<p>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!</p>
<p>That famous photograph is a result of <strong>9 hours</strong> of staring with a <strong>2.4 meter</strong> (corrected per @pela coment) diameter pupil (mirror). Your pupil is ~300 times smaller than that in diameter and a 100,000 times (=300<sup>2</sup>) 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.</p>
<p>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.</p>
<p>[Edit: PS. Please see @pela's excellent detailed answer. Earendel cannot be seen at all at visible wavelengths]</p>

CC BY-SA 4.0
2022-04-12T19:18:00.197Z
And even then, imaging devices such as cameras integrate the photons over time, but our eyes/brains don't.
Jim421616
CC BY-SA 4.0
2022-04-12T19:33:50.373Z
@Jim421616 This is interesting. It's about single photon detection, though I imagine it's a very noisy process. https://www.nature.com/articles/nature.2016.20282
Roger Wood
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2022-04-12T21:05:48.990Z
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.
pela
CC BY-SA 4.0
2022-04-12T21:08:15.990Z
Btw I seem to remember having heard that frogs are able to detect single photons. Crazy frogs…
pela
CC BY-SA 4.0
2022-04-12T21:37:53.510Z
@pela thanks, I don't know why google popped up with 4.2 m.  I'll make the corrections. My answer is misleading.
Roger Wood
CC BY-SA 4.0
2022-04-12T21:54:45.353Z
@pela I get one photon per year when I recalculate?
Roger Wood
CC BY-SA 4.0
2022-04-12T21:55:09.553Z
@pela  I gather the photon efficiency of the human eye is only a few percent - maybe frogs are much better.  I assume owls and other nocturnal creatures are among the best
Roger Wood
CC BY-SA 4.0
2022-04-13T00:04:53.493Z
Your calculation is completely correct if you assume that Hubble collected 100 photons in 9 hour. But it collected somewhat more photons: The flux is of the order of 10 nJy. Converting that to number of photons in 9 hours yields ~20,000 photons. Wrt. the nocturnal creatures, you're probably right :)
pela
CC BY-SA 4.0
2022-04-13T00:06:46.073Z
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.
pela
CC BY-SA 4.0
2022-04-13T10:02:34.970Z
Human eyes (appropriately adapted to darkness) are not much worse than the best eyes in nature. 5 or 6 yellow-greenish photons are detectable.
fraxinus
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2022-04-13T11:20:44.757Z
@user253751 Yes, googling "human eye detect single photons" return quite a few interesting reads. Cool! :)
pela
CC BY-SA 4.0
2022-04-13T21:37:40.350Z
0

