Global Frame of Reference in General RelativityHow do frames of reference work in general relativity, and are...

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Global Frame of Reference in General Relativity

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Global Frame of Reference in General Relativity


How do frames of reference work in general relativity, and are they described by coordinate systems?Whose reference frame to use for $d theta$ near a black hole?Ray tracing in General RelativityGlobal symmetries of spacetime and general covariancePhysical interpretation of Kruskal-Szekeres coordinatesIs my interpretation of these observers correct in General Relativity?Line element in Kruskal coordinatesVelocity of a photon according to a stationary observer in Schwarzschild metricCoordinate transformations in general relativityWhat is the apparent radius of the photon sphere in isotropic coordinates?






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$begingroup$


People have been saying that a global frame of reference does not exist in General Relativity.



However, from Wikipedia:




In Schwarzschild coordinates ${displaystyle (t,r,theta ,phi )}$ the Schwarzschild metric (or equivalently, the line element for proper time) has the form
$${displaystyle g=-c^{2},{dtau }^{2}=-left(1-{frac {r_{mathrm {s} }}{r}}right)c^{2},dt^{2}+left(1-{frac {r_{mathrm {s} }}{r}}right)^{-1},dr^{2}+r^{2}g_{Omega },}
$$

where ${displaystyle g_{Omega }}{displaystyle g_{Omega }}$ is the metric on the two sphere, i.e. ${displaystyle g_{Omega }=left(dtheta ^{2}+sin ^{2}theta ,dvarphi ^{2}right)}.$



–"Schwarzschild metric", Wikipedia




Isn't $left(t,r,theta,phiright)$ representing the spherical coordinates of an observer, looking at the black-hole from infinity?










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  • $begingroup$
    related: physics.stackexchange.com/questions/458854/…
    $endgroup$
    – Ben Crowell
    58 mins ago










  • $begingroup$
    Isn't $left(t,r,theta,phiright)$ representing the spherical coordinates of an observer, looking at the black-hole from infinity? The short answer is no. What makes you think it does?
    $endgroup$
    – Ben Crowell
    58 mins ago


















4












$begingroup$


People have been saying that a global frame of reference does not exist in General Relativity.



However, from Wikipedia:




In Schwarzschild coordinates ${displaystyle (t,r,theta ,phi )}$ the Schwarzschild metric (or equivalently, the line element for proper time) has the form
$${displaystyle g=-c^{2},{dtau }^{2}=-left(1-{frac {r_{mathrm {s} }}{r}}right)c^{2},dt^{2}+left(1-{frac {r_{mathrm {s} }}{r}}right)^{-1},dr^{2}+r^{2}g_{Omega },}
$$

where ${displaystyle g_{Omega }}{displaystyle g_{Omega }}$ is the metric on the two sphere, i.e. ${displaystyle g_{Omega }=left(dtheta ^{2}+sin ^{2}theta ,dvarphi ^{2}right)}.$



–"Schwarzschild metric", Wikipedia




Isn't $left(t,r,theta,phiright)$ representing the spherical coordinates of an observer, looking at the black-hole from infinity?










share|cite|improve this question









New contributor



The Last StyleBender is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$














  • $begingroup$
    related: physics.stackexchange.com/questions/458854/…
    $endgroup$
    – Ben Crowell
    58 mins ago










  • $begingroup$
    Isn't $left(t,r,theta,phiright)$ representing the spherical coordinates of an observer, looking at the black-hole from infinity? The short answer is no. What makes you think it does?
    $endgroup$
    – Ben Crowell
    58 mins ago














4












4








4





$begingroup$


People have been saying that a global frame of reference does not exist in General Relativity.



