# Tag Info

## Hot answers tagged de-sitter-spacetime

9

The reason why the AdS/CFT correspondence is interesting is not that AdS space is supposed to describe our universe, which, as you have correctly pointed out, would lead to conflicts with experiments. In the context of the correspondence, a four-dimensional (conformal) field theory is mapped to a string theory living in an $AdS_5\times S^5$ space, although ...

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There is an important difference between a Schwarzild event horizon and a cosmological event horizon: The latter is unique to each point in Space, just like the Hubble Sphere. From some symmetry considerations it should be fairly clear, that a universe in which no matter could cross any event horizon would have to be a static universe. From beyond the Black ...

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We are already living in a nearly empty de Sitter space - the cosmological constant already represents 73% of the energy density in the Universe - and the Universe won't experience any qualitative change in the future: the percentage will just approach 100%. However, once the space may be approximated as an empty de Sitter space, all moments of time are ...

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A de Sitter universe is a cosmological solution to Einstein's field equations of General Relativity which is named after Willem de Sitter. It models the universe as spatially flat and neglects ordinary matter, so the dynamics of the universe are dominated by the cosmological constant, thought to correspond to dark energy in our universe or the inflaton field ...

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Dear inflation, the answer is No, one can't define an asymptotic S-matrix in de Sitter space. See e.g. section V.A. in this paper by Bousso: http://arxiv.org/abs/hep-th/0412197 The problem is that the Penrose diagram of a de Sitter space starts and ends with a horizontal line. So at the beginning, the observer is forbidden - by causality - to set up ...

