How far could you see from a planet

[StrangeCoin]

How far could you see, on average, from a planet that was surrounded with infinite number of identical stars like the Sun, randomly (uniformly) distributed with their average distance of 10 light years apart? In other words, at what point of distance will such distribution tend to "close in" and block the view to any further away stars. Thanks.

 

[micromass]

Countably infinite or uncountably infinite?

Do we treat stars as points?

 

[StrangeCoin]

Countably infinite or uncountably infinite?

I didn't know the first one existed. If I have to pick, then whatever is more infinite, I think the second one.

Do we treat stars as points?

Not sure, if necessary I guess. What's the alternative and how they produce different results?

 

[micromass]

I didn't know the first one existed. If I have to pick, then whatever is more infinite, I think the second one.

Then I can't answer it since I have no idea how to do probability with uncountably many objects in the way you mean. I'm not even sure if the 10 light year condition can ever be satisfied.

Picking countable infinity has problem too since the answer is then automatically infinity. Unless you don't treat stars a points but as little spheres.

Also, I'm not sure what your 10 light years condition actually means. Do you mean (naively) that the average of all distances between ALL individual stars is 10 light years? Then you probably won't have a uniform distribution.

About a uniform distribution. Do you assume space to be infinite and that the stars can literally pick any place in space with the same probability? I'm afraid this is impossible.

 

[Stephen Tashi]

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

 

[StrangeCoin]

Then I can't answer it since I have no idea how to do probability with uncountably many objects in the way you mean. I'm not even sure if the 10 light year condition can ever be satisfied.

You know what I mean, please re-define it in whatever way is the most appropriate. As for the distribution, maybe if instead say there is on average 10 stars per any 100 LY cube volume of space?

Picking countable infinity has problem too since the answer is then automatically infinity. Unless you don't treat stars a points but as little spheres.

Whichever is more "realistic", spheres we need I'd say.

About a uniform distribution. Do you assume space to be infinite and that the stars can literally pick any place in space with the same probability? I'm afraid this is impossible.

Yes, infinite universe with infinite stars, it's just a hypothetical scenario. Surely the stars will "close in" and block the view completely to further away stars before the distance reaches infinity, so infinity shouldn't bother us really.

 

[Dr J L]

I'm no scientist, but It seems to me you would have to take into account perception, of distance before you can determine starting point. Because each angle will be a different distance. The probabilities will be different due to perception.

 

[Matterwave]

So, I'm going to make the assumption that you mean that in every 1000 (10^3) cubic light years you have roughly 1 star the size of the Sun.

We can make some estimates based on this (of course a better answer requires more details).

A star the size of the sun of radius R located at distance D obscures an area of the sky that is basically going to be proportional to $\frac{\pi R^2}{4\pi D^2}=\frac{1}{4}\frac{R^2}{D^2}$. This is how many steradians of the sky you will be obscuring. The sky has a total of 4pi steradians.

A lower bound for the distance can be found if the stars are not allowed to "overlap" in the sky. In this case, you need that:

$$\sum_i \frac{1}{4}\frac{R^2}{D_i^2}=4\pi$$

In a spherical shell of thickness $dD$ (assumed small), located a distance $D$ from you, there are roughly $\frac{4\pi D^2 dD}{1000 ly^3}$ stars in this shell.

So we should integrate out to find:

$$4\pi=\int_0^x \frac{1}{4}\frac{R^2}{D^2}\frac{4\pi D^2}{1000 ly^3} dD$$

This gives:

$$4\pi=\frac{R^2 x}{1000 ly^3}$$

Substituting the radius of the Sun, we find:

$$x=\frac{(4\pi)(1000 ly^3)}{R^2}\approx 2.3\times 10^{18}ly$$

This is way larger than our current observable universe (50 million times larger).

 

[StrangeCoin]

I'm no scientist, but It seems to me you would have to take into account perception, of distance before you can determine starting point. Because each angle will be a different distance. The probabilities will be different due to perception.

