It looks like you're using an Ad Blocker.

Thank you.

Some features of ATS will be disabled while you continue to use an ad-blocker.

# Ladies and gentlemen, I will now diminish the size of the Galaxy.

page: 1
15
share:

posted on Aug, 19 2014 @ 08:56 AM
Hi. This is my 50th thread on ATS. And for the occasion, I've decided to post something really, really ground breaking.

I've used the inverse square formula for a great part of my life, but I always felt that there was something a bit wrong with it. After years of pondering, I've recently found out what was wrong. So, for my hemicentennial (nerdy wink) thread, I will go on and post my most audacious thread ever. I will demonstrate how the so-called "inverse square" law might not be as accurate as it would seem. The implication of such demonstration is that the computed distances to far away stars in the Galaxy (and in the Andromeda galaxy), which are dependent on an "inverse square" luminosity law, would be over-estimated - leading to the conclusion that these stars would in reality be closer, and thus a smaller galaxy - and/or a closer Andromeda galaxy.

So, on this Tuesday morning, I take a deep breath, and with a slight tremble in my hand (after all, I may be wrong), I write this audacious post.

******

So, here we go. This post is based upon three postulates:

-postulate 1: "ideal" surfaces are rather rare in the real world,

-postulate 2: a star's apparent luminosity is directly related to its size in one's sky (the closer one is to such star, the bigger the latter will seem, and thus the brighter it will shine),

-postulate 3: stars are more accurately modelled as spheres than as squares.

******

Let us explore postulate 1.

To compute the apparent luminosity of a star, a formula, F=L/A, was implemented. This "mainstream" way to compute apparent luminosity tells us how much photons hit each square meters of a surface around a source of light.

The problem with this F=L/A formula (I will tell you more about the maths behind this formula later) is, it models the receiving surface as hollow sphere stretched all around an infinitively small source of light. It models the receiving surface as an ideally exposed surface, with all of its points at equal distance from the source of light. Furthermore, it goes on and counts the number of photons hitting a squared meter of this ideal, warped surface.

I say problem, because as I will now demonstrate to you, ideal surfaces are actually quite rare in the real world.

The Earth, for instance, is spherical in shape. Almost none of its surface matches an "ideal" surface. Now imagine what would happen if we were to place a powerful source of light at, say, 18,000 km from Earth.

Notice how many rays of light are modelled to touch an "ideal" surface, when in truth they encounter more angled surfaces (and thus are spread across more area), or miss the Earth altogether.

Now this, of course, is a rather exaggerated version of the phenomena - but even at smaller scale such discrepancy will survive. Because, want it or not, not all surfaces are shaped so that all of their points are at the same distance to a light source. This has a direct impact on the F=L/A formula's ability to model apparent luminosity - that is, the luminosity which a place perceives from a source of light.

******

Now, postulate 2.

If the Sun was closer, its apparent size in our sky would be much greater, it would cover more sky with its luminous disk, and, thus, it would shine brighter to us.

Instead of modelling the star as a point, and the receiver as a surface, I make the proposition of modelling the star as a sphere instead, and the receiver as point-like (like an eye or something).

In 2 dimensions, the equation I use, (d/r)57.295779, gives the angular diameter (in degrees) which one will perceive from a star, d being the star's diameter and r being its distance from us.

If this star would be, say, 1 AU large, and if we were to be standing at 10 AU from it, then it would cover 5.7 degrees of our sky ((1/10)*57.295779 = 5.7295779). If we were standing at 57.295779 AU from it, then we would perceive it as being exactly 1 degree large: (1/57.295779)*57.295779 = 1.

Thus, its apparent luminosity will be 5.7295779/360 th its full luminosity in the first instance, and 1/360 th its full luminosity in the second instance.

