This is not a homework question. I want some help in being able to clearly perceive the expansion of the universe and the consequences thereof.

We know that the universe is expanding in an accelerated manner. It is my understanding that space alone cannot expand without affecting the local time. Thus as per GTR, depending on the curvature of space-time and the energy density, time too shall suffer a change due to the changing space.

Assuming that there is no curvature of space-time and no mass or energy contained in a chunk of space (assuming the zero point energy of empty space as zero) between two galaxies A and B, let us say that the local time at a point P in this space goes as $t_0$. If we on Earth could somehow observe this point and measure the time at P as $t$,

  1. When we talk about the rate of expansion for example as in Hubble's Law, do we follow the comoving coordinates or the proper coordinates?

  2. How would $t$ and $t_0$ relate to each other if the rate of expansion is uniform and accelerating respectively?

  3. How much is 1s for the comoving observer at the inflation period in terms of the usual 1s now on Earth?


1 Answer 1

  1. Hubble's law $$ v=H_0 d,$$ relates the recession velocity $v$ of a distant object to it's physical distance $d$. Today, the physical distance coincides, by definiton, to the comoving distance $\chi \equiv d/(1+z)$, but if you want to know how fast two galaxies at a redshift of, say, $z=1$, you must plug in their physical distance at that time (and use $H(z)$, the Hubble parameter at that time, rather than $H_0$). Depending on how you measure their mutual distance, there are various ways to calculate this.

  2. When we observe a galaxy at redshift $z$, general relativity predicts that we see its time dilated by a factor $1+z$. That is, some physical process which takes a proper time $t_0$, will be observed to take a time $$ \boxed{t = t_0(1+z),} $$ which is the formula you're after.

    A good example of this is the time it takes for the brightness of a supernova to decline; a supernova at $z=1$ declines half as fast as a local (i.e. $z=0$) supernova (see e.g. Goldhaber et al. 2001).

    Calculating the distance to that supernova, or the age of the Universe at the time it's observed, is a bit more complicated, involving a model of the expansion rate of the Universe (see e.g. this question about the Friedmann model).

  3. Taking inflation to end when the Universe was $t\sim10^{-32}\,\mathrm{s}$ old, the corresponding redshift was of the order$^\dagger$ $z\sim4\times10^{25}$, so if we could observe physical processes there, they'd be dilated by a factor $\sim4\times10^{25}$, so 1 second would, at this dilation, take a 3 billion years. However, already 1 second after inflation, the expansion rate of the Universe had already slowed down so much that the corresponding redshift, and hence dilation factor, was "just" $\sim4\times10^9$. So the next second would only, from our point of view, seem to take $\sim140$ years.

Note that these considerations don't really have anything to do with the accelerated expansion, but is just a consequence of the expansion.

Note also that special relativity predicts the same time dilation; that is, if galaxies were flying through space rather than just being carried along with expanding space, we would observe the same relation between redshift and time dilation. However, the special relativistic interpretation of redshifts is incompatible with the relation between redshifts and magnitudes of supernova, and can be ruled out at $23\sigma$ (Davis & Lineweaver 2004).

$^\dagger$This can be calculated as follows: Until the Universe was $t_\mathrm{eq}\sim50\,000\,\mathrm{yr}$ old, corresponding to a redshift of $z_\mathrm{eq}\simeq3400$ (Planck Collaboration et al. 2018) its dynamics was dominated by the energy density of radiation, causing the size (scale factor $a=1/(1+z)$) to evolve with time as $a\propto t^{1/2}$. Calculating backward from $t_\mathrm{eq}$ yields $a\sim10^{-26}$ at $t\sim10^{-32}\,\mathrm{s}$, corresponding to $z\sim10^{26}$.

  • $\begingroup$ Downvoted? 🤷‍♂️ $\endgroup$
    – pela
    Sep 5, 2022 at 8:34

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