How far could you see from a planet

[mfb]

Clumping as it happens in our universe is a small effect - even for the largest scales, superclusters, the solid angle coverage of the stars is small compared to their size in the sky, so overlap within a cluster is rare.

The real star density is even lower than assumed in the first post - okay, this is partially canceled by some very large stars.@FactChecker: We don't live in 2 dimensions, you need the squared radius and star spacing.

Assume every (10 ly)^3 cube has one star (a random distribution gives the same result). A light ray (or imaginary "sight ray"), going through this cube, has a chance of $\frac{pi r^2}{(10ly)^2}$ to hit the star with the radius r. Therefore, as expectation value, we have to go through $\frac{(10ly)^2}{pi r^2}$ cubes, for a distance of $\frac{(10ly)^3}{pi r^2}$. Plugging in the solar radius, this gives 59*10¹⁵ light years. This is still way beyond the observable universe.

The large difference to Matterwave is the difference between the expectation value (this post) and the maximal distance (Matterwave's calculation). The overlap is just a factor of ~2.

@StrangeCoin: The stars in those image are all smaller than a pixel - you just see the finite resolution of the telescope.

I'm just here to learn. But learning start from perception within any mathematical equation on earth. If you were say on the moon or any where that does not have gravitational, pull then the perception would be different. Perception starts at any stand point, doesn't matter if on the ground or on Mars or on the moon. The variables will always be different unless in a double blind study. Plus when you are in water the perception will be different because the gravitational pull is different.

Sorry, but that does not make any sense. Gravity is not relevant here. And gravity does not change if you are in water.

 

[FactChecker]

@FactChecker: We don't live in 2 dimensions, you need the squared radius and star spacing.

Assume every (10 ly)^3 cube has one star (a random distribution gives the same result). A light ray (or imaginary "sight ray"), going through this cube, has a chance of $\frac{pi r^2}{(10ly)^2}$ to hit the star with the radius r. Therefore, as expectation value, we have to go through $\frac{(10ly)^2}{pi r^2}$ cubes, for a distance of $\frac{(10ly)^3}{pi r^2}$. Plugging in the solar radius, this gives 59*10¹⁵ light years. This is still way beyond the observable universe.

The large difference to Matterwave is the difference between the expectation value (this post) and the maximal distance (Matterwave's calculation). The overlap is just a factor of ~2.

I see your point. I had grossly underestimated the difference between 3 and 1 dimensional average distance. I guess this is the same point that @Matterwave was making.

 

[StrangeCoin]

For our actual universe, you will, on average, see the CMBR before you see a star.

How would you know it's CMBR and not some distant galaxies?

The universe is mostly empty space. Take a look at the Hubble deep field. It's mostly blackness.

In my universe it looks like Christmas.

http://en.wikipedia.org/wiki/Hubble_Ultra-Deep_Field

 

[StrangeCoin]

Perhaps Olbers figured something out: http://en.wikipedia.org/wiki/Olbers'_paradox

You know, that actually looks like much better (clearer) way to think about it.

Suppose these shells are rather thin and the stars do not overlap within their shell, so the 10 stars in the first shell are 100% visible. Then, for example, shell two would be 98% visible, shell three 95% visible, and so on.

The first question is how much this visibility percentage actually drops per each successive shell, does it ever reach 0% visibility, and if so at what shell would that be?

The second question is like the first one, only we make the stars always spawn in pairs with every two of them right next to each other. Will the result be the same?

The third question is also like the first and second one, but now we make the stars spawn in clusters randomly ranging from 1 alone to 10 of them right next to each other. Will the result be the same?


I bet 1,000 points the second result will be the same as the first one, but the third result will point to some closer shell where visibility drops to zero percent. Place your bets, please.

 

[Matterwave]

How would you know it's CMBR and not some distant galaxies?




In my universe it looks like Christmas.

http://en.wikipedia.org/wiki/Hubble_Ultra-Deep_Field

And this is what the sky looks like when a large portion of the view is starlight:

I'm going to go out on a limb and say that the Hubble picture is mostly black.

I'll make one last attempt to make my point, and then I'm out. The universe is mostly empty space. Let's look at the ACTUAL average density of stars in the observable universe. The observable universe is ~45 billion light years in radius. http://en.wikipedia.org/wiki/Observable_universe

It has ~ 10^22-10^24 stars in it. http://en.wikipedia.org/wiki/Observable_universe#Extrapolation_from_number_of_stars Let's use the upper limit of 10^24.

The average number of stars per cubic light year is:

$$n=\frac{10^24}{4\pi(45,000,000,000ly)^3/3}\approx3\cdot10^{-9}/ly^3$$

This is roughly 1 star per every 400,000,000 cubic ly, for an average distance between stars at 700 light years. 400,000 times more sparse than your estimate.

Using YOUR much more densely packed universe, we STILL get a ridiculously long distance before your line of sight hits a star on average.