However, from Wikipedia:




In Schwarzschild coordinates ${displaystyle (t,r,theta ,phi )}$ the Schwarzschild metric (or equivalently, the line element for proper time) has the form
$${displaystyle g=-c^{2},{dtau }^{2}=-left(1-{frac {r_{mathrm {s} }}{r}}right)c^{2},dt^{2}+left(1-{frac {r_{mathrm {s} }}{r}}right)^{-1},dr^{2}+r^{2}g_{Omega },}
$$

where ${displaystyle g_{Omega }}{displaystyle g_{Omega }}$ is the metric on the two sphere, i.e. ${displaystyle g_{Omega }=left(dtheta ^{2}+sin ^{2}theta ,dvarphi ^{2}right)}.$



–"Schwarzschild metric", Wikipedia




Isn't $left(t,r,theta,phiright)$ representing the spherical coordinates of an observer, looking at the black-hole from infinity?










share|cite|improve this question









New contributor



The Last StyleBender is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$




People have been saying that a global frame of reference does not exist in General Relativity.



However, from Wikipedia:




In Schwarzschild coordinates ${displaystyle (t,r,theta ,phi )}$ the Schwarzschild metric (or equivalently, the line element for proper time) has the form
$${displaystyle g=-c^{2},{dtau }^{2}=-left(1-{frac {r_{mathrm {s} }}{r}}right)c^{2},dt^{2}+left(1-{frac {r_{mathrm {s} }}{r}}right)^{-1},dr^{2}+r^{2}g_{Omega },}
$$

where ${displaystyle g_{Omega }}{displaystyle g_{Omega }}$ is the metric on the two sphere, i.e. ${displaystyle g_{Omega }=left(dtheta ^{2}+sin ^{2}theta ,dvarphi ^{2}right)}.$



–"Schwarzschild metric", Wikipedia




Isn't $left(t,r,theta,phiright)$ representing the spherical coordinates of an observer, looking at the black-hole from infinity?







general-relativity reference-frames coordinate-systems






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edited 6 hours ago









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  • $begingroup$
    related: physics.stackexchange.com/questions/458854/…
    $endgroup$
    – Ben Crowell
    58 mins ago










  • $begingroup$
    Isn't $left(t,r,theta,phiright)$ representing the spherical coordinates of an observer, looking at the black-hole from infinity? The short answer is no. What makes you think it does?
    $endgroup$
    – Ben Crowell
    58 mins ago


















  • $begingroup$
    related: physics.stackexchange.com/questions/458854/…
    $endgroup$
    – Ben Crowell
    58 mins ago










  • $begingroup$
    Isn't $left(t,r,theta,phiright)$ representing the spherical coordinates of an observer, looking at the black-hole from infinity? The short answer is no. What makes you think it does?
    $endgroup$
    – Ben Crowell
    58 mins ago
















$begingroup$
related: physics.stackexchange.com/questions/458854/…
$endgroup$
– Ben Crowell
58 mins ago




$begingroup$
related: physics.stackexchange.com/questions/458854/…
$endgroup$
– Ben Crowell
58 mins ago












$begingroup$
Isn't $left(t,r,theta,phiright)$ representing the spherical coordinates of an observer, looking at the black-hole from infinity? The short answer is no. What makes you think it does?
$endgroup$
– Ben Crowell
58 mins ago




$begingroup$
Isn't $left(t,r,theta,phiright)$ representing the spherical coordinates of an observer, looking at the black-hole from infinity? The short answer is no. What makes you think it does?
$endgroup$
– Ben Crowell
58 mins ago










3 Answers
3






active

oldest

votes


















3














$begingroup$


People have been saying that a global frame of reference does not exist in General Relativity.




Can you give an example? I don't really know what this means.



There are no global inertial reference frames in general relativity, in general.




[...] Isn't (t,r,θ,ϕ) representing the spherical coordinates of an observer, looking at the black-hole from infinity?




The word "observer" is used in totally different ways in special relativity, general relativity, and quantum mechanics. If you assume they're related, you'll get confused.



In special relativity, "observer" usually means "inertial reference frame."



In general relativity, "observer" usually means a scientist making observations from a particular vantage point (a particular location in spacetime).



