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.


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.

Thanks in advance friends.

enter image description here enter image description here

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    $\begingroup$ 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 $\endgroup$
    – Jim
    Jan 7, 2015 at 20:02
  • $\begingroup$ @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? $\endgroup$ Jan 7, 2015 at 21:18
  • $\begingroup$ 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 $\endgroup$
    – Jim
    Jan 8, 2015 at 14:17

1 Answer 1


The "teardrop" shape is perfectly compatible with a flat universe. Remember that even though the trajectory looks curved in the x/t-plot, the photons travelling along the lightcone are only moving in the x-direction.

Think of a photon sent out towards us from some distance right after the big bang. At first even though the photon is moving towards us, the space in between us and the photon is expanding faster than the speed of light so the distance to the photon actually increases. However, as the expansion of the universe slows down, eventually the photon catches up with the expansion of space and reaches us.

I think some of your confusion stems from the fact that it looks like the photon is directed away from us at first, and then somehow changes direction and eventually reaches us. This is a mistake! The photon is always directed towards us, it is just the rapid expansion of space right after the big bang that makes the distance to the photon increase faster than it can approach us.

  • $\begingroup$ I don't deny they are compatible, but your explanation doesn't utilize two dimensions of space which is what we require to measure curvature (redshift can be measured with one dimension of space). My plot (although only x and t are labeled) was meant to be a (x,y,t) graph. Perhaps you could elaborate more on the meaning of triangle I illustrated on the teardrop and explain to me why that is not defined as any sort of curvature. Other than that thank you for your input. $\endgroup$ Jan 23, 2015 at 16:44
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    $\begingroup$ The curvature on the trangle is in the time dimension, not in the spatial dimension. If you plot the trajectories of photons in the pure spatial dimensions they would (or at least could) be entirely straight. Let me note that when cosmologists say that the universe is (at least almost) flat on large scales, we mean only that space is flat. The spacetime as a whole is still curved. $\endgroup$
    – Ihle
    Jan 23, 2015 at 18:31
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    $\begingroup$ Just want to say that I considered everything you said and I believe I understand it now completely. Thanks again! $\endgroup$ Feb 20, 2015 at 14:27

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