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Is there a star over my head?
Is there a theoretical maximum size limit for a star?Is there any planet/star bigger than VY Canis Majoris?Is there any practical use for astronomy?Is there a publicly available Gliese Star ListAre there good animations of the movement of proximate stars over long time periods?How do star densities work?How could lithium burning take place in a quasi-star?Are there any known asteroids with average density similar to that of Earth's?
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Say I'm standing up straight, and I draw a straight line from my core through the top of my head (perpendicular to the ground). What is the probability that that line intersects with a star?
EDIT: I'm not trying to exclude any stars. This should include stars which we've observed and stars which we haven't yet observed but can predict due to other things we've determined (like the overall star density of the universe). Also it should include all stars regardless of naked eye magnitude limit.
star universe density hypothetical
New contributor
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add a comment |
$begingroup$
Say I'm standing up straight, and I draw a straight line from my core through the top of my head (perpendicular to the ground). What is the probability that that line intersects with a star?
EDIT: I'm not trying to exclude any stars. This should include stars which we've observed and stars which we haven't yet observed but can predict due to other things we've determined (like the overall star density of the universe). Also it should include all stars regardless of naked eye magnitude limit.
star universe density hypothetical
New contributor
$endgroup$
1
$begingroup$
Presumably you mean a naked eye magnitude star ? As the magnitude limit increases towards fainter stars, the probability is going to get very close to 1...
$endgroup$
– astrosnapper
8 hours ago
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@astrosnapper that's not obvious because of the finite age of the universe.
$endgroup$
– Steve Linton
7 hours ago
1
$begingroup$
Related: Olbers' paradox
$endgroup$
– Mike G
6 hours ago
add a comment |
$begingroup$
Say I'm standing up straight, and I draw a straight line from my core through the top of my head (perpendicular to the ground). What is the probability that that line intersects with a star?
EDIT: I'm not trying to exclude any stars. This should include stars which we've observed and stars which we haven't yet observed but can predict due to other things we've determined (like the overall star density of the universe). Also it should include all stars regardless of naked eye magnitude limit.
star universe density hypothetical
New contributor
$endgroup$
Say I'm standing up straight, and I draw a straight line from my core through the top of my head (perpendicular to the ground). What is the probability that that line intersects with a star?
EDIT: I'm not trying to exclude any stars. This should include stars which we've observed and stars which we haven't yet observed but can predict due to other things we've determined (like the overall star density of the universe). Also it should include all stars regardless of naked eye magnitude limit.
star universe density hypothetical
star universe density hypothetical
New contributor
New contributor
edited 7 hours ago
user68873
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asked 8 hours ago
user68873user68873
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1
$begingroup$
Presumably you mean a naked eye magnitude star ? As the magnitude limit increases towards fainter stars, the probability is going to get very close to 1...
$endgroup$
– astrosnapper
8 hours ago
$begingroup$
@astrosnapper that's not obvious because of the finite age of the universe.
$endgroup$
– Steve Linton
7 hours ago
1
$begingroup$
Related: Olbers' paradox
$endgroup$
– Mike G
6 hours ago
add a comment |
1
$begingroup$
Presumably you mean a naked eye magnitude star ? As the magnitude limit increases towards fainter stars, the probability is going to get very close to 1...
$endgroup$
– astrosnapper
8 hours ago
$begingroup$
@astrosnapper that's not obvious because of the finite age of the universe.
$endgroup$
– Steve Linton
7 hours ago
1
$begingroup$
Related: Olbers' paradox
$endgroup$
– Mike G
6 hours ago
1
1
$begingroup$
Presumably you mean a naked eye magnitude star ? As the magnitude limit increases towards fainter stars, the probability is going to get very close to 1...
$endgroup$
– astrosnapper
8 hours ago
$begingroup$
Presumably you mean a naked eye magnitude star ? As the magnitude limit increases towards fainter stars, the probability is going to get very close to 1...
$endgroup$
– astrosnapper
8 hours ago
$begingroup$
@astrosnapper that's not obvious because of the finite age of the universe.
$endgroup$
– Steve Linton
7 hours ago
$begingroup$
@astrosnapper that's not obvious because of the finite age of the universe.
$endgroup$
– Steve Linton
7 hours ago
1
1
$begingroup$
Related: Olbers' paradox
$endgroup$
– Mike G
6 hours ago
$begingroup$
Related: Olbers' paradox
$endgroup$
– Mike G
6 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
In short: no one knows for sure, but currently it looks that the probability is 1.
Longer: On our current understanding, the Universe is probably infinite in space. This depends on the recent WMAP satellite results, which have shown a zero curvature of the Universe below measurement precision. The other two options were a positive curvature (thus, we would live a 4D sphere), or a negative:
If the curvature is exactly zero (the last option on the picture), or it is negative, and the Universe doesn't have some exotic topology, then it is infinite.
And an infinite Universe has infinite many stars, thus it doesn't matter, where do you see, somewhere you will find a star.
However, most likely you have no option to actually see it - it is nearly surely over the cosmological horizon, thus there is no way to get any information from it, or interact with it in any sense, due to the expansion of the Universe. Note, the currently accelerating expansion continuously reduces even the count of the stars inside the cosmological horizon.
Without an universal expansion, the whole sky would be filled with stars and it would be so light than the Sun (Olbers paradoxon).
If you count only the stars beside the cosmological horizon, then the probability is very small. The typical size of the stars is in the order of $approx$ 1million km, and they are some light years away from each other ($approx 10^13$ km). They are $approx 10^7$ times more away from each other than their diameter. And even this calculation doesn't count that most space of the Universe isn't filled with any galaxy - the galaxies are disc-like objects around 20 times more far from each other than their diameter. On these reasons, I estimate the probability to $approx underlineunderline10^-17$.
$endgroup$
add a comment |
$begingroup$
Probably, maybe.
