<h3>Probability without lensing</h3>
<p>The probability of any one photon of an ensemble of isotropically emitted photons is indeed proportional to ratio between the area of the detector, and the area of the &quot;surface&quot; of a sphere centered on the source. Note, however, that a galaxy seen 13 billion years (Gyr) back in time does <em>not</em> lie 13 billion lightyears (Glyr) away, because the Universe has expanded while the photon was traveling. When NASA says &quot;13 Glyr away&quot;, it is an (in my opinion misunderstood) attempt not to confuse people.</p>
<p>The correct distance requires a model of the expansion. For the sake of this exercise, let's assume a galaxy seen 13 Gyr back in time. Such a galaxy has been redshifted by <span class="math-container">$z\simeq6.8$</span>, and its current distance is <span class="math-container">$d=28.5\,\mathrm{Glyr}$</span> (!). If the galaxy weren't gravitationally lensed, the probability would then be
<span class="math-container">$$
\begin{array}{rcl}
P_\mathrm{no\,lens} &amp; = &amp; \frac{A_\mathrm{JWST}}{4\pi d^2} \\
  &amp; \simeq &amp; \frac{25.4\,\mathrm{m}^2}{9\times10^{53}\,\mathrm{m}^2} \\
  &amp; \simeq &amp; 3\times10^{-53}.
\end{array}
$$</span></p>
<h3>Magnification factor</h3>
<p>The probability for a lensed galaxy is equal to the &quot;normal&quot; probability, multiplied by the <em>magnification factor</em> <span class="math-container">$\mu$</span> which is the factor by which the measured flux is increased. This factor is typically of order a few, or a few tens. Large magnifications are of the order 100, and in the case of <a href="https://astronomy.stackexchange.com/questions/48993/is-there-a-better-explanation-of-this-picture-showing-the-very-distant-star-ear">Earendel</a>, the magnification is up to 100,000 (depending on which model you trust most).</p>
<p>So, if you consider a galaxy that has been lensed by a factor <span class="math-container">$\mu=10$</span>, then the probability of any of the photons emitted by that galaxy hitting JWST is of the order of
<span class="math-container">$$
P_\mathrm{lens} = \mu P_\mathrm{no\,lens}\simeq 3\times10^{-52}.
$$</span></p>
<h3>How many photons does JWST catch?</h3>
<p>That's a very small number, but luckily galaxies emit many, many photons every second. A bright galaxy in these epochs would have a luminosity of, say, <span class="math-container">$L_\mathrm{gal} = 10^{10}$</span> times that of our Sun, or <span class="math-container">$4\times10^{36}\,\mathrm{W}$</span>. For an order-of-magnitude estimate, let's assume that most is emitted in the optical, with an average wavelength of <span class="math-container">$\lambda=600\,\mathrm{nm}$</span>, corresponding to an energy of <span class="math-container">$E_\mathrm{ph}=h\nu=hc/\lambda\simeq3\times10^{-19}\,\mathrm{J}$</span>.</p>
<p>Hence, each second the galaxy emits
<span class="math-container">$$
\dot{n}_\mathrm{ph} = \frac{L_\mathrm{gal}}{E_\mathrm{ph}} \simeq 10^{55}\,\mathrm{photons\,per\,second}.
$$</span></p>
<p>Because of the expansion of the Universe, the rate at which we receive individual photons here is a factor of <span class="math-container">$(1+z)$</span> smaller than the rate at which they're emitted (thanks @jawheele for catching this).</p>
<p>Hence, in this case with <span class="math-container">$\mu=10$</span>, <span class="math-container">$z=6.8$</span>, and <span class="math-container">$L_\mathrm{gal}=10^{10}\,L_\odot$</span>, JWST would catch
<span class="math-container">$$
n_\mathrm{ph} = \frac{\dot{n}_\mathrm{ph}}{1+z} P_\mathrm{lens} \sim 400\,\mathrm{photons\,per\,second}.
$$</span></p>
<p>Note that galaxies do not in general emit photons isotropically. In the case of dusty disk galaxies, they may emit quite a lot more in the face-on direction than in the edge-on direction.</p>

