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You absolutely are seeing individual stars in the Triangulum Galaxy (M33) in the zoomable image. The brighter "blobs" are a combination of foreground stars (i.e., stars in our own galaxy) and star clusters in M33, but the fainter points are almost all stars in M33 (a few may faint foreground stars or background galaxies and quasars).

This is no more mysterious than the fact that you can see individual (nearby, bright) stars with your naked eye when you go outside at night. The angular resolving power of your eyes (or of small telescopes) compared to the angular diameters of even nearby stars is terrible; nonetheless, you can still see them.

The key thing to understand is that objects which are too small to resolve will be blurred by the telescope/lens/eye (plus atmospheric turbulence if you're not in space) into a point spread function [https://en.wikipedia.org/wiki/Point_spread_function]. The light from each star is thus spread out over multiple pixels in a pattern which is basically the same for all the stars, varying only in total brightness and thus detectability; for really bright stars, the light in this pattern can be traced out over a significant fraction of the entire image. This includes features like the diffraction spikes; we don't see these for stars in M33 only because they're too faint to register.

This, exactly. And it should be noted that even with a theoretical ideal PSF, point sources would still have non-zero-sized images simply because the recording surface (film, retina, CMOS sensor) has discrete sensor elements. A point source will always be at least a single "pixel" wide in the image, it just needs to be bright enough to stand out against whatever else is projected onto that pixel.

If I'm not mistaken, ideal optics would focus a point source onto a single pixel. A blurred image would spread that point source over more pixels and a sharpening filter might be able to not only fix the blur, but locate the point source within the that single pixel. Yes? No?

I'm imagining this being similar to how adding the right kind of noise and oversampling can tease out smaller signals than one might think possible in a noise free sampling system.

Am I off base here?

No, because a pixel is not a point. Even if not-perfect optics blur the point a bit, it can still be one pixel or less. Ideal optics might be able to focus the light much smaller than one pixel, but that won't change the resulting image because the pixel sensor will pick up the same amount of light.

"A Pixel Is Not A Little Square"

http://alvyray.com/Memos/CG/Microsoft/6_pixel.pdf

And elementary photosensitive units (whether silver halide grains on photographic film, individual sensor elements in a CMOS or CCD image sensor, or photoreceptive cells in a mammalian retina) are not points. They have a nonzero area and turn photons into signal irrespective of where they hit. That article, while very informative, is not relevant. Sensels are not pixels.

Optics don't know about pixels. To a first approximation, they are continuous. Whatever discretization happens after that, it's not relevant to the optics in any way. A sharp lens can greatly outresolve a low-resolution sensor; the bottleneck is the sensor in that case. Conversely, a high-resolution sensor may not be of much use if the lens is soft - but in this case you can at least in theory use deconvolution to recover detail if you know a good approximation of the PSF of the lens. This is not uncommon when working with scientific instruments, actually, but pretty rare in the case of consumer photography equipment.

Its worth mentioning that VY Canis Majoris has a mass of "only" 17 Solar masses (+-8). This means it has a very low density.

"Despite the mass and very large size, VY CMa has an average density of 5.33 to 8.38 mg/m3 (0.00000533 to 0.00000838 kg/m3), it is over 100,000 times less dense than Earth's atmosphere at sea level (1.2 kg/m3). " [0]

Often these huge but low density objects are incorrectly rendered like our Sun with a "hard" surface. I think it's a pity because they are even weirder and more fascinating objects than that.

[0] https://en.wikipedia.org/wiki/VY_Canis_Majoris

The Mote in God's Eye had an interesting description of a star like that, including a starship going fairly deep into it and mentioning that the transition was barely noticable.

How is that possible? And how can that be determined? I'm both intrigued and in disbelief. Does it not collapse due to internal pressure from fusion reactions? Is that just heat, or radiation pressure too? Have I just answered my own questions?

It's important to note that the average density is low, it doesn't mean it's uniformly that dense. I cannot speak for this star in particular, but a lot of variable stars expand and contract so fast that material actually bounces off of it sending out a shell that may or may not re-coalesce or continue going off into space. The shell continues to glow due to the heat it already had, and the heat it receives from the main part of the star.

Their atmosphere can be thought of as 'hot near-vacuum'.

As long as there is enough light collected objects should be visible, even when in theory they would cover only a fraction of a pixel. We can rather see different stars because there's vast space between them which means they will not blend/blur into each other.

Imagine you have a camera 2.7m ly away from a star that is 2 billion km across, and it's the only significant light source in that direction.

