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I want a way to track how the composition of the FRW spacetime changes with time if we take a perfect fluid with only dust and radiation and allow some of that dust to decay to radiation.

The total energy density would be $\rho=\rho^{dust}+\rho^{rad}$, which scale as $a^{-3}$ and $a^{-4}$, respectively.

I am looking for a quantity that describes the proportion of dust vs radiation. If we took $$\frac{\rho^{dust}}{\rho}$$ that would seem to suffice but the problem is the radiation included in $\rho$ loses density at $a^{-4}$, so this quantity depends on the scale factor. So that quantity will change with the scale factor even if dust does not decay.

I suppose we could write $$\frac{\rho^{dust}}{\rho^{dust}+\rho^{rad}a}$$ but that seems contrived. Maybe that's the right quantity?

One other route i've tried: the dust is represented by the non-zero trace portion of the SEM tensor, which I could define as $D_{\mu\nu}$. Then contract this with the observer's 4-velocity so $D_{\mu\nu}U^\mu U^\nu = \rho^{dust}$. But that still changes with scale factor, and multiplying by $\sqrt{-g}$ turns it into a scalar density, which I'm not super clear on how to deal with.

Summary of what i'm trying to do: define some sort of quantity that changes only when dust decays. I want this quantity not to change with the scale factor, or at least change in such a way that the decay of dust can be separated.

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The standard expression for the rate at which the energy density in dust is lost to particle decay, which can be found in cosmology papers in a variety of contexts, is

$$\frac{\mathrm{d}\rho}{\mathrm{d}t}\equiv \frac{\partial\rho}{\partial t}+3H\rho=a^{-3}\frac{\partial}{\partial t}\left(\rho a^3\right),$$

where $\rho$ is the energy density of the dust, $H=\dot a/a$ is the Hubble rate, and $a$ is the scale factor. The generally covariant form of this expression is

$$\frac{\mathrm{d}\rho}{\mathrm{d}t}\equiv U_\mu \nabla_\nu T^{\mu\nu},$$

where $U$ is the 4-velocity of the dust, $\nabla$ is the covariant derivative, and $T$ is the stress-energy tensor of the dust. The former expression is valid only in the cosmological average, but the latter is valid locally in any context.

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  • $\begingroup$ 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? $\endgroup$ Commented Aug 8, 2023 at 4:11
  • $\begingroup$ Is that because locally we do not see expansion, and so it is not possible to separate changes to density from decay vs. kinematics? $\endgroup$ Commented Aug 8, 2023 at 4:13
  • $\begingroup$ @tertius Are you perhaps missing the partial derivative term in the covariant derivative? $\endgroup$
    – Sten
    Commented Aug 8, 2023 at 4:31
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    $\begingroup$ @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. $\endgroup$
    – Sten
    Commented Aug 8, 2023 at 13:50
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    $\begingroup$ @tertius For example, equation 9 of arxiv.org/abs/2006.16182 $\endgroup$
    – Sten
    Commented Aug 8, 2023 at 13:59

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