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?


1 Answer 1


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.

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    $\begingroup$ 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) $\endgroup$ Commented Jul 12, 2022 at 23:02
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    $\begingroup$ @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. $\endgroup$
    – jawheele
    Commented Jul 13, 2022 at 15:13
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    $\begingroup$ @pela Can't edit my old comment now, but of course I meant a division by $1+z$ $\endgroup$
    – jawheele
    Commented Jul 13, 2022 at 15:22
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    $\begingroup$ @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. $\endgroup$
    – jawheele
    Commented Jul 13, 2022 at 16:45
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    $\begingroup$ @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. $\endgroup$
    – pela
    Commented Jul 13, 2022 at 17:08

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