Anyone keen to photograph some candles, for scientific experiment?

[fjrabon]

Idk, I haven't clicked on a single link in the thread. I thought it was supposed to be someone taking a picture of a candle, not a big debate about physics and astrology that near literally has nothing to do with pushing the shutter release and enjoying the result...

We did that a long time ago. It was in a different thread that got locked though. His original theory that candles get dimmer as they get farther away was disproven with a photograph I took of a lamp. However, he more or less just ignored that and now seems to have a theory that while it doesn't happen for candles, it does happen for stars.

 

[Helen B]

I have addressed your question by explaining that resolution independence does not mean summing all the intensities into one number.

There is no such thing as 'resolution-independent'. I am not talking about your imaginary dictionary,

That's a bit of an odd thing to say. Resolution independent means that the result doesn't depend on the resolution chosen. Both words are in real dictionaries, and they have a meaning, so I don't need to use an imaginary dictionary.

I am referring to mathematics and equations used in original treatment of Olbers' paradox. And there they do sum up all the intensity into one number called 'total intensity'. Here is how it goes:

Olbers' Paradox

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)

Now let us ask just what amount of energy such a shell will send to the Earth. Since the shell's thickness is small, it is reasonable to assume that the entire shell is at a distance "r" from the earth. The energy, E, emitted by any source at distance r, produces an intensity, "I", over a given area, A, on the Earth of (inverse square law)

I = E/4p r2 (4)

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 or

T = IN (5)

Substituting the value of N previously calculated into the above, we find that

T = tnE (6)

We notice at once that the total energy received from any chosen shell does not depend upon its distance from us (no r in the above equation). The total energy received from all the shells is the sum of the contributions of each shell. If there are M shells this total is

S = tnEM (7)

But there is an infinite number of shells and so the total intensity on the earth must be infinite. Therefore, the nighttime sky should be blindingly bright!



--//--

How do you expect to deduce whether the brightness is uniform or not if your result is a single number?

Well, it's a trivial exercise to show that the above reasoning can be applied to any resolution, which in this case would mean any solid angle element (δ&#937. (This is normal scientific reasoning.) The above result is independent of the solid angle you choose. The method can be used to show that no matter what solid angle you select, the same result will happen. It's blindingly obvious. It would be literally blinding in this case, because as we have already discussed, the above ignores obstruction (see (9) in post #192 above, or the next paragraph in the article you part copied-and-pasted). The same method, adjusted to take obstruction into account, can also be used to show that the brightness of any solid angle element of the sky has the brightness of a nearby star - ie that all possible selections of δΩ however large or small (resolution independent) have the luminous emittance of a star surface and the sky is thus uniformly bright.

Is there any point in me wasting any more time on this?

 

[jake337]

 

[amolitor]

Ok, then.

Let us assume that the inverse-square law applies throughout.

I will actually describe a universe with an extremely regular distribution of an infinite number of stars, for which the night sky is uniformly bright. This can be converted to one with a random distribution of stars as follows:

- adjust the distance of each star from the viewer by a random degree, while simultaneously adjusting the size and brightness of the star so that its appearance to the viewer is unchanged. I.E. if you move it farther away, make it bigger, and if you move it closer, make it smaller. If you started with identical stars, this randomizes the size and intensity of the stars, as a little bonus.
- Now observe that since the sky is uniformly bright, I can take, for example, any two 1 degree by 1 degree squares in the sky, and swap them, without changing the appearance of the sky. This is effectively rotating stars around. The stars within the square are at varying distances, but we're rotating them around the viewer, maintaining the original distances. Randomly select pairs of squares of random dimension, and swap them. This "shuffles" the rotational position of stars. Do this as much as you like, until a suitable degree of randomness has been achieved.

So, it will suffice to show a regular distribution of an infinite number of identical stars, which produces a uniform sky brightness.

----

For the regular distribution of stars let us make an observation. I can fill any region of the sky to uniform brightness with a finite number of identical stars as follows:

- select a distance from the viewer, any distance will do
- fill the region with your stars, spacing them very close together in a regular grid, this will leave gaps, of course
- fill the gaps with a second layer of stars behind the first grid of stars. You may use the same size and intensity of star. We know that placing them further away makes them, like a candle, smaller, but not less bright.
- you might need a third or fourth layer, honestly, but it's a small number of layers to fill 100% of the gaps, if you packed the stars together pretty tightly to start with. You're just covering a piece of paper with overlapping circles, in effect.

-----

Now, select half of the night sky. Fill it as indicated above. You have used a finite number of stars.

Whatever sky remains unfilled, select half of that, and fill it in with a finite number of the same size and intensity of stars as the previous step, but placed twice as far away as the previous collection of stars. They are farther away, but will, like a candle, appear the same brightness . This will take about 2x as many stars as the previous step, despite the region being half the size.

