# Why do some lights captured by the Webb telescope have rays and others don't?

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?

• Wikipedia has some info about Point Spread Functions. You need to be familiar with the Fourier transform to properly understand this stuff. Jul 13 at 8:49
• Actually, there are 8 rays. Jul 13 at 12:04
• @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. Jul 13 at 17:30
• @fraxinus; you are right; I was assuming that the causes of the 6 bright ones was different from the 2 dimmer ones. Jul 13 at 17:32
• Because 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. Jul 14 at 11:32

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.

• Should probably also mention why they don't appear (as obviously) around extended sources. Jul 13 at 6:21
• @shaihorowitz, it looks like your last sentence is missing a
– AnoE
Jul 13 at 11:49
• @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 Jul 13 at 18:08
• @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. Jul 14 at 16:34
• @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. Jul 15 at 9:08

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.

• Your 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. Jul 15 at 9:14
• @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 Jul 15 at 9:17
• Deconvolution is difficult in the presence of noise. Forward modelling is the method of choice. Jul 15 at 9:20
• 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. Jul 16 at 1:16

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.

• For extended sources such as galaxies the spikes are however also smeared out and thus less visible, even for bright galaxies.
– rfl
Jul 13 at 7:08
• @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. Jul 13 at 7:27
• I 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. Jul 13 at 7:59
• But 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. Jul 13 at 12:24
• en.wikipedia.org/wiki/Surface_brightness Jul 13 at 19:13

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".)

• +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. Jul 14 at 16:17

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.

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.

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))

If light spreads spherically, then why do we see these light sources spread in just four (perpendicular) angles on images?

So in addition to the other answers, the "spikes" can be caused by:

1. reflections or diffraction inside the camera from brighter sources are visible in darker areas

2. scratches/scoring on the glass that light propagates through

Why do lights appear like straight lines on a windshield of a car? (becomes more prominent at sunset and night)