From the Penrose diagram of de Sitter space, we see it has a future and past conformal boundary, and they are both spacelike. So, does de Sitter space admit an asymptotic S-matrix? Sure, in the usual coordinates which only cover half of the full space, we don't have an S-matrix because it's not causally complete, but in a coordinate system which covers the entire space, why not? Just because we don't have a globally timelike Killing vector field doesn't mean we can't have an S-matrix.
Subscribe to:
Post Comments (Atom)
-
This image from NASA illustrates drag coefficients for several shapes: It is generally accepted that some variation of the teardrop/airfoil...
-
The gravitation formula says F=Gm1m2r2,so if the mass of a bob increases then the torque on it should also increase...
-
As the title says. It is common sense that sharp things cut, but how do they work at the atomical level? Answer For organic matter, such a...
-
Small vessels generally lean into a turn, whereas big vessels lean out. Why do ships lean to the outside, but boats lean to the inside of a ...
-
I'm sitting in a room next to some totally unopened cans of carbonated soft drinks (if it matters — the two affected cans are Coke Zero...
-
What exactly are the spikes, or peaks and valleys, caused by in pictures such as these Wikipedia states that "From the point of view of...
-
Problem Statement: Imagine a spherical ball is dropped from a height h, into a liquid. What is the maximum average height of the displaced...
No comments:
Post a Comment