APOD: Webb's First Deep Field (2022 Jul 13)

[Chris Peterson]

Infinity anyone?

Who's to say, through curved space, we aren't looking at ourselves at our distant past.

Not possible. The edge of the observable universe represents a horizon we can't see past.

[Chris Peterson]

Nice. What surprised me most is that the James Webb's images from its hexagonal mirrors still produces rectangular images with perfect 90 degree corners. Unlike my binoculars that has round lenses and it produces a round image. The explanation was for the old round camera lenses producing square images was because that is how the photo negative film was made. And of course science text books were rectangular and demanded rectangular photos. But nowadays when paper books are on the way out and most images are processed by computer there is no reason to still cut the round or hexagonal images into squares.

Like essentially all optical systems, the image produced by the telescope is round. The images as we see them are rectangular because the sensor is rectangular (like film in most cases), clipping out a section of the full image. The shape of the mirror or its segments doesn't matter.

[johnnydeep]

Nice. What surprised me most is that the James Webb's images from its hexagonal mirrors still produces rectangular images with perfect 90 degree corners. Unlike my binoculars that has round lenses and it produces a round image. The explanation was for the old round camera lenses producing square images was because that is how the photo negative film was made. And of course science text books were rectangular and demanded rectangular photos. But nowadays when paper books are on the way out and most images are processed by computer there is no reason to still cut the round or hexagonal images into squares.

Aren't the round images simply automatically cropped by the film or sensor into the middle rectangular portion? And CCD sensors are also rectangular as far as I know. Are you saying why not have round sensors that precisely match the round image created by the lens? (I assume Webb still ultimately uses round lenses to focus the light from the array of hexagonal mirrors.)

[VictorBorun]

Ok, there is WAY too much stuff to talk about in this FIRST image from Webb. Hubble is now just a pile of garbage in comparison

A technical question: I don't understand why this linked-to pic of the spectra of two very similar looking lensed images proves they are one and the same galaxy:

Click to view full size image

Why couldn't the two similar spectra be from two separate but still very similar galaxies at the same distance?

I take it they mean the same redshift, hence roughly the same distance.
And the same chemistry also counts, yes.
And both seem to lie in the same smooth arc going around the midground massive cluster in the first place

[VictorBorun]

These images are, I suppose, false-colour representations of infrared data. How much has the spectrum been shifted upwards - two octaves? ten?

there is a legend with RGB color codes for near infrared (NIR) and for mid-infrared (MIR)



Full Res, 4537 X 4630 here

"W" means wide band filter, the number is in .01 μm
Those wide bands are:
near infrared:

mid-infrared:

[Keyman]

https://imgs.xkcd.com/comics/the_best_camera.png

[Jim Leff]

These images are, I suppose, false-colour representations of infrared data.

Which, in turn, seems to contradict the rather fussy replies I got earlier this year when I asked ( http://asterisk.apod.com/viewtopic.php ... 30#p320630) whether Webb output could be usefully depicted in visible spectrum.

The answer, apparently, should have been “Yes, absolutely, no problem”

[Fred the Cat]

If JWST looks far enough back, would it begin to see a set of the same first galaxies with the CMB beyond - where ever it is pointing?

Maybe that’s the obvious “point” I hadn’t imagined to be a JWST mission.

[johnnydeep]

If JWST looks far enough back, would it begin to see a set of the same first galaxies with the CMB beyond - where ever it is pointing?

Maybe that’s the obvious “point” I hadn’t imagined to be a JWST mission.

Hmm. I presume you mean that because back when the universe was a mere 380,000 years old (limited by the time of "last scattering"), it was much smaller. But even then, it would only see the galaxies on the edges of the limiting sphere, which even at a young 380,000 years old, would still be much larger than any single galaxy. Presumably. So, just how big was the universe at 380,000 years young?

[johnnydeep]

Ok, there is WAY too much stuff to talk about in this FIRST image from Webb. Hubble is now just a pile of garbage in comparison

A technical question: I don't understand why this linked-to pic of the spectra of two very similar looking lensed images proves they are one and the same galaxy:

Click to view full size image

Why couldn't the two similar spectra be from two separate but still very similar galaxies at the same distance?

I take it they mean the same redshift, hence roughly the same distance.
And the same chemistry also counts, yes.
And both seem to lie in the same smooth arc going around the midground massive cluster in the first place

Still not convinced. Can't two different galaxies at the same distance show pretty much the same spectra (i.e., "chemistry")?