@pela: concerning the UV vs. IR photons: if IIUC, the IR photons that Hubble sees are UV photons that were not absorbed, but became IR by the time they got to us because of the Red Shift. Whether there are any visible photons from Earendil, I don't know; extreme UV might be shifted into our visible range, I would think.
Mike Maxwell
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2022-04-13T22:26:06.017Z
@MikeMaxwell You're right that near- and mid-UV photons may escape Earendel's host galaxy, and that they will be IR when they arrive at Earth. But photons below the ionization threshold at 917 Å are absorbed through ionization. Photons of longer wavelengths may escape the host galaxy, but will likely be scattered out of the line of sight through the Lyman α transition. So the probability of detecting photons bluer than 1216 Å × (1+z) ≈ 0.9 µm is very small, and indeed the fluxes in the optical filters are consistent with being zero.
pela
CC BY-SA 4.0
2022-04-12T18:57:54.860Z
2022-04-15T21:49:16.637Z
2022-04-15T21:49:16.637Z
Roger Wood
<p>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.</p>
<h3>Earendel is not visible to the human eye, not even in principle</h3>
<h5>Not theoretically</h5>
<p>The human eye can only detect light in the wavelength range <span class="math-container">$\lambda \simeq [400,700]\,\mathrm{nm}$</span>, which astronomers like to call <span class="math-container">$[4000,7000]\,\mathrm{Å}$</span>. Because Earendel is seen at a redshift of <span class="math-container">$z=6.2$</span>, in order to be visible when the light reaches you, it must have been a factor <span class="math-container">$(1+z)$</span> 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 <a href="https://en.wikipedia.org/wiki/Reionization" rel="noreferrer">reionization</a> 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.</p>
<p>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 <a href="https://en.wikipedia.org/wiki/Gunn%E2%80%93Peterson_trough" rel="noreferrer">Gunn-Peterson trough</a>.</p>
<p>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.</p>
<h5>Not observationally</h5>
<p>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, <a href="https://www.nature.com/articles/s41586-022-04449-y" rel="noreferrer">Welch et al. (2022)</a> (I marked them green).</p>
<p><a href="https://i.sstatic.net/pm2CV.png" rel="noreferrer"><img src="https://i.sstatic.net/pm2CV.png" alt="flux" /></a></p>
<h3>What if we pretend it's visible?</h3>
<p>We could pretend that humans <em>are</em> 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.</p>
<h5>Flux density</h5>
<p>As marked in red in the table, the detected fluxes are of the order of 20–50 nJy. A &quot;nJy&quot;, or nano-Jansky, is a measure of flux <em>density</em>, i.e. flux per frequency bin, and is equal to <span class="math-container">$10^{-32}\,\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-2}\,\mathrm{Hz}^{-1}$</span>.</p>
<h5>Flux</h5>
<p>Let's assume that the far-UV radiation that we're pretending not to be absorbed would have a similar flux density of <span class="math-container">$f_\nu = 30\,\mathrm{nJy}$</span>. The visible range of [4000–7000] Å corresponds to a frequency range of <span class="math-container">$[4.3\text{–}7.5]\times10^{14}\,\mathrm{Hz}$</span>, i.e. <span class="math-container">$\Delta\nu = 3.2\times10^{14}\,\mathrm{Hz}$</span>. Then the total flux piercing your eye is
<span class="math-container">$$
F = f_\nu\Delta\nu = 9.6\times10^{-17}\,\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-2}.
$$</span></p>
<h5>Power</h5>
<p>In the dark, your pupil expands to roughly 8 mm in diameter, so its area is <span class="math-container">$A_\mathrm{eye}\sim50\,\mathrm{mm}^{2}$</span>. The total power entering your pupil is then
<span class="math-container">$$
P = F A_\mathrm{eye}=4.8\times10^{-17}\,\mathrm{erg}\,\mathrm{s}^{-1}.
$$</span></p>
<h5>Number of photons</h5>
<p>The typical energy of a visible photon can be taken as that of a green photon with <span class="math-container">$\lambda\sim5500\,\mathrm{Å}$</span>, i.e. <span class="math-container">$E_\mathrm{ph} = h\nu = hc/\lambda = 3.6\times10^{-12}\,\mathrm{erg}$</span>. Hence, the number rate of photons entering the eye is
<span class="math-container">$$
\frac{dn}{dt} = \frac{P}{E_\mathrm{ph}} = 1.3\times10^{-5}\,\mathrm{s}^{-1}.
$$</span></p>
<h5>Waiting time</h5>
<p>Thus, the time you have to wait between each photon will, on average, be
<span class="math-container">$$
dt = \frac{1}{dn/dt}\sim1\,\mathrm{day}.
$$</span></p>

CC BY-SA 4.0
2022-04-15T20:46:58.490Z
-super answer. Thanks very much for putting that together!
Roger Wood
CC BY-SA 4.0
2022-04-15T21:46:06.383Z
Thanks for inspiring me, @RogerWood.
pela
CC BY-SA 4.0
2022-04-13T23:46:45.213Z
2022-04-13T23:46:45.213Z
pela
<p>Since a telescope is just a <em>very powerful eye</em>, 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.</p>
<p>Or is the distance so minimal that it's not even worth considering?</p>
<p>It's just a theoretical question, since I assume I will never see Earendel from my balcony.</p>

CC BY-SA 4.0
2022-04-13T15:39:28.787Z
Well, a light-year is the distance light travels in a year. I assume Earendel is 12.9 billion light years away,
rbhat
CC BY-SA 4.0
2022-04-13T20:26:00.890Z
Isn't Earendel, a.k.a. [WHL0137-LS](https://en.wikipedia.org/wiki/WHL0137-LS), currently believed to be a star, not a galaxy?
Andrew Morton
<earendel-whl0137-ls>
Earendel galaxy, 12.9 billion light-years and the human eye?
rbhat
# Earendel galaxy, 12.9 billion light-years and the human eye?