Space is mostly space. From here on out, do whatever you want.

 

[StrangeCoin]

The average number of stars per cubic light year is:

Can you explain a bit your equation, regarding solid angles you were talking about?

Is this what you are calculating, kind of "shadows" stars cast on background stars? Will that equation work on that example from Olbers' paradox?

Using YOUR much more densely packed universe, we STILL get a ridiculously long distance before your line of sight hits a star on average.

I don't understand what is supposed to be the significance of getting ridiculously long distance, or not. I find all the distances equally interesting, especially how they vary in relation to different density distributions and clustering, which unfortunately we still don't know.

 

[mfb]

How would you know it's CMBR and not some distant galaxies?

Both spectrum and angular distribution do not match for galaxies. In addition, those galaxies would have appeared and vanished all at the same time suddenly and show many more weird effects. No, that does not work.

In my universe it looks like Christmas.

Did you see my comment about angular resolution?

The first question is how much this visibility percentage actually drops per each successive shell, does it ever reach 0% visibility, and if so at what shell would that be?

It drops by a tiny amount (10*pi*(radius of sun)^2 / (4 pi (10 light years)^2), and this value has been calculated before. As average value, you never reach zero. In a static, infinite universe (which is not the universe we live in) you would eventually reach zero just by random chance. Even with your unrealistic high star concentration, this is so far away that it has no relevance in our universe.

The second question is like the first one, only we make the stars always spawn in pairs with every two of them right next to each other. Will the result be the same?

The shape of objects (circle, two circles, square, whatever) does not change the result.

The third question is also like the first and second one, but now we make the stars spawn in clusters randomly ranging from 1 alone to 10 of them right next to each other. Will the result be the same?

Yes.

I bet 1,000 points the second result will be the same as the first one, but the third result will point to some closer shell where visibility drops to zero percent. Place your bets, please.

This is not a betting game, it is mathematics.

I don't understand what is supposed to be the significance of getting ridiculously long distance, or not. I find all the distances equally interesting, especially how they vary in relation to different density distributions and clustering, which unfortunately we still don't know.

If the distance is a million times the largest distance we can see in our universe, it shows that we see mostly no stars if we look in an arbitrary direction (=the universe we live in). If the distance would be 1/10 of the observable universe, we would see stars everywhere (=not what we see).
This is a significant difference.

 

[StrangeCoin]

It drops by a tiny amount (10*pi*(radius of sun)^2 / (4 pi (10 light years)^2), and this value has been calculated before. As average value, you never reach zero.

Looks more like geometry than statistics. What probability equation is that based on?

(The third question is also like the first and second one, but now we make the stars spawn in clusters randomly ranging from 1 alone to 10 of them right next to each other. Will the result be the same?)

Yes.

So we disagree.

 

[StrangeCoin]

It drops by a tiny amount (10*pi*(radius of sun)^2 / (4 pi (10 light years)^2), and this value has been calculated before.

Is this what you are calculating?

 

[mfb]

Looks more like geometry than statistics. What probability equation is that based on?

This is indeed just geometry. There is no statistics needed at this step.

Is this what you are calculating?

50% visible? Where does that value come from?


You seem to have some goal for this thread in mind which I don't see. Can you explain what your real question is? "How far can we see" has been answered multiple times now.

 

[StrangeCoin]

This is indeed just geometry. There is no statistics needed at this step.

What's the next step?

50% visible? Where does that value come from?

Can you not see the image I posted? The percentage comes from the practical implementation of the problem as represented by that image. I simply projected "shadows" of the stars in the first shell over the second shell and counted the stars left outside those shadow areas. That's what I believe your equation is evaluating, I don't see what else could it be.

You seem to have some goal for this thread in mind which I don't see. Can you explain what your real question is? "How far can we see" has been answered multiple times now.

I'm not satisfied with the answer. Even if I accept the answer is correct, which I can't because I'm not sure I understand it, I would still be doubtful of the scope of its application. So I'm trying to verify it by applying it to different scenarios.

Goal? I enjoy solving problems I have no any prior knowledge about, so I learn new things. For me it's not about getting an answer, but about understanding it, and talking about it. This is not a homework, it's genuine interest for the purpose of satisfying intellectual hunger and also an entertainment, kind of like solving a crosswords puzzle. Isn't that exactly why you and everyone else is here on this forum?

 

[FactChecker]

The OP was asking about the mathematics of infinite sun-like stars randomly distributed with an average of 10 LY separation. The answer is that every Line Of Sight (LOS) will eventually be blocked by a star. The probability of a star blocking any LY section on the LOS is independent of distance from observer. The calculations for different shells is only confusing the mathematics.

 

[StrangeCoin]

The probability of a star blocking any LY section on the LOS is independent of distance from observer.

