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

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

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

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.

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