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In the image of the SMACS 0723 galaxy 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?

DrMcCleod
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1 Answers1

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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, 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 @jawheele 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.

pela
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    Can't comment on the science much (not my field at all) but this paper 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 Jul 12 '22 at 23:02
  • @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 Jul 13 '22 at 10:29
  • 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 Jul 13 '22 at 11:44
  • @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 Jul 13 '22 at 14:33
  • 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 Jul 13 '22 at 14:44
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    @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 Jul 13 '22 at 15:13
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    @pela Can't edit my old comment now, but of course I meant a division by $1+z$ – jawheele Jul 13 '22 at 15:22
  • @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 Jul 13 '22 at 15:24
  • @jawheele Oh man, of course, thanks! Edited! :) – pela Jul 13 '22 at 15:29
  • @SpaceTrucker Yes, it makes quite a difference. Both of the distances [us→lens] and [lens→source] enter the lensing equation. – pela Jul 13 '22 at 16:03
  • @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 Jul 13 '22 at 16:08
  • @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 Jul 13 '22 at 16:22
  • @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 Jul 13 '22 at 16:36
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    @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 Jul 13 '22 at 16:45
  • @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 Jul 13 '22 at 16:58
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    @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 Jul 13 '22 at 17:08
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