<p><strong>tl;dr</strong>: Your field of view would cover roughly one square centimeter of the sky at that time, and you would observe roughly 50 billionths of the observable Universe.</p>

<hr>

<h3>You can't really…</h3>

<p>With photons, you will never be able to see further back than to <a href="https://en.wikipedia.org/wiki/Recombination_(cosmology)" rel="noreferrer">recombination</a>, when the Universe was 380,000 yr old, because until then, the free electrons made is opaque to radiation.</p>

<p>With neutrinos, you can (in principle) see back to 1 sec after Big Bang, but to "see" all the way back to inflation, you will probably need gravitational waves.</p>

<h3>…but let's assume you can anyway</h3>

<p>Anyway, lets's do a thought experiment and assume that you have a RLT (Ridiculously Large Telescope) with a field of view (FoV) of the HST. What fraction of the Universe can you see? The answer might surprise you:</p>

<p>The farther an object is from you, the smaller an angle it spans on the sky, i.e. the smaller it looks. This is true for birds and planets and even for the nearest galaxies. However, for distant galaxies a strange effect counteracts this: Because of 1) the finite speed of light and 2) the expansion of the Universe, distant galaxies were closer to you when they emitted the light you see today, and therefore spanned a larger angle. Hence, if you compare galaxies of the same physical size, they look smaller and smaller until around 15 Glyr (giga-lightyears), after which they'll look larger and larger.</p>

<p>The Hubble Ultra Deep Field has FoV of 2.8 by 2.5 arcminutes. The region of the Universe that this FoV spans is thus largest at a distance of 15 Glyr, where it span $4.8\times4.3\,\mathrm{Mlyr}^2$ (i.e. square-mega-lightyears). But then your FoV begins to span an increasingly smaller region. At recombination, it spans only $34\times30\,\mathrm{klyr}^2$ — in other words, if a Milky Way-sized galaxy were present there (it weren't), it would be larger than the FoV.</p>

<p>It you could look all the way back to 4 hours after Big Bang, your FoV would span roughly a square-lightyear. Three minutes after BB, when all the Universe's hydrogen and helium had just been created, it would span $\sim0.1\times0.1\,\mathrm{lyr}^2$. A few microseconds after BB, it would span roughly a square-AU (the distance from Earth to Sun). Approximately $10^{-22}\,\mathrm{s}$ after BB, it would span a square kilometer.</p>

<p>And at $10^{-32}\,\mathrm{s}$ after BB, your FoV would span $0.8\times0.7\,\mathrm{cm}^2$, i.e. roughly one square-centimeter!</p>

<h3>Fraction of the Universe observed</h3>

<p>The part of the observable Universe that you would observe is the volume "behind" your FoV. That can be obtained by integrating the area along the distance, but an easier way is to simply note the fact that, if your FoV were the whole sky, you'd observe the whole Universe. Since your FoV covers a fraction of the sky of
$$
\begin{array}\\
f &amp; = &amp; \frac{2.8'\times2.5'}{\mathrm{full\,sky}}\\
  &amp; = &amp; \frac{8\times10^{-4}\,\mathrm{rad}\,\,\times\,\,7\times10^{-4}\,\mathrm{rad}}{4\pi\,\mathrm{rad}^2}\\
  &amp; = &amp; 5\times10^{-8},
\end{array}
$$
this is the fraction of the Universe that you would observe.</p>

<p>At $t=10^{-32}\,\mathrm{s}$, the part of the Universe later became our observable Universe <em>today</em>, was about ten meters in radius. Another way to calculate the fraction is to realize that the above mentioned observed square-centimeter roughly comprises a fraction $f$ of the surface area of a sphere with a radius of ten meters.</p>

CC BY-SA 4.0
2018-05-13T21:23:42.143Z
Very cool answer!!
userLTK
CC BY-SA 4.0
2018-05-13T22:35:55.850Z
Thank you.  When I aim my RLT at the 10^-32 seconds with my gravitational waves, I’m viewing a square centimeter.  This is a square centimeter of the 10 light year diameter universe, correct?
Michael James
CC BY-SA 4.0
2018-05-14T13:18:02.343Z
@MichaelJames I'm not sure what you refer to by "10 lyr". The radius of what today has become our observable Universe was, at t = 1e-32 s, roughly 10 _meters_. An observer at t = 1e-32 s (may the gods help him), would have an observable Universe of ~5e-33 cm. But none of these bear any special significance to the cm$^2$ that you'd observe with your RLT.
pela
CC BY-SA 4.0
2018-05-14T13:23:08.360Z
The _volume_ fraction of the observable Universe that you would see is the covered area as a function of distance, integrated over distance. I can calculate that later, but don't have the time right now.
pela
CC BY-SA 4.0
2018-05-14T13:24:53.810Z
Btw, if you know Python, I made a script that calculates the properties of the Universe at a given time. You can find it at [github.com/githyankipela/Universe-timeline](https://github.com/githyankipela/Universe-timeline).
pela
CC BY-SA 4.0
2018-05-14T13:25:33.700Z
Excellent!   I’ll give it a forking.
Michael James
CC BY-SA 4.0
2018-05-14T14:51:57.953Z
@MichaelJames: 5e-8 is the answer. See edit :)
pela
CC BY-SA 4.0
2018-05-13T21:13:28.183Z
2018-05-14T15:47:49.513Z
2018-05-14T15:47:49.513Z
pela
<p>Apologies in advance for my over-exposure to pop science.  I want to propose a telescope powerful enough to collect imagery from 10^-32 seconds after the “Big Bang”, or as soon as the end of the inflationary epoch.  With light coming from the universe when roughly 10 ly in diameter, along with the expansion to our current size, I would only see the tiniest fraction of light from so long ago with a telescope looking at that far of a distance.  </p>