There are at least two ways of answering the question. One is to ask what were your coordinates when you wrote the question and exactly what time it was. Then we'll need to draw a line in a model to see what you hit and whether any of those hits are stars. This assumes a complete map, which is a problem. The answer is different for everyone on Earth and changes constantly. It becomes the right question if we are in a starship. Given the vastness of space, it’s probably better to ask “How far until we hit something.”
The other answer is about probability. How often is a star directly overhead? I'll suggest one way to reason about it. There seem to be a lot of limiting factors. I'll point out a few of those, too.
First, a gut check. Our Sun is directly overhead for a good area of the Earth at all times. The sun is relatively near, so it’s coverage is special. That trillions of billions of other stars have the rest of the planet covered nonetheless seems likely.
An excellent detail of this question is whether the line you are imagining intersects with a star. I take this to mean whether the abstract line passes through any part of the star's mass, not just its center of mass or other centers.
The odds are that we are not at the center of the Universe, if "center of Universe" even has any meaning. It can be argued (it is argued) that we are at the center of the observable universe, essentially because we're looking in all directions with the same limited gear. So we can imagine a giant sphere of observability, just to give this problem some space. Imagine ourselves as a grain of sand floating at the center of a big balloon. In truth, the grain of sand is far too big in proportion to any real balloon, but imagine we are in the dead center of a balloon on an impossibly small grain.
For the balloon's dimensions, consider a sphere with a radius of 4, where the units are $1.1times 10^26$ meters. The surface of that sphere is going to be $4pi r^2$, or $64pi$ square units. If we prefer not to talk in terms with a "$pi$" mixed in, it's roughly 200 of these large square units.
Imagine that this is the area that we are gazing upon from inside the center of the balloon, sitting on our microscopic and impossibly concentric grain of sand. We can only see half of the area at once (even less, really), but we are spinning around. So we can canvas the entire inside surface of the balloon over the course of the day.
So there we are, on this spec of sand, looking at the part of the balloon we can see. One of us has a laser pointer that we can use point to different parts of the balloon and talk about them. In fact, it might be fun to imagine the laser pointer having a kind of "light pen" mode that we can use to draw inscriptions on the surface of the balloon. Plastering your name across the night sky would make for quite a show. For the sake of the illustration, you have to imagine these props having metaphysical properties. We aren't really concerned with the light pen. It's just to imagine that we're drawing lines.
Now imagine that we tried to place inside of the balloon, at scale, all of the stuff of the observable universe, or, for the sake of the question, just the stars. We would put everything inside the balloon precisely where it would be relative to our vantage point.
Now we can go through, one at a time, and consider each star individually. Each time we examine a star, we could draw the line from us to it with our laser pointer. We could use the light pen to trace the outline of the star with the laser pointer, inscribing a little circle on the surface of the balloon behind it. Every time we did this with a particular star, we would add a circle on the balloon to build up a flat map of the stars. We could process each star, one by one, and eliminate each star until the balloon is empty again. It's just us, looking back up at the map we made.
Now let's say the balloon was originally red and our light pen was drawing in green. Let's also say that the green circles we drew were colored in, filled with green. After we've processed all the stars, we've got green dots all over the inside of the balloon. The size of each green dot would first be a function of the size of the star. Bigger stars would tend to draw relatively bigger circles on the map.
This analogy is imperfect in many ways. It is imperfect here in an important respect. If you imagine that we are tracing the stars with a circular motion in the hand, which is natural, then we'll be distorting the map. The angle of the light pen in hand as we made a circular motion would be projected across a great distance. That map would be interesting for other reasons, but we're trying to identify just the areas that are on line with us, stars we are "under." We want the real size of the star to be on the map, not a size relative to the distance between us and it.
To stay true, we'll have to imagine that our map simply has a circle on it whose center is on line with us and the star it represents. The size of the star's circle is its actual size. Our sun is roughly 1.39 million kilometers across, so the circle it draws would have this diameter on our map. This is the area of points that would, regardless of distance, carry a line all the way between them and us to make a candidate for a star being "overhead".
The answer to whether at least one star is probably overhead at a given time is, in one way of thinking, the proportion of red and green on the map. How much of the entire map is green? That's roughly how likely we are to be on line with a star at any time.
If we want to keep going on this line of probability, this would be the time to get the average size of every observable star, calculate an average diameter, multiply it by the number of stars, and have an estimated area. This will be wildly off because we've flattened three or four dimensions into two and didn't account for overlap. Unfortunately, the overlap of what's overhead does not appear to be consistent. Note that when looking up at the night sky we can see the Milky Way, of which we are a part.
Also, to get those averages, you'd have to have really thoroughly indexed the observable Universe. A lot of people have been working on that for a long time, but it's very large. So if we had enough data to have reasonably good averages for things like the size of a star, we may as well forget the averages and make the actual map. We'd take care of overlapping circles that way, too. While we're at it, forget the map entirely. Just have the GPS in your phone feed your position on the globe into a model that will draw the line and check everything above you. It's the real problem we started with, just taking into appreciation that the vastness of the cosmos is so overwhelmingly large that the computation required to check what's overhead may have a shorter radius than the radius of the observable universe. Even a database to hold records of every star and its position is impossibly large.
I also read lately that the universe might be (these are guesses and arguments) at least 250 times larger than what we can observe. I have also read that the earth is flat. Maybe the universe goes on infinitely. Reasoning about that will have similar boundary conditions.
Your best bet is to actually feed your location into a model and limit the model so you can get a reasonably fast calculation. Change the question to: “What is the nearest star on this line, given a spacial and computational boundary?” You'll have to accept that somewhere beyond what can be calculated, even beyond what can be seen, there may still be a star.
New contributor
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1
$begingroup$
Welcome on the Astronomy SE! Look how nice-looking formulas I inserted into your post. It is because we have some Latex support. Type$4pi r^2$
and you will get $4pi r^2$.
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– peterh
4 hours ago
$begingroup$
Very nice. Thank you!