Doctor but not scientist, a doctor of love I guess. Doc, I think it's misleading to look at it from the observer point of view. It would perhaps be more insightful if we look at it from the point of view of a photon that needs to travel all the way to our little planet without crashing into a star on it's way.

Or perhaps better yet, we could look at it from above, from birds-view perspective. Imagine a wooden board with randomly placed nails and you have a small ball that needs go through the nails from one side of the board to another. Field of view and perspective doesn't really matter, the chance for the ball to pass the nails depends only on the distribution of the nails and the distance the ball needs to travel, plus the size of the nails and the size of the ball of course.

 

[Dr J L]

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.

Observation starts with a thought that creates a mathematical equation. That equation needs to be put to test. Doesn't matter the size of object or the velocity it is the starting point as to where things end up. To start from the observer to the observee to get reasoning and deduction to make a mathematical equation. So basically science has not figured out yet that the universe is infinite and they need a starting point. That point can be on Earth or on the moon or in water. I don't want to get off topic, cause I'm here to learn. Basically in the human body same as any planet or star there is random energy, that science has yet to explain.

 

[StrangeCoin]

A lower bound for the distance can be found if the stars are not allowed to "overlap" in the sky. In this case, you need that:

I think not allowing them to overlap imposes unnatural constraints on the starting premise that the distribution is random. Valuable example in any case.

This is way larger than our current observable universe (50 million times larger).

Thank you for that. It's beyond me though, so could you please tell us something about how it scales, how that distance varies with the stars distribution density, what is their relation?

 

[Hornbein]

I think not allowing them to overlap imposes unnatural constraints on the starting premise that the distribution is random. QUOTE]


That's true. Each shell of stars will reduce non-star space by a percentage, not linearly. So non-star space will never reach 0%, though it can get arbitrarily close.

 

[Matterwave]

I think not allowing them to overlap imposes unnatural constraints on the starting premise that the distribution is random. Valuable example in any case.

This is true. But putting in a truly random distribution means additionally calculating the probability that a star farther away is obscured in view by a closer star in the distribution. No simple way to implement this come to mind. In addition, stars are NOT randomly distributed, they are clumped together in galaxies and galaxy clusters and super clusters and super structures, etc. So, it's not like the starting premise was a realistic one anyways.

Thank you for that. It's beyond me though, so could you please tell us something about how it scales, how that distance varies with the stars distribution density, what is their relation?

You can see this in the final answer:

$$x=\frac{4\pi\bar{D}^3}{R^2}$$

Where $\bar{D}$ is the average distance between stars. So if the average distance between stars was 2 times closer, then $x$ would be smaller by a factor of 8.

The largeness of $x$ reflects the fact that space is mostly empty space. The average distance between stars is much much larger than the average size of a star (assumed to be the Sun), thus, each star only obstructs a tiny amount of your field of view. This can be readily seen in the night sky. The observable universe is several tens of billions of light years in diameter. In all this space, there are >200 billion galaxies on average each containing more than a hundred billion stars. Of all of these stars and all of this distance, the night sky is still dark.

 

[StrangeCoin]

That's true. Each shell of stars will reduce non-star space by a percentage, not linearly. So non-star space will never reach 0%, though it can get arbitrarily close.

When you say "space" in that context it seems to me we are not talking about the same thing. I'm really talking about a chance, some probability percentage, rather than anything geometrical.

x   x        x   x    x     x   xx    x     x
  x       x          x     x        x      x
    x    x    xx        x  xx          x  
  x                   x x       x    xx    x
     x    x    x               x      x
  x           x      xx  x                 x             [o] cyclops eye
x     x x     x  x          x    x
            x          x        x      x      x
    x x   x x           x  xx            x  
 x             x x       x    x    x
     xx       x       x x        x      x
  x       x       x           x         xx  x

This is how I see the problem, in 2D. All the 'x' are shooting little balls, as the distance to x increases, the chance for its balls to reach the eye, decreases. Now, if that chance does not drop to zero before the distance reaches infinity, then that means the distribution was not random to start with. Would that be true, can we make such claim about "random"?

