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Jul 18, 2021 at 13:14 vote accept wheelix
Jul 11, 2021 at 16:45 vote accept wheelix
Jul 11, 2021 at 16:45
Jul 11, 2021 at 16:45 vote accept wheelix
Jul 11, 2021 at 16:45
Jul 11, 2021 at 16:45 vote accept wheelix
Jul 11, 2021 at 16:45
Jul 11, 2021 at 16:45 vote accept wheelix
Jul 11, 2021 at 16:45
Jul 11, 2021 at 16:45 vote accept wheelix
Jul 11, 2021 at 16:45
Jul 11, 2021 at 16:45 vote accept wheelix
Jul 11, 2021 at 16:45
Jul 11, 2021 at 16:45 vote accept wheelix
Jul 11, 2021 at 16:45
Jul 11, 2021 at 16:45 vote accept wheelix
Jul 11, 2021 at 16:45
Jul 11, 2021 at 16:45 vote accept wheelix
Jul 11, 2021 at 16:45
Jul 11, 2021 at 16:45 vote accept wheelix
Jul 11, 2021 at 16:45
Jul 11, 2021 at 16:45 vote accept wheelix
Jul 11, 2021 at 16:45
Jul 11, 2021 at 16:45 vote accept wheelix
Jul 11, 2021 at 16:45
Jul 11, 2021 at 16:45 vote accept wheelix
Jul 11, 2021 at 16:45
Jul 11, 2021 at 16:45 vote accept wheelix
Jul 11, 2021 at 16:45
Jul 11, 2021 at 16:45 vote accept wheelix
Jul 11, 2021 at 16:45
Jul 11, 2021 at 16:44 vote accept wheelix
Jul 11, 2021 at 16:45
Jul 9, 2021 at 15:07 history edited ProfRob CC BY-SA 4.0
deleted 2 characters in body
Jul 9, 2021 at 14:58 history edited ProfRob CC BY-SA 4.0
added 174 characters in body
Jul 9, 2021 at 14:52 history edited ProfRob CC BY-SA 4.0
added 174 characters in body
Jul 9, 2021 at 14:39 comment added wheelix Would this imply a relation like $\Omega_m = \Omega_{m,0} a^{-3}$ is erroneous and instead should be $\Omega_m = \Omega_{m,0} a^{-3} (H_0^2/H^2)$?
Jul 9, 2021 at 14:28 comment added wheelix Just to clarify, does $(\Omega_{\Lambda} + \Omega_m + \Omega_r) = 1$ mean that $H^2 = H_0^2 (\Omega_{\Lambda} + \Omega_m + \Omega_r)$ is not true, and in fact the $H_0^2$ should be $H^2$?
Jul 9, 2021 at 6:41 history answered ProfRob CC BY-SA 4.0