New answers tagged

2

I'll answer the question in the title. No, ($n$-dimensional) de Sitter does not have a preferred frame. The reason is that it is a maximally symmetric spacetime (with isometry group $O(n,1)$*). Similarly to Minkowski space which is also maximally symmetric, there is no preferred frame because for every two points there exists an isometry transformation that ...


1

Firstly let's find a general geodesic equation (this part is not actually related to the question but you asked for it in the comments). Taking $\phi=const$ and $\theta=\pi/2$ the Lagrangian becomes $$ L=(1-r^2/l^2)\dot{t}^2-\frac{1}{(1-r^2/l^2)}\dot{r}^2$$ Where $l$ is the de sitter horizon. Substituting this in the Euler Lagrange equation we get two ...


2

Your force is correct, that is also the expression for $\ddot{r}$ in the real Schwarzschild De Sitter metric when you set the first proper time derivatives of the spatial coordinates to zero: The geodesic equation gives the radial component of the 4-acceleration (in natural units): $$ \ddot{r}= \color{gray}{ \frac{\left(\Lambda r^3-3\right) \dot{r}^2}{r \...


0

Our Universe will to become a de Sitter space, and so the entropy of this 'particle horizon at infinity' is equal in magnitude to the inverse cosmological constant of our Universe. This is known in the 'lore' of quantum gravity, see accepted answer here Entropy of an empty universe In natural units, if you divide the de Sitter entropy by $Λ$ the result ...


Top 50 recent answers are included