A 2-D version of the "mainstream" F=L/A formula actually correlates these results in a 2 D universe. In the F=L/A formula, F is the "Flux" value, L is the original luminosity of the infinitively small light source, and A is the area of the perfect surface. Except that in 2 D, a sphere's area becomes a circle's circumference, so in 2 D the formula would actually become, F=L/(6.283185*r). For a surface 10 AU away from a source of light with a Luminosity equal to 360 (one photon per degrees), F gives an output of 5.7295779, and for a surface 57.295779 AU away it gives 1.

So far, so good.

But watch what happens when we transfer all this into 3 dimensions:

In 3 dimensions, not only do we perceive the angular diameter of the star, but we also perceive the area of its disk:

This is important to consider when we compute the star's apparent luminosity. Now to find the area of a disk, we need to multiply the square of its diameter by 0.785398. As such, a disk with a diameter of 5.7 degrees will appear to have an area of 25.783095 square degrees in the sky. Similarly, a 1-degree large disk will cover 0.785398 of a square degree in your sky.

But this is where the mainstream F=L/A formula gives a different result. After setting everything on 3 dimensions, one gets the result of 32.828063 for 10 AU and the result of 1 for 57.295779 AU.

Why is there such a discrepancy? The answer is hidden in our own equation...

******

Remember when I told you that to find the area of a disk one needs to multiply the square of its diameter by 0.785398? Well, watch what would happen if I were to model stars as squares instead of disks. To find the area of a square, one only needs to put the width of the square at the power of 2. At 10 AU, 5.7295779 degrees would thus become 32.828063 square degrees - a perfect match with the mainstream formula! And at 57.295779 AU, 1 degree would stay 1 square degree - once again, perfect fit with the F=L/A formula.

The mainstream formula is equivalent to modelling stars as if they were squares instead of spheres!

It seems the "inverse square law" has the side-effect of taking stars for squares... at least, that much is evident when the formula is compared with stellar shape properties computations.

postulate 3: Since a square covers 21.4602% percent more area in the sky than a circle with the same diameter, its brightness would follow proportionally. Modelling stars as if they were squares would have the result of over-estimating their apparent luminosity by 21.4602%, leading to erroneous distance calculation of stars across the Galaxy. And, if a very bright star has to actually be closer than expected to shine as we observe it to do - then, so would those stars at the other edge of the Galaxy. And, thus, the Galaxy... would actually be smaller!

With this I concludes my 50th original post in ATS.

I hope it was much thought-provoking.

posted on Aug, 19 2014 @ 09:10 AM
Interesting. Do they actually estimate distance based on luminosity as you suggested?
If they do, shouldn't you be putting yourself in touch with atro physicists? Just a thought, it sounds like they have some serious remodelling to do, if what you say is true.

posted on Aug, 19 2014 @ 09:10 AM
Nice thought provoking thread even though I didn't understand most of it.

edit on 19-8-2014 by Briefcase because: (no reason given)

posted on Aug, 19 2014 @ 09:14 AM
Errr except you forgot the part where earth has an atmosphere, once light hits the atmosphere it can get directed in all sorts of directions, for example when we see a sunset, or a sunrise - technically it's not rising at all - it's an optical illusion, it's the suns reflection off our atmosphere, those faint left over, or coming photons are being bounced around the planet, even going as far as bouncing off the moon on the other side.

And gravity doesn't allow for square objects at such a large scale like a star or planet, it would have to be man made, but I guess that too would eventually turn into a sphere.

posted on Aug, 19 2014 @ 09:16 AM

originally posted by: Briefcase
Interesting. Do they actually estimate distance based on luminosity as you suggested?

In many cases, yes.

To measure the distance to the Andromeda Galaxy:

Curtis noticed that these novae were, on average, 10 magnitudes fainter than those that occurred elsewhere in the sky. As a result he was able to come up with a distance estimate of 500,000 light-years (3.2×1010 AU).

source: en.wikipedia.org...