[MarkBour]

I like the blob nearly at the bottom, a little right of center:

It probably already has a name.
If not, I'd propose calling it the Anglerfish Galaxy Group

Capture3.PNG (100.94 KiB) Viewed 5509 times

[AVAO]

The level of detail in the WEBB images is absolutely stunning. What surprises me is the fact that the strongly redshifted galaxies from the early universe - that are now visible compared to HUBBLE - many of these are already highly structured.

...

jac berne (flickr)

Hi all, long time, no see.

The comparison tool here is great: https://johnedchristensen.github.io/WebbCompare/

I notice that the really red object left of, and a bit lower from the brightest star is completely not visible in the Hubble image. Why is it SO red?

I'm only seeing the comparison slider tool showing for the Carina nebula image. Are the others there too, but just not working for me perhaps?

...it usually helps, if you switch your browser. Edge should work ...

[Chris Peterson]

These images are, I suppose, false-colour representations of infrared data.

Which, in turn, seems to contradict the rather fussy replies I got earlier this year when I asked ( http://asterisk.apod.com/viewtopic.php ... 30#p320630) whether Webb output could be usefully depicted in visible spectrum.

The answer, apparently, should have been “Yes, absolutely, no problem”

I don't think that's what you asked. It looks like you were asking whether the results could be represented as something that would look approximately like what we'd see in visible light without redshift.

[johnnydeep]

Hi all, long time, no see.

The comparison tool here is great: https://johnedchristensen.github.io/WebbCompare/

I notice that the really red object left of, and a bit lower from the brightest star is completely not visible in the Hubble image. Why is it SO red?

I'm only seeing the comparison slider tool showing for the Carina nebula image. Are the others there too, but just not working for me perhaps?

...it usually helps, if you switch your browser. Edge should work ...

Thanks. IE and Edge both work. For some reason, my chosen browser Vivaldi, based on Chromium, doesn't. How annoying.

[Fred the Cat]

If JWST looks far enough back, would it begin to see a set of the same first galaxies with the CMB beyond - where ever it is pointing?

Maybe that’s the obvious “point” I hadn’t imagined to be a JWST mission.

Hmm. I presume you mean that because back when the universe was a mere 380,000 years old (limited by the time of "last scattering"), it was much smaller. But even then, it would only see the galaxies on the edges of the limiting sphere, which even at a young 380,000 years old, would still be much larger than any single galaxy. Presumably. So, just how big was the universe at 380,000 years young?

Who knows? I expect the universe has as many twists and turns as a coiled extension cord. Getting it straight becomes a matter of uncoiling its story and that’s " knot" my strongpoint.

[Chris Peterson]

If JWST looks far enough back, would it begin to see a set of the same first galaxies with the CMB beyond - where ever it is pointing?

Maybe that’s the obvious “point” I hadn’t imagined to be a JWST mission. :|

Hmm. I presume you mean that because back when the universe was a mere 380,000 years old (limited by the time of "last scattering"), it was much smaller. But even then, it would only see the galaxies on the edges of the limiting sphere, which even at a young 380,000 years old, would still be much larger than any single galaxy. Presumably. So, just how big was the universe at 380,000 years young?

The (largely consensus) cosmology model tells us the radius of the Universe has increased by about 10,000 times since then. So it was still a very, very large universe compared to the size of a galaxy.

[Jim Leff]

These images are, I suppose, false-colour representations of infrared data.

Which, in turn, seems to contradict the rather fussy replies I got earlier this year when I asked ( http://asterisk.apod.com/viewtopic.php ... 30#p320630) whether Webb output could be usefully depicted in visible spectrum.

The answer, apparently, should have been “Yes, absolutely, no problem”

I don't think that's what you asked. It looks like you were asking whether the results could be represented as something that would look approximately like what we'd see in visible light without redshift.

Chris, your expertise knows no bounds. You are correcting me re: my own intention with my own question.

I was aware at the time that you'd failed to grok what I was asking, and that you were resistant to my clarification (no harm, no foul, we all appreciate your answers). And now I salute your staunch steadfastness in maintaining that my question was precisely what you originally decided it was. Backbone is good.

But I'd meekly offer the proposal that I may (or may not!) be the higher authority on my own intentions. I do realize this creates a zero sum dilemma.

