Illustrating Olbers' paradox

Of course, but you are missing my point. I picked one atom on your retina and worked out how many photons are hitting it each second. The atom chosen was arbitrary (except that we want it to be one which is exposed to the sky), so every sky-exposed atom in your retina is subject to the same argument, so every atom on your retina receives as many photons per second as it would if the Sun filled your field of view.

(One minor correction: S should have been the solid angle of sky visible to the atom, rather than the solid angle subtended by the pupil at the atom, otherwise the effect of the lens is not taken into account. In effect I described a pinhole camera instead of a human eye. Fortunately this does not affect the conclusion as just stated.)

 

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The sensor you are referring is the one used in the earlier version of my argument, which (as I already said) I have ditched in favour of the updated version I posted later (in the portion of the linked post following the second set of quote tags). The reason for the update is that the first version, while correct, relies on infinitesimal quantities, and to properly justify those I would have to make an argument almost identical to the later version anyway (with extra steps involving taking the limit as the field of view tends to zero, and then defining a differential photon flux per solid angle of sky).

So, let's see what the equivalent statement is in the second version of the argument. Here it is, along with its justification which explicitly mentions the inverse square law (I've added some bolding):

Let the star be a distance r away. If I represents the average amount of photons per unit time we'd get from that fraction of a star's surface if it were a unit distance away, then I/r[sup]2[/sup] (here we see the inverse square law) tells us the average number of photons per unit time we'll receive from it in reality.

However, if A represents the solid angle which that fraction of the star's surface would have occupied in our sky at a unit distance away, then by basic geometry A/r[sup]2[/sup] is the solid angle it takes up in reality. So, the number of photons we receive per unit time per unit solid angle from that small patch of sky is just I/A, and we see that the distance to the star has cancelled out. The number of photons we receive per unit solid angle per unit time from the fraction of star surface does not depend on how far away the star is!​

A couple of questions:

1. Are you familiar with the notion of solid angle, or am I using a concept which has no meaning to you?

2. Assuming you are OK with solid angles, do you have any issues with the validity of the argument given in those two paragraphs?

 

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Our universe does not satisfy the assumptions of Olber's paradox, so why does this matter?

Using your suggestion of plotting single-pixel stars using additive blending, the nth layer will have K[sub]1[/sub]n[sup]2[/sup] stars in it (for some K[sub]1[/sub]), and each will have a greyscale value K[sub]2[/sub]/n[sup]2[/sup] (for some K[sub]2[/sub]). Once you get so far out that K[sub]1[/sub]n[sup]2[/sup] is enormous compared to the number of pixels in your bitmap, you will be plotting onto each pixel multiple times. In fact, if your bitmap contains M pixels, you will be plotting on each pixel an average of K[sub]1[/sub]n[sup]2[/sup]/M times, so the average greyscale value of each pixel in the nth layer will be (K[sub]1[/sub]n[sup]2[/sup]/M)(K[sub]2[/sub]/n[sup]2[/sup]), which is simply K[sub]1[/sub]K[sub]2[/sub]/M. Note that this is independent of n!

(The same average applies to the layers with small n, incidentally, as you should easily be able to see. However, the small-n layers will appear to contain individual stars separated by expanses of black, while the large-n layers will tend towards a uniform grey with some random noise on top.)

That is for one layer. If you go ahead and plot all of the layers, with additive blending, the average greyscale value of each pixel will be K[sub]1[/sub]K[sub]2[/sub]/M multiplied by the number of layers, which is infinite.

 

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I can expand on the previous post a little. Consider again just a single layer, the nth one.

The number of times a particular pixel gets plotted on is a random variable with a binomial distribution with K[sub]1[/sub]n[sup]2[/sup] trials and a probability of success per trial of 1/M.

The binomial distribution has a mean of K[sub]1[/sub]n[sup]2[/sup]/M and standard deviation of n((K[sub]1[/sub]/M)(1 - 1/M))[sup]1/2[/sup]. Bearing in mind that each plotting of a pixel increases its greyscale level by K[sub]2[/sub]/n[sup]2[/sup], The appearance of the bitmap tends towards a uniform grey with a small amount of noise on top:

\( \displaystyle \textrm{mean greyscale level} = \frac{K_1 K_2}{M} \quad;\qquad \textrm{standard deviation} = \frac{K_2}{n}\sqrt{\frac{K_1}{M}(1 - \frac{1}{M})} . \)
​

Note that the standard deviation tends to zero as n goes to infinity, i.e. the grey becomes completely uniform in this limit.

 

Uh, what? You're not serious, are you? Asking a good question is the starting point of an investigation, not the end. To end the investigation, you have to answer it (with support, of course). So please: try to answer it. I'll help you with it if you have trouble.

