objection to the result of Step 1 is bizarre mostly because you are arguing against yourself. But at the same time, you seem to think that the method for Step 1 actually aplies to other steps: You are confused about what actually happens in each step. Specifically, generating the first image in Step 1 requires making assumptions about the camera and exposure specs and size and distance of the stars in each shell. If you did that when you generated the image, I don't know, but since you keep bringing it up, I can only assume you didn't -- and perhaps didn't actually generate the image. But once that first image is generated, there is no further use for the deails of the camera. Each next image is generated from the previous image, not generated from scratch from the camera details.

Your objection to the result of Step 3 comes from your failure to read/understand/absorb what I did. You keep making claims (asking leading questions) about what I did that clearly demonstrate you don't know what I did. That's on you: I explained what I did and you should go back and reread it and ask honest questions about the details until you understand what happened. Through understanding what I did in Step 3, you'll come to learn how the inverse square law works, because it is integral to the process.

Your objection to being able to assemble the shells and add the brightness values is pretty disappointing since we've discussed the issue a number of times and it isn't that hard to understand. Pictures of galaxies and our naked-eye views of the Milky Way show stars that are not individually visible. The galaxies become visible as clouds because two or more stars will send light to the same pixel and their light adds-up to visibility. And that's exactly the same as what happens when you assemble the shells in Olbers' and your paradox.

If you recall, btr tried to explain to you what happens for a one-pixel detector, but you weren't interested in listening to it. As it turns-out, that's one of the more important pieces of this that you are missing. Zoom-in on any particular pixel in your 250x250 picture and you have a one-pixel detector and his analysis applies to it exactly.

 

Stop talking about the image. The image is the result. You don't get the result by looking at some arbitrary result. You get the result by calculating the other side of the equation first.

You are not getting it, you are looking in the wrong direction. Stop looking at the image and look into the sky. First use the camera, then calculate the image. How is it not clear to you that is the only one proper way to model this situation?

So logically, the first thing you need to know is how much total energy is received by one square meter surface area per one second time interval. Why is that so hard for you? Insufficient information? It doesn't compute? What's the problem?

 

I welcome you both to this thread where I debunk QM's paranormal mysteries:
http://www.sciforums.com/showthread.php?141815-Bell-s-theorem-spooky-lucky-streaks-!

That's right, it is me, the Amazing Myth Buster! But you call me MR. B. Because, amazing, I already know that I am.
- (Paraphrasing Mr. Bobinsky from "Coraline" animated film, 2009)

 

It is clear to me. But since you provided the image, you should have already done this. If you did it wrong, that's your failure, nto mine. So please tell me: Where did you get the image? Did you calculate it?

Arright, in any case, I'm already reasonably sure that the answer is that you didn't calculate it, you found it somewhere on the internet. And since you ignored the rest of my post about my method, I must conclude that either you didn't read it or you did read it, realize you're wrong, and are dodging the issue. So let me see if I can back-into how it was calculated so we can stop making this an issue (yes, I'm sure you'll just jump to something else and keep ignoring the actual analysis, but...).

Let's assume you used the same camera as I use: a Meade DSI III, which uses the Sony ExView ICX285AL CCD chip. It has a quantum efficiency of about 60% and a well depth of 16,000 electrons. What that means is that 60% of photons of a certain energy (centered on the sun's spectrum) knock-loose electrons on each pixel, up to a maximum of 16,000 electrons to saturate the image. The spectrum is centered at about 500 nm, so that yields approximately 6.6E-10 Joules of energy to saturate your entire 250x250 pixel piece of the sensor. Since we previously agreed that we could call it a 1 second exposure, that's 6.6E-10 Watts. The pixels are 6.45 um across, so that's 1.6 mm of chip or 0.000254 w/sq meter.

Now, here's where it gets a bit tricky: Your stars are a peak of only half the saturation value, but spread-out to cover more than one pixel. As it turns-out, the total "brightness" is 1000 for the first shell, which is 4x saturation if it had been focused on one pixel, but since it wasn't, all is captured. So that works out to 0.0010 w/sq m.

I use several telescopes, but the one most suited for this is a C11 (0.05 sq meters), so that's 20,700x the area of the chip, which gives us now 4.8E-8 w/ sq meter that the star is sending to earth.

Since the Sun is 90 million miles away and sends us 1,000 watts/ sq meter, that equates to a sun-sized star, about 2.2 light years away. Not too bad of a first estimate, considering the closest star is sun-sized and about 4.4 light years away (actually two similar stars in a binary system).

