# How is $\Omega_0 = 1$ when the characteristic "teardrop" past light cone seems to admit curvature?

Introduction: The top graphic is just one I pulled from a page describing the process of detecting cosmic curvature. The second graphic is one I drew up to illustrate my misunderstanding.

My assumptions are these:

1) The characteristic "teardrop" past light cone is a correct representation of our observations.

2) Curvature is measured by the angle between two converging photons.

3) WMAP measurements of $\Omega_0 = 1$ are correct and accurate.

Question:

1) How is it conceivable that $\Omega_0 = 1$ from WMAP measurements if the teardrop past light cone admits initially parallel yet eventually converging photons?

It seems as if $\Omega_0 > 1$.

2) Photons live on the surface of light cones or teardrops and there is clearly some degree of curvature in the early universe as displayed by the graphic. If $\Omega_0 = 1$ then what exactly is meant by curvature if the curvature in the bottom graphic does not contribute to $\Omega_0$.

Keep in mind that the bottom graphic shows two dimensions of space and one of time. I have done my reading on FWR metrics to a reasonable extent and I am still lost with this so could one of you fine PSE users please show me specifically what I am misunderstanding and provide context, math or intuition.

• keep in mind the image of the triangle on the three manifolds would be light paths in non-expanding space whereas the last figure shows light paths in expanding spacetime. There's no time component in the first three images; you're looking at the path only through 2D space, not through 1D space and time like in the last figure
– Jim
Jan 7, 2015 at 20:02
• @Jim Thank you for your comment. I agree with everything you said but I suppose my problem is less with the diagrams and more with the observed value of $\Omega$. I agree that the bottom graphic is not completely analogous to the top unless our universe wasnt expanding and the above manifolds correctly represent the path photons take...We woukd still measure curvature...right? My point is that all of the MBR we receive from the horizon takes the path of the teardrop yet we measure no curvature? How is this possible? Jan 7, 2015 at 21:18
• My point was that it takes the teardrop path through time. The path through space is not teardrop. Curvature would make the path through space a teardrop
– Jim
Jan 8, 2015 at 14:17