Diffraction Effects and Artifacts in Telescopes like the JWST

[Devin-M]

Can you tell me how you measured the size of the images to be the same as your earlier claim was they are not finite but extend off the image?

https://www.cyberphysics.co.uk/topics/light/A_level/difraction.htm

 

[Devin-M]

Suppose we have 2 equal intensity (photon per second) point sources with different wavelengths. The detectable diffraction spikes of the redder point source will be larger in pixel radius. Or with long enough integration time and low enough noise, both will extend outside the image frame.

 

[Devin-M]

 

[andrew s 1905]

If there are no photons at all, then of course there will be no photons detected.

But if there's even a dim source, The Central Limit Theorem disagrees with you.

Just like rolling a die (as in "dice") that has an ever so slightly greater chance of landing on a particular number compared to any other number, the discrepancy can be determined with enough rolls. Even if the imperfection is smaller, it can be determined with a greater number of rolls.

In the case of a dim object viewed from a telescope, if the source's photon flux is greater than its surrounding background, and the photon's wavelengths are within the bandwidth of the receiver (within the filter's/sensor's bandwidth), and if the statistics of the system are stationary (i.e., we're not talking about a dynamical system such as a one-off flash, or something changing its behavior in an aperiodic fashion), then the photons can be detected with sufficient integration.

For a given exposure time of subframes, the pixel value of interest can be treated as a random variable with a mean (i.e, "average" value) and standard deviation. The standard deviation of the pixel value is the result of all the noise sources combined.

We can estimate the true mean by summing together the pixel values of multiple subframes, and then dividing by N, the number of subrames in the ensemble. (in other words, take the average value of the pixel).

What does that do to the standard deviation, you might ask? That is, the standard deviation caused by the combination of all noise sources after summing multiple subframes together?

The Central Limit Theorem shows that standard deviation of the averaged ensemble tends toward zero as N increases. Specifically by a factor of \frac{1}{\sqrt{N}}.

Similarly, if instead of stacking, you wanted to take a longer exposure (and are not at risk of saturation), with exposure time T, the time averaged noise (per unit signal) decreases by a factor of \frac{1}{\sqrt{T}} for all noise sources except the read noise, and then \frac{\mathrm{read \ noise}}{T} is added on as final step. [Edit: I'm admittedly kind of sloppy here. The units of time here are not seconds, but rather the fraction of some fixed time interval such as that used for individual subframes described above.]

The implication here is that by increasing total integration time, the estimated mean approaches the true mean with arbitrarily close precision, as total integration time increases.

Of course there may be practical limitations in any real world system. Of course. But saying that it's not possible, even in principle, is incorrect.

I am saying it is For All Practical Purposes not possible. I don't recall saying it was impossible in principle but given the finite life of stars I am inclined to think it is.

In your calculation there will come a point along the diffraction spike where the diffraction pattern will fall below the sky general background so your assumption is invalid.

We shall just have to disagree on this.

Regards Andrew

 

[Devin-M]

Forget stars, suppose we have 2 satellites within the field of view, each with a monochrome laser point source. The red laser has photon count per second of 10k and the blue laser 10k+1. Which has the larger diffraction spike— the brighter blue satellite or the dimmer red?

 

[sophiecentaur]

In your calculation there will come a point along the diffraction spike where the diffraction pattern will fall below the sky general background so your assumption is invalid.

This is the numb of the problem . Even the JWST is working in marginal conditions, even if the margins have been changed a lot. If they were operating the scope within those margins then they would be wasting many billions of dollars.
@collinsmark seems to be insisting that the limited model of his maths is all that needs to be considered but, in the limited situation of a 16 bit sensor and the presence of many other interfering low level sources there is a very real limit to how far the maths will follow reality. This is not a problem but it's what limits what we can see.

 

[collinsmark]

I am saying it is For All Practical Purposes not possible. I don't recall saying it was impossible in principle

Oh. "For All Practical Purposes" is what you meant by "FAPP" when you said:

the probability of detecting a photon is FAPP zero

I see now that "FAPP" is an acronym. That's the first time I've ever heard that word being used as an acronym. I thought you were shouting an expletive to emphasize your point. (To pound home your point, so to speak.)

I'm sorry. My bad. I think we are in agreement then. For practical reasons, yes, there are limitations. Of course.

By the way, if you didn't know, "fapp" has a much more commonly used colloquial meaning in the contemporary English language (i.e., slang) that I won't repeat here. 'Figure I should point that out so you know.

In your calculation there will come a point along the diffraction spike where the diffraction pattern will fall below the sky general background so your assumption is invalid.

Not necessarily, because sky background alone can be treated as a form of noise. The variation (and standard deviation) of a patch of sky background in an otherwise boring patch of sky tends toward zero with increased total integration time. That means that even an arbitrarily small blip above the background is detectable. Sure it might not have been detected in any one, given image. But such blips are detectable with sufficient total integration time.

 

[BillTre]

I see now that "FAPP" is an acronym. That's the first time I've ever heard that word being used as an acronym. I thought you were shouting an expletive to emphasize your point.

I certainly prefer it when a not obvious acronym is used that it is clearly spelled out in its first usage. It if only used once, lose the acronym, and do the work to make your communication clear.

 

[sophiecentaur]

The variation (and standard deviation) of a patch of sky background in an otherwise boring patch of sky tends toward zero with increased total integration time.

But the stars are stationary and will have their own images after a long exposure - not a grey background.

 

[collinsmark]

This is the numb of the problem . Even the JWST is working in marginal conditions, even if the margins have been changed a lot. If they were operating the scope within those margins then they would be wasting many billions of dollars.
@collinsmark seems to be insisting that the limited model of his maths is all that needs to be considered but, in the limited situation of a 16 bit sensor and the presence of many other interfering low level sources there is a very real limit to how far the maths will follow reality. This is not a problem but it's what limits what we can see.

I've said this before, but I'll try to say it again in different words. The bit depth of the sensor hardware is not the end-all-be-all of the overall bit depth that the camera is capable of achieving. You can increase the effective bit depth to some degree (sometimes to a very significant degree) though the process of stacking multiple subframes.

I think it's best now that I explain with examples. I'll start with some real-world examples and finish with a hypothetical, yet extreme example.

