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If all stars rotate, why was there a theory developed that requires non-rotating stars?
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According to Penrose's research, a non-rotating star would end up, after gravitational collapse, as a perfectly spherical black hole. However, every star in the universe has some kind of angular momentum.
Why even bother doing that research if that won't ever happen in the universe and does it have any implications for the future of astrophysics?
black-hole astrophysics
New contributor
urban pečoler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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show 4 more comments
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According to Penrose's research, a non-rotating star would end up, after gravitational collapse, as a perfectly spherical black hole. However, every star in the universe has some kind of angular momentum.
Why even bother doing that research if that won't ever happen in the universe and does it have any implications for the future of astrophysics?
black-hole astrophysics
New contributor
urban pečoler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
6
$begingroup$
Would you mind providing more information about the research, e.g. linking to a paper about it?
$endgroup$
– HDE 226868♦
Aug 21 at 13:58
20
$begingroup$
Frictionless spherical cows are useful abstractions too...
$endgroup$
– Beanluc
2 days ago
4
$begingroup$
I suppose it's the solution to a simplified model of reality as a first step? That's not unusual in science...
$endgroup$
– Peter A. Schneider
2 days ago
7
$begingroup$
"However, every star in the universe" You've checked them all have you?
$endgroup$
– TripeHound
2 days ago
4
$begingroup$
"All models are wrong, but some are useful"
$endgroup$
– llama
2 days ago
|
show 4 more comments
$begingroup$
According to Penrose's research, a non-rotating star would end up, after gravitational collapse, as a perfectly spherical black hole. However, every star in the universe has some kind of angular momentum.
Why even bother doing that research if that won't ever happen in the universe and does it have any implications for the future of astrophysics?
black-hole astrophysics
New contributor
urban pečoler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
According to Penrose's research, a non-rotating star would end up, after gravitational collapse, as a perfectly spherical black hole. However, every star in the universe has some kind of angular momentum.
Why even bother doing that research if that won't ever happen in the universe and does it have any implications for the future of astrophysics?
black-hole astrophysics
black-hole astrophysics
New contributor
urban pečoler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
urban pečoler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Community♦
1
1
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asked Aug 21 at 11:50
urban pečolerurban pečoler
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urban pečoler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Check out our Code of Conduct.
6
$begingroup$
Would you mind providing more information about the research, e.g. linking to a paper about it?
$endgroup$
– HDE 226868♦
Aug 21 at 13:58
20
$begingroup$
Frictionless spherical cows are useful abstractions too...
$endgroup$
– Beanluc
2 days ago
4
$begingroup$
I suppose it's the solution to a simplified model of reality as a first step? That's not unusual in science...
$endgroup$
– Peter A. Schneider
2 days ago
7
$begingroup$
"However, every star in the universe" You've checked them all have you?
$endgroup$
– TripeHound
2 days ago
4
$begingroup$
"All models are wrong, but some are useful"
$endgroup$
– llama
2 days ago
|
show 4 more comments
6
$begingroup$
Would you mind providing more information about the research, e.g. linking to a paper about it?
$endgroup$
– HDE 226868♦
Aug 21 at 13:58
20
$begingroup$
Frictionless spherical cows are useful abstractions too...
$endgroup$
– Beanluc
2 days ago
4
$begingroup$
I suppose it's the solution to a simplified model of reality as a first step? That's not unusual in science...
$endgroup$
– Peter A. Schneider
2 days ago
7
$begingroup$
"However, every star in the universe" You've checked them all have you?
$endgroup$
– TripeHound
2 days ago
4
$begingroup$
"All models are wrong, but some are useful"
$endgroup$
– llama
2 days ago
6
6
$begingroup$
Would you mind providing more information about the research, e.g. linking to a paper about it?
$endgroup$
– HDE 226868♦
Aug 21 at 13:58
$begingroup$
Would you mind providing more information about the research, e.g. linking to a paper about it?
$endgroup$
– HDE 226868♦
Aug 21 at 13:58
20
20
$begingroup$
Frictionless spherical cows are useful abstractions too...
$endgroup$
– Beanluc
2 days ago
$begingroup$
Frictionless spherical cows are useful abstractions too...
$endgroup$
– Beanluc
2 days ago
4
4
$begingroup$
I suppose it's the solution to a simplified model of reality as a first step? That's not unusual in science...
$endgroup$
– Peter A. Schneider
2 days ago
$begingroup$
I suppose it's the solution to a simplified model of reality as a first step? That's not unusual in science...
$endgroup$
– Peter A. Schneider
2 days ago
7
7
$begingroup$
"However, every star in the universe" You've checked them all have you?
$endgroup$
– TripeHound
2 days ago
$begingroup$
"However, every star in the universe" You've checked them all have you?
