Abdul,

what you have written is most impressive.  I must admit that I don't comprehend all of it, especially the calculaltions.  I want to re-read it however. 

You are to be congratulated on what you have already accomplished.  I forsee you becoming a prominent name in the space science field.

Here’s how you can visualise this "constant" from your own backyard without going into outer space 

Let’s say it’s midday now, right at this minute, and you’re standing out in the garden at the back of your house admiring all the beautiful plants around you.

Now if someone were to switch off the Sun right at this minute, as if it were just an ordinary light bulb, the sky up above would go dark, right? And the stars would come out, as if nightfall had suddenly descended across the land? Sure thing! It’s how things are for us after sunset each and every single evening.

Now imagine the earth beneath your feet gradually crumbled away and disappeared into nothingness. If you now look down beneath your feet, way down into the distance, you’ll see stars much as you would when you looked up. You would have stars all around you, way up above your head, across over to your right, over to your left and going way into the distance beneath your feet... You’d be surrounded by a 360-degree sphere of starry pinpricks of light from all across the universe. In this imaginary scenario, you have entered a perpetual interstellar night that one would encounter on every single journey going outward from our Solar System in any direction... in a rocket ship sailing toward the immensely distant stars.

If now you held out the palm of your hand in front of you, could you see it solely by the naked light from all those tiny pinpricks of stars? To answer this question, you have to evaluate "Ahad’s constant"!

Now if someone were to switch off the Sun right at this minute, as if it were just an ordinary light bulb, the sky up above would go dark, right? And the stars would come out, as if nightfall had suddenly descended across the land? Sure thing! It’s how things are for us after sunset each and every single evening.

If  I were to occupy a particular place in space where the sun were switched off like a light bulb, the immediate reflective objects would not display themselves because they are not being illuminated by the Sun any longer, thus creating a whole different view than if we left the sun "on". The position of the planets are contantly changing and thus creating a slightly different lighting as well. Also you are assuming that the person occupying the space without sun is not carrying a pocket flashlight which throws the whole thing off once used.  Hmmm...... The constant we know for sure is that there is no constant. Hmmmm..........I like you, you make my brain work. This is fun. What else you got?!

J.

This is a fun brain teaser!!!

Here are the twenty closest stars to the Sun, as they are positioned in 3D space relative to one another:-

Consider the brighter and more luminous ones amongst them on this chart: Sol (which is our Sun), Procyon, Sirius and Alpha Centauri.

Now, ask an astronomer: what is the net amount of natural light the sky is going to produce when I'm sailing about mid-way between: (a) Sol and Alpha Centauri, (b) Sirius and Procyon, (c) Alpha Centauri and Sirius, (d) Alpha Centauri and Procyon, etc...?

Whatever answer he/she will give you will be exactly the same for each of those four interstellar journeys (a), (b), (c) and (d),... That is indeed a "constant". Does the confusion stop here - lol...?

Isn't there a constant already in use by astronomers which is for a patch a sky devoid of visible stars and galaxies  and expressed in magnitudes per unit of square area?  In photometry, it is essential to know your sky background brightness before assessing small scale variations in variable objects.  This might only apply though for objects seen through our atmosphere from observatory Earth.  I have no idea whether observations carried out in space are treated similarly.

Time to do some research.

Pierre

Most impressive. The calculations went a bit over my head but I can see were you are going =). ..:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" />

Okay, here is a worked example

Consider these three bright stars, which twinkle prominently in our evening skies during early summer each year:-

Vega, Altair and Deneb

Their apparent visual magnitudes are:-

Vega = 0.03, Altair = 0.77, Deneb = 1.25

The combined brightnesses of these three stars could be worked out as follows.

Let m1=0.03, m2=0.77, m3=1.25

Now, by equation (1) above, Li/L1 = 10^[0.4*(M1-Mi)], we have:-

L2/L1 = 10^[0.4*(m1-m2)] = 10^[0.4*(0.03-0.77)] = 0.5058 (to 4 decimal places)

Similarly, L3/L1 = 10^[0.4*(m1-m3)] = 10^[0.4*(0.03-1.25)] = 0.3251

Finally, by equation (2) above, Mc = m1 - 2.5*Log10(1+L2/L1+L3/L1) = 0.03 - 2.5*Log10(1+0.5058+0.3251) = -0.63

Hence, the combined magnitude of all three stars comes to a net value of -0.63.

Now suppose we placed ourselves a couple of light years beyond the neighborhood of our Solar System, into deep interstellar space. From that vantage point, if we continued adding the magnitudes of every single star visible across the entire cosmic sky in this manner, the total luminosity would eventually tend "asymptotically" towards a final figure of circa -6.5 magnitudes.

Comparing this brightness with that of the Full Moon (which shines at approximately magnitude -12.7 in the night skies of planet Earth), the magnitude difference is (-6.5) - (-12.7) = 6.2. This equates to a brightness ratio of 10^(0.4*6.2) = 302. Hence, the total luminosity of the interstellar night sky amounts to circa 1/300th of a Full Moon’s worth ("Ahad’s constant") .

That actually makes sense. I’m starting to understand the idea of Ahad’s constant. ..:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" />

I’m doing another night shift so I got lots of time to study these things

The apparent magnitude of the Sun as seen from Earth is -26.7. As we head outward from our Solar System, the Sun’s apparent brightness will fall off according to the inverse square law. The absolute magnitude of a star is its apparent magnitude when viewed from a standard distance of 10 parsecs (32.616 light years). We know from direct measurements, the Sun has an absolute magnitude of +4.76.

The formula for working out the apparent visual magnitude 'm' of a star whose absolute magnitude is 'M', as seen from a distance of 'd' light years is given by:-

m = M - [5 - 5 * log10(d / 3.2616)] ............. Equation (3)

Using the above equation, it is possible to chart the fall off in apparent magnitude of the Sun with increasing distance, a bit like this:-

The average distance of the Earth from the Sun is 149.6 million kilometers. This yardstick distance is referred to by astronomers as an ‘astronomical unit’ (AU). Light from the Sun takes 8 minutes and 18 seconds to cross 149.6 million km of space to reach us here on Earth. In other words, that’s how long it takes to cross 1 AU. Hence, by simple arithmetic, there are 63,240 AUs in 1 light year. At a distance of some 11,500 AUs (0.182 light-year) from the Sun, its apparent magnitude would be (by equation 3 above):

m = +4.76 - [5 - 5*log10(0.182/3.2616)] = -6.5

Therefore, that is the distance where the Sun’s light output exactly matches that of the surrounding cosmic night sky’s total integrated magnitude of -6.5 ("Ahad’s constant").

It is thus possible to draw an imaginary sphere around the Sun, within which the Sun would remain the most supreme source of light, relative to the universe’s total background illumination:

Beyond the outer fringes of this theoretical sphere, our Sun would become overpowered by the collective light from the rest of the universe... Where the Sun would no longer be the most supreme source of light and heat to an interstellar traveller. It marks the edge of the Sun’s effective light dominion.

Please leave your valuable comments after reading.

http://www.openmindsblogspot.com/HOME/tabid/36/Default.aspx 

Flux equations for Ahad's constant derivation, their result and the light dominion sphere whose radius is the famous "Ahad radius"  were published in the Mathaba News Network here. Comments from a wide cross-section of the world public were collated and posted here.