Here's an image of Mars that took with my backyard telescope, a couple of years ago:

The color resolution (bit depth) seems very smooth, does it not? However, the image was taken with an 8 bit sensor! The sensor only had 256 levels. Yet look at the final image. The final image is way, way higher level resolution than 8 bits. How did I do that? Integration. The final image was composed by integrating thousands of individual subframes.

Check out this Hubble Ultra Deep Field image (I didn't take this one; Hubble Space Telescope [HST] did):

HST pointed to very, very unpopulated (unpopulated by nearby stars, nebula, etc.) patch of sky, taking many individual, 20 minute subframes and stacking them to achieve this final image.

One might say, "that's impossible, all those galaxies would just blend into the background sky glow." Well, no, it's not impossible. As HST shows us here, it is possible. As I've said in a previous post, the standard deviation of the background sky glow tends toward zero with increased total integration time, and thus the background sky glow can be subtracted out.

One might also say, "that's impossible. The detail in objects so dim would be less than 1 ADU of the bit-depth of HST's sensor." Sure, some of the detail was less than 1 ADU of the sensor for a single, 20 minute exposure, but HST gained more bit depth by integrating many individual sub-exposures.

The image was taken with 4 different filters. All subexposures were approximately 20 minutes each. For the two shorter wavelength filters, 112 individual subframes were stacked for each filter. For the higher wavelength filters, 288 individual subframes were stacked for each filter.

That makes for a total integration time of over 11 days.

And no, integrating a whole 11 days worth of subframes to produce the "Hubble Ultra Deep Field" image, instead of settling on a single, 20 minute exposure, is not a waste of many billions of dollars.

I'm not just pulling math out of my butt. This is how real-world science is done. Right here.

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Now for a hypothetical, extreme example. Consider a 1 bit camera. Each pixel can represent either an on or off.

For the purposes of this example, assume that the camera has a high quantum efficiency and the camera is operating near unity gain, ~1 \mathrm{e^-}/ADU. Also, for this hypothetical example, assume the sensor's read noise is small.

Now, put the camera on a tripod and point it at your favorite sleeping kitten, where there are both bright and dark regions (maybe the cat is sleeping in a ray of sunlight from the window). Adjust the exposure time such that some pixels are consistenly black over many different subframes, some pixlels are consistently white over many different subfames, and the rest of the pixels randomly alternate between black and white from one exposure to the next, to some varying extent from pixel location to pixel location.

Now take and record 255 separate subexposures. You'll find that when analyzing the data, in the really dark regions, some pixels are black in all 255 subframes. But a few pixels are white in 1 of the 255 frames. Moving to a slightly brighter region, there are pixel locations that are white in 2 of the 255 frames. On the really bright regions, some pixels are white in all frames. But some pixels nearby are white in only 254 of the 255 frames. Others nearby are white in only 253 of the frames. In the regions that are neutral brightness, the number of white pixels seem to be consistently around 46 out of the 255 subframes.

Now sum (or average, if you store your data in floating point format) each pixel location over all 255 subframes. Blam! you've got yourself an image with a bit depth of 8 bits. You started with a camera with only 1 bit, and now you have an 8 bit image. There's black, there's white, and 256 levels of grey, total (0 to 255).

Sure, this particular image suffers quite a bit from shot noise, but you can reduce the shot noise by integrating further, and producing an image with a bit depth greater than 8 bits as a byproduct. You'll even find that by doing so, you can eke out more detail in the shadows that were previously, consistently all black.

Isn't math neat?

 

[collinsmark]

But the stars are stationary and will have their own images after a long exposure - not a grey background.

When I said "stationary," before, I meant that in a statistical sense. I mean a stochastic system that has random properties, but such that the statistics governing the randomness are not changing. In other words, things in the process can change randomly, but the statistics do not change.
https://en.wikipedia.org/wiki/Stationary_process

So long as the star's true brightness is statistically stationary, the standard deviation of its brightness (and that of any of its diffraction artifacts) in the integrated image tend to zero as total integration time increases.

Yes, this could present issues for Cepheid variable stars, depending on the star's period and the integration strategy.

 

[Devin-M]

I tried another diffraction test with 6 spikes, but only so-so results. My exposure times were limited by thin clouds and Bortle 6 light pollution. This is a composite of around 120 exposures of Polaris, about 15 seconds per exposure at 6400iso with wire over the lens to create a 6 spike diffraction pattern.

 

[andrew s 1905]

see now that "FAPP" is an acronym. That's the first time I've ever heard that word being used as an acronym. I thought you were shouting an expletive to emphasize your point. (To pound home your point, so to speak.)

My apologies it is was not my intention to shout. It is used a lot on this forum, which is where I came across it, so did not spell it out.
Again sorry for the confusion.

Regards Andrew

 

[sophiecentaur]

So long as the star's true brightness is statistically stationary, the standard deviation of its brightness (and that of any of its diffraction artifacts) in the integrated image tend to zero as total integration time increases.

This is true but hardly relevant in a situation where a single exposure is made. There is a limit to how much you can give it if you want to avoid burn out of the brightest stars.
Then, you are stuck with the 16 bit limit unless you use stacking, which is a non linear process, so all simple bets are off.

This thread has gone in circles so we really don't know what we agree about and what we disagree about but everyone seems reasonably well informed. As far as I can see, we started off with the fact that bright star images seem to have 'longer' spikes than dim ones. Somewhere in the discussion we have explained why and now we're half in and out of theory and practice. Aren't we done?

 

[Devin-M]

Well since the 1st maxima of an airy disc of red light is larger radius than the 1st maxima for blue light of equal intensities, if we’re imaging laser point sources I wonder if the spikes would be longer for a red laser compared to an equal intensity blue laser. With the question posed this way we don’t have to worry about sky background or sensor sensitivity.

When I look at this test photo I did last night, to my eye the tips of the spikes are red, then a bit inward we have green, and further inward still we have blue.

 

[sophiecentaur]

I wonder if the spikes would be longer for a red laser compared to an equal intensity blue laser.

They will have a 'length' roughly proportional to the wavelength (also the spacing of any 'dots'. That is if everything else is controlled.
But images of sharp objects in space are mostly not monochromatic so we're really looking at the colour fringing at either end of the spikes and differences hard to discern.