To measure the diameter of the Milky Way:

how that luminosity changes over distance, when it moves further away you can calculate that distance by determining how much dimmer the flashlight is.

source: www.universetoday.com...

posted on Aug, 19 2014 @ 09:17 AM

Errr except you forgot the part where earth has an atmosphere

Actually these come as an add-on to our formulas. Notice that the mainstream F=L/A doesn't account for an atmosphere neither. This doesn't change the fact that the inverse square law might be flawed.

originally posted by: strongfp
And gravity doesn't allow for square objects at such a large scale like a star or planet, it would have to be man made, but I guess that too would eventually turn into a sphere.

My point exactly, mate.

edit on 19-8-2014 by swanne because: (no reason given)

posted on Aug, 19 2014 @ 09:24 AM
Did you get the idea from Space Balls the movie when they went traveling at light speed and then faster into plaid ? LoL. JK

I see why you think this would be a better way.

posted on Aug, 19 2014 @ 09:27 AM

originally posted by: Thiaoouba Prophecy
Did you get the idea from Space Balls the movie when they went traveling at light speed and then faster into plaid ? LoL.

Hehe, that was one great movie

But actually, I got it by attempting to observe the last solar eclipse. I wanted to make a pinhole in a cardboard so to observe the sun, and that's when I realized that a hole shaped as a square would allow more light through than a hole shaped as a circle with the same diameter.

posted on Aug, 19 2014 @ 10:39 AM
Good thread there mate. Lets see how this pans out

posted on Aug, 19 2014 @ 10:54 AM

Isn't it true that a circle can be quantized by smaller squares and as the observation distance gets longer, the difference between a circle quantized by squares and the geometric circle approaches zero?

So the 2D representation is the linearized version of the 3D representation? The linearized version is effective at a minimum distance of your choosing.

Is that germane to your point?

posted on Aug, 19 2014 @ 11:11 AM

originally posted by: InverseLookingGlass

Isn't it true that a circle can be quantized by smaller squares and as the observation distance gets longer, the difference between a circle quantized by squares and the geometric circle approaches zero?

So the 2D representation is the linearized version of the 3D representation? The linearized version is effective at a minimum distance of your choosing.

Is that germane to your point?

What you just said reminded me of this guy!

He can draw pi, and use fractals to make any object, that we see. The thing is tho, it's made up of pure triangles.

posted on Aug, 19 2014 @ 01:45 PM

Interesting thread to say the least . Does the atmosphere matter as buddy boy up there mentions? I tend to feel as though of heard those sentiments of his before.

posted on Aug, 19 2014 @ 06:20 PM

originally posted by: swanne
Instead of modelling the star as a point, and the receiver as a surface, I make the proposition of modelling the star as a sphere instead, and the receiver as point-like (like an eye or something).

apod.nasa.gov...
Neither of those models is good. You can see Betelgeuse isn't a point in this photo, which rules out the model on the left:

Explanation: Betelgeuse (sounds a lot like "beetle juice"), a red supergiant star about 600 light years distant, is seen in this Hubble Space Telescope image - the first direct picture of the surface of a star other than the Sun.

We can estimate the size of Betelgeuse from that photo, but a more accurate measurement of star size is done with interferometry:

Stellar Diameters and Temperatures IV. Predicting Stellar Angular Diameters

The number of stellar angular diameter measurements has greatly increased over the past few years due to innovations and developments in the field of long baseline optical interferometry (LBOI). We use a collection of high-precision angular diameter measurements for nearby, main-sequence stars to develop empirical relations that allow the prediction of stellar angular sizes as a function of observed photometric color. These relations are presented for a combination of 48 broad-band color indices. We empirically show for the first time a dependence on metallicity to these relations using Johnson \$(B-V)\$ and Sloan \$(g-r)\$ colors. Our relations are capable of predicting diameters with a random error of less than 5% and represent the most robust and empirical determinations to stellar angular sizes to date.

So the correct model of course would be the star as a sphere, and the receiving device not as a point but as the actual shape of the receiving device, which is often a round mirror for a telescope (not exactly a point, in fact the larger the better because then it can gather more light).