In special relativity it happens to be the case that if a scientist is moving inertially, there's a unique (up to time translation) global inertial reference frame in which they're at rest at the origin. Often people call that "their" coordinate system, though this is very misleading since you can just as well use any other coordinate system to describe their motion if you want to. But that unique inertial frame does exist, and it can be defined mathematically. There's no analogous special coordinate system for a moving scientist in general relativity. You might, in some cases, find it useful to use coordinates in which the scientist is at rest at the origin, but that doesn't uniquely define what the coordinates look like away from the origin. And often that coordinate system isn't very useful anyway, because the metric has a complicated form with respect to it, and a simpler form with respect to some other coordinate system in which the scientist isn't at rest.



Schwarzschild coordinates (t,r,θ,ϕ) are just coordinates in which the Schwarzschild metric has a simple mathematical form. They don't represent any observer, in the special-relativistic or general-relativistic sense. You can use Schwarzschild coordinates to determine what an observer (general-relativistic) will see. You can also use totally different coordinates to work out what the same observer will see, and if you did it correctly you'll get the same answer, because the underlying physics is independent of the coordinates.






share|cite|improve this answer









$endgroup$























    1














    $begingroup$

    when people talk about 'local frame of references' they are basically talking about charts, which are diffeomorphisms between an open subset of the manifold and open subset of Minkowski. A global frame of reference is actually an atlas made by the union of all charts that cover spacetime



    A chart is basically a local coordinate bidirectional function between the target space and the coordinate space, an atlas is a collection of such charts with some overlap between them.



    To drive the point home, notice that if you wanted to make a single 2D coordinate system on the sphere, you are either going to find points that behave "oddly" like the poles, for which the function is not entirely bidirectional. You can address this by adding more coordinate systems, like one for the northern hemisphere and another for the southern hemisphere, which can be both parametrized as a simple circle with the south and north pole as centers, and in this case these charts will overlap on the region of the equator. This illustrates that not all spaces can be covered with a single coordinate system, but if you allow more than one, you can consider the union of all those coordinate systems as a collection of local maps, an Atlas. When someone talks about 'global frame of references' in general geometries, they are actually referring to these structures






    share|cite|improve this answer











    $endgroup$















    • $begingroup$
      Is it okay if you can explain it using less Mathematical language? And can you address whether a global frame of reference exist or not?
      $endgroup$
      – The Last StyleBender
      9 hours ago










    • $begingroup$
      @lurscher so if we are talking about spacetime outside event horizon then the Schwarzschild metric is a chart which covers the outside spacetime righteously but to cover the whole manifold then there are 4 different chart for 4 different region of spacetime as seen in Kruskal coordinates. So can we say the whole manifold of schwarzschild sptm can be covered in one chart of Kruskal coordinate.
      $endgroup$
      – aitfel
      7 hours ago



















    0














    $begingroup$

    Depends on the metric and topology. Euclidian metric for exame you can have a global coordinate system.






    share|cite|improve this answer









    $endgroup$















    • $begingroup$
      Why is the Euclidean metric relevant to a question abour general relativity?
      $endgroup$
      – PM 2Ring
      7 hours ago














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    3 Answers
    3






    active

    oldest

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    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3














    $begingroup$


    People have been saying that a global frame of reference does not exist in General Relativity.




    Can you give an example? I don't really know what this means.



    There are no global inertial reference frames in general relativity, in general.




    [...] Isn't (t,r,θ,ϕ) representing the spherical coordinates of an observer, looking at the black-hole from infinity?




    The word "observer" is used in totally different ways in special relativity, general relativity, and quantum mechanics. If you assume they're related, you'll get confused.



    In special relativity, "observer" usually means "inertial reference frame."



    In general relativity, "observer" usually means a scientist making observations from a particular vantage point (a particular location in spacetime).