 

[Devin-M]

https://en.m.wikipedia.org/wiki/Airy_disk

 

[collinsmark]

This is true but hardly relevant in a situation where a single exposure is made. There is a limit to how much you can give it if you want to avoid burn out of the brightest stars.
Then, you are stuck with the 16 bit limit unless you use stacking, which is a non linear process, so all simple bets are off.

This thread has gone in circles so we really don't know what we agree about and what we disagree about but everyone seems reasonably well informed. As far as I can see, we started off with the fact that bright star images seem to have 'longer' spikes than dim ones. Somewhere in the discussion we have explained why and now we're half in and out of theory and practice. Aren't we done?

I may have been contributing to this thread trending in different direcitons because I like this thread and don't wish it to become a source of misinformation. So when there is a false or misleading claim I can't in good conscience just sit by and let that false claim go unchallenged. And it seems there are many in this thread, which is why, I assume, it's taking twists and turns.

Since you bring up the history of the particular circles, I'll try to consolidate them.

It may have started somewhere around the time of:

Of course the spacing will be the same but a fainter image will not produce such long spikes because lower amplitude dots are not recordable.

This is subtly incorrect. Yes, maybe in that particular image the dots were not recorded. In that image. Sure. But the dots are not the byproduct of the camera or any part of the imaging hardware. The dots are the result of diffraction/interference of the optics including the Bahtinov mask in this case. And those diffraction dots are, in fact, recordable. So to say that the dots are not recordable isn't true. Therefore I objected.

---
Then immediately there there was some misinformation about the properties of Full Width Half Maximum (FWHM) by another poster that has since been recanted, so I won't repeat them here. All of that got sorted out. But that did take a few posts to get through.
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Then there is another:

Your data near the peak can be considered to be linear but it eventually ends up being non-linear

This is untrue because the spikes themselves are linear all the way down to the quantum level (and the wavefunction of quantum mechanics [QM] is completely linear as far as anybody can tell for certain).

And if you're not talking about the linearity of the spikes themselves, but limitations of the camera sensor, the linearity of the camera sensor can be made linear with arbitrarily high precision by either by increasing the exposure time if saturation is not a concern, or by stacking. (And thus at this point the process of stacking entered into the discussion).

Also in the same post, we had this, separate separate idea that's not correct:

The bit depth of the cameras appears to be 16, which corresponds to a maximum relative magnitude of about 12, at which point the artefacts will disappear. That implies a star that is exposed at the limit will only display artefacts down to a magnitude of 12, relative to the correctly exposed star.

That claim is false. The 16 bit depth of the sensor is not the limiting factor. In later posts I explained why, complete with examples and some practical mathematics to achieve increased effective bit depth. But this 16 bit sensor seems to repeatedly brought up as an insurmountable limitation, even though in and of itself, it is not. Of course there are real-world, practical limitations regarding the dimmest things are possible for JWST to resolve. But this least significant step of the 16 bit ADC, by itself isn't one of them.

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I'll mention that at some point somewhere around there I had a miscommunication with another poster about FAPPs and fapping, but that ended up getting all sorted out, so I won't rehash that the details here.
---

With this next post I briefly thought we were in agreement about all of this stuff I just discussed above:

OK, I already accepted that the sensor itself is linear over a wide range and that stacking will allow a substantial increase in effective bit depth by averaging out random noise. I also know that the maths of diffraction goes on and on, as far down as you like.

So for a moment I thought that we had resolved the disagreements.

But I guess not, because soon after we have:

It's very misleading because, in a real image there is not a point source and also there are a number of other sources in the vicinity of the low level parts of a diffraction spike. this constitutes a 'floor' which can be significantly above the least significant step in the ADC.

Which again repeats the false claim that the least significant step in the ADC is some sort of insurmountable fundamental limit. It's not.

Furthermore, something new is mentioned about requiring a point source and overlapping patterns. Diffraction applies to all sources, not just point sources. Even if the orientation of JWST is such that the image of a dimmer star lies within the diffraction spike of a brighter star, it is still theoretically possible to resolve the brightness of the dimmer star, it just might take more total integration time and knowledge about JWST's diffraction characteristcs (which scientists are aware of). Will the brightness of the dimmer star necessarily be recorded in any given image? No. But it is recordable. I personally don't take serious issue with any of this until it's brought back to the ADC. Yes, overlapping diffraction patterns complicate matters for sure, but let's keep the ADC out of it.

Then this came out the blue:

So we agree on that. The only way to increase the effective number of bits is by using multiple images and you say there's no stacking.

Boldface mine.

I never said there was no stacking! Of course there's stacking! There's always at least some stacking. In the JWST image that posted which shows the stacking/overlap, I can see regions of at least 6, maybe 7 overlaps (could be more) and the majority are around 4 or more overlaps.

And here again with the 16 bit thing:

The linearity failure at low levels can destroy recorded spikes and 16 bits is where linearity fails. Some spikes are never there in 16 bit images.

I'm repeating myself. But 16 bits is not where linearity fails. It may be the case for a single subframe, but there's always at least a little stacking which increases the bit depth in the resulting image. And if science dictates, and more details is necessary, JWST could always can be told to increase its total integration time on that particular target in question and stack more subframes.

And regarding the overlap of diffraction patterns (such as diffraction spikes of brighter stars overlapping the central spot of the diffraction pattern of a dimmer star), yes that makes things more complicated, but not insurmountable. And none of it fundamentally limited by the "16 bit" aspect of the sensor.

And then this post broke my heart:

@collinsmark seems to be insisting that the limited model of his maths is all that needs to be considered but, in the limited situation of a 16 bit sensor and the presence of many other interfering low level sources there is a very real limit to how far the maths will follow reality. This is not a problem but it's what limits what we can see.

Good grief. Where to start. Not only does the tone of that post start off as snide and condescending, it's wrong. It's wrong in two ways:

1. Of course there's more to it than just the basic tenets, theorems, and implementation of information theory statistics, and mathematics. and I never said there wasn't anything more to it. But just because there is more to it, those simplistic, basic, theoretical ideas are essential considerations for this topic, and cannot be ignored if further understanding is to be achieved.
2. There we go again with the 16 bit sensor limitation. Implying that this as a fundamental, insurmountable limitation is wrong.
3. We can follow the maths all the way down to the quantum level with incredible precision. And this is true, even as applied to JWST images. As a matter of fact it's crucial, since in some cases we are talking about individual photons and their paths which are subject to diffraction and self interference patterns (not dissimilar to the double slit experiment). Implying that the maths and physics don't track reality at that level or above is wrong.

