On the images captured by Webb telescope one can see some lights with 6 rays, but most others don't have any. One would expect the optics to transform all light sources at infinity in the same manner. What causes these differences?
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4Wikipedia has some info about Point Spread Functions. You need to be familiar with the Fourier transform to properly understand this stuff. – PM 2Ring Jul 13 '22 at 08:49
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22Actually, there are 8 rays. – fraxinus Jul 13 '22 at 12:04
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4@PM2Ring, I am familiar with FT of course; this is a basic stuff everybody interested in Math or Physics or Engineering knows. And I understand how aperture could cause such rays. What I don't understand is why in the same picture with the same optics SOME point sources project with rays and OTHERS don't. – Michael Jul 13 '22 at 17:30
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3@fraxinus; you are right; I was assuming that the causes of the 6 bright ones was different from the 2 dimmer ones. – Michael Jul 13 '22 at 17:32
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2Because of overload. That bright object (nearby star?) has saturated the detector at its centre by orders of magnitude, and that's why its diffraction artifacts are so visible. It's the faint objects which are of interest. Note bottom centre and top right, there are other objects bright enough to cause visible diffraction artifacts but less extensively. – nigel222 Jul 14 '22 at 11:32
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@fraxinus technically, there are 12 rays, but 4 of them are hidden behind larger rays – Topcode Jul 14 '22 at 18:03
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The indispensable Scott Manley has a video about the images, including an explanation of the diffraction patterns, starting at 05:53. with some visual presentation looking as if they come from the Space Telescope Science Institute via twitter, https://mobile.twitter.com/SpaceTelescope. – Peter - Reinstate Monica Jul 15 '22 at 12:40
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To see the same effect in real life , try to look at a point source (a bulb) with your eyes squeezed.. you are gonna see two of those lines... – Ankit Jul 15 '22 at 16:50
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As the others have super perfectly explained, those are diffraction spikes. I can't really add much to that, but I found this excellent image explaining the spikes of the James Webb, so I though I could share it, in case you are curious about the particular shape: https://stsci-opo.org/STScI-01G6933BG2JKATWE1MGT1TCPJ9.png – Elkhantar Jul 15 '22 at 17:14
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In what lit scenario would you not expect your lens to show that effect? – Robbie Goodwin Jul 15 '22 at 19:11
7 Answers
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These are diffraction spikes. They are an interference pattern caused by the the arms and shape of the telescope. 
They occur around whatever is bright enough in the image which in this case is all the stars that are within the milky way. These stars show up super bright because the Webb is trying to look for super dim objects in the deep field.
The spikes very near the star (horizontal and two diagonal ones) are the three arms and the inverted image of those 3 arms.
You can think of them as a type of "shadow" the telescope is casting.
The larger lines which form the 6 points are caused by the non circular shape of Webb also causing an interference pattern.
The geometry of the arms was chosen so that the diagonal spikes from the arms line up with the spikes from the shape of the telescope. This was done to minimize the effect.
It is worth noting that all the galaxies also have diffraction spikes but they are much dimmer due to the amount of light and diffuse due to the non point like nature of the extended objects. In particular the interference pattern is a type of Point Spread Function. As Prof Rob explains this is a type of Fourier Transform done on the light due to the missing modes from the shape and arms of the telescope. A more spread out object exhibts these Point Spread Functions much less because the missing information is recovered from a slightly different angle.
