- #1
fab13
- 312
- 6
hello,
Friedmann equations lead to with cosmological constant =0 :
Omega_m -1= k/(H0^2 * R0^2) where H0 is Hubble constant and R0 the radius equal to 1 here.
If Omega_m > 1, we have necessairly k=1 but rho_m is determined by :
rho_m= rho_crit *(1+k/(H0^2 * R0^2)) (eq n°1)
I mean we cannot choose any value of rho_m, only one which is above (eq n°1) .
Moreover, if we take rho_crit= 5 protons mass and H0= 71 km/s/Mpc ( H0=71000/(3*10^(22)) s^(-1)) , rho_m takes a very big value with k=1.
Same case for Omega_m < 1, eq n°1 gives a negative value for rho_m with k=-1 and HO=71000(3*10^(22)) s^(-1).
From i know, k is an integer (-1,0,1), so, how can we choose k=1 and a free value for rho_m, and in the same time, verify the equation n°1 above ? the question is the same for k=-1.
Regards
Friedmann equations lead to with cosmological constant =0 :
Omega_m -1= k/(H0^2 * R0^2) where H0 is Hubble constant and R0 the radius equal to 1 here.
If Omega_m > 1, we have necessairly k=1 but rho_m is determined by :
rho_m= rho_crit *(1+k/(H0^2 * R0^2)) (eq n°1)
I mean we cannot choose any value of rho_m, only one which is above (eq n°1) .
Moreover, if we take rho_crit= 5 protons mass and H0= 71 km/s/Mpc ( H0=71000/(3*10^(22)) s^(-1)) , rho_m takes a very big value with k=1.
Same case for Omega_m < 1, eq n°1 gives a negative value for rho_m with k=-1 and HO=71000(3*10^(22)) s^(-1).
From i know, k is an integer (-1,0,1), so, how can we choose k=1 and a free value for rho_m, and in the same time, verify the equation n°1 above ? the question is the same for k=-1.
Regards