$endgroup$
– TripeHound
2 days ago
4
4
$begingroup$
"All models are wrong, but some are useful"
$endgroup$
– llama
2 days ago
$begingroup$
"All models are wrong, but some are useful"
$endgroup$
– llama
2 days ago
|
show 4 more comments
5 Answers
5
active
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$begingroup$
All models are approximations, we judge a model on how useful it is.
Understanding the collapse of a non-rotating star to a black hole gives insight into the nature of gravitational collapse. Much of the physics of collapse does not depend on spin. The formation of an event horizon, for example.
Models can be refined, and in this case, considering rotation leads to further insight, and a non-spherically symmetric structure with multiple singular horizons.
All models are necessarily simplifications. But the non-rotating model is still useful.
$endgroup$
add a comment |
$begingroup$
In a similar way, we could ask...
No beams can be exactly 1 meter long. No beams can be exactly
straight. The material making up a beam cannot be truly isotropic.
So why should we bother calculating the stress in a 1 meter straight
beam having isotropic material?
Because knowing how to perform this calculation is a building block for doing more complex calculations.
The non-rotating black hole calculation also provides a limiting solution. The solution for a spinning star's collapse will approach this solution as the spin approaches zero.
Similarly, Newton told us that as external forces approach zero, the path of a moving object will approach a straight line. This is useful to know even though there is no place in our universe that doesn't have gravitational influence.
$endgroup$
7
$begingroup$
Assume a spherical cow...
$endgroup$
– RonJohn
2 days ago
5
$begingroup$
I'm not sure if the metre is still defined against a standard, but if so, there is one stick that is exactly 1 metre long (by definition). Perhaps not entirely relevant to your point though.
$endgroup$
– Roland Heath
2 days ago
13
$begingroup$
@RolandHeath It hasn't been since 1960.
$endgroup$
– Graipher
2 days ago
2
$begingroup$
+1, but is it obvious that the non-rotating case is a limiting solution? A priori there might be global (topological?) effects that come into play as the angular momentum density grows towards infinity just before a singularity forms.
$endgroup$
– Henning Makholm
2 days ago
1
$begingroup$
@James: My point is that a collapsing star with low but nonzero angular momentum has to go through a phase where its angular momentum density diverges to infinity during the collapse -- whereas a star with zero angular momentum can have zero angular momentum density during its entire collapse. That might (at least a priori) give rise to a qualitative difference that is not respected by the limiting process.
$endgroup$
– Henning Makholm
yesterday
|
show 3 more comments
$begingroup$
Another consideration is that the physics that describe a rotating black hole was much harder to develop.
The maths describing the Schwarzschild (uncharged, non-spinning) black hole was developed in 1916. This was expanded to charged, non-spinning black holes in 1918 (The Reissner–Nordström metric)
It wasn't until 1963 that the Kerr metric for uncharged spinning black holes was developed. Two years later, the most general form, the Kerr-Newman metric was found.
I wouldn't fancy waiting 47 years for a more accurate black hole model to be developed before doing any meaningful work in the field.
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Also note that the pure Schwarzschild solution is static: it's eternal, not formed by collapse, and it's the only object in an otherwise empty universe. But it's still a useful solution, despite these unnatural simplifications.
$endgroup$
– PM 2Ring
yesterday
add a comment |
$begingroup$
Our sun's rotation period is 24.47 days at the equator and almost 38 days at the poles, our planet's rotational period is 23h 56m 4.098,903,691s. Use of Schwarzschild equations for either case isn't exact.
If you used the equation for non-rotating objects to calculate the time at the altitude of GPS satellites (~ 20,200 km or 12,550 miles) then you would be off by 38,636 nanoseconds per day. A Julian year is defined as 365.25 days of exactly 86,400 seconds (SI base unit), totalling exactly 31,557,600 seconds in the Julian astronomical year. The Gregorian calendar year (400 year average) is 365.2425 days.
Multiplying 365.2425 x 38,636 = 14,111,509.23 nanoseconds, that's 0.0141 seconds per year. If being off by that amount isn't of any concern to you then you can use the easier equation, such as for calculations involving the star HR 1362 which has a rotational period that is 306.9 ± 0.4 days.
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add a comment |
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You're right: all stars rotate. The only reason I can think of why astrophysicists make calculations for a non-rotating star or black hole is that it makes their calculations a bit easier. Although all stars rotate, some rotate much faster than others, and their masses vary too, so there is a wide degree of uncertainty which is reduced by calculating for a star that does not rotate.
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4
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How certain can we be that all stars rotate? There are a lot of stars and many many possible (theoretical) interactions that would slow rotation.
