Metric for 2D de Sitter?
$begingroup$
What is the correct metric to use for two dimensional de Sitter? If one starts with the following metric, which looks similar to de Sitter in 4 dimensions:
$$ds^2 = -dt^2 + e^{2H t} dx^2,$$
one can calculate $R = 2H^2$, and $R_{00} = -H^2$, which gives the $Lambda = 0$, which is not the solution one is looking for. What should be the correct metric to use for the same?
general-relativity differential-geometry metric-tensor de-sitter-spacetime
edited Dec 10 '18 at 5:46
Qmechanic♦
103k121851177
asked Dec 10 '18 at 1:59
Michael WilliamsMichael Williams
11810
$endgroup$
add a comment |
$begingroup$
What is the correct metric to use for two dimensional de Sitter? If one starts with the following metric, which looks similar to de Sitter in 4 dimensions:
$$ds^2 = -dt^2 + e^{2H t} dx^2,$$
one can calculate $R = 2H^2$, and $R_{00} = -H^2$, which gives the $Lambda = 0$, which is not the solution one is looking for. What should be the correct metric to use for the same?
general-relativity differential-geometry metric-tensor de-sitter-spacetime
edited Dec 10 '18 at 5:46
Qmechanic♦
103k121851177
asked Dec 10 '18 at 1:59
Michael WilliamsMichael Williams
11810
$endgroup$
add a comment |
$begingroup$
What is the correct metric to use for two dimensional de Sitter? If one starts with the following metric, which looks similar to de Sitter in 4 dimensions:
$$ds^2 = -dt^2 + e^{2H t} dx^2,$$
one can calculate $R = 2H^2$, and $R_{00} = -H^2$, which gives the $Lambda = 0$, which is not the solution one is looking for. What should be the correct metric to use for the same?
general-relativity differential-geometry metric-tensor de-sitter-spacetime
edited Dec 10 '18 at 5:46
Qmechanic♦
103k121851177
asked Dec 10 '18 at 1:59
Michael WilliamsMichael Williams
11810
$endgroup$
What is the correct metric to use for two dimensional de Sitter? If one starts with the following metric, which looks similar to de Sitter in 4 dimensions:
$$ds^2 = -dt^2 + e^{2H t} dx^2,$$
one can calculate $R = 2H^2$, and $R_{00} = -H^2$, which gives the $Lambda = 0$, which is not the solution one is looking for. What should be the correct metric to use for the same?
general-relativity differential-geometry metric-tensor de-sitter-spacetime
general-relativity differential-geometry metric-tensor de-sitter-spacetime
edited Dec 10 '18 at 5:46
Qmechanic♦
103k121851177
asked Dec 10 '18 at 1:59
Michael WilliamsMichael Williams
11810
edited Dec 10 '18 at 5:46
Qmechanic♦
103k121851177
asked Dec 10 '18 at 1:59
Michael WilliamsMichael Williams
11810
edited Dec 10 '18 at 5:46
Qmechanic♦
103k121851177
edited Dec 10 '18 at 5:46
Qmechanic♦
103k121851177
edited Dec 10 '18 at 5:46
Qmechanic♦
103k121851177
103k121851177
asked Dec 10 '18 at 1:59
Michael WilliamsMichael Williams
11810
asked Dec 10 '18 at 1:59
Michael WilliamsMichael Williams
11810
asked Dec 10 '18 at 1:59
Michael WilliamsMichael Williams
11810
11810
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
In two-dimensional spacetime, the Einstein tensor $R_{ab}-frac{1}{2}g_{ab}R$ is identically zero , which explains why you get $Lambda=0$.
In any number $D$ of spacetime dimensions, including $D=2$, de Sitter spacetime can be constructed like this. Start with the $D+1$ dimensional Minkowski metric
$$
-(mathrm dX^0)^2+sum_{k=1}^D(mathrm dX^k)^2.
tag{1}
$$
The submanifold defined by the condition
$$
sum_{k=1}^D(X^k)^2=L^2+(X^0)^2
tag{2}
$$
is $D$-dimensional de Sitter spacetime. The length parameter $L$ is related to the cosmological constant $Lambda$ by
$$
Lambda = frac{(D-2)(D-1)}{2L^2}.
tag{3}
$$
This is equation (4) in "Les Houches Lectures on de Sitter Space". Setting $D=2$ recovers your result $Lambda=0$.
By the way, equations (13)-(14) in the same paper show how to derive the de Sitter metric in the form
$$
-mathrm dt^2+e^{2t}sum_{k=1}^{D-1}(mathrm dx^k)^2
$$
starting from equations (1)-(2). For $D=2$, this reduces to the form shown in the question.
edited Dec 10 '18 at 8:25
Lorenzo B.
1765
answered Dec 10 '18 at 3:53
Dan YandDan Yand
9,20611536
$endgroup$
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%2f446259%2fmetric-for-2d-de-sitter%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
$begingroup$
In two-dimensional spacetime, the Einstein tensor $R_{ab}-frac{1}{2}g_{ab}R$ is identically zero , which explains why you get $Lambda=0$.
In any number $D$ of spacetime dimensions, including $D=2$, de Sitter spacetime can be constructed like this. Start with the $D+1$ dimensional Minkowski metric
$$
-(mathrm dX^0)^2+sum_{k=1}^D(mathrm dX^k)^2.
tag{1}
$$
The submanifold defined by the condition
$$
sum_{k=1}^D(X^k)^2=L^2+(X^0)^2
tag{2}
$$
is $D$-dimensional de Sitter spacetime. The length parameter $L$ is related to the cosmological constant $Lambda$ by
$$
Lambda = frac{(D-2)(D-1)}{2L^2}.
tag{3}
$$
This is equation (4) in "Les Houches Lectures on de Sitter Space". Setting $D=2$ recovers your result $Lambda=0$.