I didn't really follow the part about modeling stars as squares, but we don't do that. Perhaps you've misinterpreted the inverse square diagram showing squares; the inverse square law can apply to circles of increasing size at increasing distance too, but the squares are drawn because they make it easier to show the area calculation.

posted on Aug, 19 2014 @ 06:53 PM

Congratulations on your well done 50th thread. Red shift blue shift et al has always made me ask myself when distances are measured.... what effect does the expanding universe have on the objects of our interest and what effect does the interstellar medium of gas and dust do to our measurements ? At this stage (we think) we do pretty good with what we have to work with, however, no doubt there will be adjustments in the future... S&F

posted on Aug, 20 2014 @ 09:05 AM

Neither of those models is good. (...) So the correct model of course would be the star as a sphere, and the receiving device not as a point but as the actual shape of the receiving device

Perhaps. But I wonder how much more complex the formula would then be (especially if you consider that many receiving devices, such as the eyes or a mirror, work by focusing light unto a point), and I wonder if this complexity is justified when you compare the relatively small size of the receiving device relative to the astronomical size of stars and their sheer distances. I think that even though receiving devices might not be perfect points, approximating them as points is nevertheless a much more realistic model than suggesting that stars are points instead.

I didn't really follow the part about modeling stars as squares, but we don't do that.

So did I thought at first. But the math is clear - although mainstream don't directly model stars as squares, their result is still nevertheless equivalent to modelling stars as squares. The results are the same than those we get by modelling the angular area of a star in the sky as a square instead of a disk.

posted on Aug, 20 2014 @ 09:09 AM

You'd need an infinite number of squares to render a circle perfectly. Besides, as the distance would grow, the ratio between the area of a square and the area of a circle won't change in itself.

That's why some of us prefer to say that a circle is exactly 0.78539816339 a square (in area).

posted on Aug, 20 2014 @ 09:43 AM
Forgive me for diverging from the thread topic for one moment, but......my understanding is that the correct pronunciation (if indeed there is one) for Betelgeuse is "bettle-gurz".

posted on Aug, 20 2014 @ 01:20 PM
I can assure that the mainstream doesn't model stars at squares. Area in the case of stars is propagated by solid angle, or a Steradian.

The inverse square law is extremely easy to measure in the lab, and can be analyzed with corrections for detector shape. I know this because Iv done it and checked it as a sanity check while doing optical calibrations for a measurement for my PhD. turned out to be very accurate. Corrections for the difference in distance across the detector from the source was also made, but found to be already within the 0.001% level for my setup (distances here where of the order of 0.8 m so fairly small)

Corrections were performed analytically using a brute force integration of area over a spherical surface, and by solid angle calculations.

Now, for stars and planets, our telescopes do not resolve stellar surfaces, only the rare few as already been pointed out have ever been imaged, and not that well either. (still an amazing achievement)

Over stellar distances astronomers have a few tricks up their sleeves, many stars can have their distances determined via parallax, and of those we can determine spectral type, apparent luminosity, for some if they are in binaries or trinaries you can also get mass. While it might sound like a circular argument, using the inverse square law to prove the inverse square law... it really isn't quite the case. If you can find two stars with identical mass and very close/the same special type. They will have the same absolute luminosity. The only factor that is different between them will be the distance.

This has been done, and no discrepancy has been found that could be caused by the inverse square law being out of whack

posted on Aug, 20 2014 @ 01:31 PM

originally posted by: Mogget
Forgive me for diverging from the thread topic for one moment, but......my understanding is that the correct pronunciation (if indeed there is one) for Betelgeuse is "bettle-gurz".
correct would be Bay-Tail-Goisay

posted on Aug, 20 2014 @ 01:48 PM

I gave you a S&F for the thread. I understood everything you said, but unfortunately my level of astrophysics isn't advanced enough to refute it. I do commend you on the math heaviness and accuracy though. Usually threads like this, people start throwing out estimated numbers then try to prove their point by arguing through those made up numbers. I'm glad you used real examples with real computations. However, I still feel the inverse-square law isn't as flawed as you think. I just wish I could explain why.

new topics

top topics

15