And then, since my last post, there a brand new incorrect claim:

This is true but hardly relevant in a situation where a single exposure is made. There is a limit to how much you can give it if you want to avoid burn out of the brightest stars.
Then, you are stuck with the 16 bit limit unless you use stacking, which is a non linear process, so all simple bets are off.

Gah! The stacking process is not nonlinear. With the exception of cosmic ray/hot pixel rejection and removal, (which might be considered part of the stacking process), it's linear in the

• astrophotography sense: being performed well before the "stretch" or "curves" are applied (stacking is done in the process flow immediately after calibration frames are applied to the raw data from the sensor)
• the mathematical sense: \mathcal{O}(x + y) = \mathcal{O}(x) + \mathcal{O}(y),
• and any other sense that I can think of.

Cosmic ray/hot pixel identification, rejection and removal fit in nicely as part of the stacking processes. But this isn't a necessary part of stacking. You could still do stacking without it.

Doing just the stacking alone, the mathematical operation is linear. As a matter of fact, it's mathematically equivalent (statistically speaking) to taking a longer exposure with unlimited bit depth (zero risk of causing saturation beyond that of the subframes), at the expense of additional read noise.

And this addition of read noise does not make the process nonlinear. (Any more than the addition operator is nonlinear, which it isn't.) Nor does this read noise present an insurmountable limitation if the read noise is uncorrelated/(statistically) stationary. The overall effect of the read noise can be reduced via the Central Limit Theorem, just like any other source of uncorrelated, (statistically) stationary noise source.

You're right to be concerned about saturation during a single exposure. Saturation is a primary consideration for determining the exposure length of subframes. After that, enough frames are stacked to bring about the necessary detail and signal to noise ratio (SNR) determined by the scientific needs of the object being imaged. If the science requires more effective bit depth or better SNR, more frames are stacked.

The read noise penalty for stacking is small, although it's not negligible, so it's something that is considered and dealt with. But it's in no way an insurmountable obstacle. And to the point of the last claim, there's nothing inherently nonlinear about it.

------------------------------------------------

Summary:

There are real, true, insurmountable limitations to JWST's capabilities, both theoretical and practical. I've never said otherwise. But things like the finite bit depth of the hardware sensor and the residual, thermal glow of background space are not among them.

Things such as the overlap between the diffraction pattern of one source over another: That makes things complicated (but can be dealt with). Scientists analyzing data must take diffraction into account all the time, with every image. But they do that. That's part of the process. But nothing about that is insurmountable.

There are many, many things that make gathering scientific data from JWST or any other telescope complicated. But not all of them are insurmountable. Many obstacles brought up recently in this thread, while they may have their own complications, and require careful considerations, are not truly limitations. JWST, science and math have ways of getting around them.

So what are some of the true limitations? There are many. Here are a few biggies:

• Time. Time spent gathering data on anyone given target means less time for gathering data for other targets. JWST has a finite lifespan.
• Aperture. While JWST's primary mirror may be large, it does limit the angular resolution that it can resolve. It also limits its light gathering rate (although this one points back to the previous bullet).
• Stochastic, yet (statistically) nonstationary, dim sources. If a source is changing in a non-predictable way, and those changes are occurring fast enough given how bright it is, JWST may not be able to reliably detect those changes in detail.

And while there could be more, nowhere on that list is the 16 bit thing. And stacking is not inherently nonlinear.

 

[sophiecentaur]

@collinsmark I'm sorry but your post is far too long - unnecessarily so for anyone to bother to read thoroughly. With respect, is says little more than "I (i.e. you) was right all the time."

You took exception to a comment that diffraction spikes are all different lengths. Can you show me an image full of stars with equal (displayed) spikes all over it? Perhaps the actual wording of the statement implied something to you that was not intended. No one has suggested that diffraction theory is wrong; diffraction assumes linearity.

Your list of bullet points demonstrates that you weren't actually reading what people had written and you are inconsistent. You started off say there isn't stacking and now you say there is stacking. You talk about the central limit theorem but that involves unlimited exposure time, which will, of course, saturate a sensor. You hop from 16 to 32 bit quantisation and you ignore any effect of nearby stars. 16bits (any sampling and quantising aamof) introduces non-linearity.
You state that stacking is inherently linear but a linear sum of many images is a very crude form of stacking. Each pixel of a stacked image can use and process (e.g. median) selected values from all available pixels. I can't see that is linear.
We are stuck with FAPP and the simple maths fails at some stage.

 

[collinsmark]

Your list of bullet points demonstrates that you weren't actually reading what people had written

@collinsmark I'm sorry but your post is far too long - unnecessarily so for anyone to bother to read thoroughly.

@sophiecentaur, Who's not reading what other people have written? Could you clarify that? Do you even read what you write?!

You took exception to a comment that diffraction spikes are all different lengths. Can you show me an image full of stars with equal (displayed) spikes all over it?

Sure. Right here. @Devin-M was kind enough to generate one with direct experiment. Those two stars are of different brightness (as acquired directly from the camera). But when adjusting for the intensity, it's quite clear that the diffraction patterns are of the same size.

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Edit:

Perhaps the actual wording of the statement implied something to you that was not intended. No one has suggested that diffraction theory is wrong; diffraction assumes linearity.

My objection was your statement where you said the spikes were not "detectable" (italics mine). I don't have an objection to saying that they were not detected in a particular image. But they are in fact detectable.
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You started off say there isn't stacking and now you say there is stacking.

I never said JWST does not use stacking. I kindly request that you do not put words in my mouth. I've never said that in my life. Not on PF, not anywhere.

That said, I'd be tickled to know where you think I said that. Which post and where. Show me a quote, if you would be so kind.

You talk about the central limit theorem but that involves unlimited exposure time, which will, of course, saturate a sensor.

No it doesn't. It essentially says, among other things, that the standard deviation of a set of averaged trials decreases by a factor approaching \frac{1}{\sqrt{N}} compared to the standard deviation of the original set of trials. It says absolutely nothing about saturating sensors.