Come, now. you were doing so well earlier today, answering several questions in a row correctly and without evasion. Let's try to keep that good flow going. Please try to answer the questions I asked. Please try to answer your question.

 

Someone super-glued the pointer hand on your measuring scale.

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Would your atom conclude these two patches of sky are equally bright? Would its conclusion be correct?

 

Your solid angle is always inside angular diameter? Does that still takes into account different angular diameters of different stars? There is no however, "patch of sky" is not the same thing as "patch of any star". Patch of sky is wide and constant field of view, what you have is variable and infinitely narrow. You have mistaken "field of view" with "line of sight".

 

Are you saying patches of sky similar to Hubble Deep Field would be impossible in Olbers' paradox? It matters as it invalidates the logic of the paradox.

Perhaps with infinite exposure time, and then, even my ass would look as bright as the shiniest star on the sky. What a pleasing thought.

 

You're falling backwards. While your definition of "brightness" exists only in your head, you agreed that the second picture captures as may photons as the first. And the third as many again. And therefore you must agree that the sum of the three is three times as many as the first shell. You need to follow that pattern where it leads, not just stop after one step and declare victory. So please: go back and answer the questions in my previous post and yours.

 

Can you determine the point of saturation without knowing duration of exposure?

 

You keep talking about exposure time - it's lack of relevance is not going to change no matter how many times you ask.

 

Data insufficient, there is no answer. A car is traveling 90 m/s, how far did it go? Can you answer it without knowing some time interval? No. It simply does not compute. Just the same you can not talk about total intensity without exposure time because it has units per unit time.

Don't know what formula to use. I don't mind the number at all, whatever you say. That much I trust you.

 

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White is brighter then grey no matter how much of grey there is. If anyone is reading this be sure they are thinking you are crazy to even suggest otherwise. It's not even funny any more.

It leads no where. Without time nothing leads anywhere.

 

Try harder. If you don't understand the math, make an effort to learn it. I don't know how old you are, but seeing as how you are old enough to know how to type, you should be able to comprehend the math of this simple problem.

Wrong comparison: A car is traveling 90 m/s: How fast is it going if we watch it for 10 seconds? (90 m/s). How fast is it going if we watch it for 5 seconds? (still 90 m/s). How fast is it going if we watch it for 1 second? (still 90 m/s).

In the cases of power (J/s) and photon flux (photons/s/area) and the speed of the car m/s are all independent of the time over which they are watched.

But again: even if they weren't, it still wouldn't matter because we aren't changing the time interval:
How much faster is a car that is going 180 m/s moving than one going 90 m/s? Does it matter how long you watch them for?

You don't know what formula to use? Really?
-The second shell has twice as many stars as the first.
-The third shell has 4x as many stars as the first.
-the fourth shell has 8x as many stars as the first.

You can't figure out an equation from that pattern? Ok...Well, I guess I'll spoon-feed you a bit, but really, you should be able to do this junior-high level math yourself. It's pretty glaring for you to make jokes about a Nobel prize while being unable to comprehend simple math. Don't you care how that makes you look? How does it make you feel? Anyway:

s=a*2^(n-1)
A is the starting number of stars (10) and "n" is the shell number. So the first shell is:
s=10*2^(1-1)=10
And the 50th shell is:
s=10*2^(50-1)= 5,629,499,534,213,120 stars
But your picture is only 250x250 = 62,500 pixels. So what happens?

If there's enough of the grey layered on top of each other, it becomes white. What you need to demonstrate or at least understand is how bright the image gets when you layer all of those layers on top of each other. You know each one adds more light to the image -- so how can you be so sure that adding light over and over and over will never turn the image white? Answer: you're just guessing.

You already agreed to the pattern - now you're taking it back? Are you just purposely just trying to make this difficult to calculate to try to prevent anyone from doing the calculations to prove you wrong (while also providing no calculations to back your claim? Well sorry, but your math skills are so bad there's no way you can make it complicated enough to confusing others without first confusing yourself -- which you appear to already have done.

Ok, try this: for the sake of argument, say the camera and imaging system used to take the first image had a shutter speed of one second. But remember, it's just a rendering, not an image: you can't generate it in reality because it must somehow have the other shells blocked (unless the universe is just that small that that's the only shell). Similarly, the second picture is of just the second shell, so it must somehow have the first and every shell above the 2nd blocked. But the resulting picture when you add them - if you lived in a universe two shells wide - is what you'd still get with a one second exposure.

Regardless, adding each shell adds 1x to the amount of photons captured for our hypothetical and always the same one second exposure.

 

I assure you. See for yourself.

The question was: how far did it go? "Distance traveled" is under question, let's call it "level of saturation" and let's limit it to, say from 0 to 255.

You should then be able to calculate how much power per second would hit a pixel in Olbers' paradox universe?

