<p>According to <a href="https://en.wikipedia.org/wiki/Apparent_magnitude" rel="nofollow">this table</a>, Hubble is good down to around magnitude 31.5. What with digital stacking of exposures, that number has got to reflect a point at which detector noise is equal to incoming signal.
At the distance of M31, 778000pc, that brightness corresponds to an <a href="https://en.wikipedia.org/wiki/Absolute_magnitude" rel="nofollow">absolute magnitude</a> of M = 31.5 -5((log778000)-1) = 7.
That's about as bright as <a href="http://www.astro.wisc.edu/~dolan/constellations/extra/nearest.html" rel="nofollow">Epsilon Indus</a>, a K3 class star, <a href="https://en.wikipedia.org/wiki/K-type_main-sequence_star" rel="nofollow">0.45 to 0.8 times the mass of the Sun</a>.</p>

<p>If Hubble is seeing that deep, it's a little surprising how <em>few</em> stars it picked up in that image of our neighboring galaxy.</p>

CC BY-SA 3.0
2015-08-13T13:50:03.430Z
2015-08-13T20:43:28.860Z
2015-08-13T20:43:28.860Z
Wayfaring Stranger
<p>This is most likely not noise.</p>

<p>Photographic noise is generally going to be limited to roughly single pixels. This is because much of it is generated at the individual pixel level (often a single color, so not even an entire pixel) on the sensor. There is also some relationship to this and overheating - I'm not familiar with the cooling systems of the Hubble, but presumably it's sophisticated enough to prevent serious noise from overheating.</p>

<p>When examining the photo, you can see that these are not individual pixels but rounded seemingly point-source light. It's hard to say whether this is an accurate representation of their actual angular diameter, as they may have acquired some of that rounding by diffraction. Regardless, this suggests that these are not artifacts from the sensor but are actual light sources. For a long exposure photograph, this also means that they are a persistent light source. The colors that they are composed of also fits the colors we would expect to see from stars.</p>

<p>Unless there has been some very heavy editing done with the intention of representing noise as stars, I think it's fairly safe to say that these are all stars or other star-like light sources.</p>

CC BY-SA 3.0
2015-08-13T13:50:24.553Z
It's got to be stacked exposures. Stacking reduces noise.
Wayfaring Stranger
CC BY-SA 3.0
2015-08-13T01:04:42.883Z
2015-08-13T01:04:42.883Z
Mitch Goshorn
<p>So there is this <a href="http://cdn.spacetelescope.org/archives/images/screen/heic1502a.jpg" rel="noreferrer">huge picture</a> by Hubble.</p>

<p>Better yet, with the <a href="http://www.spacetelescope.org/images/heic1502a/zoomable/" rel="noreferrer">zoom tool</a></p>

<p>And if you zoom to the maximum level, you can see a seemingly uniformly distributed dots of orange and blue. Are those actually stars and maybe galaxies behind them and not just camera noise?</p>

<p>Furthermore I'm trying to get a grasp on how large portion of the sky this photo represents.</p>

<blockquote>
  <p>At 3.4, the apparent magnitude of the Andromeda Galaxy is one of the brightest of any Messier objects, making it visible to the naked eye on moonless nights even when viewed from areas with moderate light pollution. Although it appears more than six times as wide as the full Moon when photographed through a larger telescope, only the brighter central region is visible to the naked eye or when viewed using binoculars or a small telescope.</p>
</blockquote>

<p>Let's say this is about the half of the diameter of the galaxy so this photo is about ~3 Moons wide, is that correct?</p>

CC BY-SA 3.0
2015-08-13T06:00:36.640Z
I don't know the precise size of Andromeda, especially since it's outer edges grow rather dim, but a number of Moon and Andromeda photos have been created recently and I'd say that Andromeda, if we could see all of it, would at least 3, proboably 4-5 moons across in terms of amount of sky it takes up.   (Imagine what Andromeda might look like in 3-4 billion years as it gets much closer to the Milky way.  For more, Google image, Moon and Andromeda as there's a number of them.   Here's one:  http://stargazerslounge.com/uploads/monthly_09_2013/post-6762-137922477212.jpg
userLTK
CC BY-SA 3.0
2015-08-13T06:11:15.173Z
Small point to add, but I found this conversation interesting and related.   Stars look like bright points of light, so they can be photographed from some distance.   Here's a discussion why distant stars are easier to photograph than Pluto, which I found interesting:   http://www.reddit.com/r/explainlikeimfive/comments/1jmijg/eli5_why_we_can_take_detailed_photos_of_galaxies/    I agree with Mitch Goshorn's answer.  I've heard, those are actually stars.  Hubble has a pretty unobstructed view of Andromeda and the photo captures millions of individual starts.
userLTK
<galaxy><photography><hubble-telescope>
Are those noise or stars?
Firsh - justifiedgrid.com
# Are those noise or stars?