$endgroup$
– Valorum
Aug 21 at 22:11
$begingroup$
No one has found one yet. I suspect it would cause a sensation if one were discovered.
$endgroup$
– Michael Walsby
Aug 21 at 22:26
1
$begingroup$
@Valorum Yeah, I was thinking about a stellar collision where the stars are rotating in opposite directions. If the rotational energy is exactly opposite you'll get a non-rotating result. Very unlikely, not utterly impossible--thus it probably will happen somewhere, someday.
$endgroup$
– Loren Pechtel
2 days ago
1
$begingroup$
@LorenPechtel The rotational momentum needs to be exactly equal. I think that counts as utterly impossible.
$endgroup$
– Martin Bonner
2 days ago
$begingroup$
@Valorum Because the chance for "zero" angular momentum approaches 0 much faster than the amount of stars grow with "sample size".
$endgroup$
– paul23
8 hours ago
|
show 1 more comment
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
All models are approximations, we judge a model on how useful it is.
Understanding the collapse of a non-rotating star to a black hole gives insight into the nature of gravitational collapse. Much of the physics of collapse does not depend on spin. The formation of an event horizon, for example.
Models can be refined, and in this case, considering rotation leads to further insight, and a non-spherically symmetric structure with multiple singular horizons.
All models are necessarily simplifications. But the non-rotating model is still useful.
$endgroup$
add a comment |
$begingroup$
All models are approximations, we judge a model on how useful it is.
Understanding the collapse of a non-rotating star to a black hole gives insight into the nature of gravitational collapse. Much of the physics of collapse does not depend on spin. The formation of an event horizon, for example.
Models can be refined, and in this case, considering rotation leads to further insight, and a non-spherically symmetric structure with multiple singular horizons.
All models are necessarily simplifications. But the non-rotating model is still useful.
$endgroup$
add a comment |
$begingroup$
All models are approximations, we judge a model on how useful it is.
Understanding the collapse of a non-rotating star to a black hole gives insight into the nature of gravitational collapse. Much of the physics of collapse does not depend on spin. The formation of an event horizon, for example.
Models can be refined, and in this case, considering rotation leads to further insight, and a non-spherically symmetric structure with multiple singular horizons.
All models are necessarily simplifications. But the non-rotating model is still useful.
$endgroup$
All models are approximations, we judge a model on how useful it is.
Understanding the collapse of a non-rotating star to a black hole gives insight into the nature of gravitational collapse. Much of the physics of collapse does not depend on spin. The formation of an event horizon, for example.
Models can be refined, and in this case, considering rotation leads to further insight, and a non-spherically symmetric structure with multiple singular horizons.
All models are necessarily simplifications. But the non-rotating model is still useful.
answered Aug 21 at 12:19
James KJames K
37.8k2 gold badges66 silver badges128 bronze badges
37.8k2 gold badges66 silver badges128 bronze badges
add a comment |
add a comment |
$begingroup$
In a similar way, we could ask...
No beams can be exactly 1 meter long. No beams can be exactly
straight. The material making up a beam cannot be truly isotropic.
So why should we bother calculating the stress in a 1 meter straight
beam having isotropic material?
Because knowing how to perform this calculation is a building block for doing more complex calculations.
The non-rotating black hole calculation also provides a limiting solution. The solution for a spinning star's collapse will approach this solution as the spin approaches zero.
Similarly, Newton told us that as external forces approach zero, the path of a moving object will approach a straight line. This is useful to know even though there is no place in our universe that doesn't have gravitational influence.
$endgroup$
7
$begingroup$
Assume a spherical cow...
$endgroup$
– RonJohn
2 days ago
5
$begingroup$
I'm not sure if the metre is still defined against a standard, but if so, there is one stick that is exactly 1 metre long (by definition). Perhaps not entirely relevant to your point though.
$endgroup$
– Roland Heath
2 days ago
13
$begingroup$
@RolandHeath It hasn't been since 1960.
$endgroup$
– Graipher
2 days ago
2
$begingroup$
+1, but is it obvious that the non-rotating case is a limiting solution? A priori there might be global (topological?) effects that come into play as the angular momentum density grows towards infinity just before a singularity forms.
$endgroup$
– Henning Makholm
2 days ago
1
$begingroup$
@James: My point is that a collapsing star with low but nonzero angular momentum has to go through a phase where its angular momentum density diverges to infinity during the collapse -- whereas a star with zero angular momentum can have zero angular momentum density during its entire collapse. That might (at least a priori) give rise to a qualitative difference that is not respected by the limiting process.
$endgroup$
– Henning Makholm
yesterday
|
show 3 more comments
$begingroup$
In a similar way, we could ask...
