Schwarzchild Radius of the Universe












1












$begingroup$


According to the Wiki on the Rs, the Rs of the observable universe is 13.7BLY.
https://en.wikipedia.org/wiki/Schwarzschild_radius
(The observable universe's mass has a Schwarzschild radius of approximately 13.7 billion light-years.[7][8])



The reference for this statement is:



https://arxiv.org/abs/1008.0933 and the Encyclopedia of Distances



Can someone please explain this to me... Is this simply because to get into the non-observable portion of the universe, you have to go faster than the speed of light?






astronomy







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    According to the Wiki on the Rs, the Rs of the observable universe is 13.7BLY.
    https://en.wikipedia.org/wiki/Schwarzschild_radius
    (The observable universe's mass has a Schwarzschild radius of approximately 13.7 billion light-years.[7][8])



    The reference for this statement is:



    https://arxiv.org/abs/1008.0933 and the Encyclopedia of Distances



    Can someone please explain this to me... Is this simply because to get into the non-observable portion of the universe, you have to go faster than the speed of light?






    astronomy







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      According to the Wiki on the Rs, the Rs of the observable universe is 13.7BLY.
      https://en.wikipedia.org/wiki/Schwarzschild_radius
      (The observable universe's mass has a Schwarzschild radius of approximately 13.7 billion light-years.[7][8])



      The reference for this statement is:



      https://arxiv.org/abs/1008.0933 and the Encyclopedia of Distances



      Can someone please explain this to me... Is this simply because to get into the non-observable portion of the universe, you have to go faster than the speed of light?






      astronomy







      share|cite|improve this question











      $endgroup$




      According to the Wiki on the Rs, the Rs of the observable universe is 13.7BLY.
      https://en.wikipedia.org/wiki/Schwarzschild_radius
      (The observable universe's mass has a Schwarzschild radius of approximately 13.7 billion light-years.[7][8])



      The reference for this statement is:



      https://arxiv.org/abs/1008.0933 and the Encyclopedia of Distances



      Can someone please explain this to me... Is this simply because to get into the non-observable portion of the universe, you have to go faster than the speed of light?






      astronomy




      astronomy






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 1 hour ago







      Rick

















      asked 6 hours ago









      RickRick

      620315




      620315






















          1 Answer
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          5












          $begingroup$

          In this paper, the author begins by defining the radius of the observable universe as the radius of the Hubble sphere $r_{HS}=frac{c}{H_0}$, where $H_0$ is the Hubble constant. He then assumes that the universe is a homogeneous and isotropic collection of matter with density $rhoapprox rho_c$, where $rho_c=frac{3H^2}{8pi G}$ is the critical density of the universe at which the curvature of space is zero.



          Since he assumed that the universe is homogeneous and isotropic, the author uses the classical definition of density $rho=frac{3M}{4pi r_{HS}^3}$, where $M$ is the total mass of the observable universe, and with a bit of algebraic manipulation comes up with $r_{HS}=frac{2GM}{c^2}$. The author then asserts that $r_{HS}$ is the Schwarzschild radius of the universe, because what he came up with looks like the formula for a Schwarzschild radius.



          This is where the big problem is: the conditions that the author assumed in the beginning are not compatible with the conditions that admit the definition of a Schwarzschild radius. The Schwarzschild solution of the Einstein field equations requires that all of the mass of the universe is concentrated in a physical singularity at $r=0$, and the rest is vacuum. The author assumes essentially the exact opposite: that the mass of the universe is as spread out as possible, so that none of it is concentrated anywhere, there is no vacuum, and the universe has uniform density. As such, calling this a Schwarzschild radius doesn't really make sense, as it has nothing to do with the Schwarzschild solution besides sharing a superficial similarity in how we express their definitions. Just because he calls it a Schwarzschild radius doesn't mean that it is one.



          The moral of the story: though finding similar expressions in different contexts can often be a useful tool to guide intuition, it doesn't actually prove any connection, and isn't a substitute for an actual proof.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            probably_someone is still being kind ... the paper's author does not seem to understand even the basics of Einstein's formulation of general relativity ... the OP should just ignore this paper
            $endgroup$
            – Paul Young
            5 hours ago












          • $begingroup$
            It did not make any sense to me either which is why I posted the question. Thanks for the confirmation...
            $endgroup$
            – Rick
            1 hour ago












          Your Answer





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          active

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          active

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          5












          $begingroup$

          In this paper, the author begins by defining the radius of the observable universe as the radius of the Hubble sphere $r_{HS}=frac{c}{H_0}$, where $H_0$ is the Hubble constant. He then assumes that the universe is a homogeneous and isotropic collection of matter with density $rhoapprox rho_c$, where $rho_c=frac{3H^2}{8pi G}$ is the critical density of the universe at which the curvature of space is zero.