 

[StrangeCoin]

This is true. But putting in a truly random distribution means additionally calculating the probability that a star farther away is obscured in view by a closer star in the distribution. No simple way to implement this come to mind. In addition, stars are NOT randomly distributed, they are clumped together in galaxies and galaxy clusters and super clusters and super structures, etc. So, it's not like the starting premise was a realistic one anyways.

The thing about it is that, perhaps somewhat contrary to intuition, the more they obscure each other, the more they "clump" or "cluster" together, the number of visible stars would actually become less, not more.

So then the maximum "clump" factor would be if all the stars were in-line one right after another, and then only the closest star would be visible. It would be interesting to see results if we could include this "clump" factor into the equation.

I was searching the internet to find something related to this problem, but I miserably failed, likely due to my inability to recognize the problem expressed in some unrelated example. "Poisson distribution" kept showing in the search though, so whatever is that about might hold a solution, or at least point to some other, maybe more practical, way to think about it.

So if the average distance between stars was 2 times closer, then x would be smaller by a factor of 8.

Thanks.

 

[Matterwave]

If you input clumping of the stars, then obviously the answer will depend on which direction you look in. If you tend to look in a direction with a lot of stars, then you won't be able to see as far, if you look in a direction with few stars, then you can look much farther.

 

[StrangeCoin]

If you input clumping of the stars, then obviously the answer will depend on which direction you look in. If you tend to look in a direction with a lot of stars, then you won't be able to see as far, if you look in a direction with few stars, then you can look much farther.

We are looking for some 'average' number then. But, "direction with a lot of stars" and "direction with fewer stars" doesn't really make sense in relation to infinity, nor random distribution. Every direction should on average hold the same number of stars, some grouped closer, some grouped further away, but at some point in infinity every line of sight is supposed to go through about equal amount of stars.

 

[Matterwave]

We are looking for some 'average' number then. But, "direction with a lot of stars" and "direction with fewer stars" doesn't really make sense in relation to infinity, nor random distribution. Every direction should on average hold the same number of stars, some grouped closer, some grouped further away, but at some point in infinity every line of sight is supposed to go through about equal amount of stars.

Well, if you can come up with a model for that, let me know.

 

[StrangeCoin]

Well, if you can come up with a model for that, let me know.

I don't expect to come up with it, but rather to find it. Someone must have thought of something like this before. If I only knew what to look for, or at least recognize it when I find it.

What do you think about this:
- Expected time to completely cover a square with randomly placed smaller squares
http://math.stackexchange.com/quest...a-square-with-randomly-placed-smaller-squares

 

[StrangeCoin]

There is a way to define this problem and get the answer right out of the question. If we say that distribution of the stars is such where along each line of sight there is a certain random chance per light year that a star would be found, then that's also the answer to the question of how long on average would it take for a line of sight to be blocked by a star.

So, there should be a way to transform "density distribution" definition from the original question into this "probability distribution" factor, which would also be the answer. There must be some direct relation between those two.

 

[marcus]

Coin, there is enough space between the stars and galaxies that you can look back through them to a time BEFORE STARS were formed.
The ancient light of the Background was emitted by hot gas around year 370,000 long before any clouds of gas had time to contract and form stars. We see that ancient light coming from pretty much every direction in the sky, so it can't be being effectively blocked.

Space may be infinite in extent (we don't know whether finite or not) but time since start of expansion is finite.

 

[FactChecker]

Countably infinite or uncountably infinite?

Spacing them 10 light years apart makes them countable.

 

[StrangeCoin]

Coin, there is enough space between the stars and galaxies that you can look back through them to a time BEFORE STARS were formed.
The ancient light of the Background was emitted by hot gas around year 370,000 long before any clouds of gas had time to contract and form stars. We see that ancient light coming from pretty much every direction in the sky, so it can't be being effectively blocked.

Space may be infinite in extent (we don't know whether finite or not) but time since start of expansion is finite.

Could we describe distribution of the stars in our universe with some probability density function maybe? Sort of, what is the probability for any random line of sight to intersect a star per one light year? Or a bit simpler version, how far there is a star 'on average'?