CC BY-SA 4.0
2022-07-12T23:02:13.323Z
Can't comment on the science much (not my field at all) but [this paper](https://www.aanda.org/articles/aa/full_html/2015/10/aa25497-14/aa25497-14.html) on high redshift (z>2.5) galaxies has estimated luminosities and spectral energy distributions which could help baseline the above (although the numbers above seem pretty reasonable assumptions)
astrosnapper
CC BY-SA 4.0
2022-07-13T10:29:38.587Z
@astrosnapper Thanks, yes, a more precise estimate would be to integrate over the galaxy's SED/spectrum (possibly using higher-
z
 values than that 2.5–6.4 of Fangting et al.), but since both the magnification factor and the total luminosity of the assumed galaxy are rather arbitrary (albeit realistic) numbers anyway, I leave that as an exercise for the reader :)
pela
CC BY-SA 4.0
2022-07-13T11:44:50.510Z
Does it make a (big) difference when the lensing occurs? By the time the light reaches the cluster causing the lensing effect the cosmic expansion didn't progress as far as it does by now.
SpaceTrucker
CC BY-SA 4.0
2022-07-13T14:33:14.033Z
@Pela Small point in your last number: the frequency of photon emission versus photon detection would be redshifted in precisely the same way that the photons themselves are (the expansion of the universe expands the distance between successively emitted photons), so you need a division by $z$ in your final quantity of photons received per second.
jawheele
CC BY-SA 4.0
2022-07-13T14:44:01.110Z
Why would the probability be based on where the galaxy is _now_ vs. where it was when the photon was emitted? Or am I missing something about the meaning of your probability without lensing section and that's not what you're saying?
reirab
CC BY-SA 4.0
2022-07-13T15:13:19.660Z
@reirab The $~10^{55}$ photons emitted in the second together with those that JWST sees now have all travelled the same 28.5 billion lightyear distance at the current moment-- that is, those $10^{55}$ photons are evenly spread across a sphere of radius 28.5 billion light years *right now*. The proportion of them detected by JWST is thus the proportion of that sphere's area that JWST covers. Our distance from the galaxy at the time of emission is immaterial, then, because our detector isn't collecting light spread across a sphere at the time of emission, but at the time of detection.
jawheele
CC BY-SA 4.0
2022-07-13T15:22:26.290Z
@pela Can't edit my old comment now, but of course I meant a division by $1+z$
jawheele
CC BY-SA 4.0
2022-07-13T15:24:54.130Z
@jawheele I get that our distance from the galaxy at the time of emission is immaterial, but why wouldn't the _position_ of the galaxy at the time of emission be material? And why would the position of the galaxy _now_ be material? That was the source of my question.
reirab
CC BY-SA 4.0
2022-07-13T15:29:27.717Z
@jawheele Oh man, of course, thanks! Edited! :)
pela
CC BY-SA 4.0
2022-07-13T16:03:15.957Z
@SpaceTrucker Yes, it makes quite a difference. Both of the distances [us→lens] and [lens→source] enter the [lensing equation](https://en.wikipedia.org/wiki/Gravitational_lensing_formalism).
pela
CC BY-SA 4.0
2022-07-13T16:08:10.503Z
@reirab Perhaps it's easier to see from the point of view of the source: Emit 1e55 photons and wait $x$ years. The photons are then spread out over a surface with a radius equal to how far they travel in $x$ years. And how far is that? The answer is the distance $d$ calculated above.
pela
CC BY-SA 4.0
2022-07-13T16:22:51.590Z