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Yes. The reason is the following. For any diagonal, 4-D spacetime, such as what you are considering, the de Sitter-Schwarzschild metric, the Riemann tensor has 20 independent components. To do General Relativity, one must obtain the Ricci tensor which is obtained from contracting the Riemann tensor to obtain 10 independent components: $R_{ab} - ... 3 The positive energy theorem talks about the lower bound on the total energy/mass, like the ADM mass. To be able to define such a concept of the total energy/mass in general relativity, one needs some asymptotic region respecting a time-translational symmetry. That's the region where the gravitational potential (something like the deviation of$g_{00}$from ... 3 The motivation for this construction is explained here: He first gives the example of a space of constant positive curvature - the 3-sphere, given by taking a flat Euclidean space of one higher dimension (4) and restricting to the subspace$(x_1, x_2, x_3, x_4)$s.t. $$x_1^2+x_2^2+x_3^2+x_4^2=a^2$$ for some$a$. The metric on the sphere is just the ... 3 The answer is trivially yes. When large AdS blackholes radiate, the Hawking radiation is 'reflected' at infinity, and you eventually end up with a sort of thermodynamic equilibrium state where the blackhole becomes stable and eternal as it reabsorbs its radiation. The details are a little bit more involved, but suffice it to say nothing like that happens ... 3 Well it just so happens that Mitchell answered this question in a private email, I'm posting the relevant part here for the greater good. The Wick rotation in this case transforms a local patch of dS space into AdS space, or vice versa. Think of the difference between concave and convex, a valley and a hill. Then you should imagine that we have ... 3 That's a good question. These issues have an important philosophical core. But science is ultimately not about philosophy. The robust propositions made by science are about calculations and proofs obtained from empirically validated theories. So it is appropriate to view your question as a scientific one. Then there may actually be a pretty good reason, ... 3 One may always choose any coordinates on a spacetime manifold or any other manifold, for that matter. That's not only a simple mathematical insight but also a cornerstone of the general theory of relativity. In fact, GR starts with the postulate that all (non-singular etc.) coordinate systems are as good as any other coordinate systems and the basic laws of ... 2 I find it helpful to visualize de Sitter space-time using Felix Klein's approach to geometry; begin with projective space and pick a polarity that transforms trivially under the congruence group of the geometry. To get 3+1 de Sitter space-time one starts with a 4-d projective space that is modelled as the rays of a 5-d vector space$V_{5}$; a point in de ... 2 What you are describing is the static coordinate system for the de Sitter space, in which the metric could be written as $$ds^2 = -\left(1-\frac{r^2}{\alpha^2}\right)dt^2 + \left(1-\frac{r^2}{\alpha^2}\right)^{-1}dr^2 + r^2 d\Omega_{2}^2.$$ This is a static universe (not just 'more or less'). We see that at$r=\alpha$the metric has a cosmological horizon. ... 2 One should be careful not to get confused, because in cosmology we can define two horizons. The first is called the cosmic event horizon, defined as follows: if a galaxy outside the event horizon emits light today, it will never reach us. That is, we will never observe in the future what that galaxy looks like today (although we might observe today what the ... 2 The full statement seems to be:$T_{dS}\sim\frac{1}{R}\sim \sqrt\Lambda \implies non-SUSY$In a de Sitter universe, that the temperature (at the horizon) is inversely proportional to radius (distance to horizon) and proportional to the square root of the cosmological constant implies breaking of super symmetry. See for example Temperature at horizon in de ... 2 The definition of conformal time is actually $$a\,d\eta = dt\Leftrightarrow d\eta=\frac{dt}{a}$$ which gives you the correct result. 2 Couchyam's answer is for undeformed de Sitter space - empty if you will. So no orbits in that. What you are asking is if one put an object the mass of the earth in a de Sitter space, could objects the mass of the moon orbit the earth thing? It would seem that the earth like mass would follow a prescribed path, but since its a rule in General Relativity ... 2 I'll assume you are asking about geodesics in de Sitter (If the objects contribute to the energy-momentum tensor they perturb the metric and perform orbits). De Sitter space can be thought of as the set of points in 4+1-dimensional Minkowski space that satisfy $$X_1^2+X_2^2+X_3^2+X_4^2=1+X_0^2.$$ Cross sections of constant$X_0$are 3-spheres, so you ... 1 The short answer is your integration is both wrong and futile and your mistake was ignoring the other three parameters. For a static dS metric we have $$x_{0}=\sqrt{H^{-2}-r^{2}}\sinh(Ht)$$$$x_{1}=\sqrt{H^{-2}-r^{2}}\cosh(Ht)$$ $$x_{2}=r\cos\theta$$ $$x_{3}=r\sin\theta\cos\phi$$ $$x_{4}=r\sin\theta\sin\phi.$$ and the metric can be written as ... 1 According to commentator "bn" at The Chalkboards of Quantum Mechanics Professors (via Wired), this one is mostly about holography, how much information can be contained on a 2-D screen of a certain size. The diamonds are referring to domains of causal dependence. The square brackets are commutator brackets (Canonical commutation relation - Wikipedia), and ... 1 The master thesis by J. Hartong, On problems in de Sitter spacetime physics has some detailed explanation from Eq 2.5 to 2.6. I was able to get the final answer following his notes. Two important intermediate steps are:$\partial_aZ=-l^{-2}(\frac{X^d(y)}{X^d(x)}X_a(x)-X_a(y))$and$(\partial Z)^2=l^{-2}(1-Z^2)$,$\nabla^a(\partial_a Z)=-l^{-2}DZ$You ... 1 See L. Rodriguez and T. Yildirim, Class. Quantum Grav. 27, 155003 (2010), arXiv:1003.0026. Section 2.3 has the Schwarzschild-dS calculation. Lets define$f(r)=1-\frac{2M}{r}-\frac{r^2}{L^2}$Radius of the horizon is given by the largest real root of f(r)=0 But of course$L\rightarrow \infty$is still important. Once you obtain the energy momentum ... 1 Yes, the cosmic horizons are observer-dependent. After all, spaces like de Sitter space are maximally symmetric which means that all of their points are equally good as all other points. There can't be a privileged submanifold. This observer-dependence doesn't lead to any information loss even if one assumes that there is no physics beyond the cosmic ... 1 Everything Thriveth says in his answer is true. However, I can't help feeling it doesn't quite answer the question. It is indeed correct that no matter can ever pass beyond the cosmic horizon, when considered from the point of view of us, the observers for whom it exists. In this respect it's exactly analogous to a black hole's horizon, in that (again from ... 1 In my understanding, the cosmic horizon is a notion relative to the observer. Let us take two physicists outside of a black hole that they can both observe. They would agree on its horizon. But if the physicists are not at the same place in the universe, they would not agree about the cosmic horizon. Parts of universe will be beyond the cosmic horizon for ... 1 Rie. Like your question. I've been stewing on this for 5 yrs now, but have gotten nowhere. Suggestions. For moral support on the CC, see Carlo Rovelli's great paper:http://arxiv.org/abs/1002.3966. For physics(CC only, no dSS) see Beck: http://arxiv.org/abs/0810.0752 The electron & therefore QED must be involved. Dirac's 1935 paper was first to ... 1 The entropy is simply $$S = k_B\cdot \ln N$$ where$N$is the number of macroscopically indistinguishable microstates. If they transform as a representation of a group, then$N\$ is the dimension of this representation. The relevant Hilbert space must be a unitary representation of the isometry groups: the isometries have to preserve the sesquilinear norm ...

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