The following blue-cone Wikipedia diagram confuses me.

At any point of cosmological time the encircling horizontal lines in the diagram are of finite circumference. That is indicative of a closed model of the universe.

Queries: 1. Why does the author use a closed model of the universe to explain his point?

2.Can we conclude "It is also possible for a distance to exceed the speed of light times the age of the universe, which means that light from one part of space generated near the beginning of the Universe........." if we draw the same diagram on a flat sheet of paper[instead of using the cone we take a flat surface],remembering that the null geodesics are always straight lines in the flat spacetime context?

  • $\begingroup$ It is possible that you might be better off asking the second part of your question as its own separate post. $\endgroup$ – David Z Dec 5 '11 at 22:32

The author of the diagram explains very well how and why he drew the diagram this way:


In particular it states:"I deliberately cut off the embedding short of a full circle to emphasize that space does not loop back on itself (or, if it does, not at a distance governed by the arbitrary parameter R)".

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If you look at the top view of the cone in the Wikipedia article you linked to, you'll notice that it's not actually a closed cone. So this diagram represents the evolution of a 1D open curve in a timelike slice of spacetime, not the whole universe. It doesn't say anything about instantaneous large-scale structure.

Even if the cone were closed, you could still look at it as representing the evolution of a 1D closed curve in spacetime, which does not necessarily have to wrap around the universe. That way, the diagram could be accurate even if the universe is open.

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  • $\begingroup$ The blue-cone space-time diagram has a distinct positive curvature which is indicative of a closed cosmological model of the universe. $\endgroup$ – Anamitra Palit Dec 22 '11 at 22:25
  • $\begingroup$ Um... no it doesn't. In fact the curvature becomes negative toward the top of the diagram. $\endgroup$ – David Z Dec 23 '11 at 5:25
  • $\begingroup$ If you go round the cone at some height[some moment in cosmological time] you have a positive curvature. From bottom to top you have a negative curvature. If exact cancellation[leading to zero curvature] occurs at each and every point we are supposed to have a flat surface. The other alternative would be to assume that there are points where the net curvature is either positive or negative[ a minor language editing has been performed] $\endgroup$ – Anamitra Palit Dec 23 '11 at 9:53
  • $\begingroup$ @AnamitraPalit: you are trying to read the curvature of a 3D space from a 1D slice, there's not enough information to do that. There are slices of tori which look identical to slices of a sphere (locally). $\endgroup$ – Stephen McAteer Jan 7 '12 at 1:56

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