No beams can be exactly 1 meter long. No beams can be exactly
straight. The material making up a beam cannot be truly isotropic.
So why should we bother calculating the stress in a 1 meter straight
beam having isotropic material?
Because knowing how to perform this calculation is a building block for doing more complex calculations.
The non-rotating black hole calculation also provides a limiting solution. The solution for a spinning star's collapse will approach this solution as the spin approaches zero.
Similarly, Newton told us that as external forces approach zero, the path of a moving object will approach a straight line. This is useful to know even though there is no place in our universe that doesn't have gravitational influence.
$endgroup$
7
$begingroup$
Assume a spherical cow...
$endgroup$
– RonJohn
2 days ago
5
$begingroup$
I'm not sure if the metre is still defined against a standard, but if so, there is one stick that is exactly 1 metre long (by definition). Perhaps not entirely relevant to your point though.
$endgroup$
– Roland Heath
2 days ago
13
$begingroup$
@RolandHeath It hasn't been since 1960.
$endgroup$
– Graipher
2 days ago
2
$begingroup$
+1, but is it obvious that the non-rotating case is a limiting solution? A priori there might be global (topological?) effects that come into play as the angular momentum density grows towards infinity just before a singularity forms.
$endgroup$
– Henning Makholm
2 days ago
1
$begingroup$
@James: My point is that a collapsing star with low but nonzero angular momentum has to go through a phase where its angular momentum density diverges to infinity during the collapse -- whereas a star with zero angular momentum can have zero angular momentum density during its entire collapse. That might (at least a priori) give rise to a qualitative difference that is not respected by the limiting process.
$endgroup$
– Henning Makholm
yesterday
|
show 3 more comments
$begingroup$
In a similar way, we could ask...
No beams can be exactly 1 meter long. No beams can be exactly
straight. The material making up a beam cannot be truly isotropic.
So why should we bother calculating the stress in a 1 meter straight
beam having isotropic material?
Because knowing how to perform this calculation is a building block for doing more complex calculations.
The non-rotating black hole calculation also provides a limiting solution. The solution for a spinning star's collapse will approach this solution as the spin approaches zero.
Similarly, Newton told us that as external forces approach zero, the path of a moving object will approach a straight line. This is useful to know even though there is no place in our universe that doesn't have gravitational influence.
$endgroup$
In a similar way, we could ask...
No beams can be exactly 1 meter long. No beams can be exactly
straight. The material making up a beam cannot be truly isotropic.
So why should we bother calculating the stress in a 1 meter straight
beam having isotropic material?
Because knowing how to perform this calculation is a building block for doing more complex calculations.
The non-rotating black hole calculation also provides a limiting solution. The solution for a spinning star's collapse will approach this solution as the spin approaches zero.
Similarly, Newton told us that as external forces approach zero, the path of a moving object will approach a straight line. This is useful to know even though there is no place in our universe that doesn't have gravitational influence.
answered Aug 21 at 14:35
JamesJames
5682 silver badges8 bronze badges
5682 silver badges8 bronze badges
7
$begingroup$
Assume a spherical cow...
$endgroup$
– RonJohn
2 days ago
5
$begingroup$
I'm not sure if the metre is still defined against a standard, but if so, there is one stick that is exactly 1 metre long (by definition). Perhaps not entirely relevant to your point though.
$endgroup$
– Roland Heath
2 days ago
13
$begingroup$
@RolandHeath It hasn't been since 1960.
$endgroup$
– Graipher
2 days ago
2
$begingroup$
+1, but is it obvious that the non-rotating case is a limiting solution? A priori there might be global (topological?) effects that come into play as the angular momentum density grows towards infinity just before a singularity forms.
$endgroup$
– Henning Makholm
2 days ago
1
$begingroup$
@James: My point is that a collapsing star with low but nonzero angular momentum has to go through a phase where its angular momentum density diverges to infinity during the collapse -- whereas a star with zero angular momentum can have zero angular momentum density during its entire collapse. That might (at least a priori) give rise to a qualitative difference that is not respected by the limiting process.
$endgroup$
– Henning Makholm
yesterday
|
show 3 more comments
7
$begingroup$
Assume a spherical cow...
$endgroup$
– RonJohn
2 days ago
5
$begingroup$
I'm not sure if the metre is still defined against a standard, but if so, there is one stick that is exactly 1 metre long (by definition). Perhaps not entirely relevant to your point though.
$endgroup$
– Roland Heath
2 days ago
13
$begingroup$
@RolandHeath It hasn't been since 1960.
$endgroup$
– Graipher
2 days ago
2
$begingroup$
+1, but is it obvious that the non-rotating case is a limiting solution? A priori there might be global (topological?) effects that come into play as the angular momentum density grows towards infinity just before a singularity forms.