$endgroup$
– keparo
4 hours ago
add a comment |
$begingroup$
According to Olbers, of paradox fame, if the universe is infinite a line of sight in any direction should eventually reach a star. Why then was the night sky so dark when in theory it should be bright as day? Leaving that particular question aside, we have no proof that the universe is infinite, but it is sufficiently large that a line in any direction should sooner or later reach the surface of a star. Whether the line in question would have to travel only tens of light years to reach the star or many billions depends on where you are standing and at what particular moment you choose to draw the line. If you happened to be on the equator at the right time of year and the right time of day, the line might only have to travel a little over eight light minutes to reach a star. In the universe, as opposed to on paper, you cannot have really straight lines, only an approximation, because the universe is curved and any line defining device, a laser beam perhaps, would be acted on by a force and bent.
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$begingroup$
In short: no one knows for sure, but currently it looks that the probability is 1.
Longer: On our current understanding, the Universe is probably infinite in space. This depends on the recent WMAP satellite results, which have shown a zero curvature of the Universe below measurement precision. The other two options were a positive curvature (thus, we would live a 4D sphere), or a negative:
If the curvature is exactly zero (the last option on the picture), or it is negative, and the Universe doesn't have some exotic topology, then it is infinite.
And an infinite Universe has infinite many stars, thus it doesn't matter, where do you see, somewhere you will find a star.
However, most likely you have no option to actually see it - it is nearly surely over the cosmological horizon, thus there is no way to get any information from it, or interact with it in any sense, due to the expansion of the Universe. Note, the currently accelerating expansion continuously reduces even the count of the stars inside the cosmological horizon.
Without an universal expansion, the whole sky would be filled with stars and it would be so light than the Sun (Olbers paradoxon).
If you count only the stars beside the cosmological horizon, then the probability is very small. The typical size of the stars is in the order of $approx$ 1million km, and they are some light years away from each other ($approx 10^13$ km). They are $approx 10^7$ times more away from each other than their diameter. And even this calculation doesn't count that most space of the Universe isn't filled with any galaxy - the galaxies are disc-like objects around 20 times more far from each other than their diameter. On these reasons, I estimate the probability to $approx underlineunderline10^-17$.
$endgroup$
add a comment |
$begingroup$
In short: no one knows for sure, but currently it looks that the probability is 1.
Longer: On our current understanding, the Universe is probably infinite in space. This depends on the recent WMAP satellite results, which have shown a zero curvature of the Universe below measurement precision. The other two options were a positive curvature (thus, we would live a 4D sphere), or a negative:
If the curvature is exactly zero (the last option on the picture), or it is negative, and the Universe doesn't have some exotic topology, then it is infinite.
And an infinite Universe has infinite many stars, thus it doesn't matter, where do you see, somewhere you will find a star.
However, most likely you have no option to actually see it - it is nearly surely over the cosmological horizon, thus there is no way to get any information from it, or interact with it in any sense, due to the expansion of the Universe. Note, the currently accelerating expansion continuously reduces even the count of the stars inside the cosmological horizon.
Without an universal expansion, the whole sky would be filled with stars and it would be so light than the Sun (Olbers paradoxon).
If you count only the stars beside the cosmological horizon, then the probability is very small. The typical size of the stars is in the order of $approx$ 1million km, and they are some light years away from each other ($approx 10^13$ km). They are $approx 10^7$ times more away from each other than their diameter. And even this calculation doesn't count that most space of the Universe isn't filled with any galaxy - the galaxies are disc-like objects around 20 times more far from each other than their diameter. On these reasons, I estimate the probability to $approx underlineunderline10^-17$.
$endgroup$
add a comment |
$begingroup$
In short: no one knows for sure, but currently it looks that the probability is 1.
Longer: On our current understanding, the Universe is probably infinite in space. This depends on the recent WMAP satellite results, which have shown a zero curvature of the Universe below measurement precision. The other two options were a positive curvature (thus, we would live a 4D sphere), or a negative:
If the curvature is exactly zero (the last option on the picture), or it is negative, and the Universe doesn't have some exotic topology, then it is infinite.
And an infinite Universe has infinite many stars, thus it doesn't matter, where do you see, somewhere you will find a star.
However, most likely you have no option to actually see it - it is nearly surely over the cosmological horizon, thus there is no way to get any information from it, or interact with it in any sense, due to the expansion of the Universe. Note, the currently accelerating expansion continuously reduces even the count of the stars inside the cosmological horizon.
Without an universal expansion, the whole sky would be filled with stars and it would be so light than the Sun (Olbers paradoxon).
If you count only the stars beside the cosmological horizon, then the probability is very small. The typical size of the stars is in the order of $approx$ 1million km, and they are some light years away from each other ($approx 10^13$ km). They are $approx 10^7$ times more away from each other than their diameter. And even this calculation doesn't count that most space of the Universe isn't filled with any galaxy - the galaxies are disc-like objects around 20 times more far from each other than their diameter. On these reasons, I estimate the probability to $approx underlineunderline10^-17$.
$endgroup$
In short: no one knows for sure, but currently it looks that the probability is 1.
Longer: On our current understanding, the Universe is probably infinite in space. This depends on the recent WMAP satellite results, which have shown a zero curvature of the Universe below measurement precision. The other two options were a positive curvature (thus, we would live a 4D sphere), or a negative:
If the curvature is exactly zero (the last option on the picture), or it is negative, and the Universe doesn't have some exotic topology, then it is infinite.
And an infinite Universe has infinite many stars, thus it doesn't matter, where do you see, somewhere you will find a star.
However, most likely you have no option to actually see it - it is nearly surely over the cosmological horizon, thus there is no way to get any information from it, or interact with it in any sense, due to the expansion of the Universe. Note, the currently accelerating expansion continuously reduces even the count of the stars inside the cosmological horizon.