Continue, filling in half of what remains at each step. This will require an infinite number of steps. Each region filled will require more stars, but each step requires only a finite number of stars, and produces a region of the same brightness as the previous region.

When you are done, you will have uniform sky brightness and an infinite number of stars, in an extremely regular pattern

Shuffle these stars as suggested at the beginning of this post, to produce a random distribution of that infinitude of stars, without disturbing the uniform sky brightness.

Now you are done.

Whatever reasoning you have to demonstrate that the inverse square law produces a dark night sky, it will fail in this universe. Since it fails in this case, it must be logically incorrect.

 

[rexbobcat]

How did something so simply ten into a physics lesson?

Regardless of mathematical equations (eww). A light source will always remain at the same brightness from its origin unless there is something in the air to impede its light. :/

Why is this so complicated?

 

[fjrabon]

the ultimate issue here seems to be that Tris views the inverse square law as a sort of magic law that just holds. Sort of like the way that pi is 3.14159265...... What I think he doesn't get, which we've been trying to explain to him, is why the inverse square law holds. If he really understood why the inverse square law holds, he'd see that this whole thing he's been talking about is less than a puff of smoke. It's a misconception based on a half understanding of the inverse square law.

He's sort of latched on to the idea that things get dimmer as they get farther away, but sort of ignored the 'because their light spreads out over greater space' aspect. And then he's flat out ignoring that when light from two different sources combine, their 'brightness' is additive. He also seems to have a hard time understanding the distinction from the amount of illumination provided by a point source to a reference point (what the inverse square law is talking about) and the object's apparent brightness when focused upon (WHICH THE INVERSE SQUARE LAW DOESN'T ADDRESS)

I think the simplest way I can address this fundamental issue is this Tris:

The inverse square law has nothing to do with how bright an object looks to your eye. It was never meant to, that's not what it's used for in physics. The inverse square law is used to show the amount of TOTAL light provided by a point source to a given amount of area in a reference plane. ie it only measures the total amount of light an object provides, not how bright a focused dot 'looks'.

We've all tried to explain this to him like 1239847 times in the course of these 3 or 4 threads, so I don't expect that misconception to change any time soon.​

 

[jwbryson1]

.

 

[christop]

The right side is not correct. It's an AVERAGE of all 16 pixels, but if you truly had 1x1 image that single pixel would collect all the light and have summed brightness of all the pixels on the left. The same would be true for all the other rows. Single pixel image would always be as bright as the sum of all the pixels from the left, all the light would end up at that single pixel.

It is exactly the illumination a low-resolution sensor would produce. Sensors are calibrated to produce an illumination based on not only how much light is collected by its sensels but also on the area of the sensels. Larger sensels (lower resolution) obviously receive more light than smaller sensels if all else is equal. So the light-to-illumination conversion mechanism (amplifier circuitry or what have you) must take the sensel's area into account when computing the sensel's illumination. In my example the light received by the larger sensel must be amplified by 1/16 relative to the small sensel.

What's most concerning is that you are now claiming that lower resolution makes the image "brighter". You previously claimed the opposite. So which is it? Does lower resolution make the image brighter or dimmer? (Hint: it's neither). I can see why you're confused.

You should have all the stars be the same size (point sources), where further ones would be less bright.

Yes, I learned my lesson and I appreciate it. But there is only few dozen stars that have "resolvable" angular size, so for general case scenarios you should make all the stars be point light sources.

Why? Stars are not point sources. Stars have finite size. The small sensels in my example are part of an "ideal" high-resolution imaging sensor which is capable of resolving each individual star. The low-resolution sensor (on the right) cannot resolve each star individually so we can estimate the size of a star from the illumination of a sensel (the top right sensel's illumination is 25% so we can estimate the angular size and therefore the distance of the star).

I have no idea what point you are addressing. Illumination would add up just by having a single pixel instead of many pixels.

Exactly. Illumination adds up.

You got uniform brightness when you averaged the image from the left into only one pixel. That's not regardless of resolution, that is what automatically comes with ONE PIXEL resolution. And you are neglecting little fluctuations in brightness of individual stars, as if that would make some difference, but you are completely ignoring further away stars would be dimmer than closer stars. These are not candles any more, stars are point light sources and they get dimmer with the distance

You apparently didn't notice that all stars were individually resolved in the left image (the high-resolution sensor) and that the illumination of all 16 sensels in the bottom-left image is still uniform. I could have added an even higher-resolution sensor image which would also have been uniform illumination. Perhaps I should have added sensel grid lines to make it more obvious that each sensel is still an individual, independent sensel.

 

[pgriz]

@ Helen B: Regarding

Is there any point in me wasting any more time on this?

