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

a reply to: swanne

-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),

Before I get any more involved, I have a problem with postulate two.

The luminosity, magnitude and size of stars, though there is some correlation with distance, is not as you describe. The luminosity of stars varies depending on the type of the star.

Also, the appearance of a star appearing larger in the sky does not mean that it is closer. For instance, a very large star can be much further away than a small star, but because of how large it is it can appear to be at the same distance as the small star.

Another example; I can hold up a pencil eraser in front of my eye with the moon in the background. They both appear to be the same size, but only to my point of reference. We all know that a pencil eraser is not the size of the moon.

So, before I address your theory any further, can you please explain postulate 2 in further depth?

I don't think the calculus of 'infinitesimal' has quite clicked for you op, as much as I appreciate your effort.

So here's a question, how does this relate to objects for which we know an exact distance? Most notably we can measure planets by reflected light and we know very precisely as to how long it will take to send an object to Venus, Mars, or Sol because we've done it and measured it. If these formulas are off, shouldn't these distances also be reduced which would impact our ability to send objects there?

a reply to: PansophicalSynthesis

So, before I address your theory any further, can you please explain postulate 2 in further depth?

Don't worry, I did already took all this into consideration. Watch:

Example 1: a star with diameter 1 AU (giant star) and luminosity of 425 suns will look as if it was only 5.7 degrees large in the sky of an observer 10 AU away, and its apparent luminosity will thus be 5.7/360 th its full luminosity in 2 d (which means 6.73 suns) and will be 25.783100/360th its full luminosity in 3 d since you include its angular area (which means, apparent luminosity will equal 30.43 suns).

Example 2: a star with diameter 1 AU (giant star) and luminosity of 100 suns will look as if it was only 5.7 degrees large in the sky of an observer 10 AU away, and its apparent luminosity will thus be 5.7/360 th its full luminosity in 2 d (which means 1.58 suns) and will be 25.783100/360th its full luminosity in 3 d since you include its angular area (which means, apparent luminosity will equal 7.16 suns).

Example 3: a star with diameter 10 AU (giant star) and luminosity of 1,000 suns will look as if it was only 5.7 degrees large in the sky of an observer 100 AU away, and its apparent luminosity will thus be 5.7/360 th its full luminosity in 2 d (which means 15.83 suns) and will be 25.783100/360th its full luminosity in 3 d since you include its angular area (which means, apparent luminosity will equal 71.61 suns).

The apparent luminosity depends on the absolute luminosity of the star, but if the apparent size of the star in your sky gets incredibly small because of your distance, then its apparent luminosity will diminish accordingly.

a reply to: ErosA433

Nice of you to join the conversation; I know your free time is limited, I sincerely appreciate you taking the time to share your input.

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

Hm... What was the detector's shape in this experiment you describe? Were all points of the detector's surface at equal distance to the source of light? And, was the source of light infinitively small in shape?

Over stellar distances astronomers have a few tricks up their sleeves, many stars can have their distances determined via parallax

Indeed, but when the star is at the other side of the Galaxy, parallax becomes extremely tricky to determine. We have trouble determining Deneb's (uncertainty 500 light-years), and that's a star only 1840 light-years from us at most.

Now the size of the galaxy has been claimed to be 100,000 light years. You just can't determine the parallax of stars 100,000 light-years away, the uncertainty would be too great.

a reply to: swanne

-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),

Okay, so you need to change the language here, because what your sentence is saying is false. I addressed this in my initial reply.

I'll get around to your calculations later, though from the looks of your thought pattern in postulate 2, I must currently presume that your calculations also fail to account for relative size, distance and luminosity.

originally posted by: swanne
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.

Can you elaborate on how you're getting the result of 32.828063? In your 2D example, you give a luminosity of 360/2pi*r. What luminosity and area are you using to get 32.828063 in 3D?

a reply to: PansophicalSynthesis

though from the looks of your thought pattern in postulate 2, I must currently presume that your calculations also fail to account for relative size, distance and luminosity.

And I am telling you that you are presuming wrong.

In my last post I exposed exactly what my train of thought was. If you read it back carefully - you'll see that luminosity is included in the formula I use, along with diameter and distance. Basically, the whole of my last post could be simplified as this formula:

L((0.785398((d/r)57.295779)^2)/360)

Where L is the intrinsical luminosity of the star, d is the diameter of the star, and r is the distance to the star.
Not because I didn't mention the luminosity in postulate 2 means that I have excluded it. Of course I've included it - I simply thought that it would be so self-evident that it needed not mentioning in the postulate. In which case I would present my most sincere apologies if doing such thing only caused confusion.

a reply to: nataylor

What luminosity and area are you using to get 32.828063 in 3D?

Luminosity L (for the maintream, F=L/A formula) of the star in 2 d was indeed here set at 360 (as an example).

In 3 d, the luminosity L would become 41252.961249 (spherical area of 360, 360 being the circumference in 2 d). The receiving surface is modelled (by mainstream) as a sphere stretching all around the star. The area of such a sphere is dependent upon its radius r. For a radius of 10 units, the area of this said sphere will be equal to 4pi*r^2, or, here, 1256.637061. Since the mainstream formula is F=L/A, and since here L = 41252.961249 & A = 1256.637061, then F = 41252.961249/1256.637061, which equals 32.828063.

These values above and in the OP are, of course, an example amongst so many available, and all these values were set as such for comparison purpose only.