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The image is very nice, but the explanation of brightness is a bit confused: > But, of course, the galaxy isn’t nearly this bright. You need a dark sky to see

The image is very nice, but the explanation of brightness is a bit confused: > But, of course, the galaxy isn’t nearly this bright. You need a dark sky to see it, and, even then, it’s a barely visible fuzzy patch of light. In order to appear as bright as in the image above, the Andromeda galaxy would need to be closer. If it were close enough to look so bright, it would appear even bigger on our sky’s dome. Things closer by do not appear brighter, they merely appear bigger. The Andromeda galaxy will never appear as bright as in that image, not matter how close it is. As things get closer they do, of course, cast more light on us, but that light also comes from a larger solid angle, which means when we look at them, the brightness (per area) remains constant. It's why your computer monitor doesn't look any dimmer as you walk away. It just gets smaller in your field of view. This breaks down for pointlike objects, like single stars - anything small or far away enough to look like a point to our eye cannot get any smaller, so changes in distance are only apparent as changes in brightness. But as the image demonstrates, the Andromeda galaxy is much, much larger than a pointlike object to the naked eye, and so this does not apply (it doesn't matter that the individual light sources are pointlike stars here - for a galaxy they average out due to density, but even if they didn't, as individual points get brighter the space between them gets larger so it all still cancels out when we look at overall apparent brightness for the whole group).

ordu 5y ago
The explanation is confusing indeed because closer means not only "brighter" but also closer. But we can try to overcome this confusion by a thought experiment: lets compress Andromeda galaxy by a factor of k and divide distance by a factor of k. This way it would be k times close but it would have the same angular size. And now lets figure what the k need to be to make Andromeda this bright. Does such k exists? I believe it does. I'm not sure would it work, because my physics knowledge is limited, but if I'm right then it is a non-confusing way to interpret the confusing statement "needs to be closer to be brighter".
Archelaos 5y ago
> The Andromeda galaxy will never appear as bright as in that image, not matter how close it is. As things get closer they do, of course, cast more light on us, but that light also comes from a larger solid angle, which means when we look at them, the brightness (per area) remains constant. The expressions "look so bright" and "appear even larger" refer to a change in our perception of Andromeda, i.e., the resolution of your eyes. In the distance, the bright objects blur with the dark background, so that we do not perceive a lot of structure. Closer in, the contrast increases so that we can distinguish brighter from darker structures. This is analogous to the fact that with binoculars we can see stars that we cannot see with the naked eye. When a distant object gets closer, the area that the object occupies in our field of view becomes larger. So while its brightness under the same angle remains the same, its overall brightsness increases. Think of someone approaching with a torchlight from 100 meter to one centimeter from your eye. This is what can be seen in the eight pictures in the article. The little Andromeda spot and its halo on the first picture is as bright as Andromeda in average in the subsequent pictures; only that it becomes larger. The 7th and 8th picture show the core of Andromeda and the Milkyway together and without the darker surroundings. This is the reason why the overall brightness of the pictures is higher. (Or in the binoculars analogy: Don't look with binoculars into the sun!)
mannykannot 5y ago
We do have a close-up view of a similar galaxy - our own. While the milky way, which is a part of our galaxy that does not resolve as individual stars, is not exactly bright, it appears quite a lot brighter than Andromeda. (It would appear brighter still if not partially obscured by dust in the galactic plane, which we are viewing edge-on. In particular, it blocks most of the galactic core.)
globular-toast 5y ago
What about the inverse-square law?
- marcan_42 5y ago
  Double the distance, you get 1/4 the light. Double the distance, you get 1/4 the solid angle. The surface brightness of the object in your field of view therefore stays constant.
  - globular-toast 5y ago
    This took me a surprisingly long time to accept even though it seems intuitive based on the monitor example. After thinking about it for a while I understand it now. The inverse-square law applies to a point source. As you move further away the intensity drops off for each point, but the points get closer together in your field of view to compensate. One weird thing is that many of the stars in the sky would be as bright or brighter than the Sun, but their "disc" is just so tiny that we perceive it as being dim.
    - marcan_42 5y ago
      Yup, all stars other than the Sun are below our ability to resolve as anything other than a point, so they do indeed look dimmer than the Sun to us. The inverse square law always applies, but it relates to the total amount of light received by a constant area observer (or density at a pointlike observer). Once you're projecting the image of the source onto a plane with a lens, that doesn't map to surface brightness. Another place where the inverse square law breaks down is with beams of light, in the near field - consider something like a laser beam. While it will expand, and eventually follow the law once you get far enough, it very much doesn't at close range. And there's an interesting parallel here: to an observer, a collimated laser beam looks like a perfect pointlike object at infinity (modulo the diffraction limit, of course). Another way to see it is that focusing a beam of light in a certain way creates a virtual emitter much further away, and so you need to add that to the distance you plug into the inverse square law.
dreamcompiler 5y ago
If we reverse the situation the analysis is clearly flawed. The Sun is not a particularly bright star in an absolute sense, but it's extremely bright to us. If you moved it a few hundred AU farther away, it would become difficult to distinguish from the background stars. Clearly distance makes a difference in apparent brightness. One thing that happens is that once the star fills a 2D area in the sky its brightness stops falling off at 1/r^2 and falls off at 1/1 (no falloff) in the near field where the area is a significant fraction of the overall field.
- marcan_42 5y ago
  > If we reverse the situation the analysis is clearly flawed. The Sun is not a particularly bright star in an absolute sense, but it's extremely bright to us. If you moved it a few hundred AU farther away, it would become difficult to distinguish from the background stars. You missed my point about pointlike objects. If you moved the sun 1 AU further away, it would of course cast 1/4 of the total light on the Earth, but as an object in the sky it would have the same surface brightness as it does now. It would just look smaller. I'm talking about the appearance of objects in the sky to a camera-like observer. Once the distance increases to the point where the human eye cannot resolve it beyond a single point, then brightness starts diminishing with distance. This is the case for all single stars other than the sun. This is not the case for aggregate objects like galaxies that are large enough to see as non-points with the naked eye, like Andromeda, regardless of whether they are composed of individual pointlike stars or not.