[Chris Peterson]

Which, in turn, seems to contradict the rather fussy replies I got earlier this year when I asked ( http://asterisk.apod.com/viewtopic.php ... 30#p320630) whether Webb output could be usefully depicted in visible spectrum.

The answer, apparently, should have been “Yes, absolutely, no problem”

I don't think that's what you asked. It looks like you were asking whether the results could be represented as something that would look approximately like what we'd see in visible light without redshift.

Chris, your expertise knows no bounds. You are correcting me re: my own intention with my own question.

I was aware at the time that you'd failed to grok what I was asking, and that you were resistant to my clarification (no harm, no foul, we all appreciate your answers). And now I salute your staunch steadfastness in maintaining that my question was precisely what you originally decided it was. Backbone is good.

But I'd meekly offer the proposal that I may (or may not!) be the higher authority on my own intentions. I do realize this creates a zero sum dilemma.

I'm not correcting you. I'm pointing out what I thought you asked. And on review, it still seems that way. So I guess I don't know what you were asking. But the replies were all reasonable.

[rstevenson]

Nice. What surprised me most is that the James Webb's images from its hexagonal mirrors still produces rectangular images with perfect 90 degree corners. Unlike my binoculars that has round lenses and it produces a round image. The explanation was for the old round camera lenses producing square images was because that is how the photo negative film was made. And of course science text books were rectangular and demanded rectangular photos. But nowadays when paper books are on the way out and most images are processed by computer there is no reason to still cut the round or hexagonal images into squares.

The answer is simple, I think. The mirrors focus light into one or more of the collectors, and they are rectangular. Here’s a rectangular pic of one of them…

Rob

[FLPhotoCatcher]

Nice. What surprised me most is that the James Webb's images from its hexagonal mirrors still produces rectangular images with perfect 90 degree corners. Unlike my binoculars that has round lenses and it produces a round image. The explanation was for the old round camera lenses producing square images was because that is how the photo negative film was made. And of course science text books were rectangular and demanded rectangular photos. But nowadays when paper books are on the way out and most images are processed by computer there is no reason to still cut the round or hexagonal images into squares.

The answer is simple, I think. The mirrors focus light into one or more of the collectors, and they are rectangular. Here’s a rectangular pic of one of them… E86C69B2-BFED-42FA-83A9-F0D08ADEDC10.jpeg

Rob

Why didn't they use a sensor that better matched the actual view, or at least use a bigger one, with no reduction is resolving power? That seems like a great way to get significantly more "free" data.

[Chris Peterson]

Nice. What surprised me most is that the James Webb's images from its hexagonal mirrors still produces rectangular images with perfect 90 degree corners. Unlike my binoculars that has round lenses and it produces a round image. The explanation was for the old round camera lenses producing square images was because that is how the photo negative film was made. And of course science text books were rectangular and demanded rectangular photos. But nowadays when paper books are on the way out and most images are processed by computer there is no reason to still cut the round or hexagonal images into squares.

The answer is simple, I think. The mirrors focus light into one or more of the collectors, and they are rectangular. Here’s a rectangular pic of one of them… E86C69B2-BFED-42FA-83A9-F0D08ADEDC10.jpeg

Rob

Why didn't they use a sensor that better matched the actual view, or at least use a bigger one, with no reduction is resolving power? That seems like a great way to get significantly more "free" data.

What makes you think the sensors don't cover all of the actual view? That they aren't matched in both pixel size and area to the optics?

[Ann]

These images are, I suppose, false-colour representations of infrared data. How much has the spectrum been shifted upwards - two octaves? ten?

Oh wow. You'll have to do with my very shaky infrared "expertise". (Yeah, using the word "expertise" here is a joke.)

Anyway...

JWST User Documentation wrote:

JWST NIRCam offers 29 bandpass filters in the short wavelength (0.6–2.3 μm) and long wavelength (2.4–5.0 μm) channels.

The JWST NIRCam is the camera detecting the shortest infrared wavelengths. There is another camera too, which I think is called MIRI (Mid Infrared Instrument) (but you'll have to google, because I'm too lazy), which detects longer, mid infrared wavelengths.

