Since a telescope is just a very powerful eye, if it took the light of Earendel 12.9 billion years for it to reach the Hubble telescope, how long would it take for that same light to reach a human eye? I guess the light from Earendel would need to travel even more so that I can see it.

Or is the distance so minimal that it's not even worth considering?

It's just a theoretical question, since I assume I will never see Earendel from my balcony.

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    $\begingroup$ Do you understand why it took 12.9 billion years? $\endgroup$
    – user253751
    Apr 13, 2022 at 10:20
  • $\begingroup$ Well, a light-year is the distance light travels in a year. I assume Earendel is 12.9 billion light years away, $\endgroup$
    – rbhat
    Apr 13, 2022 at 15:39
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    $\begingroup$ So it's 12900000000 light-years away from the Hubble telescope. Is it more than 12900000000 light-years away from your eyeballs? (Hint: The Hubble telescope is approximately 0.0000000005 light-years away from your eyeballs) $\endgroup$
    – user253751
    Apr 13, 2022 at 15:48
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    $\begingroup$ Isn't Earendel, a.k.a. WHL0137-LS, currently believed to be a star, not a galaxy? $\endgroup$ Apr 13, 2022 at 20:26

2 Answers 2


Hubble is in low Earth orbit which is almost the same as being on the Earth's surface. So the time difference will be a just a few tens of milliseconds depending on where you are and where Hubble is around the curve of the Earth's surface. Yes, - definitely negligible on the scale of 10.9 billion years!

That famous photograph is a result of 9 hours of staring with a 2.4 meter (corrected per @pela coment) diameter pupil (mirror). Your pupil is ~300 times smaller than that in diameter and a 100,000 times (=3002) smaller in area, so to see the same number of photons, you would have to stare at the same point in the sky for about 100,000 times longer which is about 100 years. That photo may have as few as 100 photons correspnding to Earendel, so if you stare at the right spot in the sky for one year, your eye may eventually catch one photon that came from Earendel.

So if you're intent on seeing Earendel from your balcony you will need a very big pot of coffee to keep you awake and I'd recommend some warm clothing in case it gets chilly.

[Edit: PS. Please see @pela's excellent detailed answer. Earendel cannot be seen at all at visible wavelengths]

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    $\begingroup$ And even then, imaging devices such as cameras integrate the photons over time, but our eyes/brains don't. $\endgroup$
    – Jim421616
    Apr 12, 2022 at 19:18
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    $\begingroup$ Good answer, but minor nitpicking: Hubble's diameter is 2.4 m, not 4.2 m, and the pupil expands to somewhat larger size in the dark (up to 8-9 mm), so the area ratio is more like 1e5. Moreover, your estimate of the number of visible photons is a few orders of magnitude to low — if I've calculated correctly, your eye should receive one photon every 2-3 days. $\endgroup$
    – pela
    Apr 12, 2022 at 21:05
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    $\begingroup$ Btw I seem to remember having heard that frogs are able to detect single photons. Crazy frogs… $\endgroup$
    – pela
    Apr 12, 2022 at 21:08
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    $\begingroup$ That said, the photons are all infrared. There are no visible photons, since in the rest frame of Earendil, they were emitted in the ultraviolet and hence absorbed by neutral hydrogen in the host galaxy or close to it. $\endgroup$
    – pela
    Apr 13, 2022 at 0:06
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    $\begingroup$ @pela Humans have also been shown to be able to detect single photons with accuracy greater than if they were guessing (which is not the same as being able to actually see a star that you receive single photons from) $\endgroup$
    – user253751
    Apr 13, 2022 at 10:19

Roger Wood answers perfectly how the extra distance from Hubble to us is negligible, and estimates how much longer than Hubble you would need to look in order to detect a single photon with your eye. I'd just like to provide a little insight into exactly how long time you would have to wait to detect that photon.

Earendel is not visible to the human eye, not even in principle

Not theoretically

The human eye can only detect light in the wavelength range $\lambda \simeq [400,700]\,\mathrm{nm}$, which astronomers like to call $[4000,7000]\,\mathrm{Å}$. Because Earendel is seen at a redshift of $z=6.2$, in order to be visible when the light reaches you, it must have been a factor $(1+z)$ shorter when it was emitted, i.e. in the range ~500–1000 Å. This is in the far/extreme ultraviolet. But at that epoch — when the Universe was less than a billion years old — although reionization was about to complete there was still plenty of neutral gas. Hence, virtually everything below the ionization threshold at 917 Å was absorbed close to, if not inside, Earendel's host galaxy.

Photons of longer wavelengths may escape the host, but as they're redshifted they eventually reach the wavelength 1216 Å which corresponds to the Lyman α transition between hydrogen's ground and first excited state. Neutral hydrogen thus scatters the photons out of the line of sight, effectively absorbing them from our observations. This is the so-called Gunn-Peterson trough.

Hence, from a galaxy this far, only photons that are emitted redward of 1216 Å have a fair chance of reaching us, at which point they'll be 1216 Å × (1+z) ≈ 0.9 µm or longer, which is in the infrared (IR) region, invisible to the human eye.

Not observationally

Indeed, all observations of Earendel taken in optical filters are consistent with having zero flux, as seen in this table from the paper presenting Earendel, Welch et al. (2022) (I marked them green).


What if we pretend it's visible?

We could pretend that humans are able to detect IR photons, or we could pretend that the emitted UV photons weren't absorbed. As the former would require us to define an arbitrary range in which we could detect the photons, let's go with the latter.

Flux density

As marked in red in the table, the detected fluxes are of the order of 20–50 nJy. A "nJy", or nano-Jansky, is a measure of flux density, i.e. flux per frequency bin, and is equal to $10^{-32}\,\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-2}\,\mathrm{Hz}^{-1}$.


Let's assume that the far-UV radiation that we're pretending not to be absorbed would have a similar flux density of $f_\nu = 30\,\mathrm{nJy}$. The visible range of [4000–7000] Å corresponds to a frequency range of $[4.3\text{–}7.5]\times10^{14}\,\mathrm{Hz}$, i.e. $\Delta\nu = 3.2\times10^{14}\,\mathrm{Hz}$. Then the total flux piercing your eye is $$ F = f_\nu\Delta\nu = 9.6\times10^{-17}\,\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-2}. $$


In the dark, your pupil expands to roughly 8 mm in diameter, so its area is $A_\mathrm{eye}\sim50\,\mathrm{mm}^{2}$. The total power entering your pupil is then $$ P = F A_\mathrm{eye}=4.8\times10^{-17}\,\mathrm{erg}\,\mathrm{s}^{-1}. $$

Number of photons

The typical energy of a visible photon can be taken as that of a green photon with $\lambda\sim5500\,\mathrm{Å}$, i.e. $E_\mathrm{ph} = h\nu = hc/\lambda = 3.6\times10^{-12}\,\mathrm{erg}$. Hence, the number rate of photons entering the eye is $$ \frac{dn}{dt} = \frac{P}{E_\mathrm{ph}} = 1.3\times10^{-5}\,\mathrm{s}^{-1}. $$

Waiting time

Thus, the time you have to wait between each photon will, on average, be $$ dt = \frac{1}{dn/dt}\sim1\,\mathrm{day}. $$

  • $\begingroup$ -super answer. Thanks very much for putting that together! $\endgroup$
    – Roger Wood
    Apr 15, 2022 at 20:46
  • $\begingroup$ Thanks for inspiring me, @RogerWood. $\endgroup$
    – pela
    Apr 15, 2022 at 21:46

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