You hop from 16 to 32 bit quantisation and you ignore any effect of nearby stars. 16bits (any sampling and quantising aamof) introduces non-linearity.

The input is already quantized. From the very beginning. It starts with individual, quantized photons. From there all merely count them. We sum them. Sometimes we divide the results by a number. Nothing about any of that is nonlinear.

We can put those numbers in a 16 bit register or a 32 bit register. We can store those numbers in a file using 16 bit format or 32 bit floating point format. It doesn't matter.

Saturation is nonlinear, but that doesn't happen when stacking. That is one of the primary motivations to use stacking in the first place: it avoids saturation and nonlinearity. Stacking is about as linear as linear can get.

You state that stacking is inherently linear but a linear sum of many images is a very crude form of stacking. Each pixel of a stacked image can use and process (e.g. median) selected values from all available pixels. I can't see that is linear.

What do you think stacking is?

Averaging (as in "mean" -- not median) pixels from different subframes is what stacking is. The addition operator is a very linear operator. Division (by the number of elements summed) is a linear operator. What's not linear about that?!

We are stuck with FAPP and the simple maths fails at some stage.

Where does the math fail? Show me, please.

How might you suggest informing astronomers that use JWST, Hubble (HST), and pretty much any telescope around the world, that their stacking algorithms -- algorithms that they've been using for decades -- are all failures? Do you propose invalidating the countless academic papers that relied on astronomical data that invariably was produced, in part, using same general mathematical principles and theorems discussed here?

 

[andrew s 1905]

Two stars in the image have been manipulated to have equal lengths but other stars in the field show none at all! Regards Andrew

 

[collinsmark]

Two stars in the image have been manipulated to have equal lengths but other stars in the field show none at all! Regards Andrew

The image was manipulated so they have roughly equal brightness (even though they originally were not). The adjustment was done to make it an apples to apples comparison. The equal sizes naturally fell out of that.

Here is the original image before the stars were adjusted for equal brightness. (But without adjusting for brightness, this is an apples to oranges comparison).

[Edit: I forgot to mention the credit: Image is courtesy of a post by @Devin-M ]

 

[sophiecentaur]

How might you suggest informing astronomers that use JWST, Hubble (HST), and pretty much any telescope around the world, that their stacking algorithms -- algorithms that they've been using for decades -- are all failures? Do you propose invalidating the countless academic papers that relied on astronomical data that invariably was produced, in part, using same general mathematical principles and theorems discussed here?

This is a perfect straw man argument. "Failures"? I am simply pointing out that the world of digital (discrete) sampling is inherently non-linear and that, for example, stacking using the median value of samples introduces more non-linearity. (Did I suggest that's a bad thing?) If you use any well known amateur astro software (Say Nebulosity) there are several options of stacking algorithms. One of them is based on the median of the pixel values. (I could ask you whether you know what stacking is.)

this is an apples to oranges comparison)

You would deny that it's a so-called fair test if we are discussing visibility?
If you could only afford a four bit ADC for the sensor, would you still be able to tinker with the two different parts of the image and squeeze out two equal spike lengths? Once you fall below the minimum sig bit, you have lost that information for ever.

Where does the math fail?

The above shows failure - that's just a practical issue that the basic maths doesn't consider. Is the word 'failure' too judgmental for you? I can't think of a better description for what happens in practice.

 

[Devin-M]

In this JWST image, the dimmer stars diffraction spikes (circled red) are the same length as the brighter star’s diffraction spikes, because both extend all the way to the edge of the image frame:

With long enough integration time this will visibly & detectably be the case for every single star in the image, demonstrating the spikes are the same length but different brightness.

These spikes are caused by edges and obstructions in the optical train, not the sensor or the star. See below, I used wires in front of the lens to create a 6 spike diffraction pattern.

 

[Devin-M]

In this image I took of 2 different brightness stars, adjusted to have equal image intensity (in the 1st image), we’re only seeing the first 3 orders of diffraction on both stars. With equal image intensity both have the same shape size and appearance despite having different apparent observable brightness.

This diffraction / interference pattern was created by placing a window screen in front of the lens...

On the other hand, here is a laser passing through a pinhole (a pinhole which is equivalent to a telescope’s aperature), and we’re seeing no less than 27 orders of diffraction…

https://en.m.wikipedia.org/wiki/Airy_disk

 

[Devin-M]

Also, due to shot noise, the diffraction pattern doesn't form from the inside out, but rather across the entire image frame, somewhat randomly...

 

[collinsmark]

I am simply pointing out that the world of digital (discrete) sampling is inherently non-linear

No, it is not nonlinear when the input to the system is already discrete, such as discrete photons. We're not sampling some continuous, analog signal here. We're merely counting photons. That can be done linearly.

and that, for example, stacking using the median value of samples introduces more non-linearity. (Did I suggest that's a bad thing?) If you use any well known amateur astro software (Say Nebulosity) there are several options of stacking algorithms. One of them is based on the median of the pixel values. (I could ask you whether you know what stacking is.)

Using the median values is not a good way of stacking the bulk of the data for typical purposes if you care about the fine detail of the signal. It has its niche applications though. Using the median value as the stacking is a good way of quickly wiping out noise if you don't mind taking out the fine minutiae of the signal along with it. So there are some fringe purposes for using the median (quick noise reduction where the signal detail isn't a big concern), but it's not typical.

That said, In a typical staking algorithm, median values are commonly only used for bad pixel/cosmic ray detection and elimination. Statistical outliers relative to the median (rather than the mean) in a particular subframe are replaced with median values, or simply rejected from the stacking altogether, for that subframe. So the median is important in that way. (But only for cosmic ray/bad pixel rejection -- otherwise the mean is used.)

Using the median as a reference for pixel rejection, followed by using the mean as the end result of the stacking is a very good way of eliminating bad pixels and cosmic rays, while maintaining the statistical minutia in the signal.

And yes, before you ask, this bad pixel/cosmic ray rejection is nonlinear. That's the one aspect of this that's nonlinear. But cosmic rays and bad pixels are not stationary (statistically) anyway, and almost certainly not part of the signal, so it's a good idea to get rid of them, linearity be damned. But only a tiny proportion of pixels are rejected in this way, so it doesn't negatively affect the final image all that much.

