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Hubble deep field image at maximum resolution shows a few of the furthest galaxies very small & faint, only just visible, but suggests that there is more to see beyond if only the resolution were high enough. I know the age of the universe is 'only' about 14 billion years, but if Hubble can only show galaxies 13.2 billion light years away, how do we know the radius of the universe is 47 billion light years?

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    $\begingroup$ For clarity, this is the radius of the observable universe, not entire Universe. $\endgroup$ – Sir Cumference Jun 13 at 22:08
  • $\begingroup$ Also see astronomy.stackexchange.com/questions/3635/… and the links therein. $\endgroup$ – PM 2Ring Jun 14 at 6:22
  • $\begingroup$ @SirCumference and to add to that clarity, "observable" in this case means "detectable" in the broadest sense, rather than being limited to observing electromagnetic radiation ("light"). $\endgroup$ – Chappo Says Reinstate Monica Jun 17 at 4:17
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The observable universe's edge, by definition, is the farthest part of the Universe that light could have reached us from. Let's say light from the observable universe's edge has just reached us. The longest it could have possibly travelled for would be 13.8 billion years, since the Universe is only 13.8 billion years old.

As such, one might assume that the light came from 13.8 billion light years away, and hence the observable universe's radius should be 13.8 billion light years. The problem is that this overlooks a key phenomenon: the expansion of the universe.

The location from where the light had been emitted has grown increasingly distant from us. In other words, by the time the photon has reached us and we are able to see the object emitting it, the object has moved much farther than when it emitted the photon. This is why we are able to see objects of distances far beyond 13.8 billion light years.

So to work out the actual numbers, let's suppose an object emitted a photon at the beginning of the universe $t=0$, and it just reached us today ($t_0 = 13.8 \ Gyr$). Evidently, this object lies at the edge of the observable universe, so finding its distance allows us to find the observable universe's radius.

Let's consider a given time $t$ in the past. Suppose our photon travelled for a small interval of time $dt$; it follows that the distance it just travelled away from its source would be $dx = c \ dt$. Due to the expansion of the universe, that $dx$ will expand over time; compared to today, it was a factor $a(t) = \frac{dx(t)}{dx(t_0)}$ smaller (this is called the scale factor at time $t$). Hence, the distance it gained from its source grew to be $dx(t_0) = \frac{dx(t)}{a(t)} = \frac{c \ dt}{a(t)}$ by today.

At every time $t$, it increased by an amount $dx$. So if the photon reached us at time $t_0 = 13.8 Gyr$ after it was emitted, then the total distance it gained from its source is the sum of these changes: $$r=\int_0^{t_0} dx(t_0) = \int_0^{t_0} \frac{c}{a(t)} dt \approx 46 \ Gly$$ Which is therefore the distance of the object from us.

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  • $\begingroup$ If we can see objects far beyond 13.8b.l.yr,see them with what? I have it on good authority that Hubble can only see 13.2 b.l.yr. And if an improved version of Hubble in the relatively near future were to see something 16 b.l.yr away,would that put the cat among the pigeons? $\endgroup$ – Michael Walsby Jun 14 at 8:49
  • $\begingroup$ @MichaelWalsby The observable universe is essentially the region of the universe that could be theoretically observed by an ideal telescope. More usefully, its edge contains the farthest objects that could ever interact with us in any way, e.g. via gravity (which travels in waves at speed $c$), etc. Beyond the edge, any events in the remaining Universe have not reached us yet, and have exactly zero effect on us; by extension, no information can even theoretically be determined about them until they enter the OU. $\endgroup$ – Sir Cumference Jun 14 at 9:14
  • $\begingroup$ @MichaelWalsby Worth clarifying is that the observable universe's radius is something determined by spacecrafts that analyze fundamental properties of the Universe, e.g. Planck and WMAP; Hubble's images are of little use for that. With the measurements from these spacecrafts, we can predict $a(t)$ and use the process I described. $\endgroup$ – Sir Cumference Jun 14 at 9:30
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    $\begingroup$ @MichaelWalsby Hubble can't only see 13.2 Glyr. That number is the light travel time, multiplied by the speed of light, and is used only in popular science because then people don't start asking "how can something be farther away than 13.8 Glyr?". The most distant galaxies observed at z ~ 11, and the current distance to those galaxies are at ~32 Glyr. The regions that emitted the CMB we see today are at ~45 Glyr, and the theoretical limit — as Sir Cumference discusses — is the 46.3 Glyr you've heard. $\endgroup$ – pela Jun 14 at 11:37
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So 47 billion light years is the radius of the observable universe, not the entire universe. What the observable universe refers to is

physical limit created by the speed of light itself. Because no signals can travel faster than light, any object farther away from us than light could travel in the age of the universe simply cannot be detected, as the signals could not have reached us yet. Wikipedia

Light couldn't reach us if it were outside of this radius and thus objects emitting it would be undetectable to us.

The radius of the observable universe was calculated by the process Sir Cumference describes.

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  • $\begingroup$ I'm still no wiser as to how we know that the radius of the observable universe is 47 billion light years. $\endgroup$ – Michael Walsby Jun 13 at 23:07
  • $\begingroup$ @Michael Walsby that is plugging the number we can access into the standard model. No model no distance. Different model different distance. $\endgroup$ – Alchimista Jun 14 at 12:03

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