    In special relativity it happens to be the case that if a scientist is moving inertially, there's a unique (up to time translation) global inertial reference frame in which they're at rest at the origin. Often people call that "their" coordinate system, though this is very misleading since you can just as well use any other coordinate system to describe their motion if you want to. But that unique inertial frame does exist, and it can be defined mathematically. There's no analogous special coordinate system for a moving scientist in general relativity. You might, in some cases, find it useful to use coordinates in which the scientist is at rest at the origin, but that doesn't uniquely define what the coordinates look like away from the origin. And often that coordinate system isn't very useful anyway, because the metric has a complicated form with respect to it, and a simpler form with respect to some other coordinate system in which the scientist isn't at rest.



    Schwarzschild coordinates (t,r,θ,ϕ) are just coordinates in which the Schwarzschild metric has a simple mathematical form. They don't represent any observer, in the special-relativistic or general-relativistic sense. You can use Schwarzschild coordinates to determine what an observer (general-relativistic) will see. You can also use totally different coordinates to work out what the same observer will see, and if you did it correctly you'll get the same answer, because the underlying physics is independent of the coordinates.






    share|cite|improve this answer









    $endgroup$




















      3














      $begingroup$


      People have been saying that a global frame of reference does not exist in General Relativity.




      Can you give an example? I don't really know what this means.



      There are no global inertial reference frames in general relativity, in general.




      [...] Isn't (t,r,θ,ϕ) representing the spherical coordinates of an observer, looking at the black-hole from infinity?




      The word "observer" is used in totally different ways in special relativity, general relativity, and quantum mechanics. If you assume they're related, you'll get confused.



      In special relativity, "observer" usually means "inertial reference frame."



      In general relativity, "observer" usually means a scientist making observations from a particular vantage point (a particular location in spacetime).



      In special relativity it happens to be the case that if a scientist is moving inertially, there's a unique (up to time translation) global inertial reference frame in which they're at rest at the origin. Often people call that "their" coordinate system, though this is very misleading since you can just as well use any other coordinate system to describe their motion if you want to. But that unique inertial frame does exist, and it can be defined mathematically. There's no analogous special coordinate system for a moving scientist in general relativity. You might, in some cases, find it useful to use coordinates in which the scientist is at rest at the origin, but that doesn't uniquely define what the coordinates look like away from the origin. And often that coordinate system isn't very useful anyway, because the metric has a complicated form with respect to it, and a simpler form with respect to some other coordinate system in which the scientist isn't at rest.



      Schwarzschild coordinates (t,r,θ,ϕ) are just coordinates in which the Schwarzschild metric has a simple mathematical form. They don't represent any observer, in the special-relativistic or general-relativistic sense. You can use Schwarzschild coordinates to determine what an observer (general-relativistic) will see. You can also use totally different coordinates to work out what the same observer will see, and if you did it correctly you'll get the same answer, because the underlying physics is independent of the coordinates.






      share|cite|improve this answer









      $endgroup$


















        3














        3










        3







        $begingroup$


        People have been saying that a global frame of reference does not exist in General Relativity.




        Can you give an example? I don't really know what this means.



        There are no global inertial reference frames in general relativity, in general.




        [...] Isn't (t,r,θ,ϕ) representing the spherical coordinates of an observer, looking at the black-hole from infinity?




        The word "observer" is used in totally different ways in special relativity, general relativity, and quantum mechanics. If you assume they're related, you'll get confused.



        In special relativity, "observer" usually means "inertial reference frame."



        In general relativity, "observer" usually means a scientist making observations from a particular vantage point (a particular location in spacetime).



        In special relativity it happens to be the case that if a scientist is moving inertially, there's a unique (up to time translation) global inertial reference frame in which they're at rest at the origin. Often people call that "their" coordinate system, though this is very misleading since you can just as well use any other coordinate system to describe their motion if you want to. But that unique inertial frame does exist, and it can be defined mathematically. There's no analogous special coordinate system for a moving scientist in general relativity. You might, in some cases, find it useful to use coordinates in which the scientist is at rest at the origin, but that doesn't uniquely define what the coordinates look like away from the origin. And often that coordinate system isn't very useful anyway, because the metric has a complicated form with respect to it, and a simpler form with respect to some other coordinate system in which the scientist isn't at rest.