<p>Assuming my telescope has an identical view field as the Hubble Space Telescope, what fraction of the total volume of the universe at end of the inflationary epoch would I actually be viewing?</p>

<hubble-telescope><early-universe>
Portion of universe visible if gathering image from inflationary epoch
Michael James
# Portion of universe visible if gathering image from inflationary epoch

(Asked by: username_0 on 2018-05-13)

Apologies in advance for my over-exposure to pop science. I want to propose a telescope powerful enough to collect imagery from 10^-32 seconds after the “Big Bang”, or as soon as the end of the inflationary epoch. With light coming from the universe when roughly 10 ly in diameter, along with the expansion to our current size, I would only see the tiniest fraction of light from so long ago with a telescope looking at that far of a distance.

Assuming my telescope has an identical view field as the Hubble Space Telescope, what fraction of the total volume of the universe at end of the inflationary epoch would I actually be viewing?

**Tags:** hubble-telescope, early-universe


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## 1 answers

### Answer by username_1 on 2018-05-13. Score: 11 | Accepted answer

**tl;dr**: Your field of view would cover roughly one square centimeter of the sky at that time, and you would observe roughly 50 billionths of the observable Universe.

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### You can't really…

With photons, you will never be able to see further back than to [recombination](https://en.wikipedia.org/wiki/Recombination_(cosmology)), when the Universe was 380,000 yr old, because until then, the free electrons made is opaque to radiation.

With neutrinos, you can (in principle) see back to 1 sec after Big Bang, but to "see" all the way back to inflation, you will probably need gravitational waves.

### …but let's assume you can anyway

Anyway, lets's do a thought experiment and assume that you have a RLT (Ridiculously Large Telescope) with a field of view (FoV) of the HST. What fraction of the Universe can you see? The answer might surprise you:

The farther an object is from you, the smaller an angle it spans on the sky, i.e. the smaller it looks. This is true for birds and planets and even for the nearest galaxies. However, for distant galaxies a strange effect counteracts this: Because of 1) the finite speed of light and 2) the expansion of the Universe, distant galaxies were closer to you when they emitted the light you see today, and therefore spanned a larger angle. Hence, if you compare galaxies of the same physical size, they look smaller and smaller until around 15 Glyr (giga-lightyears), after which they'll look larger and larger.

The Hubble Ultra Deep Field has FoV of 2.8 by 2.5 arcminutes. The region of the Universe that this FoV spans is thus largest at a distance of 15 Glyr, where it span $4.8\times4.3\,\mathrm{Mlyr}^2$ (i.e. square-mega-lightyears). But then your FoV begins to span an increasingly smaller region. At recombination, it spans only $34\times30\,\mathrm{klyr}^2$ — in other words, if a Milky Way-sized galaxy were present there (it weren't), it would be larger than the FoV.

It you could look all the way back to 4 hours after Big Bang, your FoV would span roughly a square-lightyear. Three minutes after BB, when all the Universe's hydrogen and helium had just been created, it would span $\sim0.1\times0.1\,\mathrm{lyr}^2$. A few microseconds after BB, it would span roughly a square-AU (the distance from Earth to Sun). Approximately $10^{-22}\,\mathrm{s}$ after BB, it would span a square kilometer.

And at $10^{-32}\,\mathrm{s}$ after BB, your FoV would span $0.8\times0.7\,\mathrm{cm}^2$, i.e. roughly one square-centimeter!

### Fraction of the Universe observed

The part of the observable Universe that you would observe is the volume "behind" your FoV. That can be obtained by integrating the area along the distance, but an easier way is to simply note the fact that, if your FoV were the whole sky, you'd observe the whole Universe. Since your FoV covers a fraction of the sky of
$$
\begin{array}\\
f & = & \frac{2.8'\times2.5'}{\mathrm{full\,sky}}\\
& = & \frac{8\times10^{-4}\,\mathrm{rad}\,\,\times\,\,7\times10^{-4}\,\mathrm{rad}}{4\pi\,\mathrm{rad}^2}\\
& = & 5\times10^{-8},
\end{array}
$$
this is the fraction of the Universe that you would observe.

At $t=10^{-32}\,\mathrm{s}$, the part of the Universe later became our observable Universe *today*, was about ten meters in radius. Another way to calculate the fraction is to realize that the above mentioned observed square-centimeter roughly comprises a fraction $f$ of the surface area of a sphere with a radius of ten meters.

> **Comments:**
> **username_2**: Very cool answer!!  
> **username_0**: Thank you.  When I aim my RLT at the 10^-32 seconds with my gravitational waves, I’m viewing a square centimeter.  This is a square centimeter of the 10 light year diameter universe, correct?  
> **username_1**: @MichaelJames I'm not sure what you refer to by "10 lyr". The radius of what today has become our observable Universe was, at t = 1e-32 s, roughly 10 _meters_. An observer at t = 1e-32 s (may the gods help him), would have an observable Universe of ~5e-33 cm. But none of these bear any special significance to the cm$^2$ that you'd observe with your RLT.  
> **username_1**: The _volume_ fraction of the observable Universe that you would see is the covered area as a function of distance, integrated over distance. I can calculate that later, but don't have the time right now.  
> **username_1**: Btw, if you know Python, I made a script that calculates the properties of the Universe at a given time. You can find it at [github.com/githyankiusername_1/Universe-timeline](https://github.com/githyankiusername_1/Universe-timeline).  
> **username_0**: Excellent!   I’ll give it a forking.  
> **username_1**: @MichaelJames: 5e-8 is the answer. See edit :)  

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