Without an universal expansion, the whole sky would be filled with stars and it would be so light than the Sun (Olbers paradoxon).
If you count only the stars beside the cosmological horizon, then the probability is very small. The typical size of the stars is in the order of $approx$ 1million km, and they are some light years away from each other ($approx 10^13$ km). They are $approx 10^7$ times more away from each other than their diameter. And even this calculation doesn't count that most space of the Universe isn't filled with any galaxy - the galaxies are disc-like objects around 20 times more far from each other than their diameter. On these reasons, I estimate the probability to $approx underlineunderline10^-17$.
edited 4 hours ago
answered 5 hours ago
peterhpeterh
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$begingroup$
Probably, maybe.
There are at least two ways of answering the question. One is to ask what were your coordinates when you wrote the question and exactly what time it was. Then we'll need to draw a line in a model to see what you hit and whether any of those hits are stars. This assumes a complete map, which is a problem. The answer is different for everyone on Earth and changes constantly. It becomes the right question if we are in a starship. Given the vastness of space, it’s probably better to ask “How far until we hit something.”
The other answer is about probability. How often is a star directly overhead? I'll suggest one way to reason about it. There seem to be a lot of limiting factors. I'll point out a few of those, too.
First, a gut check. Our Sun is directly overhead for a good area of the Earth at all times. The sun is relatively near, so it’s coverage is special. That trillions of billions of other stars have the rest of the planet covered nonetheless seems likely.
An excellent detail of this question is whether the line you are imagining intersects with a star. I take this to mean whether the abstract line passes through any part of the star's mass, not just its center of mass or other centers.
The odds are that we are not at the center of the Universe, if "center of Universe" even has any meaning. It can be argued (it is argued) that we are at the center of the observable universe, essentially because we're looking in all directions with the same limited gear. So we can imagine a giant sphere of observability, just to give this problem some space. Imagine ourselves as a grain of sand floating at the center of a big balloon. In truth, the grain of sand is far too big in proportion to any real balloon, but imagine we are in the dead center of a balloon on an impossibly small grain.
For the balloon's dimensions, consider a sphere with a radius of 4, where the units are $1.1times 10^26$ meters. The surface of that sphere is going to be $4pi r^2$, or $64pi$ square units. If we prefer not to talk in terms with a "$pi$" mixed in, it's roughly 200 of these large square units.
Imagine that this is the area that we are gazing upon from inside the center of the balloon, sitting on our microscopic and impossibly concentric grain of sand. We can only see half of the area at once (even less, really), but we are spinning around. So we can canvas the entire inside surface of the balloon over the course of the day.
So there we are, on this spec of sand, looking at the part of the balloon we can see. One of us has a laser pointer that we can use point to different parts of the balloon and talk about them. In fact, it might be fun to imagine the laser pointer having a kind of "light pen" mode that we can use to draw inscriptions on the surface of the balloon. Plastering your name across the night sky would make for quite a show. For the sake of the illustration, you have to imagine these props having metaphysical properties. We aren't really concerned with the light pen. It's just to imagine that we're drawing lines.
Now imagine that we tried to place inside of the balloon, at scale, all of the stuff of the observable universe, or, for the sake of the question, just the stars. We would put everything inside the balloon precisely where it would be relative to our vantage point.
Now we can go through, one at a time, and consider each star individually. Each time we examine a star, we could draw the line from us to it with our laser pointer. We could use the light pen to trace the outline of the star with the laser pointer, inscribing a little circle on the surface of the balloon behind it. Every time we did this with a particular star, we would add a circle on the balloon to build up a flat map of the stars. We could process each star, one by one, and eliminate each star until the balloon is empty again. It's just us, looking back up at the map we made.
Now let's say the balloon was originally red and our light pen was drawing in green. Let's also say that the green circles we drew were colored in, filled with green. After we've processed all the stars, we've got green dots all over the inside of the balloon. The size of each green dot would first be a function of the size of the star. Bigger stars would tend to draw relatively bigger circles on the map.
This analogy is imperfect in many ways. It is imperfect here in an important respect. If you imagine that we are tracing the stars with a circular motion in the hand, which is natural, then we'll be distorting the map. The angle of the light pen in hand as we made a circular motion would be projected across a great distance. That map would be interesting for other reasons, but we're trying to identify just the areas that are on line with us, stars we are "under." We want the real size of the star to be on the map, not a size relative to the distance between us and it.
To stay true, we'll have to imagine that our map simply has a circle on it whose center is on line with us and the star it represents. The size of the star's circle is its actual size. Our sun is roughly 1.39 million kilometers across, so the circle it draws would have this diameter on our map. This is the area of points that would, regardless of distance, carry a line all the way between them and us to make a candidate for a star being "overhead".
The answer to whether at least one star is probably overhead at a given time is, in one way of thinking, the proportion of red and green on the map. How much of the entire map is green? That's roughly how likely we are to be on line with a star at any time.
If we want to keep going on this line of probability, this would be the time to get the average size of every observable star, calculate an average diameter, multiply it by the number of stars, and have an estimated area. This will be wildly off because we've flattened three or four dimensions into two and didn't account for overlap. Unfortunately, the overlap of what's overhead does not appear to be consistent. Note that when looking up at the night sky we can see the Milky Way, of which we are a part.
Also, to get those averages, you'd have to have really thoroughly indexed the observable Universe. A lot of people have been working on that for a long time, but it's very large. So if we had enough data to have reasonably good averages for things like the size of a star, we may as well forget the averages and make the actual map. We'd take care of overlapping circles that way, too. While we're at it, forget the map entirely. Just have the GPS in your phone feed your position on the globe into a model that will draw the line and check everything above you. It's the real problem we started with, just taking into appreciation that the vastness of the cosmos is so overwhelmingly large that the computation required to check what's overhead may have a shorter radius than the radius of the observable universe. Even a database to hold records of every star and its position is impossibly large.