(Asked by: username_0 on 2022-04-12)

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.

**Tags:** earendel-whl0137-ls

> **Comments:**
> **username_0**: Well, a light-year is the distance light travels in a year. I assume Earendel is 12.9 billion light years away,  
> **username_1**: Isn't Earendel, a.k.a. [WHL0137-LS](https://en.wikipedia.org/wiki/WHL0137-LS), currently believed to be a star, not a galaxy?  

---

## 2 answers

### Answer by username_2 on 2022-04-12. Score: 20

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 @username_4 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 @username_4's excellent detailed answer. Earendel cannot be seen at all at visible wavelengths]

> **Comments:**
> **username_3**: And even then, imaging devices such as cameras integrate the photons over time, but our eyes/brains don't.  
> **username_2**: @username_3 This is interesting. It's about single photon detection, though I imagine it's a very noisy process. https://www.nature.com/articles/nature.2016.20282  
> **username_4**: 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.  
> **username_4**: Btw I seem to remember having heard that frogs are able to detect single photons. Crazy frogs…  
> **username_2**: @username_4 thanks, I don't know why google popped up with 4.2 m.  I'll make the corrections. My answer is misleading.  
> **username_2**: @username_4 I get one photon per year when I recalculate?  
> **username_2**: @username_4  I gather the photon efficiency of the human eye is only a few percent - maybe frogs are much better.  I assume owls and other nocturnal creatures are among the best  
> **username_4**: Your calculation is completely correct if you assume that Hubble collected 100 photons in 9 hour. But it collected somewhat more photons: The flux is of the order of 10 nJy. Converting that to number of photons in 9 hours yields ~20,000 photons. Wrt. the nocturnal creatures, you're probably right :)  
> **username_4**: 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.  
> **username_5**: Human eyes (appropriately adapted to darkness) are not much worse than the best eyes in nature. 5 or 6 yellow-greenish photons are detectable.  
> **username_4**: @user253751 Yes, googling "human eye detect single photons" return quite a few interesting reads. Cool! :)  
> **username_6**: 0  @username_4: concerning the UV vs. IR photons: if IIUC, the IR photons that Hubble sees are UV photons that were not absorbed, but became IR by the time they got to us because of the Red Shift. Whether there are any visible photons from Earendil, I don't know; extreme UV might be shifted into our visible range, I would think.  
> **username_4**: @username_6 You're right that near- and mid-UV photons may escape Earendel's host galaxy, and that they will be IR when they arrive at Earth. But photons below the ionization threshold at 917 Å are absorbed through ionization. Photons of longer wavelengths may escape the host galaxy, but will likely be scattered out of the line of sight through the Lyman α transition. So the probability of detecting photons bluer than 1216 Å × (1+z) ≈ 0.9 µm is very small, and indeed the fluxes in the optical filters are consistent with being zero.  

---

### Answer by username_4 on 2022-04-13. Score: 6

username_2 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](https://en.wikipedia.org/wiki/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](https://en.wikipedia.org/wiki/Gunn%E2%80%93Peterson_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)](https://www.nature.com/articles/s41586-022-04449-y) (I marked them green).

[![flux](https://i.sstatic.net/pm2CV.png)](https://i.sstatic.net/pm2CV.png)

### 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 $10^{-32}\,\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-2}\,\mathrm{Hz}^{-1}$.

##### Flux

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
$$
F = f\_\nu\Delta\nu = 9.6\times10^{-17}\,\mathrm{erg}\,\mathrm{s}^{-1}\,\mathrm{cm}^{-2}.
$$

##### Power

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
$$
dt = \frac{1}{dn/dt}\sim1\,\mathrm{day}.
$$

> **Comments:**
> **username_2**: -super answer. Thanks very much for putting that together!  
> **username_4**: Thanks for inspiring me, @username_2.  

---

