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How does Hubble's constant resembles age of universe? Isn't universe getting old each day? How can a constant be a reciprocal of age of universe? Hubble's value must be variable, isn't it?


marked as duplicate by dmckee Jun 25 '13 at 3:25

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$H$, sometimes called the Hubble "constant" should actually be called properly: the Hubble parameter. It is a function of the scale factor $a(t)$ of the Universe or of the redshift $z=1/a-1$ and depends on the cosmological parameters. It gives the rate of change of the size of the Universe with respect to its size.

Through the Friedmann equations it is equal to $$H^2(a)=(\dot a(t)/a(t))^2=\frac{8\pi G}{3}\rho-\frac{kc^2}{a^2}$$ Where $\rho, \, k$ are the density and spatial curvature parameters of the Universe, respectively. If the Universe is flat, i.e. it has the critical density, then $k=0$. Since the Universe is made up of different species of matter and energy and their density evolves in a different way with time, one must write them separately in the density term present in the above equation.

If one defines: $$\Omega_i\equiv \frac{\rho_i}{\rho_\text{crit}}=\frac{8\pi G\rho_i}{3H^2},$$ where the index $i$ stands for dark matter, radiation, curvature, baryonic matter, neutrinos or cosmological constant one can write $$H^2(a)=H^2_0(\Omega_ra^{-4}+\Omega_ma^{-3}+\Omega_{\Lambda}+\Omega_ka^{-2}),$$ where $H_0$ is the Hubble parameter today, equivalent to the Hubble "constant". $r$ stands for radiation and neutrinos, $m$ for dark matter and baryons, $k$ is the spatial curvature and $\Lambda$ represents the cosmological constant.

One can see clearly from the above equation that the evolution of the Hubble parameter depends on the contents of the Universe. From the definition of $H$ one obtains: $da=a\cdot H(a)\cdot dt$ and therefore one can integrate this to obtain the age of the Universe:

$$t_\text{age}=\int_0^1 \frac{da}{aH(a)}$$

$a=1$ is defined to be today and $a=0$ the size of the Universe at the Big Bang.

  • $\begingroup$ +1. Just to emphasize for the OP's sake: It is something of a coincidence that the reciprocal of $H_0$ matches the actual age of the universe, given by the above integral, so well. This is not necessarily true at other epochs. $\endgroup$ – user10851 Jun 24 '13 at 17:07

The Hubble-constant gives the rate at which the universe expands at the present. Since the expansion of the universe hasn't been constant this is indeed not the case. So yes, the Hubble-constant is actually a time-depandant value (the name constant dates back to the time people believed the universe was static).

The behaviour in time of the hubble constant H yields whether the expansion speeds up of slows down. One can make models for the universe. The simpelest model is the Robertson-Walker universe in which one can derive the Friedmann-equations.

These Friedmann equations give the time-evolution of the Hubble-constant in this universe. If you'd plug in certain values for the amount of radiation, matter and vacuum-energy these will give the dynamics (expansion/contration) of the modeled universe.

I don't know what the actual situation is of the cosmological models, but I believe it's the same method of working.

  • $\begingroup$ can you give me logical answer ? $\endgroup$ – dibesh lamichhane Jun 24 '13 at 17:09
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    $\begingroup$ No the Hubble constant is not a constant, it varies with time. The Hubble-constant contains information of the age of the universe via the formula @Santiago Casas has given. $\endgroup$ – Nick Jun 24 '13 at 17:20

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