It's unpleasing to me. I'm not numbers-guy, I'm visionary!

What indeed. What was exposure time?

I know infinity should be enough, but for any lesser number I can't be sure as you haven't told me how much energy is received per unit time per pixel area. That's at least you can do.

I'm telling you total intensity is a function of time. You disagree?

Are you saying every pixel receives equal amount of photons from every star no matter how far away it is?

 

This is your digression so I'd like to let you take it where you want to go with it, but I can't because you've already arrived at gibberish:

"Power per second" is gibberish: power itself is already in units of per second (energy per second). You can work with power (like speed) or energy (like speed times time) but you can't work with "power per second" - there is no such thing.

Look, you've been going on about this for two days without saying why. Please explain what, exactly, the problem is with exposure time -- are you thinking that we are trying to trick you by changing it without you noticing? We aren't.

Oy; visual perhaps, but certainly not visionary. Do you have any idea what it looks like for you to claim (even kidding) a Nobel Prize when you don't even know simple math? But fair enough; looks like I'm going to have to walk you through the entire thing, even to the point of generating the simulated pictures for you as well.

Lets call it 1 second. So you tell me: what happens when you try to take a picture of 5.6 quintillion stars when you only have 62,500 pixels?

I've done quite a lot already. You haven't said why you think this is relevant, but fine:

Your hypothetical "photon" is 1 level in the "level of saturation" scale of 0-255. Your exposure is 1 second. Now what?

I do disagree. For something to be a function of something else, that something else has to be a variable. The exposure time is not a variable here, it is kept constant. What is changing is the number of shells and thus the number of photons that are collected in that never-changing 1-second exposure.

No, I'm saying each shell adds the same number of photons to the image as the last.

So I've done the calculations and made the simulations. But before I show them to you, I want to make sure you understand the logic better. I want you to explain why you think the exposure time matters. I want you to answer the question of how you display 5 quintilion stars when you only have 62,500 pixels. I need you to demonstrate you are making an effort to THINK about this.

 

True, it's energy. Is it possible to numerically express how much energy per second is hitting a single pixel in Olbers' paradox universe? Is this amount of energy per second equal for every pixel?

The same thing when you try to take a picture of Hubble Deep Field with 1 second exposure, mostly black, and the only thing you will see are the stars in the first 4-5 shells. If you want to see more, increase exposure time, or increase pixel sensitivity.

'Total intensity received' is a variable, it varies with exposure time. Therefore, total intensity received is a function of time. What happens when instead of 1 second we set exposure time to such short value it would make our Sun look almost completely dark?

I did not say exposure time is variable, I said total intensity is variable which varies with exposure time.

Each shell adds the same number of photons to the image...

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... does that make each shell appear equally bright?

Do you acknowledge there is a color-brightness contrast between white and grey square?

Quintilion-shmintilion stars and exposure time is addressed above, and before. That's all we are talking about for a while. I explained several times why and how it does not compute. If you want to prove me wrong just show us what you got already.

 

At this point, I suspect that you are mentally ill for continuing to press this particular point. The question at hand is about number of photons reaching a certain area and clearly a large patch of grey can reflect more photons than a very small patch of white. It is clear that you do not know much about astronomy or optics, but if you do not grasp this last point, then perhaps you should just trust astronomers and move on to another hobby. You might also want some professional help or even massage therapy to help you relax.

 

Yes and yes.

That's not what I was asking, but perhaps it wasn't clear, so let me try again: Do you agree that each pixel detects light from many, many stars?

It is only a variable if we actually change it: so why would we change the exposure to make the sun look completely dark?

Would you like to try again with that sentence? It is a self-contradiction. You just said exposure time is not a variable and then said it is a variable. Do you still not know what a "variable" is? A variable is something that varies. So if total intensity varies as exposure time varies, then both total intensity varies and exposure time varies. Both are variables. You should look up the words "independent variable" and "dependent variable" so I don't have to explain which is which to you.

But again: why do we care about this? Why do you want exposure time to vary?

I'm going to take that as agreement that each shell does indeed add the same number of photons to the image even though it isn't very assertively worded. Please correct me if I misunderstood your intent.

That isn't the right question. The right question is whether the sum of all shells is equally bright -- or even brighter.

No, not under any formal definition. But I do recognize that under your made-up definition there is a brightness difference. Guess what: it doesn't matter. You are so wrong that your twisted definition isn't even relevant, since Olbers' paradox shows that the whole night sky should be as bright as the picture on your left (actually, even brighter -- we'll get to that at the end).

The basic problem here is even after looking at all the pieces of the logic, sitting there waiting to be assembled, you can't assemble them. You're still just guessing about what happens if you assemble the pictures (the shells).

No it isn't. You keep making claims about it varying without ever saying why you think it should or if you think we are saying it should.

 

deleted -- DP