15

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 ...


9

A static spacetime normally means there is an irrotational global timelike Killing vector, and this isn't the case for the de Sitter geometry so the de Sitter geometry wouldn't normally be described as static. However the de Sitter metric can be written using static coordinates: $$ ds^2 = -\left(1 - \frac{\Lambda}{3}r^2\right)dt^2 + \frac{dr^2}{\left(1 - \...


8

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 ...


7

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 ...


7

Let $\mathbb{R}^{1,3}$ be the four dimensional real vector space with Minkowsky metric (quadratic form) $Q(x) = x_0^2 - x_1^2 - x_2^2 - x_3^2$ for each vector $x = (x_0,x_1,x_2,x_3) \, \in \, \mathbb{R}^{1,3}$. Define an isomorphism between $\mathbb{R}^{1,3}$ and the following four dimensional vector space of hermitian matrices $$R({1,3}) = \left\{X = \begin{...


7

The static chart written in John Rennie's answer covers only part of de Sitter spacetime (and this fact is related to the presence of a cosmological horizon where $g_{00}$ vanishes giving rise to an apparent singularity). There is no static global chart, or equivalently, there is no global timelike Killing vector with an orthogonal spacelike 3-surface. ...


5

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} - \frac{1}{...


5

Those are two different curvatures you are talking about. First, you can talk about curvature of the spacetime i.e. treating one temporal and three spatial coordinates on equal footing. Then de Sitter spacetime has constant spacetime curvature, it's basically 4d hyperboloid. Realistic cosmological solutions also all have some spacetime curvature that ...


5

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{...


4

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 ...


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

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 ...


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 ...


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

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. ...


3

I haven't read the paper by Danielsson and Van Riet, but they seem to be in good company here. Recently, four prominent string theorists wrote a paper that suggested (even metastable) De Sitter space might actually belong to the swampland (be impossible to realize in string theory): https://arxiv.org/abs/1806.08362. In this paper they formulated a ...


2

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 ...


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

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 ...


2

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 ...


2

Events are always undergoing acceleration as they evolve forward in the time dilated continuum. Therefore, when we look out into space beyond the solar system, and back in time, we are also looking down a time dilation gradient into slower time. The observer’s invariant relative rate of time is always faster than that in frames in the perceived past, and we ...


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 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 ...


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 ...


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

To assist with your problem, I will describe the general approach, when we don't necessarily know what we're looking for. As you are aware, finding the Killing vectors $X^\mu$ requires solving, $$\nabla_\mu X_\nu + \nabla_\nu X_\mu = 0$$ which is an over-determined system of differential equations. As you know, these Killing vectors are precisely those for ...


2

The fact that a bulk theory is dual to a field theory on the boundary is really nothing special. Maxwell's equations are dual to the constraint field theory on boundary conditions. The problem is that the field theory is typically some weird composite, for instance a combination of an initial and boundary condition for Maxwell will give $div \vec{B}$ on the ...


2

I think most of the questions are a little bit subjective and often the answer is currently not known. So I will just give my opinion. Why then are people are so interested in AdS/CFT from the point of view of a definition of quantum gravity? Since quantum gravity is not very well understood in general I would say any tool that helps you to analyze it ...


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