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$$H(a_\mathrm{eq})=H_0 \sqrt{\Omega_\mathrm{m}a_\mathrm{eq}^{-3}+\Omega_\mathrm{r}a_\mathrm{eq}^{-4}+\Omega_\Lambda}.$$ But $\Omega_\mathrm{m}a_\mathrm{eq}^{-3}=\Omega_\mathrm{r}a_\mathrm{eq}^{-4}$ at …

The Hubble rate is $$H(a) = H_0\sqrt{\Omega_\mathrm{m}a^{-3}+\Omega_\mathrm{r}a^{-4}+\Omega_\Lambda}$$ at the scale factor $a$, where $\Omega_\mathrm{m}$, $\Omega_\mathrm{r}$, and $\Omega_\Lambda$ are …

Deep in radiation domination ($y\ll 1$), the two solutions are $\delta=1$ and $\delta=\ln a$. Physically, these solutions describe the effects of particle drift in the complete absence of peculiar for …

The definition of $\phi$ has been changed, so that it sources peculiar gravitational forces rather than the total gravitational force. That is: $\nabla^2\phi=4\pi Ga^2\bar{\rho}\delta$ This potentia …

In this context, $\eta \to \eta + \pi(\eta,\mathbf{x})$ shouldn't be viewed as a gauge transformation. If you were to make that transformation in an unperturbed FLRW spacetime, you would end up with t …