Observable universe radius for distant observers
$begingroup$
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 Mar 9 at 10:11
Peter4075Peter4075
1,25932042
$endgroup$
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 Mar 9 at 10:11
Peter4075Peter4075
1,25932042
$endgroup$
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 Mar 9 at 10:11
Peter4075Peter4075
1,25932042
$endgroup$
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 Mar 9 at 10:11
Peter4075Peter4075
1,25932042
asked Mar 9 at 10:11
Peter4075Peter4075
1,25932042
asked Mar 9 at 10:11
Peter4075Peter4075
1,25932042
asked Mar 9 at 10:11
Peter4075Peter4075
1,25932042
asked Mar 9 at 10:11
Peter4075Peter4075
1,25932042
1,25932042
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 Mar 9 at 11:55
Anders SandbergAnders Sandberg
9,66221428
$endgroup$
$begingroup$
Shouldn't that be $r=cint_{0}^{t}frac{dt}{a(t)}$?
$endgroup$
– Peter4075
Mar 9 at 15:47
$begingroup$
@Peter4075 - I prefer to use a different variable of integration than the variable upper limit to limit confusion. Some texts use $t'$ but that looks like a derivative, and $tau$ looks like a proper time.
$endgroup$
– Anders Sandberg
Mar 9 at 16:25
$begingroup$
Fine, and thanks for your answer.
$endgroup$
– Peter4075
Mar 9 at 17:20
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "151"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f465439%2fobservable-universe-radius-for-distant-observers%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 Mar 9 at 11:55
Anders SandbergAnders Sandberg
9,66221428
$endgroup$
$begingroup$
Shouldn't that be $r=cint_{0}^{t}frac{dt}{a(t)}$?
$endgroup$
– Peter4075
Mar 9 at 15:47
$begingroup$
@Peter4075 - I prefer to use a different variable of integration than the variable upper limit to limit confusion. Some texts use $t'$ but that looks like a derivative, and $tau$ looks like a proper time.
$endgroup$
– Anders Sandberg
Mar 9 at 16:25
$begingroup$
Fine, and thanks for your answer.
$endgroup$
– Peter4075
Mar 9 at 17:20
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 Mar 9 at 11:55
Anders SandbergAnders Sandberg
9,66221428
$endgroup$
$begingroup$
Shouldn't that be $r=cint_{0}^{t}frac{dt}{a(t)}$?
$endgroup$
– Peter4075
Mar 9 at 15:47
$begingroup$
@Peter4075 - I prefer to use a different variable of integration than the variable upper limit to limit confusion. Some texts use $t'$ but that looks like a derivative, and $tau$ looks like a proper time.
$endgroup$
– Anders Sandberg
Mar 9 at 16:25
$begingroup$
Fine, and thanks for your answer.
$endgroup$
– Peter4075
Mar 9 at 17:20
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 Mar 9 at 11:55
Anders SandbergAnders Sandberg
9,66221428
$endgroup$
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 Mar 9 at 11:55
Anders SandbergAnders Sandberg
9,66221428
answered Mar 9 at 11:55
Anders SandbergAnders Sandberg
9,66221428
answered Mar 9 at 11:55
Anders SandbergAnders Sandberg
9,66221428
answered Mar 9 at 11:55
Anders SandbergAnders Sandberg
9,66221428
9,66221428
$begingroup$
Shouldn't that be $r=cint_{0}^{t}frac{dt}{a(t)}$?
$endgroup$
– Peter4075
Mar 9 at 15:47
$begingroup$
@Peter4075 - I prefer to use a different variable of integration than the variable upper limit to limit confusion. Some texts use $t'$ but that looks like a derivative, and $tau$ looks like a proper time.
$endgroup$
– Anders Sandberg
Mar 9 at 16:25
$begingroup$
Fine, and thanks for your answer.
$endgroup$
– Peter4075
Mar 9 at 17:20
add a comment |
$begingroup$
Shouldn't that be $r=cint_{0}^{t}frac{dt}{a(t)}$?
$endgroup$
– Peter4075
Mar 9 at 15:47
$begingroup$
@Peter4075 - I prefer to use a different variable of integration than the variable upper limit to limit confusion. Some texts use $t'$ but that looks like a derivative, and $tau$ looks like a proper time.
$endgroup$
– Anders Sandberg
Mar 9 at 16:25
$begingroup$
Fine, and thanks for your answer.
$endgroup$
– Peter4075
Mar 9 at 17:20
$begingroup$
Shouldn't that be $r=cint_{0}^{t}frac{dt}{a(t)}$?
$endgroup$
– Peter4075
Mar 9 at 15:47
$begingroup$
Shouldn't that be $r=cint_{0}^{t}frac{dt}{a(t)}$?
$endgroup$
– Peter4075
Mar 9 at 15:47
$begingroup$
@Peter4075 - I prefer to use a different variable of integration than the variable upper limit to limit confusion. Some texts use $t'$ but that looks like a derivative, and $tau$ looks like a proper time.
$endgroup$
– Anders Sandberg
Mar 9 at 16:25
$begingroup$
@Peter4075 - I prefer to use a different variable of integration than the variable upper limit to limit confusion. Some texts use $t'$ but that looks like a derivative, and $tau$ looks like a proper time.
$endgroup$
– Anders Sandberg
Mar 9 at 16:25
$begingroup$
Fine, and thanks for your answer.
$endgroup$
– Peter4075
Mar 9 at 17:20
$begingroup$
Fine, and thanks for your answer.
$endgroup$
– Peter4075
Mar 9 at 17:20
add a comment |
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f465439%2fobservable-universe-radius-for-distant-observers%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