 

[FactChecker]

On any section on a line of sight the size of the stars and the average distance between them would give a probability that a star would block that line on that section. If you allow the stars to be positioned randomly, with an average distance between stars, this would probably be a Poisson process. There would be an average rate of stars being on the line of sight and it could be calculated fairly easily.

Just a rough back-of-the-envelope to get things started: sun diameter ~ 1.4x10^9 meters; 10 light years ~ 10^17 meters. So let's say the odds of a star being on a 10 light-year section are about 1.4x10^9 / 10^17 = 1.4x10^-8. The average distance to a star that blocks the line of sight is about 1/(1.4x10^-8) * 10 light years ~ 7x10^8 light years.

 

[StrangeCoin]

On any section on a line of sight the size of the stars and the average distance between them would give a probability that a star would block that line on that section. If you allow the stars to be positioned randomly, with an average distance between stars, this would probably be a Poisson process. There would be an average rate of stars being on the line of sight and it could be calculated fairly easily.

Just a rough back-of-the-envelope to get things started: sun diameter ~ 1.4x10^9 meters; 10 light years ~ 10^17 meters. So let's say the odds of a star being on a 10 light-year section are about 1.4x10^9 / 10^17 = 1.4x10^-8. The average distance to a star that blocks the line of sight is about 1/(1.4x10^-8) * 10 light years ~ 7x10^8 light years.

That's great, it's just that we practically defined the answer when we defined the question, in a more practical way than we are "supposed" to. I mean, we need a way to input different types of "star density" into the equation, with different units, such as geometrical density per unit area. Or better yet to somehow include the "clump" factor to that probability distribution, so we can simulate different types of universes, more natural vis-a-vis clustering, rather than just analyzing uniform "white-noise" type of random distributions.

 

[Matterwave]

On any section on a line of sight the size of the stars and the average distance between them would give a probability that a star would block that line on that section. If you allow the stars to be positioned randomly, with an average distance between stars, this would probably be a Poisson process. There would be an average rate of stars being on the line of sight and it could be calculated fairly easily.

Just a rough back-of-the-envelope to get things started: sun diameter ~ 1.4x10^9 meters; 10 light years ~ 10^17 meters. So let's say the odds of a star being on a 10 light-year section are about 1.4x10^9 / 10^17 = 1.4x10^-8. The average distance to a star that blocks the line of sight is about 1/(1.4x10^-8) * 10 light years ~ 7x10^8 light years.

I don't think this is right. The farther away something is, the smaller it must appear. It is the square of the radius and distance that matters because the surface area obscured by a star and the total surface area of the celestial sphere at some distance are related to the square of the radius of the star and the distance to the star respectively.

If you square each and divide, you get the same answer I got earlier.

 

[StrangeCoin]

I don't think this is right. The farther away something is, the smaller it must appear.

How can appearance matter? Look at it from the 3rd person perspective, from above. There is a straight line from an eye to every star, and whether this line will be intercepted by some other star in between does not depend on the field of view or magnification, nor on any other subjective or optical property. The chance of interception depends solely on distance and the function of density distribution of the stars in the universe, plus (average) size of the stars.

 

[Matterwave]

How can appearance matter? Look at it from the 3rd person perspective, from above. There is a straight line from an eye to every star, and whether this line will be intercepted by some other star in between does not depend on the field of view or magnification, nor on any other subjective or optical property. The chance of interception depends solely on distance and the function of density distribution of the stars in the universe, plus (average) size of the stars.

Whether your line of sight ends in a star or not depends on the relative portion of the sky that the star is obstructing. The farther away the star is, the less portion of the sky that star is obstructing. The ratio with which a star obstructs your view is $\frac{R^2}{4D^2}$. As such, we need to take the square of the radius to distance ratio, not the linear radius to distance ratio. This is the same as in my previous analysis. Did you not read it?