That's probably a bit on the dense side, so a more reasonable estimate of the density of our nearby stars would probably require adjustment by a factor of 100. So all we have to do there is just go back and say the exposure time was 100 seconds instead of 1 second, then we can call the distance 44 light years.

Now we have a reality-matched explanation for how you might have generated your image of the first shell.

So I went pretty fast there...I'm sure you understood very little of it, so let me know what you need explained in more detail and if you ask nicely enough, maybe I'll explain it to you. You should notice, though, that that analysis had no impact on the first image that you already produced or the final image that I already produced. We're just backing-into the specs of how you generated - or should have generated but probably didn't - the first picture.

 

That looks like a fairly incoherent stream of consciousness set of posts and I'm really not interested. If you want people to respond, I suggest gathering your thoughts together and posting more coherently.

 

It was not calculated, it is an illustration of Olbers' paradox description from Wikipedia, with arbitrary number of stars. You insisted on using "my" image but ignored my image has pixel sensitivity and was under light exposure for some non-infinite amount of time greater than zero.

That's the stuff, finally. What you call "1.6mm chip", is analogous to film "negative" in mechanical cameras? How did you arrive to 6.6E-10 Joules of energy?

What does that mean? What properties and variables are you talking about?

How does it turn out the total "brightness" is 1000 for the first shell? What equation did you use?

What point are you making? Are you calculating something, what is it? Please name the properties and variables you are talking about by their actual names used in physics and use mathematics symbols to describe the relationship between those properties and variables.

 

Basically, yes. Digital cameras use CCD or CMOS chips. You can see what they look like/read how they work here:
http://en.wikipedia.org/wiki/Charge-coupled_device

You quoted the entire passage -- which part, specifically, did you not understand? Do you not understand how to calculate the energy of a photon given its wavelength? Do you not understand how a CCD works? What quantum efficiency is?

Zoom-in on one of the stars in your picture in any photo editor. You can see they are blobs of one or two dozen pixels. Use the eyedropper function to see what the luminance value is for each pixel.

I explained that already: I added-up the brightness (luminance) value of every pixel that makes-up one star. (and that sentence was missing a word: that's for each star in the first shell, not the total light captured in the shell).

What you actually asked for is starting with camera specs, exposure length and a hypothetical structure of the universe and generating the first image. Since we already have the first image and I did all the rest of the work and don't want to discard it and start over, I did the problem backwards, starting with the first shell image and backing-into the information that generated it. The problem works the same forwards and backwards.

In any case, I'm not inclined to turn that entire thing into a thousand word technical paper for your benefit. If there are specific pieces of the process you don't understand, ask specific questions about them and I'll answer. But if you're asking me to teach you a college level course on astronomy optics and photography, you're basically acknowledging you don't have the technical knowledge required to make the claims you have made, which is my entire point here anyway. Admitting you are way out of your depth here would be a big victory for you (yes, you, not me) and a starting point for actually learning. If you made such an admission, demonstrating you are actually willing to learn, I'd probably be willing to walk you through the entire thing step by step.

 

I'm asking you to show the equation and numbers you used to get 6.6E-10 Joules of energy.

You still haven't calculated the 1st thing you need to know: total energy received. What brightness are you talking about if you still have no clue how much light impacted the chip during the exposure time?

I asked for exactly the opposite. Numbers and equations, nothing else matters.

 

Which ones do you need? All of them? Did you understand anything I wrote? Do you understand how to calculate the energy of a photon given its wavelength?

[sigh] In my opinion, watts per square meter is more relevant because it is independent of telescope/chip size, but if you want watts, you can quite easily calculate it from the information I gave you: You have watts per square meter and you have the size of the telescope. Do you honestly need me to spoon-feed you even this simplest of calculations? At some point here, you are going to need to start doing some of the work yourself.

[edit: Note that since this is an intermediate calculation, you can actually calculate it two different ways, one starting with what the detector saw and one starting by defining the properties of the star. I've shown you both.]

Edit3: actually, since the number is carried through the calc for a while, there are at least three ways. Another is via the image analysis, with chip specs.

 

1.) Energy of one 500nm photon
E= h*v/L = (6.63e-34 * 2.99e+8) / 500e-9 = 3.97e-19 J

2,) Number of photons required to fully saturate one pixel
required for white or RfW = (well depth / efficiency) * 100
RfW = 26666.67 photons

3.) Energy required to fully saturate one pixel
E(RfW) = RfW * E(500nm) = 26666.67 * 3.97e-19 J = 1.06e-14 J

4.) Energy required to fully saturate the whole image
E(white image) = E(RfW) * (250*250) = 6.63e-10 J


That's all you need to do, then you have to say nothing more. It benefits you to make sure you know what are you talking about, and it benefits me, as well as a potential gentle reader, to understand what hat did you pull those numbers from and easier verify. I expect that you, as a teacher, would know the importance and benefits of "showing your work". Don't disappoint.