I also read lately that the universe might be (these are guesses and arguments) at least 250 times larger than what we can observe. I have also read that the earth is flat. Maybe the universe goes on infinitely. Reasoning about that will have similar boundary conditions.
Your best bet is to actually feed your location into a model and limit the model so you can get a reasonably fast calculation. Change the question to: “What is the nearest star on this line, given a spacial and computational boundary?” You'll have to accept that somewhere beyond what can be calculated, even beyond what can be seen, there may still be a star.
New contributor
$endgroup$
1
$begingroup$
Welcome on the Astronomy SE! Look how nice-looking formulas I inserted into your post. It is because we have some Latex support. Type$4pi r^2$
and you will get $4pi r^2$.
$endgroup$
– peterh
4 hours ago
$begingroup$
Very nice. Thank you!
$endgroup$
– keparo
4 hours ago
add a comment |
$begingroup$
Probably, maybe.
There are at least two ways of answering the question. One is to ask what were your coordinates when you wrote the question and exactly what time it was. Then we'll need to draw a line in a model to see what you hit and whether any of those hits are stars. This assumes a complete map, which is a problem. The answer is different for everyone on Earth and changes constantly. It becomes the right question if we are in a starship. Given the vastness of space, it’s probably better to ask “How far until we hit something.”
The other answer is about probability. How often is a star directly overhead? I'll suggest one way to reason about it. There seem to be a lot of limiting factors. I'll point out a few of those, too.
First, a gut check. Our Sun is directly overhead for a good area of the Earth at all times. The sun is relatively near, so it’s coverage is special. That trillions of billions of other stars have the rest of the planet covered nonetheless seems likely.
An excellent detail of this question is whether the line you are imagining intersects with a star. I take this to mean whether the abstract line passes through any part of the star's mass, not just its center of mass or other centers.
The odds are that we are not at the center of the Universe, if "center of Universe" even has any meaning. It can be argued (it is argued) that we are at the center of the observable universe, essentially because we're looking in all directions with the same limited gear. So we can imagine a giant sphere of observability, just to give this problem some space. Imagine ourselves as a grain of sand floating at the center of a big balloon. In truth, the grain of sand is far too big in proportion to any real balloon, but imagine we are in the dead center of a balloon on an impossibly small grain.
For the balloon's dimensions, consider a sphere with a radius of 4, where the units are $1.1times 10^26$ meters. The surface of that sphere is going to be $4pi r^2$, or $64pi$ square units. If we prefer not to talk in terms with a "$pi$" mixed in, it's roughly 200 of these large square units.
Imagine that this is the area that we are gazing upon from inside the center of the balloon, sitting on our microscopic and impossibly concentric grain of sand. We can only see half of the area at once (even less, really), but we are spinning around. So we can canvas the entire inside surface of the balloon over the course of the day.
So there we are, on this spec of sand, looking at the part of the balloon we can see. One of us has a laser pointer that we can use point to different parts of the balloon and talk about them. In fact, it might be fun to imagine the laser pointer having a kind of "light pen" mode that we can use to draw inscriptions on the surface of the balloon. Plastering your name across the night sky would make for quite a show. For the sake of the illustration, you have to imagine these props having metaphysical properties. We aren't really concerned with the light pen. It's just to imagine that we're drawing lines.
Now imagine that we tried to place inside of the balloon, at scale, all of the stuff of the observable universe, or, for the sake of the question, just the stars. We would put everything inside the balloon precisely where it would be relative to our vantage point.
Now we can go through, one at a time, and consider each star individually. Each time we examine a star, we could draw the line from us to it with our laser pointer. We could use the light pen to trace the outline of the star with the laser pointer, inscribing a little circle on the surface of the balloon behind it. Every time we did this with a particular star, we would add a circle on the balloon to build up a flat map of the stars. We could process each star, one by one, and eliminate each star until the balloon is empty again. It's just us, looking back up at the map we made.
Now let's say the balloon was originally red and our light pen was drawing in green. Let's also say that the green circles we drew were colored in, filled with green. After we've processed all the stars, we've got green dots all over the inside of the balloon. The size of each green dot would first be a function of the size of the star. Bigger stars would tend to draw relatively bigger circles on the map.
This analogy is imperfect in many ways. It is imperfect here in an important respect. If you imagine that we are tracing the stars with a circular motion in the hand, which is natural, then we'll be distorting the map. The angle of the light pen in hand as we made a circular motion would be projected across a great distance. That map would be interesting for other reasons, but we're trying to identify just the areas that are on line with us, stars we are "under." We want the real size of the star to be on the map, not a size relative to the distance between us and it.
To stay true, we'll have to imagine that our map simply has a circle on it whose center is on line with us and the star it represents. The size of the star's circle is its actual size. Our sun is roughly 1.39 million kilometers across, so the circle it draws would have this diameter on our map. This is the area of points that would, regardless of distance, carry a line all the way between them and us to make a candidate for a star being "overhead".
The answer to whether at least one star is probably overhead at a given time is, in one way of thinking, the proportion of red and green on the map. How much of the entire map is green? That's roughly how likely we are to be on line with a star at any time.
If we want to keep going on this line of probability, this would be the time to get the average size of every observable star, calculate an average diameter, multiply it by the number of stars, and have an estimated area. This will be wildly off because we've flattened three or four dimensions into two and didn't account for overlap. Unfortunately, the overlap of what's overhead does not appear to be consistent. Note that when looking up at the night sky we can see the Milky Way, of which we are a part.
Also, to get those averages, you'd have to have really thoroughly indexed the observable Universe. A lot of people have been working on that for a long time, but it's very large. So if we had enough data to have reasonably good averages for things like the size of a star, we may as well forget the averages and make the actual map. We'd take care of overlapping circles that way, too. While we're at it, forget the map entirely. Just have the GPS in your phone feed your position on the globe into a model that will draw the line and check everything above you. It's the real problem we started with, just taking into appreciation that the vastness of the cosmos is so overwhelmingly large that the computation required to check what's overhead may have a shorter radius than the radius of the observable universe. Even a database to hold records of every star and its position is impossibly large.