And keep in mind, without stacking, you couldn't even do that. There are no statistical outliers between subframes with only a single subframe. You'd be stuck with that bad pixel or cosmic ray. Trying to eliminate cosmic rays and bad pixels with only a single subframe is not nearly as robust.

I would love to discuss stacking algorithms such as the ins and outs of the Drizzle algorithm. (Probably in a different thread, though.) Drizzle is an advanced stacking algorithm that does everything I've described in this thread about stacking, plus quite a bit more. Many modern implementations of Drizzle are flexible, so if you wanted to keep it simple, such as the methods I discussed in this thread, you could. But you can also use its full power so long as the subframes are dithered during acquisition (commanding the telescope to point in a slightly [or even not so slightly] different direction and/or orientation between subframes).

The Drizzle algorithm is a core component in JWST's pipeline (see the section on "Imaging combination"), It's the algorithm that JWST uses for stacking.

I could discuss stacking algorithms all day. It's something I find fascinating and something I am interested in. Understanding Drizzle requires a firm grasp of basic stacking principles first. But unless we're all on the same page there, it would be like discussing Riemannian geometry without a firm grasp of Euclidean geometry. Something tells me you wouldn't be interested in reading my posts anyway though. Suffice it to say Drizzle involves stacking.

You would deny that it's a so-called fair test if we are discussing visibility?

Sure, as long as the data is scaled accordingly, equalizing the relative brightness, so that it becomes an apples-to-apples comparison, even visually.

Once again, I have no qualms with saying that spikes were not detected in some particular image. What I object to is the claim that spikes are not detectable.

If you could only afford a four bit ADC for the sensor, would you still be able to tinker with the two different parts of the image and squeeze out two equal spike lengths?

Would I "still"? I don't understand, why would I even tinker with two different parts of the same image in the first place? I wouldn't do that. I would be concerned with corresponding pixels between different subframes, not different pixels in the same subframe.

----

All else being equal, if I switched from a 16 bit ADC to 4 bit ADC, the smart thing to do is reduce the subframe exposure time to avoid saturation in the subframes. To compensate, a greater number of subframes would be taken. That includes a some extra subframes to compensate for the extra read noise. The total integration time would be a little longer, but other that it would be a statistically equivalent system.

Once you fall below the minimum sig bit, you have lost that information for ever.

No, because I've compensated for that by choosing the exposure time per subframe and number of subframes accordingly. Nothing is lost, statistically speaking.

The above shows failure - that's just a practical issue that the basic maths doesn't consider. Is the word 'failure' too judgmental for you?

"Fails" is the word that you chose, not me.

And no, there isn't any failure if you divide up the necessary total integration time between the subframe length and number of subframes wisely, which is something that you would need to do anyway. 'Unless you choose them like an idiot, and then there's failure.

 

[Devin-M]

Suppose we have 3 different monochrome laser sources of varying intensity but the same wavelength shining through a diffraction grating and we take 3 exposures of consistent exposure length and consistent processing... Since the photons are discrete & arrive randomly, we end up with three diffraction patterns, all the same angular and pixel dimension size, but varying only in brightness...

^The three lasers were different brightnesses, exposure times were the same, and all the diffraction patterns ended up the same size with identical image processing...

 

[sophiecentaur]

In this JWST image, the dimmer stars diffraction spikes (circled red) are the same length as the brighter star’s diffraction spikes, because both extend all the way to the edge of the image frame:

I think you would agree that the spikes get progressively dimmer, the further out (that's what we find with diffraction). So, if every star in that image has detectable spikes extending to the edge, there must be stars available with images that are as bright as the dimmest end of the dimmest spike. @collinsmark seems to imply saying that those 'dimmest' stars will still have spikes going out to the edge? In which case, those are not the dimmest stars available. I really don't see how anyone can argue with that logic.

I don't understand how one can say that the single photon quantisation (a quantum efficiency of 100%) is the same as the smallest step of the sensor ADC. The photon energy corresponds to the range of the frequencies involved but the digital value is what it is.

varying only in brightness..

Yes and if the light level is reduced, you can lose the faintest fringe altogether.

 

[Devin-M]

Yes and if the light level is reduced, you can lose the faintest fringe altogether.

Not necessarily. Count 10 photons from a bright source that passes through a pin hole and only 2 from a dim source. Same exposure time and the sensor is an array of single photon counters.

Suppose the 10 photons from the bright source arrive by random chance within the 0th, 3rd, 5th and 6th diffraction orders.

The 2 photons from the dim source arrive by chance within the 8th and 12th diffraction orders. The dimmer source's spikes in this case are longer than the bright source's spikes.

http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/grating.html

 

[collinsmark]

I don't understand how one can say that the single photon quantisation (a quantum efficiency of 100%) is the same as the smallest step of the sensor ADC. The photon energy corresponds to the range of the frequencies involved but the digital value is what it is.

I'll try to help you understand how that works, using examples. I'll use JWST my first example, and my own camera as a second example:

JWST:

First, you can find the specs here:
https://jwst-docs.stsci.edu/jwst-ne...detector-overview/nircam-detector-performance

Note the "Gain (e-/ADU)" and the "Quantum efficiency (QE)" in Table 1.

As you can see, JWST uses two different gain settings, one for short wavelengths and another for long.

The Quantum Efficiency (QE) is naturally dependent on the wavelength.

So let's say that were presently using JWST's F405M filter, which is a long wavelength filter with a passband at 4 microns.

Going back to Table 1, we see that
Gain: 1.82 [e-/ADU]
QE: 90% [e-/photon]We can convert that to photons by dividing the two:
\frac{1.82 \ \mathrm{[e^- /ADU]}}{0.9 \ [e^- / \mathrm{photon}]} = 2.02 \ \mathrm{photons/ADU}

So for when JWST is using its F405M filter, it takes on average about 2 photons to trigger an ADU increment.

My Deep Sky Camera: ZWO ASI6200MM-Pro:

You can find the specs here:
https://astronomy-imaging-camera.com/product/asi6200mm-pro-mono

So, for example, if I'm using my Hydrogen-alpha (Hα) filter, which as a wavelength of 656.3 nm, we can see that the QE is about 60%.

