Schwarzchild Radius of the UniverseIs the “far” universe expanding more quickly?Does the math work out for there being enough time for the formation of the heavier elements and their distribution as seen in today's universe?How is the universe expanding?What would happen in the final days of the universe?How much of the universe is observable at visible wavelengths?What's the point of looking at distances beyond $13,7$ billion light years?How long was the universe radiation dominated?physical meaning of dark matter virial radiusWhat happens in the event that the cooling radius is shorter than the virial radius of a Cold Dark Matter Halo?The Cosmic Microwave Background Paradox
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Schwarzchild Radius of the Universe
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Schwarzchild Radius of the Universe
Is the “far” universe expanding more quickly?Does the math work out for there being enough time for the formation of the heavier elements and their distribution as seen in today's universe?How is the universe expanding?What would happen in the final days of the universe?How much of the universe is observable at visible wavelengths?What's the point of looking at distances beyond $13,7$ billion light years?How long was the universe radiation dominated?physical meaning of dark matter virial radiusWhat happens in the event that the cooling radius is shorter than the virial radius of a Cold Dark Matter Halo?The Cosmic Microwave Background Paradox
$begingroup$
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
asked 8 hours ago
RickRick
620315
$endgroup$
add a comment |
$begingroup$
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
asked 8 hours ago
RickRick
620315
$endgroup$
add a comment |
$begingroup$
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
asked 8 hours ago
RickRick
620315
$endgroup$
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
asked 8 hours ago
RickRick
620315
asked 8 hours ago
RickRick
620315
edited 3 hours ago
Rick
asked 8 hours ago
RickRick
620315
asked 8 hours ago
RickRick
620315
asked 8 hours ago
RickRick
620315
620315
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In this paper, the author begins by defining the radius of the observable universe as the radius of the Hubble sphere $r_HS=fraccH_0$, 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 $rho_c=frac3H^28pi G$ 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 $rho=frac3M4pi r_HS^3$, where $M$ is the total mass of the observable universe, and with a bit of algebraic manipulation comes up with $r_HS=frac2GMc^2$. 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.
answered 7 hours ago
probably_someoneprobably_someone
18.8k12960
$endgroup$
$begingroup$
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
7 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
3 hours ago
add a comment |
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1 Answer
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$begingroup$
In this paper, the author begins by defining the radius of the observable universe as the radius of the Hubble sphere $r_HS=fraccH_0$, 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 $rho_c=frac3H^28pi G$ 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 $rho=frac3M4pi r_HS^3$, where $M$ is the total mass of the observable universe, and with a bit of algebraic manipulation comes up with $r_HS=frac2GMc^2$. 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.
answered 7 hours ago
probably_someoneprobably_someone
18.8k12960
$endgroup$
$begingroup$
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
7 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
3 hours ago
add a comment |
$begingroup$
In this paper, the author begins by defining the radius of the observable universe as the radius of the Hubble sphere $r_HS=fraccH_0$, 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 $rho_c=frac3H^28pi G$ 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 $rho=frac3M4pi r_HS^3$, where $M$ is the total mass of the observable universe, and with a bit of algebraic manipulation comes up with $r_HS=frac2GMc^2$. 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.
answered 7 hours ago
probably_someoneprobably_someone
18.8k12960
$endgroup$
$begingroup$
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
7 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
3 hours ago
add a comment |
$begingroup$
In this paper, the author begins by defining the radius of the observable universe as the radius of the Hubble sphere $r_HS=fraccH_0$, 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 $rho_c=frac3H^28pi G$ 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 $rho=frac3M4pi r_HS^3$, where $M$ is the total mass of the observable universe, and with a bit of algebraic manipulation comes up with $r_HS=frac2GMc^2$. 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.
answered 7 hours ago
probably_someoneprobably_someone
18.8k12960
$endgroup$
In this paper, the author begins by defining the radius of the observable universe as the radius of the Hubble sphere $r_HS=fraccH_0$, 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 $rho_c=frac3H^28pi G$ 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 $rho=frac3M4pi r_HS^3$, where $M$ is the total mass of the observable universe, and with a bit of algebraic manipulation comes up with $r_HS=frac2GMc^2$. 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.
answered 7 hours ago
probably_someoneprobably_someone
18.8k12960
answered 7 hours ago
probably_someoneprobably_someone
18.8k12960
answered 7 hours ago
probably_someoneprobably_someone
18.8k12960
answered 7 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
7 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
3 hours ago
add a comment |
$begingroup$
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
7 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
3 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
7 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
7 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
3 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
3 hours ago
add a comment |
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