Hubble's law states that $v=Hx$

Age of the universe is calculated by $T= \frac{x}{v} = \frac{1}{H}$

but the velocity is not constant; it changes with distance, so I think that this equation cannot be applied because simply the velocity is NOT uniform

$v$ should have been replaced with $\frac{\mathrm{d}x}{\mathrm{d}t}$, so it gives this differential equation $\frac{\mathrm{d}x}{\mathrm{d}t}=Hx$.

Nevertheless, when this differential equation is solved, and we substitute for initial condition ($x=0$ when $t=0$), it appears that there are no solutions, because it will be an exponential function, which can't take the value $0$ ever.

What is wrong with my understanding?

  • $\begingroup$ The universe has had different expansion profiles throughout its history, the equation we use isn't so simple as the one you used $\endgroup$ – Jim Jul 10 '14 at 18:34
  • $\begingroup$ related: physics.stackexchange.com/q/104153 $\endgroup$ – Kyle Oman Jul 10 '14 at 18:36
  • $\begingroup$ I know that H changes, but assuming H is constant is a reasonable approximation, you should know that 1/H estimates the age of the universe to be 13.7 which is very close to the actual value 13.8 $\endgroup$ – user3613971 Jul 10 '14 at 18:49
  • $\begingroup$ $1/H_0=14.6 Gyr$ give or take. Only slightly longer than the age of Methuselah star at 14.4 (+/- 0.8) Gyr $\endgroup$ – Jim Jul 10 '14 at 18:57

You're actually pretty close to the correct method for estimating the age of the Universe, but $H$ is not constant with time, it is $H=H(t)$.

One of the many ways of writing the equation to solve is:

$$t(z) = \frac{1}{H_0}\int_z^\infty \frac{dz}{(1+z)E(z)}$$

Here $z$ is redshift; $z\rightarrow\infty$ at the Big Bang, and $z=0$ now, so if you integrate from $0$ to $\infty$ you get $t(0)$ , the age of the Universe.

$E(z)$ is a function describing the relative content of the Universe at different redshifts, one example of $E(z)$ is the one for the $\rm\Lambda CDM$ model:

$$E(z) = \sqrt{\Omega_{\Lambda,0}+(1-\Omega_0)(1+z)^2+\Omega_{m,0}(1+z)^3+\Omega_{r,0}(1+z)^4}$$

The $\Omega$ are density parameters, which you can read more about here or here. The $0$ subscripts indicate that these should be the values measured at $z=0$, so these are constants and you can write the age of the Universe in terms of them (or you could, if that integral had an analytic solution - I don't suggest trying to solve it by hand, unless you make some simplifying assumptions or approximations first!).

The reason that $t(0)\approx1/H_0$ works reasonably well is that the Universe has been expanding at very nearly the rate it is expanding at now for the majority of the time since the Big Bang. So approximating $v$ as constant at the present value actually gets you pretty close to the right answer.

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  • $\begingroup$ Your $E(z)$ is for a cosmological constant model. For generic dark energy, you should have $\Omega_{\Lambda,0}exp[-3\int_0^1\frac{da'}{a'}(1+w(a'))]$ $\endgroup$ – Jim Jul 10 '14 at 18:51
  • $\begingroup$ Thanks. This answered a part of my question which is why my way of solving doesn't give a solution. $\endgroup$ – user3613971 Jul 10 '14 at 18:51
  • $\begingroup$ another part is why x is just divided by v? and surprisingly it gives a close number which 13.7 ?? according to the approximation, v is not constant, only H is constant, so how he could do that? $\endgroup$ – user3613971 Jul 10 '14 at 18:54
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    $\begingroup$ @Jim basic $\rm \Lambda CDM$ is the cosmological constant model. I don't see any reason to complicate this further with the DE equation of state. $\endgroup$ – Kyle Oman Jul 10 '14 at 18:57
  • $\begingroup$ @user3613971 Added a paragraph to answer why assuming a constant $v$ gives nearly the right answer. $\endgroup$ – Kyle Oman Jul 10 '14 at 19:02

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