But if I set my camera gain to 0, corresponding to around 0.8 [e-/ADU], that gives

\frac{0.8 \ \mathrm{[e^- /ADU]}}{0.6 \ [e^- / \mathrm{photon}]} = 1.33 \ \mathrm{photons/ADU}

So with this gain setting and when using the Hα filter, it takes on average about 1.33 photons to increment an ADU.

(I like to set my camera gain to something higher for better read noise, but I'll just use a gain setting of 0 here for the example.)

Had I instead set the gain setting to around 25, which corresponds to somewhere close to 0.6 [e-/ADU], that would correspond to roughly 1 photon/ADU on average.

 

[sophiecentaur]

@collinsmark thanks for that information. It's interesting that the efficiency seems to be so high. However, isn't there something fundamental that means having gain too high will just increase the effect of shot noise wasting one bit of bit depth?
I do more or less take your point about equating a lsb to the arrival of one (or two) photons if the gain is suitable.

The 2 photons from the dim source arrive by chance within the 8th and 12th diffraction orders. The dimmer source's spikes in this case are longer than the bright source's spikes.

I don't understand how the result of such an arbitrary set of events would be relevant to the statistics of real events producing an identifiable spike pattern on a regular basis. Your instance doesn't produce a spike; it triggers a single pixel. You seem to suggest that a bright star would not be expected to produce a 'clearer' pattern of photons arriving at the sensor than a (very) dim one. I know that quantum effects / statistics of small numbers can sometimes give surprising results but normal principles of signal to noise ratio start to kick in at low numbers. Relative numbers of 'events' still tend to correlate with the continuum of analogue values that diffraction integrals give you.

With equal image intensity both have the same shape size and appearance despite having different apparent observable brightness.

That should be no surprise to anyone. That large image of many stars shows long spikes for many stars that are not actually near saturation but the fainter ones do not have identifiable spikes. When the absolute value of the spike pattern from a low intensity star falls below 1 increment then probability of it causing a hit gets less and less. I really don't see why this isn't obvious.

 

[Devin-M]

Here's a red diffraction pattern extending to the edge of the telescope's image circle:

Here's a blue diffraction pattern extending to the edge of the telescope's imaging circle:

Blue Channel:

Red Channel:

Here's both superimposed in the same viewing frame:

The center of the red diffraction pattern has an RGB value of 226,0,0 and the center of the blue diffraction pattern has an RGB value of 0,0,150... Which diffraction pattern is "larger?"

 

[collinsmark]

@collinsmark thanks for that information. It's interesting that the efficiency seems to be so high. However, isn't there something fundamental that means having gain too high will just increase the effect of shot noise wasting one bit of bit depth?

@sophiecentaur Yes, the gain setting does affect the noise, but it's a bit counterintuitive, and might not be what you think.

• Firstly, it's mostly all a matter of read noise. I'll get to shot noise in a second, but read noise is affected more by the gain setting.
• The read noise contribution is actually lower with higher gain, not the other way around. So if you want to lower your sensor's noise, and you're not at risk of saturation, increase the gain.
• The gain value does little to nothing to the pixel detection or the electron well. It really doesn't make the sensor more "sensitive," as one might think. Rather, in most cases it merely affects the ADC operation, which is on the back-end of the pixel, not the front.

Yeah, I know, that might sound counterintuitive that increasing the gain lowers the noise. Especially if you're a typical photographer. This same thing applies to a DSLR camera. Higher ISO settings actually produce less noise, not more.

If you tell that to a photographer friend, they will yell back, "no way! When I take a photo with higher ISO it always comes back noisier!"

But what they're forgetting, is that when they increase the sensor gain, they naturally compensate by lowering the exposure: either reducing the exposure time (a.k.a., shutter speed), lower the aperture (a.k.a., f/stop), or both. Lowering the exposure reduces the number of photons accumulated, and that is what increases the noise: in the form of shot noise, since there's less signal. And for that matter, the signal to noise ratio (SNR) is reduced because, well, there's less signal. But this entire test is an apples-to-oranges comparison. Too many variables.

To make an apples-to-apples comparison, you need to lock down the exposure (f/stop and shutter speed) and compare pictures with only changing the gain setting (ISO). If you take a photo of a dim object (to avoid saturation even at high gain) and analyze the results with both gains, you'll find the one with high gain actually has the higher SNR!

Read noise vs. gain characteristics are typically nonlinear and vary quite a bit from camera model to camera model. Look at this plot showing the read noise vs. gain for my new planetary camera, ZWO ASI585MC. (Notice that it's different than the camera from my last post. These characteristics do vary quite a bit by camera model.

One thing that seems to be true in every camera I've ever seen is while although the read noise vs gain may be nonlinear, it's always decreasing monotonically: higher gain means lower read noise.

Yeah, if you're like me, this is difficult to wrap one's head around. I've created an analogy that might help.

Imagine that you keep a pile of marshmallows (representing electrons) in a silo (the well). And you have a measuring stick (the ADC) that coverts the physical height of the marshmallow pile to number (the output), representing the height.

A rabid badger (representing the source of read noise) is always grabbing the measuring stick and yanking it back and forth and all over, so it's difficult to get a steady reading off of the measuring stick.

Now here's the neat thing: You have the power change the size of the measuring stick and the rabid badger. Increasing the "gain" setting, has the effect of shrinking the measuring stick, and the badger gets smaller too, to some extent.

This however, has some ramifications:

• By increasing the gain (and thus shrinking the measuring stick), the stick can no longer measure as high. The maximum height of the measuring stick is reduced. In other words, it's dynamic range (DR) is reduced.
• Higher gain settings shrink the size of the rabid badger too! The smaller rabid badger doesn't do as much shaking in reference to the actual, physical height of the marshmallow stack. In other words, the important measure: the amount of shaking of the stick in units of marshmallow height, is reduced.
• The noise of the output value (i.e., the final output number) might change with greater ADU numbers, but not relatively so. If you divide the standard deviation of the output number by the mean of the output number, you'll find the result is less when the gain setting is higher.
• The size of the rabid badger does not necessarily shrink the same amount as the measuring stick. If you halve the size of the stick, you don't necessarily halve the size of the rabid badger, but it does get get smaller.
• Notice that changing the gain setting really has nothing to do with the marshmallows, silo, or anything that comes before that.

