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Due to the space expansion, the universe should act as a giant telescope. The farther objects are from us (beyond the redshift of $z\approx 1.5$), the larger (not smaller!) they should appear in the sky. Per Wikipedia:

an object "behind" another of the same size, beyond a certain redshift (roughly z=1.5), appears larger on the sky

This optical effect is explained by the Angular Diameter Distance decreasing with distance in an expanding universe, whether open, closed, or flat:

enter image description here

The smaller the angular diameter distance, the larger the object appears in the sky. Therefore, galaxies of the same size should appear the smallest in the sky at the redshift of about $z\approx 1.5$. The galaxies that are closer to us should naturally appear larger, because they are closer, but, counterintuitively, the galaxies that are farther away from us also should appear larger due to the optical effects of the space expansion. Thus the universe should act as a giant telescope magnifying distant objects the more the further they are.

However, when we look at the most distant objects discovered by the Hubble telescope with the redshift ranging from $z=8.6$ to $z=11.9$, they all appear extremely small on the background of other also very distant (but not as extremely distant) Ultra Deep Field galaxies:

enter image description here

What is the explanation of why more distant galaxies appear the smallest on the Ultra Deep Field image while they should appear the largest according to the angular diameter distance formula?

$$d_A=\cfrac{c}{H_0 q^2_0} \cfrac{(zq_0+(q_0 -1)(\sqrt{2q_0 z+1}-1))}{(1+z)^2}$$

Qmechanic
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safesphere
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  • It seems to me that you are assuming that the UDF is not enlarged. How big do you think the galaxies should appear in the sky without such enlargement? – Kyle Kanos Aug 14 '18 at 11:43
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    @KyleKanos The linear size of the Hubble UDF image is about one tenth of the diameter of the moon or 11.5 square arc minutes. It would take 13 million such images to cover the entire sky. This animation shows the relative magnification of the image: https://m.youtube.com/watch?v=K5ZbrDJYP-c - The question is about relative sizes of the galaxies depending on their distances, not about the absolute magnification. – safesphere Aug 14 '18 at 14:40
  • i know how big the UDF is, your mentioning it is a non sequitur. I believe I do understand what your question is, that's why I made the comment. You seem to think that the UDF is smaller than it should, I'm really asking why you think that. – Kyle Kanos Aug 14 '18 at 14:46
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    @KyleKanos No, I am asking about relative sizes of more and less distant galaxies on the same UDF image, not about the size of the entire UDF. The most distant galaxies shown in little squares with the $z$ indicated next to them should, according to the ADD formula, appear larger than (at least some of) the less distant (unmarked) galaxies on the same image, but this does not seem to be the case. Essentially I wonder if the ADD concept as described has any experimental proof, whether on this image or on any other image. – safesphere Aug 14 '18 at 15:21
  • so insert 'galaxies within the' in front of UDF and you get the same idea. Does the ADD formula say that the objects appear larger than they should or that each further object appears larger than nearer objects? – Kyle Kanos Aug 14 '18 at 15:32
  • @KyleKanos The latter. For any two distant ($z\gt 1.5$) objects of the same physical size, the one that appears smaller in the sky is closer to us. – safesphere Aug 14 '18 at 17:11
  • Seems to me there's a fundamental flaw in the question. If, as you say, "The smaller the angular diameter distance, the larger the object appears in the sky", then these $z=8, 9, 10$ galaxies should appear smaller than everything except $z < 0.2$ (or so) galaxies, per the chart you gave. So then the question becomes whether the Hubble UDF contains $z < 0.2$ galaxies, which is pretty clearly "no". – Allure Dec 20 '19 at 00:33
  • @Allure On the chart, higher points correspond to a smaller size visible in the sky. A galaxy of the same size appears the smallest in the sky at $z\approx 1.5$. At a larger $z$ (farther away), the same galaxy should appear larger. So the high $z$ galaxies should appear larger on the photo than the lower $z$ galaxies, but they do not. There is no flaw in the question. This matter is just very counterintuitive and can be confusing. – safesphere Dec 20 '19 at 07:34
  • @safesphere At https://scitechdaily.com/spin-doctors-vast-cosmic-filaments-and-galaxy-rotation-video/ , there's discussion of a recent study relating the orientation of galaxies to the rotation axes of galactic filaments (with the axes of rotation of the larger galaxies being orthogonal to the rotation axis of the nearest filament, while those of the smaller galaxies aren't), which might help to explain the theory/observation divergence you've noticed. Nikodem Poplawski's torsion-based cosmology, which implies scale reductions between local iterations of a multiverse, might also explain it. – Edouard Jan 20 '20 at 18:51
  • There's only a 360 degrees in a circle. Likewise there's only so many steradians in a sphere. So I don't see how the whole sky can be magnified, which seems to be what you talking about. To magnify something is to make it (or rather its image) subtend a larger angle at the eye. "Appear larger" is vague. – Matthew Christopher Bartsh Aug 10 '22 at 05:26
  • @MatthewChristopherBartsh Yes, "appears larger" in the Wikipedia article (as of 4 years ago) referred to a larger angle at the eye for more distant galaxies compared to those closer to us of the same physical size. – safesphere Aug 10 '22 at 06:02

1 Answers1

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Two reasons, both to do with comparing like with like.

  1. Measuring the angular diameter of a galaxy is not a trivial matter. Where do you consider the edge of the galaxy to be? Whilst the angular diameter distance decreases at high redshift, the luminosity distance increases as $(1+z)^2 d_A$ and the flux from a galaxy decreases as $(1+z)^{-4} d_A^{-2}$, so very distant galaxies get faint very quickly. The dynamic range achievable in the imaging of "galaxies" at $z\sim 10$ is extremely limited, so that essentially only their bright nuclei or brightest starburst regions are seen above the background.

  2. It is unlikely that galaxies in the present day, or even at $z \sim 2$ are anything like the galaxies observed at $z>6$. The highest redshift objects are either hugely powerful quasars (at $6<z<8$) at where most of the light comes from a compact nucleus, or from starbursts in small proto-galaxy fragments that eventually merge to form the larger galaxies seen in the universe today.

For example, the record holder (or it was at some point), GN-z11 at $z=11.1$ was found to have a half-light radius of only 0.6 kpc (about 20 times smaller than the Milky Way) but to have a star formation rate about 20 times higher than the Milky Way (Oesch et al. 2016).

ProfRob
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  • I see your points on the limited precision. Still it would be an interesting field of research. It would also apply to structures larger than galaxies. For example, if the geometry is as described, more distant (and therefore larger) structures should appear not more, but less uniform. Thanks for your answer! +1 – safesphere May 03 '20 at 17:01
  • @safesphere the CMB is an example. The spatial frequency of the first acoustic peak gives the expansion rate of the universe and it is (shall we say) within 10% of the value estimated from the more local universe. – ProfRob May 03 '20 at 18:43