Observable universe radius for distant observers
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The radius of the observable universe is about 46 Gly. Is that figure true for all current observers in our universe? Is it true if the universe is finite or infinite, flat or curved?
cosmology space-expansion observable-universe
asked 3 hours ago
Peter4075Peter4075
1,25232042
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$begingroup$
The radius of the observable universe is about 46 Gly. Is that figure true for all current observers in our universe? Is it true if the universe is finite or infinite, flat or curved?
cosmology space-expansion observable-universe
asked 3 hours ago
Peter4075Peter4075
1,25232042
$endgroup$
add a comment |
$begingroup$
The radius of the observable universe is about 46 Gly. Is that figure true for all current observers in our universe? Is it true if the universe is finite or infinite, flat or curved?
cosmology space-expansion observable-universe
asked 3 hours ago
Peter4075Peter4075
1,25232042
$endgroup$
The radius of the observable universe is about 46 Gly. Is that figure true for all current observers in our universe? Is it true if the universe is finite or infinite, flat or curved?
cosmology space-expansion observable-universe
cosmology space-expansion observable-universe
asked 3 hours ago
Peter4075Peter4075
1,25232042
asked 3 hours ago
Peter4075Peter4075
1,25232042
asked 3 hours ago
Peter4075Peter4075
1,25232042
asked 3 hours ago
Peter4075Peter4075
1,25232042
asked 3 hours ago
Peter4075Peter4075
1,25232042
1,25232042
add a comment |
add a comment |
1 Answer
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If the universe has a FLRW metric, then there is a cosmological time $t$ that all observers at rest relative to CMB or the matter in the universe will experience at the same rate. This is true regardless of the curvature and whether the universe is infinite or merely unbounded.
The radius of the observable universe (in co-moving coordinates) is calculated by integrating $$r=c int_0^t frac{du}{a(u)}$$ (where $a(t)$ is the scale factor) from the start to the present cosmological time. All observers with the same $t$ will agree on $r$.
The slightly conceptually tricky part is defining "all current observers". We can define current observers to mean "all observers at rest relative to the matter or CMB that see the same scale factor $a(t)=1$ as us (i.e. have the same cosmological time $t$)" and get a well-defined slice of constant $t$ across the space-time manifold. This is less arbitrary and problematic than talking about simultaneity in special relativity, where there is nothing to compare to and no real simultaneity (everybody have their own present-time slices across spacetime, all equally valid). In a homogeneous and isotropic cosmology there is a frame of reference that is shared.
answered 1 hour ago
Anders SandbergAnders Sandberg
9,61221428
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1 Answer
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1 Answer
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$begingroup$
If the universe has a FLRW metric, then there is a cosmological time $t$ that all observers at rest relative to CMB or the matter in the universe will experience at the same rate. This is true regardless of the curvature and whether the universe is infinite or merely unbounded.
The radius of the observable universe (in co-moving coordinates) is calculated by integrating $$r=c int_0^t frac{du}{a(u)}$$ (where $a(t)$ is the scale factor) from the start to the present cosmological time. All observers with the same $t$ will agree on $r$.
The slightly conceptually tricky part is defining "all current observers". We can define current observers to mean "all observers at rest relative to the matter or CMB that see the same scale factor $a(t)=1$ as us (i.e. have the same cosmological time $t$)" and get a well-defined slice of constant $t$ across the space-time manifold. This is less arbitrary and problematic than talking about simultaneity in special relativity, where there is nothing to compare to and no real simultaneity (everybody have their own present-time slices across spacetime, all equally valid). In a homogeneous and isotropic cosmology there is a frame of reference that is shared.
answered 1 hour ago
Anders SandbergAnders Sandberg
9,61221428
$endgroup$
add a comment |
$begingroup$
If the universe has a FLRW metric, then there is a cosmological time $t$ that all observers at rest relative to CMB or the matter in the universe will experience at the same rate. This is true regardless of the curvature and whether the universe is infinite or merely unbounded.
The radius of the observable universe (in co-moving coordinates) is calculated by integrating $$r=c int_0^t frac{du}{a(u)}$$ (where $a(t)$ is the scale factor) from the start to the present cosmological time. All observers with the same $t$ will agree on $r$.
The slightly conceptually tricky part is defining "all current observers". We can define current observers to mean "all observers at rest relative to the matter or CMB that see the same scale factor $a(t)=1$ as us (i.e. have the same cosmological time $t$)" and get a well-defined slice of constant $t$ across the space-time manifold. This is less arbitrary and problematic than talking about simultaneity in special relativity, where there is nothing to compare to and no real simultaneity (everybody have their own present-time slices across spacetime, all equally valid). In a homogeneous and isotropic cosmology there is a frame of reference that is shared.
answered 1 hour ago
Anders SandbergAnders Sandberg
9,61221428
$endgroup$
add a comment |
$begingroup$
If the universe has a FLRW metric, then there is a cosmological time $t$ that all observers at rest relative to CMB or the matter in the universe will experience at the same rate. This is true regardless of the curvature and whether the universe is infinite or merely unbounded.
The radius of the observable universe (in co-moving coordinates) is calculated by integrating $$r=c int_0^t frac{du}{a(u)}$$ (where $a(t)$ is the scale factor) from the start to the present cosmological time. All observers with the same $t$ will agree on $r$.
The slightly conceptually tricky part is defining "all current observers". We can define current observers to mean "all observers at rest relative to the matter or CMB that see the same scale factor $a(t)=1$ as us (i.e. have the same cosmological time $t$)" and get a well-defined slice of constant $t$ across the space-time manifold. This is less arbitrary and problematic than talking about simultaneity in special relativity, where there is nothing to compare to and no real simultaneity (everybody have their own present-time slices across spacetime, all equally valid). In a homogeneous and isotropic cosmology there is a frame of reference that is shared.
answered 1 hour ago
Anders SandbergAnders Sandberg
9,61221428
$endgroup$
If the universe has a FLRW metric, then there is a cosmological time $t$ that all observers at rest relative to CMB or the matter in the universe will experience at the same rate. This is true regardless of the curvature and whether the universe is infinite or merely unbounded.
The radius of the observable universe (in co-moving coordinates) is calculated by integrating $$r=c int_0^t frac{du}{a(u)}$$ (where $a(t)$ is the scale factor) from the start to the present cosmological time. All observers with the same $t$ will agree on $r$.
The slightly conceptually tricky part is defining "all current observers". We can define current observers to mean "all observers at rest relative to the matter or CMB that see the same scale factor $a(t)=1$ as us (i.e. have the same cosmological time $t$)" and get a well-defined slice of constant $t$ across the space-time manifold. This is less arbitrary and problematic than talking about simultaneity in special relativity, where there is nothing to compare to and no real simultaneity (everybody have their own present-time slices across spacetime, all equally valid). In a homogeneous and isotropic cosmology there is a frame of reference that is shared.
answered 1 hour ago
Anders SandbergAnders Sandberg
9,61221428
answered 1 hour ago
Anders SandbergAnders Sandberg
9,61221428
answered 1 hour ago
Anders SandbergAnders Sandberg
9,61221428
answered 1 hour ago
Anders SandbergAnders Sandberg
9,61221428
9,61221428
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