Some of the increased read noise (at lower gain) might arguably be the result of higher quantization at lower gain settings (i.e., posterization). But I'm lead to believe that while posterization might play a role in explaining that, there is more to it.

Whatever the case, here's a word of advice: if you're far from saturation, and for what ever reason you don't want to or can't increase the exposure time, increase the gain instead to maximize SNR.

 

[sophiecentaur]

Which diffraction pattern is "larger?"

The colour images could be misleading because our sensitivity / perception of colours is a 'human quality. Personally, I can't see the blue fringes at all clearly on my phone. An astrophotographer, using a monochrome camera would be better placed to come to a meaningful conclusion. If the monochrome images do not follow the maths then there's something wrong, once they've been normalised. If there's a ratio, of two in the wavelengths then there should be a factor of two in the spacing (for small angles, at least).

 

[Devin-M]

Blue Red Channel:

Red Channel:

Assume they both go all the way to the edge but the same color. One is dimmer, which is “larger?”

 

[Devin-M]

Here's some proof light can diffract across an entire sensor on the JWST...

Or even 2 sensors...

 

[sophiecentaur]

which is “larger?”

Sorry, I don't get where you are coming from. If they both 'go to the edge' then the edge defines their extent. What does this demonstrate?
The theoretical curve never goes to zero.

Here's some proof light can diffract across an entire sensor on the JWST..

The word "proof" implies that there is some doubt. You notice the star centre is well burnt out; there is no limit to how far out the spike could go. The (reflecting) struts supporting the reflector are the long line sources and the pattern from a line will be of the form (sin(x)/x)² in the direction normal to the strut. Their reflecting area is pretty significant. The envelope falls off as 1/x², which takes a lot of peaks before it starts to disappear.

 

[pinball1970]

This is latest.

Dwarf galaxy WLM from NIRCAM. It is near and looks very spread out.
I do not think there is an image from Andromeda as yet, that is the nearest? So the details from that will be impressive too. Individual stars from the galaxy.

https://blogs.nasa.gov/webb/2022/11/09/beneath-the-night-sky-in-a-galaxy-not-too-far-away/

A video zoom in.

https://webbtelescope.org/contents/media/videos/01GH1WV11D7J5KPTT50ZQQ5TBA

 

[Devin-M]

Followup question. Above I’ve taken the same image of the same star and split the image into just the red light (red channel) on the left and just the blue light (blue channel) on the right. We can see the apparent position of the dimmer stars with no visible diffraction spikes are the same in both images (bottom left, top right, and the one just below the right side diffraction spike), but the points of maximum brightness of the diffraction spikes are in different positions depending on color.

Does it mean that only photons of the same color/ wavelength interfere with each other?

For example suppose I use a red monochrome laser as a point source viewed through my aperture, and note the distances of the maxima from the apparent source. Next I aim at a star (broadband) and take an exposure sufficiently long to see the same number of maxima, and look at the red channel. Will the red spots be same distance from the apparent position in both cases, even though in the star/broadband case there are additional wavelengths of light traveling through the aperature? Will the blue wavelengths for example change the apparent positions of the red interference maxima?

 

[sophiecentaur]

Does it mean that only photons of the same color/ wavelength interfere with each other?

The mechanism of interference relies on two (or more) sources of precisely the same wavelength. If a red photon is detected at the same time as a blue photon then the interference patterns associated with each wavelength will be totally different; they are independent.

I suggest you read around about the mechanism of interference / diffraction and you will find that photons do not come into it. It is always described in terms of waves. (I think this has already been pointed out in this thread.)

 

[Devin-M]

Is it true that the dark areas in the spikes are destructive interference and this is only possible because starlight has both spatial and temporal coherence? (In other words, if a point source did not have spatial and temporal coherence would the dark areas in the spikes vanish?

 

[sophiecentaur]

How could the spatial coherence not be perfect for a point? Do you know what coherence means?

 

[Devin-M]

Well from the below source, light from the sun isn’t coherent, but light from the stars is apparently, so presumably light from the sun would be coherent if viewed from a great distance, but I’m not sure why a non coherent source becomes coherent when viewed from increasing distance.

https://www.mr-beam.org/en/blogs/news/kohaerentes-licht

 

[sophiecentaur]

the light from a star is hardly monochromatic so how can it be “coherent’? Diffraction fringes are not visible unless you filter the light.
But coherence is a quantity with a continuum of values; everything from excellent to zero.

 

[Devin-M]

I thought destructive interference was causing the dark regions in the spikes, but isn’t coherence a prerequisite for destructive interference?

 

[sophiecentaur]

I thought destructive interference was causing the dark regions in the spikes, but isn’t coherence a prerequisite for destructive interference?

Variations, not just minima require some degree of coherence but it's not just on/off. As coherence decreases, so the patterns become less distinct. You will get a hold of this better if you avoid taking single comments that you hear / read and try to follow the details of the theory. It may be that the argument you heard relates to twinkling stars. Fair enough but the details count and a statement about one situation may not cover all cases.

 

[Devin-M]

In this video he’s able to obtain a destructive interference pattern with just sunlight & a double slit… (3:29)

 

[sophiecentaur]

In this video he’s able to obtain a destructive interference pattern with just sunlight & a double slit… (3:29)

And your point is?
The equipment he uses is introducing some coherence with a collimating slit which is causing the formation of diffraction effects. You don't only get diffraction effects with a high quality laser.
And I think we have gone past the stage, in this discussion, where we should be quoting links showing ripples on water. To discuss the limitations imposed on the basic calculations of diffraction, you need to introduce the concept and effects of coherence.
The coherence length of any source is the mean length over which all the individual bursts of light exist - very long for a good laser, short for a discharge tube and even shorter from a hot object. Interference will only occur when the path differences involved are shorter than the coherence length. That will be about boresight when you use sunlight.

 

[mfb]

Starlight is coherent over the size of your setup. That includes sunlight from a given direction - but an unfiltered Sun will come from an angle of half a degree, usually that makes the blob of sunlight much wider than any interference patterns.

Does it mean that only photons of the same color/ wavelength interfere with each other?

Light is linear so you can always treat interference as something that happens photon by photon (the double-slit experiment has been done with single-photon sources, and also with electrons). If each photon only interferes with itself then it's obvious why different colors (different wavelengths) can have different patterns.