(Asked by: username_0 on 2015-08-12)

So there is this [huge picture](http://cdn.spacetelescope.org/archives/images/screen/heic1502a.jpg) by Hubble.

Better yet, with the [zoom tool](http://www.spacetelescope.org/images/heic1502a/zoomable/)

And if you zoom to the maximum level, you can see a seemingly uniformly distributed dots of orange and blue. Are those actually stars and maybe galaxies behind them and not just camera noise?

Furthermore I'm trying to get a grasp on how large portion of the sky this photo represents.

> At 3.4, the apparent magnitude of the Andromeda Galaxy is one of the brightest of any Messier objects, making it visible to the naked eye on moonless nights even when viewed from areas with moderate light pollution. Although it appears more than six times as wide as the full Moon when photographed through a larger telescope, only the brighter central region is visible to the naked eye or when viewed using binoculars or a small telescope.

Let's say this is about the half of the diameter of the galaxy so this photo is about ~3 Moons wide, is that correct?

**Tags:** galaxy, photography, hubble-telescope

> **Comments:**
> **username_1**: I don't know the precise size of Andromeda, especially since it's outer edges grow rather dim, but a number of Moon and Andromeda photos have been created recently and I'd say that Andromeda, if we could see all of it, would at least 3, proboably 4-5 moons across in terms of amount of sky it takes up.   (Imagine what Andromeda might look like in 3-4 billion years as it gets much closer to the Milky way.  For more, Google image, Moon and Andromeda as there's a number of them.   Here's one:  http://stargazerslounge.com/uploads/monthly_09_2013/post-6762-137922477212.jpg  
> **username_1**: Small point to add, but I found this conversation interesting and related.   Stars look like bright points of light, so they can be photographed from some distance.   Here's a discussion why distant stars are easier to photograph than Pluto, which I found interesting:   http://www.reddit.com/r/explainlikeimfive/comments/1jmijg/eli5_why_we_can_take_detailed_photos_of_galaxies/    I agree with username_3's answer.  I've heard, those are actually stars.  Hubble has a pretty unobstructed view of Andromeda and the photo captures millions of individual starts.  

---

## 2 answers

### Answer by username_3 on 2015-08-13. Score: 5 | Accepted answer

This is most likely not noise.

Photographic noise is generally going to be limited to roughly single pixels. This is because much of it is generated at the individual pixel level (often a single color, so not even an entire pixel) on the sensor. There is also some relationship to this and overheating - I'm not familiar with the cooling systems of the Hubble, but presumably it's sophisticated enough to prevent serious noise from overheating.

When examining the photo, you can see that these are not individual pixels but rounded seemingly point-source light. It's hard to say whether this is an accurate representation of their actual angular diameter, as they may have acquired some of that rounding by diffraction. Regardless, this suggests that these are not artifacts from the sensor but are actual light sources. For a long exposure photograph, this also means that they are a persistent light source. The colors that they are composed of also fits the colors we would expect to see from stars.

Unless there has been some very heavy editing done with the intention of representing noise as stars, I think it's fairly safe to say that these are all stars or other star-like light sources.

> **Comments:**
> **username_2**: It's got to be stacked exposures. Stacking reduces noise.  

---

### Answer by username_2 on 2015-08-13. Score: 4

According to [this table](https://en.wikipedia.org/wiki/Apparent_magnitude), Hubble is good down to around magnitude 31.5. What with digital stacking of exposures, that number has got to reflect a point at which detector noise is equal to incoming signal.
At the distance of M31, 778000pc, that brightness corresponds to an [absolute magnitude](https://en.wikipedia.org/wiki/Absolute_magnitude) of M = 31.5 -5((log778000)-1) = 7.
That's about as bright as [Epsilon Indus](http://www.astro.wisc.edu/~dolan/constellations/extra/nearest.html), a K3 class star, [0.45 to 0.8 times the mass of the Sun](https://en.wikipedia.org/wiki/K-type_main-sequence_star).

If Hubble is seeing that deep, it's a little surprising how *few* stars it picked up in that image of our neighboring galaxy.


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