@pela But that sphere is centered on where said photons were emitted, no? Not the current location (x years later) of the thing that emitted them? This is why I was trying to figure out the relevance of, "Note, however, that a galaxy seen 13 billion years (Gyr) back in time does not lie 13 billion lightyears (Glyr) away, because the Universe has expanded while the photon was traveling." Why does it matter where the galaxy is now? This is the part that's not clear to me. Did you mean instead that the point in space from which the photon was emitted does not lie 13 Glyr away?
reirab
CC BY-SA 4.0
2022-07-13T16:36:39.590Z
@reirab It sounds like your confusion is sourced from a common misunderstanding of the nature of the universe's expansion. The galaxy has not moved (up to rounding errors). It is in virtually the same position it was 13 billion years ago, according to the only meaningful notion of "position" available. Nevertheless, the notion of distance in the universe has stretched such that it is now some $1+z \approx 7.7$ times farther away than it was then. The relevant sphere in each spatial slice of constant cosmological time is centered on the galaxy's single, fixed position in the spatial slices.
jawheele
CC BY-SA 4.0
2022-07-13T16:45:19.543Z
@reirab Even if the galaxy is moving (e.g., has some relativistically significant peculiar velocity with respect to the CMB, which is generally unlikely), the above reasoning holds exactly when applied to the position in the spatial slices of the point of emission: that point of emission is fixed in the spatial slices and is the center of the spheres of interest, but our distance to that *fixed* point of emission has still grown by a factor of $\sim7.7$ since emission.
jawheele
CC BY-SA 4.0
2022-07-13T16:58:46.447Z
@jawheele Yes, that was indeed the source of my confusion. Thanks. It was not at all clear to me from originally reading the answer that where the galaxy is now and where it was 13 billion years ago would be reasonably interchangeable for purposes of the calculation.
reirab
CC BY-SA 4.0
2022-07-13T17:08:56.250Z
@reirab jawheele explains well, but with the risk of creating more confusing, I can add that, while a galaxy seen 13 Gyr back in time is _today_ 28.5 Glyr away, its distance from the (progenitor of the) Milky Way at the _time of emission_ was only 3.6 Glyr.
pela
CC BY-SA 4.0
2022-07-14T13:15:41.647Z
Comments are not for extended discussion; this conversation has been [moved to chat](https://chat.stackexchange.com/rooms/137781/discussion-on-answer-by-pela-what-is-the-probability-of-a-photon-from-one-of-the).
called2voyage
CC BY-SA 4.0
2022-07-12T22:10:46.027Z
2022-07-14T13:51:25.653Z
2022-07-14T13:51:25.653Z
pela
<p>In the <a href="https://www.jwst.nasa.gov/content/webbLaunch/assets/images/firstImages/image1-DeepField/STSCI-J-p22035a-4000px.jpg" rel="nofollow noreferrer">image of the SMACS 0723 galaxy</a> from the James Webb space telescope can be seen distorted galaxies whose light has taken about 13 billion years to reach the telescope.</p>
<p>On the face of it, the probability of a particular photon launched from one of those galaxies hitting the JWST detector would be the surface area of the JWST mirror array divided by the surface area of a sphere of radius 13 billion light years, <strong>BUT</strong> those galaxy images are distorted via the gravitational lens effect of the galaxy that they have passed around which would presumably affect things.</p>
<p>So, the question is, given the non-straight path of the photon and the expansion of space since it was emitted, what is the probability that a photon emitted from one of the distant, red-shifted, distorted galaxies in the image above would end up striking the JWST detector?</p>