I also read lately that the universe might be (these are guesses and arguments) at least 250 times larger than what we can observe. I have also read that the earth is flat. Maybe the universe goes on infinitely. Reasoning about that will have similar boundary conditions.
Your best bet is to actually feed your location into a model and limit the model so you can get a reasonably fast calculation. Change the question to: “What is the nearest star on this line, given a spacial and computational boundary?” You'll have to accept that somewhere beyond what can be calculated, even beyond what can be seen, there may still be a star.
New contributor
$endgroup$
1
$begingroup$
Welcome on the Astronomy SE! Look how nice-looking formulas I inserted into your post. It is because we have some Latex support. Type$4pi r^2$
and you will get $4pi r^2$.
$endgroup$
– peterh
4 hours ago
$begingroup$
Very nice. Thank you!
$endgroup$
– keparo
4 hours ago
add a comment |
$begingroup$
Probably, maybe.
There are at least two ways of answering the question. One is to ask what were your coordinates when you wrote the question and exactly what time it was. Then we'll need to draw a line in a model to see what you hit and whether any of those hits are stars. This assumes a complete map, which is a problem. The answer is different for everyone on Earth and changes constantly. It becomes the right question if we are in a starship. Given the vastness of space, it’s probably better to ask “How far until we hit something.”
The other answer is about probability. How often is a star directly overhead? I'll suggest one way to reason about it. There seem to be a lot of limiting factors. I'll point out a few of those, too.
First, a gut check. Our Sun is directly overhead for a good area of the Earth at all times. The sun is relatively near, so it’s coverage is special. That trillions of billions of other stars have the rest of the planet covered nonetheless seems likely.
An excellent detail of this question is whether the line you are imagining intersects with a star. I take this to mean whether the abstract line passes through any part of the star's mass, not just its center of mass or other centers.
The odds are that we are not at the center of the Universe, if "center of Universe" even has any meaning. It can be argued (it is argued) that we are at the center of the observable universe, essentially because we're looking in all directions with the same limited gear. So we can imagine a giant sphere of observability, just to give this problem some space. Imagine ourselves as a grain of sand floating at the center of a big balloon. In truth, the grain of sand is far too big in proportion to any real balloon, but imagine we are in the dead center of a balloon on an impossibly small grain.
For the balloon's dimensions, consider a sphere with a radius of 4, where the units are $1.1times 10^26$ meters. The surface of that sphere is going to be $4pi r^2$, or $64pi$ square units. If we prefer not to talk in terms with a "$pi$" mixed in, it's roughly 200 of these large square units.
Imagine that this is the area that we are gazing upon from inside the center of the balloon, sitting on our microscopic and impossibly concentric grain of sand. We can only see half of the area at once (even less, really), but we are spinning around. So we can canvas the entire inside surface of the balloon over the course of the day.
So there we are, on this spec of sand, looking at the part of the balloon we can see. One of us has a laser pointer that we can use point to different parts of the balloon and talk about them. In fact, it might be fun to imagine the laser pointer having a kind of "light pen" mode that we can use to draw inscriptions on the surface of the balloon. Plastering your name across the night sky would make for quite a show. For the sake of the illustration, you have to imagine these props having metaphysical properties. We aren't really concerned with the light pen. It's just to imagine that we're drawing lines.
Now imagine that we tried to place inside of the balloon, at scale, all of the stuff of the observable universe, or, for the sake of the question, just the stars. We would put everything inside the balloon precisely where it would be relative to our vantage point.
Now we can go through, one at a time, and consider each star individually. Each time we examine a star, we could draw the line from us to it with our laser pointer. We could use the light pen to trace the outline of the star with the laser pointer, inscribing a little circle on the surface of the balloon behind it. Every time we did this with a particular star, we would add a circle on the balloon to build up a flat map of the stars. We could process each star, one by one, and eliminate each star until the balloon is empty again. It's just us, looking back up at the map we made.
Now let's say the balloon was originally red and our light pen was drawing in green. Let's also say that the green circles we drew were colored in, filled with green. After we've processed all the stars, we've got green dots all over the inside of the balloon. The size of each green dot would first be a function of the size of the star. Bigger stars would tend to draw relatively bigger circles on the map.
This analogy is imperfect in many ways. It is imperfect here in an important respect. If you imagine that we are tracing the stars with a circular motion in the hand, which is natural, then we'll be distorting the map. The angle of the light pen in hand as we made a circular motion would be projected across a great distance. That map would be interesting for other reasons, but we're trying to identify just the areas that are on line with us, stars we are "under." We want the real size of the star to be on the map, not a size relative to the distance between us and it.
To stay true, we'll have to imagine that our map simply has a circle on it whose center is on line with us and the star it represents. The size of the star's circle is its actual size. Our sun is roughly 1.39 million kilometers across, so the circle it draws would have this diameter on our map. This is the area of points that would, regardless of distance, carry a line all the way between them and us to make a candidate for a star being "overhead".
The answer to whether at least one star is probably overhead at a given time is, in one way of thinking, the proportion of red and green on the map. How much of the entire map is green? That's roughly how likely we are to be on line with a star at any time.
If we want to keep going on this line of probability, this would be the time to get the average size of every observable star, calculate an average diameter, multiply it by the number of stars, and have an estimated area. This will be wildly off because we've flattened three or four dimensions into two and didn't account for overlap. Unfortunately, the overlap of what's overhead does not appear to be consistent. Note that when looking up at the night sky we can see the Milky Way, of which we are a part.
