Metric for 2D de Sitter?
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♦
102k121831154
asked Dec 10 '18 at 1:59
Michael Williams
11810
add a comment |
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♦
102k121831154
asked Dec 10 '18 at 1:59
Michael Williams
11810
add a comment |
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♦
102k121831154
asked Dec 10 '18 at 1:59
Michael Williams
11810
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♦
102k121831154
asked Dec 10 '18 at 1:59
Michael Williams
11810
edited Dec 10 '18 at 5:46
Qmechanic♦
102k121831154
asked Dec 10 '18 at 1:59
Michael Williams
11810
edited Dec 10 '18 at 5:46
Qmechanic♦
102k121831154
edited Dec 10 '18 at 5:46
Qmechanic♦
102k121831154
edited Dec 10 '18 at 5:46
Qmechanic♦
102k121831154
102k121831154
asked Dec 10 '18 at 1:59
Michael Williams
11810
asked Dec 10 '18 at 1:59
Michael Williams
11810
asked Dec 10 '18 at 1:59
Michael Williams
11810
11810
add a comment |
add a comment |
1 Answer
1
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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 Yand
7,4971931
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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votes
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 Yand
7,4971931
add a comment |
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 Yand
7,4971931
add a comment |
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 Yand
7,4971931
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 Yand
7,4971931
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 Yand
7,4971931
answered Dec 10 '18 at 3:53
Dan Yand
7,4971931
answered Dec 10 '18 at 3:53
Dan Yand
7,4971931
7,4971931
add a comment |
add a comment |
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