$endgroup$
– Henning Makholm
2 days ago
1
$begingroup$
@James: My point is that a collapsing star with low but nonzero angular momentum has to go through a phase where its angular momentum density diverges to infinity during the collapse -- whereas a star with zero angular momentum can have zero angular momentum density during its entire collapse. That might (at least a priori) give rise to a qualitative difference that is not respected by the limiting process.
$endgroup$
– Henning Makholm
yesterday
7
7
$begingroup$
Assume a spherical cow...
$endgroup$
– RonJohn
2 days ago
$begingroup$
Assume a spherical cow...
$endgroup$
– RonJohn
2 days ago
5
5
$begingroup$
I'm not sure if the metre is still defined against a standard, but if so, there is one stick that is exactly 1 metre long (by definition). Perhaps not entirely relevant to your point though.
$endgroup$
– Roland Heath
2 days ago
$begingroup$
I'm not sure if the metre is still defined against a standard, but if so, there is one stick that is exactly 1 metre long (by definition). Perhaps not entirely relevant to your point though.
$endgroup$
– Roland Heath
2 days ago
13
13
$begingroup$
@RolandHeath It hasn't been since 1960.
$endgroup$
– Graipher
2 days ago
$begingroup$
@RolandHeath It hasn't been since 1960.
$endgroup$
– Graipher
2 days ago
2
2
$begingroup$
+1, but is it obvious that the non-rotating case is a limiting solution? A priori there might be global (topological?) effects that come into play as the angular momentum density grows towards infinity just before a singularity forms.
$endgroup$
– Henning Makholm
2 days ago
$begingroup$
+1, but is it obvious that the non-rotating case is a limiting solution? A priori there might be global (topological?) effects that come into play as the angular momentum density grows towards infinity just before a singularity forms.
$endgroup$
– Henning Makholm
2 days ago
1
1
$begingroup$
@James: My point is that a collapsing star with low but nonzero angular momentum has to go through a phase where its angular momentum density diverges to infinity during the collapse -- whereas a star with zero angular momentum can have zero angular momentum density during its entire collapse. That might (at least a priori) give rise to a qualitative difference that is not respected by the limiting process.
$endgroup$
– Henning Makholm
yesterday
$begingroup$
@James: My point is that a collapsing star with low but nonzero angular momentum has to go through a phase where its angular momentum density diverges to infinity during the collapse -- whereas a star with zero angular momentum can have zero angular momentum density during its entire collapse. That might (at least a priori) give rise to a qualitative difference that is not respected by the limiting process.
$endgroup$
– Henning Makholm
yesterday
|
show 3 more comments
$begingroup$
Another consideration is that the physics that describe a rotating black hole was much harder to develop.
The maths describing the Schwarzschild (uncharged, non-spinning) black hole was developed in 1916. This was expanded to charged, non-spinning black holes in 1918 (The Reissner–Nordström metric)
It wasn't until 1963 that the Kerr metric for uncharged spinning black holes was developed. Two years later, the most general form, the Kerr-Newman metric was found.
I wouldn't fancy waiting 47 years for a more accurate black hole model to be developed before doing any meaningful work in the field.
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$begingroup$
Also note that the pure Schwarzschild solution is static: it's eternal, not formed by collapse, and it's the only object in an otherwise empty universe. But it's still a useful solution, despite these unnatural simplifications.
$endgroup$
– PM 2Ring
yesterday
add a comment |
$begingroup$
Another consideration is that the physics that describe a rotating black hole was much harder to develop.
The maths describing the Schwarzschild (uncharged, non-spinning) black hole was developed in 1916. This was expanded to charged, non-spinning black holes in 1918 (The Reissner–Nordström metric)
It wasn't until 1963 that the Kerr metric for uncharged spinning black holes was developed. Two years later, the most general form, the Kerr-Newman metric was found.
I wouldn't fancy waiting 47 years for a more accurate black hole model to be developed before doing any meaningful work in the field.
$endgroup$
$begingroup$
Also note that the pure Schwarzschild solution is static: it's eternal, not formed by collapse, and it's the only object in an otherwise empty universe. But it's still a useful solution, despite these unnatural simplifications.
$endgroup$
– PM 2Ring
yesterday
add a comment |
$begingroup$
Another consideration is that the physics that describe a rotating black hole was much harder to develop.
The maths describing the Schwarzschild (uncharged, non-spinning) black hole was developed in 1916. This was expanded to charged, non-spinning black holes in 1918 (The Reissner–Nordström metric)
It wasn't until 1963 that the Kerr metric for uncharged spinning black holes was developed. Two years later, the most general form, the Kerr-Newman metric was found.
