Why does $n = -3(1+ \omega)$? [closed]

Question

$n$ comes from how the density of any component of the universe scales as the universe expands in the FLRW metric. For matter, $n=-3$ since the mass of matter is conserved.

$\omega$ comes from the equation of state for any component in the universe.

$$ p = \omega ρ $$

Pressure equals omega times energy density. For matter $\omega = 0$, matter, often termed dust, has no associated pressure. And you can see how the formula in the question works.

$$ -3 = -3(1+0) $$ For dark energy, $n = 0$ and $\omega = -1$. That also works $$ 0 = -3(1 + -1) $$ But I can't for the life of me see why $n$ and $\omega$ should even be related, let alone by this formula.

Answer 1

It comes from the continuity equation.

The continuity equation reads \begin{equation} \dot{\rho} = -3 H (\rho + p) \end{equation} Using the equation of state$^\star$, $p=w\rho$, this equation becomes \begin{equation} \frac{d\rho}{dt} = - 3 \frac{da}{dt} (1+w)\frac{\rho}{a} \end{equation} Cancelling the $dt$'s$^\dagger$, and rearranging, we get the equation \begin{equation} \frac{d\rho}{\rho} = -3(1+w)\frac{da}{a} \end{equation} which has the solution \begin{equation} \rho = \rho_0 a^{-3(1+w)} \end{equation} where $\rho=\rho_0$ when $a=1$.

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$^\star$ For some reason (I honestly suspect because of sloppy handwriting), there are multiple notations used for the equation of state parameter. I was taught that the symbol is $w$, but I have also seen $\omega$ and $\varpi$ used.

$^\dagger$ More formally, you can rerarrange the equation so the left hand side is \begin{equation} \frac{d\rho/dt}{da/dt} = \frac{d\rho}{da} \end{equation} which works so long as $\rho$ and $a$ are monotonic functions of time. (Which they are in the $\Lambda$CDM model).

Answer 2

Essentially, it's energy conservation -- see e.g. https://arxiv.org/abs/gr-qc/9712019, Chapter 8 (in particular, the part around Eqns. (8.15) to (8.23)).