<galaxy><james-webb-space-telescope><photons>
What is the probability of a photon from one of these galaxies hitting the James Webb detector?
DrMcCleod
# What is the probability of a photon from one of these galaxies hitting the James Webb detector?

(Asked by: username_0 on 2022-07-12)

In the [image of the SMACS 0723 galaxy](https://www.jwst.nasa.gov/content/webbLaunch/assets/images/firstImages/image1-DeepField/STSCI-J-p22035a-4000px.jpg) from the James Webb space telescope can be seen distorted galaxies whose light has taken about 13 billion years to reach the telescope.

On the face of it, the probability of a particular photon launched from one of those galaxies hitting the JWST detector would be the surface area of the JWST mirror array divided by the surface area of a sphere of radius 13 billion light years, **BUT** those galaxy images are distorted via the gravitational lens effect of the galaxy that they have passed around which would presumably affect things.

So, the question is, given the non-straight path of the photon and the expansion of space since it was emitted, what is the probability that a photon emitted from one of the distant, red-shifted, distorted galaxies in the image above would end up striking the JWST detector?

**Tags:** galaxy, james-webb-space-telescope, photons


---

## 1 answers

### Answer by username_1 on 2022-07-12. Score: 27 | Accepted answer

### Probability without lensing

The probability of any one photon of an ensemble of isotropically emitted photons is indeed proportional to ratio between the area of the detector, and the area of the "surface" of a sphere centered on the source. Note, however, that a galaxy seen 13 billion years (Gyr) back in time does *not* lie 13 billion lightyears (Glyr) away, because the Universe has expanded while the photon was traveling. When NASA says "13 Glyr away", it is an (in my opinion misunderstood) attempt not to confuse people.

The correct distance requires a model of the expansion. For the sake of this exercise, let's assume a galaxy seen 13 Gyr back in time. Such a galaxy has been redshifted by $z\simeq6.8$, and its current distance is $d=28.5\,\mathrm{Glyr}$ (!). If the galaxy weren't gravitationally lensed, the probability would then be
$$
\begin{array}{rcl}
P\_\mathrm{no\,lens} & = & \frac{A\_\mathrm{JWST}}{4\pi d^2} \\
& \simeq & \frac{25.4\,\mathrm{m}^2}{9\times10^{53}\,\mathrm{m}^2} \\
& \simeq & 3\times10^{-53}.
\end{array}
$$

### Magnification factor

The probability for a lensed galaxy is equal to the "normal" probability, multiplied by the *magnification factor* $\mu$ which is the factor by which the measured flux is increased. This factor is typically of order a few, or a few tens. Large magnifications are of the order 100, and in the case of [Earendel](https://astronomy.stackexchange.com/questions/48993/is-there-a-better-explanation-of-this-picture-showing-the-very-distant-star-ear), the magnification is up to 100,000 (depending on which model you trust most).

So, if you consider a galaxy that has been lensed by a factor $\mu=10$, then the probability of any of the photons emitted by that galaxy hitting JWST is of the order of
$$
P\_\mathrm{lens} = \mu P\_\mathrm{no\,lens}\simeq 3\times10^{-52}.
$$

### How many photons does JWST catch?

That's a very small number, but luckily galaxies emit many, many photons every second. A bright galaxy in these epochs would have a luminosity of, say, $L\_\mathrm{gal} = 10^{10}$ times that of our Sun, or $4\times10^{36}\,\mathrm{W}$. For an order-of-magnitude estimate, let's assume that most is emitted in the optical, with an average wavelength of $\lambda=600\,\mathrm{nm}$, corresponding to an energy of $E\_\mathrm{ph}=h\nu=hc/\lambda\simeq3\times10^{-19}\,\mathrm{J}$.

Hence, each second the galaxy emits
$$
\dot{n}\_\mathrm{ph} = \frac{L\_\mathrm{gal}}{E\_\mathrm{ph}} \simeq 10^{55}\,\mathrm{photons\,per\,second}.
$$

Because of the expansion of the Universe, the rate at which we receive individual photons here is a factor of $(1+z)$ smaller than the rate at which they're emitted (thanks @username_4 for catching this).

Hence, in this case with $\mu=10$, $z=6.8$, and $L\_\mathrm{gal}=10^{10}\,L\_\odot$, JWST would catch
$$
n\_\mathrm{ph} = \frac{\dot{n}\_\mathrm{ph}}{1+z} P\_\mathrm{lens} \sim 400\,\mathrm{photons\,per\,second}.
$$

Note that galaxies do not in general emit photons isotropically. In the case of dusty disk galaxies, they may emit quite a lot more in the face-on direction than in the edge-on direction.