I wouldn't fancy waiting 47 years for a more accurate black hole model to be developed before doing any meaningful work in the field.
$endgroup$
Another consideration is that the physics that describe a rotating black hole was much harder to develop.
The maths describing the Schwarzschild (uncharged, non-spinning) black hole was developed in 1916. This was expanded to charged, non-spinning black holes in 1918 (The Reissner–Nordström metric)
It wasn't until 1963 that the Kerr metric for uncharged spinning black holes was developed. Two years later, the most general form, the Kerr-Newman metric was found.
I wouldn't fancy waiting 47 years for a more accurate black hole model to be developed before doing any meaningful work in the field.
answered Aug 21 at 22:55
IngolifsIngolifs
2,4391 gold badge10 silver badges26 bronze badges
2,4391 gold badge10 silver badges26 bronze badges
$begingroup$
Also note that the pure Schwarzschild solution is static: it's eternal, not formed by collapse, and it's the only object in an otherwise empty universe. But it's still a useful solution, despite these unnatural simplifications.
$endgroup$
– PM 2Ring
yesterday
add a comment |
$begingroup$
Also note that the pure Schwarzschild solution is static: it's eternal, not formed by collapse, and it's the only object in an otherwise empty universe. But it's still a useful solution, despite these unnatural simplifications.
$endgroup$
– PM 2Ring
yesterday
$begingroup$
Also note that the pure Schwarzschild solution is static: it's eternal, not formed by collapse, and it's the only object in an otherwise empty universe. But it's still a useful solution, despite these unnatural simplifications.
$endgroup$
– PM 2Ring
yesterday
$begingroup$
Also note that the pure Schwarzschild solution is static: it's eternal, not formed by collapse, and it's the only object in an otherwise empty universe. But it's still a useful solution, despite these unnatural simplifications.
$endgroup$
– PM 2Ring
yesterday
add a comment |
$begingroup$
Our sun's rotation period is 24.47 days at the equator and almost 38 days at the poles, our planet's rotational period is 23h 56m 4.098,903,691s. Use of Schwarzschild equations for either case isn't exact.
If you used the equation for non-rotating objects to calculate the time at the altitude of GPS satellites (~ 20,200 km or 12,550 miles) then you would be off by 38,636 nanoseconds per day. A Julian year is defined as 365.25 days of exactly 86,400 seconds (SI base unit), totalling exactly 31,557,600 seconds in the Julian astronomical year. The Gregorian calendar year (400 year average) is 365.2425 days.
Multiplying 365.2425 x 38,636 = 14,111,509.23 nanoseconds, that's 0.0141 seconds per year. If being off by that amount isn't of any concern to you then you can use the easier equation, such as for calculations involving the star HR 1362 which has a rotational period that is 306.9 ± 0.4 days.
$endgroup$
add a comment |
$begingroup$
Our sun's rotation period is 24.47 days at the equator and almost 38 days at the poles, our planet's rotational period is 23h 56m 4.098,903,691s. Use of Schwarzschild equations for either case isn't exact.
If you used the equation for non-rotating objects to calculate the time at the altitude of GPS satellites (~ 20,200 km or 12,550 miles) then you would be off by 38,636 nanoseconds per day. A Julian year is defined as 365.25 days of exactly 86,400 seconds (SI base unit), totalling exactly 31,557,600 seconds in the Julian astronomical year. The Gregorian calendar year (400 year average) is 365.2425 days.
Multiplying 365.2425 x 38,636 = 14,111,509.23 nanoseconds, that's 0.0141 seconds per year. If being off by that amount isn't of any concern to you then you can use the easier equation, such as for calculations involving the star HR 1362 which has a rotational period that is 306.9 ± 0.4 days.
$endgroup$
add a comment |
$begingroup$
Our sun's rotation period is 24.47 days at the equator and almost 38 days at the poles, our planet's rotational period is 23h 56m 4.098,903,691s. Use of Schwarzschild equations for either case isn't exact.
If you used the equation for non-rotating objects to calculate the time at the altitude of GPS satellites (~ 20,200 km or 12,550 miles) then you would be off by 38,636 nanoseconds per day. A Julian year is defined as 365.25 days of exactly 86,400 seconds (SI base unit), totalling exactly 31,557,600 seconds in the Julian astronomical year. The Gregorian calendar year (400 year average) is 365.2425 days.
Multiplying 365.2425 x 38,636 = 14,111,509.23 nanoseconds, that's 0.0141 seconds per year. If being off by that amount isn't of any concern to you then you can use the easier equation, such as for calculations involving the star HR 1362 which has a rotational period that is 306.9 ± 0.4 days.