How many pixels would you expect the image of the star to be? It can't be less than 1. You'd still see the individual star even though it is a lot smaller than 1 pixel. Imperfect optics mean it will likely appear larger than 1 pixel.

Easily. Which focal length/field of view lens is on that camera?

I would have thought all these stars and galaxies are at optical-infinity?

True, and focal length is in fact defined (or at least thought of) under the assumption of parallel light rays coming in from an optically infinite distance. https://en.wikipedia.org/wiki/Focal_length

Yes? The sensor is not at optical infinity however, and the focal length determines the field of view you get. Quite clearly, one pixel of a camera with 0.001 degree angular resolution and the one of 10 degree FOV give you very different results.

Conversely, how many stars are contributing to that pixel?

2 billion km = 2e12m, not 2e9

The fact that individual stars can be seen doesn't necessarily mean the telescope has a resolution equal to the angular size of this star. A point source will be apparent even if it isn't resolved. Otherwise we could never see stars with our naked eye.

I was wondering about this as well, haven't ever found a good answer.

https://earthsky.org/tonight/can-we-see-stars-outside-our-mi...

Only addresses visible stars, which are all in our Galaxy. The thing is Andromeda actually covers more of the night sky than the moon, so I kinda wonder if an instrument could resolve individual stars. I don't know a whole lot about that other than Airy discs, which would suggest no, stars are too close to each other in angular terms.

There are quite a lot of star clusters in M33/Triangulum (e.g. https://arxiv.org/abs/1402.3029), so some of the bright dots may not be individual stars. There are stars which are bright enough to dominate the light within a PSF (point spread function, or resolved region size). They will, of course, be physically much smaller than that size.

In the article, the image is stated to be 665 megapixels and the width of the galaxy is given as 60,000 light years.

Under the simplifying assumptions that the image is square and encompasses the exact galaxy width, that's about 22 trillion km (or ~11,000 VY Canis Majoris) per pixel. So I guess we're not seeing individual stars (edit: in the target galaxy) - or my math is also off somewhere.

Well the stated resolution of Hubble is about 3 orders of magnitude smaller than necessary to resolve VY per my likely flawed calculation, so we're roughly within an order of magnitude of each other.

Either that or all those stars are uuuuuge

It doesn't matter if the star is smaller than the pixel. You can see them individually as long as you can resolve the distance between them.

Those are definitely individual stars. They might between us and the Triangulum galaxy.

The ones with diffraction spikes are foreground stars. The others are in the Triangulum galaxy.

You only need to resolve the distance between stars, not the stars themselves.

Those blobs are individual stars, but not necessarily stars in Triangulum...

Platelets aren't red. Still very interesting.

Lol i thought there was something wrong with it when i wrote it down.

Platlets are about a tenth of the size of erythrocytes. Sometimes up to one fifth the size, for big ones.

Horizons on earth start at maybe 25 or 30 miles away.

Cars are about 10 feet across.

But you’re saying an 8 (maybe 6) micrometer erythrocyte, less than 10 feet away (on a car window), can eclipse a half a millimeter graphite pencil lead at a distance of 86Km (53 miles)?

Also, a meter is a million micrometers. A millimeter is a thousand micrometers. You would need 100 copies of a 10 micrometer object, to span a millimeter. Fifty, to span half.

Are you saying VY Canis Majoris is the erythrocyte to Triangulum’s pencil lead? As it eclipses said object at 53 miles?

Or are you saying the erythrocyte is fixed onto a house window, and an erythrocyte can occlude the direct observation of an actual star in the sky (the biggest one we know of), even at 53 miles, which is over, beyond and even twice past any terrestrial horizon, and would need to be on a window in a skyscraper taller than anything ever built?

EDIT: Sorry, the Burj Kalifa is 2,700 feet tall, so its spire would peek over the horizon, even at distances up to around 60 miles. So, I need to update my mental trivia model. We actually have built something that big, but it only happened about ten years ago.

https://en.wikipedia.org/wiki/Burj_Khalifa

> Or are you saying the erythrocyte is fixed onto a house window, and an erythrocyte can occlude the direct observation of an actual star in the sky (the biggest one we know of), even at 53 miles, which is over, beyond and even twice past any terrestrial horizon, and would need to be on a window in a skyscraper taller than anything ever built?

Yes. Also, I'm pretty sure your objections here are handled by "(flat, obviously)". The point is to make an analogy to sizes and distances on a human scale, not to make a point about the geometry of Earth.

The pencil lead and the car are distractions, that lead the reader to confuse possible relative size analogies, with frames of direct observation.