> **Comments:**
> **username_2**: Can't comment on the science much (not my field at all) but [this paper](https://www.aanda.org/articles/aa/full_html/2015/10/aa25497-14/aa25497-14.html) on high redshift (z>2.5) galaxies has estimated luminosities and spectral energy distributions which could help baseline the above (although the numbers above seem pretty reasonable assumptions)  
> **username_1**: @username_2 Thanks, yes, a more precise estimate would be to integrate over the galaxy's SED/spectrum (possibly using higher- z  values than that 2.5–6.4 of Fangting et al.), but since both the magnification factor and the total luminosity of the assumed galaxy are rather arbitrary (albeit realistic) numbers anyway, I leave that as an exercise for the reader :)  
> **username_3**: Does it make a (big) difference when the lensing occurs? By the time the light reaches the cluster causing the lensing effect the cosmic expansion didn't progress as far as it does by now.  
> **username_4**: @username_1 Small point in your last number: the frequency of photon emission versus photon detection would be redshifted in precisely the same way that the photons themselves are (the expansion of the universe expands the distance between successively emitted photons), so you need a division by $z$ in your final quantity of photons received per second.  
> **username_5**: Why would the probability be based on where the galaxy is _now_ vs. where it was when the photon was emitted? Or am I missing something about the meaning of your probability without lensing section and that's not what you're saying?  
> **username_4**: @username_5 The $~10^{55}$ photons emitted in the second together with those that JWST sees now have all travelled the same 28.5 billion lightyear distance at the current moment-- that is, those $10^{55}$ photons are evenly spread across a sphere of radius 28.5 billion light years *right now*. The proportion of them detected by JWST is thus the proportion of that sphere's area that JWST covers. Our distance from the galaxy at the time of emission is immaterial, then, because our detector isn't collecting light spread across a sphere at the time of emission, but at the time of detection.  
> **username_4**: @username_1 Can't edit my old comment now, but of course I meant a division by $1+z$  
> **username_5**: @username_4 I get that our distance from the galaxy at the time of emission is immaterial, but why wouldn't the _position_ of the galaxy at the time of emission be material? And why would the position of the galaxy _now_ be material? That was the source of my question.  
> **username_1**: @username_4 Oh man, of course, thanks! Edited! :)  
> **username_1**: @username_3 Yes, it makes quite a difference. Both of the distances [us→lens] and [lens→source] enter the [lensing equation](https://en.wikipedia.org/wiki/Gravitational_lensing_formalism).  
> **username_1**: @username_5 Perhaps it's easier to see from the point of view of the source: Emit 1e55 photons and wait $x$ years. The photons are then spread out over a surface with a radius equal to how far they travel in $x$ years. And how far is that? The answer is the distance $d$ calculated above.  
> **username_5**: @username_1 But that sphere is centered on where said photons were emitted, no? Not the current location (x years later) of the thing that emitted them? This is why I was trying to figure out the relevance of, "Note, however, that a galaxy seen 13 billion years (Gyr) back in time does not lie 13 billion lightyears (Glyr) away, because the Universe has expanded while the photon was traveling." Why does it matter where the galaxy is now? This is the part that's not clear to me. Did you mean instead that the point in space from which the photon was emitted does not lie 13 Glyr away?  
> **username_4**: @username_5 It sounds like your confusion is sourced from a common misunderstanding of the nature of the universe's expansion. The galaxy has not moved (up to rounding errors). It is in virtually the same position it was 13 billion years ago, according to the only meaningful notion of "position" available. Nevertheless, the notion of distance in the universe has stretched such that it is now some $1+z \approx 7.7$ times farther away than it was then. The relevant sphere in each spatial slice of constant cosmological time is centered on the galaxy's single, fixed position in the spatial slices.  
> **username_4**: @username_5 Even if the galaxy is moving (e.g., has some relativistically significant peculiar velocity with respect to the CMB, which is generally unlikely), the above reasoning holds exactly when applied to the position in the spatial slices of the point of emission: that point of emission is fixed in the spatial slices and is the center of the spheres of interest, but our distance to that *fixed* point of emission has still grown by a factor of $\sim7.7$ since emission.  
> **username_5**: @username_4 Yes, that was indeed the source of my confusion. Thanks. It was not at all clear to me from originally reading the answer that where the galaxy is now and where it was 13 billion years ago would be reasonably interchangeable for purposes of the calculation.  
> **username_1**: @username_5 username_4 explains well, but with the risk of creating more confusing, I can add that, while a galaxy seen 13 Gyr back in time is _today_ 28.5 Glyr away, its distance from the (progenitor of the) Milky Way at the _time of emission_ was only 3.6 Glyr.  
> **username_6**: Comments are not for extended discussion; this conversation has been [moved to chat](https://chat.stackexchange.com/rooms/137781/discussion-on-answer-by-username_1-what-is-the-probability-of-a-photon-from-one-of-the).  

---

