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

$$n$$ 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.

• I don't think this question should be closed for needing details or clarity; at least to a cosmologist I understand what the question is asking. I think the OP could edit the question to define all their symbols more clearly to improve the question, though. Dec 16, 2022 at 12:53

The continuity equation reads $$$$\dot{\rho} = -3 H (\rho + p)$$$$ Using the equation of state$$^\star$$, $$p=w\rho$$, this equation becomes $$$$\frac{d\rho}{dt} = - 3 \frac{da}{dt} (1+w)\frac{\rho}{a}$$$$ Cancelling the $$dt$$'s$$^\dagger$$, and rearranging, we get the equation $$$$\frac{d\rho}{\rho} = -3(1+w)\frac{da}{a}$$$$ which has the solution $$$$\rho = \rho_0 a^{-3(1+w)}$$$$ where $$\rho=\rho_0$$ when $$a=1$$.
$$^\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$$ More formally, you can rerarrange the equation so the left hand side is $$$$\frac{d\rho/dt}{da/dt} = \frac{d\rho}{da}$$$$ which works so long as $$\rho$$ and $$a$$ are monotonic functions of time. (Which they are in the $$\Lambda$$CDM model).