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Aug 8, 2023 at 13:59 comment added Sten @tertius For example, equation 9 of arxiv.org/abs/2006.16182
Aug 8, 2023 at 13:50 comment added Sten @tertius Your calculation is right, except that it doesn't have to equal zero (that's only if the energy of the dust alone is conserved). If you're not sure how $a^{-3}\frac{\partial}{\partial t}\left(\rho a^3\right)$ comes out of that, try working backwards, expanding the derivative with the product rule.
Aug 8, 2023 at 8:03 comment added perchlorious Do you happen to have a paper or reference for the first equation. Interested in reading how they use it.
Aug 8, 2023 at 4:31 comment added Sten @tertius Are you perhaps missing the partial derivative term in the covariant derivative?
Aug 8, 2023 at 4:13 comment added perchlorious Is that because locally we do not see expansion, and so it is not possible to separate changes to density from decay vs. kinematics?
Aug 8, 2023 at 4:11 comment added perchlorious Thanks. This is exactly what I was looking for. One follow up: the first equation clearly accounts for changes to the dust density from both kinematical expansion as well as decay. The second equation, which I just wrote out, seems to only reproduce the kinematical changes, not decay changes. The $\mu=0$ component is $d\rho/dt = -3H\rho$, when $U_0 = 1$. Have I missed something?
Aug 8, 2023 at 3:43 history answered Sten CC BY-SA 4.0