Probably not, but I do very much appreciate your clear (at least to some of us) explanations. Once we're past this particular conrundum, we can probably start addressing the question of how many angels can dance on a head of a pin. But that may devolve into a discussion of the footprint size of each class of angel, and the variation in pin head sizes.

 

[tris_d]

That's a bit of an odd thing to say. Resolution independent means that the result doesn't depend on the resolution chosen. Both words are in real dictionaries, and they have a meaning, so I don't need to use an imaginary dictionary.

We are talking about an image of the night sky here, formed by human eyes or a camera, and why does it look the way it looks. Your result being a single scalar number is not "resolution independent", it is COMPLETELY WITHOUT any resolution, and when talking about images that's a nonsense to start with.

Well, it's a trivial exercise to show that the above reasoning can be applied to any resolution, which in this case would mean any solid angle element (δ&#937. (This is normal scientific reasoning.) The above result is independent of the solid angle you choose. The method can be used to show that no matter what solid angle you select, the same result will happen. It's blindingly obvious. It would be literally blinding in this case, because as we have already discussed, the above ignores obstruction (see (9) in post #192 above, or the next paragraph in the article you part copied-and-pasted). The same method, adjusted to take obstruction into account, can also be used to show that the brightness of any solid angle element of the sky has the brightness of a nearby star - ie that all possible selections of δΩ however large or small (resolution independent) have the luminous emittance of a star surface and the sky is thus uniformly bright.

Solid angle has nothing to do with resolution or distribution. Distribution involves a sequence, that is more than one number. And your result is a single number, it can not have any distribution, it's "uniform" just by being a single. It's a farce.

Don't you realize if you take "resolution independent" approach and calculate total intensity of either of those two images above you get a single number that would make you believe those images are both bright and uniform, which is obviously not the case if you take resolution into account. Do you understand now why such mathematics is wrong and unsuitable to deduce anything about any distribution or actual brightness?

 

[amolitor]

Tris, I have no trouble following Helen's posts.

I cannot follow your posts at all, they appear to simply be strings on non-sequiturs with some keywords sprinkled in. I say "appear" here quite deliberately -- I pass, in this post, no judgement on whether your remarks are correct or not. My point here is to note that if you plan to write a paper about this, you need to work on your expository skills. A publishable paper exists to communicate ideas to other people, and your writing on this forum simply doesn't accomplish that. As a people, I can state firmly that you have succeeded in communicating no ideas at all to me. My sense is that you have not communicated successfully with anyone else in this forum, either.

You should consider what to do about that, before sitting down to write your paper.

 

[tris_d]

Ok, then.

Let us assume that the inverse-square law applies throughout.

I will actually describe a universe with an extremely regular distribution of an infinite number of stars, for which the night sky is uniformly bright. This can be converted to one with a random distribution of stars as follows:

- adjust the distance of each star from the viewer by a random degree, while simultaneously adjusting the size and brightness of the star so that its appearance to the viewer is unchanged. I.E. if you move it farther away, make it bigger, and if you move it closer, make it smaller.

That's wrong. If we adjust the night to look like a pink unicorn then when you look above at the night sky you see a pink unicorn.

If you started with identical stars, this randomizes the size and intensity of the stars, as a little bonus.
- Now observe that since the sky is uniformly bright, I can take, for example, any two 1 degree by 1 degree squares in the sky, and swap them, without changing the appearance of the sky. This is effectively rotating stars around. The stars within the square are at varying distances, but we're rotating them around the viewer, maintaining the original distances. Randomly select pairs of squares of random dimension, and swap them. This "shuffles" the rotational position of stars. Do this as much as you like, until a suitable degree of randomness has been achieved.

So, it will suffice to show a regular distribution of an infinite number of identical stars, which produces a uniform sky brightness.

No. Since the night sky looks like a pink unicorn, I can take, for example, any two 1 degree by 1 degree squares in the sky, and swap them, without changing the appearance of the sky.

Whatever reasoning you have to demonstrate that the inverse square law produces a dark night sky, it will fail in this universe. Since it fails in this case, it must be logically incorrect.

Don't you have anything better to do?

 

[amolitor]

Are you actually under the impression that repeating back fragments of my remarks, and sprinkling them with snarky bits, constitutes a rebuttal?

This is, at least, the second time you've replied to me in this fashion. As a published scientist, I can assure you that it's not a good road to getting a paper into print.

 

[The_Traveler]

I am quite pleased I didn't get caught up in this thread. but have clumped trid_d along with others in this group referred to below.

 

[unpopular]

I say we just abandon ship and ignore all future liter which tris casts.

I still say he's a genuine physicist just trying to **** with us.

---

amolitor - just curious, what field of science did you study?