shai horowitz
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18Should probably also mention why they don't appear (as obviously) around extended sources. – Kyle Oman Jul 13 '22 at 06:21
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1It's probably because dot-like sources line-of-sight is hidden behind arms meeting point (black plate at the top), so in this case diffraction becomes a dominating process. For extended objects (like galaxies), bigger part of light goes round-about this plate, so line-of-sight is less important to them. If that plate (and arms) could be made fully transparent,- I suspect diffraction pattern should vanish at all. – Agnius Vasiliauskas Jul 13 '22 at 07:38
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1@shai horowitz You have it the wrong way round, the major, evenly spaced, six-pointed pattern is from the mirror shape and the smaller lines are from the struts: https://webbtelescope.org/contents/media/images/01G529MX46J7AFK61GAMSHKSSN – smernst Jul 13 '22 at 08:20
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thanks for the comments are corrections. I have edited the answer to reflect them. – shai horowitz Jul 13 '22 at 10:12
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1I understand that aperture and such could cause diffraction rays. What I don't understand why the rays appear for SOME point sources at infinity, but not OTHERS. For astronomers the distance to a Milky Way star and to a distance cluster of galaxy is different, but for optics they are the same: both at field distance infinity. If the optics, including the arms of the telescope and its mirror shape, acts in identical manner on all light objects, why SOME got rays and OTHERS didn't? – Michael Jul 13 '22 at 17:53
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2@Michael if they were truly at infinity then they would be points. we wouldn't be able to see any features. I can see spiral arms and many features so the galaxies are clearly spanning some small angle of the sky. this angle makes the diffraction spikes more diffuse but again of primary importance is that these objects are orders of magnitude apart in terms of brightness – shai horowitz Jul 13 '22 at 18:08
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2@Michael, if you want to get really technical, look up the definition of radiance. Even if the telescope receives more total light from a certain distant galaxy than it receives from a certain star in our own galaxy, the radiance of the star can be much higher because of its apparent, nearly infinitessimal size in the sky. The image of the star is a single point, much brighter than any point in the image of the galaxy, and its diffraction pattern is equally bright and sharp as compared to the galaxy. – Solomon Slow Jul 14 '22 at 16:34
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2@Michael I suppose the foreground stars are orders of magnitude brighter than the far-away galaxies. A printed or displayed image cannot communicate that information: There is only one white, so you are naturally left wondering: Why does one white point have diffraction spikes but the other one does not? But the actual photon density, if you want, from foreground objects is so high (a thousand times higher? a million times higher? visible to the naked eye!?) that the pretty minimal light from diffraction, arms etc. is still bright enough to visibly register; not so with the dim sources. – Peter - Reinstate Monica Jul 15 '22 at 09:08
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Any image is the convolution of the intrinsic image of an object and the point spread function of the telescope, which is (in broad terms) the Fourier Transform of its aperture and any obscuring supporting structure for the secondary mirror.
Even the nearest stars appear as point sources to JWST, so their images are essentially just the Fourier Transform of the aperture/telescope. The tiled, hexagonal structure and secondary mirror support of JWST means that this looks like a 6-pointed object, with the brightness of these "diffraction spikes" decreasing rapidly with distance from the true position of the source. The main 6-pointed pattern is due to the hexagonal shape of the primary mirror, combined with the three-strut secondary support. There is also a horizontal spikelet that is just due to the mirror support (e.g., see here).
In contrast, most galaxies are not point sources, they are resolved into images of finite size by JWST. These are also convolved with the 6-spike point spread function, but the finite size of the sources blurs it out so it can't be obviously seen in the final image. i.e. The spikes contain some small fraction of the brightness and are very narrow. In a galaxy image, this small fraction is spread across many pixels and becomes much less obvious.
In pictures, another factor in how prominent the diffraction spikes appear (around stars) will depend to some extent on how brightness in the image is scaled with respect to true brightness. Often the images are scaled to enhance the appearance of fainter stars/galaxies and this makes the diffraction spikes more prominent, especially around the brightest stars in an image.
ProfRob
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2Your referral to Fourier Transform is just a "metaphor" in the sense that the effect of the optics amount to what one could compute with FT, right? Because if the image was actually computed one could probably take geometry and arms etc. into account and compensate for them computationally. And for some reason I wonder why one cannot post-process an image that way: The geometry of the mirror is known, so one could remove the known "erroneous" light coming from diffraction of the bright sources. It probably masks some low-amplitude real data which is lost forever, but still. – Peter - Reinstate Monica Jul 15 '22 at 09:14
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@Peter-ReinstateMonica The effects are not due to geometric optics alone. I don't know the details, but when dealing with a diffraction pattern I normally start with a Fourier transform – ProfRob Jul 15 '22 at 09:17
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2Deconvolution is difficult in the presence of noise. Forward modelling is the method of choice. – ProfRob Jul 15 '22 at 09:20
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Note if you look close at the upper right part, you can see what look to be a couple of galaxies that have fuzzier and less well-formed spikes with them. – The_Sympathizer Jul 16 '22 at 01:16
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Every source of light has their spikes in the image. Even extended galaxies have broad and washy spikes.