By the way, equations (13)-(14) in the same paper show how to derive the de Sitter metric in the form
$$
-mathrm dt^2+e^{2t}sum_{k=1}^{D-1}(mathrm dx^k)^2
$$
starting from equations (1)-(2). For $D=2$, this reduces to the form shown in the question.
edited Dec 10 '18 at 8:25
Lorenzo B.
1765
answered Dec 10 '18 at 3:53
Dan YandDan Yand
9,20611536
$endgroup$
add a comment |
$begingroup$
In two-dimensional spacetime, the Einstein tensor $R_{ab}-frac{1}{2}g_{ab}R$ is identically zero , which explains why you get $Lambda=0$.
In any number $D$ of spacetime dimensions, including $D=2$, de Sitter spacetime can be constructed like this. Start with the $D+1$ dimensional Minkowski metric
$$
-(mathrm dX^0)^2+sum_{k=1}^D(mathrm dX^k)^2.
tag{1}
$$
The submanifold defined by the condition
$$
sum_{k=1}^D(X^k)^2=L^2+(X^0)^2
tag{2}
$$
is $D$-dimensional de Sitter spacetime. The length parameter $L$ is related to the cosmological constant $Lambda$ by
$$
Lambda = frac{(D-2)(D-1)}{2L^2}.
tag{3}
$$
This is equation (4) in "Les Houches Lectures on de Sitter Space". Setting $D=2$ recovers your result $Lambda=0$.
By the way, equations (13)-(14) in the same paper show how to derive the de Sitter metric in the form
$$
-mathrm dt^2+e^{2t}sum_{k=1}^{D-1}(mathrm dx^k)^2
$$
starting from equations (1)-(2). For $D=2$, this reduces to the form shown in the question.
edited Dec 10 '18 at 8:25
Lorenzo B.
1765
answered Dec 10 '18 at 3:53
Dan YandDan Yand
9,20611536
$endgroup$
add a comment |
$begingroup$
In two-dimensional spacetime, the Einstein tensor $R_{ab}-frac{1}{2}g_{ab}R$ is identically zero , which explains why you get $Lambda=0$.
In any number $D$ of spacetime dimensions, including $D=2$, de Sitter spacetime can be constructed like this. Start with the $D+1$ dimensional Minkowski metric
$$
-(mathrm dX^0)^2+sum_{k=1}^D(mathrm dX^k)^2.
tag{1}
$$
The submanifold defined by the condition
$$
sum_{k=1}^D(X^k)^2=L^2+(X^0)^2
tag{2}
$$
is $D$-dimensional de Sitter spacetime. The length parameter $L$ is related to the cosmological constant $Lambda$ by
$$
Lambda = frac{(D-2)(D-1)}{2L^2}.
tag{3}
$$
This is equation (4) in "Les Houches Lectures on de Sitter Space". Setting $D=2$ recovers your result $Lambda=0$.
By the way, equations (13)-(14) in the same paper show how to derive the de Sitter metric in the form
$$
-mathrm dt^2+e^{2t}sum_{k=1}^{D-1}(mathrm dx^k)^2
$$
starting from equations (1)-(2). For $D=2$, this reduces to the form shown in the question.
edited Dec 10 '18 at 8:25
Lorenzo B.
1765
answered Dec 10 '18 at 3:53
Dan YandDan Yand
9,20611536
$endgroup$
In two-dimensional spacetime, the Einstein tensor $R_{ab}-frac{1}{2}g_{ab}R$ is identically zero , which explains why you get $Lambda=0$.
In any number $D$ of spacetime dimensions, including $D=2$, de Sitter spacetime can be constructed like this. Start with the $D+1$ dimensional Minkowski metric
$$
-(mathrm dX^0)^2+sum_{k=1}^D(mathrm dX^k)^2.
tag{1}
$$
The submanifold defined by the condition
$$
sum_{k=1}^D(X^k)^2=L^2+(X^0)^2
tag{2}
$$
is $D$-dimensional de Sitter spacetime. The length parameter $L$ is related to the cosmological constant $Lambda$ by
$$
Lambda = frac{(D-2)(D-1)}{2L^2}.
tag{3}
$$
This is equation (4) in "Les Houches Lectures on de Sitter Space". Setting $D=2$ recovers your result $Lambda=0$.
By the way, equations (13)-(14) in the same paper show how to derive the de Sitter metric in the form
$$
-mathrm dt^2+e^{2t}sum_{k=1}^{D-1}(mathrm dx^k)^2
$$
starting from equations (1)-(2). For $D=2$, this reduces to the form shown in the question.
edited Dec 10 '18 at 8:25
Lorenzo B.
1765
answered Dec 10 '18 at 3:53
Dan YandDan Yand
9,20611536
edited Dec 10 '18 at 8:25
Lorenzo B.
1765
edited Dec 10 '18 at 8:25
Lorenzo B.
1765
edited Dec 10 '18 at 8:25
Lorenzo B.
1765
1765
answered Dec 10 '18 at 3:53
Dan YandDan Yand
9,20611536
answered Dec 10 '18 at 3:53
Dan YandDan Yand
9,20611536
answered Dec 10 '18 at 3:53
Dan YandDan Yand
9,20611536
9,20611536
add a comment |
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%2f446259%2fmetric-for-2d-de-sitter%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