$endgroup$
Our sun's rotation period is 24.47 days at the equator and almost 38 days at the poles, our planet's rotational period is 23h 56m 4.098,903,691s. Use of Schwarzschild equations for either case isn't exact.
If you used the equation for non-rotating objects to calculate the time at the altitude of GPS satellites (~ 20,200 km or 12,550 miles) then you would be off by 38,636 nanoseconds per day. A Julian year is defined as 365.25 days of exactly 86,400 seconds (SI base unit), totalling exactly 31,557,600 seconds in the Julian astronomical year. The Gregorian calendar year (400 year average) is 365.2425 days.
Multiplying 365.2425 x 38,636 = 14,111,509.23 nanoseconds, that's 0.0141 seconds per year. If being off by that amount isn't of any concern to you then you can use the easier equation, such as for calculations involving the star HR 1362 which has a rotational period that is 306.9 ± 0.4 days.
answered 2 days ago
RobRob
1,6821 gold badge5 silver badges18 bronze badges
1,6821 gold badge5 silver badges18 bronze badges
add a comment |
add a comment |
$begingroup$
You're right: all stars rotate. The only reason I can think of why astrophysicists make calculations for a non-rotating star or black hole is that it makes their calculations a bit easier. Although all stars rotate, some rotate much faster than others, and their masses vary too, so there is a wide degree of uncertainty which is reduced by calculating for a star that does not rotate.
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4
$begingroup$
How certain can we be that all stars rotate? There are a lot of stars and many many possible (theoretical) interactions that would slow rotation.
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– Valorum
Aug 21 at 22:11
$begingroup$
No one has found one yet. I suspect it would cause a sensation if one were discovered.
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– Michael Walsby
Aug 21 at 22:26
1
$begingroup$
@Valorum Yeah, I was thinking about a stellar collision where the stars are rotating in opposite directions. If the rotational energy is exactly opposite you'll get a non-rotating result. Very unlikely, not utterly impossible--thus it probably will happen somewhere, someday.
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– Loren Pechtel
2 days ago
1
$begingroup$
@LorenPechtel The rotational momentum needs to be exactly equal. I think that counts as utterly impossible.
$endgroup$
– Martin Bonner
2 days ago
$begingroup$
@Valorum Because the chance for "zero" angular momentum approaches 0 much faster than the amount of stars grow with "sample size".
$endgroup$
– paul23
8 hours ago
|
show 1 more comment
$begingroup$
You're right: all stars rotate. The only reason I can think of why astrophysicists make calculations for a non-rotating star or black hole is that it makes their calculations a bit easier. Although all stars rotate, some rotate much faster than others, and their masses vary too, so there is a wide degree of uncertainty which is reduced by calculating for a star that does not rotate.
$endgroup$
4
$begingroup$
How certain can we be that all stars rotate? There are a lot of stars and many many possible (theoretical) interactions that would slow rotation.
$endgroup$
– Valorum
Aug 21 at 22:11
$begingroup$
No one has found one yet. I suspect it would cause a sensation if one were discovered.
$endgroup$
– Michael Walsby
Aug 21 at 22:26
1
$begingroup$
@Valorum Yeah, I was thinking about a stellar collision where the stars are rotating in opposite directions. If the rotational energy is exactly opposite you'll get a non-rotating result. Very unlikely, not utterly impossible--thus it probably will happen somewhere, someday.
$endgroup$
– Loren Pechtel
2 days ago
1
$begingroup$
@LorenPechtel The rotational momentum needs to be exactly equal. I think that counts as utterly impossible.
$endgroup$
– Martin Bonner
2 days ago
$begingroup$
@Valorum Because the chance for "zero" angular momentum approaches 0 much faster than the amount of stars grow with "sample size".
$endgroup$
– paul23
8 hours ago
|
show 1 more comment
$begingroup$
You're right: all stars rotate. The only reason I can think of why astrophysicists make calculations for a non-rotating star or black hole is that it makes their calculations a bit easier. Although all stars rotate, some rotate much faster than others, and their masses vary too, so there is a wide degree of uncertainty which is reduced by calculating for a star that does not rotate.
$endgroup$
You're right: all stars rotate. The only reason I can think of why astrophysicists make calculations for a non-rotating star or black hole is that it makes their calculations a bit easier. Although all stars rotate, some rotate much faster than others, and their masses vary too, so there is a wide degree of uncertainty which is reduced by calculating for a star that does not rotate.
answered Aug 21 at 12:21
Michael WalsbyMichael Walsby
9211 silver badge6 bronze badges
9211 silver badge6 bronze badges
4
$begingroup$
How certain can we be that all stars rotate? There are a lot of stars and many many possible (theoretical) interactions that would slow rotation.
$endgroup$
– Valorum
Aug 21 at 22:11
$begingroup$
No one has found one yet. I suspect it would cause a sensation if one were discovered.