Ultimately we need watts per pixel, but first we need watts pet total sensor area, thus watts per meter will do as pixel size is defined. The problem is you don't have any watts, yet. What you calculated above are not watts, it is a constant pixel property with units in joules, it does not tell you how bright the sky is. You first need to look towards the sky and see how much total energy would hit the sensor surface in one second interval of exposure time. Stars first, image second, always remember.

 

Now prepare to eat your own soup, your favorite "do you see where this goes?", and it goes as follows... Exposure time is 1 second, number of photons per second received from a star in the first shell is 100,000 photons.

Code:

 1st shell star photons received per second = 100,000 
 2nd shell star photons received per second =  25,000 
 3rd shell star photons received per second =   6,250
 4th shell star photons received per second =   1,562.5
 5th shell star photons received per second =     390.6
 6th shell star photons received per second =      97.7
 7th shell star photons received per second =      24.6
 8th shell star photons received per second =       6.2
 9th shell star photons received per second =       1.5


Do you see where this goes?

10th shell star photons received per second =       0.3
11th shell star photons received per second =       0.09
12th shell star photons received per second =       0.02
13th shell star photons received per second =       0.006
14th shell star photons received per second =       0.002
15th shell star photons received per second =       0.0004
16th shell star photons received per second =       0.00009
17th shell star photons received per second =       0.00002
18th shell star photons received per second =       0.000006
19th shell star photons received per second =       0.000002
20th shell star photons received per second =       0.0000004


Zero is unusual number...

And finally we arrive here, just like in the real world:

Please Register or Log in to view the hidden image!

 

The revelation of irrefutable truth.

Code:

 1st shell star photons received per second = 100,000 
 2nd shell star photons received per second =  25,000 
 3rd shell star photons received per second =   6,250
 4th shell star photons received per second =   1,562.5
 5th shell star photons received per second =     390.6
 6th shell star photons received per second =      97.7
 7th shell star photons received per second =      24.6
 8th shell star photons received per second =       6.2
 9th shell star photons received per second =       1.5
---------------------------------------------------------
10th shell star photons received per second =       0.3
11th shell star photons received per second =       0.09
12th shell star photons received per second =       0.02
13th shell star photons received per second =       0.006
14th shell star photons received per second =       0.002

While the stars in the first shell overexposed the sensor, the stars after the 5th shell are barely visible, if at all. And then, the 10th shell, 0.3 photons per second per star. Huh? It means the stars in the 10th shells are not only very dim, but they are as black as it gets (less than 6 photons per star), and from 70% of them we received not even a single photon. One second we receive some photons from 30% of them, but the next second it's going to be some other 30%, so less than a second frame rate (exposure time) of the human eye, makes them all practically invisible. Mystery solved.

 

[sigh] I had some hope that after calling me "teacher" in post #175 and actually starting to do calculations on your own that we might have turned a corner, but now it appears not. Quick clarification on that though:

You misunderstand how teaching/learning works. As the student, you are responsible for showing your work and as the teacher, I merely show you where you err.

Anyway, it is very interesting that you are able to re-create my claculations, while still displaying an apparent inability grasp the simple logic by which they were constucted. It implies trolling: that you are capable of more than you are giving.

Huh? You did all those other calcs and now you're hung up on how I went from Joules and seconds to Watts? Really? Or did you just not read the whole thing -- you say I don't have Watts, when I clearly do.

Why didn't you account for the fact that the shells/image/pixels contain more than one star in your calculations? Why didn't you account for the fact that you are looking at all the shells at once? These are fairly obvious parts of the puzzle, discussed multiple times.

How, precisely, did you generate that image? It does not appear to me to be a derivative of the other images -- the star arrangement is different.

 

At no point you considered the energy coming from the stars, you were only talking about sensor properties. How much total energy is received by the sensor surface per one second time interval?

I accounted for everything. It's indisputably irrefutable truth. Give it up!

My butler did it, pixel by pixel, painstakingly so I couldn't bear to watch.

 

Read the rest of the post. It's in there.

You did not calculate how many photons hit the sensor.

I'm guessing by your switch to trolling that you have figured out that you are in error.