2 added 790 characters in body edited Oct 26 '18 at 18:00 Hugo V 1,58066 silver badges1616 bronze badges Hubble's constant is not actually a constant. It is true the Hubble parameter is constant in regards to position and velocity of objects in the Universe at a certain time, meaning that for any measurement you get that $$v \propto d$$, where v is the velocity of the object away from us and d is its distance, Hubble's constant being exactly this proportionality factor. But this only means that for any given time the paremeter is the same for all objects, meaning that if you "look' at the sky at any given time you will find such a paremeter such that $$v \propto d$$ for all objects. This does not mean that the paremeter is constant over time, meaning that for two different timetimes you still get that $$v \propto d$$, but with a different proportionality factor, so the rate of expansion actually changes. The parameter is an estimate of the age of the universe, because the actual calculation of the age of the universe is: $$t_0=\frac {F( \Omega_m, \Omega_{\Lambda},...)}{H_0}$$ Where $$F$$ is a function of the relative densities of matter ($$\Omega_m$$), of dark energy ($$\Omega_{\Lambda}$$)... For a proper calculation this needs to be taken into account, but for an approximate calculation it is fine to take $$F \approx 1$$, because that is actually close to the current accepted value of $$F$$, according to the accepted model, the $$\Lambda$$CDM. The issue with the parameter becoming constant, it won't, although it will assymptotically approach a fixed value. In spite of that, accounting for the changes on $$F$$, it would still be possible to calculate the age of the universe that way. Hubble's constant is not actually a constant. It is true the Hubble parameter is constant in regards to position and velocity of objects in the Universe at a certain time, meaning that for any measurement you get that $$v \propto d$$, where v is the velocity of the object away from us and d is its distance, Hubble's constant being exactly this proportionality factor. But this only means that for any given time the paremeter is the same for all objects, meaning that if you "look' at the sky at any given time you will find such a paremeter such that $$v \propto d$$ for all objects. This does not mean that the paremeter is constant over time, meaning that for two different time you still get that $$v \propto d$$, but with a different proportionality factor, so the rate of expansion actually changes. Hubble's constant is not actually a constant. It is true the Hubble parameter is constant in regards to position and velocity of objects in the Universe at a certain time, meaning that for any measurement you get that $$v \propto d$$, where v is the velocity of the object away from us and d is its distance, Hubble's constant being exactly this proportionality factor. But this only means that for any given time the paremeter is the same for all objects, meaning that if you "look' at the sky at any given time you will find such a paremeter such that $$v \propto d$$ for all objects. This does not mean that the paremeter is constant over time, meaning that for two different times you still get that $$v \propto d$$, but with a different proportionality factor, so the rate of expansion actually changes. The parameter is an estimate of the age of the universe, because the actual calculation of the age of the universe is: $$t_0=\frac {F( \Omega_m, \Omega_{\Lambda},...)}{H_0}$$ Where $$F$$ is a function of the relative densities of matter ($$\Omega_m$$), of dark energy ($$\Omega_{\Lambda}$$)... For a proper calculation this needs to be taken into account, but for an approximate calculation it is fine to take $$F \approx 1$$, because that is actually close to the current accepted value of $$F$$, according to the accepted model, the $$\Lambda$$CDM. The issue with the parameter becoming constant, it won't, although it will assymptotically approach a fixed value. In spite of that, accounting for the changes on $$F$$, it would still be possible to calculate the age of the universe that way. 1 answered Oct 26 '18 at 14:31 Hugo V 1,58066 silver badges1616 bronze badges Hubble's constant is not actually a constant. It is true the Hubble parameter is constant in regards to position and velocity of objects in the Universe at a certain time, meaning that for any measurement you get that $$v \propto d$$, where v is the velocity of the object away from us and d is its distance, Hubble's constant being exactly this proportionality factor. But this only means that for any given time the paremeter is the same for all objects, meaning that if you "look' at the sky at any given time you will find such a paremeter such that $$v \propto d$$ for all objects. This does not mean that the paremeter is constant over time, meaning that for two different time you still get that $$v \propto d$$, but with a different proportionality factor, so the rate of expansion actually changes.