        Schwarzschild coordinates (t,r,θ,ϕ) are just coordinates in which the Schwarzschild metric has a simple mathematical form. They don't represent any observer, in the special-relativistic or general-relativistic sense. You can use Schwarzschild coordinates to determine what an observer (general-relativistic) will see. You can also use totally different coordinates to work out what the same observer will see, and if you did it correctly you'll get the same answer, because the underlying physics is independent of the coordinates.






        share|cite|improve this answer









        $endgroup$




        People have been saying that a global frame of reference does not exist in General Relativity.




        Can you give an example? I don't really know what this means.



        There are no global inertial reference frames in general relativity, in general.




        [...] Isn't (t,r,θ,ϕ) representing the spherical coordinates of an observer, looking at the black-hole from infinity?




        The word "observer" is used in totally different ways in special relativity, general relativity, and quantum mechanics. If you assume they're related, you'll get confused.



        In special relativity, "observer" usually means "inertial reference frame."



        In general relativity, "observer" usually means a scientist making observations from a particular vantage point (a particular location in spacetime).



        In special relativity it happens to be the case that if a scientist is moving inertially, there's a unique (up to time translation) global inertial reference frame in which they're at rest at the origin. Often people call that "their" coordinate system, though this is very misleading since you can just as well use any other coordinate system to describe their motion if you want to. But that unique inertial frame does exist, and it can be defined mathematically. There's no analogous special coordinate system for a moving scientist in general relativity. You might, in some cases, find it useful to use coordinates in which the scientist is at rest at the origin, but that doesn't uniquely define what the coordinates look like away from the origin. And often that coordinate system isn't very useful anyway, because the metric has a complicated form with respect to it, and a simpler form with respect to some other coordinate system in which the scientist isn't at rest.



        Schwarzschild coordinates (t,r,θ,ϕ) are just coordinates in which the Schwarzschild metric has a simple mathematical form. They don't represent any observer, in the special-relativistic or general-relativistic sense. You can use Schwarzschild coordinates to determine what an observer (general-relativistic) will see. You can also use totally different coordinates to work out what the same observer will see, and if you did it correctly you'll get the same answer, because the underlying physics is independent of the coordinates.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 7 hours ago









        benrgbenrg

        2,2887 silver badges18 bronze badges




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            1














            $begingroup$

            when people talk about 'local frame of references' they are basically talking about charts, which are diffeomorphisms between an open subset of the manifold and open subset of Minkowski. A global frame of reference is actually an atlas made by the union of all charts that cover spacetime



            A chart is basically a local coordinate bidirectional function between the target space and the coordinate space, an atlas is a collection of such charts with some overlap between them.



            To drive the point home, notice that if you wanted to make a single 2D coordinate system on the sphere, you are either going to find points that behave "oddly" like the poles, for which the function is not entirely bidirectional. You can address this by adding more coordinate systems, like one for the northern hemisphere and another for the southern hemisphere, which can be both parametrized as a simple circle with the south and north pole as centers, and in this case these charts will overlap on the region of the equator. This illustrates that not all spaces can be covered with a single coordinate system, but if you allow more than one, you can consider the union of all those coordinate systems as a collection of local maps, an Atlas. When someone talks about 'global frame of references' in general geometries, they are actually referring to these structures






            share|cite|improve this answer











            $endgroup$















            • $begingroup$
              Is it okay if you can explain it using less Mathematical language? And can you address whether a global frame of reference exist or not?
              $endgroup$
              – The Last StyleBender
              9 hours ago










            • $begingroup$
              @lurscher so if we are talking about spacetime outside event horizon then the Schwarzschild metric is a chart which covers the outside spacetime righteously but to cover the whole manifold then there are 4 different chart for 4 different region of spacetime as seen in Kruskal coordinates. So can we say the whole manifold of schwarzschild sptm can be covered in one chart of Kruskal coordinate.
              $endgroup$
              – aitfel
              7 hours ago
















            1














            $begingroup$

            when people talk about 'local frame of references' they are basically talking about charts, which are diffeomorphisms between an open subset of the manifold and open subset of Minkowski. A global frame of reference is actually an atlas made by the union of all charts that cover spacetime



            A chart is basically a local coordinate bidirectional function between the target space and the coordinate space, an atlas is a collection of such charts with some overlap between them.