So anyway. If I understand things correctly, and 0.6 μm is the same thing as 600 nm, then the shortest wavelength that JWST can detect corresponds to this visible color: ███

So this is the shortest wavelength that JWST can detect, and any emission at this wavelength will be mapped as blue ███. (The color sample I just showed you corresponds to a wavelength of 441 nm, which probably qualifies as the height of "blue-ness" that the human eye can detect.) The difference in wavelength between 600 nm and 441 nm is not so large, only some "160 steps on the wavelength meter" (or how do you put that in words?).

All right. But please note that 2.3 μm, 2300 nm, also counts as a short wavelength to JWST, and this wavelength will, perhaps, be mapped as this visible color: ███ (This color sample corresponds to 550 nm, which is a wavelength that is sometimes used by Hubble as a green filter.)

Let's assume that 2300 nm is mapped as "green", or 550 nm, by JWST. The difference between 550 nm and 2300 nm corresponds to "1750 steps on the wavelength meter". That's a lot.

2300 nm is way, way off the optical scale of wavelengths that the human eye can detect. After all, when Hubble takes pictures in the infrared, it is usually no more infrared than - if I remember correctly - 814 nm. 2300 nm is almost three times longer than the longest infrared wavelength than Hubble can see.

To get an idea of wavelengths and color, go to this page. Scroll right as far as you can get to see where the human ability to detect longer and longer wavelengths fizzles out. The scale ends at 780 nm, which corresponds to a dark brown color to the human eye. Beyond 780 nm, the human sensitivity will quickly drop to zero. 2300 nm, the longest wavelength that the JWST NIRCam classifies as "short", is almost exactly three times longer than the longest, darkest brown wavelength that is reasonably visible to the human eye.

Go to this page to see a bit more about the JWST NIRCam filters and how their wavelengths are mapped.

Ann

[FLPhotoCatcher]

The answer is simple, I think. The mirrors focus light into one or more of the collectors, and they are rectangular. Here’s a rectangular pic of one of them… E86C69B2-BFED-42FA-83A9-F0D08ADEDC10.jpeg

Rob

Why didn't they use a sensor that better matched the actual view, or at least use a bigger one, with no reduction is resolving power? That seems like a great way to get significantly more "free" data.

What makes you think the sensors don't cover all of the actual view? That they aren't matched in both pixel size and area to the optics?

As Rob said, "The answer is simple, I think. The mirrors focus light into one or more of the collectors, and they [the collectors] are rectangular."
I think the incoming light would be round-ish, and the collector is rectangle. I have no proof one way or the other though.

[zmon21]

in stephan's quintet, NGC 7320 is 7 times closer than the other four galaxies, however they look almost the same size. as size is inversely proportional to the square root, the other galaxies are roughly 49 times bigger than NGC 7320.  diameter of NGC 7320 is 30,415.05 light years. then the other four galaxies diameters are around 1.5 million lys. is this right?

[Chris Peterson]

Why didn't they use a sensor that better matched the actual view, or at least use a bigger one, with no reduction is resolving power? That seems like a great way to get significantly more "free" data.

What makes you think the sensors don't cover all of the actual view? That they aren't matched in both pixel size and area to the optics?

As Rob said, "The answer is simple, I think. The mirrors focus light into one or more of the collectors, and they [the collectors] are rectangular."
I think the incoming light would be round-ish, and the collector is rectangle. I have no proof one way or the other though.

The corrected field at the focal plane is round. So yes, if you used a round detector, you could get more coverage than using a square one. But those pixels would hardly be "free". In fact, there's really no way to make a round detector because of the way that rows and columns are read. And cutting a round section from a semiconductor crystal? Really difficult! So what you'd need to do is make an oversized square detector, and it would have its corners outside the well corrected area of the focal plane. You could either use those pixels at reduced resolution, or ignore them. (Most ordinary cameras have worse resolution in the corners because of this.) And, of course, that larger sensor with more pixels would be MUCH more expensive. So they really are using the most effective, practical method.

And, of course, there's the processing to consider. All of the algorithms are based on the Cartesian coordinate system, because that's how the pixels are arranged. If you had a circular image, you'd still need to embed it in a square matrix, with a bunch of pixels outside the circle that would have to be accounted for, but also ignored. Unless you invented an entirely new sort of detector that somehow had the pixels arranged in a circular array... and developed all new analysis software that worked in polar coordinates.

(Years ago I worked a little with a very exotic image sensor, that was hexagonal with hexagonal pixels. It was developed for a camera in a missile that was intended for tracking a target. Very specialized and very unsuitable for ordinary imaging. And very expensive.)