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Friedmann equation relates Hubble constant to Universe age as 1/H0 times mass density correction factor adding a few percent. What would Hubble have observed 7 billion years ago? The same value? if so what is special about the present moment that 1/H0 fits the current age so closely? Or would the earlier Hubble have measured a bigger constant corresponding to the shorter age? If so the Friedmann equation must be wrong.


marked as duplicate by John Rennie spacetime Jul 2 '18 at 11:10

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  • $\begingroup$ A quick (inaccurate) calculation using this numerical solution of the Friedmann equation finds the Hubble constant to be $H = \dot{a}/a = (0.6619901-0.6615104)/(6.01-5.99)/0.6612529$ $ = 0.03627205264 \text{ Gyrs}^{-1}$ $ \approx 35 \text{ km/s/Mpc}^{-1}$. As such $1/H_0 \approx 27.6\text{ Gyrs}$, a wildly different value from the expected $7.4 \text{ Gyrs}$. This is because $1/H_0$ is only a good approximation when $H$ increases similarly to an exponential curve. However, in the early universe this wasn't so, so $1/H_0$ is a bad approximation. $\endgroup$ – Beta Decay Jul 2 '18 at 11:48
  • $\begingroup$ The behavior of the Hubble parameter over time depends on the cosmological model. The fact that the Hubble time today is close to the universe age in the Friedman model is indeed one of many arguments against the correctness of this model. $\endgroup$ – safesphere Jul 2 '18 at 12:40