How did you arrive to that conclusion? Surely if you walk a straight path across a random mine field you will have a higher chance to step on a mine if the mine field is 100 meters long than if it was 10.

 

[mfb]

What's the next step?

Multiplying the fractions of unoccupied space for each shell.

Can you not see the image I posted?

I can, but a shell that blocks 50% of the solid angle is an impractical approach. Your stars are way too huge and nearby to match your conditions, yet even the star density in our universe. Also, images are a good way to visualize things, but not to replace calculations here.

I would still be doubtful of the scope of its application.

The application: we can see the cosmic microwave background.

How did you arrive to that conclusion? Surely if you walk a straight path across a random mine field you will have a higher chance to step on a mine if the mine field is 100 meters long than if it was 10.

Yes, but for each meter the chance to hit a mine is the same.

 

[StrangeCoin]

(10*pi*(radius of sun)^2 / (4 pi (10 light years)^2)

10*pi*(radius of sun)^2

Is this circle area multiplied by 10 stars in the first shell? How can this describe density if the number of stars and their total coverage area is not related per some defined field of view area?


4 pi (10 light years)^2

Where did 4pi come from? How can this describe density if there is no defined number of stars and their total coverage area per some defined field of view area?


In other words...

INPUT shell 1:
field of view = 30 degrees
1st shell number of stars = 10
1st shell star radius = sun radius
distance to 1st shell = 10 light years

INPUT shell 2:
field of view = 30 degrees
2nd shell number of stars = 40
2nd shell star radius = sun radius
distance to 2nd shell = 20 light years
----

OUTPUT:
number of stars visible in the 2nd shell = ??

So what I'm saying is that I don't see that equation takes all the necessary input parameters into account.

 

[FactChecker]

How did you arrive to that conclusion? Surely if you walk a straight path across a random mine field you will have a higher chance to step on a mine if the mine field is 100 meters long than if it was 10.

Yes, but any section of the line of sight that is one light year long has the same probability of a star being on it as any other section that is one light year long. So the probability is just proportional to length, and sections of equal length have equal probabilities.

 

[StrangeCoin]

Yes, but any section of the line of sight that is one light year long has the same probability of a star being on it as any other section that is one light year long. So the probability is just proportional to length, and sections of equal length have equal probabilities.

Probability is inversely proportional to length, yes, and it is also inversely proportional to the size and number of the stars per unit volume of space.

However, I don't think we can get probability from purely geometrical relations, I believe we need to wrap that up into some probability equation.

The other thing is that the equation suggested as the answer deals with solid angles and field of view, some sort of perspective projections, but we see in the first sentence above those things are not really a part of the equation.

 

[mfb]

10*pi*(radius of sun)^2

Is this circle area multiplied by 10 stars in the first shell? How can this describe density if the number of stars and their total coverage area is not related per some defined field of view area?

Take the whole field of view all around you. A sphere with a radius of 10 light years has an area of 4 pi (10 light years)^2

field of view = 30 degrees
1st shell number of stars = 10

Now you modify the star density - but that does not change the general idea, you just get some other prefactor.

OUTPUT:
number of stars visible in the 2nd shell = ??

The expected number of visible stars is the total number of stars multiplied by the fraction of empty space in the first shell. If you don't choose an arbitrary 30 degrees field but take the full shell, the fraction occupied by the first shell is:

(10*pi*(radius of sun)^2 / (4 pi (10 light years)^2)

That's such a simple concept, I'm running out of ideas how to explain this in even more different ways.

 

[FactChecker]

The other thing is that the equation suggested as the answer deals with solid angles and field of view, some sort of perspective projections, but we see in the first sentence above those things are not really a part of the equation.

As the distance increases, the arc blocked by one star decreases, but also the arc not blocked decreases by exactly the same factor. Since the stars are infinite and remain spaced on average at 10 light years, there are always enough stars to keep the percentage of blocked sky at a fixed average no matter how far away the "shell" being calculated is. So the odds of a particular line of sight being blocked is the same in the hundredth light year distance as it is in the first light year distance.

 

[StrangeCoin]

Take the whole field of view all around you. A sphere with a radius of 10 light years has an area of 4 pi (10 light years)^2

All the terms in the equation are properties of the 1st shell, and there is no any information about the 2nd shell?

So visibility of the 2nd shell does not depend on its distance from the 1st shell, nor it depends on the distance to the observer?

 

[mfb]

All the terms in the equation are properties of the 1st shell, and there is no any information about the 2nd shell?

The thing you quoted is for the first shell, but the second shell just gets an additional factor of 4 both in the numerator and denominator, and those cancel.

So visibility of the 2nd shell does not depend on its distance from the 1st shell, nor it depends on the distance to the observer?

The shells are chosen to be equidistant in this approach, otherwise it is messy to implement a constant star density. The visible fraction of the second shell (= the fraction of shell 1 that is not blocked!) does not depend on the distance of the shell, right.