# Why is the Hubble time same with age of the universe assuming ideally constant expansion?

Practically, the Hubble time does not perfectly same with age of the universe. But that's considered to coincide when the rate of expansion is constant.

However, the constant rate does not mean constant speed.

Suppose that the Hubble constant is 1 m/s/m, that's not changed along the time.

Then, the object far from 10 m is receded by 10 m/s.

Because, the object was slower in past, the moving time must be more than 1 s.

But, the inverse of the Hubble constant is 1 s. So, the moving time of the object is more than the age of the universe. Is it possible?

• Does this answer your question? Why can we trust Hubble Time if the rate of expansion is not constant? Commented Dec 6, 2023 at 15:53
• @AlexRobinson Thank you, but that's not my question. I think the Hubble time is not same with the age of the universe even if the rate of expansion is ideally constant.
– XX X
Commented Dec 6, 2023 at 15:59

The present expansion of the universe follows Hubble's law $$\frac{da}{dt} = H a\ ,$$ where $$H_0$$ is the Hubble parameter and $$a$$ represents a scale factor (perhaps the distance between two galaxies).

The age of the universe will only be $$H^{-1}$$ at any epoch if the expansion rate itself, $$da/dt$$ is constant. This would imply a Hubble parameter varying as $$H=H_0a^{-1}$$, where $$H_0$$ is the present Hubble parameter at a scale factor of $$a=1$$, and would be the case for an unrealistic universe with just curvature and no matter, radiation or dark energy, since the evolution of $$H$$ is governed by $$H^2 = H_0^2 \left( \frac{\Omega_r}{a^4} + \frac{\Omega_M}{a^3} + \frac{\Omega_k}{a^2} + \Omega_{\Lambda}\right)\ .$$ Here, $$H_0$$ is the Hubble parameter now, $$a(t)$$ is the scale factor of the universe, $$\Omega_r$$ is the current (i.e. $$a=1$$) ratio of the radiation density to the critical density and $$\Omega_M$$, $$\Omega_k$$ and $$\Omega_{\Lambda}$$ are the equivalent densities for the matter (baryonic and dark), curvature and (constant) vacuum energy densities.

In other, more realistic universes then $$H$$ changes in different and more complex ways and the age of the universe is some multiple of $$H_0^{-1}$$, and not exactly $$H^{-1}$$ at other times. That is because the Hubble parameter changes and it changes in a way that is not just proportional to $$a^{-1}$$.

Your idea of a universe with a Hubble parameter that does not change with time would be a universe that has always been dominated by dark energy. In this case, the expansion is exponential; $$H = H_0 \Omega_\Lambda^{1/2}$$ and rolling the clock back, the separation between two objects would never be zero, but would obey $$a = \exp[H_0 \Omega_\Lambda^{1/2}(t-t_0)] \ ,$$ and the age of the universe would be infinite. This is also unrealistic, since whilst dark energy has a fixed density, matter and radiation densities were much bigger when the universe was smaller.

The age of the universe actually is quite close to $$H_0^{-1}$$ because of the "cosmic coincidence" that we live in a universe where the matter and dark energy densities are similar. The decelerating effects of the matter are cancelled out, and now exceeded, by the dark energy, resulting in an $$a$$ versus $$t$$ relationship which has a tangent ($$da/dt= H_0$$) at the current epoch ($$a=1$$) that can be extrapolated back to $$a=0$$ at roughly the correct time of the big bang. See Why can we trust Hubble Time if the rate of expansion is not constant? .

• Thank you. I finally know why my model is unrealistic. And maybe, I had misunderstood the constant expansion.
– XX X
Commented Dec 7, 2023 at 4:07
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