Also, to get those averages, you'd have to have really thoroughly indexed the observable Universe. A lot of people have been working on that for a long time, but it's very large. So if we had enough data to have reasonably good averages for things like the size of a star, we may as well forget the averages and make the actual map. We'd take care of overlapping circles that way, too. While we're at it, forget the map entirely. Just have the GPS in your phone feed your position on the globe into a model that will draw the line and check everything above you. It's the real problem we started with, just taking into appreciation that the vastness of the cosmos is so overwhelmingly large that the computation required to check what's overhead may have a shorter radius than the radius of the observable universe. Even a database to hold records of every star and its position is impossibly large.
I also read lately that the universe might be (these are guesses and arguments) at least 250 times larger than what we can observe. I have also read that the earth is flat. Maybe the universe goes on infinitely. Reasoning about that will have similar boundary conditions.
Your best bet is to actually feed your location into a model and limit the model so you can get a reasonably fast calculation. Change the question to: “What is the nearest star on this line, given a spacial and computational boundary?” You'll have to accept that somewhere beyond what can be calculated, even beyond what can be seen, there may still be a star.
New contributor
$endgroup$
Probably, maybe.
There are at least two ways of answering the question. One is to ask what were your coordinates when you wrote the question and exactly what time it was. Then we'll need to draw a line in a model to see what you hit and whether any of those hits are stars. This assumes a complete map, which is a problem. The answer is different for everyone on Earth and changes constantly. It becomes the right question if we are in a starship. Given the vastness of space, it’s probably better to ask “How far until we hit something.”
The other answer is about probability. How often is a star directly overhead? I'll suggest one way to reason about it. There seem to be a lot of limiting factors. I'll point out a few of those, too.
First, a gut check. Our Sun is directly overhead for a good area of the Earth at all times. The sun is relatively near, so it’s coverage is special. That trillions of billions of other stars have the rest of the planet covered nonetheless seems likely.
An excellent detail of this question is whether the line you are imagining intersects with a star. I take this to mean whether the abstract line passes through any part of the star's mass, not just its center of mass or other centers.
The odds are that we are not at the center of the Universe, if "center of Universe" even has any meaning. It can be argued (it is argued) that we are at the center of the observable universe, essentially because we're looking in all directions with the same limited gear. So we can imagine a giant sphere of observability, just to give this problem some space. Imagine ourselves as a grain of sand floating at the center of a big balloon. In truth, the grain of sand is far too big in proportion to any real balloon, but imagine we are in the dead center of a balloon on an impossibly small grain.
For the balloon's dimensions, consider a sphere with a radius of 4, where the units are $1.1times 10^26$ meters. The surface of that sphere is going to be $4pi r^2$, or $64pi$ square units. If we prefer not to talk in terms with a "$pi$" mixed in, it's roughly 200 of these large square units.
Imagine that this is the area that we are gazing upon from inside the center of the balloon, sitting on our microscopic and impossibly concentric grain of sand. We can only see half of the area at once (even less, really), but we are spinning around. So we can canvas the entire inside surface of the balloon over the course of the day.
So there we are, on this spec of sand, looking at the part of the balloon we can see. One of us has a laser pointer that we can use point to different parts of the balloon and talk about them. In fact, it might be fun to imagine the laser pointer having a kind of "light pen" mode that we can use to draw inscriptions on the surface of the balloon. Plastering your name across the night sky would make for quite a show. For the sake of the illustration, you have to imagine these props having metaphysical properties. We aren't really concerned with the light pen. It's just to imagine that we're drawing lines.
Now imagine that we tried to place inside of the balloon, at scale, all of the stuff of the observable universe, or, for the sake of the question, just the stars. We would put everything inside the balloon precisely where it would be relative to our vantage point.
Now we can go through, one at a time, and consider each star individually. Each time we examine a star, we could draw the line from us to it with our laser pointer. We could use the light pen to trace the outline of the star with the laser pointer, inscribing a little circle on the surface of the balloon behind it. Every time we did this with a particular star, we would add a circle on the balloon to build up a flat map of the stars. We could process each star, one by one, and eliminate each star until the balloon is empty again. It's just us, looking back up at the map we made.
Now let's say the balloon was originally red and our light pen was drawing in green. Let's also say that the green circles we drew were colored in, filled with green. After we've processed all the stars, we've got green dots all over the inside of the balloon. The size of each green dot would first be a function of the size of the star. Bigger stars would tend to draw relatively bigger circles on the map.
This analogy is imperfect in many ways. It is imperfect here in an important respect. If you imagine that we are tracing the stars with a circular motion in the hand, which is natural, then we'll be distorting the map. The angle of the light pen in hand as we made a circular motion would be projected across a great distance. That map would be interesting for other reasons, but we're trying to identify just the areas that are on line with us, stars we are "under." We want the real size of the star to be on the map, not a size relative to the distance between us and it.
To stay true, we'll have to imagine that our map simply has a circle on it whose center is on line with us and the star it represents. The size of the star's circle is its actual size. Our sun is roughly 1.39 million kilometers across, so the circle it draws would have this diameter on our map. This is the area of points that would, regardless of distance, carry a line all the way between them and us to make a candidate for a star being "overhead".
The answer to whether at least one star is probably overhead at a given time is, in one way of thinking, the proportion of red and green on the map. How much of the entire map is green? That's roughly how likely we are to be on line with a star at any time.
If we want to keep going on this line of probability, this would be the time to get the average size of every observable star, calculate an average diameter, multiply it by the number of stars, and have an estimated area. This will be wildly off because we've flattened three or four dimensions into two and didn't account for overlap. Unfortunately, the overlap of what's overhead does not appear to be consistent. Note that when looking up at the night sky we can see the Milky Way, of which we are a part.
Also, to get those averages, you'd have to have really thoroughly indexed the observable Universe. A lot of people have been working on that for a long time, but it's very large. So if we had enough data to have reasonably good averages for things like the size of a star, we may as well forget the averages and make the actual map. We'd take care of overlapping circles that way, too. While we're at it, forget the map entirely. Just have the GPS in your phone feed your position on the globe into a model that will draw the line and check everything above you. It's the real problem we started with, just taking into appreciation that the vastness of the cosmos is so overwhelmingly large that the computation required to check what's overhead may have a shorter radius than the radius of the observable universe. Even a database to hold records of every star and its position is impossibly large.