Why are they not always visible then?
The only difference is the surface brightness of the sources. The point spread function of the telescope aperture spreads the source brightness always over the same pattern, regardless of its brightness. This point spread function has a gigantic main lobe that contains almost all of the intensity and the spike intensity is lower by many orders of magnitude. Therefore, only bright sources gather enough detector counts in the spikes to make them discernible over the surrounding intensity.
For less bright sources, the spike intensity is just so low, that we cannot see them.
tobalt
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3For extended sources such as galaxies the spikes are however also smeared out and thus less visible, even for bright galaxies. – rfl Jul 13 '22 at 07:08
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3@rfl ... which is what I wrote in the second sentence. Anyway, I inserted a rhetorical question to better section my huge wall-of-text and follow along its twisted logic. – tobalt Jul 13 '22 at 07:27
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3I upvoted this answer because I think it concisely gets to the key difference: colloquially speaking, the "brightness", but perhaps more appropriately, the "surface brightness" or the incident flux. When taking flux into consideration, the extended nature of the sources is indeed relevant, but a pedantic reader might also consider other factors such as contrast or the sensitivity of the instrument. Hopefully this helps improve the answer. – Alwin Jul 13 '22 at 07:59
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1But it's the surface brightness that matters, not the brightness. The surface brightness of a star is the same as ther Sun. It is many, many orders of magnitude larger than that of any galaxy, except perhaps a quasar. – ProfRob Jul 13 '22 at 12:24
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@Alwin ah I see. This is what I meant. I'll edit it. So the apparent brightness then is indeed the integrated surface brightness over the extent of the object.. i.e. the half moon has the same surface brightness as the full moon, but only half the apparent brightness. I incorrectly thought that surface brightness meant the absolute (instead of apparent) surface brightness. – tobalt Jul 13 '22 at 19:35
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As tobalt wrote, it's all about surface brightness. But the crucial bit, which is not apparent enough, is just how much brighter these stars appear than the further away galaxies. If JWST had taken the picture with lower exposure it could have captured this more faithfully, perhaps like this:
...but that wouldn't be very useful for scientific purposes and certainly not look as inspiring.
Due to a combination of overexposure of the sensor itself, and post-processing, we see much more information in the actual published image – but at the price of discarding the direct information about the foreground stars' brightness, and also of blowing the intensity of their diffraction spikes completely out of proportion. (Which does of course provide a way of estimating the true brightness again, as well as a visual impression of their "brilliance".)
leftaroundabout
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5+1 Actually this is the only answer that goes straight to the point. The stars are grossly oversaturated, so even their relatively weak diffraction is apparent. – dominecf Jul 14 '22 at 16:17
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On the images captured by Webb telescope one can see some lights with 6 rays, but most others don't have any
They have, simply bigger blobs of light (galaxies, etc) outshines diffraction pattern. Take a look at some section of pic
which was zoomed-in, then edge detection algorithm was performed and white balance was automatically corrected.
Agnius Vasiliauskas
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Every object in this photo is displaying "spikes." The question begins with the false premise that "some...have rays and others don't."
Some rays are bright, while some rays are dim and diffused. The intensity of the rays depend on the brightness of the light source. They also depend upon the relative diameter in the photo of the light source. Larger sources such as galaxies have diffused rays, seeming to not display rays to the naked eye observer.
Closer examination of the photo show every light source to be producing the rays.
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I would like to add a few things to the nice answer by @profrob, because I recently had a similar question:
The "cross" pattern is due to light reflecting off other surfaces in one of two ways. Most likely it's either an artifact of your camera imaging system where reflections or diffraction inside the camera from bright sources are visible in darker fields (https://petapixel.com/2018/05/19/the-physics-behind-sunbursts-and-how-it-can-help-you-focus-your-photos/), ...or it's due to scratches/scoring on a window you're looking through (Why do lights appear like straight lines on a windshield of a car? (becomes more prominent at sunset and night))
So in addition to the other answers, the "spikes" can be caused by:
reflections or diffraction inside the camera from brighter sources are visible in darker areas
scratches/scoring on the glass that light propagates through
Árpád Szendrei
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