 

[StrangeCoin]

Whether your line of sight ends in a star or not depends on the relative portion of the sky that the star is obstructing. The farther away the star is, the less portion of the sky that star is obstructing. The ratio with which a star obstructs your view is $\frac{R^2}{4D^2}$. As such, we need to take the square of the radius to distance ratio, not the linear radius to distance ratio.

You didn't really disagree. You are talking about some specific field of view and counting stars that would fit into it. I am talking about general, or on average, distance at which the stars will have zero chance for their photons to reach us, and the field of view is simply not a part of this equation.

This is the same as in my previous analysis. Did you not read it?

I thought you defined 360 degrees field of view, so it seemed that equation is "general" in the same way I was thinking about it. But with your latest explanation I think we are not talking about the same thing anymore.

 

[marcus]

... what is the probability for any random line of sight to intersect a star per one light year? Or a bit simpler version, how far there is a star 'on average'?

For all practical purposes I believe the answer to the first question is ZERO probability.

The second question is how far is there a star on a random line of sight. Since there is almost never a star on a random line of sight (in the real universe we see) the answer to this question would be UNDEFINED.

Lines of sight comprise a set in the 4D spacetime called the LIGHT CONE. Rays in the past light cone go back to before there were stars. e.g. to the emission of a reddish glow from hot gas (which has become our circa 1 mm wavelength ancient light background.
The vast majority of rays in the past light cone do not intersect a star. Or so I think anyway .

 

[Matterwave]

You didn't really disagree. You are talking about some specific field of view and counting stars that would fit into it. I am talking about general, or on average, distance at which the stars will have zero chance for their photons to reach us, and the field of view is simply not a part of this equation.

I thought you defined 360 degrees field of view, so it seemed that equation is "general" in the same way I was thinking about it. But with your latest explanation I think we are not talking about the same thing anymore.

I'm saying that if you were to pay attention to the derivation, you would see that the answer to your second question "how far do we see on average before we see a star" is actually (approximately) the answer I gave you in my post #8 (assuming roughly random distribution of stars). The question you asked is really just another way of asking the question I answered.

My post #8 tells you how far you go before roughly all of the sky is covered by stars if they didn't overlap. This is just the approximate converse to the statement of how far you have to look on average before your line of sight hits a star.

@Marcus: I think the OP is talking about a purely mathematical question, and he is not talking about the current universe, but a theoretical infinite and static universe.

 

[StrangeCoin]

I'm saying that if you were to pay attention to the derivation, you would see that the answer to your second question "how far do we see on average before we see a star" is actually (approximately) the answer I gave you in my post #8 (assuming roughly random distribution of stars). The question you asked is really just another way of asking the question I answered.

I'm not sure 4Pi should be a part of the equation. The number I'm talking about does not depend on the field of view or solid angle of the stars in the foreground. The number I'm talking about is solely a function of general density distribution and average star diameter, nothing else. Are you sure we talking about the same number?

@Marcus: I think the OP is talking about a purely mathematical question, and he is not talking about the current universe, but a theoretical infinite and static universe.

I'm talking about both. Hopefully we'll end up with some equation that would be applicable to our actual universe as well.

 

[Matterwave]

I'm talking about both. Hopefully we'll end up with some equation that would be applicable to our actual universe as well.

For our actual universe, you will, on average, see the CMBR before you see a star. The universe is mostly empty space. Take a look at the Hubble deep field. It's mostly blackness.

 

[StrangeCoin]

For all practical purposes I believe the answer to the first question is ZERO probability.

http://cdn4.sci-news.com/images/2013/12/image_1590-Million-Stars-Catalogue.jpg
http://www.sci-news.com/astronomy/science-catalogue-stars-galaxies-01590.html

Looks more like 83% to me.

 

[FactChecker]

The OP asked a question with specific assumptions. The question was simple and the answer is fairly easy to estimate. My estimate is 7x10^8 light years. Since the diameter of the observable universe is much larger (93x10^9), that result would make the night sky bright. So the assumptions do not fit the real universe (with clustering due to gravity, limited number of stars, etc.) Still, a simple model may serve a purpose if only to show how much should be added.

 

[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.