Quarlep said:
I want to know Universe density according to this equation( ##k=-1##) ?
##H^2(t)-8πρG/3=-k/a^2(t)##
##ρ_U=ρ_m+p_r##
##ρ_U##=Universe density
##ρ_m##=Matter density
##p_r##=Radiation density
Too many parameters, too few equations. Also, it depends a bit upon your conventions. There are two commonly-used conventions:
1. ##k = {-1, 0, 1}##. With this convention, the scale parameter ##a## becomes the radius of curvature.
2. ##a(now) = 1##. This is usually the easiest. With this convention, ##k## can take on any number, and represents the amount of spatial curvature today (when ##a = 1##).
One way to make things easier to deal with is to use the concept of density fractions. With density fractions, I can rewrite:
H^2(t) = {8\pi G \over 3} \left(\rho_m + \rho_r\right) - {k \over a^2}
as:
H^2(t) = const \left({\Omega_m \over a^3} + {\Omega_r \over a^4} + {\Omega_k \over a^2}\right)
Here I've introduced a constant on the left, and a series of ##\Omega## constants. The equation becomes simplest if we require than when ##a = 1##, ##\Omega_m + \Omega_r + \Omega_k = 1##. In this case, when ##a = 1##, we get:
H^2(now) = const
Where this is the same constant that goes out in front, so our constant is just ##H_0^2##, and the Friedmann equation becomes:
H^2(t) = H_0^2 \left({\Omega_m \over a^3} + {\Omega_r \over a^4} + {\Omega_k \over a^2}\right)
So now we have four unknowns, but only one equation. That's not enough to give an answer unless three of the unknowns are set. What's done in practice is to take some measurable quantity which is a function of ##H(t)##:
F(data) = f(H(t))
We then take lots of data points, and use computer simulations to figure out which choices for the constants ##H_0, \Omega_m, \Omega_r, \Omega_k,## and ##\Omega_\Lambda## fit the data most closely (that last one stems from the cosmological constant).