ordu 5y ago

The explanation is confusing indeed because closer means not only "brighter" but also closer. But we can try to overcome this confusion by a thought experiment: lets compress Andromeda galaxy by a factor of k and divide distance by a factor of k. This way it would be k times close but it would have the same angular size. And now lets figure what the k need to be to make Andromeda this bright. Does such k exists? I believe it does. I'm not sure would it work, because my physics knowledge is limited, but if I'm right then it is a non-confusing way to interpret the confusing statement "needs to be closer to be brighter".

Archelaos 5y ago

> The Andromeda galaxy will never appear as bright as in that image, not matter how close it is. As things get closer they do, of course, cast more light on us, but that light also comes from a larger solid angle, which means when we look at them, the brightness (per area) remains constant. The expressions "look so bright" and "appear even larger" refer to a change in our perception of Andromeda, i.e., the resolution of your eyes. In the distance, the bright objects blur with the dark background, so that we do not perceive a lot of structure. Closer in, the contrast increases so that we can distinguish brighter from darker structures. This is analogous to the fact that with binoculars we can see stars that we cannot see with the naked eye. When a distant object gets closer, the area that the object occupies in our field of view becomes larger. So while its brightness under the same angle remains the same, its overall brightsness increases. Think of someone approaching with a torchlight from 100 meter to one centimeter from your eye. This is what can be seen in the eight pictures in the article. The little Andromeda spot and its halo on the first picture is as bright as Andromeda in average in the subsequent pictures; only that it becomes larger. The 7th and 8th picture show the core of Andromeda and the Milkyway together and without the darker surroundings. This is the reason why the overall brightness of the pictures is higher. (Or in the binoculars analogy: Don't look with binoculars into the sun!)

mannykannot 5y ago

We do have a close-up view of a similar galaxy - our own. While the milky way, which is a part of our galaxy that does not resolve as individual stars, is not exactly bright, it appears quite a lot brighter than Andromeda. (It would appear brighter still if not partially obscured by dust in the galactic plane, which we are viewing edge-on. In particular, it blocks most of the galactic core.)

globular-toast 5y ago

What about the inverse-square law?

marcan_42 5y ago
Double the distance, you get 1/4 the light. Double the distance, you get 1/4 the solid angle. The surface brightness of the object in your field of view therefore stays constant.
- globular-toast 5y ago
  This took me a surprisingly long time to accept even though it seems intuitive based on the monitor example. After thinking about it for a while I understand it now. The inverse-square law applies to a point source. As you move further away the intensity drops off for each point, but the points get closer together in your field of view to compensate. One weird thing is that many of the stars in the sky would be as bright or brighter than the Sun, but their "disc" is just so tiny that we perceive it as being dim.
  - marcan_42 5y ago
    Yup, all stars other than the Sun are below our ability to resolve as anything other than a point, so they do indeed look dimmer than the Sun to us. The inverse square law always applies, but it relates to the total amount of light received by a constant area observer (or density at a pointlike observer). Once you're projecting the image of the source onto a plane with a lens, that doesn't map to surface brightness. Another place where the inverse square law breaks down is with beams of light, in the near field - consider something like a laser beam. While it will expand, and eventually follow the law once you get far enough, it very much doesn't at close range. And there's an interesting parallel here: to an observer, a collimated laser beam looks like a perfect pointlike object at infinity (modulo the diffraction limit, of course). Another way to see it is that focusing a beam of light in a certain way creates a virtual emitter much further away, and so you need to add that to the distance you plug into the inverse square law.

dreamcompiler 5y ago

If we reverse the situation the analysis is clearly flawed. The Sun is not a particularly bright star in an absolute sense, but it's extremely bright to us. If you moved it a few hundred AU farther away, it would become difficult to distinguish from the background stars. Clearly distance makes a difference in apparent brightness. One thing that happens is that once the star fills a 2D area in the sky its brightness stops falling off at 1/r^2 and falls off at 1/1 (no falloff) in the near field where the area is a significant fraction of the overall field.

marcan_42 5y ago
> If we reverse the situation the analysis is clearly flawed. The Sun is not a particularly bright star in an absolute sense, but it's extremely bright to us. If you moved it a few hundred AU farther away, it would become difficult to distinguish from the background stars. You missed my point about pointlike objects. If you moved the sun 1 AU further away, it would of course cast 1/4 of the total light on the Earth, but as an object in the sky it would have the same surface brightness as it does now. It would just look smaller. I'm talking about the appearance of objects in the sky to a camera-like observer. Once the distance increases to the point where the human eye cannot resolve it beyond a single point, then brightness starts diminishing with distance. This is the case for all single stars other than the sun. This is not the case for aggregate objects like galaxies that are large enough to see as non-points with the naked eye, like Andromeda, regardless of whether they are composed of individual pointlike stars or not.