Shape of de Sitter Universe: Is Hyperboloid Misleading?

  • Level: Undergrad 
  • Thread starter Thread starter gerald V
  • Start date Start date
  • Tags Tags
    Shape Universe
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
2 replies · 2K views
gerald V
Messages
67
Reaction score
3
TL;DR
How is the de Sitter universe best depicted?
I am confused about the shape of the de Sitter universe. The Misner-Thorne-Wheeler says it can be regarded as the submanifold given by ##-x_1^2 + x_2^2 + x_3^2 +x_4^2 + x_5^2 = k## of a flat space with lineelement ##\mbox{d}s^2 = -\mbox{d}x_1^2 + \mbox{d}x_2^2 + \mbox{d}x_3^2 +\mbox{d}x_4^2 + \mbox{d}x_5^2## (I am aware that there are generalizations with more dimensions as well as with more mixed signs, but this is not my point). ##k## is a constant which can be positive or negative. One oftenly sees figures depicting this de Sitter universe (two dimensions supressed) as a nice hyperboloid. In the following I only regard two degrees of freedom. The line element shall read ##\mbox{d}s^2 = -\mbox{d}x^2 + \mbox{d}y^2##, and the equation ##-x^2 + y^2 = k## shall define a 1-dimensional submanifold. If depicted on a sheet of paper, this equation yields a hyperbola. But a sheet of paper has euclidean symmetry, not pseudoeuclidean. If one wants to take into account the pseudoeuclidean metric of the embedding space, then one has to do the tricks familiar from sketches for Special Relativity, with length contraction and so on. But if one does so, the „hyperbola“ looks like a circle on a sheet of paper, right? What else could it look like?Is my conclusion right? So istn’t the depiction of the de Sitter universe as a hyperboloid completely misleading? Wouldn’t it be more appropriate to depict it as 4-sphere? I am aware that somebody might argue that this question has no answer, because actually there be no embedding space for our universe - what sounds to me a bit like an evasion.

Thank you very much in advance.
 
Physics news on Phys.org
Usually the de Sitter Robertson-Walker spacetime is the "vacuum" solution of the Friedman equations with positive cosmological constant. It can have all three curvature values ##k=\pm 1##, ##k=0## of the FLRW metric.
 
vanhees71 said:
It can have all three curvature values ##k=\pm 1##, ##k=0## of the FLRW metric.
To be clear, these are three different coordinate charts on the same spacetime; they are not three different spacetime geometries that all share the same name ("de Sitter"). It's also worth noting that not all of those charts cover the entire spacetime.
 
  • Like
Likes   Reactions: cianfa72 and vanhees71