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 .