I've done some research on this site and others and I'm having a bit of trouble.


Why is the observable universe so big?

Does the speed of light change?


From my understanding, FLRW/GR explains that the universe is expanding at a fixed rate, only dependent on the distance objects are apart. This rate is something like (21km/s)/lightyear. And we've come to show that this can even violate the speed of light at large distances, because the space itself is changing not the matter within it. I also hear the universe is 13.7 billion years old.

I took the radius of the observable universe, which should be in theory the furthest object from us in space (46,500,000,000 lightyears) and divided it by the expansion of the universe to get about 2.2 billion km/s. If something is moving away from us that fast we definitely won't see any more light from it, but there is still some coming because of what was sent before it was so far away.

How then do we know the size of the universe is as big as it is? And how do we really know how far away everything is when time is being distorted here as well, or even how old the universe is?

One other thing, if you take the speed of light and divide that by the expansion rate, you get about 14k lightyears being the max distance something can actually be from us for the light to still eventually reach us. Does this mean the things we now say are 46.5 billion light years from us were only 14k lightyears from us when they emitted that light?

  • 1
    $\begingroup$ Have you read en.wikipedia.org/wiki/Distance_measures_(cosmology) ? $\endgroup$
    – PM 2Ring
    Commented Jan 23, 2019 at 2:06
  • $\begingroup$ I have now, and that does get me thinking, we're measuring all of this in lightyears, and we use red shift to calculate how long it's been moving. Is it possible that the speed of light or the amount of red shift is what is actually changing? $\endgroup$ Commented Jan 23, 2019 at 2:11
  • $\begingroup$ I didnt understand the $21km/s/ly$ ? Are you sure thats correct ? $\endgroup$
    – seVenVo1d
    Commented Jan 23, 2019 at 3:57
  • $\begingroup$ What do you mean by "this can even violate the speed of light at large distances," ? $\endgroup$
    – seVenVo1d
    Commented Jan 23, 2019 at 4:48

1 Answer 1


I would like to give you a general answer related to your questions.

The Hubble Parameter is $70km/s/Mpc$. We can measure the expansion rate $(H)$ as an observable using the Hubble's Law (for $v<<c$) ($z=v_{rec}/c=Hr/c$) where $r$ is the proper distance to the galaxies. ($0<z<0.5$)

$H$ also appears in the Friedmann Equation.

$$H^2=\frac {8\pi G } {c^2}\epsilon-\frac {\kappa c^2} {R^2_0a^2(t)}$$

So in perspective of General Relativity, we can also see that the Hubble Parameter is related to the energy density-curvature of the universe.

The Hubble Law can be written as, $$V_{rec}=Hr$$ if we make $V_{rec}=c$ we get,

$c=Hr$ and from here $$r=c/H_0$$

this distance is the radius of the Hubble sphere, and the radius is approximately $4300Mpc$.

So this means that after this distance the objects "seems" to move faster than the speed of light. But it does not violate any kind of law. The reason is the objects are not actually moving faster than the speed of light. But the space itself expands. Also, the calculation in cosmology is based on FLRW metric from the GR and not SR.

I'll quote,

General relativity was specifically derived to be able to predict motion when global inertial frames were not available. Galaxies that are receding from us superluminally are at rest locally (their peculiar velocity, $v_{pec} = 0$ and motion in their local inertial frames remains well described by special relativity. They are in no sense catching up with photons $v_{pec} = c$. Rather, the galaxies and the photons are both receding from us at recession velocities greater than the speed of light.


We can measure the observable universe redius by using the FLRW metric,

Taking the $z=0$ for now and $z=\infty$ as for $t=0$, the radius of the observable universe is,

$$d_{particle \,horizon}=c\eta(z)=cH_0^{-1}\int_{z=0}^{\infty}\frac {dz} {E(z)}=46.9\, \text{Billion light year}$$

and the age of the universe can be calculated as $$t=1/H_0=4.41\times 10^{17}s = 13.9 \,\text {Billion year}$$ Also the observation suggests that these are indeed the true values. (There are also error bars in these measurements by I 'll ignore them for now)

Another point I would like to mention is that the radius of the observable radius is actually defined as the particle horizon in cosmology.

The particle horizon is the distance traveled by light from $t=0$ to $t=now$ (including the expansion factor) to us. Hence the furthest thing we can see is the CMB radiation at z=1100. Particle Horizon is the natural horizon where we can see the furthest distance.

Edit: How can see objects further away then the Hubble Radius ?

The main reason is the $H$ actually decreases (however approaches to a certain value) in time so Hubble radius increases. Which means that once were the photons at the region of $v_{rec}>c$ can pass to a region where $v_{rec}<c$. And when they pass that region we case we can see the distant galaxies.

  • $\begingroup$ Thank you for the well thought out answer, people like you are why this site is great. I understand most of what you are saying, the math makes a lot more sense now, but I still have one problem with this. How does light ever make it to us if the space it is moving into is expanding faster than the speed of it moving through it? $\endgroup$ Commented Jan 23, 2019 at 14:41
  • $\begingroup$ @SethWeinstein I would suggest you to read Sectlon 3.3 In the referance that I gave you. $\endgroup$
    – seVenVo1d
    Commented Jan 23, 2019 at 15:26
  • $\begingroup$ @SethWeinstein I edited my post maybe it helps more $\endgroup$
    – seVenVo1d
    Commented Jan 23, 2019 at 16:55

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