I also read lately that the universe might be (these are guesses and arguments) at least 250 times larger than what we can observe. I have also read that the earth is flat. Maybe the universe goes on infinitely. Reasoning about that will have similar boundary conditions.
Your best bet is to actually feed your location into a model and limit the model so you can get a reasonably fast calculation. Change the question to: “What is the nearest star on this line, given a spacial and computational boundary?” You'll have to accept that somewhere beyond what can be calculated, even beyond what can be seen, there may still be a star.
New contributor
edited 3 hours ago
New contributor
answered 5 hours ago
keparokeparo
1214 bronze badges
1214 bronze badges
New contributor
New contributor
1
$begingroup$
Welcome on the Astronomy SE! Look how nice-looking formulas I inserted into your post. It is because we have some Latex support. Type$4pi r^2$
and you will get $4pi r^2$.
$endgroup$
– peterh
4 hours ago
$begingroup$
Very nice. Thank you!
$endgroup$
– keparo
4 hours ago
add a comment |
1
$begingroup$
Welcome on the Astronomy SE! Look how nice-looking formulas I inserted into your post. It is because we have some Latex support. Type$4pi r^2$
and you will get $4pi r^2$.
$endgroup$
– peterh
4 hours ago
$begingroup$
Very nice. Thank you!
$endgroup$
– keparo
4 hours ago
1
1
$begingroup$
Welcome on the Astronomy SE! Look how nice-looking formulas I inserted into your post. It is because we have some Latex support. Type
$4pi r^2$
and you will get $4pi r^2$.$endgroup$
– peterh
4 hours ago
$begingroup$
Welcome on the Astronomy SE! Look how nice-looking formulas I inserted into your post. It is because we have some Latex support. Type
$4pi r^2$
and you will get $4pi r^2$.$endgroup$
– peterh
4 hours ago
$begingroup$
Very nice. Thank you!
$endgroup$
– keparo
4 hours ago
$begingroup$
Very nice. Thank you!
$endgroup$
– keparo
4 hours ago
add a comment |
$begingroup$
According to Olbers, of paradox fame, if the universe is infinite a line of sight in any direction should eventually reach a star. Why then was the night sky so dark when in theory it should be bright as day? Leaving that particular question aside, we have no proof that the universe is infinite, but it is sufficiently large that a line in any direction should sooner or later reach the surface of a star. Whether the line in question would have to travel only tens of light years to reach the star or many billions depends on where you are standing and at what particular moment you choose to draw the line. If you happened to be on the equator at the right time of year and the right time of day, the line might only have to travel a little over eight light minutes to reach a star. In the universe, as opposed to on paper, you cannot have really straight lines, only an approximation, because the universe is curved and any line defining device, a laser beam perhaps, would be acted on by a force and bent.
$endgroup$
add a comment |
$begingroup$
According to Olbers, of paradox fame, if the universe is infinite a line of sight in any direction should eventually reach a star. Why then was the night sky so dark when in theory it should be bright as day? Leaving that particular question aside, we have no proof that the universe is infinite, but it is sufficiently large that a line in any direction should sooner or later reach the surface of a star. Whether the line in question would have to travel only tens of light years to reach the star or many billions depends on where you are standing and at what particular moment you choose to draw the line. If you happened to be on the equator at the right time of year and the right time of day, the line might only have to travel a little over eight light minutes to reach a star. In the universe, as opposed to on paper, you cannot have really straight lines, only an approximation, because the universe is curved and any line defining device, a laser beam perhaps, would be acted on by a force and bent.
$endgroup$
add a comment |
$begingroup$
According to Olbers, of paradox fame, if the universe is infinite a line of sight in any direction should eventually reach a star. Why then was the night sky so dark when in theory it should be bright as day? Leaving that particular question aside, we have no proof that the universe is infinite, but it is sufficiently large that a line in any direction should sooner or later reach the surface of a star. Whether the line in question would have to travel only tens of light years to reach the star or many billions depends on where you are standing and at what particular moment you choose to draw the line. If you happened to be on the equator at the right time of year and the right time of day, the line might only have to travel a little over eight light minutes to reach a star. In the universe, as opposed to on paper, you cannot have really straight lines, only an approximation, because the universe is curved and any line defining device, a laser beam perhaps, would be acted on by a force and bent.
$endgroup$
According to Olbers, of paradox fame, if the universe is infinite a line of sight in any direction should eventually reach a star. Why then was the night sky so dark when in theory it should be bright as day? Leaving that particular question aside, we have no proof that the universe is infinite, but it is sufficiently large that a line in any direction should sooner or later reach the surface of a star. Whether the line in question would have to travel only tens of light years to reach the star or many billions depends on where you are standing and at what particular moment you choose to draw the line. If you happened to be on the equator at the right time of year and the right time of day, the line might only have to travel a little over eight light minutes to reach a star. In the universe, as opposed to on paper, you cannot have really straight lines, only an approximation, because the universe is curved and any line defining device, a laser beam perhaps, would be acted on by a force and bent.
answered 5 hours ago
Michael WalsbyMichael Walsby
9991 silver badge6 bronze badges
9991 silver badge6 bronze badges
add a comment |
add a comment |
user68873 is a new contributor. Be nice, and check out our Code of Conduct.
user68873 is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
Presumably you mean a naked eye magnitude star ? As the magnitude limit increases towards fainter stars, the probability is going to get very close to 1...
$endgroup$
– astrosnapper
8 hours ago
$begingroup$
@astrosnapper that's not obvious because of the finite age of the universe.
$endgroup$
– Steve Linton
7 hours ago
1
$begingroup$
Related: Olbers' paradox
$endgroup$
– Mike G
6 hours ago