          Since he assumed that the universe is homogeneous and isotropic, the author uses the classical definition of density $rho=frac{3M}{4pi r_{HS}^3}$, where $M$ is the total mass of the observable universe, and with a bit of algebraic manipulation comes up with $r_{HS}=frac{2GM}{c^2}$. The author then asserts that $r_{HS}$ is the Schwarzschild radius of the universe, because what he came up with looks like the formula for a Schwarzschild radius.



          This is where the big problem is: the conditions that the author assumed in the beginning are not compatible with the conditions that admit the definition of a Schwarzschild radius. The Schwarzschild solution of the Einstein field equations requires that all of the mass of the universe is concentrated in a physical singularity at $r=0$, and the rest is vacuum. The author assumes essentially the exact opposite: that the mass of the universe is as spread out as possible, so that none of it is concentrated anywhere, there is no vacuum, and the universe has uniform density. As such, calling this a Schwarzschild radius doesn't really make sense, as it has nothing to do with the Schwarzschild solution besides sharing a superficial similarity in how we express their definitions. Just because he calls it a Schwarzschild radius doesn't mean that it is one.



          The moral of the story: though finding similar expressions in different contexts can often be a useful tool to guide intuition, it doesn't actually prove any connection, and isn't a substitute for an actual proof.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            probably_someone is still being kind ... the paper's author does not seem to understand even the basics of Einstein's formulation of general relativity ... the OP should just ignore this paper
            $endgroup$
            – Paul Young
            5 hours ago












          • $begingroup$
            It did not make any sense to me either which is why I posted the question. Thanks for the confirmation...
            $endgroup$
            – Rick
            1 hour ago
















          5












          $begingroup$

          In this paper, the author begins by defining the radius of the observable universe as the radius of the Hubble sphere $r_{HS}=frac{c}{H_0}$, where $H_0$ is the Hubble constant. He then assumes that the universe is a homogeneous and isotropic collection of matter with density $rhoapprox rho_c$, where $rho_c=frac{3H^2}{8pi G}$ is the critical density of the universe at which the curvature of space is zero.



          Since he assumed that the universe is homogeneous and isotropic, the author uses the classical definition of density $rho=frac{3M}{4pi r_{HS}^3}$, where $M$ is the total mass of the observable universe, and with a bit of algebraic manipulation comes up with $r_{HS}=frac{2GM}{c^2}$. The author then asserts that $r_{HS}$ is the Schwarzschild radius of the universe, because what he came up with looks like the formula for a Schwarzschild radius.



          This is where the big problem is: the conditions that the author assumed in the beginning are not compatible with the conditions that admit the definition of a Schwarzschild radius. The Schwarzschild solution of the Einstein field equations requires that all of the mass of the universe is concentrated in a physical singularity at $r=0$, and the rest is vacuum. The author assumes essentially the exact opposite: that the mass of the universe is as spread out as possible, so that none of it is concentrated anywhere, there is no vacuum, and the universe has uniform density. As such, calling this a Schwarzschild radius doesn't really make sense, as it has nothing to do with the Schwarzschild solution besides sharing a superficial similarity in how we express their definitions. Just because he calls it a Schwarzschild radius doesn't mean that it is one.



          The moral of the story: though finding similar expressions in different contexts can often be a useful tool to guide intuition, it doesn't actually prove any connection, and isn't a substitute for an actual proof.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            probably_someone is still being kind ... the paper's author does not seem to understand even the basics of Einstein's formulation of general relativity ... the OP should just ignore this paper
            $endgroup$
            – Paul Young
            5 hours ago












          • $begingroup$
            It did not make any sense to me either which is why I posted the question. Thanks for the confirmation...
            $endgroup$
            – Rick
            1 hour ago














          5












          5








          5





          $begingroup$

          In this paper, the author begins by defining the radius of the observable universe as the radius of the Hubble sphere $r_{HS}=frac{c}{H_0}$, where $H_0$ is the Hubble constant. He then assumes that the universe is a homogeneous and isotropic collection of matter with density $rhoapprox rho_c$, where $rho_c=frac{3H^2}{8pi G}$ is the critical density of the universe at which the curvature of space is zero.



