这是一份UCL伦敦大学学院PHAS00032作业代写的成功案
The density parameter is a very useful way of specifying the density of the Universe. Let’s start with the Friedmann equation again. Recalling that $H=\dot{a} / a$, it reads
$$
H^{2}=\frac{8 \pi G}{3} \rho-\frac{k}{a^{2}} .
$$
For a given value of $H$, there is a special value of the density which would be required in order to make the geometry of the Universe flat, $k=0$. This is known as the critical density $\rho_{\mathrm{c}}$, which we see is given by
$$
\rho_{\mathrm{c}}(t)=\frac{3 H^{2}}{8 \pi G} .
$$
Note that the critical density changes with time, since $H$ does. Since we know the present value of the Hubble constant [at least in terms of $h$ defined in equation (6.1)], we can compute the present critical density. Since $G=6.67 \times 10^{-11} \mathrm{~m}^{3} \mathrm{~kg}^{-1} \mathrm{sec}^{-2}$, and converting megaparsecs to metres using conversion factors quoted on page xiv, it is
$$
\rho_{\mathrm{c}}\left(t_{0}\right)=1.88 \mathrm{~h}^{2} \times 10^{-26} \mathrm{~kg} \mathrm{~m}^{-3} .
$$
PHAS00032 COURSE NOTES ：
In the same way that it is useful to express the density as a fraction of the critical density, it is convenient to define a density parameter for the cosmological constant as
$$
\Omega_{\Lambda}=\frac{\Lambda}{3 H^{2}} .
$$
Although $\Lambda$ is a constant, $\Omega_{\Lambda}$ is not since $H$ varies with time. Repeating the steps used to write the Friedmann equation in the form of equation (6.9), we then find
$$
\Omega+\Omega_{\Lambda}-1=\frac{k}{a^{2} H^{2}} .
$$
The condition to have a flat Universe, $k=0$, generalizes to
$$
\Omega+\Omega_{\Lambda}=1 .
$$