            To drive the point home, notice that if you wanted to make a single 2D coordinate system on the sphere, you are either going to find points that behave "oddly" like the poles, for which the function is not entirely bidirectional. You can address this by adding more coordinate systems, like one for the northern hemisphere and another for the southern hemisphere, which can be both parametrized as a simple circle with the south and north pole as centers, and in this case these charts will overlap on the region of the equator. This illustrates that not all spaces can be covered with a single coordinate system, but if you allow more than one, you can consider the union of all those coordinate systems as a collection of local maps, an Atlas. When someone talks about 'global frame of references' in general geometries, they are actually referring to these structures






            share|cite|improve this answer











            $endgroup$















            • $begingroup$
              Is it okay if you can explain it using less Mathematical language? And can you address whether a global frame of reference exist or not?
              $endgroup$
              – The Last StyleBender
              9 hours ago










            • $begingroup$
              @lurscher so if we are talking about spacetime outside event horizon then the Schwarzschild metric is a chart which covers the outside spacetime righteously but to cover the whole manifold then there are 4 different chart for 4 different region of spacetime as seen in Kruskal coordinates. So can we say the whole manifold of schwarzschild sptm can be covered in one chart of Kruskal coordinate.
              $endgroup$
              – aitfel
              7 hours ago














            1














            1










            1







            $begingroup$

            when people talk about 'local frame of references' they are basically talking about charts, which are diffeomorphisms between an open subset of the manifold and open subset of Minkowski. A global frame of reference is actually an atlas made by the union of all charts that cover spacetime



            A chart is basically a local coordinate bidirectional function between the target space and the coordinate space, an atlas is a collection of such charts with some overlap between them.



            To drive the point home, notice that if you wanted to make a single 2D coordinate system on the sphere, you are either going to find points that behave "oddly" like the poles, for which the function is not entirely bidirectional. You can address this by adding more coordinate systems, like one for the northern hemisphere and another for the southern hemisphere, which can be both parametrized as a simple circle with the south and north pole as centers, and in this case these charts will overlap on the region of the equator. This illustrates that not all spaces can be covered with a single coordinate system, but if you allow more than one, you can consider the union of all those coordinate systems as a collection of local maps, an Atlas. When someone talks about 'global frame of references' in general geometries, they are actually referring to these structures






            share|cite|improve this answer











            $endgroup$



            when people talk about 'local frame of references' they are basically talking about charts, which are diffeomorphisms between an open subset of the manifold and open subset of Minkowski. A global frame of reference is actually an atlas made by the union of all charts that cover spacetime



            A chart is basically a local coordinate bidirectional function between the target space and the coordinate space, an atlas is a collection of such charts with some overlap between them.



            To drive the point home, notice that if you wanted to make a single 2D coordinate system on the sphere, you are either going to find points that behave "oddly" like the poles, for which the function is not entirely bidirectional. You can address this by adding more coordinate systems, like one for the northern hemisphere and another for the southern hemisphere, which can be both parametrized as a simple circle with the south and north pole as centers, and in this case these charts will overlap on the region of the equator. This illustrates that not all spaces can be covered with a single coordinate system, but if you allow more than one, you can consider the union of all those coordinate systems as a collection of local maps, an Atlas. When someone talks about 'global frame of references' in general geometries, they are actually referring to these structures







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            edited 8 hours ago

























            answered 9 hours ago









            lurscherlurscher

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            • $begingroup$
              Is it okay if you can explain it using less Mathematical language? And can you address whether a global frame of reference exist or not?
              $endgroup$
              – The Last StyleBender
              9 hours ago