$endgroup$
– Michael Walsby
Aug 21 at 22:26
1
$begingroup$
@Valorum Yeah, I was thinking about a stellar collision where the stars are rotating in opposite directions. If the rotational energy is exactly opposite you'll get a non-rotating result. Very unlikely, not utterly impossible--thus it probably will happen somewhere, someday.
$endgroup$
– Loren Pechtel
2 days ago
1
$begingroup$
@LorenPechtel The rotational momentum needs to be exactly equal. I think that counts as utterly impossible.
$endgroup$
– Martin Bonner
2 days ago
$begingroup$
@Valorum Because the chance for "zero" angular momentum approaches 0 much faster than the amount of stars grow with "sample size".
$endgroup$
– paul23
8 hours ago
|
show 1 more comment
4
$begingroup$
How certain can we be that all stars rotate? There are a lot of stars and many many possible (theoretical) interactions that would slow rotation.
$endgroup$
– Valorum
Aug 21 at 22:11
$begingroup$
No one has found one yet. I suspect it would cause a sensation if one were discovered.
$endgroup$
– Michael Walsby
Aug 21 at 22:26
1
$begingroup$
@Valorum Yeah, I was thinking about a stellar collision where the stars are rotating in opposite directions. If the rotational energy is exactly opposite you'll get a non-rotating result. Very unlikely, not utterly impossible--thus it probably will happen somewhere, someday.
$endgroup$
– Loren Pechtel
2 days ago
1
$begingroup$
@LorenPechtel The rotational momentum needs to be exactly equal. I think that counts as utterly impossible.
$endgroup$
– Martin Bonner
2 days ago
$begingroup$
@Valorum Because the chance for "zero" angular momentum approaches 0 much faster than the amount of stars grow with "sample size".
$endgroup$
– paul23
8 hours ago
4
4
$begingroup$
How certain can we be that all stars rotate? There are a lot of stars and many many possible (theoretical) interactions that would slow rotation.
$endgroup$
– Valorum
Aug 21 at 22:11
$begingroup$
How certain can we be that all stars rotate? There are a lot of stars and many many possible (theoretical) interactions that would slow rotation.
$endgroup$
– Valorum
Aug 21 at 22:11
$begingroup$
No one has found one yet. I suspect it would cause a sensation if one were discovered.
$endgroup$
– Michael Walsby
Aug 21 at 22:26
$begingroup$
No one has found one yet. I suspect it would cause a sensation if one were discovered.
$endgroup$
– Michael Walsby
Aug 21 at 22:26
1
1
$begingroup$
@Valorum Yeah, I was thinking about a stellar collision where the stars are rotating in opposite directions. If the rotational energy is exactly opposite you'll get a non-rotating result. Very unlikely, not utterly impossible--thus it probably will happen somewhere, someday.
$endgroup$
– Loren Pechtel
2 days ago
$begingroup$
@Valorum Yeah, I was thinking about a stellar collision where the stars are rotating in opposite directions. If the rotational energy is exactly opposite you'll get a non-rotating result. Very unlikely, not utterly impossible--thus it probably will happen somewhere, someday.
$endgroup$
– Loren Pechtel
2 days ago
1
1
$begingroup$
@LorenPechtel The rotational momentum needs to be exactly equal. I think that counts as utterly impossible.
$endgroup$
– Martin Bonner
2 days ago
$begingroup$
@LorenPechtel The rotational momentum needs to be exactly equal. I think that counts as utterly impossible.
$endgroup$
– Martin Bonner
2 days ago
$begingroup$
@Valorum Because the chance for "zero" angular momentum approaches 0 much faster than the amount of stars grow with "sample size".
$endgroup$
– paul23
8 hours ago
$begingroup$
@Valorum Because the chance for "zero" angular momentum approaches 0 much faster than the amount of stars grow with "sample size".
$endgroup$
– paul23
8 hours ago
|
show 1 more comment
urban pečoler is a new contributor. Be nice, and check out our Code of Conduct.
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Would you mind providing more information about the research, e.g. linking to a paper about it?
$endgroup$
– HDE 226868♦
Aug 21 at 13:58
20
$begingroup$
Frictionless spherical cows are useful abstractions too...
$endgroup$
– Beanluc
2 days ago
4
$begingroup$
I suppose it's the solution to a simplified model of reality as a first step? That's not unusual in science...
$endgroup$
– Peter A. Schneider
2 days ago
7
$begingroup$
"However, every star in the universe" You've checked them all have you?
$endgroup$
– TripeHound
2 days ago
4
$begingroup$
"All models are wrong, but some are useful"
$endgroup$
– llama
2 days ago