          Since he assumed that the universe is homogeneous and isotropic, the author uses the classical definition of density $rho=frac{3M}{4pi r_{HS}^3}$, where $M$ is the total mass of the observable universe, and with a bit of algebraic manipulation comes up with $r_{HS}=frac{2GM}{c^2}$. The author then asserts that $r_{HS}$ is the Schwarzschild radius of the universe, because what he came up with looks like the formula for a Schwarzschild radius.



          This is where the big problem is: the conditions that the author assumed in the beginning are not compatible with the conditions that admit the definition of a Schwarzschild radius. The Schwarzschild solution of the Einstein field equations requires that all of the mass of the universe is concentrated in a physical singularity at $r=0$, and the rest is vacuum. The author assumes essentially the exact opposite: that the mass of the universe is as spread out as possible, so that none of it is concentrated anywhere, there is no vacuum, and the universe has uniform density. As such, calling this a Schwarzschild radius doesn't really make sense, as it has nothing to do with the Schwarzschild solution besides sharing a superficial similarity in how we express their definitions. Just because he calls it a Schwarzschild radius doesn't mean that it is one.



          The moral of the story: though finding similar expressions in different contexts can often be a useful tool to guide intuition, it doesn't actually prove any connection, and isn't a substitute for an actual proof.






          share|cite|improve this answer









          $endgroup$



          In this paper, the author begins by defining the radius of the observable universe as the radius of the Hubble sphere $r_{HS}=frac{c}{H_0}$, where $H_0$ is the Hubble constant. He then assumes that the universe is a homogeneous and isotropic collection of matter with density $rhoapprox rho_c$, where $rho_c=frac{3H^2}{8pi G}$ is the critical density of the universe at which the curvature of space is zero.



          Since he assumed that the universe is homogeneous and isotropic, the author uses the classical definition of density $rho=frac{3M}{4pi r_{HS}^3}$, where $M$ is the total mass of the observable universe, and with a bit of algebraic manipulation comes up with $r_{HS}=frac{2GM}{c^2}$. The author then asserts that $r_{HS}$ is the Schwarzschild radius of the universe, because what he came up with looks like the formula for a Schwarzschild radius.



          This is where the big problem is: the conditions that the author assumed in the beginning are not compatible with the conditions that admit the definition of a Schwarzschild radius. The Schwarzschild solution of the Einstein field equations requires that all of the mass of the universe is concentrated in a physical singularity at $r=0$, and the rest is vacuum. The author assumes essentially the exact opposite: that the mass of the universe is as spread out as possible, so that none of it is concentrated anywhere, there is no vacuum, and the universe has uniform density. As such, calling this a Schwarzschild radius doesn't really make sense, as it has nothing to do with the Schwarzschild solution besides sharing a superficial similarity in how we express their definitions. Just because he calls it a Schwarzschild radius doesn't mean that it is one.



          The moral of the story: though finding similar expressions in different contexts can often be a useful tool to guide intuition, it doesn't actually prove any connection, and isn't a substitute for an actual proof.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 5 hours ago









          probably_someoneprobably_someone

          18.8k12960




          18.8k12960












          • $begingroup$
            probably_someone is still being kind ... the paper's author does not seem to understand even the basics of Einstein's formulation of general relativity ... the OP should just ignore this paper
            $endgroup$
            – Paul Young
            5 hours ago












          • $begingroup$
            It did not make any sense to me either which is why I posted the question. Thanks for the confirmation...
            $endgroup$
            – Rick
            1 hour ago


















          • $begingroup$
            probably_someone is still being kind ... the paper's author does not seem to understand even the basics of Einstein's formulation of general relativity ... the OP should just ignore this paper
            $endgroup$
            – Paul Young
            5 hours ago












          • $begingroup$
            It did not make any sense to me either which is why I posted the question. Thanks for the confirmation...
            $endgroup$
            – Rick
            1 hour ago
















          $begingroup$
          probably_someone is still being kind ... the paper's author does not seem to understand even the basics of Einstein's formulation of general relativity ... the OP should just ignore this paper
          $endgroup$
          – Paul Young
          5 hours ago






          $begingroup$
          probably_someone is still being kind ... the paper's author does not seem to understand even the basics of Einstein's formulation of general relativity ... the OP should just ignore this paper
          $endgroup$
          – Paul Young
          5 hours ago














          $begingroup$
          It did not make any sense to me either which is why I posted the question. Thanks for the confirmation...
          $endgroup$
          – Rick
          1 hour ago




          $begingroup$
          It did not make any sense to me either which is why I posted the question. Thanks for the confirmation...
          $endgroup$
          – Rick
          1 hour ago


















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