            • $begingroup$
              @lurscher so if we are talking about spacetime outside event horizon then the Schwarzschild metric is a chart which covers the outside spacetime righteously but to cover the whole manifold then there are 4 different chart for 4 different region of spacetime as seen in Kruskal coordinates. So can we say the whole manifold of schwarzschild sptm can be covered in one chart of Kruskal coordinate.
              $endgroup$
              – aitfel
              7 hours ago


















            • $begingroup$
              Is it okay if you can explain it using less Mathematical language? And can you address whether a global frame of reference exist or not?
              $endgroup$
              – The Last StyleBender
              9 hours ago










            • $begingroup$
              @lurscher so if we are talking about spacetime outside event horizon then the Schwarzschild metric is a chart which covers the outside spacetime righteously but to cover the whole manifold then there are 4 different chart for 4 different region of spacetime as seen in Kruskal coordinates. So can we say the whole manifold of schwarzschild sptm can be covered in one chart of Kruskal coordinate.
              $endgroup$
              – aitfel
              7 hours ago
















            $begingroup$
            Is it okay if you can explain it using less Mathematical language? And can you address whether a global frame of reference exist or not?
            $endgroup$
            – The Last StyleBender
            9 hours ago




            $begingroup$
            Is it okay if you can explain it using less Mathematical language? And can you address whether a global frame of reference exist or not?
            $endgroup$
            – The Last StyleBender
            9 hours ago












            $begingroup$
            @lurscher so if we are talking about spacetime outside event horizon then the Schwarzschild metric is a chart which covers the outside spacetime righteously but to cover the whole manifold then there are 4 different chart for 4 different region of spacetime as seen in Kruskal coordinates. So can we say the whole manifold of schwarzschild sptm can be covered in one chart of Kruskal coordinate.
            $endgroup$
            – aitfel
            7 hours ago




            $begingroup$
            @lurscher so if we are talking about spacetime outside event horizon then the Schwarzschild metric is a chart which covers the outside spacetime righteously but to cover the whole manifold then there are 4 different chart for 4 different region of spacetime as seen in Kruskal coordinates. So can we say the whole manifold of schwarzschild sptm can be covered in one chart of Kruskal coordinate.
            $endgroup$
            – aitfel
            7 hours ago











            0














            $begingroup$

            Depends on the metric and topology. Euclidian metric for exame you can have a global coordinate system.






            share|cite|improve this answer









            $endgroup$















            • $begingroup$
              Why is the Euclidean metric relevant to a question abour general relativity?
              $endgroup$
              – PM 2Ring
              7 hours ago
















            0














            $begingroup$

            Depends on the metric and topology. Euclidian metric for exame you can have a global coordinate system.






            share|cite|improve this answer









            $endgroup$















            • $begingroup$
              Why is the Euclidean metric relevant to a question abour general relativity?
              $endgroup$
              – PM 2Ring
              7 hours ago














            0














            0










            0







            $begingroup$

            Depends on the metric and topology. Euclidian metric for exame you can have a global coordinate system.






            share|cite|improve this answer









            $endgroup$



            Depends on the metric and topology. Euclidian metric for exame you can have a global coordinate system.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 9 hours ago









            Leo KovacicLeo Kovacic

            1017 bronze badges




            1017 bronze badges















            • $begingroup$
              Why is the Euclidean metric relevant to a question abour general relativity?
              $endgroup$
              – PM 2Ring
              7 hours ago


















            • $begingroup$
              Why is the Euclidean metric relevant to a question abour general relativity?
              $endgroup$
              – PM 2Ring
              7 hours ago
















            $begingroup$
            Why is the Euclidean metric relevant to a question abour general relativity?
            $endgroup$
            – PM 2Ring
            7 hours ago




            $begingroup$
            Why is the Euclidean metric relevant to a question abour general relativity?
            $endgroup$
            – PM 2Ring
            7 hours ago











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